Dataset Viewer
ID
stringlengths 8
10
| Year
int64 1.98k
2.02k
| Problem Number
int64 1
15
| Question
stringlengths 37
2.66k
| Answer
int64 0
997
| Part
stringclasses 2
values | Solution
sequencelengths 1
25
|
|---|---|---|---|---|---|---|
1983-I-1 | 1,983 | 1 | Let $x$ , $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_xw=24$ , $\log_y w = 40$ and $\log_{xyz}w=12$ . Find $\log_zw$ . | 60 | null | [
"The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms. $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. If we now convert everything to a power of $120$, it will be easy to isolate $z$ and $w$. $x^{120}=w^5$, $y^{120}=w^3$, and $(xyz)^{120}=w^{10}$. With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\\log_zw=060.",
"First we'll convert everything to exponential form. $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. The only expression containing $z$ is $(xyz)^{12}=w$. It now becomes clear that one way to find $\\log_z w$ is to find what $x^{12}$ and $y^{12}$ are in terms of $w$. Taking the square root of the equation $x^{24}=w$ results in $x^{12}=w^{\\frac{1}{2}}$. Raising both sides of $y^{40}=w$ to the $\\frac{12}{40}$th power gives $y^{12}=w^{\\frac{3}{10}}$. Going back to $(xyz)^{12}=w$, we can substitute the $x^{12}$ and $y^{12}$ with $w^{1/2}$ and $w^{3/10}$, respectively. We now have $w^{1/2}w^{3/10}z^{12}=w$. Simplifying, we get $z^{60}=w$. So our answer is $060.",
"Applying the change of base formula, \\begin{align*} \\log_x w = 24 &\\implies \\frac{\\log w}{\\log x} = 24 \\implies \\frac{\\log x}{\\log w} = \\frac 1 {24} \\\\ \\log_y w = 40 &\\implies \\frac{\\log w}{\\log y} = 40 \\implies \\frac{\\log y}{\\log w} = \\frac 1 {40} \\\\ \\log_{xyz} w = 12 &\\implies \\frac{\\log {w}}{\\log {xyz}} = 12 \\implies \\frac{\\log x +\\log y + \\log z}{\\log w} = \\frac 1 {12} \\end{align*} Therefore, $\\frac {\\log z}{\\log w} = \\frac 1 {12} - \\frac 1 {24} - \\frac 1{40} = \\frac 1 {60}$. Hence, $\\log_z w = 060.",
"Since $\\log_a b = \\frac{1}{\\log_b a}$, the given conditions can be rewritten as $\\log_w x = \\frac{1}{24}$, $\\log_w y = \\frac{1}{40}$, and $\\log_w xyz = \\frac{1}{12}$. Since $\\log_a \\frac{b}{c} = \\log_a b - \\log_a c$, $\\log_w z = \\log_w xyz - \\log_w x - \\log_w y = \\frac{1}{12}-\\frac{1}{24}-\\frac{1}{40}=\\frac{1}{60}$. Therefore, $\\log_z w = 060.",
"If we convert all of the equations into exponential form, we receive $x^{24}=w$, $y^{40}=w$, and $(xyz)^{12}=w$. The last equation can also be written as $x^{12}y^{12}z^{12}=w$. Also note that by multiplying the first two equations, we get, $x^{24}y^{40}= w^{2}$. Taking the square root of this, we find that $x^{12}y^{20}=w$. Recall, $x^{12}y^{12}z^{12}=w$. Thus, $z^{12}= y^{8}$. Also recall, $y^{40}=w$. Therefore, $z^{60}$ = $y^{40}$ = $w$. So, $\\log_z w$ = $060. -Dhillonr25, Bobbob",
"Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \\Rightarrow z^3=y^2.$ We are looking for $\\log_z w,$ which by substitution, is $\\log_{y^{\\frac{2}{3}}} y^{40} = 40 \\div \\frac{2}{3} =60 ~coolmath2017"
] |
1983-I-2 | 1,983 | 2 | Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$ . | 15 | null | [
"It is best to get rid of the absolute values first. Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$. Adding these together, we find that the sum is equal to $30-x$, which attains its minimum value (on the given interval $p \\leq x \\leq 15$) when $x=15$, giving a minimum of $015.",
"Let $p$ be equal to $15 - \\varepsilon$, where $\\varepsilon$ is an almost neglectable value. Because of the small value $\\varepsilon$, the domain of $f(x)$ is basically the set ${15}$. plugging in $15$ gives $\\varepsilon + 0 + 15 - \\varepsilon$, or $15$, so the answer is $015"
] |
1983-I-3 | 1,983 | 3 | What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$ ? | 20 | null | [
"If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with. Instead, we substitute $y$ for $x^2+18x+30$, so that the equation becomes $y=2\\sqrt{y+15}$. Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second root is extraneous since $2\\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$, we get $-6=6$, which is obviously false). Hence we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$, $x^2+18x+30=10 \\Longrightarrow x^2+18x+20=0.$ Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \\cdot 1 \\cdot 20 = 244$, which is positive. Thus by Vieta's formulas, the product of the real roots is simply $020.",
"We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \\[(x^2+ 18x + 45) - 2\\sqrt{x^2+18x+45} - 15 = 0.\\] Letting $n = \\sqrt{x^2+18x+45}$, we have $n^2-2n-15 = 0 \\Longrightarrow (n-5)(n+3) = 0$. Because the square root of a real number can't be negative, the only possible $n$ is $5$. Substituting that in, we have \\[\\sqrt{x^2+18x+45} = 5 \\Longrightarrow x^2 + 18x + 45 = 25 \\Longrightarrow x^2+18x+20=0.\\] Reasoning as in Solution 1, the product of the roots is $020.",
"Begin by completing the square on both sides of the equation, which gives \\[(x+9)^2-51=2\\sqrt{(x+3)(x+15)}\\] Now by substituting $y=x+9$, we get $y^2-51=2\\sqrt{(y-6)(y+6)}$, or \\[y^4-106y^2+2745=0\\] The solutions in $y$ are then \\[y=x+9=\\pm3\\sqrt{5},\\pm\\sqrt{61}\\] Turns out, $\\pm3\\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \\[\\left(\\sqrt{61}-9\\right)\\left(-\\sqrt{61}-9\\right)=81-61=020\\] By difference of squares.",
"We are given the equation \\[x^2+18x+30=2\\sqrt{x^2+18x+45}\\] Squaring both sides yields \\[(x^2+18x+30)^2=4(x^2+18x+45)\\] \\[(x^2+18x+30)^2=4(x^2+18x+30+15)\\] \\[(x^2+18x+30)^2=4(x^2+18x+30)+60\\] \\[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\\] Substituting $y=x^2+18x+30$ yields \\[y^2-4y-60=0\\] \\[(y+6)(y-10)=0\\] Thus $y=x^2+18x+30=-6,10$. However if $y=-6$, the left side of the equation \\[x^2+18x+30=2\\sqrt{x^2+18x+45}\\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \\[x^2+18x+30=10\\] \\[x^2+18x+20=0\\] Since the discriminant $\\sqrt{18^2-4\\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $020. ~ Nafer"
] |
1983-I-4 | 1,983 | 4 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy] | 26 | null | [
"Because we are given a right angle, we look for ways to apply the Pythagorean Theorem. Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$. Let $OE=x$ and $OD=y$. We're trying to find $x^2+y^2$. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(D--O--F--B,dashed); dot(O); dot(A); dot(B); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$B$\",B,S); label(\"$C$\",C,SE); label(\"$D$\",D,NE); label(\"$E$\",F,SW); [/asy] Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$. Thus, $\\left(\\sqrt{50}\\right)^2 = y^2 + (6-x)^2$, and $\\left(\\sqrt{50}\\right)^2 = x^2 + (y+2)^2$. We solve this system to get $x = 1$ and $y = 5$, such that the answer is $1^2 + 5^2 = 026.",
"We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$. We know how long $OA$ and $AB$ are, so if we can find $\\cos \\angle OAB$, we'll be in good shape. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); dot(O); dot(A); dot(B); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$B$\",B,S); label(\"$C$\",C,SE); [/asy] We can find $\\cos \\angle OAB$ using angles $OAC$ and $BAC$. First we note that by Pythagoras, \\[AC = \\sqrt{AB^2 + BC^2} = \\sqrt{36 + 4} = \\sqrt{40} = 2 \\sqrt{10}.\\] If we let $M$ be the midpoint of $AC$, that mean that $AM = \\sqrt{10}$. Since $\\triangle OAC$ is isosceles ($OA = OC$ from the definition of a circle), $M$ is also the foot of the altitude from $O$ to $AC.$ [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); pair M = (A+C)/2; draw(P); draw(O--C--A--cycle); draw(O--M, dashed); draw(rightanglemark(O,M,A,25)); dot(O); dot(A); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$M$\",M,SSW); label(\"$C$\",C,SE); label(\"$\\sqrt{50}$\", (O+A)/2, NW); label(\"$\\sqrt{10}$\", (A+M)/2, E); [/asy] It follows that $OM = \\sqrt{40} = 2 \\sqrt{10}$. Therefore \\begin{align*} \\cos \\angle OAC = \\frac{\\sqrt{10}}{\\sqrt{50}} &= \\frac{1}{\\sqrt{5}}, \\\\ \\sin \\angle OAC = \\frac{2 \\sqrt{10}}{\\sqrt{50}} &= \\frac{2}{\\sqrt{5}}. \\end{align*} Meanwhile, from right triangle $ABC,$ we have \\begin{align*} \\cos \\angle BAC = \\frac{6}{\\sqrt{40}} &= \\frac{3}{\\sqrt{10}}, \\\\ \\sin \\angle BAC = \\frac{2}{\\sqrt{40}} &= \\frac{1}{\\sqrt{10}}. \\end{align*} This means that by the angle subtraction formulas, \\begin{align*} \\cos \\angle OAB &= \\cos (\\angle OAC - \\angle BAC) \\\\ &= \\cos \\angle OAC \\cos \\angle BAC + \\sin \\angle OAC \\sin \\angle BAC \\\\ &= \\frac{1}{\\sqrt{5}} \\cdot \\frac{3}{\\sqrt{10}} + \\frac{2}{\\sqrt{5}} \\cdot \\frac{1}{\\sqrt{10}} \\\\ &= \\frac{5}{5 \\sqrt{2}} = \\frac{1}{\\sqrt{2}}. \\end{align*} Now we have all we need to use the law of cosines on $\\triangle OAB.$ This tells us that \\begin{align*} OB^2 &= AO^2 + AB^2 - 2 AO \\cdot AB \\cdot \\cos \\angle OAB \\\\ &= 50 + 36 - 2 \\cdot 5 \\sqrt{2} \\cdot 6 \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= 86 - 2 \\cdot 5 \\cdot 6 \\\\ &= 26. \\end{align*}",
"We'll use the law of cosines. Let $O$ be the center of the circle; we wish to find $OB$. We know how long $OA$ and $AB$ are, so if we can find $\\cos \\angle OAB$, we'll be in good shape. [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); dot(O); dot(A); dot(B); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$B$\",B,S); label(\"$C$\",C,SE); [/asy] We can find $\\cos \\angle OAB$ using angles $OAC$ and $BAC$. First we note that by Pythagoras, \\[AC = \\sqrt{AB^2 + BC^2} = \\sqrt{36 + 4} = \\sqrt{40} = 2 \\sqrt{10}.\\] If we let $M$ be the midpoint of $AC$, that mean that $AM = \\sqrt{10}$. Since $\\triangle OAC$ is isosceles ($OA = OC$ from the definition of a circle), $M$ is also the foot of the altitude from $O$ to $AC.$ [asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); pair M = (A+C)/2; draw(P); draw(O--C--A--cycle); draw(O--M, dashed); draw(rightanglemark(O,M,A,25)); dot(O); dot(A); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$M$\",M,SSW); label(\"$C$\",C,SE); label(\"$\\sqrt{50}$\", (O+A)/2, NW); label(\"$\\sqrt{10}$\", (A+M)/2, E); [/asy] It follows that $OM = \\sqrt{40} = 2 \\sqrt{10}$. Therefore \\begin{align*} \\cos \\angle OAC = \\frac{\\sqrt{10}}{\\sqrt{50}} &= \\frac{1}{\\sqrt{5}}, \\\\ \\sin \\angle OAC = \\frac{2 \\sqrt{10}}{\\sqrt{50}} &= \\frac{2}{\\sqrt{5}}. \\end{align*} Meanwhile, from right triangle $ABC,$ we have \\begin{align*} \\cos \\angle BAC = \\frac{6}{\\sqrt{40}} &= \\frac{3}{\\sqrt{10}}, \\\\ \\sin \\angle BAC = \\frac{2}{\\sqrt{40}} &= \\frac{1}{\\sqrt{10}}. \\end{align*} This means that by the angle subtraction formulas, \\begin{align*} \\cos \\angle OAB &= \\cos (\\angle OAC - \\angle BAC) \\\\ &= \\cos \\angle OAC \\cos \\angle BAC + \\sin \\angle OAC \\sin \\angle BAC \\\\ &= \\frac{1}{\\sqrt{5}} \\cdot \\frac{3}{\\sqrt{10}} + \\frac{2}{\\sqrt{5}} \\cdot \\frac{1}{\\sqrt{10}} \\\\ &= \\frac{5}{5 \\sqrt{2}} = \\frac{1}{\\sqrt{2}}. \\end{align*} Now we have all we need to use the law of cosines on $\\triangle OAB.$ This tells us that \\begin{align*} OB^2 &= AO^2 + AB^2 - 2 AO \\cdot AB \\cdot \\cos \\angle OAB \\\\ &= 50 + 36 - 2 \\cdot 5 \\sqrt{2} \\cdot 6 \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= 86 - 2 \\cdot 5 \\cdot 6 \\\\ &= 26. \\end{align*}",
"Mark the midpoint $M$ of $AC$. Then, drop perpendiculars from $O$ to $AB$ (with foot $T_1$), $M$ to $OT_1$ (with foot $T_2$), and $M$ to $AB$ (with foot $T_3$). [asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {\"O\",\"$T_1$\",\"B\",\"C\",\"M\",\"A\",\"$T_3$\",\"M\",\"$T_2$\"}; for(int i=0;i<a.length;++i) { dl(n[i],a[i],dir(degrees(a[i],false) ) ); draw(a[(i-1)%a.length]--a[i]); }; dot(a); draw(a[5]--a[1]); draw(a[0]--a[3]); draw(a[0]--a[4]); draw(a[0]--a[2]); draw(a[0]--a[5]); draw(a[5]--a[2]--a[3]--cycle,blue+linewidth(0.7)); draw(a[0]--a[8]--a[7]--cycle,blue+linewidth(0.7)); [/asy] First notice that by computation, $OAC$ is a $\\sqrt {50} - \\sqrt {40} - \\sqrt {50}$ isosceles triangle, so $AC = MO$. Then, notice that $\\angle MOT_2 = \\angle T_3MO = \\angle BAC$. Therefore, the two blue triangles are congruent, from which we deduce $MT_2 = 2$ and $OT_2 = 6$. As $T_3B = 3$ and $MT_3 = 1$, we subtract and get $OT_1 = 5,T_1B = 1$. Then the Pythagorean Theorem tells us that $OB^2 = 026.",
"Draw segment $OB$ with length $x$, and draw radius $OQ$ such that $OQ$ bisects chord $AC$ at point $M$. This also means that $OQ$ is perpendicular to $AC$. By the Pythagorean Theorem, we get that $AC=\\sqrt{(BC)^2+(AB)^2}=2\\sqrt{10}$, and therefore $AM=\\sqrt{10}$. Also by the Pythagorean theorem, we can find that $OM=\\sqrt{50-10}=2\\sqrt{10}$. Next, find $\\angle BAC=\\arctan{\\left(\\frac{2}{6}\\right)}$ and $\\angle OAM=\\arctan{\\left(\\frac{2\\sqrt{10}}{\\sqrt{10}}\\right)}$. Since $\\angle OAB=\\angle OAM-\\angle BAC$, we get \\[\\angle OAB=\\arctan{2}-\\arctan{\\frac{1}{3}}\\]\\[\\tan{(\\angle OAB)}=\\tan{(\\arctan{2}-\\arctan{\\frac{1}{3}})}\\]By the subtraction formula for $\\tan$, we get\\[\\tan{(\\angle OAB)}=\\frac{2-\\frac{1}{3}}{1+2\\cdot \\frac{1}{3}}\\]\\[\\tan{(\\angle OAB)}=1\\]\\[\\cos{(\\angle OAB)}=\\frac{1}{\\sqrt{2}}\\]Finally, by the Law of Cosines on $\\triangle OAB$, we get \\[x^2=50+36-2(6)\\sqrt{50}\\frac{1}{\\sqrt{2}}\\]\\[x^2=026.\\]",
"Draw segment $OB$ with length $x$, and draw radius $OQ$ such that $OQ$ bisects chord $AC$ at point $M$. This also means that $OQ$ is perpendicular to $AC$. By the Pythagorean Theorem, we get that $AC=\\sqrt{(BC)^2+(AB)^2}=2\\sqrt{10}$, and therefore $AM=\\sqrt{10}$. Also by the Pythagorean theorem, we can find that $OM=\\sqrt{50-10}=2\\sqrt{10}$. Next, find $\\angle BAC=\\arctan{\\left(\\frac{2}{6}\\right)}$ and $\\angle OAM=\\arctan{\\left(\\frac{2\\sqrt{10}}{\\sqrt{10}}\\right)}$. Since $\\angle OAB=\\angle OAM-\\angle BAC$, we get \\[\\angle OAB=\\arctan{2}-\\arctan{\\frac{1}{3}}\\]\\[\\tan{(\\angle OAB)}=\\tan{(\\arctan{2}-\\arctan{\\frac{1}{3}})}\\]By the subtraction formula for $\\tan$, we get\\[\\tan{(\\angle OAB)}=\\frac{2-\\frac{1}{3}}{1+2\\cdot \\frac{1}{3}}\\]\\[\\tan{(\\angle OAB)}=1\\]\\[\\cos{(\\angle OAB)}=\\frac{1}{\\sqrt{2}}\\]Finally, by the Law of Cosines on $\\triangle OAB$, we get \\[x^2=50+36-2(6)\\sqrt{50}\\frac{1}{\\sqrt{2}}\\]\\[x^2=026.\\]",
"We use coordinates. Let the circle have center $(0,0)$ and radius $\\sqrt{50}$; this circle has equation $x^2 + y^2 = 50$. Let the coordinates of $B$ be $(a,b)$. We want to find $a^2 + b^2$. $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$, respectively, both lie on the circle. From this we obtain the system of equations $a^2 + (b+6)^2 = 50$ $(a+2)^2 + b^2 = 50$ After expanding these terms, we notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$. after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$. Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta. Solving, we get $a=5$ and $b=-1$, so the distance is $a^2 + b^2 = 026.",
"[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); draw(O--C); dot(O); dot(A); dot(B); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$B$\",B,S); label(\"$C$\",C,SE); [/asy] I will use the law of cosines in triangle $\\triangle OAC$ and $\\triangle OBC$. $AC = \\sqrt{AB^2 + BC^2} = \\sqrt{6^2 + 2^2} = 2 \\sqrt{10}$ $\\cos \\angle ACB = \\frac{2}{2\\sqrt{10}} = \\frac{1}{\\sqrt{10}}$ $\\cos \\angle ACO = \\frac{AC^2+OC^2-OA^2}{2 \\cdot AC \\cdot OC} = \\frac{(2\\sqrt{10})^2+(\\sqrt{50})^2-(\\sqrt{50})^2}{2 \\cdot 2\\sqrt{10} \\cdot \\sqrt{50}} = \\frac{1}{\\sqrt{5}}$ $\\sin \\angle ACB = \\sqrt{1-\\cos^2 \\angle ACB} = \\sqrt{1-(\\frac{1}{\\sqrt{10}})^2} = \\frac{3}{\\sqrt{10}}$ $\\sin \\angle ACO = \\sqrt{1-\\cos^2 \\angle ACO} = \\sqrt{1-(\\frac{1}{\\sqrt{5}})^2} = \\frac{2}{\\sqrt{5}}$ $\\cos \\angle OCB = \\cos (\\angle ACB - \\angle ACO) = \\cos \\angle ACB \\cdot \\cos \\angle ACO + \\sin \\angle ACB \\cdot \\sin \\angle ACO = \\frac{1}{\\sqrt{10}} \\cdot \\frac{1}{\\sqrt{5}} + \\frac{3}{\\sqrt{10}} \\cdot \\frac{2}{\\sqrt{5}} = \\frac{7}{5\\sqrt{2}}$ $OB^2 = OC^2 + BC^2 - 2 \\cdot OC \\cdot BC \\cdot \\cos \\angle OCB = (\\sqrt{50})^2 + 2^2 - 2 \\cdot \\sqrt{50} \\cdot 2 \\cdot \\frac{7}{5\\sqrt{2}} = 50 + 4 - 28 = 026 ~isabelchen",
"[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r); pair D=(A.x,0),F=(0,B.y); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); draw(P); draw(C--B--O--A--B); draw(O--B); draw(A--C); draw(O--C); dot(O); dot(A); dot(B); dot(C); label(\"$O$\",O,SW); label(\"$A$\",A,NE); label(\"$B$\",B,S); label(\"$C$\",C,SE); [/asy] I will use the law of cosines in triangle $\\triangle OAC$ and $\\triangle OBC$. $AC = \\sqrt{AB^2 + BC^2} = \\sqrt{6^2 + 2^2} = 2 \\sqrt{10}$ $\\cos \\angle ACB = \\frac{2}{2\\sqrt{10}} = \\frac{1}{\\sqrt{10}}$ $\\cos \\angle ACO = \\frac{AC^2+OC^2-OA^2}{2 \\cdot AC \\cdot OC} = \\frac{(2\\sqrt{10})^2+(\\sqrt{50})^2-(\\sqrt{50})^2}{2 \\cdot 2\\sqrt{10} \\cdot \\sqrt{50}} = \\frac{1}{\\sqrt{5}}$ $\\sin \\angle ACB = \\sqrt{1-\\cos^2 \\angle ACB} = \\sqrt{1-(\\frac{1}{\\sqrt{10}})^2} = \\frac{3}{\\sqrt{10}}$ $\\sin \\angle ACO = \\sqrt{1-\\cos^2 \\angle ACO} = \\sqrt{1-(\\frac{1}{\\sqrt{5}})^2} = \\frac{2}{\\sqrt{5}}$ $\\cos \\angle OCB = \\cos (\\angle ACB - \\angle ACO) = \\cos \\angle ACB \\cdot \\cos \\angle ACO + \\sin \\angle ACB \\cdot \\sin \\angle ACO = \\frac{1}{\\sqrt{10}} \\cdot \\frac{1}{\\sqrt{5}} + \\frac{3}{\\sqrt{10}} \\cdot \\frac{2}{\\sqrt{5}} = \\frac{7}{5\\sqrt{2}}$ $OB^2 = OC^2 + BC^2 - 2 \\cdot OC \\cdot BC \\cdot \\cos \\angle OCB = (\\sqrt{50})^2 + 2^2 - 2 \\cdot \\sqrt{50} \\cdot 2 \\cdot \\frac{7}{5\\sqrt{2}} = 50 + 4 - 28 = 026 ~isabelchen",
"Notice that $50=5^2+5^2=7^2+1^2$, and by the size of the diagram, it seems reasonable that $OA$ represents $5^2+5^2$, and $OC$ means the $7^1+1^2$, and indeed, the values work ($7-5=2$ and $5+1=6$), so $OB^2=5^2+1^2=026 Note: THIS IS NOT A RELIABLE SOLUTION, as diagrams on the tests are not usually drawn to scale."
] |
1983-I-5 | 1,983 | 5 | Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have? | 4 | null | [
"Solution 1 One way to solve this problem is by substitution. We have $x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$ Hence observe that we can write $w=x+y$ and $z=xy$. This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$. Because we want the largest possible $w$, let's find an expression for $z$ in terms of $w$. $w^2-7=2z \\implies z=\\frac{w^2-7}{2}$. Substituting, $w^3-21w+20=0$, which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division). The largest possible solution is therefore $x+y=w=004",
"One way to solve this problem is by substitution. We have $x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$ Hence observe that we can write $w=x+y$ and $z=xy$. This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$. Because we want the largest possible $w$, let's find an expression for $z$ in terms of $w$. $w^2-7=2z \\implies z=\\frac{w^2-7}{2}$. Substituting, $w^3-21w+20=0$, which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division). The largest possible solution is therefore $x+y=w=004",
"An alternate way to solve this is to let $x=a+bi$ and $y=c+di$. Because we are looking for a value of $x+y$ that is real, we know that $d=-b$, and thus $y=c-bi$. Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up. $(a+bi)^2+(c-bi)^2=7+0i$ $(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$ Looking at the imaginary part of that equation, $2ab-2cb=0$, so $a=c$, and $x$ and $y$ are actually complex conjugates. Looking at the real part of the equation and plugging in $a=c$, $2a^2-2b^2=7$, or $2b^2=2a^2-7$. Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$, which equals $10$ (ignoring the odd powers of $i$, since they would not result in something in the form of $10+0i$): $a^3+3a(bi)^2+a^3+3a(-bi)^2=10$ $2a^3-6ab^2=10$ Since we know that $2b^2=2a^2-7$, it can be plugged in for $b^2$ in the above equation to yield: $2a^3-3a(2a^2-7)=10$ $-4a^3+21a=10$ $4a^3-21a+10=0$ Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested. From the Rational Roots theorem, $a=10, a=5, a=\\frac{5}{2}$ all fail, but $a=2$ does work. Thus, the real part of both numbers is $2$, and their sum is $004",
"Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$, such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$. Substituting $c=\\frac{b^2-7}{2}$, we deduce that $b^3-21b-20=0$, whose roots are $-4$, $-1$, and $5$. Since $-b$ is the sum of the roots and is maximized when $b=-4$, the answer is $-(-4)=004",
"$x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \\implies xy = 7 - \\frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$. Letting $p = x+y$, we have that $p^3 - 21p + 20 = 0$. Testing $p = 1$, we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \\implies p = -5, 1, 4 \\implies 4"
] |
1983-I-6 | 1,983 | 6 | Let $a_n=6^{n}+8^{n}$ . Determine the remainder on dividing $a_{83}$ by $49$ . | 35 | null | [
"Solution 1 Firstly, we try to find a relationship between the numbers we're provided with and $49$. We notice that $49=7^2$, and both $6$ and $8$ are greater or less than $7$ by $1$. Thus, expressing the numbers in terms of $7$, we get $a_{83} = (7-1)^{83}+(7+1)^{83}$. Applying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403\\cdot7^{81}+\\cdots + 83\\cdot7)$. We realize that all of these terms are divisible by $49$ except the final term. After some quick division, our answer is $035 This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"Firstly, we try to find a relationship between the numbers we're provided with and $49$. We notice that $49=7^2$, and both $6$ and $8$ are greater or less than $7$ by $1$. Thus, expressing the numbers in terms of $7$, we get $a_{83} = (7-1)^{83}+(7+1)^{83}$. Applying the Binomial Theorem, half of our terms cancel out and we are left with $2(7^{83}+3403\\cdot7^{81}+\\cdots + 83\\cdot7)$. We realize that all of these terms are divisible by $49$ except the final term. After some quick division, our answer is $035 This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"Since $\\phi(49) = 42$ (see Euler's totient function), Euler's Totient Theorem tells us that $a^{42} \\equiv 1 \\pmod{49}$ where $\\text{gcd}(a,49) = 1$. Thus $6^{83} + 8^{83} \\equiv 6^{2(42)-1}+8^{2(42)-1}$ $\\equiv 6^{-1} + 8^{-1} \\equiv \\frac{8+6}{48}$ $\\equiv \\frac{14}{-1}\\equiv 035 This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"As the value of $a$ is obviously $6^{83}+8^{83}$ we look for a pattern with others. With a bit of digging, we discover that $6^n+6^m$ where $m$ and $n$ are odd is equal to $35 This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"As the value of $a$ is obviously $6^{83}+8^{83}$ we look for a pattern with others. With a bit of digging, we discover that $6^n+6^m$ where $m$ and $n$ are odd is equal to $35 This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"$6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\\ldots-8^{81}6+8^{82})$ Becuase $7|(6+8)$, we only consider $6^{82}-6^{81}8+\\ldots-8^{81}6+8^{82} \\pmod{7}$ $6^{82}-6^{81}8+\\ldots-8^{81}6+8^{82} \\equiv (-1)^{82} - (-1)^{81}+ \\ldots - (-1)^1 + 1 = 83 \\equiv 6 \\pmod{7}$ $6^{83} + 8^{83} \\equiv 14 \\cdot 6 \\equiv 035 This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"Repeat the steps of taking modulo $49$ after reducing the exponents over and over again until you get a residue of $49,$ namely $35.$ This bashing takes a lot of time but it isn’t too bad. ~peelybonehead",
"Repeat the steps of taking modulo $49$ after reducing the exponents over and over again until you get a residue of $49,$ namely $35.$ This bashing takes a lot of time but it isn’t too bad. ~peelybonehead"
] |
1983-I-7 | 1,983 | 7 | Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator? | 57 | null | [
"We can use complementary counting, by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$. Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place. We will place $A$, $B$, and $C$ with and without the restriction. There are $22$ places to put $A$, followed by $21$ places to put $B$, and $20$ places to put $C$ after $A$ and $B$. Hence, there are $22\\cdot21\\cdot20$ ways to place $A, B, C$ in between these people with restrictions. Without restrictions, there are $22$ places to put $A$, followed by $23$ places to put $B$, and $24$ places to put $C$ after $A$ and $B$. Hence, there are $22\\cdot23\\cdot24$ ways to place $A,B,C$ in between these people without restrictions. Thus, the desired probability is $1-\\frac{22\\cdot21\\cdot20}{22\\cdot23\\cdot24}=1-\\frac{420}{552}=1-\\frac{35}{46}=\\frac{11}{46}$, and the answer is $11+46=057.",
"There are $(25-1)! = 24!$ configurations for the knights about the table (since configurations that are derived from each other simply by a rotation are really the same, and should not be counted multiple times). There are ${3\\choose 2} = 3$ ways to pick a pair of knights from the trio, and there are $2! = 2$ ways to determine in which order they are seated. Since these two knights must be together, we let them be a single entity, so there are $(24-1)! = 23!$ configurations for the entities. However, this overcounts the instances in which the trio sits together; when all three knights sit together, then two of the pairs from the previous case are counted. However, we only want to count this as one case, so we need to subtract the number of instances in which the trio sits together (as a single entity). There are $3! = 6$ ways to determine their order, and there are $(23-1)! = 22!$ configurations. Thus, the required probability is $\\frac{2 \\times 3 \\times 23! - 6 \\times 22!}{24!} = \\frac{11}{46}$, and the answer is $057.",
"Number the knights around the table from $1$ to $25$. The total number of ways to pick the knights is \\[{25\\choose 3} = \\frac{25\\cdot24\\cdot23}{3\\cdot2\\cdot1} = 25\\cdot23\\cdot4\\] There are two possibilities: either all three sit next to each other, or two sit next to each other and one is not sitting next to the other two. Case 1: All three sit next to each other. In this case, you are picking $(1,2,3)$, $(2,3,4)$, $(3,4,5)$, $(4,5,6)$, ...,$(23,24,25)$, $(24,25,1)$, $(25,1,2)$. This makes $25$ combinations. Case 2: Like above, there are $25$ ways to pick the pair of knights sitting next to each other. Once a pair is picked, you cannot pick either of the two adjacent knights. (For example, if you pick $(5,6)$, you may not pick 4 or 7.) Thus, there are $25-4=21$ ways to pick the third knight, for a total of $25\\cdot21$ combinations. Thus, you have a total of $25 + (25\\cdot21) = 25\\cdot22$ allowable ways to pick the knights. The probability is $\\frac{25\\cdot22}{25\\cdot23\\cdot4} = \\frac{11}{46}$, and the answer is $057.",
"Pick an arbitrary spot for the first knight. Then pick spots for the next two knights in order. Case 1: The second knight sits next to the first knight. There are $2$ possible places for this out of $24$, so the probability of this is $\\frac{1}{12}$. We do not need to consider the third knight. Case 2: The second knight sits two spaces apart from the first knight. There are $2$ possible places for this out of $24$, so the probability is $\\frac{1}{12}$. Then there are $3$ places out of a remaining $23$ for the third knight to sit, so the total probability for this case is $\\frac{1}{12} \\times \\frac{3}{23}$. Case 3: The second knight sits three or more spaces apart from the first knight. There are $20$ possible places for this out of $24$, so the probability is $\\frac{5}{6}$. Then there are $4$ places to put the last knight out of $23$, so the total probability for this case is $\\frac{5}{6}\\times\\frac{4}{23}$. Now we add the probabilities to get the total: \\[\\frac{1}{12}+\\frac{1}{12}\\times\\frac{3}{23}+\\frac{5}{6}\\times\\frac{4}{23}=\\frac{1}{12}\\times\\frac{1}{23}\\left(23+3+40\\right)=\\frac{66}{12\\times 23}\\] \\[=\\frac{6\\times 11}{6\\times 2 \\times 23}=\\frac{11}{46}\\] so the answer is $057.",
"We simplify this problem by using complementary counting and fixing one knight in place. Then, either a knight can sit two spaces apart from the fixed knight, or a knight can sit more than two spaces apart from the fixed knight. The probability is then $\\frac{24\\left(23\\right)-\\left[2\\left(20\\right)+20\\left(19\\right)\\right]}{24\\left(23\\right)}=\\frac{11}{46}$, so the answer is $11+46=057.",
"Let $K_1, K_2, K_3$ be the knights in clockwise order. Let $A$ be the distance between $K_1$ and $K_2$, $B$ the distance between $K_2$ and $K_3$ and $C$ the distance between $K_3$ and $K_1$. $A + B + C = 25$ and $A, B, C \\geq 1$. In order to use stars and bars the numbers must be greater than or equal to 0 instead of 1, so we define $A_1 = A - 1, B_1 = B - 1, C_1 = C - 1$. $A_1 + B_1 + C_1 = 22$, so by stars and bars there are $\\binom{22 + 3 - 1}{3 - 1} = 276$ possibilities. The condition is not satisfied if $A, B, C \\geq 2$, so we can use complementary counting. Let $A_2 = A - 2, B_2 = B - 2, C_2 = C - 2$. $A_2 + B_2 + C_2 = 19$, and by stars and bars there are $\\binom{19 + 3 - 1}{3 - 1} = 210$ possibilities. This means there are $276 - 210 = 66$ possibilities where the condition is satisfied, so the probability is $\\frac{11}{46}$, resulting in $057.",
"Let $K_1, K_2, K_3$ be the knights in clockwise order. Let $A$ be the distance between $K_1$ and $K_2$, $B$ the distance between $K_2$ and $K_3$ and $C$ the distance between $K_3$ and $K_1$. $A + B + C = 25$ and $A, B, C \\geq 1$. In order to use stars and bars the numbers must be greater than or equal to 0 instead of 1, so we define $A_1 = A - 1, B_1 = B - 1, C_1 = C - 1$. $A_1 + B_1 + C_1 = 22$, so by stars and bars there are $\\binom{22 + 3 - 1}{3 - 1} = 276$ possibilities. The condition is not satisfied if $A, B, C \\geq 2$, so we can use complementary counting. Let $A_2 = A - 2, B_2 = B - 2, C_2 = C - 2$. $A_2 + B_2 + C_2 = 19$, and by stars and bars there are $\\binom{19 + 3 - 1}{3 - 1} = 210$ possibilities. This means there are $276 - 210 = 66$ possibilities where the condition is satisfied, so the probability is $\\frac{11}{46}$, resulting in $057.",
"There are $\\binom{25}{3}=\\frac{25 \\cdot 24 \\cdot 23}{3 \\cdot 2}=2300$ ways to chose $3$ knights out of $25$ knights. To ensure at least $2$ adjacent knights are chosen, first choose $1$ of the $25$ pairs of adjacent knights. After choosing the adjacent pair of knights, there are $23$ ways to choose the third knight. There are $25$ choices of $3$ knights sitting consecutively, which are counted twice. For example, choosing $(1,2)$ first, then choosing $3$ is the same as choosing $(2,3)$ first, then choosing $1$. Therefore, there are $25 \\cdot 23 -25=550$ ways to choose $3$ knights where at least $2$ of the $3$ had been sitting next to each other. \\[P=\\frac{550}{2300}=\\frac{11}{46}\\] The answer is $11+46=\\textbf{057} ~isabelchen",
"There are two possible scenarios satisfying this condition, one where two knights sit next to each other, while the third knight is lonely, and the other is where all three knights sit next to each other. On the numerator will be the total number of ways to position the knights, and on the denominator will be the number of ways to choose three knights. Denominator: $\\binom{25}{3}$ ways to choose three knights Numerator: $\\implies$ $Case_1:$ Two knights are next to each other, third is lonely :( $25$ ways to situate two knights, $21$ ways for third knight $\\implies$ $Case_2:$ All three knights are next to each other $25$ ways So, we have $P = \\frac{21 \\cdot 25+25}{\\binom{25}{3}}$ $= \\frac{22 \\cdot 25 \\cdot 6}{25 \\cdot 24 \\cdot 3}$ $= \\frac{11}{46}$, and our answer is $11+46=057 ~mineric (formatted shalomkeshet)"
] |
1983-I-8 | 1,983 | 8 | What is the largest $2$ -digit prime factor of the integer $n = {200\choose 100}$ ? | 61 | null | [
"Solution 1 Expanding the binomial coefficient, we get ${200 \\choose 100}=\\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \\le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such prime is $061 ~ Nafer",
"Expanding the binomial coefficient, we get ${200 \\choose 100}=\\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \\le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such prime is $061 ~ Nafer",
"The only way a single prime number will be left out after all the cancelation will have to satisfy the condition that out of all the multiples of the prime number we want to find, 2 of the multiples will have to be between 100 and 200, $61$ is good for this solution (if you know your primes) because $61 \\cdot 2$ and $61 \\cdot 3$ both lie in the interval $(100, 200)$. The reason for this is because if 2 copies of the prime $p$ we want to find are on the numerator and one copy on the denominator (AKA it contains $2p$ and also $\\dfrac{1}{p}$), there will only be 1 copy of the prime we want to find when all the cancellation is made when reducing the original fraction (only $p$ is left after cancellation). ~ $Michaellin16$ (formatted shalomkeshet) Solution 2: Clarification of Solution 1 We know that \\[{200\\choose100}=\\frac{200!}{100!100!}\\] Since $p<100$, there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\\frac{200!}{100!100!}$. (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,) So basically, $p$ is the largest prime number such that \\[\\left \\lfloor\\frac{200}{p}\\right \\rfloor>3\\] Since $p<\\frac{200}{3}=66.66...$, the largest prime value for $p$ is $p=61 ~ Nafer",
"The only way a single prime number will be left out after all the cancelation will have to satisfy the condition that out of all the multiples of the prime number we want to find, 2 of the multiples will have to be between 100 and 200, $61$ is good for this solution (if you know your primes) because $61 \\cdot 2$ and $61 \\cdot 3$ both lie in the interval $(100, 200)$. The reason for this is because if 2 copies of the prime $p$ we want to find are on the numerator and one copy on the denominator (AKA it contains $2p$ and also $\\dfrac{1}{p}$), there will only be 1 copy of the prime we want to find when all the cancellation is made when reducing the original fraction (only $p$ is left after cancellation). ~ $Michaellin16$ (formatted shalomkeshet) Solution 2: Clarification of Solution 1 We know that \\[{200\\choose100}=\\frac{200!}{100!100!}\\] Since $p<100$, there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\\frac{200!}{100!100!}$. (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,) So basically, $p$ is the largest prime number such that \\[\\left \\lfloor\\frac{200}{p}\\right \\rfloor>3\\] Since $p<\\frac{200}{3}=66.66...$, the largest prime value for $p$ is $p=61 ~ Nafer",
"We know that \\[{200\\choose100}=\\frac{200!}{100!100!}\\] Since $p<100$, there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\\frac{200!}{100!100!}$. (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,) So basically, $p$ is the largest prime number such that \\[\\left \\lfloor\\frac{200}{p}\\right \\rfloor>3\\] Since $p<\\frac{200}{3}=66.66...$, the largest prime value for $p$ is $p=61 ~ Nafer"
] |
1983-I-9 | 1,983 | 9 | Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$ . | 12 | null | [
"Let $y=x\\sin{x}$. We can rewrite the expression as $\\frac{9y^2+4}{y}=9y+\\frac{4}{y}$. Since $x>0$, and $\\sin{x}>0$ because $0< x<\\pi$, we have $y>0$. So we can apply AM-GM: \\[9y+\\frac{4}{y}\\ge 2\\sqrt{9y\\cdot\\frac{4}{y}}=12\\] The equality holds when $9y=\\frac{4}{y}\\Longleftrightarrow y^2=\\frac49\\Longleftrightarrow y=\\frac23$. Therefore, the minimum value is $012 is attainable by the Intermediate Value Theorem).",
"We can rewrite the numerator to be a perfect square by adding $-\\dfrac{12x \\sin x}{x \\sin x}$. Thus, we must also add back $12$. This results in $\\dfrac{(3x \\sin x-2)^2}{x \\sin x}+12$. Thus, if $3x \\sin x-2=0$, then the minimum is obviously $12$. We show this possible with the same methods in Solution 1; thus the answer is $012.",
"Let $y = x\\sin{x}$ and rewrite the expression as $f(y) = 9y + \\frac{4}{y}$, similar to the previous solution. To minimize $f(y)$, take the derivative of $f(y)$ and set it equal to zero. The derivative of $f(y)$, using the Power Rule, is $f'(y)$ = $9 - 4y^{-2}$ $f'(y)$ is zero only when $y = \\frac{2}{3}$ or $y = -\\frac{2}{3}$. It can further be verified that $\\frac{2}{3}$ and $-\\frac{2}{3}$ are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative $f''(y)=8y^{-3}$ is positive. However, since $x \\sin{x}$ is always positive in the given domain, $y = \\frac{2}{3}$. Therefore, $x\\sin{x}$ = $\\frac{2}{3}$, and the answer is $\\frac{(9)(\\frac{2}{3})^2 + 4}{\\frac{2}{3}} = 012.",
"Let $y = x\\sin{x}$ and rewrite the expression as $f(y) = 9y + \\frac{4}{y}$, similar to the previous solution. To minimize $f(y)$, take the derivative of $f(y)$ and set it equal to zero. The derivative of $f(y)$, using the Power Rule, is $f'(y)$ = $9 - 4y^{-2}$ $f'(y)$ is zero only when $y = \\frac{2}{3}$ or $y = -\\frac{2}{3}$. It can further be verified that $\\frac{2}{3}$ and $-\\frac{2}{3}$ are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative $f''(y)=8y^{-3}$ is positive. However, since $x \\sin{x}$ is always positive in the given domain, $y = \\frac{2}{3}$. Therefore, $x\\sin{x}$ = $\\frac{2}{3}$, and the answer is $\\frac{(9)(\\frac{2}{3})^2 + 4}{\\frac{2}{3}} = 012.",
"As above, let $y = x\\sin{x}$. Add $\\frac{12y}{y}$ to the expression and subtract $12$, giving $f(x) = \\frac{(3y+2)^2}{y} - 12$. Taking the derivative of $f(x)$ using the Chain Rule and Quotient Rule, we have $\\frac{\\text{d}f(x)}{\\text{d}x} = \\frac{6y(3y+2)-(3y+2)^2}{y^2}$. We find the minimum value by setting this to $0$. Simplifying, we have $6y(3y+2) = (3y+2)^2$ and $y = \\pm{\\frac{2}{3}} = x\\sin{x}$. Since both $x$ and $\\sin{x}$ are positive on the given interval, we can ignore the negative root. Plugging $y = \\frac{2}{3}$ into our expression for $f(x)$, we have $\\frac{(3(\\frac{2}{3})+2)^2}{y}-12 = \\frac{16}{\\left(\\frac{2}{3}\\right)}-12 = 012.",
"As above, let $y = x\\sin{x}$. Add $\\frac{12y}{y}$ to the expression and subtract $12$, giving $f(x) = \\frac{(3y+2)^2}{y} - 12$. Taking the derivative of $f(x)$ using the Chain Rule and Quotient Rule, we have $\\frac{\\text{d}f(x)}{\\text{d}x} = \\frac{6y(3y+2)-(3y+2)^2}{y^2}$. We find the minimum value by setting this to $0$. Simplifying, we have $6y(3y+2) = (3y+2)^2$ and $y = \\pm{\\frac{2}{3}} = x\\sin{x}$. Since both $x$ and $\\sin{x}$ are positive on the given interval, we can ignore the negative root. Plugging $y = \\frac{2}{3}$ into our expression for $f(x)$, we have $\\frac{(3(\\frac{2}{3})+2)^2}{y}-12 = \\frac{16}{\\left(\\frac{2}{3}\\right)}-12 = 012.",
"Set $\\frac{9x^2\\sin^2 x + 4}{x\\sin x}$ equal to $y$. Then multiply by $x\\sin x$ on both sides to get $9x^2\\sin^2 x + 4 = y\\cdot x\\sin x$. We then subtract $yx\\sin x$ from both sides to get $9x^2\\sin^2 x + 4 - yx\\sin x = 0$. This looks like a quadratic so set $z= x\\sin x$ and use quadratic equation on $9z^2 - yz + 4 = 0$ to see that $z = \\frac{y\\pm\\sqrt{y^2-144}}{18}$. We know that $y$ must be an integer and as small as it can be, so $y$ = 12. We plug this back in to see that $x\\sin x = \\frac{2}{3}$ which we can prove works using methods from solution 1. This makes the answer $012 -awesomediabrine",
"Seeing that we need to minimize, we think inequalities, and seeing squares, we think RMS-AM-GM-HM. From this inequality, we know that $\\sqrt{\\frac{(3x\\sin x)^2+2^2}{2}} \\geq \\sqrt{(3x\\sin x)(2)}$, with equality holding when $3x\\sin x=2$. From this inequality, we can see the following: \\begin{align*} \\sqrt{\\frac{(3x\\sin x)^2+2^2}{2}} \\geq \\sqrt{(3x\\sin x)(2)} \\\\ \\frac{9x^2\\sin^2x+4}{2} \\geq 6x\\sin x \\\\ \\frac{9x^2\\sin^2x+4}{x\\sin x} \\geq 12 \\end{align*} We can prove that the equality condition is possible as in Solution $1$. Thus, our answer is $012."
] |
1983-I-10 | 1,983 | 10 | The numbers $1447$ , $1005$ and $1231$ have something in common: each is a $4$ -digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | 432 | null | [
"Suppose that the two identical digits are both $1$. Since the thousands digit must be $1$, only one of the other three digits can be $1$. This means the possible forms for the number are $11xy,\\qquad 1x1y,\\qquad1xy1$ Because the number must have exactly two identical digits, $x\\neq y$, $x\\neq1$, and $y\\neq1$. Hence, there are $3\\cdot9\\cdot8=216$ numbers of this form. Now suppose that the two identical digits are not $1$. Reasoning similarly to before, we have the following possibilities: $1xxy,\\qquad1xyx,\\qquad1yxx.$ Again, $x\\neq y$, $x\\neq 1$, and $y\\neq 1$. There are $3\\cdot9\\cdot8=216$ numbers of this form. Thus the answer is $216+216=432.",
"Consider a sequence of $4$ digits instead of a $4$-digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\\frac{1}{10}$. This means we can find all possible sequences with one digit repeated twice, and then divide by $10$. If we let the three distinct digits of the sequence be $a, b,$ and $c$, with $a$ repeated twice, we can make a table with all possible sequences: \\[\\begin{tabular}{ccc} aabc & abac & abca \\\\ baac & baca & \\\\ bcaa && \\\\ \\end{tabular}\\] There are $6$ possible sequences. Next, we can see how many ways we can pick $a$, $b$, and $c$. This is $10(9)(8) = 720$, because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $432 as our answer.",
"We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated. Case 1: There are $9$ choices for the hundreds digit (it cannot be $1$), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\\cdot8\\cdot7=504$ numbers. Case 2: One of the three numbers must be $1$, and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$, and there are $9$ choices (any number except for $1$) to pick the other number that is repeated, so a total of $3\\cdot9=27$ numbers. Case 3: We will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times. When $1$ is repeated $3$ times, then one of the digits is not $1$. There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\\cdot3=27$ numbers. When a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number. So in Case 3 there are $27+9=36$ numbers Case 4: There is only $1$ number: $1111$. There are a total of $1000$ $4$-digit numbers that begin with $1$ (from $1000$ to $1999$), so by complementary counting you get $1000-(504+27+36+1)=432 numbers. Solution by Kinglogic",
"We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated. Case 1: There are $9$ choices for the hundreds digit (it cannot be $1$), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\\cdot8\\cdot7=504$ numbers. Case 2: One of the three numbers must be $1$, and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$, and there are $9$ choices (any number except for $1$) to pick the other number that is repeated, so a total of $3\\cdot9=27$ numbers. Case 3: We will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times. When $1$ is repeated $3$ times, then one of the digits is not $1$. There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\\cdot3=27$ numbers. When a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number. So in Case 3 there are $27+9=36$ numbers Case 4: There is only $1$ number: $1111$. There are a total of $1000$ $4$-digit numbers that begin with $1$ (from $1000$ to $1999$), so by complementary counting you get $1000-(504+27+36+1)=432 numbers. Solution by Kinglogic",
"Let us proceed by casework. Case 1: We will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$, which is $9$ options. For the last digit, their are $8$ possibilities since it can be neither $1$ or the other number that is chosen. The final outcome is ${3}\\choose{2}$ $* 9 * 8 = 216$ possibilities for this case. Case 2: The last case will be the amount of numbers that have two identical digits thare are $1$. There are ${3}\\choose{1}$ places to pick the $1$. For the other $2$ digits, there are $9$ and $8$ options respectively. Thus, we have $3 * 9 * 8 = 216$. Summing the two cases, we get $216 + 216 = 432.",
"Let us proceed by casework. Case 1: We will count the amount of numbers that have two identical digits that are not one. The thousands digit is fixed, and we are choosing two spots to hold two identical digits that are chosen from $0, 2-9$, which is $9$ options. For the last digit, their are $8$ possibilities since it can be neither $1$ or the other number that is chosen. The final outcome is ${3}\\choose{2}$ $* 9 * 8 = 216$ possibilities for this case. Case 2: The last case will be the amount of numbers that have two identical digits thare are $1$. There are ${3}\\choose{1}$ places to pick the $1$. For the other $2$ digits, there are $9$ and $8$ options respectively. Thus, we have $3 * 9 * 8 = 216$. Summing the two cases, we get $216 + 216 = 432."
] |
1983-I-11 | 1,983 | 11 | The solid shown has a square base of side length $s$ . The upper edge is parallel to the base and has length $2s$ . All other edges have length $s$ . Given that $s=6\sqrt{2}$ , what is the volume of the solid? [asy] import three; size(170); pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); label("A",A, S); label("B",B, S); label("C",C, S); label("D",D, S); label("E",E,N); label("F",F,N); [/asy] | 288 | null | [
"Solution 1 First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$. [asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); label(\"$12\\sqrt{2}$\",(E+F)/2,N); label(\"$6\\sqrt{2}$\",(A+B)/2,S); label(\"6\",(3*s/2,s/2,3),ENE); [/asy] Next, we complete t he figure into a triangular prism, and find its volume, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=288. ~skibidi solver",
"First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$, and one of the legs is $3\\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$. [asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; pen d = linewidth(0.65); pen l = linewidth(0.5); currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(B--Ba--Ca--C,dashed+d); draw(A--Aa--Da--D,dashed+d); draw(E--(E.x,E.y,0),dashed+l); draw(F--(F.x,F.y,0),dashed+l); draw(Aa--E--Da,dashed+d); draw(Ba--F--Ca,dashed+d); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,S); label(\"D\",D,NE); label(\"E\",E,N); label(\"F\",F,N); label(\"$12\\sqrt{2}$\",(E+F)/2,N); label(\"$6\\sqrt{2}$\",(A+B)/2,S); label(\"6\",(3*s/2,s/2,3),ENE); [/asy] Next, we complete t he figure into a triangular prism, and find its volume, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$. Now, we subtract off the two extra pyramids that we included, whose combined volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=288. ~skibidi solver",
"[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label(\"A\",A,(-1,-1,0)); label(\"B\",B,( 2,-1,0)); label(\"C\",C,( 1, 1,0)); label(\"D\",D,(-1, 1,0)); label(\"E\",E,(0,0,1)); label(\"F\",F,(0,0,1)); label(\"G\",G,(0,0,-1)); label(\"H\",H,(0,0,-1)); [/asy] Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = 288. ~skibidi solver",
"We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\\sqrt{2}-\\sqrt{2}h$ since it decreases linearly with respect to $h$, and $l=6\\sqrt{2}+\\sqrt{2}h$ since it similarly increases linearly with respect to $h$. Now we solve:\\[\\int_0^6(6\\sqrt{2}-\\sqrt{2}h)(6\\sqrt{2}+\\sqrt{2}h)\\ \\text{d}h =\\int_0^6(72-2h^2)\\ \\text{d}h=72(6)-2\\left(\\frac{1}{3}\\right)\\left(6^3\\right)=288. ~skibidi solver",
"Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \\sqrt{6}$, and that little portion that hangs out has a length of $3\\sqrt2$. This is a triangular pyramid with a base of $3\\sqrt6, 3\\sqrt6, 3\\sqrt2$, and a height of $3\\sqrt{2}$. Since there are two of these, we can compute the sum of the volumes of these two to be $72$. Now we are left with a triangular prism with a base of dimensions $3\\sqrt6, 3\\sqrt6, 3\\sqrt2$ and a height of $6\\sqrt2$. We can compute the volume of this to be 216, and thus our answer is $288. ~skibidi solver",
"From solution 1, the height of the solid is $6$. Construct a triangular prism with base ABCD, and with the height of the solid. The volume of this triangular prism is $( 6\\sqrt{2} )^2 \\cdot 6 \\cdot 1/2 = 216$ Notice that the solid is symetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a regular tetrahedron. This tetrahedron has sidelength of $6\\sqrt{2}$ and since the formula for the volume of a regular tetrahedron is $\\frac{s^3}{6\\sqrt{2}}$ the volume of this tetrahedron is $72$. $216 + 72 = 288. ~skibidi solver"
] |
1983-I-12 | 1,983 | 12 | Diameter $AB$ of a circle has length a $2$ -digit integer (base ten). Reversing the digits gives the length of the perpendicular chord $CD$ . The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$ . Pdfresizer.com-pdf-convert-aimeq12.png | 65 | null | [
"Let $AB=10x+y$ and $CD=10y+x$. It follows that $CO=\\frac{AB}{2}=\\frac{10x+y}{2}$ and $CH=\\frac{CD}{2}=\\frac{10y+x}{2}$. Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on $2CO$, $2OH$ and $2CH$, we deduce \\[(2OH)^2=\\left(10x+y\\right)^2-\\left(10y+x\\right)^2=99(x+y)(x-y)\\] Because $OH$ is a positive rational number and $x$ and $y$ are integral, the quantity $99(x+y)(x-y)$ must be a perfect square. Hence either $x-y$ or $x+y$ must be a multiple of $11$, but as $x$ and $y$ are different digits, $1+0=1 \\leq x+y \\leq 9+9=18$, so the only possible multiple of $11$ is $11$ itself. However, $x-y$ cannot be 11, because both must be digits. Therefore, $x+y$ must equal $11$ and $x-y$ must be a perfect square. The only pair $(x,y)$ that satisfies this condition is $(6,5)$, so our answer is $065.)"
] |
1983-I-13 | 1,983 | 13 | For $\{1, 2, 3, \ldots, n\}$ and each of its nonempty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract succesive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+1=5$ and for $\{5\}$ it is simply $5$ . Find the sum of all such alternating sums for $n=7$ . | 448 | null | [
"Solution 1 Let $S$ be a non- empty subset of $\\{1,2,3,4,5,6\\}$. Then the alternating sum of $S$, plus the alternating sum of $S \\cup \\{7\\}$, is $7$. This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S\\cup \\{7\\}$. Because there are $2^{6}=64$ of these pairs of sets, the sum of all possible subsets of our given set is $64 \\cdot 7$, giving an answer of $448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Let $S$ be a non- empty subset of $\\{1,2,3,4,5,6\\}$. Then the alternating sum of $S$, plus the alternating sum of $S \\cup \\{7\\}$, is $7$. This is because, since $7$ is the largest element, when we take an alternating sum, each number in $S$ ends up with the opposite sign of each corresponding element of $S\\cup \\{7\\}$. Because there are $2^{6}=64$ of these pairs of sets, the sum of all possible subsets of our given set is $64 \\cdot 7$, giving an answer of $448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Consider a given subset $T$ of $S$ that contains $7$; then there is a subset $T'$ which contains all the elements of $T$ except for $7$, and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$, their signs are opposite. Thus the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 \\cdot 2^6 = 448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Consider a given subset $T$ of $S$ that contains $7$; then there is a subset $T'$ which contains all the elements of $T$ except for $7$, and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$, their signs are opposite. Thus the sum of the alternating sums of $T$ and $T'$ is equal to 7. There are $2^6$ subsets containing 7, so our answer is $7 \\cdot 2^6 = 448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Denote the desired total of all alternating sums of an $n$-element set as $S_n$. We are looking for $S_7$. Notice that all alternating sums of an $n$-element set are also alternating sums of an $n+1$-element set. However, when we go from an $n$ to $n+1$ element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the $n$-element set. There are $2^n$ subsets of an $n+1$-element set that includes the new element, giving us the relationship $S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)$. When $n = 6$, we therefore get $S_ 7 = 2^6(7) = 448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of $7 \\cdot 64=448$. If we look at when 6 appears, which it also does 64 times, whether it comes as positive or negative depends on the presence of 7. Half of the subsets with 6 have 7 resulting in subtracting 6 each time, while the other half does not have 7 adding 6 each time, so these contributions of sixes cancel each other out giving a net gain of 0. The same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because $0=(1-1)^{n}=\\binom{n}{0}-\\binom{n}{1}+\\binom{n}{2}-...\\binom{n}{n}$ via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be $448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Let $\\mathbb{N}_n := \\{1, 2, 3, \\dots n\\}$. Let the alternating sum of a certain subset of $S$ of $\\mathbb{N}_n$ be $\\xi(S),$ and let \\[\\mathcal{A}(\\mathbb{N}_n) := \\sum_{S \\subseteq \\mathbb{N}_n} \\xi(S).\\] We see that \\[\\mathcal{A}(\\mathbb{N}_n) = \\sum_{S \\subseteq \\mathbb{N}_n} \\xi(S) = \\sum_{n \\in S, S \\subseteq \\mathbb{N}_n} \\xi(S) + \\sum_{S \\subseteq \\mathbb{N}_{n-1}} \\xi(S) = \\mathcal{A}(\\mathbb{N}_{n-1}) + \\sum_{n \\in S, S \\subseteq \\mathbb{N}_n} \\left( n - \\xi(S - \\{n\\}) \\right),\\] as if $n \\in S,$ $n$ is the largest element in $S.$ Now, we know that \\[\\sum_{n \\in S, S \\subseteq \\mathbb{N}_n} n - \\xi(S - \\{n\\}) = \\sum_{S \\subseteq \\mathbb{N}_{n-1}} n - \\xi(S) = n \\cdot 2^{n-1} - \\sum_{S \\subseteq \\mathbb{N}_{n-1}} \\xi(S) = n \\cdot 2^{n-1} - \\mathcal{A}(\\mathbb{N}_{n-1}),\\] so \\[\\mathcal{A}(\\mathbb{N}_{n}) = n \\cdot 2^{n-1}.\\] Thus, our answer (which is the $n = 7$ case) is $\\mathcal{A}(\\mathbb{N}_{7}) = 7 \\cdot 2^6 = 448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Computing the first few $S_n$ gives, $S_1$ $=$ $1$, $S_2$ $=$ $4$, $S_3$ $=$ $12$, and $S_4$ $=$ $32$. From this, we think the formula for $S_n$ must be an exponential function in terms of $2^n$ because the amount of subsets that a set with $n$ elements has is $2^n$ and this is a question about subsets so we try $S_n$ = $2^n(x)$. Trying this formula for the first few $S_n$, gives away $x$ to be $n/2$ or more simply $S_n$ = $2^{n-1}(n)$ (try proving this by induction) . Substuiting $n = 7$ gives the requested answer of $S_ 7 = 2^6(7) = 448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian",
"Computing the first few $S_n$ gives, $S_1$ $=$ $1$, $S_2$ $=$ $4$, $S_3$ $=$ $12$, and $S_4$ $=$ $32$. From this, we think the formula for $S_n$ must be an exponential function in terms of $2^n$ because the amount of subsets that a set with $n$ elements has is $2^n$ and this is a question about subsets so we try $S_n$ = $2^n(x)$. Trying this formula for the first few $S_n$, gives away $x$ to be $n/2$ or more simply $S_n$ = $2^{n-1}(n)$ (try proving this by induction) . Substuiting $n = 7$ gives the requested answer of $S_ 7 = 2^6(7) = 448. Sidenote This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you would like to see why the formula holds, please observe Solution 3. -ThemathCanadian"
] |
1983-I-14 | 1,983 | 14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$ . [asy]size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); draw(O2--O1); dot(O1); dot(O2); draw(Q--R); label("$Q$",Q,NW); label("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);[/asy] | 130 | null | [
"Firstly, notice that if we reflect $R$ over $P$, we get $Q$. Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$, we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$, and with radius $6$) that intersects circle $A$ at $Q$. The rest is just finding lengths, as follows. B is reflected like so Since $P$ is the midpoint of segment $BC$, $AP$ is a median of $\\triangle ABC$. Because we know $AB=12$, $BP=PC=6$, and $AP=8$, we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get $AC = \\sqrt{56}$. The Kite is formed Now we have a kite $AQCP$ with $AQ=AP=8$, $CQ=CP=6$, and diagonal $AC=\\sqrt{56}$, and all we need is the length of the other diagonal $PQ$. The easiest way it can be found is with the Pythagorean Theorem. Let $2x$ be the length of $PQ$. Then \\[\\sqrt{36-x^2} + \\sqrt{64-x^2} = \\sqrt{56}.\\] Solving this equation, we find that $x^2=\\frac{65}{2}$, so $PQ^2 = 4x^2 = 130",
"[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"$A$\",a,S); label(\"$B$\",b,S); label(\"$M$\",m,NE); label(\"$N$\",n,NE); label(\"$P$\",p,N); label(\"$Q$\",q,NW); label(\"$R$\",r,E); label(\"$12$\",(14,0),SW); label(\"$6$\",(23,0),S); [/asy] Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=130 ~happypi31415",
"[asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label(\"$A$\",a,S); label(\"$B$\",b,S); label(\"$M$\",m,NE); label(\"$N$\",n,NE); label(\"$P$\",p,N); label(\"$Q$\",q,NW); label(\"$R$\",r,E); label(\"$12$\",(14,0),SW); label(\"$6$\",(23,0),S); [/asy] Draw additional lines as indicated. Note that since triangles $AQP$ and $BPR$ are isosceles, the altitudes are also bisectors, so let $QM=MP=PN=NR=x$. Since $\\frac{AR}{MR}=\\frac{BR}{NR},$ triangles $BNR$ and $AMR$ are similar. If we let $y=BN$, we have $AM=3BN=3y$. Applying the Pythagorean Theorem on triangle $BNR$, we have $x^2+y^2=36$. Similarly, for triangle $QMA$, we have $x^2+9y^2=64$. Subtracting, $8y^2=28\\Rightarrow y^2=\\frac72\\Rightarrow x^2=\\frac{65}2\\Rightarrow QP^2=4x^2=130 ~happypi31415",
"Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}\\left(\\frac{x}{16}\\right)$ and $\\cos^{-1}\\left(\\frac{x}{12}\\right)$. So we have $\\cos^{-1}\\left(\\frac{x}{16}\\right)+\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)=180^{\\circ}-\\cos^{-1}\\left(\\frac{x}{12}\\right).$ Taking the cosine of both sides, and simplifying using the addition formula for $\\cos$ as well as the identity $\\sin^{2}{x} + \\cos^{2}{x} = 1$, gives $x^2=130 ~happypi31415",
"Let $QP=PR=x$. Angles $QPA$, $APB$, and $BPR$ must add up to $180^{\\circ}$. By the Law of Cosines, $\\angle APB=\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)$. Also, angles $QPA$ and $BPR$ equal $\\cos^{-1}\\left(\\frac{x}{16}\\right)$ and $\\cos^{-1}\\left(\\frac{x}{12}\\right)$. So we have $\\cos^{-1}\\left(\\frac{x}{16}\\right)+\\cos^{-1}\\left(\\frac{{-11}}{24}\\right)=180^{\\circ}-\\cos^{-1}\\left(\\frac{x}{12}\\right).$ Taking the cosine of both sides, and simplifying using the addition formula for $\\cos$ as well as the identity $\\sin^{2}{x} + \\cos^{2}{x} = 1$, gives $x^2=130 ~happypi31415",
"Let $QP = PR = x$. Extend the line containing the centers of the two circles to meet $R$, and to meet the other side of the large circle at a point $S$. The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$. The length of the diameter of the larger circle is $16$. Thus by Power of a Point in the circle passing through $Q$, $R$, and $S$, we have $x \\cdot 2x = 10 \\cdot (10+16) = 260$, so $x^2 = 130 ~happypi31415",
"Let $QP = PR = x$. Extend the line containing the centers of the two circles to meet $R$, and to meet the other side of the large circle at a point $S$. The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$. The length of the diameter of the larger circle is $16$. Thus by Power of a Point in the circle passing through $Q$, $R$, and $S$, we have $x \\cdot 2x = 10 \\cdot (10+16) = 260$, so $x^2 = 130 ~happypi31415",
"[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label(\"$A$\",a,S); label(\"$B$\",b,S); label(\"$M$\",m,NE); label(\"$N$\",n,NE); label(\"$P$\",p,N); label(\"$Q$\",q,NW); label(\"$R$\",r,E); label(\"$12$\",(14,0),SW); label(\"$T$\", t , NW); [/asy] Note that the midpoint of $AB$ is $T.$ Also, since $AM,NB$ bisect $QP$ and $PR,$ respectively, $P$ is the midpoint of $MN.$ Thus, $AM+NB=2PT.$ let $AM=a,BN=b.$ This means that $a+b=2PT.$ Using Stewart's Theorem on $\\Delta APB,$ the median $PT$ has length $\\sqrt{14}.$ Thus, $a+b=2\\sqrt{14}.$ Also, since $MP=PN$, from the Pythagorean Theorem, $8^2-a^2=6^2-b^2\\implies a^2-b^2=28.$ Thus, $a-b=\\frac{28}{2\\sqrt{14}}=\\sqrt{14}.$ We conclude that $QP=MN=\\sqrt{12^2-(a-b)^2}=\\sqrt{130}\\implies130 ~happypi31415",
"[asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label(\"$A$\",a,S); label(\"$B$\",b,S); label(\"$M$\",m,NE); label(\"$N$\",n,NE); label(\"$P$\",p,N); label(\"$Q$\",q,NW); label(\"$R$\",r,E); label(\"$12$\",(14,0),SW); label(\"$T$\", t , NW); [/asy] Note that the midpoint of $AB$ is $T.$ Also, since $AM,NB$ bisect $QP$ and $PR,$ respectively, $P$ is the midpoint of $MN.$ Thus, $AM+NB=2PT.$ let $AM=a,BN=b.$ This means that $a+b=2PT.$ Using Stewart's Theorem on $\\Delta APB,$ the median $PT$ has length $\\sqrt{14}.$ Thus, $a+b=2\\sqrt{14}.$ Also, since $MP=PN$, from the Pythagorean Theorem, $8^2-a^2=6^2-b^2\\implies a^2-b^2=28.$ Thus, $a-b=\\frac{28}{2\\sqrt{14}}=\\sqrt{14}.$ We conclude that $QP=MN=\\sqrt{12^2-(a-b)^2}=\\sqrt{130}\\implies130 ~happypi31415",
"Looking at Drawing 2 (by the way, we don't need point $R$), we set $AM=a$ and $BN=b$, and the desired length$QP=x=PR$. We know that a radius perpendicular to a chord bisects the chord, so $MP=\\frac{x}{2}$ and $PN=\\frac{x}{2}$. Draw line $AP$ and $PB$, and we see that they are radii of Circles $A$ and $B$, respectively. We can write the Pythagorean relationships $a^2+\\left(\\frac{x}{2}\\right)^2=8^2$ for triangle $AMP$ and $b^2+\\left(\\frac{x}{2}\\right)^2=6^2$ for triangle $BNP$. We also translate segment $MN$ down so that $N$ coincides with $B$, and form another right triangle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$, the longer leg is the same as $MN=x$, and the hypotenuse is $AB=12$. We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$. Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$, and solve), you find that $x=\\sqrt{130}$, so $x^2 = 130 ~happypi31415",
"Looking at Drawing 2 (by the way, we don't need point $R$), we set $AM=a$ and $BN=b$, and the desired length$QP=x=PR$. We know that a radius perpendicular to a chord bisects the chord, so $MP=\\frac{x}{2}$ and $PN=\\frac{x}{2}$. Draw line $AP$ and $PB$, and we see that they are radii of Circles $A$ and $B$, respectively. We can write the Pythagorean relationships $a^2+\\left(\\frac{x}{2}\\right)^2=8^2$ for triangle $AMP$ and $b^2+\\left(\\frac{x}{2}\\right)^2=6^2$ for triangle $BNP$. We also translate segment $MN$ down so that $N$ coincides with $B$, and form another right triangle. From that triangle, you can see that the shorter leg is on the left side, having length $a-b$, the longer leg is the same as $MN=x$, and the hypotenuse is $AB=12$. We can write the Pythagorean relationship $(a-b)^2+x^2=12^2$. Solving the system of 3 unknowns and 3 equations (One of the best ways to do it is to solve for $a$ in the first equation and $b$ in the second equation, and substitute into the third equation, get an equation only in terms of $x$, and solve), you find that $x=\\sqrt{130}$, so $x^2 = 130 ~happypi31415",
"The centers are collinear, you can prove it (but that is already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from $P$ to $AB.$ You then have 2 separate segments, separated by the foot of the altitude of $P$. Call them $a$ and $b$ respectively. Call the measure of the foot of the altitude of $P$ $h$. You then have 3 equations: \\[(1)\\quad a+b=12\\] (this is given by the fact that the distance between the centers is 12. \\[(2)\\quad a^2+h^2=64\\]. This is given by the fact that P is on the circle with radius 8. \\[(3)\\quad b^2+h^2=36\\]. This is given by the fact that P is on the circle with radius 6. Subtract (3) from (2) to get that $a^2-b^2=28$. As per (1), then you have $a-b=\\frac{7}{3}$ (4). Add (1) and (4) to get that $2a=\\frac{43}{3}$. Then substitute into (1) to get $b=\\frac{29}{6}$. Substitute either a or b into (2) or (3) to get $h=\\sqrt{455}{6}$. Then to get $PQ=PR$ it is just $\\sqrt{(b+6)^2+h^2}=\\sqrt{\\frac{65^2}{6^2}+\\frac{455}{6^2}}=\\sqrt{\\frac{4680}{36}}=\\sqrt{130}$. $PQ^2=130 ~happypi31415",
"We use coordinate geometry to approach this problem. Let the center of larger circle be the origin $O_1$, the smaller circle be $O_2$, and the x-axis be $O_1O_2$. Hence, we can get the two circle equations: $x^2+y^2 = 64$ and $(x-12)^2+y^2=36$. Let point $P$ be $(a, b)$. Noting that it lies on both circles, we can plug the coordinates into both equations: $a^2 +b^2 = 64$ $(a-12)^2+ b^2 \\Rightarrow a^2-24a+144+b^2 = 64$ Substituting $a^2+b^2 = 64$ into equation 2 and solving for $a$, we get $a = \\frac{43}{6}$. The problem asks us to find $QP^2$, which is congruent to $PR^2$. Using the distance formula for $P(a, b)$ and $R(18, 0)$ (by Solution 7's collinear proof), we get $PR^2 = (18-a)^2 +b^2$. Using $a^2+b^2 = 64$, we find that $b^2 = \\frac{455}{36}$. Plugging the variables $a$ and $b^2$ in, we get $PR^2 = QP^2 = 130 ~happypi31415",
"We use coordinate geometry to approach this problem. Let the center of larger circle be the origin $O_1$, the smaller circle be $O_2$, and the x-axis be $O_1O_2$. Hence, we can get the two circle equations: $x^2+y^2 = 64$ and $(x-12)^2+y^2=36$. Let point $P$ be $(a, b)$. Noting that it lies on both circles, we can plug the coordinates into both equations: $a^2 +b^2 = 64$ $(a-12)^2+ b^2 \\Rightarrow a^2-24a+144+b^2 = 64$ Substituting $a^2+b^2 = 64$ into equation 2 and solving for $a$, we get $a = \\frac{43}{6}$. The problem asks us to find $QP^2$, which is congruent to $PR^2$. Using the distance formula for $P(a, b)$ and $R(18, 0)$ (by Solution 7's collinear proof), we get $PR^2 = (18-a)^2 +b^2$. Using $a^2+b^2 = 64$, we find that $b^2 = \\frac{455}{36}$. Plugging the variables $a$ and $b^2$ in, we get $PR^2 = QP^2 = 130 ~happypi31415",
"Let the center of the circle with radius $8$ be $A,$ and let the center of the one with radius $6$ be $B.$ Also, let $QP = PR = x.$ Using law of cosines on triangle $\\Delta APB,$ we have that $\\cos{\\angle{APB}} = -\\frac{{11}}{24}.$ Angle chasing gives that $\\angle{QAR} = \\angle{APB},$ so their cosines are the same. Applying law of cosines again on triangle $\\Delta QAR,$ we have $\\left(2x^2\\right)=64+324-2(8)(18)\\left(-\\frac{11}{24}\\right),$ which gives that $x^2 = 130 ~happypi31415",
"Let the center of the circle with radius $8$ be $A,$ and let the center of the one with radius $6$ be $B.$ Also, let $QP = PR = x.$ Using law of cosines on triangle $\\Delta APB,$ we have that $\\cos{\\angle{APB}} = -\\frac{{11}}{24}.$ Angle chasing gives that $\\angle{QAR} = \\angle{APB},$ so their cosines are the same. Applying law of cosines again on triangle $\\Delta QAR,$ we have $\\left(2x^2\\right)=64+324-2(8)(18)\\left(-\\frac{11}{24}\\right),$ which gives that $x^2 = 130 ~happypi31415",
"If you call X the second intersection of the two circles and the centers $O_1$, $O_2$ respectively, note that triangles $O_1 X O_2$ and $Q X R$ are similar by Spiral Sym. $P$ is the midpoint of segment $Q R$. If the midpoint of $O_1 O_2$ is $M$, then $\\frac{X M}{O_1 M} = \\frac{X P}{Q P}$. By Appolonius Median Length theorem, $X M = \\sqrt{14}$. Note that $X P$ is simply two times the height from X to $O_1O_2$, and as a result, by Heron's formula, $X P = \\frac{\\sqrt{(7)(13)(5)}}{3}$, and from our ratio, $Q P = \\sqrt{130}$. As a result, the square is $130$, and we are done. - sepehr2010",
"If you call X the second intersection of the two circles and the centers $O_1$, $O_2$ respectively, note that triangles $O_1 X O_2$ and $Q X R$ are similar by Spiral Sym. $P$ is the midpoint of segment $Q R$. If the midpoint of $O_1 O_2$ is $M$, then $\\frac{X M}{O_1 M} = \\frac{X P}{Q P}$. By Appolonius Median Length theorem, $X M = \\sqrt{14}$. Note that $X P$ is simply two times the height from X to $O_1O_2$, and as a result, by Heron's formula, $X P = \\frac{\\sqrt{(7)(13)(5)}}{3}$, and from our ratio, $Q P = \\sqrt{130}$. As a result, the square is $130$, and we are done. - sepehr2010"
] |
1983-I-15 | 1,983 | 15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ ? [asy]size(140); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy] | 175 | null | [
"Solution 1 As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\\ell$? The answer to this question is: a line $m$ parallel to $\\ell$, such that $m$ and $P$ are (1) on opposite sides of $\\ell$. and (2) at the same distance from $\\ell$. [asy] size(170); pair P = (0,0), L1 = (-3, 0), L2 = (1.5, 1.5), M1 = (-3, 1), M2 = (1.5, 2.5); pair X = .6*M1 + .4*M2, M = (P+X)/2; draw(L1--L2); draw(M1--M2); draw(P--X, dotted); dot(M); dot(\"$P$\", P, S); dot(\"$X$\", X, N); label(\"$\\ell$\", L2, E); label(\"$m$\", M2, E); [/asy] Applied to this problem, this means that $D$ is the only point that lies on both (1) the given circle, and (2) the line through $D$ parallel to $BC$. This means that $BC$ is parallel to the tangent to the given circle at $D$. If we take $O$ to be the center of the given circle, then this means that $OD$ is perpendicular to $BC$. Let $M$ be the midpoint of chord $BC,$ and let $N$ be the intersection of $OD$ with the line through $A$ parallel to $BC$. [asy] size(170); pair O = (0,0), D = (0, 5), B = (-3, 4), C = (3, 4), A = (-4, 3), EE = (4, 3); pair M = (B+C)/2, NN = (A+EE)/2; draw(circle(O, 5)); draw(O--D); draw(B--C); draw(A--EE); draw(B--O--A); dot(\"$O$\", O, SE); label(\"$D$\", D, N); label(\"$B$\", B, WNW); label(\"$A$\", A, WNW); label(\"$C$\", C, ENE); label(\"$M$\", M, NE); label(\"$N$\", NN, SE);[/asy] Since $BC = 6,$ we know that $BM = 3$; since $OB$ (a radius of the circle) is 5, we can conclude that $\\triangle BMO$ is a 3-4-5 right triangle. Since $D$ and line $AN$ are equidistant from line $BC,$ we know that $MN = 1$, and thus $ON = 3$. This makes $\\triangle ANO$ also a 3-4-5 right triangle. We're looking for $\\sin \\angle AOB$, and we can find that using the angle subtraction formula for sine. We have \\begin{align*} \\sin \\angle AOB &= \\sin(\\angle AOM - \\angle BOM) \\\\ &= \\sin \\angle AOM \\cos \\angle BOM - \\cos \\angle AOM \\sin \\angle BOM \\\\ &= \\frac{4}{5} \\cdot \\frac{4}{5} - \\frac{3}{5} \\cdot \\frac{3}{5} \\\\ &= \\frac{16 - 9}{25} = \\frac{7}{25}. \\end{align*} This is in lowest terms, so our answer is $mn = 7 \\cdot 25 = 175$. Solution 2 -Credit to Adamz for diagram- [asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label(\"$O$\",O,SW); pair M = (4,0);dot(M);label(\"$M$\",M,SE); pair N = (4,2);dot(N);label(\"$N$\",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label(\"$B$\",B,NE); pair C = (4,-3);dot(C);label(\"$C$\",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label(\"$P$\",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label(\"$A$\",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label(\"$Q$\",Q,S); pair R = (3,0); dot(R); label(\"$R$\",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label(\"$D$\",D,E); draw(A--D); [/asy]Let $A$ be any fixed point on circle $O$, and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AO$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to $BC$ at point $N$. Let $M$ be the midpoint of the chord $BC$. From right triangle $OMB$, we have $OM = \\sqrt{OB^2 - BM^2} =4$. This gives $\\tan \\angle BOM = \\frac{BM}{OM} = \\frac 3 4$. Notice that the distance $OM$ equals $PN + PO \\cos \\angle AOM = r(1 + \\cos \\angle AOM)$, where $r$ is the radius of circle $P$. Hence \\[\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5\\] (where $R$ represents the radius, $5$, of the large circle given in the question). Therefore, since $\\angle AOM$ is clearly acute, we see that \\[\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3\\] Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the subtraction formula for $\\tan$ to obtain \\[\\tan \\angle AOB =\\frac{\\tan \\angle AOM - \\tan \\angle BOM}{1 + \\tan \\angle AOM \\cdot \\tan \\angle BOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}\\] It follows that $\\sin \\angle AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, such that the answer is $7 \\cdot 25=175 ~skibidi solver",
"As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\\ell$? The answer to this question is: a line $m$ parallel to $\\ell$, such that $m$ and $P$ are (1) on opposite sides of $\\ell$. and (2) at the same distance from $\\ell$. [asy] size(170); pair P = (0,0), L1 = (-3, 0), L2 = (1.5, 1.5), M1 = (-3, 1), M2 = (1.5, 2.5); pair X = .6*M1 + .4*M2, M = (P+X)/2; draw(L1--L2); draw(M1--M2); draw(P--X, dotted); dot(M); dot(\"$P$\", P, S); dot(\"$X$\", X, N); label(\"$\\ell$\", L2, E); label(\"$m$\", M2, E); [/asy] Applied to this problem, this means that $D$ is the only point that lies on both (1) the given circle, and (2) the line through $D$ parallel to $BC$. This means that $BC$ is parallel to the tangent to the given circle at $D$. If we take $O$ to be the center of the given circle, then this means that $OD$ is perpendicular to $BC$. Let $M$ be the midpoint of chord $BC,$ and let $N$ be the intersection of $OD$ with the line through $A$ parallel to $BC$. [asy] size(170); pair O = (0,0), D = (0, 5), B = (-3, 4), C = (3, 4), A = (-4, 3), EE = (4, 3); pair M = (B+C)/2, NN = (A+EE)/2; draw(circle(O, 5)); draw(O--D); draw(B--C); draw(A--EE); draw(B--O--A); dot(\"$O$\", O, SE); label(\"$D$\", D, N); label(\"$B$\", B, WNW); label(\"$A$\", A, WNW); label(\"$C$\", C, ENE); label(\"$M$\", M, NE); label(\"$N$\", NN, SE);[/asy] Since $BC = 6,$ we know that $BM = 3$; since $OB$ (a radius of the circle) is 5, we can conclude that $\\triangle BMO$ is a 3-4-5 right triangle. Since $D$ and line $AN$ are equidistant from line $BC,$ we know that $MN = 1$, and thus $ON = 3$. This makes $\\triangle ANO$ also a 3-4-5 right triangle. We're looking for $\\sin \\angle AOB$, and we can find that using the angle subtraction formula for sine. We have \\begin{align*} \\sin \\angle AOB &= \\sin(\\angle AOM - \\angle BOM) \\\\ &= \\sin \\angle AOM \\cos \\angle BOM - \\cos \\angle AOM \\sin \\angle BOM \\\\ &= \\frac{4}{5} \\cdot \\frac{4}{5} - \\frac{3}{5} \\cdot \\frac{3}{5} \\\\ &= \\frac{16 - 9}{25} = \\frac{7}{25}. \\end{align*} This is in lowest terms, so our answer is $mn = 7 \\cdot 25 = 175$. Solution 2 -Credit to Adamz for diagram- [asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label(\"$O$\",O,SW); pair M = (4,0);dot(M);label(\"$M$\",M,SE); pair N = (4,2);dot(N);label(\"$N$\",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label(\"$B$\",B,NE); pair C = (4,-3);dot(C);label(\"$C$\",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label(\"$P$\",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label(\"$A$\",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label(\"$Q$\",Q,S); pair R = (3,0); dot(R); label(\"$R$\",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label(\"$D$\",D,E); draw(A--D); [/asy]Let $A$ be any fixed point on circle $O$, and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AO$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to $BC$ at point $N$. Let $M$ be the midpoint of the chord $BC$. From right triangle $OMB$, we have $OM = \\sqrt{OB^2 - BM^2} =4$. This gives $\\tan \\angle BOM = \\frac{BM}{OM} = \\frac 3 4$. Notice that the distance $OM$ equals $PN + PO \\cos \\angle AOM = r(1 + \\cos \\angle AOM)$, where $r$ is the radius of circle $P$. Hence \\[\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5\\] (where $R$ represents the radius, $5$, of the large circle given in the question). Therefore, since $\\angle AOM$ is clearly acute, we see that \\[\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3\\] Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the subtraction formula for $\\tan$ to obtain \\[\\tan \\angle AOB =\\frac{\\tan \\angle AOM - \\tan \\angle BOM}{1 + \\tan \\angle AOM \\cdot \\tan \\angle BOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}\\] It follows that $\\sin \\angle AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, such that the answer is $7 \\cdot 25=175 ~skibidi solver",
"-Credit to Adamz for diagram- [asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label(\"$O$\",O,SW); pair M = (4,0);dot(M);label(\"$M$\",M,SE); pair N = (4,2);dot(N);label(\"$N$\",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label(\"$B$\",B,NE); pair C = (4,-3);dot(C);label(\"$C$\",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label(\"$P$\",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label(\"$A$\",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label(\"$Q$\",Q,S); pair R = (3,0); dot(R); label(\"$R$\",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label(\"$D$\",D,E); draw(A--D); [/asy]Let $A$ be any fixed point on circle $O$, and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AO$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to $BC$ at point $N$. Let $M$ be the midpoint of the chord $BC$. From right triangle $OMB$, we have $OM = \\sqrt{OB^2 - BM^2} =4$. This gives $\\tan \\angle BOM = \\frac{BM}{OM} = \\frac 3 4$. Notice that the distance $OM$ equals $PN + PO \\cos \\angle AOM = r(1 + \\cos \\angle AOM)$, where $r$ is the radius of circle $P$. Hence \\[\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5\\] (where $R$ represents the radius, $5$, of the large circle given in the question). Therefore, since $\\angle AOM$ is clearly acute, we see that \\[\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3\\] Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the subtraction formula for $\\tan$ to obtain \\[\\tan \\angle AOB =\\frac{\\tan \\angle AOM - \\tan \\angle BOM}{1 + \\tan \\angle AOM \\cdot \\tan \\angle BOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}\\] It follows that $\\sin \\angle AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, such that the answer is $7 \\cdot 25=175 ~skibidi solver",
"This solution, while similar to Solution 2, is arguably more motivated and less contrived. Firstly, we note the statement in the problem that \"$AD$ is the only chord starting at $A$ and bisected by $BC$\" – what is its significance? What is the criterion for this statement to be true? We consider the locus of midpoints of the chords from $A$. It is well-known that this is the circle with diameter $AO$, where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor $\\frac{1}{2}$ and center $A$. Thus, the locus is the result of the dilation with scale factor $\\frac{1}{2}$ and centre $A$ of circle $O$. Let the center of this circle be $P$. Now, $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$. The rest of this problem is straightforward. Our goal is to find $\\sin \\angle AOB = \\sin{\\left(\\angle AOM - \\angle BOM\\right)}$, where $M$ is the midpoint of $BC$. We have $BM=3$ and $OM=4$. Let $R$ be the projection of $A$ onto $OM$, and similarly let $Q$ be the projection of $P$ onto $OM$. Then it remains to find $AR$ so that we can use the addition formula for $\\sin$. As $PN$ is a radius of circle $P$, $PN=2.5$, and similarly, $PO=2.5$. Since $OM=4$, we have $OQ=OM-QM=OM-PN=4-2.5=1.5$. Thus $PQ=\\sqrt{2.5^2-1.5^2}=2$. Further, we see that $\\triangle OAR$ is a dilation of $\\triangle OPQ$ about center $O$ with scale factor $2$, so $AR=2PQ=4$. Lastly, we apply the formula: \\[\\sin{\\left(\\angle AOM - \\angle BOM\\right)} = \\sin \\angle AOM \\cos \\angle BOM - \\sin \\angle BOM \\cos \\angle AOM = \\left(\\frac{4}{5}\\right)\\left(\\frac{4}{5}\\right)-\\left(\\frac{3}{5}\\right)\\left(\\frac{3}{5}\\right)=\\frac{7}{25}\\] Thus the answer is $7\\cdot25=175 ~skibidi solver",
"Let the circle have equation $x^2 + y^2 = 25$, with centre $O(0,0)$. Since $BC=6$, we can calculate (by the Pythagorean Theorem) that the distance from $O$ to the line $BC$ is $4$. Therefore, we can let $B=(3,4)$ and $C=(-3,4)$. Now, assume that $A$ is any point on the major arc BC, and $D$ any point on the minor arc BC. We can write $A=(5 \\cos \\alpha, 5 \\sin \\alpha)$, where $\\alpha$ is the angle measured from the positive $x$ axis to the ray $OA$. It will also be convenient to define $\\angle XOB = \\alpha_0$. Firstly, since $B$ must lie in the minor arc $AD$, we see that $\\alpha \\in \\left(-\\frac{\\pi}{2}-\\alpha_0,\\alpha_0\\right)$. However, since the midpoint of $AD$ must lie on $BC$, and the highest possible $y$-coordinate of $D$ is $5$, we see that the $y$-coordinate cannot be lower than $3$, that is, $\\alpha \\in \\left[\\sin^{-1}\\frac{3}{5},\\alpha_0\\right)$. Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that $P$ is the intersection point of $AD$ and $BC$, so that by the theorem, $OP$ is perpendicular to $AD$. So, if $AD$ is the only chord starting at $A$ which is bisected by $BC$, this means that $P$ is the only point on the chord $BC$ such that $OP$ is perpendicular to $AD$. Now suppose that $P=(p,4)$, where $p \\in (-3,3)$. The fact that $OP$ must be perpendicular to $AD$ is equivalent to the following equation: \\[-1 = \\left(\\text{slope of } OP\\right)\\left(\\text{slope of } AP\\right)\\] which becomes \\[-1 = \\frac{4}{p} \\cdot \\frac{5\\sin \\alpha - 4}{5\\cos \\alpha - p}\\] This rearranges to \\[p^2 - (5\\cos \\alpha)p + 16 - 20 \\sin \\alpha = 0\\] Given that this equation must have only one real root $p\\in (-3,3)$, we study the following function: \\[f(x) = x^2 - (5\\cos \\alpha)x + 16 - 20 \\sin \\alpha\\] First, by the fact that the equation $f(x)=0$ has real solutions, its discriminant $\\Delta$ must be non-negative, so we calculate \\[\\begin{split}\\Delta & = (5\\cos \\alpha)^2 - 4(16-20\\sin \\alpha) \\\\ & = 25 (1- \\sin^2 \\alpha) - 64 + 80 \\sin \\alpha \\\\ & = -25 \\sin^2 \\alpha + 80\\sin \\alpha - 39 \\\\ & = (13 - 5\\sin \\alpha)(5\\sin \\alpha - 3)\\end{split}\\] It is obvious that this is in fact non-negative. If it is actually zero, then $\\sin \\alpha = \\frac{3}{5}$, and $\\cos \\alpha = \\frac{4}{5}$. In this case, $p = \\frac{5\\cos \\alpha}{2} = 2 \\in (-3,3)$, so we have found a possible solution. We thus calculate $\\sin(\\text{central angle of minor arc } AB) = \\sin (\\alpha_0 - \\alpha) = \\frac{4}{5}\\cdot \\frac{4}{5} - \\frac{3}{5} \\cdot \\frac{3}{5} = \\frac{7}{25}$ by the subtraction formula for $\\sin$. This means that the answer is $7 \\cdot 25 = 175$. Addendum to Solution 4 Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows. Suppose that $\\Delta > 0$, which would mean that there could be two real roots of $f(x)$, one lying in the interval $(-3,3)$, and another outside of it. We also see, by Vieta's Formulas, that the average of the two roots is $\\frac{5\\cos \\alpha}{2}$, which is non-negative, so the root outside of $(-3,3)$ must be no less than $3$. By considering the graph of $y=f(x)$, which is a \"U-shaped\" parabola, it is now evident that $f(-3) > 0$ and $f(3)\\leq 0$. We can just use the second inequality: \\[0 \\geq f(3) = 25 - 15\\cos \\alpha - 20 \\sin \\alpha\\] so \\[3\\cos \\alpha + 4 \\sin \\alpha \\geq 5\\] The only way for this inequality to be satisfied is when $A=B$ (by applying the Cauchy-Schwarz inequality, or just plotting the line $3x+4y=5$ to see that point $A$ can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point $A$ lies in the half-plane above the line $3x+4y=5$, inclusive, and the half-plane below the line $-3x+4y=5$, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.) Solution 5 Let the center of the circle be $O$. Fix $B,C,$ and $A$. Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$, it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$. Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$. Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$. Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$. Clearly, $KO = \\sqrt{OC^2-KC^2} = \\sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\\frac{5}{2}$. The same applies for $FO$, which also equals $\\frac{5}{2}$. By the Pythagorean theorem, we deduce that $FL = 2$, so $EK = 2$. This is very important information! Now we know that $BE = 1$, so by Power of a Point, $AE = ED = \\sqrt{5}$. We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$, we deduce that $EO = 2\\sqrt{5}$. $EC=OC=5$, so $\\sin (CEO) = \\frac{2\\sqrt{5}}{5}$. Furthermore, since $\\sin (CEO) = \\cos(DEC)$, we know that $\\cos (DEC) = \\frac{2\\sqrt{5}}{5}$. By the law of cosines, \\[DC^2 = (\\sqrt{5})^2 + 5^2 -10\\sqrt{5} \\cdot \\frac{2\\sqrt{5}}{5} = 10\\]Therefore, $DC = \\sqrt{10} \\Longleftrightarrow BA = \\sqrt{2}$. Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$. Then, by the Pythagorean theorem, $OZ = \\frac{7\\sqrt{2}}{2}$. Thus, $\\sin (BOZ) = \\frac{\\sqrt{2}}{10}$ and $\\cos (BOZ) = \\frac{7\\sqrt{2}}{10}$. As a result, $\\sin (BOA) = \\sin (2 BOZ) = 2\\sin(BOZ)\\cos(BOZ) = \\frac{7}{25}$. $7 \\cdot 25 = 175 ~skibidi solver",
"Let the circle have equation $x^2 + y^2 = 25$, with centre $O(0,0)$. Since $BC=6$, we can calculate (by the Pythagorean Theorem) that the distance from $O$ to the line $BC$ is $4$. Therefore, we can let $B=(3,4)$ and $C=(-3,4)$. Now, assume that $A$ is any point on the major arc BC, and $D$ any point on the minor arc BC. We can write $A=(5 \\cos \\alpha, 5 \\sin \\alpha)$, where $\\alpha$ is the angle measured from the positive $x$ axis to the ray $OA$. It will also be convenient to define $\\angle XOB = \\alpha_0$. Firstly, since $B$ must lie in the minor arc $AD$, we see that $\\alpha \\in \\left(-\\frac{\\pi}{2}-\\alpha_0,\\alpha_0\\right)$. However, since the midpoint of $AD$ must lie on $BC$, and the highest possible $y$-coordinate of $D$ is $5$, we see that the $y$-coordinate cannot be lower than $3$, that is, $\\alpha \\in \\left[\\sin^{-1}\\frac{3}{5},\\alpha_0\\right)$. Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that $P$ is the intersection point of $AD$ and $BC$, so that by the theorem, $OP$ is perpendicular to $AD$. So, if $AD$ is the only chord starting at $A$ which is bisected by $BC$, this means that $P$ is the only point on the chord $BC$ such that $OP$ is perpendicular to $AD$. Now suppose that $P=(p,4)$, where $p \\in (-3,3)$. The fact that $OP$ must be perpendicular to $AD$ is equivalent to the following equation: \\[-1 = \\left(\\text{slope of } OP\\right)\\left(\\text{slope of } AP\\right)\\] which becomes \\[-1 = \\frac{4}{p} \\cdot \\frac{5\\sin \\alpha - 4}{5\\cos \\alpha - p}\\] This rearranges to \\[p^2 - (5\\cos \\alpha)p + 16 - 20 \\sin \\alpha = 0\\] Given that this equation must have only one real root $p\\in (-3,3)$, we study the following function: \\[f(x) = x^2 - (5\\cos \\alpha)x + 16 - 20 \\sin \\alpha\\] First, by the fact that the equation $f(x)=0$ has real solutions, its discriminant $\\Delta$ must be non-negative, so we calculate \\[\\begin{split}\\Delta & = (5\\cos \\alpha)^2 - 4(16-20\\sin \\alpha) \\\\ & = 25 (1- \\sin^2 \\alpha) - 64 + 80 \\sin \\alpha \\\\ & = -25 \\sin^2 \\alpha + 80\\sin \\alpha - 39 \\\\ & = (13 - 5\\sin \\alpha)(5\\sin \\alpha - 3)\\end{split}\\] It is obvious that this is in fact non-negative. If it is actually zero, then $\\sin \\alpha = \\frac{3}{5}$, and $\\cos \\alpha = \\frac{4}{5}$. In this case, $p = \\frac{5\\cos \\alpha}{2} = 2 \\in (-3,3)$, so we have found a possible solution. We thus calculate $\\sin(\\text{central angle of minor arc } AB) = \\sin (\\alpha_0 - \\alpha) = \\frac{4}{5}\\cdot \\frac{4}{5} - \\frac{3}{5} \\cdot \\frac{3}{5} = \\frac{7}{25}$ by the subtraction formula for $\\sin$. This means that the answer is $7 \\cdot 25 = 175$. Addendum to Solution 4 Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows. Suppose that $\\Delta > 0$, which would mean that there could be two real roots of $f(x)$, one lying in the interval $(-3,3)$, and another outside of it. We also see, by Vieta's Formulas, that the average of the two roots is $\\frac{5\\cos \\alpha}{2}$, which is non-negative, so the root outside of $(-3,3)$ must be no less than $3$. By considering the graph of $y=f(x)$, which is a \"U-shaped\" parabola, it is now evident that $f(-3) > 0$ and $f(3)\\leq 0$. We can just use the second inequality: \\[0 \\geq f(3) = 25 - 15\\cos \\alpha - 20 \\sin \\alpha\\] so \\[3\\cos \\alpha + 4 \\sin \\alpha \\geq 5\\] The only way for this inequality to be satisfied is when $A=B$ (by applying the Cauchy-Schwarz inequality, or just plotting the line $3x+4y=5$ to see that point $A$ can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point $A$ lies in the half-plane above the line $3x+4y=5$, inclusive, and the half-plane below the line $-3x+4y=5$, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.) Solution 5 Let the center of the circle be $O$. Fix $B,C,$ and $A$. Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$, it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$. Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$. Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$. Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$. Clearly, $KO = \\sqrt{OC^2-KC^2} = \\sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\\frac{5}{2}$. The same applies for $FO$, which also equals $\\frac{5}{2}$. By the Pythagorean theorem, we deduce that $FL = 2$, so $EK = 2$. This is very important information! Now we know that $BE = 1$, so by Power of a Point, $AE = ED = \\sqrt{5}$. We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$, we deduce that $EO = 2\\sqrt{5}$. $EC=OC=5$, so $\\sin (CEO) = \\frac{2\\sqrt{5}}{5}$. Furthermore, since $\\sin (CEO) = \\cos(DEC)$, we know that $\\cos (DEC) = \\frac{2\\sqrt{5}}{5}$. By the law of cosines, \\[DC^2 = (\\sqrt{5})^2 + 5^2 -10\\sqrt{5} \\cdot \\frac{2\\sqrt{5}}{5} = 10\\]Therefore, $DC = \\sqrt{10} \\Longleftrightarrow BA = \\sqrt{2}$. Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$. Then, by the Pythagorean theorem, $OZ = \\frac{7\\sqrt{2}}{2}$. Thus, $\\sin (BOZ) = \\frac{\\sqrt{2}}{10}$ and $\\cos (BOZ) = \\frac{7\\sqrt{2}}{10}$. As a result, $\\sin (BOA) = \\sin (2 BOZ) = 2\\sin(BOZ)\\cos(BOZ) = \\frac{7}{25}$. $7 \\cdot 25 = 175 ~skibidi solver",
"Let the center of the circle be $O$. Fix $B,C,$ and $A$. Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$, it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$. Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$. Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$. Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$. Clearly, $KO = \\sqrt{OC^2-KC^2} = \\sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\\frac{5}{2}$. The same applies for $FO$, which also equals $\\frac{5}{2}$. By the Pythagorean theorem, we deduce that $FL = 2$, so $EK = 2$. This is very important information! Now we know that $BE = 1$, so by Power of a Point, $AE = ED = \\sqrt{5}$. We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 25$, we deduce that $EO = 2\\sqrt{5}$. $EC=OC=5$, so $\\sin (CEO) = \\frac{2\\sqrt{5}}{5}$. Furthermore, since $\\sin (CEO) = \\cos(DEC)$, we know that $\\cos (DEC) = \\frac{2\\sqrt{5}}{5}$. By the law of cosines, \\[DC^2 = (\\sqrt{5})^2 + 5^2 -10\\sqrt{5} \\cdot \\frac{2\\sqrt{5}}{5} = 10\\]Therefore, $DC = \\sqrt{10} \\Longleftrightarrow BA = \\sqrt{2}$. Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$. Then, by the Pythagorean theorem, $OZ = \\frac{7\\sqrt{2}}{2}$. Thus, $\\sin (BOZ) = \\frac{\\sqrt{2}}{10}$ and $\\cos (BOZ) = \\frac{7\\sqrt{2}}{10}$. As a result, $\\sin (BOA) = \\sin (2 BOZ) = 2\\sin(BOZ)\\cos(BOZ) = \\frac{7}{25}$. $7 \\cdot 25 = 175 ~skibidi solver",
"Let I be the intersection of AD and BC. Lemma: $AI = ID$ if and only if $\\angle AIO = 90$. Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\\angle AIO = 90$, We can get $\\triangle AIO \\cong \\triangle OID$ Let be this the circle with diameter AO. Thus, we get $\\angle AIO = 90$, implying I must lie on $\\omega$. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC. Now, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C. Let Z be (0,5). Let Y be (-5,0). Let X be the center of $\\omega$. Since $\\omega$'s radius is $\\frac{5}{2}$, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \\frac{3}{5}$. $sin(BOZ) = \\frac{3}{5}$. If we let $sin(\\theta) = \\frac{3}{5}$, we can find that what we are looking for is $sin(90 - 2\\theta)$, which we can evaluate and get $\\frac{7}{25} \\implies 175 ~skibidi solver",
"Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\\overline{AO}$. Additionally, this circle must be tangent to $\\overline{BC}$. Let the center of this circle be $P$. Let $M$ be the midpoint of $BC$, $N$ be the foot of the perpendicular from $P$ to $\\overline{BM}$, and $K$ be the foot of the perpendicular from $B$ to $\\overline{AP}$. Let $x=BK$. From right triangle $BKO$, we get $KO = \\sqrt{25-x^2}$. Thus, $KP = \\sqrt{25-x^2}-\\frac52$. Since $BO = 5$, $BM = 3$, and $\\angle BMO$ is right, $MO=4$. From quadrilateral $MNPO$, we get $MN = \\sqrt{PO^2 - (MO - NP)^2} = \\sqrt{(5/2)^2 - (4 - 5/2)^2} = \\sqrt{(5/2)^2 - (3/2)^2} = 2$. Thus, $BN = 1$. Since angles $BNP$ and $BKP$ are right, we get \\[BK^2+KP^2 = BN^2 + NP^2 \\implies x^2 + \\left(\\sqrt{25-x^2}-\\frac52\\right)^2 = \\left(\\frac52\\right)^2 + 1\\] \\[25 - 5\\sqrt{25-x^2} = 1\\] \\[5\\sqrt{25-x^2} = 24\\] \\[25(25-x^2) = 24^2\\] \\[25x^2 = 25^2 - 24^2 = 49\\] \\[x = \\frac75\\] Thus, $\\sin \\angle AOB = \\frac{x}{5} = \\frac{7}{25}\\implies 175 ~skibidi solver",
"Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\\overline{AO}$. Additionally, this circle must be tangent to $\\overline{BC}$. Let the center of this circle be $P$. Let $M$ be the midpoint of $BC$, $N$ be the foot of the perpendicular from $P$ to $\\overline{BM}$, and $K$ be the foot of the perpendicular from $B$ to $\\overline{AP}$. Let $x=BK$. From right triangle $BKO$, we get $KO = \\sqrt{25-x^2}$. Thus, $KP = \\sqrt{25-x^2}-\\frac52$. Since $BO = 5$, $BM = 3$, and $\\angle BMO$ is right, $MO=4$. From quadrilateral $MNPO$, we get $MN = \\sqrt{PO^2 - (MO - NP)^2} = \\sqrt{(5/2)^2 - (4 - 5/2)^2} = \\sqrt{(5/2)^2 - (3/2)^2} = 2$. Thus, $BN = 1$. Since angles $BNP$ and $BKP$ are right, we get \\[BK^2+KP^2 = BN^2 + NP^2 \\implies x^2 + \\left(\\sqrt{25-x^2}-\\frac52\\right)^2 = \\left(\\frac52\\right)^2 + 1\\] \\[25 - 5\\sqrt{25-x^2} = 1\\] \\[5\\sqrt{25-x^2} = 24\\] \\[25(25-x^2) = 24^2\\] \\[25x^2 = 25^2 - 24^2 = 49\\] \\[x = \\frac75\\] Thus, $\\sin \\angle AOB = \\frac{x}{5} = \\frac{7}{25}\\implies 175 ~skibidi solver",
"Let the center of the circle be O. O is at (0,0). Rotate the circle so that the line BC has slope 0, and so that C is in the 1st quadrant. Since BC = 6, and B can be reflected across the y axis to become C, we can say that B is on x=-3, and C is on x=3. Since it must lie on the circle $x^2+y^2=25$, B = (-3,4) and C = (3,4). So BC is on the line y=4. Let A be at $(x,\\sqrt{25-x^2})$. Let D be at $(z,\\sqrt{25-z^2})$. Notice that the midpoint of AD lies on BC. Since BC is on the line y=4, using the midpoint formula we can say, $\\sqrt{25-x^2} + \\sqrt{25-z^2} = 8$. We treat x like a constant, since it is determined by where A lies on the circle. Notice that if the above equation is true for z=p, it is also true for z=-p. But this is impossible because the problem states that AD is the only chord starting at A which is bisected by BC. This problem is solved if z=-z=0. Thus, z=0, and D=(0,5). plugging in z=0 into $\\sqrt{25-x^2} + \\sqrt{25-z^2} = 8$, we find x=+-4. Since AB is a minor arc, we assume x=-4. A is therefore on (-4,3) and B on (-3,4). We can use complex numbers to find the angle between A and B, because arguments subtract when you divide complex numbers. $\\frac{-4+3i}{-3+4i} = \\frac{24+7i}{25}$ Which gives us $\\frac{7}{25}$. $7 \\cdot 25 = 175 ~skibidi solver",
"Let the center of the circle be O. O is at (0,0). Rotate the circle so that the line BC has slope 0, and so that C is in the 1st quadrant. Since BC = 6, and B can be reflected across the y axis to become C, we can say that B is on x=-3, and C is on x=3. Since it must lie on the circle $x^2+y^2=25$, B = (-3,4) and C = (3,4). So BC is on the line y=4. Let A be at $(x,\\sqrt{25-x^2})$. Let D be at $(z,\\sqrt{25-z^2})$. Notice that the midpoint of AD lies on BC. Since BC is on the line y=4, using the midpoint formula we can say, $\\sqrt{25-x^2} + \\sqrt{25-z^2} = 8$. We treat x like a constant, since it is determined by where A lies on the circle. Notice that if the above equation is true for z=p, it is also true for z=-p. But this is impossible because the problem states that AD is the only chord starting at A which is bisected by BC. This problem is solved if z=-z=0. Thus, z=0, and D=(0,5). plugging in z=0 into $\\sqrt{25-x^2} + \\sqrt{25-z^2} = 8$, we find x=+-4. Since AB is a minor arc, we assume x=-4. A is therefore on (-4,3) and B on (-3,4). We can use complex numbers to find the angle between A and B, because arguments subtract when you divide complex numbers. $\\frac{-4+3i}{-3+4i} = \\frac{24+7i}{25}$ Which gives us $\\frac{7}{25}$. $7 \\cdot 25 = 175 ~skibidi solver"
] |
1984-I-1 | 1,984 | 1 | Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$ , $a_2$ , $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$ . | 93 | null | [
"One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of $a_1$, then use that to calculate $a_2$ and sum another arithmetic series to get our answer. A somewhat quicker method is to do the following: for each $n \\geq 1$, we have $a_{2n - 1} = a_{2n} - 1$. We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \\ldots + (a_{98} - 1) + a_{98} = 137$. The left-hand side of this equation is simply $2(a_2 + a_4 + \\ldots + a_{98}) - 49$, so our desired value is $\\frac{137 + 49}{2} = 93.",
"If $a_1$ is the first term, then $a_1+a_2+a_3 + \\cdots + a_{98} = 137$ can be rewritten as: $98a_1 + 1+2+3+ \\cdots + 97 = 137$ $\\Leftrightarrow$ $98a_1 + \\frac{97 \\cdot 98}{2} = 137$ Our desired value is $a_2+a_4+a_6+ \\cdots + a_{98}$ so this is: $49a_1 + 1+3+5+ \\cdots + 97$ which is $49a_1+ 49^2$. So, from the first equation, we know $49a_1 = \\frac{137}{2} - \\frac{97 \\cdot 49}{2}$. So, the final answer is: $\\frac{137 - 97(49) + 2(49)^2}{2} = 93.",
"A better approach to this problem is to notice that from $a_{1}+a_{2}+\\cdots a_{98}=137$ that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be $\\frac{137-49}{2}$. Thus, if we want to find the sum of all of the even elements we simply add $49$ common differences to this giving us $\\frac{137-49}{2}+49=93.",
"We want to find the value of $a_2+a_4+a_6+a_8+\\ldots+a_{98}$, which can be rewritten as $a_1+1+a_2+2+a_3+\\ldots+a_{49}+49 \\implies a_1+a_2+a_3+\\ldots+a_{49}+\\frac{49 \\cdot 50}{2}$. We can split $a_1+a_2+a_3+\\ldots+a_{98}$ into two parts: \\[a_1+a_2+a_3+\\ldots+a_{49}\\] and \\[a_{50}+a_{51}+a_{52}+\\ldots+a_{98}\\] Note that each term in the second expression is $49$ greater than the corresponding term, so, letting the first equation be equal to $x$, we get $a_1+a_2+a_3+\\ldots+a_{98}=137=2x+49^2 \\implies x=\\frac{137-49^2}{2}$. Calculating $49^2$ by sheer multiplication is not difficult, but you can also do $(50-1)(50-1)=2500-100+1=2401$. We want to find the value of $x+\\frac{49 \\cdot 50}{2}=x+49 \\cdot 25=x+1225$. Since $x=\\frac{137-2401}{2}$, we find $x=-1132$. $-1132+1225=93. - PhunsukhWangdu",
"Since we are dealing with an arithmetic sequence, \\[a_2+a_4+a_6+a_8+\\ldots+a_{98} = 49a_{50}\\] We can also figure out that \\[a_1+a_2+a_3+\\ldots+a_{98} = a_1 + 97a_{50} = 137\\] \\[a_1 = a_{50}-49 \\Rightarrow 98a_{50}-49 = 137\\] Thus, $49a_{50} = \\frac{137 + 49}{2} = 93 ~kempwood"
] |
1984-I-2 | 1,984 | 2 | The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$ . | 592 | null | [
"Any multiple of 15 is a multiple of 5 and a multiple of 3. Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0. The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$. For this number to be divisible by 3, $a$ must be divisible by 3. We also know that $a>0$ since $n$ is positive. Thus $n$ must have at least three copies of the digit 8. The smallest number which meets these two requirements is 8880. Thus the answer is $\\frac{8880}{15} = 592.",
"Notice how $8 \\cdot 10^k \\equiv 8 \\cdot (-5)^k \\equiv 5 \\pmod{15}$ for all integers $k \\geq 2$. Since we are restricted to only the digits $8,0$, because $8\\equiv -7 \\pmod{15}$ we can't have an $8$ in the optimal smallest number. We can just 'add' fives to quickly get $15 \\equiv 0 \\pmod{15}$ to get our answer. Thus n is $80+800+8000=8880$ and $n/15=8880/15=592"
] |
1984-I-3 | 1,984 | 3 | A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$ , the resulting smaller triangles $t_{1}$ , $t_{2}$ , and $t_{3}$ in the figure, have areas $4$ , $9$ , and $49$ , respectively. Find the area of $\triangle ABC$ . [asy] size(200); pathpen=black+linewidth(0.65);pointpen=black; pair A=(0,0),B=(12,0),C=(4,5); D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("A",C,N);MP("B",A,SW);MP("C",B,SE); /* sorry mixed up points according to resources diagram. */ MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N); MP("t_2",(B+C+(B-C)*5/12+(C-B)*5/6)/2+(-0.3,0.1),WSW); MP("t_1",(A+C+(C-A)*3/4+(A-C)*5/12)/2+(0,0.15),ESE); [/asy] | 144 | null | [
"By the transversals that go through $P$, all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \\dfrac{ab\\sin C}{2}$ to show that the areas are proportional (the sides are proportional and the angles are equal) Hence, we can write the lengths of corresponding sides of the triangle as $2x,\\ 3x,\\ 7x$. Thus, the corresponding side on the large triangle is $12x$, and the area of the triangle is $12^2 = 144.",
"Alternatively, since the triangles are similar by $AA$, then the ratios between the bases and the heights of each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume $\\dfrac{base}{height} = \\dfrac{2}{1}.$ That means that the base of $t_{1}$ is 4, the base of $t_{2}$ is 6, and the base of $t_{3}$ is 14. Since the quadrilaterals underneath $t_{1}$ and $t_{2}$ are both parallelograms, and opposite sides of a parallelogram are congruent, the base of the large triangle is $4 + 14 + 6 = 24$. Therefore, the height of the entire triangle would be twelve, so therefore, the area of the large triangle is $\\dfrac{1}{2} \\cdot 24 \\cdot 12 = 144.",
"The base of $\\triangle{ABC}$ is $BC$. Let the base of $t_1$ be $x$, the base of $t_2$ be $y$, and the base of $t_3$ be $z$. Since $\\triangle{ABC}, t_1, t_2,$ and $t_3$ are all similar, the sections in $\\triangle{ABC}$ that aren't $t_1,t_2,$ or $t_3$ are all parallelograms. Hence, $BC=x+z+y$. We can relate $t_1,t_2,$ and $t_3$ by the square root of the ratio of their areas. $\\sqrt{\\frac{4}{9}}=\\frac{2}{3}$ and $\\sqrt{\\frac{4}{49}}=\\frac{2}{7}$ so $y=\\frac{3x}{2}$ and $z=\\frac{7x}{2}$. $x+\\frac{7x}{2}+\\frac{3x}{2}=6x$, so $\\triangle{ABC}$ has a base that is $6$ times $t_1$. $[\\triangle{ABC}]=36[t_1]=36 \\cdot 4=144. -PhunsukhWangdu",
"Since the three lines through $P$ are parallel to the sides, $t_1$, $t_2$, $t_3$, and $\\triangle{ABC}$ are similar by $AA$ similarity. Suppose the area of $\\triangle{ABC}$ is $x^2$, so the ratio of the base of $t_1$ to the base of $t_2$ to the base of $t_3$ to the base of $\\triangle{ABC}$ is $2:3:7:x$. Because the quadrilaterals below $t_1$ and $t_2$ are parallelograms, the base of $\\triangle{ABC}$ is equal to the sum of the bases of $t_1, t_2,$ and $t_3$. Therefore, $x$ equals $2+3+7=12$ so the area of $\\triangle{ABC}$ equals $x^2=12^2=144 -Yiyj1"
] |
1984-I-4 | 1,984 | 4 | Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$ . However, if $68$ is removed, the average of the remaining numbers drops to $55$ . What is the largest number that can appear in $S$ ? | 649 | null | [
"Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$ We are given that \\begin{align*} \\frac{s+68}{n+1}&=56, \\\\ \\frac{s}{n}&=55. \\end{align*} Clearing denominators, we have \\begin{align*} s+68&=56n+56, \\\\ s&=55n. \\end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$ The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\\cdot1=649 ~JBL (Solution) ~MRENTHUSIASM (Reconstruction)",
"Suppose that $S$ has $n$ numbers other than the $68,$ and the sum of these numbers is $s.$ We are given that \\begin{align*} \\frac{s+68}{n+1}&=56, \\\\ \\frac{s}{n}&=55. \\end{align*} Clearing denominators, we have \\begin{align*} s+68&=56n+56, \\\\ s&=55n. \\end{align*} Subtracting the equations, we get $68=n+56,$ from which $n=12.$ It follows that $s=660.$ The sum of the twelve remaining numbers in $S$ is $660.$ To maximize the largest number, we minimize the other eleven numbers: We can have eleven $1$s and one $660-11\\cdot1=649 ~JBL (Solution) ~MRENTHUSIASM (Reconstruction)",
"Suppose that $S$ has $n$ numbers other than the $68.$ We have the following table: \\[\\begin{array}{c|c|c|c} & & & \\\\ [-2.5ex] & \\textbf{Count} & \\textbf{Arithmetic Mean} & \\textbf{Sum} \\\\ \\hline & & & \\\\ [-2.5ex] \\textbf{Initial} & n+1 & 56 & 56(n+1) \\\\ \\hline & & & \\\\ [-2.5ex] \\textbf{Final} & n & 55 & 55n \\end{array}\\] We are given that \\[56(n+1)-68=55n,\\] from which $n=12.$ It follows that the sum of the remaining numbers in $S$ is $55n=660.$ We continue with the last paragraph of Solution 1 to get the answer $649 ~MRENTHUSIASM",
"Suppose that $S$ has $n$ numbers other than the $68.$ We have the following table: \\[\\begin{array}{c|c|c|c} & & & \\\\ [-2.5ex] & \\textbf{Count} & \\textbf{Arithmetic Mean} & \\textbf{Sum} \\\\ \\hline & & & \\\\ [-2.5ex] \\textbf{Initial} & n+1 & 56 & 56(n+1) \\\\ \\hline & & & \\\\ [-2.5ex] \\textbf{Final} & n & 55 & 55n \\end{array}\\] We are given that \\[56(n+1)-68=55n,\\] from which $n=12.$ It follows that the sum of the remaining numbers in $S$ is $55n=660.$ We continue with the last paragraph of Solution 1 to get the answer $649 ~MRENTHUSIASM"
] |
1984-I-5 | 1,984 | 5 | Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ . | 512 | null | [
"Use the change of base formula to see that $\\frac{\\log a}{\\log 8} + \\frac{2 \\log b}{\\log 4} = 5$; combine denominators to find that $\\frac{\\log ab^3}{3\\log 2} = 5$. Doing the same thing with the second equation yields that $\\frac{\\log a^3b}{3\\log 2} = 7$. This means that $\\log ab^3 = 15\\log 2 \\Longrightarrow ab^3 = 2^{15}$ and that $\\log a^3 b = 21\\log 2 \\Longrightarrow a^3 b = 2^{21}$. If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$, so taking the fourth root of that, $ab = 2^9 = 512.",
"We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become $\\frac{\\ln a}{\\ln 8} + \\frac{2 \\ln b}{\\ln 4} = 5$ and $\\frac{\\ln b}{\\ln 8} + \\frac{2 \\ln a}{\\ln 4} = 7$. Adding the equations and factoring, we get $(\\frac{1}{\\ln 8}+\\frac{2}{\\ln 4})(\\ln a+ \\ln b)=12$. Rearranging we see that $\\ln ab = \\frac{12}{\\frac{1}{\\ln 8}+\\frac{2}{\\ln 4}}$. Again, we pull exponents out of our logarithms to get $\\ln ab = \\frac{12}{\\frac{1}{3 \\ln 2} + \\frac{2}{2 \\ln 2}} = \\frac{12 \\ln 2}{\\frac{1}{3} + 1} = \\frac{12 \\ln 2}{\\frac{4}{3}} = 9 \\ln 2$. This means that $\\frac{\\ln ab}{\\ln 2} = 9$. The left-hand side can be interpreted as a base-2 logarithm, giving us $ab = 2^9 = 512.",
"This solution is very similar to the above two, but it utilizes the well-known fact that $\\log_{m^k}{n^k}= \\log_m{n}.$ Thus, $\\log_8a+\\log_4b^2=5 \\Rightarrow \\log_{2^3}{(\\sqrt[3]{a})^3} + \\log_{2^2}{b^2} = 5 \\Rightarrow \\log_2{\\sqrt[3]{a}} + \\log_2{b} = 5 \\Rightarrow \\log_2{\\sqrt[3]{a}b} = 5.$ Similarly, $\\log_8b+\\log_4a^2=7 \\Rightarrow \\log_2{\\sqrt[3]{b}a} = 7.$ Adding these two equations, we have $\\log_2{a^{\\frac{4}{3}}b^{\\frac{4}{3}}} = 12 \\Rightarrow ab = 2^{12\\times\\frac{3}{4}} = 2^9 = 512.",
"We can change everything to a common base, like so: $\\log_8{a} + \\log_8{b^3} = 5,$ $\\log_8{b} + \\log_8{a^3} = 7.$ We set the value of $\\log_8{a}$ to $x$, and the value of $\\log_8{b}$ to $y.$ Now we have a system of linear equations: \\[x + 3y = 5,\\] \\[y + 3x = 7.\\] Now add the two equations together then simplify, we'll get $x+y=3$. So $\\log_8{ab} = \\log_8{a} + \\log_8{b} = 3$, $ab = 8^3 = 512",
"Add the two equations to get $\\log_8 {a}+ \\log_8 {b}+ \\log_{a^2}+\\log_{b^2}=12$. This can be simplified with the log property $\\log_n {x}+\\log_n {y}=log_n {xy}$. Using this, we get $\\log_8 {ab}+ \\log_4 {a^2b^2}=12$. Now let $\\log_8 {ab}=c$ and $\\log_4 {a^2b^2}=k$. Converting to exponents, we get $8^c=ab$ and $4^k=(ab)^2$. Sub in the $8^c$ to get $k=3c$. So now we have that $k+c=12$ and $k=3c$ which gives $c=3$, $k=9$. This means $\\log_4 {a^2b^2}=9$ so $4^9=(ab)^2 \\implies ab=(2^2)^9 \\implies 2^9 \\implies 512",
"Add the equations and use the facts that $\\log{a} +\\log{b}=\\log{ab}$ and $\\log{k^n} =n\\log{k}$ to get \\[\\log_8{ab} +2\\log_4{ab}=12\\] Now use the change of base identity with base as 2: \\[\\dfrac{\\log_2{ab}}{\\log_2{8}}+\\dfrac{2\\log_2{ab}}{\\log_2{4}}=12\\] Which gives: \\[\\frac{4}{3}\\log_2{ab}=12\\] Solving gives, $ab=2^9=512",
"By properties of logarithms, we know that $\\log_8 {a}+ \\log_4 {b ^ 2} = \\log_2 {a ^ {1/3}}+ \\log_2 {b} = 5$. Using the fact that $\\log_a {b} + \\log_a {c} = log_a{b*c}$, we get $\\log_2 {a^{1/3} * b} = 5$. Similarly, we know that $\\log_2 {a * b^{1/3}} = 7$. From these two equations, we get $a^{1/3} * b = 2^5$ and $a * b^{1/3} = 2^7$. Multiply the two equations to get $a^{4/3} * b^{4/3} = 2^{12}$. Solving, we get that $a*b = 2^{12*3/4} = 2^9 =$$512.",
"Adding both of the equations, we get \\[\\log_8{ab} +2\\log_4{ab}=12\\] Furthermore, we see that $\\log_4 {ab}$ is $\\frac{3}{2}$ times $\\log_8 {ab}.$ Substituting $\\log_8 {ab}$ as $x,$ we get $x+3x=12,$ so $x=3.$ Therefore, we have $\\log_8 {ab} = 3,$ so $ab= 8^3=512 ~ math_comb01",
"Change all equations to base 64. We then get: \\[\\log_{64}(a^2) + \\log_{64}(b^6) = 5\\] and \\[\\log_{64}(b^2) + \\log_{64}(a^6) = 7.\\] Using the property \\(\\log(a) + \\log(b) = \\log(ab)\\), we get: \\[\\log_{64}(a^2b^6) = 5\\] and \\[\\log_{64}(a^6b^2) = 7.\\] Then: \\[a^2b^6 = 64^5\\] and \\[a^6b^2 = 64^7.\\] Simplifying, we have: \\[ab^3 = 8^5\\] and \\[a^3b = 8^7.\\] Substituting and solving, we get: \\[a = 8^2\\] and \\[b = 8.\\] Then: \\[ab = 8^3 = 512.\\]",
"Given: - $\\log_8 a + \\log_4 b^2 = 5$ - $\\log_8 b + \\log_4 a^2 = 7$ We set up a system by subtracting 2 from the second equation to make both equal to 5: - $\\log_8 a + \\log_4 b^2 = 5$ - $\\log_8 b + \\log_4 a^2 - 2 = 5$ Setting these equal: $\\log_8 a + \\log_4 b^2 = \\log_8 b + \\log_4 a^2 - 2$ Using the property $\\log_4 x^2 = 2\\log_4 x$: $\\log_8 a + 2\\log_4 b = \\log_8 b + 2\\log_4 a - 2$ Converting to base 2, where $\\log_8 x = \\frac{\\log_2 x}{3}$ and $\\log_4 x = \\frac{\\log_2 x}{2}$: $\\frac{\\log_2 a}{3} + 2\\cdot\\frac{\\log_2 b}{2} = \\frac{\\log_2 b}{3} + 2\\cdot\\frac{\\log_2 a}{2} - 2$ Simplifying: $\\frac{\\log_2 a}{3} + \\log_2 b = \\frac{\\log_2 b}{3} + \\log_2 a - 2$ Multiplying through by 6 to eliminate fractions: $2\\log_2 a + 6\\log_2 b = 2\\log_2 b + 6\\log_2 a - 12$ Rearranging: $2\\log_2 a - 6\\log_2 a + 6\\log_2 b - 2\\log_2 b = -12$ $-4\\log_2 a + 4\\log_2 b = -12$ $\\log_2 b - \\log_2 a = -3$ Therefore: $\\log_2\\left(\\frac{b}{a}\\right) = -3$ $\\frac{b}{a} = \\frac{1}{8}$ $b = \\frac{a}{8}$ From the first original equation: $\\log_8 a + \\log_4 b^2 = 5$ Substituting $b = \\frac{a}{8}$: $\\log_8 a + \\log_4 \\left(\\frac{a}{8}\\right)^2 = 5$ $\\log_8 a + \\log_4 \\left(\\frac{a^2}{64}\\right) = 5$ $\\log_8 a + \\log_4 a^2 - \\log_4 64 = 5$ Since $\\log_4 64 = \\log_4 4^3 = 3$: $\\log_8 a + \\log_4 a^2 - 3 = 5$ $\\log_8 a + \\log_4 a^2 = 8$ Now we have: - $\\log_8 a + \\log_4 a^2 = 8$ (derived) - $\\log_8 b + \\log_4 a^2 = 7$ (given) Subtracting: $\\log_8 a - \\log_8 b = 1$ $\\log_8\\left(\\frac{a}{b}\\right) = 1$ $\\frac{a}{b} = 8$ This confirms our earlier finding that $b = \\frac{a}{8}$, so $a = 8b$. Substituting this back into the first original equation: $\\log_8 (8b) + \\log_4 b^2 = 5$ $\\log_8 8 + \\log_8 b + \\log_4 b^2 = 5$ $1 + \\log_8 b + \\log_4 b^2 = 5$ $\\log_8 b + \\log_4 b^2 = 4$ From the second original equation: $\\log_8 b + \\log_4 (8b)^2 = 7$ $\\log_8 b + \\log_4 (64b^2) = 7$ $\\log_8 b + \\log_4 64 + \\log_4 b^2 = 7$ $\\log_8 b + 3 + \\log_4 b^2 = 7$ $\\log_8 b + \\log_4 b^2 = 4$ Thus both equations yield the same constraint: $\\log_8 b + \\log_4 b^2 = 4$ Converting to base 2: $\\frac{\\log_2 b}{3} + \\frac{2\\log_2 b}{2} = 4$ $\\frac{\\log_2 b}{3} + \\log_2 b = 4$ $\\frac{\\log_2 b + 3\\log_2 b}{3} = 4$ $\\frac{4\\log_2 b}{3} = 4$ $\\log_2 b = 3$ Therefore: $b = 2^3 = 8$ $a = 8b = 8 \\cdot 8 = 64$ $ab = 64 \\cdot 8 = 512$ The value of $ab$ is $512. ~ brandonyee",
"Let us convert everything from the first equation to log base-2. We get $\\frac{\\log_2{a}}{3} + \\frac{\\log_2{b^2}}{2} = 5$. Multiplying by $6$, we get $2\\log_2{a} + 3\\log_2{b^2} = 30$. By the properties of logarithms, $\\log_2{a^2b^6} = 30$, so $a^2b^6 = 2^{30}$. We can use the exact same process for the next equation. We have $\\log_2{b^2a^6} = 42$, so $a^6b^2 = 2^{42}$. We can now multiply our two equations of products of powers of $a$ and $b$ to get $a^8b^8 = 2^{42} \\cdot 2^{30} = 2^{72}$. Taking the $8$-th root of both sides, we get $ab = 2^{\\frac{72}{8}} = 2^9 = 512 ~idk12345678"
] |
1984-I-6 | 1,984 | 6 | Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ , $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line? | 24 | null | [
"The line passes through the center of the bottom circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle. Draw the midpoint of $\\overline{AC}$ (the centers of the other two circles), and call it $M$. If we draw the feet of the perpendiculars from $A,C$ to the line (call $E,F$), we see that $\\triangle AEM\\cong \\triangle CFM$ by HA congruency; hence $M$ lies on the line. The coordinates of $M$ are $\\left(\\frac{19+14}{2},\\frac{84+92}{2}\\right) = \\left(\\frac{33}{2},88\\right)$. Thus, the slope of the line is $\\frac{88 - 76}{\\frac{33}{2} - 17} = -24$, and the answer is $024. Remark: Notice the fact that the radius is 3 is not used in this problem; in fact changing the radius does not affect the answer.",
"First of all, we can translate everything downwards by $76$ and to the left by $14$. Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem: Two circles, each of radius $3$, are drawn with centers at $(0, 16)$, and $(5, 8)$. A line passing through $(3,0)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line? Note that this is equivalent to finding a line such that the distance from $(0,16)$ to the line is the same as the distance from $(5,8)$ to the line. Let the line be $y - ax - b = 0$. Then, we have that: \\[\\frac{|-5a + 8 - b|}{\\sqrt{a^2+1}}= \\frac{|16 - b|}{\\sqrt{a^2+1}} \\Longleftrightarrow |-5a+8-b| = |16-b|\\]We can split this into two cases. Case 1: $16-b = -5a + 8 - b \\Longleftrightarrow a = -\\frac{8}{5}$ In this case, the absolute value of the slope of the line won’t be an integer, and since this is an AIME problem, we know it’s not possible. Case 2: $b-16 = -5a + 8 - b \\Longleftrightarrow 2b + 5a = 24$ But we also know that it passes through the point $(3,0)$, so $-3a-b = 0 \\Longleftrightarrow b = -3a$. Plugging this in, we see that $2b + 5a = 24 \\Longleftrightarrow a = -24$. $24.",
"Consider the region of the plane between $x=16$ and $x=17$. The parts of the circles centered at $(14,92)$ and $(19,84)$ in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since $(17,76)$ is $8$ units below the center of the lower circle, we will have the line exit the region $8$ units above the center of the upper circle, at $(16,100)$. We then find that the slope of the line is $-24$ and our answer is $024, and the fact that certain areas are equal is evident.)",
"Consider the region of the plane between $x=16$ and $x=17$. The parts of the circles centered at $(14,92)$ and $(19,84)$ in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since $(17,76)$ is $8$ units below the center of the lower circle, we will have the line exit the region $8$ units above the center of the upper circle, at $(16,100)$. We then find that the slope of the line is $-24$ and our answer is $024, and the fact that certain areas are equal is evident.)",
"We can redefine the coordinate system so that the center of the center circle is the origin, for easier calculations, as the slope of the line and the congruence of the circles do not depend on it. $O_1=(-3, 16)$ $O_2=(0,0)$, and $O_3=(2,8)$. A line bisects a circle iff it passes through the center. Therefore, we can ignore the bottom circle because it contributes an equal area with any line. A line passing through the centroid of any plane system with two perpendicular lines of reflectional symmetry bisects it. We have defined two points of the line, which are $(0,0)$ and $(-\\frac{1}{2},12)$. We use the slope formula to calculate the slope, which is $-24$, leading to an answer of $024 Solution by a1b2",
"Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose $\\ell$ is the desired line. Draw lines $\\ell_1$ and $\\ell_2$ both parallel to $\\ell$ such that $\\ell_1$ passes through $(14,92)$ and $\\ell_2$ passes through $(19,84)$. Clearly, $\\ell$ must be the \"average\" of $\\ell_1$ and $\\ell_2$. Suppose $\\ell:=y=mx+b, \\ell_1:=y=mx+c, \\ell_2:=y=mx+d$. Then $b=76-17m, c=92-14m, d=84-19m$. So we have that \\[76-17m=\\frac{92-14m+84-19m}{2},\\] which yields $m=-24$ for an answer of $024. ~yofro",
"Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose $\\ell$ is the desired line. Draw lines $\\ell_1$ and $\\ell_2$ both parallel to $\\ell$ such that $\\ell_1$ passes through $(14,92)$ and $\\ell_2$ passes through $(19,84)$. Clearly, $\\ell$ must be the \"average\" of $\\ell_1$ and $\\ell_2$. Suppose $\\ell:=y=mx+b, \\ell_1:=y=mx+c, \\ell_2:=y=mx+d$. Then $b=76-17m, c=92-14m, d=84-19m$. So we have that \\[76-17m=\\frac{92-14m+84-19m}{2},\\] which yields $m=-24$ for an answer of $024. ~yofro"
] |
1984-I-7 | 1,984 | 7 | The function f is defined on the set of integers and satisfies $f(n)= \begin{cases} n-3 & \mbox{if }n\ge 1000 \\ f(f(n+5)) & \mbox{if }n<1000 \end{cases}$ Find $f(84)$ . | 997 | null | [
"Define $f^{h} = f(f(\\cdots f(f(x))\\cdots))$, where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \\ldots f^{y}(1004)$. $1004 = 84 + 5(y - 1) \\Longrightarrow y = 185$. So we now need to reduce $f^{185}(1004)$. Let’s write out a couple more iterations of this function: \\begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\\\ &=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\\end{align*} So this function reiterates with a period of 2 for $x$. It might be tempting at first to assume that $f(1004) = 1001$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$: \\[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=997.",
"We start by finding values of the function right under $1000$ since they require iteration of the function. \\[f(999)=f(f(1004))=f(1001)=998\\] \\[f(998)=f(f(1003))=f(1000)=997\\] \\[f(997)=f(f(1002))=f(999)=998\\] \\[f(996)=f(f(1001))=f(998)=997\\] Soon we realize the $f(k)$ for integers $k<1000$ either equal $998$ or $997$ based on its parity. (If short on time, a guess of $998$ or $997$ can be taken now.) If $k$ is even, $f(k)=997$. If $k$ is odd, $f(k)=998$. $84$ has even parity, so $f(84)=997$. The result may be rigorously shown through induction.",
"Assume that $f(84)$ is to be performed $n+1$ times. Then we have \\[f(84)=f^{n+1}(84)=f(f^n(84+5))\\] In order to find $f(84)$, we want to know the smallest value of \\[f^n(84+5)\\ge1000\\] Because then \\[f(84)=f(f^n(84+5))=(f^n(84+5))-3\\] From which we'll get a numerical value for $f(84)$. Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\\frac{n}{2}$ times, the result is greater than or equal to $1000$. In this case, the value of $n$ for $f(84)$ is $916$, because \\[84+\\frac{916}{2}\\cdot5-\\frac{916}{2}\\cdot3=1000\\Longrightarrow f^{916}(84+5))=1000\\] Thus \\[f(84)=f(f^{916}(84+5))=f(1000)=1000-3=997\\] ~ Nafer",
"Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it $f$, you can call it whatever you want) with parameter $n$ (or 84 in this case) and say if $n$ is greater than 1000, then return $n-3$. Else, return $f(f(n + 5))$. Python code: def f(n): if n >= 1000: return n - 3 else: return f(f(n + 5)) print(f(84)) Or [python]def f(n): if n < 1000: return f(f(n + 5)) else: return n-3 print(f(84))[/python] ~ NL008, Sernegeti22",
"Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it $f$, you can call it whatever you want) with parameter $n$ (or 84 in this case) and say if $n$ is greater than 1000, then return $n-3$. Else, return $f(f(n + 5))$. Python code: def f(n): if n >= 1000: return n - 3 else: return f(f(n + 5)) print(f(84)) Or [python]def f(n): if n < 1000: return f(f(n + 5)) else: return n-3 print(f(84))[/python] ~ NL008, Sernegeti22"
] |
1984-I-8 | 1,984 | 8 | The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$ . | 160 | null | [
"We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$, then $0=(r^3-1)(r^6+r^3+1)=r^9-1$. The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\\circ$ for integer $m$ (the ninth degree roots of unity). Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$. This reduces $\\theta$ to either $120^{\\circ}$ or $160^{\\circ}$. But $\\theta$ can't be $120^{\\circ}$ because if $r=\\cos 120^\\circ +i\\sin 120^\\circ$, then $r^6+r^3+1=3$. (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\\circ$ was one of them.) This leaves $\\theta=160.",
"The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$. Applying the quadratic formula gives roots $y=-\\frac{1}{2}\\pm \\frac{\\sqrt{3}i}{2}$, which have arguments of $120$ and $240,$ respectively. We can write them as $z^3 = \\cos 240^\\circ + i\\sin 240^\\circ$ and $z^3 = \\cos 120^\\circ + i\\sin 120^\\circ$. So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! For $\\cos 240^\\circ + i\\sin 240$ we have $(\\cos 240^\\circ + i\\sin 240^\\circ)^{1/3}$ $\\Rightarrow$ $\\cos 80^\\circ + i\\sin 80^\\circ, \\cos 200^\\circ + i\\sin200^\\circ,$ and $\\cos 320^\\circ + i\\sin320^\\circ.$ Similarly for $(\\cos 120^\\circ + i\\sin 120^\\circ)^{1/3}$, we have $\\cos 40^\\circ + i\\sin 40^\\circ, \\cos 160^\\circ + i\\sin 160^\\circ,$ and $\\cos 280^\\circ + i\\sin 280^\\circ.$ The only argument out of all these roots that fits the description is $\\theta = 160. ~programmeruser ~ blueballoon",
"As in Solution 2, make the substitution $u=z^3$. Then we are left with $u^2+u+1=0$, and we would have the solutions $e^{\\frac{2\\pi}{3}i},e^{\\frac{4\\pi}{3}i},e^{\\frac{8\\pi}{3}i},e^{\\frac{10}{3}\\pi}$. The latter two solutions are obtained by adding one extra revolution around the unit circle. (Notice how we omitted $e^{\\frac{6\\pi}{3}i}$, since this would yield $e^{2i\\pi}=1$ and does not satisfy the equation.) Now, we substitute back, which gives us $z^3=e^{\\frac{2\\pi}{3}i},e^{\\frac{4\\pi}{3}i},e^{\\frac{10}{3}\\pi}\\implies z=e^{\\frac{2\\pi}{9}i},e^{\\frac{4\\pi}{9}i},e^{\\frac{8\\pi}{9}i},e^{\\frac{10\\pi}{9}i}$. The only root in the range $\\frac{\\pi}{2}<\\theta<\\pi$ is achieved when $\\theta=\\frac{8\\pi}{9}=160^\\circ in this equation, and they are obtained by simply adding more revolutions around the unit circle and dividing by 3."
] |
1984-I-9 | 1,984 | 9 | In tetrahedron $ABCD$ , edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$ . These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$ . | 20 | null | [
"[asy] /* modified version of olympiad modules */ import three; real markscalefactor = 0.03; path3 rightanglemark(triple A, triple B, triple C, real s=8) { triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } path3 anglemark(triple A, triple B, triple C, real t=8 ... real[] s) { triple M,N,P[],Q[]; path3 mark; int n=s.length; M=t*markscalefactor*unit(A-B)+B; N=t*markscalefactor*unit(C-B)+B; for (int i=0; i<n; ++i) { P[i]=s[i]*markscalefactor*unit(A-B)+B; Q[i]=s[i]*markscalefactor*unit(C-B)+B; } mark=arc(B,M,N); for (int i=0; i<n; ++i) { if (i%2==0) { mark=mark--reverse(arc(B,P[i],Q[i])); } else { mark=mark--arc(B,P[i],Q[i]); } } if (n%2==0 && n!=0) mark=(mark--B--P[n-1]); else if (n!=0) mark=(mark--B--Q[n-1]); else mark=(mark--B--cycle); return mark; } size(200); import three; defaultpen(black+linewidth(0.7)); pen small = fontsize(10); triple A=(0,0,0),B=(3,0,0),C=(1.8,10,0),D=(1.5,4,4),Da=(D.x,D.y,0),Db=(D.x,0,0); currentprojection=perspective(16,-10,8); draw(surface(A--B--C--cycle),rgb(0.6,0.7,0.6),nolight); draw(surface(A--B--D--cycle),rgb(0.7,0.6,0.6),nolight); /* draw pyramid - other lines + angles */ draw(A--B--C--A--D--B--D--C); draw(D--Da--Db--cycle); draw(rightanglemark(D,Da,Db));draw(rightanglemark(A,Db,D));draw(anglemark(Da,Db,D,15)); /* labeling points */ label(\"$A$\",A,SW);label(\"$B$\",B,S);label(\"$C$\",C,S);label(\"$D$\",D,N);label(\"$30^{\\circ}$\",Db+(0,.35,0.08),(1.5,1.2),small); label(\"$3$\",(A+B)/2,S); label(\"$15\\mathrm{cm}^2$\",(Db+C)/2+(0,-0.5,-0.1),NE,small); label(\"$12\\mathrm{cm}^2$\",(A+D)/2,NW,small); [/asy] Position face $ABC$ on the bottom. Since $[\\triangle ABD] = 12 = \\frac{1}{2} \\cdot AB \\cdot h_{ABD}$, we find that $h_{ABD} = 8$. Because the problem does not specify, we may assume both $ABC$ and $ABD$ to be isosceles triangles. Thus, the height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron. So, $h = \\frac{1}{2} (8) = 4$. The volume of the tetrahedron is thus $\\frac{1}{3}Bh = \\frac{1}{3} \\cdot15 \\cdot 4 = 020.",
"It is clear that $DX=8$ and $CX=10$ where $X$ is the foot of the perpendicular from $D$ and $C$ to side $AB$. Thus $[DXC]=\\frac{ab\\sin{c}}{2}=20=5 \\cdot h \\rightarrow h = 4$ where h is the height of the tetrahedron from $D$. Hence, the volume of the tetrahedron is $\\frac{bh}{3}=15\\cdot \\frac{4}{3}=020 is the altitude of the tetrahedron.",
"It is clear that $DX=8$ and $CX=10$ where $X$ is the foot of the perpendicular from $D$ and $C$ to side $AB$. Thus $[DXC]=\\frac{ab\\sin{c}}{2}=20=5 \\cdot h \\rightarrow h = 4$ where h is the height of the tetrahedron from $D$. Hence, the volume of the tetrahedron is $\\frac{bh}{3}=15\\cdot \\frac{4}{3}=020 is the altitude of the tetrahedron.",
"Make faces $ABC$ and $ABD$ right triangles. This makes everything a lot easier. Then do everything in solution 1.",
"Make faces $ABC$ and $ABD$ right triangles. This makes everything a lot easier. Then do everything in solution 1.",
"We can use 3D coordinates. Let $A = (0, 0, 0)$ and $B = (3, 0, 0).$ WLOG, let $D = \\left(\\frac{3}{2}, 8, 0\\right)$, because the area of $\\Delta{ABD} = 12$ and the tetrahedron area won't change if we put it somewhere else with $y=8.$ To find $C$, we can again let the $x$-coordinate be $\\frac{3}{2}$ for simplicity. Note that $C$ is $10$ units away from $AB$ because the area of $\\Delta{ABC}$ is $15$. Since the angle between $ABD$ and $ABC$ is $30^\\circ$, we can form a 30-60-90 triangle between $A$, $B$, and an altitude dropped from $C$ onto face $ABD$. Since $10$ is the hypotenuse, we get $5\\sqrt{3}$ and $5$ as legs. Then $y=5\\sqrt{3}$ and $z=5$, so $C = \\left(\\frac{3}{2}, 5\\sqrt{3}, 5\\right).$ (I highly advise you to draw both the tetrahedron and 30-60-90 triangle to get a better perspective.) Now, we can move onto vectors. To find the volume of the tetrahedron, we use the formula $\\frac{1}{3}Bh.$ Letting $\\Delta{ABC}$ be the base we have $B = 15$ (from the problem statement). We need to find the distance between $D$ and $ABC$, and to do this, we should find the projection of $D$ onto face $ABC$. Note that we can simplify this to projecting $D$ onto $\\mathbf{\\overrightarrow{C}}.$ This is because we know the projection will have the same $x$-coordinate as $D$ and $C$, as both are $\\frac{3}{2}.$ Now we find $\\text{proj}_{\\mathbf{\\overrightarrow{D}}} \\mathbf{\\overrightarrow{C}}$, or plugging in our coordinates, $\\text{proj}_{\\langle\\frac{3}{2}, 5\\sqrt{3}, 5\\rangle} \\left\\langle\\frac{3}{2}, 8, 0\\right\\rangle$. Let the $x$-coordinates for both be $0$ for simplicity, because we can always add a $\\frac{3}{2}$ at the end. Using the projection formula, we get \\[\\langle 0, 6, 2\\sqrt{3}\\rangle.\\] Finally, we calculate the distance between $\\left(\\frac{3}{2}, 6, 2\\sqrt{3}\\right)$ and $D$ to be $4$. So the height is $4$, and plugging into our tetrahedron formula we get \\[\\frac{1}{3}\\cdot 15\\cdot 4 = 20.\\] -kilobyte144",
"We can use 3D coordinates. Let $A = (0, 0, 0)$ and $B = (3, 0, 0).$ WLOG, let $D = \\left(\\frac{3}{2}, 8, 0\\right)$, because the area of $\\Delta{ABD} = 12$ and the tetrahedron area won't change if we put it somewhere else with $y=8.$ To find $C$, we can again let the $x$-coordinate be $\\frac{3}{2}$ for simplicity. Note that $C$ is $10$ units away from $AB$ because the area of $\\Delta{ABC}$ is $15$. Since the angle between $ABD$ and $ABC$ is $30^\\circ$, we can form a 30-60-90 triangle between $A$, $B$, and an altitude dropped from $C$ onto face $ABD$. Since $10$ is the hypotenuse, we get $5\\sqrt{3}$ and $5$ as legs. Then $y=5\\sqrt{3}$ and $z=5$, so $C = \\left(\\frac{3}{2}, 5\\sqrt{3}, 5\\right).$ (I highly advise you to draw both the tetrahedron and 30-60-90 triangle to get a better perspective.) Now, we can move onto vectors. To find the volume of the tetrahedron, we use the formula $\\frac{1}{3}Bh.$ Letting $\\Delta{ABC}$ be the base we have $B = 15$ (from the problem statement). We need to find the distance between $D$ and $ABC$, and to do this, we should find the projection of $D$ onto face $ABC$. Note that we can simplify this to projecting $D$ onto $\\mathbf{\\overrightarrow{C}}.$ This is because we know the projection will have the same $x$-coordinate as $D$ and $C$, as both are $\\frac{3}{2}.$ Now we find $\\text{proj}_{\\mathbf{\\overrightarrow{D}}} \\mathbf{\\overrightarrow{C}}$, or plugging in our coordinates, $\\text{proj}_{\\langle\\frac{3}{2}, 5\\sqrt{3}, 5\\rangle} \\left\\langle\\frac{3}{2}, 8, 0\\right\\rangle$. Let the $x$-coordinates for both be $0$ for simplicity, because we can always add a $\\frac{3}{2}$ at the end. Using the projection formula, we get \\[\\langle 0, 6, 2\\sqrt{3}\\rangle.\\] Finally, we calculate the distance between $\\left(\\frac{3}{2}, 6, 2\\sqrt{3}\\right)$ and $D$ to be $4$. So the height is $4$, and plugging into our tetrahedron formula we get \\[\\frac{1}{3}\\cdot 15\\cdot 4 = 20.\\] -kilobyte144"
] |
1984-I-10 | 1,984 | 10 | Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$ . From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$ , John could not have determined this. What was Mary's score? (Recall that the AHSME consists of $30$ multiple choice problems and that one's score, $s$ , is computed by the formula $s=30+4c-w$ , where $c$ is the number of correct answers and $w$ is the number of wrong answers. Students are not penalized for problems left unanswered.) | 119 | null | [
"Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then \\begin{align*} s&=30+4c-w \\\\ &=30+4(c-1)-(w-4) \\\\ &=30+4(c+1)-(w+4). \\end{align*} Therefore, Mary could not have left at least five blank; otherwise, one more correct and four more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have a score above $80$, or even $30$.) It follows that $c+w\\geq 26$ and $w\\leq 3$, so $c\\geq 23$ and $s=30+4c-w\\geq 30+4(23)-3=119$. So Mary scored at least $119$. To see that no result other than $23$ right/$3$ wrong produces $119$, note that $s=119\\Rightarrow 4c-w=89$ so $w\\equiv 3\\pmod{4}$. But if $w=3$, then $c=23$, which was the result given; otherwise $w\\geq 7$ and $c\\geq 24$, but this implies at least $31$ questions, a contradiction. This makes the minimum score $119.",
"Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then \\begin{align*} s&=30+4c-w \\\\ &=30+4(c-1)-(w-4) \\\\ &=30+4(c+1)-(w+4). \\end{align*} Therefore, Mary could not have left at least five blank; otherwise, one more correct and four more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have a score above $80$, or even $30$.) It follows that $c+w\\geq 26$ and $w\\leq 3$, so $c\\geq 23$ and $s=30+4c-w\\geq 30+4(23)-3=119$. So Mary scored at least $119$. To see that no result other than $23$ right/$3$ wrong produces $119$, note that $s=119\\Rightarrow 4c-w=89$ so $w\\equiv 3\\pmod{4}$. But if $w=3$, then $c=23$, which was the result given; otherwise $w\\geq 7$ and $c\\geq 24$, but this implies at least $31$ questions, a contradiction. This makes the minimum score $119.",
"A less technical approach that still gets the job done: Pretend that the question is instead a game, where we are trying to get certain numbers by either adding $4$ or $5.$ The maximum number we can get is $70.$ The goal of the game is to find out what number we can achieve in only ONE method, while all other numbers above that can be achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score. If this is already confusing, I suggest not looking further.) For example, the number $\"21\"$ can be achieved with only $1$ method $(4+4+4+4+5).$ However, $25$, which is a larger number than $21$, can be achieved with multiple methods (e.g. $5 \\cdot 5$ or $4 \\cdot 5 + 5$), hence $21$ is not the number we are trying to find. If we make a table of adding $4$ or adding $5$, we will see we get $4, 8, 12, 16, 20,$ etc. if we add only $4$s and if we add $5$ to those numbers then we will get $9, 13, 17, 21, 25,$ etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to $20$, then there will be multiple methods (because $20 = 4 \\cdot 5 = 5 \\cdot 4$). Hence, the number we are looking for cannot be $20$ plus one of those base numbers. Instead, it must be $10$ plus that base number, because that results in the same last digit while maintaining only one method to solve. For example, if we start with $4$, the number $14$ would have only $1$ method to solve, but the number $24$ would have multiple (because $4 + 20 = 24$ and we are trying to avoid adding $20$). The largest number we see that is in our base numbers is $21.$ Hence, our maximum number is $21 + 10 = 31.$ Note that if we have the number $25$, that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding $20$ to a previous base number, which we don't want. And since the maximum number of this game is $31$, that is the number we subtract from the maximum score of $150$, so we get $150 - 31 = 119 if you want a concise method using inequalities that's probably better than this solution.",
"A less technical approach that still gets the job done: Pretend that the question is instead a game, where we are trying to get certain numbers by either adding $4$ or $5.$ The maximum number we can get is $70.$ The goal of the game is to find out what number we can achieve in only ONE method, while all other numbers above that can be achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score. If this is already confusing, I suggest not looking further.) For example, the number $\"21\"$ can be achieved with only $1$ method $(4+4+4+4+5).$ However, $25$, which is a larger number than $21$, can be achieved with multiple methods (e.g. $5 \\cdot 5$ or $4 \\cdot 5 + 5$), hence $21$ is not the number we are trying to find. If we make a table of adding $4$ or adding $5$, we will see we get $4, 8, 12, 16, 20,$ etc. if we add only $4$s and if we add $5$ to those numbers then we will get $9, 13, 17, 21, 25,$ etc. Now a key observation to getting this problem correct is that if we can add one of those previous base numbers to $20$, then there will be multiple methods (because $20 = 4 \\cdot 5 = 5 \\cdot 4$). Hence, the number we are looking for cannot be $20$ plus one of those base numbers. Instead, it must be $10$ plus that base number, because that results in the same last digit while maintaining only one method to solve. For example, if we start with $4$, the number $14$ would have only $1$ method to solve, but the number $24$ would have multiple (because $4 + 20 = 24$ and we are trying to avoid adding $20$). The largest number we see that is in our base numbers is $21.$ Hence, our maximum number is $21 + 10 = 31.$ Note that if we have the number $25$, that can be solved via multiple methods, and if we keep repeating our cycle of base numbers, we are basically adding $20$ to a previous base number, which we don't want. And since the maximum number of this game is $31$, that is the number we subtract from the maximum score of $150$, so we get $150 - 31 = 119 if you want a concise method using inequalities that's probably better than this solution.",
"Based on the value of $c,$ we construct the following table: \\[\\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} &\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&&&&&&&&&&&&& \\\\ [-2.5ex] \\boldsymbol{c} &\\boldsymbol{\\cdots}&\\boldsymbol{12}&\\boldsymbol{13}&\\boldsymbol{14}&\\boldsymbol{15}&\\boldsymbol{16}&\\boldsymbol{17}&\\boldsymbol{18}&\\boldsymbol{19}&\\boldsymbol{20}&\\boldsymbol{21}&\\boldsymbol{22}&\\boldsymbol{23}&\\boldsymbol{24}&\\boldsymbol{25}&\\boldsymbol{26}&\\boldsymbol{27}&\\boldsymbol{28}&\\boldsymbol{29}&\\boldsymbol{30} \\\\ \\hline \\hline &&&&&&&&&&&&&&&&&&& \\\\ [-2.25ex] \\boldsymbol{s_{\\min}} &\\cdots&60&65&70&75&80&85&90&95&100&105&110&115&120&125&130&135&140&145&150 \\\\ \\hline &&&&&&&&&&&&&&&&&&& \\\\ [-2.25ex] \\boldsymbol{s_{\\max}} &\\cdots&78&82&86&90&94&98&102&106&110&114&118&122&126&130&134&138&142&146&150 \\end{array}\\] For a fixed value of $c,$ note that $s_{\\min}$ occurs at $w=30-c,$ and $s_{\\max}$ occurs at $w=0.$ Moreover, all integers from $s_{\\min}$ through $s_{\\max}$ are attainable. To find Mary's score, we look for the lowest score $\\boldsymbol{s}$ such that $\\boldsymbol{s\\geq80}$ and $\\boldsymbol{s}$ is contained in exactly one interval. Let $S(c)$ denote the interval of all possible scores $s$ with $c$ correct answers. We need: $S(c)\\not\\subset S(c-1)\\cup S(c+1).$ $s\\in S(c)$ but $s\\not\\in S(c-1)\\cup S(c+1).$ It follows that the least such value of $c$ is $23,$ from which the lowest such score $s$ is $119 ~MRENTHUSIASM",
"Based on the value of $c,$ we construct the following table: \\[\\begin{array}{c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} &\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&\\hspace{5.5mm}&&&&&&&&&&&&& \\\\ [-2.5ex] \\boldsymbol{c} &\\boldsymbol{\\cdots}&\\boldsymbol{12}&\\boldsymbol{13}&\\boldsymbol{14}&\\boldsymbol{15}&\\boldsymbol{16}&\\boldsymbol{17}&\\boldsymbol{18}&\\boldsymbol{19}&\\boldsymbol{20}&\\boldsymbol{21}&\\boldsymbol{22}&\\boldsymbol{23}&\\boldsymbol{24}&\\boldsymbol{25}&\\boldsymbol{26}&\\boldsymbol{27}&\\boldsymbol{28}&\\boldsymbol{29}&\\boldsymbol{30} \\\\ \\hline \\hline &&&&&&&&&&&&&&&&&&& \\\\ [-2.25ex] \\boldsymbol{s_{\\min}} &\\cdots&60&65&70&75&80&85&90&95&100&105&110&115&120&125&130&135&140&145&150 \\\\ \\hline &&&&&&&&&&&&&&&&&&& \\\\ [-2.25ex] \\boldsymbol{s_{\\max}} &\\cdots&78&82&86&90&94&98&102&106&110&114&118&122&126&130&134&138&142&146&150 \\end{array}\\] For a fixed value of $c,$ note that $s_{\\min}$ occurs at $w=30-c,$ and $s_{\\max}$ occurs at $w=0.$ Moreover, all integers from $s_{\\min}$ through $s_{\\max}$ are attainable. To find Mary's score, we look for the lowest score $\\boldsymbol{s}$ such that $\\boldsymbol{s\\geq80}$ and $\\boldsymbol{s}$ is contained in exactly one interval. Let $S(c)$ denote the interval of all possible scores $s$ with $c$ correct answers. We need: $S(c)\\not\\subset S(c-1)\\cup S(c+1).$ $s\\in S(c)$ but $s\\not\\in S(c-1)\\cup S(c+1).$ It follows that the least such value of $c$ is $23,$ from which the lowest such score $s$ is $119 ~MRENTHUSIASM",
"Given that Mary's score is $30+4c-w$, two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$. Since it is clear that $c>1$, we must have $w<4$. In order to minimize the score, assume that $w=3$. The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order to minimize the score, assume that the number of problems left blank is $4$, making the number of correct problems $23$. Substituting, we get that $s=30+23{\\,\\cdot\\,}4-3$, so $s=119. ~JeffersonJ",
"Given that Mary's score is $30+4c-w$, two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$. Since it is clear that $c>1$, we must have $w<4$. In order to minimize the score, assume that $w=3$. The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order to minimize the score, assume that the number of problems left blank is $4$, making the number of correct problems $23$. Substituting, we get that $s=30+23{\\,\\cdot\\,}4-3$, so $s=119. ~JeffersonJ"
] |
1984-I-11 | 1,984 | 11 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ . | 106 | null | [
"First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and \"non-birch\" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.) The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making ${8\\choose5} = 56$ different ways to arrange this. There are ${12 \\choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\\frac{56}{792} = \\frac{7}{99}$. The answer is $7 + 99 = 106.",
"Let $b$, $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$s to separate the $5$ $b$s. Specifically, \\[b,n,b,n,b,n,b,n,b\\] Since we have $7$ $n$s, we are placing the extra $3$ $n$s into the $6$ intervals beside the $b$s. Now doing simple casework. If all $3$ $n$s are in the same interval, there are $6$ ways. If $2$ of the $3$ $n$s are in the same interval, there are $6\\cdot5=30$ ways. If the $n$s are in $3$ different intervals, there are ${6 \\choose 3} =20$ ways. In total there are $6+30+20=56$ ways. There are ${12\\choose5}=792$ ways to distribute the birch trees among all $12$ trees. Thus the probability equals $\\frac{56}{792}=\\frac{7}{99}\\Longrightarrow m+n=7+99=106. ~ Nafer",
"Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion. The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees. The total number of configurations is given by $\\frac{12!}{3! \\cdot 4! \\cdot 5!}$. To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE. $\\#$(configurations with at least one pair of adjacent Birch trees) $=$ $\\#$(configurations with one pair) $-$ $\\#$(configurations with two pairs) $+$ $\\#$(configurations with three pairs) $-$ $\\#$(configurations with four pairs). To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\\frac{11!}{3! \\cdot 3! \\cdot 4!}$ configurations. For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\\frac{10!}{2! \\cdot 3! \\cdot 4!}$ cases. So our second term is $\\frac{2 \\cdot 10!}{2! \\cdot 3! \\cdot 4!}$. The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\\frac{2 \\cdot 9!}{3! \\cdot 4!}$ arrangements. The final term can happen in one way (BBBBB). This gives $\\frac{8!}{3! \\cdot 4!}$ arrangements. Substituting these into our PIE expression, we find that there are $25760$ configurations with at least one pair of adjacent Birch trees. Therefore, there are a total of $\\frac{12!}{3! \\cdot 4! \\cdot 5!} - 25760 = 1960$ configurations with no adjacent Birch trees. Thus, the probability of a given configuration having no two adjacent Birch trees is given by $\\frac{1960}{\\frac{12!}{3! \\cdot 4! \\cdot 5!}} = \\frac{7}{99}$. Therefore, the desired result is given by $7+99 = 106. ~th1nq3r",
"Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion. The number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees. The total number of configurations is given by $\\frac{12!}{3! \\cdot 4! \\cdot 5!}$. To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE. $\\#$(configurations with at least one pair of adjacent Birch trees) $=$ $\\#$(configurations with one pair) $-$ $\\#$(configurations with two pairs) $+$ $\\#$(configurations with three pairs) $-$ $\\#$(configurations with four pairs). To compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\\frac{11!}{3! \\cdot 3! \\cdot 4!}$ configurations. For the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\\frac{10!}{2! \\cdot 3! \\cdot 4!}$ cases. So our second term is $\\frac{2 \\cdot 10!}{2! \\cdot 3! \\cdot 4!}$. The third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\\frac{2 \\cdot 9!}{3! \\cdot 4!}$ arrangements. The final term can happen in one way (BBBBB). This gives $\\frac{8!}{3! \\cdot 4!}$ arrangements. Substituting these into our PIE expression, we find that there are $25760$ configurations with at least one pair of adjacent Birch trees. Therefore, there are a total of $\\frac{12!}{3! \\cdot 4! \\cdot 5!} - 25760 = 1960$ configurations with no adjacent Birch trees. Thus, the probability of a given configuration having no two adjacent Birch trees is given by $\\frac{1960}{\\frac{12!}{3! \\cdot 4! \\cdot 5!}} = \\frac{7}{99}$. Therefore, the desired result is given by $7+99 = 106. ~th1nq3r",
"Here is a solution leaving out nothing. This solution is dedicated to those that are in self study and wish to learn the most they can. I will make it as elementary as possible and intuition based. Arrange first the $3$ maple and $4$ oaks as $MMMOOOO$. We then notice that for none of the $5$ birch trees to be adjacent, they must be put in between these $M$'s and $O$'s. We then see that there are $8$ spots to put these $5$ birch trees in. So we can select $5$ spots for these birch trees in $\\binom{8}{5}$. But then, we can rearrange the $M$'s and $O$'s in $7!/(3!4!)=\\binom{7}{3}$ ways. So then there are $\\binom{8}{5}\\binom{7}{3}$ valid arrangements with no given consecutive birch trees. There are then a total of $\\frac{12!}{3!4!5!}$ different total arrangements. Therefore the probability is given as $\\frac{\\binom{8}{5}\\binom{7}{3}}{\\frac{12!}{3!4!5!}}=\\frac{7}{99}$, so the answer is $7+99=106. ~th1nq3r"
] |
1984-I-12 | 1,984 | 12 | A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$ . If $x=0$ is a root for $f(x)=0$ , what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$ ? | 401 | null | [
"If $f(2+x)=f(2-x)$, then substituting $t=2+x$ gives $f(t)=f(4-t)$. Similarly, $f(t)=f(14-t)$. In particular, \\[f(t)=f(14-t)=f(14-(4-t))=f(t+10)\\] Since $0$ is a root, all multiples of $10$ are roots, and anything congruent to $4\\pmod{10}$ are also roots. To see that these may be the only integer roots, observe that the function \\[f(x) = \\sin \\frac{\\pi x}{10}\\sin \\frac{\\pi (x-4)}{10}\\] satisfies the conditions and has no other roots. In the interval $-1000\\leq x\\leq 1000$, there are $201$ multiples of $10$ and $200$ numbers that are congruent to $4 \\pmod{10}$, therefore the minimum number of roots is $401.",
"We notice that the function has reflectional symmetry across both $x=2$ and $x=7$. We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\\pm 10$ as roots. Continuing this shows that the roots are $0 \\mod 10$ or $4 \\mod 10$. There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $401 Solution by a1b2",
"We notice that the function has reflectional symmetry across both $x=2$ and $x=7$. We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\\pm 10$ as roots. Continuing this shows that the roots are $0 \\mod 10$ or $4 \\mod 10$. There are 200 positive roots and 200 negative roots. 0 is also a root, and adding these gives a result of $401 Solution by a1b2",
"Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent with them. Then find the x values for the functions that are equal to f(2) and f(7). You will notice that it starts at x=0, then it goes to x=5, x=10, etc... each f() has two possible x values, but we are only counting the total number of x values so ignore the double counted values. This means that $x = 0, \\pm 5, \\pm 10, \\pm 15... \\pm 1000$ so the answer is 400 + 1 = $401",
"Let $z$ be an arbitrary zero. If $z=2-x$, then $x=2-z$ and $2+x=4-z$. Repeat with other equation to find if $z$ is a zero then so are $4-z$ and $14-z$. From $0$, we get $4$ and $14$. Now note that applying either of these twice will return $z$, so we must apply them in an alternating fashion for distinct roots. Doing so to $4$ and $14$ returns $10$ and $-10$, respectively. A pattern will emerge of each path hitting a multiple of $10$ after $2$ moves. Hence, we will reach $\\pm 1000$ after $200$ jumps in either direction. Including zero, there are $2\\cdot200+1=401 ~N828335"
] |
1984-I-13 | 1,984 | 13 | Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$ | 15 | null | [
"Solution 1 We know that $\\tan(\\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\\tan(x+y) = \\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}$. Let $a = \\cot^{-1}(3)$, $b=\\cot^{-1}(7)$, $c=\\cot^{-1}(13)$, and $d=\\cot^{-1}(21)$. We have $\\tan(a)=\\frac{1}{3},\\quad\\tan(b)=\\frac{1}{7},\\quad\\tan(c)=\\frac{1}{13},\\quad\\tan(d)=\\frac{1}{21}$, so $\\tan(a+b) = \\frac{\\frac{1}{3}+\\frac{1}{7}}{1-\\frac{1}{21}} = \\frac{1}{2}$ and $\\tan(c+d) = \\frac{\\frac{1}{13}+\\frac{1}{21}}{1-\\frac{1}{273}} = \\frac{1}{8}$, so $\\tan((a+b)+(c+d)) = \\frac{\\frac{1}{2}+\\frac{1}{8}}{1-\\frac{1}{16}} = \\frac{2}{3}$. Thus our answer is $10\\cdot\\frac{3}{2}=015, we are left with \\[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\\] \\[= (2+i)(13+i)(21+i)\\] \\[= (25+15i)(21+i)\\] \\[= (5+3i)(21+i)\\] \\[= (102+68i)\\] \\[= (3+2i)\\] \\[= 10\\cot \\tan^{-1}\\frac{2}{3}\\] \\[= 10 \\cdot \\frac{3}{2} = 015\\]",
"We know that $\\tan(\\arctan(x)) = x$ so we can repeatedly apply the addition formula, $\\tan(x+y) = \\frac{\\tan(x)+\\tan(y)}{1-\\tan(x)\\tan(y)}$. Let $a = \\cot^{-1}(3)$, $b=\\cot^{-1}(7)$, $c=\\cot^{-1}(13)$, and $d=\\cot^{-1}(21)$. We have $\\tan(a)=\\frac{1}{3},\\quad\\tan(b)=\\frac{1}{7},\\quad\\tan(c)=\\frac{1}{13},\\quad\\tan(d)=\\frac{1}{21}$, so $\\tan(a+b) = \\frac{\\frac{1}{3}+\\frac{1}{7}}{1-\\frac{1}{21}} = \\frac{1}{2}$ and $\\tan(c+d) = \\frac{\\frac{1}{13}+\\frac{1}{21}}{1-\\frac{1}{273}} = \\frac{1}{8}$, so $\\tan((a+b)+(c+d)) = \\frac{\\frac{1}{2}+\\frac{1}{8}}{1-\\frac{1}{16}} = \\frac{2}{3}$. Thus our answer is $10\\cdot\\frac{3}{2}=015, we are left with \\[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\\] \\[= (2+i)(13+i)(21+i)\\] \\[= (25+15i)(21+i)\\] \\[= (5+3i)(21+i)\\] \\[= (102+68i)\\] \\[= (3+2i)\\] \\[= 10\\cot \\tan^{-1}\\frac{2}{3}\\] \\[= 10 \\cdot \\frac{3}{2} = 015\\]",
"Apply the formula $\\cot^{-1}x + \\cot^{-1} y = \\cot^{-1}\\left(\\frac {xy-1}{x+y}\\right)$ repeatedly. Using it twice on the inside, the desired sum becomes $\\cot (\\cot^{-1}2+\\cot^{-1}8)$. This sum can then be tackled by taking the cotangent of both sides of the inverse cotangent addition formula shown at the beginning. Solution 3 On the coordinate plane, let $O=(0,0)$, $A_1=(3,0)$, $A_2=(3,1)$, $B_1=(21,7)$, $B_2=(20,10)$, $C_1=(260,130)$, $C_2=(250,150)$, $D_1=(5250,3150)$, $D_2=(5100,3400)$, and $H=(5100,0)$. We see that $\\cot^{-1}(\\angle A_2OA_1)=3$, $\\cot^{-1}(\\angle B_2OB_1)=7$, $\\cot^{-1}(\\angle C_2OC_1)=13$, and $\\cot^{-1}(\\angle D_2OD_1)=21$. The sum of these four angles forms the angle of triangle $OD_2H$, which has a cotangent of $\\frac{5100}{3400}=\\frac{3}{2}$, which must mean that $\\cot( \\cot^{-1}3+\\cot^{-1}7+\\cot^{-1}13+\\cot^{-1}21)=\\frac{3}{2}$. So the answer is $10\\cdot\\left(\\frac{3}{2}\\right)=015, we are left with \\[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\\] \\[= (2+i)(13+i)(21+i)\\] \\[= (25+15i)(21+i)\\] \\[= (5+3i)(21+i)\\] \\[= (102+68i)\\] \\[= (3+2i)\\] \\[= 10\\cot \\tan^{-1}\\frac{2}{3}\\] \\[= 10 \\cdot \\frac{3}{2} = 015\\]",
"On the coordinate plane, let $O=(0,0)$, $A_1=(3,0)$, $A_2=(3,1)$, $B_1=(21,7)$, $B_2=(20,10)$, $C_1=(260,130)$, $C_2=(250,150)$, $D_1=(5250,3150)$, $D_2=(5100,3400)$, and $H=(5100,0)$. We see that $\\cot^{-1}(\\angle A_2OA_1)=3$, $\\cot^{-1}(\\angle B_2OB_1)=7$, $\\cot^{-1}(\\angle C_2OC_1)=13$, and $\\cot^{-1}(\\angle D_2OD_1)=21$. The sum of these four angles forms the angle of triangle $OD_2H$, which has a cotangent of $\\frac{5100}{3400}=\\frac{3}{2}$, which must mean that $\\cot( \\cot^{-1}3+\\cot^{-1}7+\\cot^{-1}13+\\cot^{-1}21)=\\frac{3}{2}$. So the answer is $10\\cdot\\left(\\frac{3}{2}\\right)=015, we are left with \\[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\\] \\[= (2+i)(13+i)(21+i)\\] \\[= (25+15i)(21+i)\\] \\[= (5+3i)(21+i)\\] \\[= (102+68i)\\] \\[= (3+2i)\\] \\[= 10\\cot \\tan^{-1}\\frac{2}{3}\\] \\[= 10 \\cdot \\frac{3}{2} = 015\\]",
"Recall that $\\cot^{-1}\\theta = \\frac{\\pi}{2} - \\tan^{-1}\\theta$ and that $\\arg(a + bi) = \\tan^{-1}\\frac{b}{a}$. Then letting $w = 1 + 3i, x = 1 + 7i, y = 1 + 13i,$ and $z = 1 + 21i$, we are left with \\[10\\cot(\\frac{\\pi}{2} - \\arg w + \\frac{\\pi}{2} - \\arg x + \\frac{\\pi}{2} - \\arg y + \\frac{\\pi}{2} - \\arg z) = 10\\cot(2\\pi - \\arg wxyz)\\] \\[= -10\\cot(\\arg wxyz).\\] Expanding $wxyz$, we are left with \\[(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)\\] \\[= (2+i)(13+i)(21+i)\\] \\[= (25+15i)(21+i)\\] \\[= (5+3i)(21+i)\\] \\[= (102+68i)\\] \\[= (3+2i)\\] \\[= 10\\cot \\tan^{-1}\\frac{2}{3}\\] \\[= 10 \\cdot \\frac{3}{2} = 015\\]"
] |
1984-I-14 | 1,984 | 14 | What is the largest even integer that cannot be written as the sum of two odd composite numbers? | 38 | null | [
"Take an even positive integer $x$. $x$ is either $0 \\bmod{6}$, $2 \\bmod{6}$, or $4 \\bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases: If $x \\ge 18$ and is $0 \\bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites. If $x\\ge 44$ and is $2 \\bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites. If $x\\ge 34$ and is $4 \\bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites. Clearly, if $x \\ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $038 is the largest possible number that is not expressible as the sum of two odd composite numbers.",
"Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$, yielding a maximal answer of 38. Since $38-25=13$, which is prime, the answer is $038.",
"Let $2n$ be an even integer. Using the Chicken McNugget Theorem on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as $n+n$. We bash each case until we find one that works.",
"Let $2n$ be an even integer. Using the Chicken McNugget Theorem on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as $n+n$. We bash each case until we find one that works.",
"The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35... For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14. If you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $38",
"The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35... For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14. If you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $38",
"Claim: The answer is $038 -Alexlikemath",
"All numbers that could possibly work must be $2 \\cdot p$ where $p$ is prime. As previous solutions stated, the maximum number that could possibly work by Chicken McNugget is $9 \\cdot 25 - 9 - 25 = 225-34 = 191$. We then bash from top to bottom: 1. $178 = 89 \\cdot 2 => 87 + 91$ - refuted 2. $166 = 83 \\cdot 2 => 81 + 85$ - refuted 3. $158 = 79 \\cdot 2 => 77 + 81$ - refuted 4. $146 = 73 \\cdot 2 => 69 + 77$ - refuted 5. $142 = 71 \\cdot 2 => 65 + 77$ - refuted 6. $134 = 67 \\cdot 2 => 65 + 69$ - refuted 7. $122 = 61 \\cdot 2 => 57 + 65$ - refuted 8. $118 = 59 \\cdot 2 => 55 + 63$ - refuted 9. $106 = 53 \\cdot 2 => 51 + 55$ - refuted 10. $94 = 47 \\cdot 2 => 45 + 49$ - refuted 11. $86 = 43 \\cdot 2 => 35 + 51$ - refuted 12. $82 = 41 \\cdot 2 => 33 + 49$ - refuted 13. $74 = 37 \\cdot 2 => 35 + 39$ - refuted 14. $62 = 31 \\cdot 2 => 27 + 35$ - refuted 15. $58 = 29 \\cdot 2 => 25 + 33$ - refuted 16. $46 = 23 \\cdot 2 => 21 + 25$ - refuted 17. $38 = 19 \\cdot 2 => = 19 + 19$ - it works! Because we did a very systematic bash as shown, we are confident the answer is $038 ~Arcticturn",
"As stated above, all numbers that could possibly work must be $2 \\cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \\cdot p$ can be expressed as ($p$+$q$)+($p$-$q$) for some positive, even $q$ less then $p$. If $p$ = $1 \\bmod{30}$, p±4 would both be composite If $p$ = $7 \\bmod{30}$, p±2 would both be composite If $p$ = $11 \\bmod{30}$, p±14 would both be composite If $p$ = $13 \\bmod{30}$, p±8 would both be composite If $p$ = $17 \\bmod{30}$, p±8 would both be composite If $p$ = $19 \\bmod{30}$, p±14 would both be composite If $p$ = $23 \\bmod{30}$, p±2 would both be composite If $p$ = $29 \\bmod{30}$, p±4 would both be composite So $p$ < 30 From here, just try all possible p and find the answer is $038 ~Mathophobia",
"As stated above, all numbers that could possibly work must be $2 \\cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \\cdot p$ can be expressed as ($p$+$q$)+($p$-$q$) for some positive, even $q$ less then $p$. If $p$ = $1 \\bmod{30}$, p±4 would both be composite If $p$ = $7 \\bmod{30}$, p±2 would both be composite If $p$ = $11 \\bmod{30}$, p±14 would both be composite If $p$ = $13 \\bmod{30}$, p±8 would both be composite If $p$ = $17 \\bmod{30}$, p±8 would both be composite If $p$ = $19 \\bmod{30}$, p±14 would both be composite If $p$ = $23 \\bmod{30}$, p±2 would both be composite If $p$ = $29 \\bmod{30}$, p±4 would both be composite So $p$ < 30 From here, just try all possible p and find the answer is $038 ~Mathophobia"
] |
1985-I-1 | 1,985 | 1 | Let $x_1=97$ , and for $n>1$ let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2 \ldots x_8$ . | 384 | null | [
"Since $x_n=\\frac{n}{x_{n-1}}$, $x_n \\cdot x_{n - 1} = n$. Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$, $x_3x_4 = 4$, $x_5x_6 = 6$ and $x_7x_8 = 8$ so \\[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\\cdot4\\cdot6\\cdot8 = 384 was completely unneeded!",
"Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel: \\[x_1x_2x_3x_4x_5x_6x_7x_8=x_1\\cdot\\dfrac{2}{x_1}\\cdot\\dfrac{3}{\\dfrac{2}{x_1}}\\cdot\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}\\cdot\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}\\cdot\\dfrac{6}{\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}}\\cdot\\dfrac{7}{\\dfrac{6}{\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}}}\\cdot\\dfrac{8}{\\dfrac{7}{\\dfrac{6}{\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}}}}\\] \\[=\\left (x_1\\cdot\\dfrac{2}{x_1} \\right )\\cdot \\left (\\dfrac{3}{\\dfrac{2}{x_1}}\\cdot\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}} \\right )\\cdot \\left (\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}\\cdot\\dfrac{6}{\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}} \\right )\\cdot \\left (\\dfrac{7}{\\dfrac{6}{\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}}}\\cdot\\dfrac{8}{\\dfrac{7}{\\dfrac{6}{\\dfrac{5}{\\dfrac{4}{\\dfrac{3}{\\dfrac{2}{x_1}}}}}}} \\right )\\] \\[=(2)\\cdot (4)\\cdot (6)\\cdot (8)=384\\]"
] |
1985-I-2 | 1,985 | 2 | When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{cm}^3$ . When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{cm}^3$ . What is the length (in cm) of the hypotenuse of the triangle? | 26 | null | [
"Let one leg of the triangle have length $a$ and let the other leg have length $b$. When we rotate around the leg of length $a$, the result is a cone of height $a$ and radius $b$, and so of volume $\\frac 13 \\pi ab^2 = 800\\pi$. Likewise, when we rotate around the leg of length $b$ we get a cone of height $b$ and radius $a$ and so of volume $\\frac13 \\pi b a^2 = 1920 \\pi$. If we divide this equation by the previous one, we get $\\frac ab = \\frac{\\frac13 \\pi b a^2}{\\frac 13 \\pi ab^2} = \\frac{1920}{800} = \\frac{12}{5}$, so $a = \\frac{12}{5}b$. Then $\\frac{1}{3} \\pi \\left(\\frac{12}{5}b\\right)b^2 = 800\\pi$ so $b^3 = 1000$ and $b = 10$ so $a = 24$. Then by the Pythagorean Theorem, the hypotenuse has length $\\sqrt{a^2 + b^2} = 026.",
"Let $a$, $b$ be the $2$ legs, we have the $2$ equations \\[\\frac{a^2b\\pi}{3}=800\\pi,\\frac{ab^2\\pi}{3}=1920\\pi\\] Thus $a^2b=2400, ab^2=5760$. Multiplying gets \\begin{align*} (a^2b)(ab^2)&=2400\\cdot5760\\\\ (ab)^3&=(2^5\\cdot3\\cdot5^2)(2^7\\cdot3^2\\cdot5)\\\\ ab&=\\sqrt[3]{2^{12}\\cdot3^3\\cdot5^3}=240 \\end{align*} Adding gets \\begin{align*} a^2b+ab^2=ab(a+b)&=2400+5760\\\\ 240(a+b)&=240\\cdot(10+24)\\\\ a+b&=34 \\end{align*} Let $h$ be the hypotenuse then \\begin{align*} h&=\\sqrt{a^2+b^2}\\\\ &=\\sqrt{(a+b)^2-2ab}\\\\ &=\\sqrt{34^2-2\\cdot240}\\\\ &=\\sqrt{676}\\\\ &=26 \\end{align*} ~ Nafer",
"Let $a$ and $b$ be the two legs of the equation. We can find $\\frac{a}{b}$ by doing $\\frac{1920\\pi}{800\\pi}$. This simplified is $\\frac{12}{5}$. We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively. Since the volume of the first cone is $800\\pi$, we use the formula for the volume of a cone and get $100\\pi x^3=800 \\pi$. Solving for $x$, we get $x=2$. Plugging in the side lengths to the Pythagorean Theorem, we get an answer of $026. ~bobthegod78",
"Let $a$ and $b$ be the two legs of the equation. We can find $\\frac{a}{b}$ by doing $\\frac{1920\\pi}{800\\pi}$. This simplified is $\\frac{12}{5}$. We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively. Since the volume of the first cone is $800\\pi$, we use the formula for the volume of a cone and get $100\\pi x^3=800 \\pi$. Solving for $x$, we get $x=2$. Plugging in the side lengths to the Pythagorean Theorem, we get an answer of $026. ~bobthegod78"
] |
1985-I-3 | 1,985 | 3 | Find $c$ if $a$ , $b$ , and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$ , where $i^2 = -1$ . | 198 | null | [
"Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$. Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$. Since $a, b$ are integers, this means $b$ is a divisor of 107, which is a prime number. Thus either $b = 1$ or $b = 107$. If $b = 107$, $3a^2 - 107^2 = 1$ so $3a^2 = 107^2 + 1$, but $107^2 + 1$ is not divisible by 3, a contradiction. Thus we must have $b = 1$, $3a^2 = 108$ so $a^2 = 36$ and $a = 6$ (since we know $a$ is positive). Thus $c = 6^3 - 3\\cdot 6 = 198. Video Solution by SpreadTheMathLove https://www.youtube.com/watch?v=mw2A1Fa7APM"
] |
1985-I-4 | 1,985 | 4 | A small square is constructed inside a square of area $1$ by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices, as shown in the figure. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$ . AIME 1985 Problem 4.png | 32 | null | [
"The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram. Using the smaller side of the parallelogram, $1/n$, as the base, where the height is 1, we find that the area of the parallelogram is $A = \\frac{1}{n}$. By the Pythagorean Theorem, the longer base of the parallelogram has length $l = \\sqrt{1^2 + \\left(\\frac{n - 1}{n}\\right)^2} = \\frac{1}{n}\\sqrt{2n^2 - 2n + 1}$, so the parallelogram has height $h = \\frac{A}{l} = \\frac{1}{\\sqrt{2n^2 - 2n + 1}}$. But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$. Solving this quadratic equation gives $n = 32.",
"Surrounding the square with area $\\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \\frac{1}{1985} = 1$, where $X$ is the area of the of the 4 triangles. We can thus use proportions to solve this problem. \\begin{eqnarray*} \\frac{GF}{BE}=\\frac{CG}{CB}\\implies \\frac{\\frac{1}{\\sqrt{1985}}}{BE}=\\frac{\\frac{1}{n}}{1}\\implies BE=\\frac{n\\sqrt{1985}}{1985} \\end{eqnarray*} Also, \\begin{eqnarray*} \\frac{BE}{1}=\\frac{EC}{\\frac{n-1}{n}}\\implies EC=\\frac{\\sqrt{1985}}{1985}(n-1) \\end{eqnarray*} Thus, \\begin{eqnarray*} 2(BE)(EC)+\\frac{1}{1985}=1\\\\ 2n^{2}-2n+1=1985\\\\ n(n-1)=992 \\end{eqnarray*} Simple factorization and guess and check gives us $32.",
"Line Segment $DE = \\frac{1}{n}$, so $EC = 1 - \\frac{1}{n} = \\frac{n-1}{n}$. Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\\frac{1}{\\sqrt{1985}}$, as it is the same length as the sides of the square. Notice that $\\triangle CEL$ is similar to $\\triangle HDE$ by $AA$ similarity. Thus, $\\frac{LC}{HE} = \\frac{EC}{DE} = n-1$, so $LC = \\frac{n-1}{\\sqrt{1985}}$. Notice that $\\triangle CEL$ is also similar to $\\triangle CDF$ by $AA$ similarity. Thus, $\\frac{FC}{EC} = \\frac{DC}{LC}$, and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$. Solving this quadratic equation yields $n =32."
] |
1985-I-5 | 1,985 | 5 | A sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$ . What is the sum of the first $2001$ terms of this sequence if the sum of the first $1492$ terms is $1985$ , and the sum of the first $1985$ terms is $1492$ ? | 986 | null | [
"The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$. Then $a_3 = b - a$, $a_4 = (b - a) - b = -a$, $a_5 = -a - (b - a) = -b$, $a_6 = -b - (-a) = a - b$, $a_7 = (a - b) - (-b) = a$ and $a_8 = a - (a - b) = b$. Since the sequence is recursively defined by the first 2 terms, after this point it must continue to repeat. Thus, in particular $a_{j + 6} = a_j$ for all $j$, and so repeating this $n$ times, $a_{j + 6n} = a_j$ for all integers $n$ and $j$. Because of this, the sum of the first 1492 terms can be greatly simplified: $1488 = 6 \\cdot 248$ is the largest multiple of 6 less than 1492, so \\[\\sum_{i = 1}^{1492} a_i = (a_{1489} + a_{1490} + a_{1491} + a_{1492})+ \\sum_{i = 1}^{1488} a_i = (a_1 + a_2 + a_3 + a_4) + \\sum_{n = 0}^{247}\\sum_{j = 1}^6 a_{6n + j}\\] \\[=(a + b + (b - a) + (-a)) + \\sum_{n = 0}^{247}\\sum_{j = 1}^6 a_j = 2b - a,\\] where we can make this last step because $\\sum_{j = 1}^6 a_j = 0$ and so the entire second term of our expression is zero. Similarly, since $1980 = 6 \\cdot 330$, $\\sum_{i = 1}^{1985} a_i = (a_1 + a_2 + a_3 + a_4 + a_5) + \\sum_{i = 1}^{1980}a_i = a + b + (b - a) + (-a) + (-b) = b - a$. Finally, $\\sum_{i = 1}^{2001}a_i = a_1 + a_2 + a_3 + \\sum_{i = 1}^{1998} a_i = a + b + (b - a) = 2b$. Then by the givens, $2b - a = 1985$ and $b - a = 1492$ so $b = 1985 - 1492 = 493$ and so the answer is $2\\cdot 493 = 986."
] |
End of preview. Expand
in Data Studio
README.md exists but content is empty.
- Downloads last month
- 91