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IMO-1959-1
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_1
|
Prove that the fraction \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).
|
[
"Denoting the greatest common divisor of \\(a, b\\) as \\((a,b)\\), we use the Euclidean algorithm:\n\n\\[\n(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1\n\\]\n\nTheir greatest common divisor is 1, so \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"Proof by contradiction:\n\nAssume that \\(\\dfrac{14n+3}{21n+4}\\) is a reducible fraction where \\(p\\) is the greatest common factor of \\(14n+3\\) and \\(21n + 4\\). Thus,\n\n\\[\n21n+4\\equiv 0\\pmod{p} \\implies 42n+8\\equiv 0\\pmod{p}\n\\]\n\nSubtracting the second equation from the first equation we get \\(1\\equiv 0\\pmod{p}\\) which is clearly absurd.\n\nHence \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"Proof by contradiction:\n\nAssume that \\(\\dfrac{14n+3}{21n+4}\\) is a reducible fraction.\n\nIf a certain fraction \\(\\dfrac{a}{b}\\) is reducible, then the fraction \\(\\dfrac{2a}{3b}\\) is reducible, too. In this case, \\(\\dfrac{2a}{3b} = \\dfrac{42n+8}{42n+9}\\).\n\nThis fraction consists of two consecutives numbers, which never share any factor. So in this case, \\(\\dfrac{2a}{3b}\\) is irreducible, which is absurd.\n\nHence \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"We notice that:\n\n\\[\n\\frac{21n+4}{14n+3} = \\frac{(14n+3)+(7n+1)}{14n+3} = 1+\\frac{7n+1}{14n+3}\n\\]\n\nSo it follows that \\(7n+1\\) and \\(14n+3\\) must be coprime for every natural number \\(n\\) for the fraction to be irreducible. Now the problem simplifies to proving \\(\\frac{7n+1}{14n+3}\\) irreducible. We re-write this fraction as:\n\n\\[\n\\frac{7n+1}{(7n+1)+(7n+1) + 1} = \\frac{7n+1}{2(7n+1)+1}\n\\]\n\nSince the denominator \\(2(7n+1) + 1\\) differs from a multiple of the numerator \\(7n+1\\) by 1, the numerator and the denominator must be relatively prime natural numbers. Hence it follows that \\(\\frac{21n+4}{14n+3}\\) is irreducible.\n\nQ.E.D",
"By Bezout's Lemma, \\(3 \\cdot (14n+3) - 2 \\cdot (21n + 4) = 1\\), so the GCD of the numerator and denominator is \\(1\\) and the fraction is irreducible.",
"To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with \\(\\frac{21n}{14n}\\) + \\(\\frac{4}{3}\\). Simplifying the fraction, we end up with. \\(\\frac{3}{2}\\). Now combining these fractions with addition as shown before in the problem, we end up with \\(\\frac{17}{6}\\). It's important to note that 17 is 1 off from being divisible by 6, and you'll see why later down this explanation. Now we expirement with some numbers. Plugging 1 in gives us \\(\\frac{25}{17}\\). Notice how both numbers are one off from being divisible by 8(25 is next to 24, 17 is next to 16). Trying 2, we end up with \\(\\frac{46}{31}\\). Again, both results are 1 off from being multiples of 15. Trying 3, we end up with \\(\\frac{67}{45}\\). Again, both end up 1 away from being multiples of 22. This is where the realization comes in that two scenarios keep reoccurring:\n\n1. Both Numerator and Denominator keep ending up 1 value away from being multiples of the same number, but\n\n2. One always ends up being a prime number. Knowing that prime numbers only factors are 1 and itself, the fraction ends up being a paradoxical expression where one prime number is always being produced, and even with larger values, like say 15, implementing it in gives us \\(\\frac{319}{199}\\), we keep ending up with results that at the end will only have a common divisor of 1.",
"Let \\(\\gcd (21n+4,14n+3)=a.\\) So for some co-prime positive integers \\(x,y\\) we have\n\n\\[\n21n+4 = ax \\qquad (1)\n\\]\n\n\\[\n14n+3 = ay \\qquad (2)\n\\]\n\nMultiplying \\((1)\\) by \\(2\\) and \\((2)\\) by \\(3\\) and then subtracting \\((1)\\) from \\((2)\\) we get\n\n\\[\n42n+9-(42n+8)=3ay-2ax\n\\]\n\n\\[\n\\implies a(3y-2x)=1.\n\\]\n\nWe must have \\(a=1\\), since \\(a\\) is an positive integer. Thus, \\(\\gcd(21n+4,14n+3)=1\\) which means the fraction is irreducible, as needed."
] |
IMO-1959-2
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_2
|
For what real values of \(x\) is
\[
\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,
\]
given (a) \(A=\sqrt{2}\), (b) \(A=1\), (c) \(A=2\), where only non-negative real numbers are admitted for square roots?
|
[
"The square roots imply that \\(x\\ge \\frac{1}{2}\\).\n\nSquare both sides of the given equation:\n\n\\[\nA^2 = \\Big( x + \\sqrt{2x - 1}\\Big) + 2 \\sqrt{x + \\sqrt{2x - 1}} \\sqrt{x - \\sqrt{2x - 1}} + \\Big( x - \\sqrt{2x - 1}\\Big)\n\\]\n\nAdd the first and the last terms to get:\n\n\\[\nA^2 = 2x + 2 \\sqrt{x + \\sqrt{2x - 1}} \\sqrt{x - \\sqrt{2x - 1}}\n\\]\n\nMultiply the middle terms, and use \\((a + b)(a - b) = a^2 - b^2\\) to get:\n\n\\[\nA^2 = 2x + 2 \\sqrt{x^2 - 2x + 1}\n\\]\n\nSince the term inside the square root is a perfect square, and by factoring 2 out, we get\n\n\\[\nA^2 = 2(x + \\sqrt{(x-1)^2})\n\\]\n\nUse the property that \\(\\sqrt{x^2}=|x|\\) to get\n\n\\[\nA^2 = 2(x+|x-1|)\n\\]\n\nCase I: If \\(x \\le 1\\), then \\(|x-1| = 1 - x\\), and the equation reduces to \\(A^2 = 2\\). This is precisely part (a) of the question, for which the valid interval is now \\(x \\in \\left[ \\frac{1}{2}, 1 \\right]\\)\n\nCase II: If \\(x > 1\\), then \\(|x-1| = x - 1\\) and we have\n\n\\[\nx = \\frac{A^2 + 2}{4} > 1\n\\]\n\nwhich simplifies to\n\n\\[\nA^2 > 2\n\\]\n\nThis tells there that there is no solution for (b), since we must have \\(A^2 \\ge 2\\)\n\nFor (c), we have \\(A = 2\\), which means that \\(A^2 = 4\\), so the only solution is \\(x=\\frac{3}{2}\\).\n\n~flamewavelight (Expanded)\n\n~phoenixfire (edited)",
"Note that the equation can be rewritten to\n\n\\[\n\\sqrt{(\\sqrt{2x-1}+1)^2} + \\sqrt{(\\sqrt{2x-1}-1)^2}=A\\sqrt{2}\n\\]\n\ni.e., \\(\\sqrt{2x-1}+1 + |\\sqrt{2x-1}-1|=A\\sqrt{2}\\).\n\nCase I: when \\(2x-1\\ge 1\\) (i.e., \\(x\\ge 1\\)), the equation becomes \\(2\\sqrt{2x-1}=\\sqrt{2}A\\). For (a), we have \\(x=1\\); for (b) we have \\(x=\\frac{3}{4}\\); for (c) we have \\(x=\\frac{3}{2}\\). Since \\(x\\ge 1\\), (b) \\(x=\\frac{3}{4}\\) is not what we want.\n\nCase II: when \\(0\\le 2x-1 <1\\) (i.e., \\(1/2\\le x <1\\)), the equation becomes \\(2=\\sqrt{2}A\\), which only works for (a) \\(A=\\sqrt{2}\\).\n\nIn summary, any \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\) is a solution for (a); there is no solution for (b); there is one solution for (c), which is \\(x=\\frac{3}{2}\\).",
"For real \\(x\\), \\(x \\ge \\frac{1}{2}\\). Also,\n\n\\[\n\\sqrt{x+\\sqrt{2x-1}} = \\frac{1}{\\sqrt{2}} |\\sqrt{2x-1}+1|\n\\]\n\n\\[\n= \\frac{1}{\\sqrt{2}} (\\sqrt{2x-1}+1),\n\\]\n\n\\[\n\\sqrt{x-\\sqrt{2x-1}} = \\frac{1}{\\sqrt{2}} |\\sqrt{2x-1}-1|\n\\]\n\nIt is easy to know:\n\n(1) When \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\),\n\n\\[\n|\\sqrt{2x-1}-1| = 1-\\sqrt{2x-1}.\n\\]\n\n(2) When \\(x \\ge 1,\\)\n\n\\[\n|\\sqrt{2x-1}-1| = \\sqrt{2x-1}-1.\n\\]\n\nLet's define\n\n\\[\nA = \\sqrt{x+\\sqrt{2x-1}} + \\sqrt{x-\\sqrt{2x-1}} =\n\\]\n\n\\[\n= \\frac{1}{\\sqrt{2}} (\\sqrt{2x-1}+1+|\\sqrt{2x-1}-1|).\n\\]\n\nHence, for \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\), \\(A = \\sqrt{2};\\) \\(x \\ge 1,\\) \\(A \\ge \\sqrt{2}.\\)\n\nTo sum up, the solution for (a) \\(x \\in \\left[\\frac{1}{2}, 1\\right]\\); (b) No solution; (c) \\(x=\\frac{3}{2}\\).\n\n~EUCLID_2003"
] |
IMO-1959-3
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_3
|
Let \(a,b,c\) be real numbers. Consider the quadratic equation in \(\cos{x}\) :
\[
a\cos ^{2}x + b\cos{x} + c = 0.
\]
Using the numbers \(a,b,c\), form a quadratic equation in \(\cos{2x}\), whose roots are the same as those of the original equation. Compare the equations in \(\cos{x}\) and \(\cos{2x}\) for \(a=4, b=2, c=-1\).
|
[
"Let the original equation be satisfied only for \\(\\cos{x}=m, \\cos{x}=n\\). Then we wish to construct a quadratic with roots \\(2m^2 -1, 2n^2 -1\\).\n\nClearly, the sum of the roots of this quadratic must be\n\n\\[\n2 (m^2 + n^2) - 2 = 2 \\left(\\frac{b^2-2ac}{a^2}\\right) - 2 = \\frac{2b^2 - 4ac - 2a^2}{a^2},\n\\]\n\nand the product of its roots must be\n\n\\[\n4m^2 n^2 - 2(m^2 + n^2) + 1 = \\frac{4c^2}{a^2}+\\frac{4ac - 2b^2}{a^2} + \\frac{a^2}{a^2} = \\frac{(a+2c)^2 - 2b^2}{a^2}\n\\]\n\nThus the following quadratic fulfils the conditions:\n\n\\[\na^2 \\cos ^2 {2x} + (2a^2 + 4ac - 2b^2)\\cos{2x} + (a+2c)^2 - 2b^2 = 0\n\\]\n\nNow, when we let \\(a=4, b=2, c= -1\\), our equations are\n\n\\[\n4 \\cos^2 {x} + 2 \\cos {x} - 1 = 0\n\\]\n\nand\n\n\\[\n16 \\cos^2 {2x} + 8 \\cos {2x} - 4 = 0,\n\\]\n\nThis simplifies to previous equation. The first root of the first equation \\(\\cos {x} = \\frac{-1 + \\sqrt{5}}{4}\\) corresponds to \\(\\cos {2x} = \\frac{-1 - \\sqrt{5}}{4}\\) and the second root of the first equation \\(\\cos {x} = \\frac{-1 - \\sqrt{5}}{4}\\) corresponds to \\(\\cos {2x} = \\frac{-1 + \\sqrt{5}}{4}\\). These are relatively well-known trigonometric values, yielding roots \\(x = \\boxed{72^\\circ, 144^\\circ}\\)."
] |
IMO-1959-4
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_4
|
Construct a right triangle with a given hypotenuse \(c\) such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.
|
[
"We denote the catheti of the triangle as \\(a\\) and \\(b\\). We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse. (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)",
"The conditions of the problem require that\n\n\\[\nab = \\frac{c^2}{4}.\n\\]\n\nHowever, we notice that twice the area of the triangle \\(abc\\) is \\(ab\\), since \\(a\\) and \\(b\\) form a right angle. However, twice the area of the triangle is also the product of \\(c\\) and the altitude to \\(c\\). Hence the altitude to \\(c\\) must have length \\(\\frac{c}{4}\\). Therefore if we construct a circle with diameter \\(c\\) and a line parallel to \\(c\\) and of distance \\(\\frac{c}{4}\\) from \\(c\\), either point of intersection between the line and the circle will provide a suitable third vertex for the triangle. Q.E.D.",
"We denote the angle between \\(b\\) and \\(c\\) as \\(\\alpha\\). The problem requires that\n\n\\[\nab = \\frac{c^2}{4},\n\\]\n\nor, equivalently, that\n\n\\[\n2 \\frac{ab}{c^2} = \\frac{1}{2}.\n\\]\n\nHowever, since \\(\\frac{a}{c} = \\sin{\\alpha};\\; \\frac{b}{c} = \\cos{\\alpha}\\), we can rewrite the condition as\n\n\\[\n2\\sin{\\alpha}\\cos{\\alpha} = \\frac{1}{2},\n\\]\n\nor, equivalently, as\n\n\\[\n\\sin{2\\alpha} = \\frac{1}{2}.\n\\]\n\nFrom this it becomes apparent that \\(2\\alpha = \\frac{\\pi}{6}\\) or \\(\\frac{5\\pi}{6}\\); hence the other two angles in the triangle must be \\(\\frac{ \\pi }{12}\\) and \\(\\frac{ 5 \\pi }{12}\\), which are not difficult to construct. Q.E.D.\n\nNote. It is not difficult to reconcile these two constructions. Indeed, we notice that the altitude of the triangle is of length \\(c \\sin{\\alpha}\\cos{\\alpha}\\), which both of the solutions set equal to \\(\\frac{c}{4}\\) .",
"If we let the legs be \\(a\\) and \\(b\\) with \\(a < b\\), then \\(c^2 = a^2 + b^2\\). Because \\(c^2 = 4 a b\\) as well, we immediately deduce via some short computations through quadratic formula that \\(a = (2 - \\sqrt{3})b\\) and \\(a = (2 + \\sqrt{3})b\\). Thus, \\(\\frac{a}{b} = \\tan 15^\\circ\\) and \\(\\frac{a}{b} = \\tan 75^\\circ\\), and so one of the angles of the triangle must be \\(15^\\circ\\) and the other must be \\(75^\\circ\\). A \\(15^\\circ\\) angle is easily constructed by bisecting a \\(30^\\circ\\) angle (which is formed by constructing the altitude of an equilateral triangle), and \\(75^\\circ\\) angle is constructed by constructing a 15 degree angle on top of the 60 degree point.",
"We draw perpendicular \\(h\\) on \\(c\\) and denote the angle between the median and \\(c\\) as \\(\\alpha\\). The area of the triangle is \\(\\frac{ab}{2}\\) or \\(\\frac{ch}{2}\\)\n\nWe can write \\(c\\) as \\(2\\sqrt{ab}\\) So we have,\n\n\\[\n\\frac{ab}{2} = \\frac{2\\sqrt{ab}h}{2}\n\\]\n\nWhich simplifies to be\n\n\\[\n\\frac{\\sqrt{ab}}{h} = \\frac{1}{2}\n\\]\n\nOr\n\n\\[\n\\sin\\alpha = \\sin\\frac{\\pi}{6}\n\\]\n\nSo, we draw \\(c\\) and from it's midpoint, we draw a segment \\(\\frac{c}{2}\\) forming an angle \\(\\frac{\\pi}{6}\\) Join the vertices to get the required triangle.",
"This is a geometric solution. We start with some observations in the first paragraph and describe the construction in the next paragraph. Consider any right triangle with hypotenuse c. Construct square with side length c on the opposite side of c from the triangle. Construct 3 more triangles congruent to the original triangle on the outside of each of the three other sides of the square in such a way that a larger square with side length a+b is formed. The area of this larger square is c^2+2ab = 1.5c^2 so a+b = (3/2)^(1/2) *c\n\nNow go back to the given c. Construct a square with side length c first. Let two adjacent corners of it be M and N. Construct the square's diagonals and let center of the square be O. Construct a circle around O with radius (3/2)^(1/2) * OM. To do so, take OM, first draw a 2:1 right triangle with OM being the short side to obtain a hypothenuse with length 3^(1/2) OM, then draw a 1:1 right triangle from that side to obtain a hypothenuse with length 6^(1/2)*OM. then bisect it for desired length. Next draw the circular locus of points X such that MXO is 45 degrees. To accomplish this, simply find the point on the perpendicular bisector of OM that is 1/(2OM) from OM. There are two such points, pick the one closer to N. Draw circle around this new point going through O and M. The intersection of the two circles is the desired third vertex of the triangle with the given hypotenuse c."
] |
IMO-1959-5
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_5
|
An arbitrary point \(M\) is selected in the interior of the segment \(AB\). The squares \(AMCD\) and \(MBEF\) are constructed on the same side of \(AB\), with the segments \(AM\) and \(MB\) as their respective bases. The circles about these squares, with respective centers \(P\) and \(Q\), intersect at \(M\) and also at another point \(N\). Let \(N'\) denote the point of intersection of the straight lines \(AF\) and \(BC\).
(a) Prove that the points \(N\) and \(N'\) coincide.
(b) Prove that the straight lines \(MN\) pass through a fixed point \(S\) independent of the choice of \(M\).
(c) Find the locus of the midpoints of the segments \(PQ\) as \(M\) varies between \(A\) and \(B\).
|
[
"## Part A\n\nSince the triangles \\(AFM, CBM\\) are congruent, the angles \\(AFM, CBM\\) are congruent; hence \\(AN'B\\) is a right angle. Therefore \\(N'\\) must lie on the circumcircles of both quadrilaterals; hence it is the same point as \\(N\\).\n\n\\[\n1IMO5A.JPG\n\\]\n\n## Part B\n\nWe observe that \\(\\frac{AM}{MB} = \\frac{CM}{MB} = \\frac{AN}{NB}\\) since the triangles \\(ABN, BCM\\) are similar. Then \\(NM\\) bisects \\(ANB\\).\n\nWe now consider the circle with diameter \\(AB\\). Since \\(ANB\\) is a right angle, \\(N\\) lies on the circle, and since \\(MN\\) bisects \\(ANB\\), the arcs it intercepts are congruent, i.e., it passes through the bisector of arc \\(AB\\) (going counterclockwise), which is a constant point.\n\n## Part C\n\nDenote the midpoint of \\(PQ\\) as \\(R\\). It is clear that \\(R\\)'s distance from \\(AB\\) is the average of the distances of \\(P\\) and \\(Q\\) from \\(AB\\), i.e., half the length of \\(AB\\), which is a constant. Therefore the locus in question is a line segment.",
"\\[\nIMO19595A.png\n\\]\n\n## Part a)\n\nNotice that \\(\\angle BAF = \\arctan (\\frac{MF}{AM}) = \\frac{MB}{AM}\\) and \\(\\angle ABC = \\arctan (\\frac{MC}{MB}) = \\frac{AM}{MB}\\).\n\n\\(\\implies\\) \\(\\angle BAF + \\angle ABC = 90^{\\circ}\\).\n\nNow notice that \\(\\angle FNB = 90^{\\circ}\\). Considering \\(\\angle BAF\\) as \\(\\theta\\), this gives \\(\\angle MBN = 90^{\\circ} - \\theta\\) and thus \\(\\angle MFN = 90^{\\circ} + \\theta\\). But notice that \\(\\angle MFA = 90^{\\circ} - \\theta\\), which means that \\(\\angle AFN = 180^{\\circ}\\). Therefore points \\(A, F, N\\) are collinear. Now \\(\\angle BNF = 90^{\\circ}\\) and \\(\\angle ANC = 90^{\\circ}\\). Therefore, \\(\\angle BNC = 180^{\\circ}\\) and thus points \\(B, N, C\\) are collinear. Therefore, AF and BC intersect at N.\n\n## Part b)\n\nConstruct the bisector of arc AB above AB. Call it X. \\(\\angle ANM = \\angle ACM = 45^{\\circ}\\). Now \\(\\angle ANB = 90^{\\circ}\\) which means N lies on the circle with AB as diameter.\n\n\\(\\implies\\) \\(\\angle ANX = \\angle ABX = 45^{\\circ} = \\angle ANM\\). Therefore since M and X are on the same side of \\(N\\), \\(NM\\) passes through \\(X\\) wherever we choose \\(M\\) on \\(AB\\).\n\n## Part c)\n\nLet the midpoint of \\(PQ\\) be \\(R\\). Let \\(G\\) be the midpoint of \\(AM\\). Let \\(F\\) be the midpoint of \\(MB\\). Let \\(I\\) be the foot of the perpendicular from \\(R\\) onto \\(AB\\). Therefore by Midpoint Theorem, \\(RI = \\frac{PG + HQ}{2} = \\frac{AG + HB}{2} = \\frac{AB}{4}\\). Therefore the distance \\(RI\\) is a constant and thus the locus is a straight line parallel to \\(AB\\) at a distance (to \\(AB\\), of course) of \\(\\frac{AB}{4}\\)."
] |
IMO-1959-6
|
https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_6
|
Two planes, \(P\) and \(Q\), intersect along the line \(p\). The point \(A\) is in the plane \(P\), and the point \({C}\) is in the plane \(Q\); neither of these points lies on the straight line \(p\). Construct an isosceles trapezoid \(ABCD\) (with \(AB\) parallel to \(DC\)) in which a circle can be inscribed, and with vertices \(B\) and \(D\) lying in the planes \(P\) and \(Q\), respectively.
|
[
"We first observe that we must have both lines \\(AB\\) (which we shall denote \\(a\\)) and \\(DC\\) (which we shall denote \\(c\\)) parallel to \\(p\\), since if one of them is not, then neither can be and they must both intersect \\(p\\) (since they are both coplanar with \\(p\\)), making them skew.\n\nNow we note since a circle can be inscribed in the trapezoid, we must have \\(AB + DC = AD + BC\\), and since the trapezoid is isosceles, this implies that each of the trapezoid's legs has length equal to the average of the lengths of the bases.\n\nWe can find this average by dropping perpendicular \\(AA'\\) to \\(c\\) such that \\(A'\\) is on \\(c\\). The average will be \\(A'C\\), which is one of the sides of the rectangle with sides on \\(a\\) and \\(c\\) with vertices at \\(A\\) and \\({C}\\).\n\nWe now draw a circle with center \\({C}\\) that contains \\(A'\\). The intersections of this circle with \\(a\\) are the two possible values of \\(B\\), from either of which it is trivial to determine the corresponding location for \\(D\\). It is worth noting that the intersection points may concur (in which case there is only one distinct possibility (a square)), or they may not occur at all. Q.E.D."
] |
IMO-1960-1
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_1
|
Determine all three-digit numbers \(N\) having the property that \(N\) is divisible by 11, and \(\dfrac{N}{11}\) is equal to the sum of the squares of the digits of \(N\).
|
[
"Let \\(N = 100a + 10b+c\\) for some digits \\(a,b,\\) and \\(c\\). Then\n\n\\[\n100a + 10b+c = 11m\n\\]\n\nfor some \\(m\\). We also have \\(m=a^2+b^2+c^2\\). Substituting this into the first equation and simplification, we get\n\n\\[\n100a+10b+c = 11a^2 +11b^2 +11c^2\n\\]\n\nFor an integer divisible by \\(11\\), the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by \\(11\\). Thus we get: \\(b = a + c\\) or \\(b = a + c - 11\\).\n\nCase \\(1\\): Let \\(b=a+c\\). We get\n\n\\[\n100a+c+10a+10c = 11a^2 +11c^2+11(a+c)^2\n\\]\n\n\\[\n10a+c = 2a^2+2ac+2c^2\n\\]\n\nSince the right side is even, the left side must also be even. Let \\(c=2q\\) for some \\(q = 0,1,2,3,4\\). Then\n\n\\[\n10a+2q=2a^2+4aq+8q^2\n\\]\n\n\\[\n5a+q=a^2+2aq+4q^2\n\\]\n\nSubstitute \\(q=0,1,2,3,4\\) into the last equation and then solve for \\(a\\).\n\nWhen \\(q=0\\), we get \\(a=5\\). Thus \\(c=0\\) and \\(b=5\\). We get that \\(N=550\\) which works.\n\nWhen \\(q=1\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nWhen \\(q=2\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nWhen \\(q=3\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nWhen \\(q=4\\), we get that \\(a\\) is not an integer. There is no \\(N\\) for this case.\n\nCase \\(2\\): Let \\(b = a + c - 11\\). We get\n\n\\[\n100a+c+10a+10c -110= 11(a^2+(a+c)^2-22(a+c)+c^2+121)\n\\]\n\n\\[\n10a+c=2a^2+2c^2+2ac-22a-22c+131\n\\]\n\n\\[\n2(a-8)^2+2(c-\\frac{23}{4})^2+2ac-\\frac{505}{8}=0\n\\]\n\nNow we test all \\(c=0\\rightarrow10\\). When \\(c=0,1,2,4,5,6,7,8,9\\), we get no integer solution to \\(a\\). Thus, for these values of \\(c\\), there is no valid \\(N\\). However, when \\(c=3\\), we get\n\n\\[\n2(a-8)^2+2(3-\\frac{23}{4})^2+6a-\\frac{505}{8}=0\n\\]\n\n\\[\n2(a-8)^2+6a-48 = 0\n\\]\n\nWe get that \\(a=8\\) is a valid solution. For this case, we get \\(a=8,b=0,c=3\\), so \\(N=803\\), and this is a valid value. Thus, the answers are \\(\\boxed{N=550,803}\\).",
"Define a ten to be all ten positive integers which begin with a fixed tens digit.\n\nWe can make a systematic approach to this:\n\nBy inspection, \\(\\dfrac{N}{11}\\) must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.\n\nFor a given ten, the sum of the squares of the digits of \\(N\\) increases faster than \\(\\dfrac{N}{11}\\), so we can have at most one number in every ten that works.\n\nWe check the first ten:\n\n\\[\n11*11=121\n\\]\n\n\\[\n1^2+2^2+1^2=4\n\\]\n\n\\[\n12*11=132\n\\]\n\n\\[\n1^2+3^2+2^2=14\n\\]\n\n11 is too small and 12 is too large, so all numbers below 11 will be too small and all numbers above 12 will be too large, so no numbers in the first ten work.\n\nWe try the second ten:\n\n\\[\n21*11=231\n\\]\n\n\\[\n2^2+3^2+1^2=14\n\\]\n\n\\[\n22*11=242\n\\]\n\n\\[\n2^2+4^2+2^2=24\n\\]\n\nTherefore, no numbers in the second ten work.\n\nWe continue, to find out that 50 and 73 are the only ones that works.\n\n\\(N=50*11=550\\), \\(N=73*11=803\\) so there are two \\(N\\) that works.",
"Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases. We end up getting \\(\\boxed{N=550,803}\\)."
] |
IMO-1960-2
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_2
|
For what values of the variable \(x\) does the following inequality hold:
\[
\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?
\]
|
[
"Set \\(x = -\\frac{1}{2} + \\frac{a^2}{2}\\), where \\(a\\ge0\\). \\(\\frac{4\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)^2}{\\left(1-\\sqrt{1+2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)}\\right)^2}<2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)+9\\)\n\nAfter simplifying, we get \\((a+1)^2<a^2+8\\)\n\nSo \\(a^2+2a+1<a^2+8\\)\n\nWhich gives \\(a<\\frac{7}{2}\\) and hence \\(-\\frac{1}{2} \\le x<\\frac{45}{8}\\).\n\nBut \\(x=0\\) makes the LHS indeterminate.\n\nSo, answer: \\(-\\frac{1}{2} \\le x<\\frac{45}{8}\\), except \\(x=0\\).",
"If \\(x \\neq 0\\), then the LHS is defined and rewrites as follows:\n\n\\begin{align*} \\frac{4x^2}{(1 - \\sqrt{2x + 1})^2} &= \\biggl(\\frac{2x}{1 - \\sqrt{2x + 1}}\\biggl)^2 \\\\ &= \\biggl( \\frac{2x}{1 - \\sqrt{2x + 1}} \\cdot \\frac{1 + \\sqrt{2x + 1}}{1 + \\sqrt{2x + 1}} \\biggl)^2 \\\\ &= (1 + \\sqrt{2x + 1})^2 \\\\ &= 2x + 2\\sqrt{2x + 1} + 2. \\end{align*}\n\nThe inequality therefore holds if and only if\n\n\\[\n2x + 2\\sqrt{2x + 1} + 2 < 2x + 9.\n\\]\n\nor\n\n\\[\n\\sqrt{2x + 1} < \\frac{7}{2}.\n\\]\n\nSo \\(2x + 1 < 49/4\\) and therefore \\(x < 45/8\\). But if \\(x < -1/2\\) then the inequality makes no sense, since \\(\\sqrt{2x + 1}\\) is imaginary. So the original inequality holds iff \\(x\\) is in \\([-1/2, 0) \\cup (0, 45/8).\\)"
] |
IMO-1960-3
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_3
|
In a given right triangle \(ABC\), the hypotenuse \(BC\), of length \(a\), is divided into \(n\) equal parts (\(n\) an odd integer). Let \(\alpha\) be the acute angle subtending, from \(A\), that segment which contains the midpoint of the hypotenuse. Let \(h\) be the length of the altitude to the hypotenuse of the triangle. Prove that:
\[
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
\]
|
[
"Using coordinates, let \\(A=(0,0)\\), \\(B=(b,0)\\), and \\(C=(0,c)\\). Also, let \\(PQ\\) be the segment that contains the midpoint of the hypotenuse with \\(P\\) closer to \\(B\\).\n\n\\[\n[asy] size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); dot(Q); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NW); label(\"P\",P,ENE); label(\"Q\",Q,NNE); draw(A--B--C--cycle); draw(A--P); draw(A--Q); [/asy]\n\\]\n\nThen, \\(P = \\frac{n+1}{2}B+\\frac{n-1}{2}C = \\left(\\frac{n+1}{2}b,\\frac{n-1}{2}c\\right)\\), and \\(Q = \\frac{n-1}{2}B+\\frac{n+1}{2}C = \\left(\\frac{n-1}{2}b,\\frac{n+1}{2}c\\right)\\).\n\nSo, \\(\\text{slope}\\)\\((PA)=\\tan{\\angle PAB}=\\frac{c}{b}\\cdot\\frac{n-1}{n+1}\\), and \\(\\text{slope}\\)\\((QA)=\\tan{\\angle QAB}=\\frac{c}{b}\\cdot\\frac{n+1}{n-1}\\).\n\nThus, \\(\\tan{\\alpha} = \\tan{(\\angle QAB - \\angle PAB)} = \\frac{(\\frac{c}{b}\\cdot\\frac{n+1}{n-1})-(\\frac{c}{b}\\cdot\\frac{n-1}{n+1})}{1+(\\frac{c}{b}\\cdot\\frac{n+1}{n-1})\\cdot(\\frac{c}{b}\\cdot\\frac{n-1}{n+1})}\\) \\(= \\frac{\\frac{c}{b}\\cdot\\frac{4n}{n^2-1}}{1+\\frac{c^2}{b^2}} = \\frac{4nbc}{(n^2-1)(b^2+c^2)}=\\frac{4nbc}{(n^2-1)a^2}\\).\n\nSince \\([ABC]=\\frac{1}{2}bc=\\frac{1}{2}ah\\), \\(bc=ah\\) and \\(\\tan{\\alpha}=\\frac{4nh}{(n^2-1)a}\\) as desired.",
"Let \\(P, Q, R\\) be points on side \\(BC\\) such that segment \\(PR\\) contains midpoint \\(Q\\), with \\(P\\) closer to \\(C\\) and (without loss of generality) \\(AC \\le AB\\). Then if \\(AD\\) is an altitude, then \\(D\\) is between \\(P\\) and \\(C\\). Combined with the obvious fact that \\(Q\\) is the midpoint of \\(PR\\) (for \\(n\\) is odd), we have\n\n\\[\n\\tan {\\angle PAR} = \\tan (\\angle RAD - \\angle PAD) = \\frac{\\frac{PR}{h}}{1 + \\frac{DP \\cdot DR}{h^2}} = \\frac{PR \\cdot h}{h^2 + DP \\cdot DR} = \\frac{PR \\cdot h}{AQ^2 - DQ^2 + DP \\cdot DR} = \\frac{PR \\cdot h}{\\frac{a^2}{4} - PQ^2} = \\frac{\\frac{a}{n} \\cdot h}{\\frac{a^2}{4} - \\frac{a^2}{4n^2}} = \\frac{4nh}{(n^2-1)a}.\n\\]",
"Let \\(\\angle ACB = x\\), and \\(\\angle ABC = 90^\\circ - x\\). Let \\(M\\) be the midpoint on the hypotenuse \\(BC\\), and \\(Q\\) and \\(P\\) be points such that \\(PQ\\) contains \\(BC\\), with \\(Q\\) closer to \\(C\\) and \\(P\\) closer to \\(B\\). The midpoint will always be in the middle of line \\(QP\\), unless \\(n\\) is even or infinite, which it is not. Given such a triangle, we can express the altitude to the hypotenuse as:\n\n\\[\nh = a \\cos(x) \\sin(x)\n\\]\n\nNext, we shall denote line \\(AM\\) as \\(f\\), where \\(AM\\) is the median to the hypotenuse. This means that line \\(AM = BM = CM\\), and as \\(BM = \\frac{a}{2}\\), we have:\n\n\\[\nf = \\frac{a}{2}\n\\]\n\nWe know that \\(\\angle MAB = 90^\\circ - x\\), and \\(\\angle MAC = x\\). This means that \\(\\angle AMB = 2x\\) and \\(\\angle AMC = 180^\\circ - 2x\\). The length of \\(QP\\) is \\(\\frac{a}{n}\\). Let \\(\\angle QAM = k\\) and \\(\\angle PAM = z\\), such that \\(\\angle QAP\\) (or \\(\\alpha\\)) equals \\(k + z\\). This means that \\(\\angle AQM = 2x - k\\), and \\(\\angle APM = 180^\\circ - 2x - z\\).\n\nAs \\(M\\) is in the middle of \\(QP\\), we have \\(QM = PM = \\frac{a}{2n}\\). Applying the sine law on triangle \\(AQM\\), we get:\n\n\\[\n\\frac{\\sin(k)}{\\frac{a}{2n}} = \\frac{\\sin(2x - k)}{\\frac{a}{2}}\n\\]\n\nSimplifying:\n\n\\[\n\\frac{2n \\sin(k)}{a} = \\frac{2 \\sin(2x - k)}{a}\n\\]\n\n\\[\nn \\sin(k) = \\sin(2x - k)\n\\]\n\nUsing the identity \\(\\sin(2x - k) = \\sin(2x) \\cos(k) - \\cos(2x) \\sin(k)\\), and since \\(\\sin(2x) = 2 \\sin(x) \\cos(x)\\), we substitute:\n\n\\[\n\\sin(2x) = \\frac{2h}{a}\n\\]\n\nThus:\n\n\\[\nn \\sin(k) = \\frac{2h}{a} \\cos(k) - \\cos(2x) \\sin(k)\n\\]\n\nNow, we know that:\n\n\\[\n\\cos(2x) = \\frac{\\sqrt{a^2 - 4h^2}}{a}\n\\]\n\nSubstituting this into the equation:\n\n\\[\nn \\sin(k) + \\sin(k) \\frac{\\sqrt{a^2 - 4h^2}}{a} = \\cos(k) \\frac{2h}{a}\n\\]\n\nFactoring out \\(\\sin(k)\\):\n\n\\[\n\\sin(k) \\left( n + \\frac{\\sqrt{a^2 - 4h^2}}{a} \\right) = \\cos(k) \\frac{2h}{a}\n\\]\n\nThus:\n\n\\[\n\\tan(k) = \\frac{2h}{an + \\sqrt{a^2 - 4h^2}}\n\\]\n\nBy performing similar steps with \\(\\tan(z)\\), we can use the addition formula for \\(\\tan(z+k)\\) to find \\(\\tan(\\alpha)\\), where \\(\\alpha = z + k\\).\n\nCourtesy of EobardThawne",
"Let \\(D\\) and \\(E\\) be points on \\(BC\\) such that \\(\\alpha=\\angle DAE\\), \\(D\\) is closer to \\(B\\) than to \\(C\\), and \\(E\\) is closer to \\(C\\) than \\(B\\). Furthermore, let \\(\\alpha_1=\\angle BAD\\) and \\(\\alpha_2=\\angle CAE\\), and let \\(\\beta=\\angle B\\) and \\(\\gamma=\\angle C\\). Applying Law of Sines on \\(\\triangle BAD\\) yields \\begin{align*} \\frac{\\sin\\alpha_1}{\\frac{a(n-1)}{2n}}&=\\frac{\\sin(180-(\\alpha_1+\\beta))}{c}\\\\ \\frac{2cn}{a(n-1)}\\sin\\alpha_1&=\\sin(\\alpha_1+\\beta)\\\\ \\frac{2cn}{a(n-1)}\\sin\\alpha_1&=\\sin\\alpha_1\\cos\\beta+\\cos\\alpha_1\\sin\\beta\\\\ \\frac{2cn}{a(n-1)}\\sin\\alpha_1&=\\sin\\alpha_1\\cdot\\frac{c}{a}+\\cos\\alpha_1\\cdot\\frac{b}{a}\\\\ \\frac{2cn}{n-1}\\sin\\alpha_1&=c\\sin\\alpha_1+b\\cos\\alpha_1\\\\ \\frac{cn+c}{n-1}\\sin\\alpha_1&=b\\cos\\alpha_1\\\\ \\tan\\alpha_1&=\\frac{b(n-1)}{c(n+1)} \\end{align*} Similarly, \\(\\tan\\alpha_2=\\frac{c(n-1)}{b(n+1)}\\). Next, we use the sum of tangents formula: \\begin{align*} \\tan(\\alpha_1+\\alpha_2)&=\\frac{\\frac{b(n-1)}{c(n+1)}+\\frac{c(n-1)}{b(n+1)}}{1-\\frac{b(n-1)}{c(n+1)}\\cdot\\frac{c(n-1)}{b(n+1)}}\\\\ &=\\frac{\\left(\\frac{b}{c}+\\frac{c}{b}\\right)\\cdot\\frac{n-1}{n+1}}{1-\\frac{(n-1)^2}{(n+1)^2}}\\\\ &=\\frac{\\left(\\frac{b}{c}+\\frac{c}{b}\\right)(n^2-1)}{(n+1)^2-(n-1)^2}\\\\ &=\\frac{\\left(\\frac{b}{c}+\\frac{c}{b}\\right)(n^2-1)}{4n}\\\\ &=\\frac{(b^2+c^2)(n^2-1)}{4nbc}\\\\ \\end{align*} Since \\(\\alpha+\\alpha_1+\\alpha_2=90^\\circ\\), we need to find \\(\\tan\\alpha=\\tan\\left(\\frac{\\pi}{2}-(\\alpha_1+\\alpha_2)\\right)\\). This is equal to \\(\\cot(\\alpha_1+\\alpha_2)\\), which is the reciprocal of \\(\\tan(\\alpha_1+\\alpha_2)\\), or \\(\\frac{4nbc}{(b^2+c^2)(n^2-1)}\\). Since \\(bc=ah\\) due to triangle areas and \\(a^2=b^2+c^2\\) due to the Pythagorean Theorem, we conclude that the expression is equal to \\(\\frac{4nah}{a^2(n^2-1)}\\), which simplifies to \\(\\frac{4nh}{(n^2-1)a}\\).\n\n~eevee9406"
] |
IMO-1960-4
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_4
|
Construct triangle \(ABC\), given \(h_a\), \(h_b\) (the altitudes from \(A\) and \(B\)), and \(m_a\), the median from vertex \(A\).
|
[
"Let \\(M_a\\), \\(M_b\\), and \\(M_c\\) be the midpoints of sides \\(\\overline{BC}\\), \\(\\overline{CA}\\), and \\(\\overline{AB}\\), respectively. Let \\(H_a\\), \\(H_b\\), and \\(H_c\\) be the feet of the altitudes from \\(A\\), \\(B\\), and \\(C\\) to their opposite sides, respectively. Since \\(\\triangle ABC\\sim\\triangle M_bM_aC\\), with \\(M_bM_a=\\frac12 AB\\), the distance from \\(M_a\\) to side \\(\\overline{AC}\\) is \\(\\frac{h_b}{2}\\).\n\nConstruct \\(AM_a\\) with length \\(m_a\\). Draw a circle centered at \\(A\\) with radius \\(h_a\\). Construct the tangent \\(l_1\\) to this circle through \\(M_a\\). \\(\\overline{BC}\\) lies on \\(l_1\\).\n\nDraw a circle centered at \\(M_a\\) with radius \\(\\frac{h_b}{2}\\). Construct the tangent \\(l_2\\) to this circle through \\(A\\). \\(\\overline{AC}\\) lies on \\(l_2\\). Then \\(C=l_1\\cap l_2\\).\n\nConstruct the line \\(l_3\\) parallel to \\(l_2\\) so that the distance between \\(l_2\\) and \\(l_3\\) is \\(h_b\\) and \\(M_a\\) lies between these lines. \\(B\\) lies on \\(l_3\\). Then \\(B=l_1\\cap l_3\\)."
] |
IMO-1960-5
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_5
|
Consider the cube \(ABCDA'B'C'D'\) (with face \(ABCD\) directly above face \(A'B'C'D'\)).
a) Find the locus of the midpoints of the segments \(XY\), where \(X\) is any point of \(AC\) and \(Y\) is any point of \(B'D'\);
b) Find the locus of points \(Z\) which lie on the segment \(XY\) of part a) with \(ZY = 2XZ\).
|
[
"Let \\(A=(0,0,2)\\), \\(B=(2,0,2)\\), \\(C=(2,0,0)\\), \\(D=(0,0,0)\\), \\(A'=(0,2,2)\\), \\(B'=(2,2,2)\\), \\(C'=(2,2,0)\\), and \\(D'=(0,2,0)\\). Then there exist real \\(x\\) and \\(y\\) in the closed interval \\([0,2]\\) such that \\(X=(x,0,2-x)\\) and \\(Y=(y,2,y)\\).\n\nThe midpoint of \\(XY\\) has coordinates \\(((x+y)/2, 1, (2-x+y)/2)\\). Let \\(a\\) and \\(b\\) be the \\(x\\)- and \\(z\\)-coordinates of the midpoint of \\(XY\\), respectively. We then have that \\(a+b=y+1\\) and \\(a-b=x-1\\), so \\(a+b\\in [1,3]\\) and \\(a-b\\in [-1,1]\\). The region of points that satisfy these inequalities is the closed square with vertices at \\((1,1,2)\\), \\((2,1,1)\\), \\((1,1,0)\\), and \\((0,1,1)\\). For every point \\(P\\) in this region, there exist unique points \\(X\\) and \\(Y\\) such that \\(P\\) is the midpoint of \\(XY\\).\n\nIf \\(Z\\in XY\\) and \\(ZY=2XZ\\), then \\(Z\\) has coordinates \\(((2x+y)/3, 2/3, (4-2x+y)/3)\\). Let \\(a\\) and \\(b\\) be the \\(x\\)- and \\(z\\)- coordinates of \\(Z\\). We then have that \\(a+b=(4/3)+(2/3)y\\) and \\(a-b=(4x-4)/3\\), and \\(a\\in (4/3,8/3)\\) and \\(b\\in (-4/3,4/3)\\). The region of points that satisfy these inequalities is the closed rectangle with vertices at \\((0,2/3,4/3)\\), \\((2/3,2/3,1)\\), \\((1,2/3,2/3)\\), and \\((4/3,2/3,0)\\). For every point \\(Z\\) in this region, there exist unique points \\(X\\) and \\(Y\\) such that \\(Z\\in XY\\) and \\(ZY=2XZ\\)."
] |
IMO-1960-6
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_6
|
Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let \(V_1\) be the volume of the cone and \(V_2\) be the volume of the cylinder.
a) Prove that \(V_1 \neq V_2\);
b) Find the smallest number \(k\) for which \(V_1 = kV_2\); for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.
|
[
"Part (a):\n\nLet \\(R\\) denote the radius of the cone, and let \\(r\\) denote the radius of the cylinder and sphere. Let \\(l\\) denote the slant height of the cone, and let \\(h\\) denote the height of the cone.\n\nConsider a plane that contains the axis of the cone. This plane will slice the cone and sphere into a circle \\(\\omega\\) inscribed in an isosceles triangle \\(T\\).\n\nThe area of \\(T\\) may be computed in two different ways:\n\n\\[\n[T] = \\frac{1}{2} \\times r \\times (2l + 2R) = r(l+R)\n\\]\n\n\\[\n[T] = \\frac{1}{2} \\times 2R \\times h = Rh\n\\]\n\nFrom this, we deduce that \\(r = \\frac{Rh}{l+R}\\).\n\nNow, we calculate our volumes:\n\n\\[\nV_1 = \\frac{1}{3}\\pi R^2 h\n\\]\n\n\\[\nV_2 = \\pi r^2 \\times 2r = 2\\pi r^3 = \\frac{2R^3 h^3}{(l+R)^3}\n\\]\n\nNow, we will compute the quantity \\(\\frac{3(l+R)^3}{\\pi R^5 h} (V_1 - V_2)\\) and prove that it is always greater than \\(0\\). Let \\(x = \\frac{h}{R}\\). Clearly, \\(x\\) can be any positive real number. Define \\(W_1 = \\frac{3(l+R)^3}{\\pi R^5 h} V_1\\) and \\(W_2 = \\frac{3(l+R)^3}{\\pi R^5 h} V_2\\). We will calculate \\(W_1\\) and \\(W_2\\) in terms of \\(x\\) and then compute the desired quantity \\(W_1 - W_2\\).\n\nIt is easy to see that:\n\n\\[\nW_1 = (\\sqrt{x^2+1} + 1)^3\n\\]\n\n\\[\nW_2 = 6x^2\n\\]\n\nNow, let \\(u = \\sqrt{x^2+1}\\). Since \\(x > 0\\), it follows that \\(u > 1\\). We now have:\n\n\\[\nW_1 = (u + 1)^3\n\\]\n\n\\[\nW_2 = 6(u^2 - 1)\n\\]\n\nDefine \\(f(u) = W_1 - W_2\\). It follows that:\n\n\\[\nf(u) = (u+1)^3 - 6(u^2 - 1)\n\\]\n\n\\[\nf(u) = u^3 - 3u^2 + 3u + 7\n\\]\n\n\\[\nf(u) = (u-1)^3 + 8 > 8 > 0\n\\]\n\nWe see that \\(f(u) > 8\\) for all allowed values of \\(u\\). Thus, \\(V_1 - V_2 > \\frac{8\\pi R^5 h}{3(l+R)^3}\\), meaning that \\(V_1 > V_2\\). We have thus proved that \\(V_1 \\ne V_2\\), as desired.\n\nPart (b):\n\nFrom our earlier work in calculating the volumes \\(V_1\\) and \\(V_2\\), we easily see that:\n\n\\[\n\\frac{V_1}{V_2} = \\frac{(u+1)^3}{6(u^2 - 1)}\n\\]\n\nRe-expressing and simplifying, we have:\n\n\\[\n\\frac{6V_1}{V_2} = \\frac{(u+1)^2}{u-1} = (u-1) + 4 + \\frac{4}{u-1}\n\\]\n\nBy the AM-GM Inequality, \\((u-1) + \\frac{4}{u-1} \\ge 4\\), meaning that \\(\\frac{V_1}{V_2} \\ge \\frac{4}{3}\\). Equality holds if and only if \\(u-1 = \\frac{4}{u-1}\\), meaning that \\(u=3\\) and \\(x = 2\\sqrt{2}\\).\n\nIf we check the case \\(x = \\frac{h}{R} = 2\\sqrt{2}\\), we may calculate \\(V_1\\) and \\(V_2\\):\n\n\\[\nV_1 = \\frac{1}{3} \\pi R^2 h = \\frac{2\\sqrt{2}}{3}\\pi R^3\n\\]\n\n\\[\nV_2 = 2\\pi r^3 = 2\\pi \\left(\\frac{R\\times 2\\sqrt{2} R}{4R}\\right)^3 = \\frac{\\pi}{\\sqrt{2}} R^3\n\\]\n\nIndeed, we have \\(\\frac{V_1}{V_2} = \\frac{4}{3}\\), meaning that our minimum of \\(k = \\frac{4}{3}\\) can be achieved.\n\nThus, we have proved that the minimum value of \\(k\\) such that \\(V_1 = kV_2\\) is \\(\\frac{4}{3}\\).\n\nNow, let \\(\\theta\\) be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following:\n\n\\[\n\\tan\\left(\\frac{\\theta}{2}\\right) = \\frac{1}{x} = \\frac{\\sqrt{2}}{4}\n\\]\n\nFrom the double-angle formula for tangent,\n\n\\[\n\\theta = \\arctan\\left(\\frac{4\\sqrt{2}}{7}\\right)\n\\]\n\nThis angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths \\(4\\sqrt{2}\\) and \\(7\\). This is straightforward, and the angle opposite the leg of length \\(4\\sqrt{2}\\) will be the desired angle \\(\\theta\\).\n\nIt follows that we have successfully constructed the desired angle \\(\\theta\\).",
"Part a):\n\nLet \\(r\\) and \\(a\\) denote the radius of the sphere and the radius of the base of the cone, respectively.\n\nConsider a plane that contains a cross section of the cone. We get a circle (the sphere) inscribed in both an isosceles triangle (the cone) and a rectangle (the cylinder). Let \\(A\\) and \\(A_1\\) be the vertices at the base of the triangle and let \\(H_1 H_2 = h\\) be the altitude through the third vertex \\(H_1\\). Let \\(\\theta\\) be the angle with arms \\(AO\\) and \\(AA_1\\), where \\(O\\) is the center of the circle; \\(0 < \\theta < \\frac{\\pi}{4}\\) .\n\nWe have \\(\\tan \\theta = \\frac{r}{a}\\) ; \\(0 < \\tan \\theta < 1\\). For the two volumes we have \\(V_2 = \\pi r^2 \\times 2r = 2\\pi r^3\\) and \\(V_1 = \\frac{\\pi a^2 \\times H_1 H_2}{3} = \\frac{\\pi a^3 \\tan (\\angle A_1 A H_1) }{3}\\) .\n\nKnowing \\(\\angle A_1 A H_1 = 2 \\times \\angle A_1 A O = 2 \\theta\\), and using the formula for a double-angle tangent, we can simplify the result:\n\n\\[\nV_1 = \\frac{\\pi a^3 \\tan (\\angle A_1 A H_1)}{3}\n\\]\n\n\\[\nV_1 = \\frac{2 \\pi r^3 \\tan \\theta}{3 \\tan^3 \\theta (1 - \\tan^2 \\theta)}\n\\]\n\n\\(V_1 = k V_2\\), where \\(k = \\frac{1}{3\\tan^2 \\theta (1 - \\tan^2 \\theta)}\\)\n\nWe see that the biquadratic equation \\(3 \\tan^4 \\theta - 3 \\tan^2 \\theta + 1 = 0\\) has no real roots for \\(\\tan \\theta\\), which means that \\(k \\neq 1\\). Therefore \\(V_1 \\neq V_2\\) .\n\nPart b):\n\nLet \\(\\tan^2 \\theta = x; 0 < x < 1\\). Consider the function \\(f(x) = 3x - 3x^2\\). The coefficient in front of \\(x^2\\) is negative, which means the function has an upper bound, i.e. k has a lower bound (since it is the reciprocal of \\(f(x)\\)). Using the vertex formula for a quadratic function, we get \\(x = -\\frac{b}{2a} = \\frac{1}{2}\\) , which gives us \\(f(x) = y_{max} = \\frac{3}{4}\\) . From there \\(k_{min} = \\frac{4}{3}\\) .\n\nFor \\(\\tan^2 \\theta\\) we have \\(\\tan^2 \\theta = \\frac{1}{2}; \\tan \\theta = +/- \\frac{\\sqrt{2}}{4}\\) . We defined \\(\\tan \\theta\\) to be in the interval (0;1), so \\(\\tan \\theta = \\frac{\\sqrt{2}}{4}\\) . That gives us \\(\\tan \\angle A_1 A H_1 = \\frac{4 \\sqrt{2}}{7}\\) . To find \\(2\\theta\\) we construct a right triangle with legs equal to \\(7\\) and \\(4\\sqrt{2}\\) using a unit length reference."
] |
IMO-1960-7
|
https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_7
|
An isosceles trapezoid with bases \(a\) and \(c\) and altitude \(h\) is given.
a) On the axis of symmetry of this trapezoid, find all points \(P\) such that both legs of the trapezoid subtend right angles at \(P\);
b) Calculate the distance of \(P\) from either base;
c) Determine under what conditions such points \(P\) actually exist. Discuss various cases that might arise.
|
[
"(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.\n\n(b) Let \\(x\\) be the distance from \\(P\\) to one of the bases; then \\(h - x\\) must be the distance from \\(P\\) to the other base. Similar triangles give \\(\\frac{x}{\\frac{a}{2}} = \\frac{\\frac{c}{2}}{h - x}\\), so \\(x^2 - hx + \\frac{ac}{4} = 0\\) and so \\(x = \\frac{h \\pm \\sqrt{h^2 - ac}}{2}.\\)\n\n(c) When \\(h^2 \\ge ac\\).\n\n[1]\n\nIn our above picture, ABCD is our trapezoid with AB=a and CD=c and its height is 'h'. AF and BE are perpendicular to CD such that AF= BE= h. XY is our axis of symmetry and it intersects with CD at a point O. Point O is our origin of reference whose coordinates are (0,0).\n\nLet our point P be on the axis of symmetry at z distance from the origin O.\n\nThe coordinates of the points A,B,C,D,E,F and P are given in the figure.\n\nNow,\n\nSlope of the line \\(PC= (z-0)/(0-c/2) = -2z/c\\) Slope of the line \\(PB= (z-h)/(0-a/2) = -2(z-h)/a\\)\n\nSince the leg BC subtends a right angle at P, the angle BPC should be a right angle. This means that the product of the slope of PC and PB is -1.\n\ni.e\n\n\\[\n4z(z-h)=-ac\n\\]\n\nor \\(z^2 - zh + ac/4= 0\\)\n\nNow, solving for z, we get, \\(z= [(h + ( h^2 - ac ) ^1/2 ]/2\\) and \\([(h - ( h^2 - ac ) ^1/2 ]/2\\)\n\nSo, z is the distance of the points from the base CD..\n\nAlso the points are possible only when , \\(h^2 - ac >= 0\\).. and doesn't exist for \\(h^2 -ac <0\\)"
] |
IMO-1961-1
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_1
|
Solve the system of equations:
\[
\begin{matrix} \quad x + y + z \!\!\! &= a \; \, \\ x^2 +y^2+z^2 \!\!\! &=b^2 \\ \qquad \qquad xy \!\!\! &= z^2 \end{matrix}
\]
where \(a\) and \(b\) are constants. Give the conditions that \(a\) and \(b\) must satisfy so that \(x, y, z\) (the solutions of the system) are distinct positive numbers.
|
[
"Note that \\(x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2\\), so the first two equations become\n\n\\[\n\\begin{matrix} \\quad (x + y) + z \\!\\!\\! &= a \\; \\; (*) \\\\ (x+y)^2 - z^2 \\!\\!\\! &=b^2 (**) \\end{matrix}\n\\]\n\n.\n\nWe note that \\((x+y)^2 - z^2 = \\Big[ (x+y)+z \\Big]\\Big[ (x+y)-z\\Big]\\), so if \\(a\\) equals 0, then \\(b\\) must also equal 0. We then have \\(x+y = -z\\); \\(xy = (x+y)^2\\). This gives us \\(x^2 + xy + y^2 = 0\\). Mutiplying both sides by \\((x-y)\\), we have \\(x^3 - y^3 = 0\\). Since we want \\(x,y\\) to be real, this implies \\(x = y\\). But \\(x^2 + x^2 + x^2\\) can only equal 0 when \\(x=0\\) (which, in this case, implies \\(y,z = 0\\)). Hence there are no positive solutions when \\(a = 0\\).\n\nWhen \\(a \\neq 0\\), we divide \\((**)\\) by \\((*)\\) to obtain the system of equations\n\n\\[\n\\begin{matrix} (x+y)+z &= a \\; \\quad \\\\ (x+y)-z &= b^2/a \\end{matrix}\n\\]\n\n,\n\nwhich clearly has solution \\(x+y = \\frac{a^2 + b^2}{2a}\\), \\(z = \\frac{a^2 - b^2}{2a}\\). In order for these both to be positive, we must have positive \\(a\\) and \\(a^2 > b^2\\). Now, we have \\(x+y = \\frac{a^2 + b^2}{2a}\\); \\(xy = \\left(\\frac{a^2 - b^2}{2a}\\right)^2\\), so \\(x,y\\) are the roots of the quadratic \\(m^2 - \\frac{a^2 + b^2}{2a}m + \\left(\\frac{a^2 - b^2}{2a}\\right)^2\\). The discriminant for this equation is\n\n\\[\n\\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2 = \\frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}\n\\]\n\n.\n\nIf the expressions \\((3a^2 - b^2), (3b^2 - a^2)\\) were simultaneously negative, then their sum, \\(2(a^2 + b^2)\\), would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when \\(3a^2 > b^2\\) and \\(3b^2 > a^2\\). But we have already replaced the first inequality with the sharper bound \\(a^2 > b^2\\). It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from \\(\\left(\\frac{a^2 + b^2}{2a}\\right)^2 > \\left(\\frac{a^2 + b^2}{2a}\\right)^2 - \\left(2\\frac{a^2 -b^2}{2a}\\right)^2\\) or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if \\(a\\) is positive and \\(3b^2 > a^2 > b^2\\). Q.E.D.",
"Obviously, \\(a=x+y+z>0\\). The third equation implies that \\(x,z,y\\) is a geometric sequence. Then let \\(x=\\frac{z}{r}\\) and \\(y=rz\\), with \\(r,z>0\\) and \\(r\\neq1\\). Then the first two equations become:\n\n\\[\n\\left(r+1+\\frac{1}{r}\\right)z=a~~~(1)\n\\]\n\nand\n\n\\[\n\\left(r^2+1+\\frac{1}{r^2}\\right)z^2=b^2~~~(2)\n\\]\n\nTaking \\(\\frac{(2)}{(1)}\\) (since \\(z>0\\)), we get:\n\n\\[\n\\frac{(r^2+1+\\frac{1}{r^2})z^2}{(r+1+\\frac{1}{r})z}=\\left(r-1+\\frac{1}{r}\\right)z=\\frac{b^2}{a}~~~(3)\n\\]\n\nWe can then take \\((1)^2-(2)\\) and \\((2)-(3)^2\\) to get:\n\n\\[\n\\left(r^2+2r+3+\\frac{2}{r}+\\frac{1}{r^2}\\right)z^2-\\left(r^2+1+\\frac{1}{r^2}\\right)z^2=2z^2\\left(r+1+\\frac{1}{r}\\right)=a^2-b^2~~~(4)\n\\]\n\nand\n\n\\[\n\\left(r^2+1+\\frac{1}{r^2}\\right)z^2-\\left(r^2-2r+3-\\frac{2}{r}+\\frac{1}{r^2}\\right)z^2=2z^2\\left(r-1+\\frac{1}{r}\\right)=b^2-\\frac{b^4}{a^2}=\\frac{b^2}{a^2}(a^2-b^2)~~~(5)\n\\]\n\nLet \\(k=r+\\frac{1}{r}\\). By AM-GM, \\(k\\ge2\\) with equality at \\(r=\\frac{1}{r}\\implies r=1\\), which is impossible. Hence, \\(k>2\\). Then, \\(\\frac{(5)}{(4)}\\) becomes:\n\n\\[\n\\frac{k-1}{k+1}=\\frac{b^2}{a^2}~(6)\\implies a^2+b^2=(a^2-b^2)k>2(a^2-b^2)\\implies3b^2>a^2\n\\]\n\nFrom the above restrictions on \\(a\\) and \\(b\\), we see that there must exist some \\(k>2\\) satisfying \\((6)\\), and hence, some \\(r>0\\neq1\\) satisfying \\((6)\\). From \\((4)\\), if \\(a^2-b^2>0\\), then there must exist some positive \\(z\\) satisfying \\((4)\\), and consequently since \\((4)\\) and \\((6)\\) are equivalent to the remaining equations, they satisfy \\((1)\\) and \\((2)\\). Hence, \\(x,y,z\\) satisfy the original system, and from the restrictions on \\(r\\) and \\(z\\), they are distinct positive reals. Hence, \\(\\boxed{a>0\\text{ and }3b^2>a^2>b^2}\\). \\(\\blacksquare\\)\n\n~rhydon516"
] |
IMO-1961-2
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_2
|
Let \(a\), \(b\), and \(c\) be the lengths of a triangle whose area is S. Prove that
\[
a^2 + b^2 + c^2 \ge 4S\sqrt{3}
\]
In what case does equality hold?
|
[
"By Heron's formula, we have\n\n\\[\nS = \\sqrt{s(s-a)(s-b)(s-c)}.\n\\]\n\nThis can be simplified to\n\n\\[\nS = \\sqrt{\\left(\\frac{a+b+c}{2}\\right)\\left(\\frac{-a+b+c}{2}\\right)\\left(\\frac{a-b+c}{2}\\right)\\left(\\frac{a+b-c}{2}\\right)}.\n\\]\n\nNext, we can factor out all of the \\(2\\)s and use a clever difference of squares:\n\n\\[\nS = \\frac{1}{4}\\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\n\\]\n\n\\[\nS = \\frac{1}{4}\\sqrt{((b+c)^2 - a^2)(a^2 - (b-c)^2)}\n\\]\n\n\\[\nS = \\frac{1}{4}\\sqrt{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}.\n\\]\n\nWe can now use difference of squares again:\n\n\\[\nS = \\frac{1}{4}\\sqrt{4b^2c^2-(b^2 + c^2 - a^2)^2}\n\\]\n\n\\[\n4S\\sqrt{3} = \\sqrt{3(4b^2c^2 - (b^2 + c^2 - a^2)^2)}\n\\]\n\nWe must prove that the RHS of this equation is less than or equal to \\(a^2 + b^2 + c^2\\).\n\nLet \\(a^2 = A\\), \\(b^2 = B\\), \\(c^2 = C\\). Then, our inequality can be reduced to\n\n\\[\nA + B + C \\geq \\sqrt{6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2}.\n\\]\n\nWe now have to prove\n\n\\[\n(A + B + C)^2 \\geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2.\n\\]\n\nWe can simplify:\n\n\\[\nA^2 + B^2 + C^2 + 2AB + 2BC + 2CA \\geq 6AB + 6BC + 6CA - 3A^2 - 3B^2 - 3C^2\n\\]\n\n\\[\n4A^2 + 4B^2 + 4C^2 \\geq 4AB + 4BC + 4CA\n\\]\n\n\\[\nA^2 + B^2 + C^2 \\geq AB + BC + CA.\n\\]\n\nFinally, we can apply AM-GM:\n\n\\[\n\\frac{A^2 + B^2}{2} \\geq AB\n\\]\n\n\\[\n\\frac{B^2 + C^2}{2} \\geq BC\n\\]\n\n\\[\n\\frac{C^2 + A^2}{2} \\geq CA\n\\]\n\nAdding these all up, we have the desired inequality\n\n\\[\nA^2 + B^2 + C^2 \\geq AB + BC + CA,\n\\]\n\nand so the proof is complete.\\(\\square\\)\n\nTo have \\(A + B + C = 4S\\sqrt{3}\\), we must satisfy\n\n\\[\n\\frac{A^2 + B^2}{2} = AB,\n\\]\n\n\\[\n\\frac{B^2 + C^2}{2} = BC,\n\\]\n\n\\[\n\\frac{C^2 + A^2}{2} = CA.\n\\]\n\nThis is only true when \\(A = B = C\\), and thus \\(a = b = c\\). Therefore, equality happens when the triangle is equilateral.\n\n~mathboy100",
"We firstly use the duality principle. \\(a=x+y~~b=x+z~~c=y+z\\) The LHS becomes \\((x+y)^2+(x+z)^2+(y+z)^2\\) and the RHS becomes \\(4\\sqrt{3}\\sqrt{(x+y+z)xyz}\\) If we use Heron's formula. By AM-GM \\(\\frac{(x+y+z)^3}{27} \\ge xyz\\) Making this substitution \\([ABC]\\) becomes \\(\\sqrt{(x+y+z)^4\\frac{1}{27}}\\) and once we take the square root of the area then our RHS becomes \\(\\frac{4\\sqrt{3}}{3\\sqrt{3}}(x+y+z)^2=\\frac{4}{3}(x+y+z)^2\\) Multiplying the RHS and the LHS by 3 we get the LHS to be \\(3((x+y)^2+(x+z)^2+(y+z)^2)=6(x^2+y^2+z^2+xy+yz+xz).\\) Our RHS becomes \\(4(x^2+y^2+z^2)+8(xy+yz+xz).\\) Subtracting \\(4(x^2+y^2+z^2)+6(xy+yz+xz)\\) we have the LHS equal to \\((2(x^2+y^2+z^2))\\) and the RHS being \\(2(xy+xz+yz)\\) If LHS \\(\\ge\\) RHS then LHS-RHS\\(\\ge 0\\) LHS-RHS=\\(2(x^2+y^2+z^2)-2(xy+xz+yz)=x^2-2xy+y^2+x^2-2x+z^2+y^2-2yz+z^2=(x-y)^2+(x-z)^2+(y-z)^2.\\) \\((x-y)^2+(x-z)^2+(y-z)^2 \\ge 0\\) by the trivial inequality so therefore, \\(a^2 + b^2 + c^2 \\ge 4S\\sqrt{3}\\) and we're done.\n\n~PEKKA",
"Let \\(\\theta\\) be the angle between edges \\(a\\) and \\(b\\) subtending \\(c\\).\n\nWe notice that \\(0\\le(a-b\\cos x)^2=a^2-2ab\\cos x+b^2\\cos^2x\\le a^2-2ab\\cos x+b^2\\) holds for all \\(a,b,x\\in\\mathbb R\\).\n\nIf we let \\(x=\\theta+30^\\circ\\),\n\n\\[\n\\begin{align*}a^2-2ab\\sin(\\theta+30^\\circ)+b^2&\\geq0\\\\ a^2-2ab(\\sin\\theta\\cos30^\\circ+\\cos\\theta\\sin30^\\circ)+b^2&\\geq0\\\\a^2-ab\\sin\\theta\\sqrt3-ab\\cos\\theta+b^2&\\geq0\\\\a^2-ab\\cos\\theta+b^2&\\geq ab\\sin\\theta\\sqrt3\\\\a^2-ab\\cos\\theta+b^2&\\geq2\\left(\\frac12ab\\sin\\theta\\right)\\sqrt3\\\\a^2-ab\\cos\\theta+b^2&\\geq2T\\sqrt3\\\\2a^2-2ab\\cos\\theta+2b^2&\\geq4T\\sqrt3\\\\a^2+b^2+(a^2-2ab\\cos\\theta+b^2)&\\geq4\\sqrt3\\ T\\\\\\boxed{a^2+b^2+c^2\\geq4\\sqrt3\\ T}\\ \\blacksquare\\end{align*}\n\\]\n\nFor equality, it is necessary for \\(a^2-2ab\\sin x+b^2=0\\implies a=b\\) and \\(\\sin x=\\sin(\\theta+30^\\circ)=1\\implies\\theta=60^\\circ\\), which implies that \\(\\triangle ABC\\) is equilateral.\n\n~ztilB",
"Again, we use Heron's formula but we also set \\(a=y+z\\), \\(b=z+x\\), \\(c=x+y\\). Then \\(S=\\sqrt{xyz(x+y+z)}\\) and\n\n\\[\n\\begin{align*} 4\\sqrt{3xyz(x+y+z)} &= \\frac{4}{3}\\sqrt{27xyz(x+y+z)} \\\\ &\\le \\frac{4}{3}(x^2+y^2+z^2+2xy+2yz+2zx) \\end{align*}\n\\]\n\nand if we set \\(A=xy+yz+zx\\) and \\(B=x^2+y^2+z^2\\) then it suffices to prove that \\(A+B \\le \\frac{2}{3}(A+2B)\\) which is equivalent to \\(A\\le B\\) and is obviously true. ~alexanderchew2021"
] |
IMO-1961-3
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_3
|
Solve the equation
\[
\cos^n{x} - \sin^n{x} = 1
\]
where \(n\) is a given positive integer.
|
[
"Since \\(cos^2x + sin^2x = 1\\), we cannot have solutions with \\(n\\ne2\\) and \\(0<|cos(x)|,|sin(x)|<1\\). Nor can we have solutions with \\(n=2\\), because the sign is wrong. So the only solutions have \\(sin (x) = 0\\) or \\(cos (x) = 0\\), and these are: \\(x =\\) multiple of \\(\\pi\\), and \\(n\\) even; \\(x\\) even multiple of \\(\\pi\\) and \\(n\\) odd; \\(x\\) = even multiple of \\(\\pi + \\frac{3\\pi}{2}\\) and \\(n\\) odd.",
"First consider \\(n=1\\). Squaring both sides yields \\(\\cos^2{x}-2\\cos{x}\\sin{x}+\\sin^2{x}=1\\), and utilizing identities yields \\(\\sin{2x}=0\\). This results in solutions \\(x=\\frac{\\pi}{2}k\\) for integers \\(k\\). After manually checking the solutions in the original equation, we conclude that only \\(x=2k\\pi\\) and \\(x=\\frac{3\\pi}{2}+2k\\pi\\) are solutions to the equation.\n\nNext, we consider \\(n=2\\). This is equivalent to \\(\\cos{2x}=1\\), so the solutions are \\(x=k\\pi\\) for integers \\(k\\). Once again, substituting into the equation results in a valid identity, so these are the only solutions.\n\nLet \\(f(x)=\\cos^n{x}-\\sin^n{x}\\). Taking the derivative yields \\(f'(x)=-n\\sin{x}\\cos{x}(\\sin^{n-2}{x}+\\cos^{n-2}{x})\\) for \\(n>2\\). We wish to consider extrema, so we find all \\(x\\)-values such that \\(f'(x)=0\\) by checking the zeros of each factor. In \\(x\\in[0,2\\pi]\\), \\(f'(x)=0\\) for even \\(n\\) when \\(x=0,\\frac{\\pi}{2},\\pi,\\frac{3\\pi}{2},2\\pi\\), and for odd \\(n\\) when \\(x=0,\\frac{\\pi}{2},\\frac{3\\pi}{4},\\pi,\\frac{3\\pi}{2},\\frac{7\\pi}{4},2\\pi\\). The differences between the two come from the \\(\\sin^{n-2}{x}+\\cos^{n-2}{x}\\) factor, reaching zeros when \\(\\tan^{n-2}{x}=-1\\). This is impossible when \\(n\\) is even, and if \\(n\\) is odd, it implies that \\(\\tan{x}=-1\\), hence the solutions.\n\nWe then calculate the value of \\(f(x)\\) at every such \\(x\\). For even \\(n\\), the \\(x\\)-values \\(0,\\frac{\\pi}{2},\\pi,\\frac{3\\pi}{2},2\\pi\\) result in \\(f(x)\\)-values of \\(1,-1,1,-1,1\\). Since these are the only possible extrema, we conclude that \\(1\\) is always a maximum and \\(-1\\) is always a minimum; thus the only solutions are the maxima, which occur at \\(x=0,\\pi,2\\pi\\) when \\(x\\in[0,2\\pi]\\); more generally, we have \\(x=k\\pi\\) for integers \\(k\\).\n\nNext, for odd \\(n\\), the \\(x\\)-values \\(x=0,\\frac{\\pi}{2},\\frac{3\\pi}{4},\\pi,\\frac{3\\pi}{2},\\frac{7\\pi}{4},2\\pi\\) yield \\(f(x)\\)-values of \\(1,-1,-2\\left(\\frac{1}{\\sqrt{2}}\\right)^n,-1,1,2\\left(\\frac{1}{\\sqrt{2}}\\right)^n,1\\). Obviously \\(x=0,\\frac{3\\pi}{2},2\\pi\\) are all solutions; we show that on our closed domain these are the only solutions. On \\(x=0\\) to \\(x=\\frac{\\pi}{2}\\), the function is always decreasing; if it were increasing at any point, an extremum would be required to change the sign of the derivative; however, there does not exist an extremum between these two points (since we have already found all the extrema!); thus only \\(x=0\\) satisfies the equation. We can repeat this argument for all values between two extrema; eventually, we can conclude that the only times that \\(f(x)=1\\) are the above (since no segment contains \\(1\\) at any point other than an endpoint; as a result, the function cannot equal \\(1\\) at any point other than an endpoint of some segment, which are the extrema). As a result, extending this to all \\(x\\), the solutions are \\(x=2k\\pi\\) and \\(x=\\frac{3\\pi}{2}+2k\\pi\\) for integers \\(k\\).\n\nSince these results match our \\(n=1\\) and \\(n=2\\) cases, we conclude that when \\(n\\) is odd, the solutions are \\(x=2k\\pi\\) and \\(x=\\frac{3\\pi}{2}+2k\\pi\\) for integers \\(k\\), and when \\(n\\) is even, the solutions are \\(x=k\\pi\\) for integers \\(k\\).\n\n~ eevee9406"
] |
IMO-1961-4
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_4
|
In the interior of triangle \(P_1P_2P_3\) a point \(P\) is given. Let \(Q_1,Q_2,Q_3\) be the intersections of \(PP_1, PP_2,PP_3\) with the opposing edges of triangle \(P_1P_2P_3\). Prove that among the ratios \(\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}\) there exists one not larger than \(2\) and one not smaller than \(2\).
|
[
"Let \\([ABC]\\) denote the area of triangle \\(ABC\\).\n\nSince triangles \\(P_1P_2P_3\\) and \\(PP_2P_3\\) share the base \\(P_2P_3\\), we have \\(\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_1}{P_1Q_1}\\).\n\nSimilarly, \\(\\frac{[PP_1P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}, \\frac{[PP_1P_2]}{[P_1P_2P_3]}=\\frac{PQ_3}{P_3Q_3}\\).\n\nAdding all of these gives \\(\\frac{[PP_1P_3]}{[P_1P_2P_3]}+\\frac{[PP_1P_2]}{[P_1P_2P_3]}+\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}+\\frac{PQ_3}{P_3Q_3}+\\frac{PQ_1}{P_1Q_1}=1\\).\n\nWe see that we must have at least one of the three fractions not greater than \\(\\frac{1}{3}\\), and at least one not less than \\(\\frac{1}{3}\\). These correspond to ratios \\(\\frac{PP_i}{PQ_i}\\) being less than or equal to \\(2\\), and greater than or equal to \\(2\\), respectively, so we are done.",
"Let \\(K_1=[P_2PP_3], K_2=[P_3PP_1],\\) and \\(K_3=[P_1PP_2].\\) Note that by same base in triangles \\(P_2PP_3\\) and \\(P_2P_1P_3,\\)\n\n\\[\n\\frac{P_1P}{PQ_1}+1=\\frac{P_1Q_1}{PQ_1}=\\frac{[P_2P_1P_3]}{[P_2PP_3]}=\\frac{K_1+K_2+K_3}{K_1}.\n\\]\n\nThus,\n\n\\[\n\\frac{P_1P}{PQ_1}=\\frac{K_2+K_3}{K_1}\n\\]\n\n\\[\n\\frac{P_2P}{PQ_2}=\\frac{K_1+K_3}{K_2}\n\\]\n\n\\[\n\\frac{P_3P}{PQ_3}=\\frac{K_1+K_2}{K_3}.\n\\]\n\nWithout loss of generality, assume \\(K_1\\leq K_2\\leq K_3.\\) Hence,\n\n\\[\n\\frac{P_1P}{PQ_1}=\\frac{K_2+K_3}{K_1}\\geq \\frac{K_1+K_1}{K_1}=2\n\\]\n\nand\n\n\\[\n\\frac{P_3P}{PQ_3}=\\frac{K_1+K_2}{K+3}\\leq \\frac{K_3+K_3}{K_3}=2,\n\\]\n\nas desired. \\(\\blacksquare\\)"
] |
IMO-1961-5
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_5
|
Construct a triangle ABC if the following elements are given: \(AC = b, AB = c\), and \(\angle AMB = \omega \left(\omega < 90^{\circ}\right)\) where M is the midpoint of BC. Prove that the construction has a solution if and only if
\[
b \tan{\frac{\omega}{2}} \le c < b
\]
In what case does equality hold?
|
[
"Prolong BA to a point D such that \\(BD = 2AB\\). Take circle through B and D such that the minor arc BD is equal to \\(2*\\omega\\) so that for points P on the major arc BD we have \\(\\angle BPD = \\omega\\). Draw a circle with center A and radius AC, and the point of intersection of this circle and the major arc BD will be C. In general there are two possibilities for C.\n\nLet X be the intersection of the arc BN and the perpendicular to the segment BN through A. For the construction to be possible we require \\(AX \\geqslant AC > AB\\). But \\(\\frac{AB}{AX} = \\tan{\\frac{\\omega}{2}}\\), so we get the condition in the question"
] |
IMO-1961-6
|
https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_6
|
Consider a plane \(\epsilon\) and three non-collinear points \(A,B,C\) on the same side of \(\epsilon\); suppose the plane determined by these three points is not parallel to \(\epsilon\). In plane \(\epsilon\) take three arbitrary points \(A',B',C'\). Let \(L,M,N\) be the midpoints of segments \(AA', BB', CC'\); Let \(G\) be the centroid of the triangle \(LMN\). (We will not consider positions of the points \(A', B', C'\) such that the points \(L,M,N\) do not form a triangle.) What is the locus of point \(G\) as \(A', B', C'\) range independently over the plane \(\epsilon\)?
|
[
"We will consider the various points in terms of their coordinates in space. We have \\(L=\\frac{A+A^\\prime}{2},M=\\frac{B+B^\\prime}{2},N=\\frac{C+C^\\prime}{2}\\). Since the centroid of a triangle is the average of the triangle's vertices, we have \\(G=\\frac{1}{3}\\left(L+M+N\\right)=\\frac{1}{6}\\left(A+B+C+A^\\prime+B^\\prime+C^\\prime\\right)\\). It is clear now that \\(G\\) is midpoint of the line segment connecting the centroid of \\(ABC\\) and the centroid of \\(A^\\prime B^\\prime C^\\prime\\). It is obvious that the centroid of \\(A^\\prime B^\\prime C^\\prime\\) can be any point on plane \\(\\epsilon\\). Thus, the locus of \\(G\\) is the plane parallel to \\(\\epsilon\\) and halfway between the centroid of \\(ABC\\) and \\(\\epsilon\\)."
] |
IMO-1962-1
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_1
|
Find the smallest natural number \(n\) which has the following properties:
(a) Its decimal representation has 6 as the last digit.
(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number \(n\).
|
[
"As the new number starts with a \\(6\\) and the old number is \\(1/4\\) of the new number, the old number must start with a \\(1\\).\n\nAs the new number now starts with \\(61\\), the old number must start with \\(\\lfloor 61/4\\rfloor = 15\\).\n\nWe continue in this way until the process terminates with the new number \\(615\\,384\\) and the old number \\(n=\\boxed{153\\,846}\\).",
"We know from the two properties that for some string \\(x\\), \\(n=10x+6\\). Let the number of digits in \\(x\\) be \\(a\\); then moving the \\(6\\) to the front would give it place value \\(10^a\\); as a result, \\(4n=6\\cdot10^a+x\\). Multiplying this by \\(10\\) gives \\(40n=6\\cdot10^{a+1}+10x\\), and subtracting the former yields \\(39n=6(10^{a+1}-1)\\), or \\(13n=2(10^{a+1}-1)\\). As a result, \\(13|(10^{a+1}-1)\\). By Fermat's Little Theorem, we know that \\(10^{12}-1\\) divides \\(13\\), so it isn't difficult to try values of \\(a+1\\) less than \\(13\\) to find the smallest such \\(a\\).\n\nEventually, we notice that \\(10^6-1\\) divides \\(13\\), so \\(a=5\\). Then \\(\\boxed{n=153846}\\), and since the number ends in \\(6\\), we know that \\(x\\) is also an integer, so this is the solution.\n\n~eevee9406"
] |
IMO-1962-2
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_2
|
Determine all real numbers \(x\) which satisfy the inequality:
\[
\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}
\]
(Note that the problem as written in the official PDF https://www.imo-official.org/year_info.aspx?year=1962 does not have the outer \(\sqrt{\ }\))
|
[
"Obviously we need \\(\\sqrt{3-x} \\geq \\sqrt{x+1}\\) for the outer square root to be defined, \\(x\\leq 3\\) for the first inner square root to be defined, and \\(x\\geq -1\\) for the second inner square root to be defined. Solving these we get that the left hand side is defined for \\(x\\in \\left[ -1,1 \\right]\\).\n\nNow obviously the function \\(f(x)=\\sqrt{\\sqrt{3-x}-\\sqrt{x+1}}\\) is continuous on \\(\\left[ -1,1 \\right]\\), with \\(f(-1)=\\sqrt 2\\) and \\(f(1)=0\\). Moreover, as \\(3-x\\) is a decreasing and \\(x+1\\) an increasing function, both \\(\\sqrt{3-x}\\) and \\(-\\sqrt{x+1}\\) are decreasing functions, and hence \\(f(x)\\) is a decreasing function. Therefore there is exactly one solution to \\(f(x)=\\dfrac{1}{2}\\).\n\nWe can now find this solution:\n\n\\[\n\\begin{align*} \\sqrt{\\sqrt{3-x}-\\sqrt{x+1}} &= \\dfrac{1}{2} \\\\ \\sqrt{3-x}-\\sqrt{x+1} &= \\dfrac{1}{4} \\\\ \\sqrt{3-x} &= \\dfrac{1}{4} + \\sqrt{x+1} \\\\ 3-x &= \\dfrac 1{16} + x+1 + \\dfrac{\\sqrt{x+1}}2 \\\\ 2 - 2x - \\dfrac 1{16} &= \\dfrac{\\sqrt{x+1}}2 \\\\ 31 - 32x &= 8\\sqrt{x+1} \\\\ 1024 x^2 - 1984x + 961 &= 64(x+1) \\\\ 1024 x^2 - 2048x + 897 &= 0 \\end{align*}\n\\]\n\n(Note the little trick in the third row: placing the square roots on opposite sides of the equation. Squaring the equation in the second row would work as well, but this way is a little more pleasant, as the one remaining square root after the squaring will essentially be one of the original two, not their product.)\n\nSolving the quadratic equation for \\(x\\), we get\n\n\\[\nx_{1,2}=\\dfrac{ 2^{11} \\pm \\sqrt{ 2^{22} - 2^{12}\\cdot 897} }{2^{11}} = 1 \\pm \\dfrac{\\sqrt{127}}{32}\n\\]\n\nThe reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than \\(1\\) is the additional solution, and \\(x=1-\\dfrac{\\sqrt{127}}{32}\\) is the root we need.\n\nHence the solutions to the given inequality are precisely the reals in the interval \\(\\boxed{ \\left[ ~ -1,\\quad 1-\\dfrac{\\sqrt{127}}{32} ~ \\right) }\\).\n\n\\[\n[asy] import graph; size(300,300,IgnoreAspect); real f(real x) {return sqrt( sqrt(3-x) - sqrt(x+1) );}; real g(real x) {return 1/2;}; draw(graph(f,-1,1),blue,\"$f(x)=\\sqrt{\\sqrt{3-x}-\\sqrt{x+1}}$\"); draw(graph(g,-1,1),red,\"$g(x)=1/2$\"); xaxis(\"$x$\",BottomTop,Ticks); yaxis(\"$y$\",LeftRight,Ticks); attach(legend(),point(E),20E,UnFill); [/asy]\n\\]"
] |
IMO-1962-3
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_3
|
Consider the cube \(ABCDA'B'C'D'\)(\(ABCD\) and \(A'B'C'D'\) are the upper and lower bases, respectively, and edges \(AA'\), \(BB'\), \(CC'\), \(DD'\) are parallel). The point \(X\) moves at constant speed along the perimeter of the square \(ABCD\) in the direction \(ABCDA\), and the point \(Y\) moves at the same rate along the perimeter of the square \(B'C'CB\) in the direction \(B'C'CBB'\). Points \(X\) and \(Y\) begin their motion at the same instant from the starting positions \(A\) and \(B'\), respectively. Determine and draw the locus of the midpoints of the segments \(XY\).
|
[
"First we prove a small lemma: If the particles \\(X\\) and \\(Y\\) move along straight lines at constant velocities, then the locus of the midpoint of \\(XY\\) is also a line. This is rather trivial, since all lines in 3D space may take the parametric form \\((at+n, bt+m, ct+p)\\), with \\(t\\) being time, and the average of two such lines must also have a linear parametric form.\n\nThe locus clearly starts at the midpoint of \\(AB'\\), or the center of face \\(ABB'A'\\). As \\(X\\) moves from \\(A\\) to \\(B\\), and \\(Y\\) moves from \\(C'\\) to \\(C\\), both \\(X\\) and \\(Y\\) move along straight lines, so the midpoint of \\(XY\\) traces out a line segment, starting at the midpoint of \\(AB'\\) and ending at the midpoint of \\(BC'\\), or the center of face \\(BCC'B'\\). This concludes the first phase of motion. A quick check reveals that the locus goes nowhere during the second and fourth phases of motion, and only moves backward on the third. Thus the locus is just the segment connecting the centers of sides \\(ABB'A'\\) and \\(BCC'B'\\).\n\nsolution is actually wrong, it's not segment it's parallelogram- author mixed the movement of points (even if he clearly understands what he is doing- just a little mistake) Here is correct soluton:\n\nAnswer: the rhombus CUVW, where U is the center of ABCD, V is the center of ABB'A, and W is the center of BCC'B'.\n\nTake rectangular coordinates with A as (0, 0, 0) and C' as (1, 1, 1). Let M be the midpoint of XY. Whilst X is on AB and Y on B'C', X is (x, 0, 0) and Y is (1, x, 1), so M is (x/2 + 1/2, x/2, 1/2) = x (1, 1/2, 1/2) + (1-x) (1/2, 0, 1/2) = x W + (1-x) V, so M traces out the line VW.\n\nWhilst X is on BC and Y is on C'C, X is (1, x, 0) and Y is ( 1, 1, 1-x), so M is (1, x/2+1/2, 1/2 - x/2) = x (1, 1, 0) + (1-x) (1, 1/2, 1/2) = x C + (1-x) W, so M traces out the line WC.\n\nWhilst X is on CD and Y is on CB, X is (1-x, 1, 0) and Y is (1, 1-x, 0), so M is (1-x/2, 1-x/2, 0) = x (1, 1, 0) + (1-x) (1/2, 1/2, 0) = x C + (1-x) U, so M traces out the line CU.\n\nWhilst X is on DA and Y is on BB', X is (0, 1-x, 0) and Y is (1, 0, x), so M is (1/2, 1/2 - x/2, x/2) = x (1/2, 0, 1/2) + (1-x) (1/2, 1/2, 0) = x V + (1-x) U, so M traces out the line UV."
] |
IMO-1962-4
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_4
|
Solve the equation \(\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1\).
|
[
"First, note that we can write the left hand side as a cubic function of \\(\\cos^2 x\\). So there are at most \\(3\\) distinct values of \\(\\cos^2 x\\) that satisfy this equation. Therefore, if we find three values of \\(x\\) that satisfy the equation and produce three different \\(\\cos^2 x\\), then we found all solutions to this cubic equation (without expanding it, which is another viable option). Indeed, we find that \\(\\frac{\\pi}2\\), \\(\\frac{\\pi}4\\), and \\(\\frac{\\pi}6\\) all satisfy the equation, and produce three different values of \\(\\cos^2 x\\), namely \\(0\\), \\(\\frac12\\), and \\(\\frac34\\). So we solve \\(\\cos^2 x = \\text{each of these}\\). Therefore, our solutions are:\n\n\\[\nx = \\frac{(2k+1)\\pi}2,\\, \\frac{(2k+1)\\pi}4,\\, \\frac{(6k+1)\\pi}6,\\, \\frac{(6k+5)\\pi}6 \\quad \\forall k\\in Z\n\\]"
] |
IMO-1962-5
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_5
|
On the circle \(K\) there are given three distinct points \(A,B,C\). Construct (using only straightedge and compass) a fourth point \(D\) on \(K\) such that a circle can be inscribed in the quadrilateral thus obtained.
|
[
"Aviso: La siguiente solución está en español\n\ngeneralidad ya que es simétrico para todas las combinaciones posibles)\n\nCircunferencia K en 2 puntos, uno de ello se nombrará con E (va a estar del mismo lado que B con respecto a AC) y el otro se nombrará F\n\niguales, abrir con el compás la distancia BE y se traza la Circunferencia de Centro en F y Radio BE, así esta nueva circunferencia corta a la circunferencia K en 2 puntos, llámese D al punto que está del mismo lado de A con respecto a la mediatriz de AC, ese es el punto pedido que cumple la"
] |
IMO-1962-6
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_6
|
Consider an isosceles triangle. Let \(r\) be the radius of its circumscribed circle and \(\rho\) the radius of its inscribed circle. Prove that the distance \(d\) between the centers of these two circles is
\[
d=\sqrt{r(r-2\rho)}
\]
.
|
[
"<geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra> Instead of an isosceles triangle, let us consider an arbitrary triangle \\(ABC\\). Let \\(ABC\\) have circumcenter \\(O\\) and incenter \\(I\\). Extend \\(AI\\) to meet the circumcircle again at \\(L\\). Then extend \\(LO\\) so it meets the circumcircle again at \\(M\\). Consider the point where the incircle meets \\(AB\\), and let this be point \\(D\\). We have \\(\\angle ADI = \\angle MBL = 90^{\\circ}, \\angle IAD = \\angle LMB\\); thus, \\(\\triangle ADI \\sim \\triangle MBL\\), or \\(\\frac {ID}{BL} = \\frac {AI} {ML} \\iff ID \\cdot ML = 2r\\rho = AI \\cdot BL\\). Now, drawing line \\(BI\\), we see that \\(\\angle BIL = \\frac {1}{2}\\angle A + \\frac {1}{2}\\angle ABC, \\angle IBL = \\frac {1}{2}\\angle ABC + \\angle CBL = \\frac {1}{2}\\angle ABC + \\frac {1}{2}\\angle A\\). Therefore, \\(BIL\\) is isosceles, and \\(IL = BL\\). Substituting this back in, we have \\(2r\\rho = AI\\cdot IL\\). Extending \\(OI\\) to meet the circumcircle at \\(P,Q\\), we see that \\(AI\\cdot IL = PI\\cdot QI\\) by Power of a Point. Therefore, \\(2r\\rho = PI \\cdot QI = (PO + OI)(QO - OI) = (r + d)(r - d)\\), and we have \\(2r\\rho = r^2 - d^2 \\iff d = \\sqrt {r(r - 2\\rho)}\\), and we are done."
] |
IMO-1962-7
|
https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_7
|
The tetrahedron \(SABC\) has the following property: there exist five spheres, each tangent to the edges \(SA, SB, SC, BC, CA, AB\), or to their extensions.
(a) Prove that the tetrahedron \(SABC\) is regular.
(b) Prove conversely that for every regular tetrahedron five such spheres exist.
|
[
"\\[\nIMO 1962 P7 01.png\n\\]\n\nPart (a)\n\nLet points \\(P_{SA}, P_{SB}, P_{SC}, P_{BC}, P_{CA}, P_{AB}\\) be the points where the smallest sphere is tangent to the edges \\(SA, SB, SC, BC, CA, AB\\) respectively.\n\nFor each face of the tetrahedron there is a circular cross section of the smallest sphere. Since that circle also needs to be tangent to the edges, then that circle is the incircle of each triangular face.\n\nFrom the properties of an incircle we know that:\n\n\\(|AP_{AB}|=|AP_{AC}|\\), \\(|BP_{BC}|=|BP_{BA}|\\), and \\(|CP_{CA}|=|CP_{CB}|\\) in \\(\\Delta ABC\\)\n\nThere is a larger circle that is tangent to \\(AB\\) and the extensions of \\(CA\\) and \\(CB\\).\n\nThis larger circle is a cross section of the sphere that's also tangent to the extension of \\(CS\\) away from \\(S\\) and the edges of \\(\\Delta ABC\\)\n\nTherefore, this larger circle and the incircle of \\(\\Delta ABC\\) are part of that same sphere.\n\nIn order for these two circles to be part of the same sphere and also tangent to line \\(AB\\), then the point of the tangent of this larger circle needs to be the same as point \\(P_{AB}\\)\n\nThe only way these to circles can share the same tangent point on edge \\(AB\\) is if \\(|AP_{AB}|=|BP_{AB}|\\)\n\nUsing the same argument with the larder circle of edge \\(BC\\) then \\(|BP_{BC}|=|CP_{BC}|\\)\n\nand with the larger circle of edge \\(AC\\) then \\(|CP_{AC}|=|AP_{AC}|\\)\n\nThis results in:\n\n\\(|AP_{AB}|=|AP_{AC}|=|BP_{BC}|=|BP_{BA}|=|CP_{CA}|=|CP_{CB}|\\) in \\(\\Delta ABC\\)\n\nwhich means that\n\n\\(|AB|=|BC|=|AC|\\) in \\(\\Delta ABC\\), thus \\(\\Delta ABC\\) is an equilteral triangle.\n\nLikewise,\n\n\\(|AB|=|BS|=|AS|\\) in \\(\\Delta ABS\\), thus \\(\\Delta ABS\\) is an equilteral triangle.\n\n\\(|AS|=|SC|=|AC|\\) in \\(\\Delta ASC\\), thus \\(\\Delta ASC\\) is an equilteral triangle.\n\n\\(|SB|=|BC|=|SC|\\) in \\(\\Delta SBC\\), thus \\(\\Delta SBC\\) is an equilteral triangle.\n\nSince all four faces are equilateral triangles, then tetrahedron \\(ABCS\\) is a regular tetrahedron.\n\nPart (b)\n\nLet's consider a regular tetrahedron in the cartesian space with center at \\((0,0,0)\\) with \\(r_c\\) as the circumradius.\n\nWe can write the coordinates for the four vertices as:\n\n\\[\nV_1=\\left( \\frac{2\\sqrt{2}}{3}r_c ,\\;0,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nV_2=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;\\frac{\\sqrt{6}}{3}r_c,\\;\\frac{-r_c}{3}\\right)\n\\]\n\n\\[\nV_3=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;-\\frac{\\sqrt{6}}{3}r_c,\\;\\frac{-r_c}{3}\\right)\n\\]\n\n\\[\nV_4=\\left(0 ,\\;0,\\;r_c\\right)\n\\]\n\nAnd the midpoints of all edges as:\n\n\\[\nM_{12}=\\frac{V_1-V_2}{2}=\\left( \\frac{\\sqrt{2}}{6}r_c ,\\;\\frac{\\sqrt{6}}{6}r_c,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{13}=\\frac{V_1-V_3}{2}=\\left( \\frac{\\sqrt{2}}{6}r_c ,\\;-\\frac{\\sqrt{6}}{6}r_c,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{23}=\\frac{V_2-V_3}{2}=\\left( -\\frac{\\sqrt{2}}{3}r_c ,\\;0,\\;-\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{14}=\\frac{V_1-V_4}{2}=\\left( \\frac{\\sqrt{2}}{3}r_c ,\\;0,\\;\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{24}=\\frac{V_2-V_4}{2}=\\left( -\\frac{\\sqrt{2}}{6}r_c ,\\;\\frac{\\sqrt{6}}{6}r_c,\\;\\frac{r_c}{3}\\right)\n\\]\n\n\\[\nM_{34}=\\frac{V_3-V_4}{2}=\\left( -\\frac{\\sqrt{2}}{6}r_c ,\\;-\\frac{\\sqrt{6}}{6}r_c,\\;\\frac{r_c}{3}\\right)\n\\]\n\nNow we calculate the following six dot products:\n\n\\[\n(V_1-V_2) \\bullet M_{12}=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left( \\frac{\\sqrt{6}}{3}r_c\\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_1-V_3) \\bullet M_{13}=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left(- \\frac{\\sqrt{6}}{3}r_c\\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_2-V_3) \\bullet M_{23}=0+\\left(- \\frac{\\sqrt{6}}{6}r_c\\right)\\left( 0\\right)+ 0=0\n\\]\n\n\\[\n(V_1-V_4) \\bullet M_{13}=\\left( 2\\frac{\\sqrt{2}}{3}r_c \\right)\\left( \\frac{\\sqrt{2}}{3}r_c \\right)+0+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{4}{9}r_c^2-\\frac{4}{9}r_c^2=0\n\\]\n\n\\[\n(V_2-V_4) \\bullet M_{13}=\\left( -\\frac{\\sqrt{2}}{3}r_c \\right)\\left( -\\frac{\\sqrt{2}}{6}r_c \\right)+\\left( \\frac{\\sqrt{6}}{3}r_c \\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{1}{9}r_c^2+\\frac{1}{3}r_c^2-\\frac{4}{9}r_c^2=0\n\\]\n\n\\[\n(V_3-V_4) \\bullet M_{13}=\\left( -\\frac{\\sqrt{2}}{3}r_c \\right)\\left( -\\frac{\\sqrt{2}}{6}r_c \\right)+\\left( -\\frac{\\sqrt{6}}{3}r_c \\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ \\left(- \\frac{4}{3}r_c\\right)\\left( \\frac{1}{3}r_c \\right)=\\frac{1}{9}r_c^2+\\frac{1}{3}r_c^2-\\frac{4}{9}r_c^2=0\n\\]\n\nSince all these dot producs equal to \\(0\\) that means that the lines from the center to each of the midpoints are all perpendicular.\n\nNow we calculate the distances from \\((0,0,0)\\) to all the midpoints:\n\n\\[\n|M_{12}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{13}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{23}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(-\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{14}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{24}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\n\\[\n|M_{34}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(\\frac{r_c}{3}\\right)^2}=\\frac{\\sqrt{3}}{3}r_c\n\\]\n\nSince all distances are all the same and all dot products are \\(0\\), then we have our first sphere at \\((0,0,0)\\) with a radius of \\(\\frac{\\sqrt{3}}{3}r_c\\)\n\nNow we will look at the other 4 spheres.\n\nLet \\(E_{ij}\\) be the point of tangent of the larger sphere on the extension of line \\(V{i}V{j}\\) in the direction of \\(V{i}\\) to \\(V{j}\\) and beyond \\(V{j}\\) for \\(i=1,2,3,4\\); \\(j=1,2,3,4\\); and \\(i \\ne j\\)\n\nSince \\(|V_j E_{ij}|=|V_j M_{ij}|=\\frac{1}{2} |V_j V_i|\\), then \\(\\frac{1}{2}(V_j-V_i)=E_{ij}-V_j\\), thus \\(E_{ij}=\\frac{3V_j-V_i}{2}\\)\n\nUsing this formula we calculate the following:\n\n\\[\nE_{41}=\\left( \\sqrt{2} r_c ,\\;0,\\;-r_c \\right)\n\\]\n\n\\[\nE_{42}=\\left( -\\frac{\\sqrt{2}}{2}r_c ,\\;\\frac{\\sqrt{6}}{2}r_c,\\;-r_c \\right)\n\\]\n\n\\[\nE_{43}=\\left( -\\frac{\\sqrt{2}}{2}r_c ,\\;-\\frac{\\sqrt{6}}{2}r_c,\\;-r_c \\right)\n\\]\n\nWe will start with the sphere below the base of the tetrahedron opposite of vertex \\(V_{4}\\) below \\(\\Delta V{1}V{2}V{3}\\)\n\nThe center of this larger sphere is at \\(C_1=(0,0,-2r_c)\\) and it is tangent at points \\(M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}\\)\n\nWe calculate the following dot products:\n\n\\[\n(V_1-V_2) \\bullet (M_{12}-C_1)=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left( \\frac{\\sqrt{6}}{3}r_c\\right)\\left( \\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_1-V_3) \\bullet (M_{13}-C_1)=\\left( -\\sqrt{2}r_c \\right)\\left( \\frac{\\sqrt{2}}{6}r_c \\right)+ \\left(- \\frac{\\sqrt{6}}{3}r_c\\right)\\left( -\\frac{\\sqrt{6}}{6}r_c \\right)+ 0=-\\frac{1}{3}r_c^2+\\frac{1}{3}r_c^2=0\n\\]\n\n\\[\n(V_2-V_3) \\bullet (M_{23}-C_1)=0+\\left(- \\frac{\\sqrt{6}}{6}r_c\\right)\\left( 0\\right)+ 0=0\n\\]\n\n\\[\n(E_{41}-V_4) \\bullet (E_{41}-C_1) = \\left( \\sqrt{2}r_c \\right)^2+0+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0\n\\]\n\n\\[\n(E_{42}-V_4) \\bullet (E_{42}-C_1) = \\left( -\\frac{\\sqrt{2}}{2}r_c \\right)^2+\\left( \\frac{\\sqrt{6}}{2}r_c \\right)^2+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0\n\\]\n\n\\[\n(E_{43}-V_4) \\bullet (E_{43}-C_1) = \\left( -\\frac{\\sqrt{2}}{2}r_c \\right)^2+\\left( -\\frac{\\sqrt{6}}{2}r_c \\right)^2+\\left( -2r_c \\right)\\left( -r_c+2r_c \\right)=2r_c^2-2r_c^2=0\n\\]\n\nSince all these dot producs equal to \\(0\\) that means that the lines from the center \\(C_1\\) to each of the point \\(M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}\\) are all perpendicular.\n\nNow we calculate the distances from \\(C_1=(0,0,-2r_c)\\) to points \\(M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}\\)\n\n\\[\n|C_1M_{12}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1M_{13}|=\\sqrt{\\left( \\frac{\\sqrt{2}}{6}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{6}r_c\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1M_{23}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{3}r_c\\right)^2+\\left(0\\right)^2+\\left(-\\frac{5}{3}r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1E_{41}|=\\sqrt{\\left( \\sqrt{2}r_c\\right)^2+\\left(0\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1E_{42}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{2}r_c\\right)^2+\\left(\\frac{\\sqrt{6}}{2}r_c\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\n\\[\n|C_1E_{43}|=\\sqrt{\\left( -\\frac{\\sqrt{2}}{2}r_c\\right)^2+\\left(-\\frac{\\sqrt{6}}{2}r_c\\right)^2+\\left(-r_c\\right)^2}=\\sqrt{3}r_c\n\\]\n\nSince all distances are all the same and all dot products are \\(0\\), then we have one of the larger spheres at \\((0,0,-2r_c)\\) with a radius of \\(\\sqrt{3}r_c\\)\n\nThen, the other three larger spheres which are the same size as the sphere with center at \\(C_1\\) are congruent and tangent to their respective sides near the other faces of the tetrahedron.\n\nand this proves that a tetrahedron with any circumradius \\(r_c\\) will have these 5 spheres, one with radius \\(\\frac{\\sqrt{3}}{3}r_c\\), at the center of the tetrahedron and the other 4 with radius \\(\\sqrt{3}r_c\\) at centers that are at a distance of \\(3r_c\\) away from any of the vertices of the tetrahedron in the direction from that vertex to the center of its opposite face.\n\nThus these five spheres exist for any regular tetrahedron.\n\n~Tomas Diaz. [email protected]"
] |
IMO-1963-1
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_1
|
Find all real roots of the equation
\[
\sqrt{x^2-p}+2\sqrt{x^2-1}=x
\]
,
where \(p\) is a real parameter.
|
[
"Assuming \\(x \\geq 0\\), square the equation, obtaining\n\n\\[\n4\\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2\n\\]\n\n. If we have \\(p + 4 \\geq 4x^2\\), we can square again, obtaining\n\n\\[\nx^2 = \\frac {(p - 4)^2}{4(4 - 2p)} \\implies x = \\pm\\frac {p - 4}{2\\sqrt {4 - 2p}}\n\\]\n\nWe must have \\(4 - 2p > 0 \\iff p < 2\\), so we have\n\n\\[\nx = \\frac {4 - p}{2\\sqrt {4 - 2p}}\n\\]\n\nHowever, this is only a solution when\n\n\\[\np + 4 \\geq 4x^2 = \\frac {(p - 4)^2}{4 - 2p} \\iff (p + 4)(4 - 2p)\\geq(p - 4)^2 \\iff 0\\geq p(3p - 4)\n\\]\n\nso we have \\(p\\geq 0\\) and \\(p \\leq \\frac {4}{3}\\)\n\nand \\(x = \\frac {4 - p}{2\\sqrt {4 - 2p}}\\)"
] |
IMO-1963-2
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_2
|
Point \(A\) and segment \(BC\) are given. Determine the locus of points in space which are the vertices of right angles with one side passing through \(A\), and the other side intersecting the segment \(BC\).
|
[
"Let \\(\\omega_1\\) be the circle with diameter \\(AB\\), and let \\(\\omega_2\\) be the circle with diameter \\(AC\\). Then the locus is simply the set of points inside either \\(\\omega_1\\) or \\(\\omega_2\\), but not both.\n\nTo see this, suppose the right angle's ray that does not pass through \\(A\\) intersects segment \\(BC\\) at \\(X\\). Then the right angle's vertex must lie on the circle with diameter \\(AX\\). So, for a particular \\(X\\), the desired locus is a circle with diameter \\(AX\\). Accounting for all possible \\(X\\), the total locus is the union of the circumferences of all circles that have a diameter \\(AX\\), where \\(X\\) is some point on \\(BC\\).\n\nAs \\(X\\) moves from \\(B\\) to \\(C\\), the motion of the circle with diameter \\(AX\\) is continuous and fluid. Any point \\(P\\) lying within \\(\\omega_1\\) but outside \\(\\omega_2\\) will eventually be intersected by this moving circle, since it went from inside to outside the circle. This applies similarly to all points inside \\(\\omega_2\\) but outside \\(\\omega_1\\). Also, the points inside both \\(\\omega_1\\) and \\(\\omega_2\\) are never intersected by this moving circle, as it always stays inside.\n\n(This proof sucks and needs some formalism)"
] |
IMO-1963-3
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_3
|
In an \(n\)-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
\[
a_1\ge a_2\ge \cdots \ge a_n.
\]
Prove that \(a_1=a_2=\cdots = a_n\).
|
[
"Let \\(a_1 = p_1p_2\\), \\(a_2 = p_2p_3\\), etc.\n\nPlot the \\(n\\)-gon on the cartesian plane such that \\(p_1p_2\\) is on the \\(x\\)-axis and the entire shape is above the \\(x\\)-axis. There are two cases: the number of sides is even, and the number of sides is odd:\n\n\\[\n\\textbf{Case 1: Even}\n\\]\n\nIn this case, the side with the topmost points will be \\(p_{\\frac{n}{2}+1}p_{\\frac{n}{2}+2}\\). To obtain the \\(y\\)-coordinate of this top side, we can multiply the lengths of the sides \\(a_1\\), \\(a_2\\), ... \\(a_{\\frac{n}{2}}\\) by the sine of the angle they make with the \\(x\\)-axis:\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = 1}^{\\frac{n}{2}}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (1)}\n\\]\n\nWe can obtain the \\(y\\)-coordinate of the top side in a different way by multiplying the lengths of the sides \\(a_{\\frac{n}{2}+1}\\), \\(a_{\\frac{n}{2}+2}\\), ... \\(a_n\\) by the sine of the angle they make with the \\(x\\)-axis to get the \\(\\emph{negated}\\) \\(y\\)-coordinate of the top side:\n\n\\[\n-y\\textrm{-coordinate} = \\sum_{k = \\frac{n}{2}+1}^{n}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = \\frac{n}{2}+1}^{n}a_k \\cdot -\\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\n= \\sum_{k = \\frac{n}{2}+1}^{n}a_k \\cdot \\sin \\frac{2\\pi(k-\\frac{n}{2}-1)}{n}\n\\]\n\n\\[\n= \\sum_{k = 1}^{\\frac{n}{2}}a_{k+\\frac{n}{2}} \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (2)}\n\\]\n\nIt must be true that \\(\\textbf{(1)} = \\textbf{(2)}\\). This implies that \\(a_k = a_{k+\\frac{n}{2}}\\) for all \\(1 \\leq k \\leq \\frac{n}{2}\\), and therefore \\(a_1=a_2=\\cdots = a_n\\).\n\n\\[\n\\textbf{Case 2: Odd}\n\\]\n\nThis case is very similar to before. We will compute the \\(y\\)-coordinate of the top point \\(p_{frac{n+3}{2}}\\) two ways:\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = 2}^{\\frac{n+1}{2}}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (3)}\n\\]\n\n\\[\n-y\\textrm{-coordinate} = \\sum_{k = \\frac{n+3}{2}}^{n}a_k \\cdot \\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\ny\\textrm{-coordinate} = \\sum_{k = \\frac{n+3}{2}}^{n}a_k \\cdot -\\sin \\frac{2\\pi(k-1)}{n}\n\\]\n\n\\[\n= \\sum_{k = \\frac{n+3}{2}}^{n}a_k \\cdot \\sin \\frac{2\\pi(n - k + 1)}{n}\n\\]\n\n\\[\n= \\sum_{k = 2}^{\\frac{n+1}{2}}a_{n-k+2} \\cdot \\sin \\frac{2\\pi(k-1)}{n}.\\textbf{ (4)}\n\\]\n\nIt must be true that \\(\\textbf{(3)} = \\textbf{(4)}\\). Then, we get \\(a_k = a_{n-k+2}\\) for all \\(2 \\leq k \\leq \\frac{n+1}{2}\\). Therefore, \\(a_2=a_3=\\cdots = a_n\\). It is trivial that \\(a_1\\) is then equal to the other values, so \\(a_1=a_2=\\cdots = a_n\\). This completes the proof. \\(\\square\\)\n\n~mathboy100",
"Define the vector \\(\\vec{v_i}\\) to equal \\(\\cos{\\left(\\frac{2\\pi}{n}i\\right)}\\vec{i}+\\sin{\\left(\\frac{2\\pi}{n}i\\right)}\\vec{j}\\). Now rotate and translate the given polygon in the Cartesian Coordinate Plane so that the side with length \\(a_i\\) is parallel to \\(\\vec{v_i}\\). We then have that\n\n\\[\n\\sum_{i=1}^{n} a_i\\vec{v_i}=\\vec{0}\\Rightarrow \\sum_{i=1}^{n} a_i\\cos{\\left(\\frac{2\\pi}{n}i\\right)} = \\sum_{i=1}^{n} a_i\\sin{\\left(\\frac{2\\pi}{n}i\\right)} =0\n\\]\n\nBut \\(a_i\\geq a_{n-i}\\) for all \\(i\\leq \\lfloor \\frac{n}{2}\\rfloor\\), so\n\n\\[\na_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)} = -a_i\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)} \\geq -a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)}\n\\]\n\nfor all \\(i\\leq \\lfloor \\frac{n}{2}\\rfloor\\). This shows that \\(a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}+a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)}\\geq 0\\), with equality when \\(a_i=a_{n-i}\\). Therefore\n\n\\[\n\\sum_{i=1}^{n} a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}=\\sum_{i=1}^{\\lfloor \\frac{n}{2}\\rfloor} a_i \\sin{\\left(\\frac{2\\pi}{n}i\\right)}+a_{n-i}\\sin{\\left(\\frac{2\\pi}{n}(n-i)\\right)} \\geq 0\n\\]\n\nThere is equality only when \\(a_i=a_{n-i}\\) for all \\(i\\). This implies that \\(a_1=a_{n-1}\\) and \\(a_2=a_n\\), so we have that \\(a_1=a_2=\\cdots =a_n\\). \\(\\blacksquare\\)"
] |
IMO-1963-4
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_4
|
Find all solutions \(x_1,x_2,x_3,x_4,x_5\) of the system
\[
\begin{eqnarray*} x_5+x_2&=&yx_1\\ x_1+x_3&=&yx_2\\ x_2+x_4&=&yx_3\\ x_3+x_5&=&yx_4\\ x_4+x_1&=&yx_5,\end{eqnarray*}
\]
where \(y\) is a parameter.
|
[
"Notice: The following words are Chinese.\n\n首先,我们可以将以上5个方程相加,得到:\n\n\\[\n2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)\n\\]\n\n当\\(x_1+x_2+x_3+x_4+x_5=0\\)时,因为\\(x_1,x_2,x_3,x_4,x_5\\)关于原方程组轮换对称,所以\n\n\\[\nx_1=x_2=x_3=x_4=x_5=0\n\\]\n\n若反之,则方程两边同除以\\((x_1+x_2+x_3+x_4+x_5)\\),得到\\(y=2\\),显然解为\n\n\\[\nx_1=x_2=x_3=x_4=x_5\n\\]\n\n综上所述,若\\(y=2\\),最终答案为\\(x_1=x_2=x_3=x_4=x_5\\),否则答案为\\(x_1=x_2=x_3=x_4=x_5=0\\)\n\nThe solution in English (translated by Google Translate):\n\nFirst of all, we can add the five equations to get:\n\n\\[\n2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)\n\\]\n\nWhen \\(x_1+x_2+x_3+x_4+x_5=0\\), Because \\(x_1,x_2,x_3,x_4,x_5\\) is symmetric in the original equations,\n\n\\[\nx_1=x_2=x_3=x_4=x_5=0\n\\]\n\nOtherwise, dividing both sides by \\((x_1+x_2+x_3+x_4+x_5\\), we get \\(y=2\\), and clearly\n\n\\[\nx_1=x_2=x_3=x_4=x_5\n\\]\n\nSummarizing, if \\(y=2\\), then the answer is of the form \\(x_1=x_2=x_3=x_4=x_5\\). Otherwise, \\(x_1=x_2=x_3=x_4=x_5=0\\).\n\n## Mistake\n\nWhile doing this question, I found out that the answer is actually wrong, \\(y\\) can equal \\(\\frac{-1-\\sqrt{5}}{2}\\) and \\(\\frac{-1+\\sqrt{5}}{2}\\) and still produce an infinite number of solutions in the form \\((n,n,-\\frac{ny}{y+1},-2ny,-\\frac{ny}{y+1})\\) where \\(n\\) is a real number and the set is cyclic (Ex: The set can correspond to \\((x_{1},x_{2},x_{3},x_{4},x_{5})\\) or \\((x_{2},x_{3},x_{4},x_{5},x_{1})\\), either works. Order matters, but not starting position.). For example, if \\(n=1\\) and \\(y=\\frac{-1+\\sqrt{5}}{2}\\) the set will be \\((1,1,\\frac{\\sqrt{5}-3}{2},1-\\sqrt{5},\\frac{\\sqrt{5}-3}{2})\\), which you can test and find out that it still works even though the set isn't symmetric.\n\nCan someone change this answer so it's correct?\n\nEdit: 亲爱的中国盆友，我找到错误了。 If \\(x_{1}+x_{2}+x_{3}+x_{4}+x_{5} = 0\\), y can be anything(不一定要轮换对称)."
] |
IMO-1963-5
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_5
|
Prove that \(\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}\).
|
[
"Because the sum of the \\(x\\)-coordinates of the seventh roots of unity is \\(0\\), we have\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\frac{8\\pi}{7}} + \\cos{\\frac{10\\pi}{7}} + \\cos{\\frac{12\\pi}{7}} + \\cos{\\frac{14\\pi}{7}} = 0\n\\]\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\frac{8\\pi}{7}} + \\cos{\\frac{10\\pi}{7}} + \\cos{\\frac{12\\pi}{7}} = -1\n\\]\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\left(-\\frac{6\\pi}{7}\\right)} + \\cos{\\left(-\\frac{4\\pi}{7}\\right)} + \\cos{\\left(-\\frac{2\\pi}{7}\\right)} = -1.\n\\]\n\nNow, we can apply \\(\\cos{x} = \\cos{\\left(-x\\right)}\\) to obtain\n\n\\[\n2\\left(\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}}\\right) = -1\n\\]\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} = -\\frac{1}{2}.\n\\]\n\nFinally, since \\(\\cos{x} = -\\cos{(\\pi-x)}\\),\n\n\\[\n\\cos{\\frac{2\\pi}{7}} - \\cos{\\frac{3\\pi}{7}} - \\cos{\\frac{\\pi}{7}} = -\\frac{1}{2}\n\\]\n\n\\[\n\\cos{\\frac{\\pi}{7}} - \\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{3\\pi}{7}} = \\frac{1}{2}\\textrm{. }\\square\n\\]\n\n~mathboy100",
"Let \\(\\cos{\\frac{\\pi}{7}}-\\cos{\\frac{2\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}=S\\). We have\n\n\\[\nS=\\cos{\\frac{\\pi}{7}}-\\cos{\\frac{2\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}=\\cos{\\frac{\\pi}{7}}+\\cos{\\frac{3\\pi}{7}}+\\cos{\\frac{5\\pi}{7}}\n\\]\n\nThen, by product-sum formulae, we have\n\n\\[\nS \\cdot 2 \\sin{\\frac{\\pi}{7}} = \\sin{\\frac{2\\pi}{7}}+\\sin{\\frac{4\\pi}{7}}-\\sin{\\frac{2\\pi}{7}}+\\sin{\\frac{6\\pi}{7}}-\\sin{\\frac{4\\pi}{7}}=\\sin{\\frac{6\\pi}{7}}=\\sin{\\frac{\\pi}{7}}\n\\]\n\nThus \\(S = 1/2\\). \\(\\blacksquare\\)",
"Let \\(a=\\sin{\\frac{\\pi}{7}}\\) and \\(b=\\cos{\\frac{\\pi}{7}}\\). From the addition formulae, we have\n\n\\[\nS=b-[b^2-a^2]+[ b(b^2-a^2)-a(2ab) ]=b-b^2+b^3+a^2(1-3b)\n\\]\n\nFrom the Trigonometric Identity, \\(a^2=1-b^2\\), so\n\n\\[\nS=b-b^2+b^3+(1-b^2)(1-3b)=4b^3-2b^2-2b+1\n\\]\n\nWe must prove that \\(S=1/2\\). It suffices to show that \\(8b^3-4b^2-4b+1=0\\).\n\nNow note that \\(\\cos{\\frac{4\\pi}{7}}=-\\cos{\\frac{3\\pi}{7}}\\). We can find these in terms of \\(a\\) and \\(b\\):\n\n\\[\n\\cos{\\frac{4\\pi}{7}}=[b^2-a^2]^2-[2ab]^2=b^4-6a^2b^2+b^4=b^4-6(1-b^2)b^2+b^4=8b^4-8b^2+1\n\\]\n\n\\[\n\\cos{\\frac{3\\pi}{7}}=b[b^2-a^2]-a[2ab]=b^3-3a^2b=b^3-3(1-b^2)b=3b-4b^3\n\\]\n\nTherefore \\(8b^4-8b^2+1=-(3b-4b^3)\\Rightarrow 8b^4+4b^3-8b^2-3b+1=0\\). Note that this can be factored:\n\n\\[\n8b^4+4b^3-8b^2-3b+1=(8b^3-4b^2-4b+1)(b+1)=0\n\\]\n\nClearly \\(b\\neq -1\\), so \\(8b^3-4b^2-4b+1=0\\). This proves the result. \\(\\blacksquare\\)",
"Let \\(\\omega=\\mathrm{cis}\\left(\\frac{\\pi}{14}\\right)\\). Thus it suffices to show that \\(\\omega+\\omega^{-1}-\\omega^2-\\omega^{-2}+\\omega^3+\\omega^{-3}=1\\). Now using the fact that \\(\\omega^k=\\omega^{14+k}\\) and \\(-\\omega^2=\\omega^9\\), this is equivalent to\n\n\\[\n\\omega+\\omega^3+\\omega^5+\\omega^7+\\omega^9+\\omega^{11}+\\omega^{13}-\\omega^7\n\\]\n\n\\[\n\\omega\\left(\\frac{\\omega^{14}-1}{\\omega^2-1}\\right)-\\omega^7\n\\]\n\nBut since \\(\\omega\\) is a \\(14\\)th root of unity, \\(\\omega^{14}=1\\). The answer is then \\(-\\omega^{7}=1\\), as desired.\n\n~yofro",
"We let \\(\\omega = \\mathrm{cis} \\ \\left(\\dfrac{2\\pi}{7} \\right)\\). We therefore have \\(w^i\\), where \\(0 \\leq i \\leq 6\\), are the \\(7^{\\text{th}}\\) roots of unity. Since \\(\\sum_{i = 0}^6 \\omega^i = 0\\), then \\(\\sum_{i = 1}^6 \\omega^i = -1\\), so \\(\\sum_{i = 1}^3 \\mathrm{Re}(\\omega^i) = -\\dfrac{1}{2}\\). Therefore, because \\(\\mathrm{cis} \\ \\alpha = \\cos \\alpha + i \\sin \\alpha, \\mathrm{Re}(\\mathrm{cis} \\ \\alpha) = \\cos \\alpha\\), so\n\n\\[\n\\cos \\left(\\dfrac{2\\pi}{7} \\right) + \\cos \\left( \\dfrac{4\\pi}{7} \\right) + \\cos \\left(\\dfrac{6\\pi}{7} \\right) = -\\dfrac{1}{2}\n\\]\n\n\\[\n\\implies -\\cos \\left(\\dfrac{2\\pi}{7} \\right) - \\cos \\left(\\dfrac{4\\pi}{7} \\right) - \\cos \\left(\\dfrac{6\\pi}{7} \\right) = \\dfrac{1}{2}\n\\]\n\nSince \\(\\cos \\alpha = - \\cos(\\pi - \\alpha)\\), we have \\(\\cos \\left(\\dfrac{\\pi}{7} \\right) - \\cos \\left(\\dfrac{2\\pi}{7} \\right) + \\cos \\left(\\dfrac{3\\pi}{7} \\right) = \\dfrac{1}{2}\\) and we are done \\(\\blacksquare\\)\n\n~Yiyj1"
] |
IMO-1963-6
|
https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_6
|
Five students, \(A,B,C,D,E\), took part in a contest. One prediction was that the contestants would finish in the order \(ABCDE\). This prediction was very poor. In fact no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order \(DAECB\). This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.
|
[
"We are given that no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so in order \\(ABCDE\\). None of them finished in that order. Also only two of them had their actual positions in \\(DAECB\\). After imposing these two conditions the list of possible outcomes is: (1)\\(CAEBD\\), (2)\\(DCAEB\\), (3)\\(DCEBA\\), (4)\\(EDACB\\). One more condition is that two disjoint pairs of students predicted to finish consecutively actually did so. Out of the above four in the list, (1) and (2) have \\(AE\\) as the correctly predicted consecutive finishers(but only 1 pair), (3) has no correctly predicted consecutive finishers. But (4) has 2 disjoint correctly predicted consecutive finishers who are \\(DA\\) and \\(CB\\). Hence, order is \\(EDACB\\)."
] |
IMO-1964-1
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_1
|
(a) Find all positive integers \(n\) for which \(2^n-1\) is divisible by \(7\).
(b) Prove that there is no positive integer \(n\) for which \(2^n+1\) is divisible by \(7\).
|
[
"We claim \\(2^n\\) is equivalent to \\(2, 4,\\) and \\(1\\) \\(\\pmod{7}\\) for \\(n\\) congruent to \\(1\\), \\(2\\), and \\(0\\) \\(\\pmod{3}\\), respectively.\n\n(a) From the statement above, only \\(n\\) divisible by \\(3\\) will work.\n\n(b) Again from the statement above, \\(2^n\\) can never be congruent to \\(-1\\) \\(\\pmod{7}\\), so there are no solutions for \\(n\\).",
"This solution is clearer and easier to understand.\n\n(1) Since we know that \\(2^n-1\\) is congruent to 0 (mod 7), we know that \\(2^n\\) is congruent to 8 mod 7, which means \\(2^n\\) is congruent to 1 mod 7.\n\nExperimenting with the residue of \\(2^n\\) mod 7:\n\n\\(n\\)=1: 2\n\n\\(n\\)=2: 4\n\n\\(n\\)=3: 1 (this is because when \\(2^n\\) is doubled to \\(2*2^n\\), the residue doubles too, but \\(4*2=8\\) is congruent to 1 (mod 7).\n\n\\(n\\)=4: 2\n\n\\(n\\)=5: 4\n\n\\(n\\)=6: 1\n\nThrough induction, we easily show that this is true since the residue doubles every time you double \\(2^n\\).\n\nSo, the residue of \\(2^n\\) mod 7 cycles in 2, 4, 1. Therefore, \\(n\\) must be a multiple of 3. Proved.\n\n(2) According to part (1), the residue of \\(2^n\\) cycles in 2, 4, 1.\n\nIf \\(2^n+1\\) is congruent to 0 mod 7, then \\(2^n\\) must be congruent to 6 mod 7, but this is not possible due to how \\(2^n\\) mod 7 cycles. Therefore, there is no solution. Proved.\n\n~hastapasta"
] |
IMO-1964-2
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_2
|
Suppose \(a, b, c\) are the sides of a triangle. Prove that
\[
a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.
\]
|
[
"Let \\(b+c-a = x\\), \\(c+a-b = y\\), and \\(a+b-c = z\\). Then, \\(a = \\frac{y+z}{2}\\), \\(b = \\frac{x+z}{2}\\), and \\(c = \\frac{x+y}{2}\\). By AM-GM,\n\n\\[\n\\frac{x+y}{2} \\geq \\sqrt{xy},\n\\]\n\n\\[\n\\frac{y+z}{2} \\geq \\sqrt{yz},\n\\]\n\n\\[\n\\textrm{and }\\frac{x+z}{2} \\geq \\sqrt{xz}.\n\\]\n\nMultiplying these equations, we have\n\n\\[\n\\frac{x+y}{2} \\cdot \\frac{y+z}{2} \\cdot \\frac{x+z}{2} \\geq xyz\n\\]\n\n\\[\n\\therefore abc \\geq (a+b-c)(b+c-a)(c+a-b).\n\\]\n\nWe can now simplify:\n\n\\[\n(a+b-c)(b+c-a)(c+a-b) \\leq abc\n\\]\n\n\\[\n(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \\leq abc\n\\]\n\n\\[\na(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \\leq abc\n\\]\n\n\\[\n-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \\leq abc\n\\]\n\n\\[\na^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \\leq abc\n\\]\n\n\\[\na^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\\le{3abc}\\textrm{. }\\square\n\\]\n\n~mathboy100",
"We can use the substitution \\(a=x+y\\), \\(b=x+z\\), and \\(c=y+z\\) to get\n\n\\[\n2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\\leq 3(x+y)(x+z)(y+z)\n\\]\n\n\\[\n2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz\n\\]\n\n\\[\nx^2y+x^2z+y^2x+y^2z+z^2x+z^2y\\geq 6xyz\n\\]\n\n\\[\n\\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\\geq xyz=\\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}\n\\]\n\nThis is true by AM-GM. We can work backwards to get that the original inequality is true.",
"Rearrange to get\n\n\\[\na(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \\ge 0,\n\\]\n\nwhich is true by Schur's inequality."
] |
IMO-1964-3
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_3
|
A circle is inscribed in a triangle \(ABC\) with sides \(a,b,c\). Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from \(\triangle ABC\). In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of \(a,b,c\)).
|
[
"Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let the points of contact of MH,FG,DE with the in circle be X,Y,Z respectively. Then perimeter of BHM = BH+HX+XM+MB=BH+HP+MR+BM=BP+BQ=2(s-b) and similar results follow!\n\n\\[\n[BHM]/[BCA] = \\text{(perimeter of BHM/perimeter of BCA)}^2 ={(2s-2b)/(a+b+c)}^2 ={(c+a-b)/(c+a+b)}^2\n\\]\n\nDenote \\([ABC]\\) the area of \\(\\triangle ABC\\) and \\((ABC)\\) the perimeter of \\(\\triangle ABC\\).\n\nThen \\(\\frac{[BHM]}{[ABC]} = \\left(\\frac{(BHM)}{(ABC)}\\right)^{2} =\\left(\\frac{c+a-b}{c+a+b}\\right)^{2}\\).\n\nSo \\([BHM]=\\left(\\frac{c+a-b}{c+a+b}\\right)^{2}\\cdot [ABC]\\).\n\nWe know, \\(r_{1}\\) is the radius of the incircle of \\(\\triangle BHM\\): \\(r_{1}= 2 \\cdot \\frac{[BHM]}{(BHM)}\\).\n\nArea of the incircle of \\(\\triangle BHM\\)\n\n\\[\n= \\pi \\cdot 4 \\cdot (\\frac{[BHM]}{(BHM)})^{2}= 4\\pi \\frac{(c+a-b)^{2}}{(c+a+b)^{4}} \\cdot ([ABC])^{2}\n\\]\n\nArea of the incircle of \\(\\triangle ABC\\):\\(4\\pi (\\frac{[ABC]}{a+b+c})^{2}\\).\n\nSum of the area of the 4 incircles:\n\n\\[\n4 \\pi ([ABC])^{2}\\left[\\frac{(c+a-b)^{2}+(b+c-a)^{2}+(c+a-b)^{2}}{(a+b+c)^{4}}\\right]+4\\pi (\\frac{[ABC]}{a+b+c})^{2}\n\\]\n\n\\[\n=4 \\pi \\frac{([ABC])^{2}}{(a+b+c)^{2}}\\left[\\frac{(c+a-b)^{2}+(b+c-a)^{2}+(c+a-b)^{2}+(a+b+c)^{2}}{(a+b+c)^{2}}\\right]\n\\]\n\n\\[\n=16 \\pi \\frac{([ABC])^{2}(a^{2}+b^{2}+c^{2})}{(a+b+c)^{4}}\n\\]"
] |
IMO-1964-4
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_4
|
Seventeen people correspond by mail with one another - each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there are at least three people who write to each other about the same topic.
|
[
"Lemma: Consider a complete graph with 6 vertices colored with 2 colors. There exists a monochromatic triangle.\n\nProof: Consider one vertex and all connections leading out from it. Call it \\(V_1\\). It has 5 edges coming out from it. By the Pigeonhole Principle, there are at least 3 of the same color. Call this color red. Call those vertices \\(V_2\\), \\(V_3\\) and \\(V_4\\). If any of the segments \\(V_2V_3\\), \\(V_2V_4\\), or \\(V_3V_4\\) are red, then we have a monochromatic triangle with vertices \\(V_1\\) and the other two that are also red. If they are all the other color, then we have a monochromatic triangle with vertices \\(V_2\\),\\(V_3\\), and \\(V_4\\). \\(\\blacksquare\\)\n\nMain Problem: Represent these people as vertices on a connected graph with 17 vertices and colored with 3 colors, one corresponding with each topic. So this problem is reduced to showing that on a connected graph with 17 vertices and colored with three colors, there exists some monochromatic triangle. Look at an arbitrary vertex. Call it \\(V_1\\). Look at the 16 other vertices that it is connected to. By the Pigeonhole Principle, there are at least 6 vertices connected to \\(V_1\\) that are all one color. Call this color 1. If any of the connections inbetween these six vertices are in color 1, then we are done. If none of them are color 1, we know that that there are only 2 colors in those 6 vertices. By Lemma 1, we know that there is a monochromatic triangle in those 6 vertices. So we are done."
] |
IMO-1964-5
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_5
|
Suppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have.
|
[
"Suppose, those five points are \\((A, B, C, D, E)\\). Now, we want to create some special structure. Let, we take the line \\(BC\\) and draw a perpendicular from \\(A\\) on \\(BC\\), andd call it \\(P_1\\). We can do this set up in \\(\\binom{5}{1}\\binom{4}{2}=30\\) ways. There will \\(30\\) such \\(P_i\\) s.\n\nNow, we will find how many other perpendiculars intersect the line. We can do this in total \\(20\\) ways. Why? See, can draw perpendiculars from \\(B\\) and \\(C\\) to other lines( we haven't counted the perpendicular from \\(B\\) to \\(AC\\) and perpendicular from \\(C\\) on \\(AB\\) , as they intersect \\(P_1\\) at the same point) in \\(5\\) ways for each. So, total \\(10\\) ways.\n\nNow, \\(5\\) perpendiculars from each \\(D\\) and \\(E\\) on the other lines except on \\(BC\\)( because in this case teh perpendiculars from \\(D\\) and \\(E\\) will be parallel to \\(P_1\\) , and so shall not intersect). So,total \\(10\\) cases. From, these two cases we get \\(P_1\\) will be intersected at at most \\(5*6*(5+5+5+5)=600\\) points.\n\nBut, as we have passed this algorithm over all the five points, we have counted each intersection points twice. So, there are total \\(\\frac{600}{2}=300\\) ways.\n\nNow, as we had excluded the orthocentres, we have to add now. There are total \\(\\binom{5}{3}=10\\) orthocentres. Also we should add those vertices as these are also point of intersection of silimar perpendiculars, there are \\(5\\) such.\n\nSo, total ways \\(300+10+5=315\\)."
] |
IMO-1964-6
|
https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_6
|
In tetrahedron \(ABCD\), vertex \(D\) is connected with \(D_0\), the centroid of \(\triangle ABC\). Lines parallel to \(DD_0\) are drawn through \(A,B\) and \(C\). These lines intersect the planes \(BCD, CAD\) and \(ABD\) in points \(A_1, B_1,\) and \(C_1\), respectively. Prove that the volume of \(ABCD\) is one third the volume of \(A_1B_1C_1D_0\). Is the result true if point \(D_o\) is selected anywhere within \(\triangle ABC\)?
|
[
"\\[\nIMO 1964 P6 01.png\n\\]\n\nLet \\(A_{2}\\) be the point where line \\(AD_{0}\\) intersects line \\(BC\\)\n\nLet \\(B_{2}\\) be the point where line \\(BD_{0}\\) intersects line \\(AC\\)\n\nLet \\(C_{2}\\) be the point where line \\(CD_{0}\\) intersects line \\(AB\\)\n\nFrom centroid properties we have:\n\n\\[\n|AA_{2}|=3|D_{0}A_{2}|\n\\]\n\n\\[\n|BB_{2}|=3|D_{0}B_{2}|\n\\]\n\n\\[\n|CC_{2}|=3|D_{0}C_{2}|\n\\]\n\nTherefore,\n\n\\[\n\\frac{|AA_{2}|}{|D_{0}A_{2}|}=3\n\\]\n\n\\[\n\\frac{|BB_{2}|}{|D_{0}B_{2}|}=3\n\\]\n\n\\[\n\\frac{|CC_{2}|}{|D_{0}C_{2}|}=3\n\\]\n\nSince \\(\\Delta D_{0}A_{2}A_{1}\\sim \\Delta AA_{2}A_{1}\\), then \\(|AA_{1}|=|DD_{0}| \\frac{|AA_{2}|}{|D_{0}A_{2}|}=3|DD_{0}|\\)\n\nSince \\(\\Delta D_{0}B_{2}B_{1}\\sim \\Delta BB_{2}B_{1}\\), then \\(|BB_{1}|=|DD_{0}| \\frac{|BB_{2}|}{|D_{0}B_{2}|}=3|DD_{0}|\\)\n\nSince \\(\\Delta D_{0}C_{2}C_{1}\\sim \\Delta CC_{2}C_{1}\\), then \\(|CC_{1}|=|DD_{0}| \\frac{|CC_{2}|}{|D_{0}C_{2}|}=3|DD_{0}|\\)\n\nSince \\(|AA_{2}|=|BB_{2}|=|CC_{2}|\\) and \\(AA_{1} \\parallel BB_{1} \\parallel CC_{1} \\parallel DD_{0}\\),\n\nthen \\(\\Delta A_{1}B_{1}C_{1}\\parallel \\Delta ABC\\), and \\(Area_{\\Delta A_{1}B_{1}C_{1}}=Area_{\\Delta ABC}\\)\n\nLet \\(h_{D}\\) be the perpendicular distance from \\(D\\) to \\(\\Delta ABC\\)\n\nLet \\(h_{\\Delta A_{1}B_{1}C_{1}}\\) be the perpendicular distance from \\(\\Delta A_{1}B_{1}C_{1}\\) to \\(\\Delta ABC\\)\n\n\\[\n\\frac{h_{\\Delta A_{1}B_{1}C_{1}}}{h_{D}}=\\frac{|AA_{2}|}{|D_{0}A_{2}|}=\\frac{|BB_{2}|}{|D_{0}B_{2}|}=\\frac{|CC_{2}|}{|D_{0}C_{2}|}=3\n\\]\n\n\\[\nh_{\\Delta A_{1}B_{1}C_{1}}=3h_{D}\n\\]\n\n\\[\n\\frac{1}{3}h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}=3\\frac{1}{3}h_{D}Area_{\\Delta ABC}\n\\]\n\n\\[\n\\frac{h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}}{3}=3\\frac{h_{D}Area_{\\Delta ABC}}{3}\n\\]\n\nSince \\(Volume_{ABCD}=\\frac{h_{D}Area_{\\Delta ABC}}{3}\\) and \\(Volume_{A_{1}B_{1}C_{1}D_{0}}=\\frac{h_{\\Delta A_{1}B_{1}C_{1}}Area_{\\Delta A_{1}B_{1}C_{1}}}{3}\\)\n\nthen, \\(Volume_{A_{1}B_{1}C_{1}D_{0}}=3Volume_{ABCD}\\)\n\nthus, \\(Volume_{ABCD}=\\frac{1}{3}Volume_{A_{1}B_{1}C_{1}D_{0}}\\)\n\nthis proves that the volume of \\(ABCD\\) is one third the volume of \\(A_1B_1C_1D_0\\)\n\nThe result is NOT true if point \\(D_o\\) is selected anywhere within \\(\\triangle ABC\\) as ratios of \\(\\frac{|AA_{2}|}{|D_{0}A_{2}|}\\), \\(\\frac{|BB_{2}|}{|D_{0}B_{2}|}\\), and \\(\\frac{|CC_{2}|}{|D_{0}C_{2}|}\\) will have values other than 3 as the point is no longer a centroid. Also, it will make the ratios \\(\\frac{|AA_{2}|}{|D_{0}A_{2}|} \\ne \\frac{|BB_{2}|}{|D_{0}B_{2}|}\\ne \\frac{|CC_{2}|}{|D_{0}C_{2}|}\\) which means that \\(\\Delta A_{1}B_{1}C_{1} \\nparallel \\Delta ABC\\) and the volume relationship will no longer hold true.\n\n~Tomas Diaz. [email protected]"
] |
IMO-1965-1
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_1
|
Determine all values \(x\) in the interval \(0\leq x\leq 2\pi\) which satisfy the inequality
\[
2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.
\]
|
[
"We shall deal with the left side of the inequality first (\\(2\\cos x \\leq \\left| \\sqrt{1+\\sin 2x} - \\sqrt{1-\\sin 2x } \\right|\\)) and the right side after that.\n\nIt is clear that the left inequality is true when \\(\\cos x\\) is non-positive, and that is when \\(x\\) is in the interval \\([\\pi/2, 3\\pi/2]\\). We shall now consider when \\(\\cos x\\) is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. \\(4\\cos^2{x}\\leq 1+\\sin 2x+1-\\sin 2x-2\\sqrt{1-\\sin^2 2x}=2-2\\sqrt{\\cos^2{2x}}\\). This inequality is equivalent to \\(2\\cos^2 x\\leq 1-\\left| \\cos 2x\\right|\\). I shall now divide this problem into cases.\n\nCase 1: \\(\\cos 2x\\) is non-negative. This means that \\(x\\) is in one of the intervals \\([0,\\pi/4]\\) or \\([7\\pi/4, 2\\pi]\\). We must find all \\(x\\) in these two intervals such that \\(2\\cos^2 x\\leq 1-\\cos 2x\\). This inequality is equivalent to \\(2\\cos^2 x\\leq 2\\sin^2 x\\), which is only true when \\(x=\\pi/4\\) or \\(7\\pi/4\\).\n\nCase 2: \\(\\cos 2x\\) is negative. This means that \\(x\\) is in one of the interavals \\((\\pi/4, \\pi/2)\\) or \\((3\\pi/2, 7\\pi/4)\\). We must find all \\(x\\) in these two intervals such that \\(2\\cos^2 x\\leq 1+\\cos 2x\\), which is equivalent to \\(2\\cos^2 x\\leq 2\\cos^2 x\\), which is true for all \\(x\\) in these intervals.\n\nTherefore the left inequality is true when \\(x\\) is in the union of the intervals \\([\\pi/4, \\pi/2)\\), \\((3\\pi/2, 7\\pi/4]\\), and \\([\\pi/2, 3\\pi/2]\\), which is the interval \\([\\pi/4, 7\\pi/4]\\). We shall now deal with the right inequality.\n\nAs above, we can square it and have it be true whenever the original right inequality is true, so we do that. \\(2-2\\sqrt{\\cos^2{2x}}\\leq 2\\), which is always true. Therefore the original right inequality is always satisfied, and all \\(x\\) such that the original inequality is satisfied are in the interval \\([\\pi/4, 7\\pi/4]\\).",
"Manipulate the inequality so that it becomes:\n\n\\[\n\\sqrt{2}\\cos x \\leq \\left| \\sqrt{\\frac{1+\\cos(\\frac{\\pi}{2}-2x)}{2}} - \\sqrt{\\frac{1-\\cos(\\frac{\\pi}{2}- 2x)}{2}} \\right| \\leq 1\n\\]\n\nInside the absolute value, we identify the half-angle formulas. However, since we do not know the sign of the resultant expression, we have to use absolute value signs since the principal square root is always positive; then the inequality becomes\n\n\\[\n\\sqrt{2}\\cos x \\leq \\left| \\left|\\cos\\left(\\frac{\\pi}{4}-x\\right)\\right| - \\left|\\sin\\left(\\frac{\\pi}{4}-x\\right)\\right|\\right| \\leq 1\n\\]\n\nwhich, considering absolute value, is the same as\n\n\\[\n\\sqrt{2}\\cos x \\leq \\left| \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\right| - \\left|\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\right| \\leq 1\n\\]\n\nSince both inner absolute values range between \\(0\\) and \\(1\\), their positive difference also ranges from \\(0\\) to \\(1\\), so the right inequality is always true. Thus we take a look at the left inequality. For \\(x\\in\\left[\\frac{\\pi}{2},\\frac{3\\pi}{2}\\right]\\), the left-hand side is never positive; however, the right-hand side is always nonnegative due to absolute value, so the inequality holds for these \\(x\\)-values. As a result, we consider the remaining two sections of the interval.\n\nFor \\(x\\in\\left[0,\\frac{\\pi}{4}\\right]\\) and \\(x\\in\\left[\\frac{7\\pi}{4},2\\pi\\right]\\), when considering the \\(-\\frac{\\pi}{4}\\), inside the absolute value, cosine is positive but sine is negative; thus our inequality becomes \\begin{align*} \\sqrt{2}\\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right) +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2} +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2}\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\sin\\left(\\frac{\\pi}{4}\\right) +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\cos\\left(\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\sin\\left(\\left(x-\\frac{\\pi}{4}\\right)+\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le |\\sin x|\\\\ \\end{align*} If \\(x\\in\\left[0,\\frac{\\pi}{4}\\right]\\), then due to the interval, \\(\\sin x\\) and \\(\\cos x\\) are both never negative; thus, \\begin{align*} \\cos x &\\le \\sin x\\\\ 1&\\le\\tan x\\\\ \\end{align*} But this is never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.\n\nIf \\(x\\in\\left[\\frac{7\\pi}{4},2\\pi\\right]\\), then due to the interval, \\(\\sin x\\) is negative and \\(\\cos x\\) is positive; thus, \\begin{align*} \\cos x &\\le -\\sin x\\\\ 1&\\le-\\tan x\\\\ -1&\\ge\\tan x\\\\ \\end{align*} But this is also never true in our interval (except at one endpoint, which is included in the interval below), so the inequality does not hold.\n\nFor \\(x\\in\\left[\\frac{\\pi}{4},\\frac{\\pi}{2}\\right]\\), inside the absolute value, when considering the \\(-\\frac{\\pi}{4}\\), both cosine and sine are positive; thus our inequality becomes \\begin{align*} \\sqrt{2}\\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2} -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\cdot\\frac{\\sqrt{2}}{2}\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right)\\cos\\left(\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\sin\\left(\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le \\left|\\cos\\left(\\left(x-\\frac{\\pi}{4}\\right)+\\frac{\\pi}{4}\\right)\\right|\\\\ \\cos x &\\le |\\cos x|\\\\ \\end{align*} This is always true, so the inequality holds.\n\nFor \\(x\\in\\left[\\frac{3\\pi}{2},\\frac{7\\pi}{4}\\right]\\), inside the absolute value, when considering the \\(-\\frac{\\pi}{4}\\), both cosine and sine are negative; thus our inequality becomes \\begin{align*} \\sqrt{2}\\cos x &\\le \\left|-\\cos\\left(x-\\frac{\\pi}{4}\\right) +\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\sqrt{2}\\cos x &\\le \\left|-\\left(\\cos\\left(x-\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right)\\right|\\\\ \\sqrt{2}\\cos x &\\le \\left|\\cos\\left(x-\\frac{\\pi}{4}\\right) -\\sin\\left(x-\\frac{\\pi}{4}\\right)\\right|\\\\ \\end{align*} This is the same as the case for \\(x\\in\\left[\\frac{\\pi}{4},\\frac{\\pi}{2}\\right]\\), so the inequality holds.\n\nCombining our findings, we find that the solutions are \\(\\boxed{x\\in\\left[\\frac{\\pi}{4},\\frac{7\\pi}{4}\\right]}\\).\n\n~eevee9406"
] |
IMO-1965-2
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_2
|
Consider the system of equations
\[
a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0
\]
\[
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0
\]
\[
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0
\]
with unknowns \(x_1\), \(x_2\), \(x_3\). The coefficients satisfy the conditions:
(a) \(a_{11}\), \(a_{22}\), \(a_{33}\) are positive numbers;
(b) the remaining coefficients are negative numbers;
(c) in each equation, the sum of the coefficients is positive.
Prove that the given system has only the solution \(x_1 = x_2 = x_3 = 0\).
|
[
"Clearly if the \\(x_i\\) are all equal, then they are equal to 0. Now let's assume WLOG that \\(x_1=0\\). If \\(x_2\\) or \\(x_3\\) is 0, then the other is clearly zero, so let's consider the case where neither are 0. \\(a_{12}\\) and \\(a_{21}\\) are negative, so exactly one of \\(x_2\\) or \\(x_3\\) is positive. Unfortunately this means that one of \\(a_{22}x_2 + a_{23}x_3\\) or \\(a_{32}x_2 + a_{33}x_3 = 0\\) is positive and the other is negative, so the equation couldn't possibly be satisfied if \\(x_2\\) or \\(x_3\\) isn't 0. We have covered the case where one of the \\(x_i\\) is 0, now let's assume that none of them are 0.\n\nIf two are positive and one is negative, then when the negative \\(x_i\\) is paired with one of the positive \\(a_i\\), the corresponding equation is negative. This is bad. If two are negative and one is positive, then when the positive \\(x_i\\) is paired with one of the positive \\(a_i\\), the corresponding equation is positive. This is also bad. Therefore the \\(x_i\\) all have the same sign.\n\nCase 1: The \\(x_i\\) are all positive. WLOG \\(x_1\\leq x_2\\leq x_3\\). Now consider the third equation, \\(a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0\\). Therefore \\(x_2(a_{31} +a_{32}+a_{33})+ a_{31}(x_1-x_2)+a_{33}(x_3-x_2)= 0\\), but all of the terms on the LHS are non-negative and the first one is positive, so this is impossible.\n\nCase 2: The \\(x_i\\) are all negative. WLOG \\(x_1\\geq x_2\\geq x_3\\). Consider the third equation, \\(a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0\\). Therefore \\(x_3(a_{31}+a_{32}+a_{33})+a_{31}(x_1-x_3)+a_{32}(x_2-x_3)=0\\), but all of the terms on the LHS are non-positive and the first one is negative, so this is impossible.\n\nTherefore at least one of the \\(x_i\\) is 0, which implies all of them are 0.",
"We will prove that the matrix\n\n\\[\na_{11}\\ \\ \\ \\ a_{12}\\ \\ \\ \\ a_{13}\n\\]\n\n\\[\na_{21}\\ \\ \\ \\ a_{22}\\ \\ \\ \\ a_{23}\n\\]\n\n\\[\na_{31}\\ \\ \\ \\ a_{32}\\ \\ \\ \\ a_{33}\n\\]\n\nhas its determinant \\(> 0.\\)\n\nThe determinant is\n\n\\[\na_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{32}a_{21} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{33}a_{12}a_{21}.\n\\]\n\nAfter a little algebraic manipulation we can rewrite this as\n\n\\[\n[a_{33}(a_{11} + a_{13}) - a_{13}(a_{31} + a_{33})](a_{21} + a_{22} + a_{23}) - (a_{21}a_{33} - a_{23}a_{31})(a_{11} + a_{12} + a_{13}) - (a_{11}a_{23} - a_{13}a_{21})(a_{31} + a_{32} + a_{33})\n\\]\n\n(or we can just verify that this is true).\n\nNote that \\(a_{11} + a_{12} + a_{13} > 0\\) implies \\(a_{11} + a_{12} > 0\\) and \\(a_{11} + a_{13} > 0.\\) The expression above is clearly \\(> 0\\). To show this in a simple way, I will just write out the sign of each factor:\n\n\\[\n[(+)(+) - (-)(+)]\\ (+) - ((-)(+) - (-)(-))\\ (+) - ((+)(-) - (-)(-))\\ (+)\n\\]\n\nso now we can see that the end expression is \\((+).\\)\n\n[Solution by pf02, November 2024]"
] |
IMO-1965-3
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_3
|
Given the tetrahedron \(ABCD\) whose edges \(AB\) and \(CD\) have lengths \(a\) and \(b\) respectively. The distance between the skew lines \(AB\) and \(CD\) is \(d\), and the angle between them is \(\omega\). Tetrahedron \(ABCD\) is divided into two solids by plane \(\varepsilon\), parallel to lines \(AB\) and \(CD\). The ratio of the distances of \(\varepsilon\) from \(AB\) and \(CD\) is equal to \(k\). Compute the ratio of the volumes of the two solids obtained.
|
[
"Let the plane meet \\(AD\\) at \\(X\\), \\(BD\\) at \\(Y\\), \\(BC\\) at \\(Z\\) and \\(AC\\) at \\(W\\). Take a plane parallel to \\(BCD\\) through \\(WX\\) and let it meet \\(AB\\) in \\(P\\).\n\n\\[\nProb 1965 3.png\n\\]\n\nSince the distance of \\(AB\\) from \\(WXYZ\\) is \\(k\\) times the distance of \\(CD\\), we have that \\(AX = k \\cdot XD\\) and hence \\(AX/AD = k/(k+1).\\) Similarly \\(AP/AB = AW/AC = AX/AD.\\) \\(XY\\) is parallel to \\(AB\\), so also \\(AX/AD = BY/BD = BZ/BC.\\)\n\nvol \\(ABWXYZ =\\) vol \\(APWX +\\) vol \\(WXPBYZ.\\) \\(APWX\\) is similar to the tetrahedron \\(ABCD.\\) The sides are \\(k/(k + 1)\\) times smaller, so vol \\(APWX = k^3/(k + 1)^3\\) vol \\(ABCD.\\)\n\nThe base of the prism \\(WXPBYZ\\) is \\(BYZ\\) which is similar to \\(BCD\\) with sides \\(k/(k + 1)\\) times smaller and hence area \\(k^2/(k + 1)^2\\) times smaller. Its height is \\(1/(k + 1)\\) times the height of \\(A\\) above \\(ABCD,\\) so vol prism \\(= 3 k^2/(k + 1)^3\\) vol \\(ABCD.\\)\n\nThus vol \\(ABWXYZ = (k^3 + 3k^2)/(k + 1)^3\\) vol \\(ABCD.\\)\n\nWe get the volume of the other piece as vol \\(ABCD\\ -\\) vol \\(ABWXYZ,\\) and hence the ratio is (after a little manipulation) \\(k^2(k + 3)/(3k + 1).\\)\n\n## Remark (added by pf02, November 2024)\n\nNote that the problem is untypically sloppy or misleading. It mentions the sizes \\(a, b, d, \\omega\\) as if they are needed. In fact, as the solution above shows, they are not needed either in expressing the result, or in solving the problem. A thorough problem solver might worry about not having used all the data in the problem.\n\nHowever, one can imagine other solutions, where these quantities would be used in the process of solving the problem. For example one could break up the doubly truncated triangular prism \\(ABWXYZ\\) into pyramids \\(APWX, PXYZW, BPZY\\). Computing the volume of each of these pyramids would require all the data in the problem. The end result should of course be the same, but the thorough problem solver would not have the uneasy feeling of not having used all the data in the problem."
] |
IMO-1965-4
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_4
|
Find all sets of four real numbers \(x_1\), \(x_2\), \(x_3\), \(x_4\) such that the sum of any one and the product of the other three is equal to \(2\).
|
[
"Let \\(P = x_1x_2x_3x_4\\) be the product of the four real numbers.\n\nThen, for \\(i = 1,2,3,4\\) we have: \\(x_i + \\prod_{j \\neq i}x_j = 2\\).\n\nMultiplying by \\(x_i\\) yields:\n\n\\(x^2_i + P = 2x_i \\Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \\Longleftrightarrow x_i = 1 \\pm t\\) where \\(t = \\pm \\sqrt{1-P} \\in \\mathbb{R}\\).\n\nIf \\(t=0\\), then we have \\((x_1,x_2,x_3,x_4)=(1,1,1,1)\\) which is a solution.\n\nSo assume that \\(t \\neq 0\\). WLOG, let at least two of \\(x_i\\) equal \\(1+t\\), and \\(x_1 \\ge x_2 \\ge x_3 \\ge x_4\\) OR \\(x_1 \\le x_2 \\le x_3 \\le x_4\\).\n\nCase I: \\(x_1 = x_2 = x_3 = x_4 = 1+t\\)\n\nThen we have:\n\n\\[\n(1+t)+(1+t)^3 = 2 \\Longleftrightarrow t^3+3t^2+4t = 0 \\Longleftrightarrow t(t^2+3t+4) = 0\n\\]\n\nWhich has no non-zero solutions for \\(t\\).\n\nCase II: \\(x_1 = x_2 = x_3 = 1+t\\) AND \\(x_4 = 1-t\\)\n\nThen we have:\n\n\\((1-t)+(1+t)^3 = 2 \\Longleftrightarrow t^3+3t^2+2t = 0\\) \\(\\Longleftrightarrow t(t+1)(t+2) = 0 \\Longleftrightarrow t \\in \\{0,-1,-2\\}\\)\n\nAND\n\n\\((1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0\\) \\(\\Longleftrightarrow -t(t-1)(t+2) = 0 \\Longleftrightarrow t \\in \\{0,1,-2\\}\\)\n\nSo, we have \\(t = -2\\) as the only non-zero solution, and thus, \\((x_1,x_2,x_3,x_4) = (-1,-1,-1,3)\\) and all permutations are solutions.\n\nCase III: \\(x_1 = x_2 = 1+t\\) AND \\(x_3 = x_4 = 1-t\\)\n\nThen we have:\n\n\\((1-t)+(1-t)(1+t)^2 = 2 \\Longleftrightarrow -t^3-t^2 = 0\\) \\(\\Longleftrightarrow -t^2(t+1) = 0 \\Longleftrightarrow t \\in \\{0,-1\\}\\)\n\nAND\n\n\\((1+t)+(1+t)(1-t)^2 = 2 \\Longleftrightarrow t^3-t^2 = 0\\) \\(\\Longleftrightarrow t^2(t-1) = 0 \\Longleftrightarrow t \\in \\{0,1\\}\\)\n\nThus, there are no non-zero solutions for \\(t\\) in this case.\n\nTherefore, the solutions are: \\((1,1,1,1)\\); \\((3,-1,-1,-1)\\); \\((-1,3,-1,-1)\\); \\((-1,-1,3,-1)\\); \\((-1,-1,-1,3)\\).",
"We have to solve the system of equations\n\n\\[\nx_1 + x_2x_3x_4 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1) \\\\ x_2 + x_1x_3x_4 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (2) \\\\ x_3 + x_1x_2x_4 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (3) \\\\ x_4 + x_1x_2x_3 = 2\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (4)\n\\]\n\nSubtract (2) from (1) and factor. We get\n\n\\((x_1 - x_2)(1 - x_3x_4) = 0\\),\n\nwhich implies \\(x_1 - x_2 = 0\\) or \\(1 - x_3x_4 = 0\\).\n\nSimilarly, subtracting (3) and then (4) from (1) and factoring, we get\n\n\\[\n(x_1 - x_3)(1 - x_2x_4) = 0 \\\\ (x_1 - x_4)(1 - x_2x_3) = 0\n\\]\n\nThey imply \\(x_1 - x_3 = 0\\) or \\(1 - x_2x_4 = 0\\), and \\(x_1 - x_4 = 0\\) or \\(1 - x_2x_3 = 0\\).\n\nWe will consider four possibilities:\n\n1. \\(x_2 = x_1, x_3 = x_1, x_4 = x_1\\)\n\n2. \\(x_2 = x_1, x_3 = x_1\\) and \\(x_2x_3 = 1\\)\n\n3. \\(x_2 = x_1\\) and \\(x_2x_3 = 1, x_2x_4 = 1\\)\n\n4. \\(x_2x_3 = 1, x_2x_4 = 1, x_3x_4 = 1\\)\n\nNote that in fact, there are four more possibilities, but they just correspond to permutations in \\(x_1, x_2, x_3, x_4\\) of cases 2. and 3., so there is no harm in not dealing with them explicitly.\n\nCase 1. Plug \\(x_2, x_3, x_4\\) in equation (1). We get \\(x_1 + x_1^3 = 2\\). This is an equation of degree \\(3\\) whose only real root is \\(x_1 = 1\\). We get the solution \\(x_1 = x_2 = x_3 = x_4 = 1\\).\n\nCase 2. Plug \\(x_2, x_3\\) into \\(x_2x_3 = 1\\), and get \\(x_1^2 = 1\\). We get \\(x_1 = 1\\) or \\(x_1 = -1\\). The first solution, \\(x_1 = 1\\), yields \\(x_2 = x_3 = 1\\), and using (4), \\(x_4 = 1\\). The second solution, \\(x_1 = -1\\), yields \\(x_2 = x_3 = -1\\), and using (4), \\(x_4 = 3\\). We get the solution \\(x_1 = x_2 = x_3 = -1, x_4 = 3.\\) Because of the permutations of \\(x_1, x_2, x_3, x_4\\), we also get the solutions \\((-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)\\).\n\nCase 3. Plug in \\(x_2x_3, x_2x_4\\) in (4) and (3), and get \\(x_4 + x_1 = 2\\) and \\(x_3 + x_1 = 2\\). Now plug \\(x_2, x_3, x_4\\) in (1), and get \\(x_1 + x_1(2 - x_1)^2 = 2\\). This equation becomes \\((x_1 - 1)^2(x_1 - 2) = 0\\). \\(x_1 = 1\\) yields \\((1, 1, 1, 1)\\) which we already know, and \\(x_1 = 2\\) yields \\((2, 2, 0, 0)\\) (or some other values, depending on where we plug \\(x_1 = 2\\)), which do not give a possible solution.\n\nCase 4. plugging \\(x_2x_3, x_2x_4, x_3x_4\\) into (4), (3), (2), we get \\(x_4 + x_1 = 2, x_3 + x_1 = 2, x_2 + x_1 = 2\\). Plugging \\(x_2, x_3, x_4\\) into (1), we get \\(x_1 + (2 - x_1)^3 = 2\\). The solutions of this equation are \\(1, 2, 3\\). Note that \\(x_1 = 1\\) and \\(x_1 = 3\\) yield solutions we already know, and \\(x_1 = 2\\) is impossible.\n\nThus, the five solutions to the problem are \\((1, 1, 1, 1), (3, -1, -1, -1), (-1, 3, -1, -1), (-1, -1, 3, -1), (-1, -1, -1, 3)\\).\n\n[Solution by pf02, November 2024]"
] |
IMO-1965-5
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_5
|
Consider \(\triangle OAB\) with acute angle \(AOB\). Through a point \(M \neq O\) perpendiculars are drawn to \(OA\) and \(OB\), the feet of which are \(P\) and \(Q\) respectively. The point of intersection of the altitudes of \(\triangle OPQ\) is \(H\). What is the locus of \(H\) if \(M\) is permitted to range over (a) the side \(AB\), (b) the interior of \(\triangle OAB\)?
|
[
"Let \\(O(0,0),A(a,0),B(b,c)\\). Equation of the line \\(AB: y=\\frac{c}{b-a}(x-a)\\). Point \\(M \\in AB : M(\\lambda,\\frac{c}{b-a}(\\lambda-a))\\). Easy, point \\(P(\\lambda,0)\\). Point \\(Q = OB \\cap MQ\\), \\(MQ \\bot OB\\). Equation of \\(OB : y=\\frac{c}{b}x\\), equation of \\(MQ : y=-\\frac{b}{c}(x-\\lambda)+\\frac{c}{b-a}(\\lambda-a)\\). Solving: \\(x_{Q}=\\frac{1}{b^{2}+c^{2}}\\left[b^{2}\\lambda+\\frac{c^{2}(\\lambda-a)b}{b-a}\\right]\\). Equation of the first altitude: \\(x=\\frac{1}{b^{2}+c^{2}}\\left[b^{2}\\lambda+\\frac{c^{2}(\\lambda-a)b}{b-a}\\right] \\quad (1)\\). Equation of the second altitude: \\(y=-\\frac{b}{c}(x-\\lambda)\\quad\\quad (2)\\). Eliminating \\(\\lambda\\) from (1) and (2):\n\n\\[\nac \\cdot x + (b^{2}+c^{2}-ab)y=abc\n\\]\n\na line segment \\(MN , M \\in OA , N \\in OB\\). Second question: the locus consists in the \\(\\triangle OMN\\).",
"This solution is a simplified version of the previous solution, it fills in some gaps. and it provides more information.\n\nJust like in the previous solution, we use analytic (coordinate) geometry, but we don't care how the axes are chosen.\n\nLet \\(A, B, O\\) have coordinates \\((a_1, a_2), (b_1, b_2), (c_1, c_2)\\), and let \\(M = (\\lambda a_1 + (1 - \\lambda b_1, \\lambda a_2 + (1 - \\lambda b_2)\\) with \\(\\lambda \\in [0, 1]\\). The idea is to just follow the degrees of the expressions and equations in \\(\\lambda, x, y\\) involved as we make the computations for obtaining the coordinates of \\(H\\), and the equation of the curve \\(H\\) is on. We will see that the equation for \\(H\\) is an equation of degree \\(1\\), so we will know that it is a line. We don't need to write out the equation explicitly.\n\nThe coordinates of \\(M\\) are expressions of degree \\(1\\) in \\(\\lambda\\).\n\nThe equation for \\(MP\\) (the perpendicular from \\(M\\) to \\(OA\\)) is an equation of degree \\(1\\) in \\(x, y\\) with constant coefficients for \\(x, y\\), and whose constant term is an expression of degree \\(1\\) in \\(\\lambda\\).\n\nThe coordinates of \\(P\\) (the foot of the perpendicular from \\(P\\) to \\(OA\\)) are expressions of degree \\(1\\) in \\(\\lambda\\).\n\nThe equation of the perpendicular from \\(P\\) to \\(OB\\) is of degree \\(1\\) in \\(x, y\\), with constant coefficients for \\(x, y\\), and whose constant term is an expression of degree \\(1\\) in \\(\\lambda\\). This corresponds to equation (2) in the above solution.\n\nSimilarly, the equation of the perpendicular from \\(Q\\) to \\(OA\\) is of degree \\(1\\) in \\(x, y\\), with constant coefficients for \\(x, y\\), and whose constant term is an expression of degree \\(1\\) in \\(\\lambda\\). This corresponds to equation (1) in the above solution.\n\nNow, in principle, we would have to solve the system of two equations (1) and (2) to obtain the coordinates of \\(H\\) as expressions of \\(\\lambda\\), and then eliminate \\(\\lambda\\) to obtain the equation in \\(x, y\\) for \\(H\\). As a shortcut, we can eliminate \\(\\lambda\\) directly from the two equations (1) and (2). Either way, the result is an equation of degree \\(1\\) in \\(x, y\\).\n\nThis tells us that the locus is on a line. We just need to specify which set of points on this line is the locus. And, we want to make the line explicit.\n\nThe previous solution, with a good amount of hand waving, tells us that the solution is \"a line segment \\(B_1A_1, B_1 \\in OA, A_1 \\in OB\\)\". (On top of the hand waving the solution uses the unhappy notation \\(M\\) for \\(B_1\\) and \\(N\\) for \\(A_1\\), which is bad because \\(M\\) has already been used!) We will do better than that.\n\nLet \\(A_1\\) be the foot of the perpendicular from \\(A\\) to \\(OB\\), and \\(B_1\\) be the foot of the perpendicular from \\(B\\) to \\(OA\\). (For this paragraph see the picture shown in Solution 3.) Consider the limit situation when \\(M = A\\). Then \\(Q = A_1\\), and \\(P = A\\). It follows that the intersection \\(H\\) of the perpendiculars from \\(P\\) to \\(OB\\) and \\(Q\\) to \\(OA\\) is \\(A_1\\). Similarly, the limit situation when \\(M = B\\) yields \\(H = B_1\\). Now it is reasonable to say that when \\(M\\) moves from \\(A\\) to \\(B\\), \\(H\\) moves from \\(A_1\\) to \\(B_1\\). So, the locus is the line segment joining the feet \\(A_1, B_1\\) of the perpendiculars in \\(\\triangle OAB\\) from \\(A, B\\). This answers question (a).\n\nFor part (b) of the problem, with a good amount of hand waving, the previous solution says \"the locus consists in the \\(\\triangle OB_1A_1\\)\". We justify this by pointing out that if \\(M\\) is inside \\(\\triangle OAB\\), then we can take the triangle \\(\\triangle OA'B'\\), such that \\(A' \\in OA\\), \\(B' \\in OB\\), \\(A'B'\\) going through \\(M\\) and parallel to \\(AB\\). Then \\(H\\) will be on the corresponding segment \\(A_1'B_1'\\) determined by the feet of the perpendiculars in \\(\\triangle OA'B'\\). Conversely, it is easy to see that any point \\(H \\in \\triangle OA_1B_1\\) is on a segment \\(A_1'B_1'\\) obtained from a triangle \\(\\triangle OA'B'\\), and \\(H\\) is obtained from a point \\(M \\in A'B'\\). This answers question (b).\n\n[Solution by pf02, October 2024]",
"This solution is elementary, it does not use analytic geometry.\n\nLet \\(A_1\\) be the foot of the perpendicular from \\(A\\) to \\(OB\\), and \\(B_1\\) be the foot of the perpendicular from \\(B\\) to \\(OA\\). Construct \\(H\\) as described in the statement of the problem.\n\n\\[\nProb 1965 5.png\n\\]\n\nWe will prove that \\(H \\in A_1B_1\\).\n\nFirst, as shown in the second picture, take \\(MQ \\perp OB\\), and let \\(H_1\\) be the intersection of \\(A_1B_1\\) with the perpendicular from \\(Q\\) to \\(OA\\). From the triangle \\(\\triangle BAA_1\\) we have \\(\\frac{AM}{MB} = \\frac{A_1Q}{QB}\\) because \\(AA_1 \\parallel MQ\\). From the triangle \\(\\triangle A_1BB_1\\) we have \\(\\frac{A_1Q}{QB} = \\frac{A_1H_1}{H_1B_1}\\) because \\(QH_1 \\parallel BB_1\\). So, \\(\\frac{AM}{MB} = \\frac{A_1H_1}{H_1B_1}\\).\n\nSecond, as shown in the third picture, take \\(MP \\perp OA\\), and let \\(H_2\\) be the intersection of \\(A_1B_1\\) with the perpendicular from \\(P\\) to \\(OB\\). By a similar argument, we have \\(\\frac{AM}{MB} = \\frac{A_1H_2}{H_2B_1}\\).\n\nIt follows that \\(H_1 = H_2\\) since the two points divide \\(A_1B_1\\) in the same ratio. This is the point \\(H\\) from the statement of the problem.\n\nThe solution to the problem is now completed by repeating the last two paragraphs from Solution 2.\n\n[Solution by pf02, October 2024]"
] |
IMO-1965-6
|
https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_6
|
In a plane a set of \(n\) points (\(n\geq 3\)) is given. Each pair of points is connected by a segment. Let \(d\) be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length \(d\). Prove that the number of diameters of the given set is at most \(n\).
|
[
"Image of problem Solution. Credits to user awe-sum.\n\n## Remarks (added by pf02, October 2024)\n\n1. As a public service, I will upload the \"Image of problem Solution\" mentioned above to this web page. That way, a reader can see the \"Solution\" immediately, without having to go to another web site, and we are not subjected to the possibility of the imgur.com website being taken down, or Imgur's parent company deciding to delete this particular image. Credits for the image are due to user awe-sum, as pointed out above.\n\n2. This \"Solution\" is presented very badly, and edited very badly. Indeed, some terms are undefined, left to the reader to make sense of (e.g. \"incident\", \"degree\"). But let us be forgiving, and let us do our best to make sense of the \"Solution\".\n\n3. The \"Solution\" is incomplete, to the point of not being a solution at all. There are some serious gaps, which raise questions which are not addressed, and a reader can not be expected to fill in the details. These are:\n\na. The author says \"... at least one point is incident to at least three diagonals\". But, it could also happen that two points are \"incident\" to two \"diagonals\" (i.e. diameters) each. The author does not address this possibility at all.\n\nb. The author says \"consider all points at distance exactly d from point Q (the green circle). We see that all of them are outside the locus, the exception being P.\" This is far from obvious. It assumes that all the other k-4 points (those points of the k given points which are not highlighted in the picture) are inside the \"locus (the blue shaded region)\". In fact, it seems to this reader that this is not necessarily true.\n\nc. Another issue with this \"Solution\" is that it assumes \\(n \\ge 4\\), while the statement of the problem has \\(n \\ge 3\\). This shortcoming is easy to fix, unlike the previous two I mentioned.\n\n4. I will give two solutions below, in the section \"Solution 2\" and \"Section 3\".",
"\\[\nProblem 1965 6 sol by awe-sum.png\n\\]",
"\\(\\mathbf{Definition:}\\) For the purpose of this proof, let us define an \\(equilateral\\ arc\\ triangle\\) as the shape we obtain when we take a triangle \\(\\triangle ABC\\) in which we replace the side \\(AB\\) by the \\(\\pi/6\\) arc of the circle centered at \\(C\\) with radius \\(CA = CB\\), going from \\(A\\) to \\(B\\), and similarly for the other two sides \\(BC\\) and \\(AC\\).\n\nSee the picture below.\n\n\\[\nProb 1965 6 fig1.png\n\\]\n\nNote that if the sides of the original equilateral triangle were of length \\(d\\), then the distance from a vertex to any point on the opposite arc is \\(d\\). Also, the distance between any two points on the arcs, such that none of them are vertices is \\(< d\\). And finally, the distance from a vertex to any point on the arcs adjacent to it (which is not another vertex) is \\(< d\\).\n\n\\(\\mathbf{Lemma:}\\) A configuration of \\(n\\) points has exactly \\(n\\) diameters if an only if \\(3\\) of the points are the vertices of an equilateral arc triangle, and the other \\(n - 3\\) are on the boundary of the triangle.\n\n\\[\nProb 1965 6 fig2.png\n\\]\n\nIn the configuration shown above, we have \\(9\\) points with \\(9\\) diameters.\n\n\\(\\mathbf{Proof\\ of\\ the\\ Lemma:}\\) It is clear that if we have a configuration as described in the lemma then there are \\(n\\) diameters. We will prove the converse by induction. We want to prove that if we have a configuration of \\(n\\) points with \\(n\\) diameters, then the points are in a configuration as described above.\n\nIf \\(n = 3\\) it is clear that the \\(3\\) points have to be the vertices of an equilateral triangle since the sides are equal (to \\(d\\)). Assume we know the statement to be true for \\(n\\), and prove it for \\(n + 1\\). Assume that we have \\(n + 1\\) points with \\(n + 1\\) diameters \\(d\\). Consider \\(n\\) of the points, and let \\(Q\\) be the remaining point. It follows from the induction hypothesis that \\(3\\) of the \\(n\\) are the vertices of an equilateral arc triangle, and the other \\(n - 3\\) are on the boundary.\n\nNow let us examine where \\(Q\\) may be. It can not be outside of the equilateral arc triangle because then the distance to at least one of the vertices would be \\(d' > d\\), and the diameter of the set would be \\(d'\\) rather than \\(d\\). It can not be inside the triangle because then all the distances from \\(Q\\) to the \\(n\\) points considered are \\(< d\\), and we would not have \\(n + 1\\) diameters, as assumed. So \\(Q\\) must be on the boundary of the equilateral arc triangle, which proves the induction step, and the lemma.\n\nNow the solution to the problem is easy to finish. Assume we have \\(n\\) points given, and the configuration has \\(m\\) diameters. I claim that we can not have \\(m > n\\). We prove this by contradiction. If we did, let us consider a set \\(S\\) consisting of only \\(n\\) of the diameters. Because of the lemma, this implies that the \\(n\\) points form a configuration as described in the lemma, i.e. \\(3\\) of them are the vertices of an equilateral arc triangle, and the other \\(n - 3\\) are on the boundary. Now consider a diameter \\(D\\) which is not in \\(S\\). It is the distance of a segment between two of the points \\(P_1, P_2\\) in the configuration. If one of \\(P_1, P_2\\) is a vertex, then this diameter \\(D\\) is in \\(S\\), which is a contradiction. If none of \\(P_1, P_2\\) are vertices, then the length of \\(P_1P_2 < d\\), so \\(D = P_1P_2\\) is not a diameter, which is again a contradiction.\n\nIt follows that \\(m \\le n\\) which proves the problem.\n\n[Solution by pf02, October 2024]",
"See Proposition 2.3 on page 19 in \"Lecture Notes: Combinatorics in the Plane\" by Torsten Ueckerdt, from March 12, 2015 at https://www.math.kit.edu/iag6/lehre/combplane2013s/media/lecture_notes.pdf ."
] |
IMO-1966-1
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https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_1
|
In a mathematical contest, three problems, \(A\), \(B\), and \(C\) were posed. Among the participants there were \(25\) students who solved at least one problem each. Of all the contestants who did not solve problem \(A\), the number who solved \(B\) was twice the number who solved \(C\). The number of students who solved only problem \(A\) was one more than the number of students who solved \(A\) and at least one other problem. Of all students who solved just one problem, half did not solve problem \(A\). How many students solved only problem \(B\)?
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[
"Let us draw a Venn Diagram.\n\n\\[\nIMO1966.png\n\\]\n\nLet \\(a\\) be the number of students solving both B and C. Then for some positive integer \\(x\\), \\(2x - a\\) students solved B only, and \\(x - a\\) students solved C only. Let \\(2y - 1\\) be the number of students solving A; then \\(y\\) is the number of students solving A only. We have by given\n\n\\[\n2y - 1 + 3x - a = 25\n\\]\n\nand\n\n\\[\ny = 3x - 2a.\n\\]\n\nSubstituting for y into the first equation gives\n\n\\[\n9x - 5a = 26.\n\\]\n\nThus, because \\(x\\) and \\(a\\) are positive integers with \\(x-a \\ge 0\\), we have \\(x = 4\\) and \\(a = 2\\). (Note that \\(x = 9\\) and \\(a = 11\\) does not work.) Hence, the number of students solving B only is \\(2x - a = 8 - 2 = \\boxed{6}.\\)"
] |
IMO-1966-2
|
https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_2
|
Let \(a\), \(b\), and \(c\) be the lengths of the sides of a triangle, and \(\alpha,\beta,\gamma\) respectively, the angles opposite these sides. Prove that if
\[
a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}),
\]
the triangle is isosceles.
|
[
"We'll prove that the triangle is isosceles with \\(a=b\\). We'll prove that \\(a=b\\). Assume by way of contradiction WLOG that \\(a>b\\). First notice that as \\(\\gamma = \\pi -\\alpha-\\beta\\) then and the identity \\(\\tan\\left(\\frac \\pi 2 - x \\right)=\\cot x\\) our equation becomes:\n\n\\[\na+b=\\cot \\frac{\\alpha +\\beta}{2}\\left(a\\tan \\alpha + b\\tan \\beta \\right)\n\\]\n\n\\[\n\\iff a\\tan\\frac{\\alpha +\\beta}{2}+b\\tan \\frac{\\alpha +\\beta}{2}=a\\tan \\alpha + b\\tan \\beta\n\\]\n\n\\[\n\\iff a\\left(\\tan \\alpha -\\tan \\frac{\\alpha +\\beta}{2}\\right)+b\\left(\\tan \\beta -\\tan \\frac{\\alpha +\\beta}{2} \\right)=0\n\\]\n\nUsing the identity \\(\\tan (A-B)=\\frac {\\tan A-\\tan B}{1+\\tan A\\tan B}\\) \\(\\iff \\tan A-\\tan B=\\tan(A-B)(1+\\tan A\\tan B)\\) and inserting this into the above equation we get:\n\n\\[\n\\iff a\\tan \\frac{\\alpha -\\beta}{2}\\left(1+\\tan \\alpha \\tan \\frac{\\alpha +\\beta}{2}\\right)+b\\tan \\frac{\\beta -\\alpha}{2}\\left(1+\\tan \\beta \\tan \\frac{\\alpha +\\beta}{2} \\right)=0\n\\]\n\n\\[\n\\underbrace{\\iff}_{\\tan -A=-\\tan A}a\\tan \\frac{\\alpha -\\beta}{2}\\left(1+\\tan \\alpha \\tan \\frac{\\alpha +\\beta}{2}\\right)-b\\tan \\frac{\\alpha -\\beta}{2}\\left(1+\\tan \\beta \\tan \\frac{\\alpha +\\beta}{2} \\right)=0\n\\]\n\n\\[\n\\iff \\tan \\frac{\\alpha -\\beta}{2}\\left(a-b+\\tan \\frac{\\alpha +\\beta}{2}(a\\tan\\alpha -b\\tan \\beta) \\right)=0\n\\]\n\nNow, since \\(a>b\\) and the definitions of \\(a,b,\\alpha,\\beta\\) being part of the definition of a triangle, \\(\\alpha >\\beta\\). Now, \\(\\pi >\\alpha -\\beta >0\\) (as \\(\\alpha+\\beta +\\gamma = \\pi\\) and the angles are positive), \\(\\tan \\frac{\\alpha -\\beta}{2}\\neq 0\\), and furthermore, \\(\\tan \\frac{\\alpha+\\beta}{2}>0\\). By all the above,\n\n\\[\n\\left(a-b+\\tan \\frac{\\alpha +\\beta}{2}(a\\tan\\alpha -b\\tan \\beta) \\right)>0\n\\]\n\nWhich contradicts our assumption, thus \\(a\\leq b\\). By the symmetry of the condition, using the same arguments, \\(a\\geq b\\). Hence \\(a=b\\).",
"First, we'll prove that both \\(\\alpha\\) and \\(\\beta\\) are acute. At least one of them has to be acute because these are angles of a triangle. We can assume that \\(\\alpha\\) is acute. We want to show that \\(\\beta\\) is acute as well. For a proof by contradiction, assume \\(\\beta \\ge \\frac{\\pi}{2}\\).\n\nFrom the hypothesis, it follows that \\((a + b) \\tan \\frac{\\alpha + \\beta}{2} = a \\tan \\alpha + b \\tan \\beta\\).\n\nFrom \\(\\alpha < \\frac{\\pi}{2} \\le \\beta\\) it follows that \\(a < b\\). So,\n\n\\[\nb \\tan \\beta = (a + b) \\tan \\frac{\\alpha + \\beta}{2} - a \\tan \\alpha > 2a \\tan \\frac{\\alpha + \\beta}{2} - a \\tan \\alpha \\ge a (2 \\tan \\left( \\frac{\\alpha}{2} + \\frac{\\pi}{4} \\right) - \\tan \\alpha) =\n\\]\n\n\\[\n2a \\left( \\frac{\\tan \\frac{\\alpha}{2} + 1}{1 - \\tan \\frac{\\alpha}{2}} - \\frac{\\tan \\frac{\\alpha}{2}}{1 - \\tan^2 \\frac{\\alpha}{2}} \\right) = 2a \\cdot \\frac{\\tan^2 \\frac{\\alpha}{2} + \\tan \\frac{\\alpha}{2} + 1} {1 - \\tan^2 \\frac{\\alpha}{2}} > 0\n\\]\n\nbecause the numerator is \\(> 0\\) (because \\(Y^2 + Y + 1 > 0\\) for any real \\(Y\\)), and the denominator is also \\(> 0\\) (because \\(\\alpha < \\frac{\\pi}{2}\\), so \\(\\tan \\frac{\\alpha}{2} < \\tan \\frac{\\pi}{4} = 1\\)).\n\nIt follows that \\(\\tan \\beta > 0\\), so it can not be that \\(\\beta \\ge \\frac{\\pi}{2}\\).\n\nNow, we will prove that \\((a + b) \\tan \\frac{\\alpha + \\beta}{2} = a \\tan \\alpha + b \\tan \\beta\\) implies \\(\\alpha = \\beta\\).\n\nReplace \\(a = \\sin \\alpha \\cdot 2R\\) and \\(b = \\sin \\beta \\cdot 2R\\) (in fact, we don't care that \\(R\\) is the radius of the circumscribed circle), and simplify by \\(2R\\). We get\n\n\\((\\sin \\alpha + \\sin \\beta) \\tan \\frac{\\alpha + \\beta}{2} = \\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta\\).\n\nThis becomes\n\n\\[\n\\left( \\sin \\frac{\\alpha + \\beta}{2} \\tan \\frac{\\alpha + \\beta}{2} \\right) \\cdot \\cos \\frac{\\alpha - \\beta}{2} = \\frac{1}{2}(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta)\n\\]\n\nWe will show that the function \\(f(x) = \\tan x \\sin x\\) is convex on the interval \\(\\left( 0, \\frac{\\pi}{2} \\right)\\). Indeed, the first derivative is \\(f'(x) = \\frac{\\sin x}{\\cos^2 x} + \\sin x\\), and the second derivative is \\(f''(x) = \\frac{\\cos^4 x - \\cos ^2 x + 2}{\\cos^3 x}\\).\n\nWe have \\(f''(x) > 0\\) on \\(\\left( 0, \\frac{\\pi}{2} \\right)\\) since the numerator is \\(> 0\\) (because \\(Y^2 - Y + 1 > 0\\) for any real \\(Y\\)), and the denominator is \\(> 0\\) on the interval \\(\\left( 0, \\frac{\\pi}{2} \\right)\\). It follows that \\(f(x) = \\tan x \\sin x\\) is convex on the interval \\(\\left( 0, \\frac{\\pi}{2} \\right)\\).\n\nUsing the convexity we have \\(f \\left( \\frac{x + y}{2} \\right) \\le \\frac{1}{2} (f(x) + f(y))\\). In our case, we have\n\n\\(\\frac{1}{2}(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta) = \\left( \\sin \\frac{\\alpha + \\beta}{2} \\tan \\frac{\\alpha + \\beta}{2} \\right) \\cdot \\cos \\frac{\\alpha - \\beta}{2} \\le \\frac{1}{2}(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta) \\cdot \\cos \\frac{\\alpha - \\beta}{2}\\).\n\nWe can simplify by \\(\\sin \\alpha \\tan \\alpha + \\sin \\beta \\tan \\beta\\) because it is positive (because both \\(\\alpha, \\beta\\) are acute!), and we get\n\n\\(1 \\le \\cos \\frac{\\alpha - \\beta}{2}\\). This is possible only when \\(\\cos \\frac{\\alpha - \\beta}{2} = 1\\), i.e. \\(\\alpha = \\beta\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1966-3
|
https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_3
|
Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.
|
[
"We will need the following lemma to solve this problem:\n\n\\(\\emph{Lemma:}\\) Let \\(MNOP\\) be a regular tetrahedron, and \\(T\\) a point inside it. Let \\(x_1, x_2, x_3, x_4\\) be the distances from \\(T\\) to the faces \\(MNO, MNP, MOP\\), and \\(NOP\\). Then, \\(x_1 + x_2 + x_3 + x_4\\) is constant, independent of \\(T\\).\n\n\\[\n\\emph{Proof:}\n\\]\n\nWe will compute the volume of \\(MNOP\\) in terms of the areas of the faces and the distances from the point \\(T\\) to the faces:\n\n\\[\n\\textrm{Volume}(MNOP) = [MNO] \\cdot x_1 \\cdot \\frac{1}{3} + [MNP] \\cdot x_2 \\cdot \\frac{1}{3} + [MOP] \\cdot x_3 \\cdot \\frac{1}{3} + [NOP] \\cdot x_4 \\cdot \\frac{1}{3}\n\\]\n\n\\[\n= [MNO] \\cdot \\frac{(x_1 + x_2 + x_3 + x_4)}{3}\n\\]\n\nbecause the areas of the four triangles are equal. (\\([ABC]\\) stands for the area of \\(\\triangle ABC\\).) Then\n\n\\[\n\\frac{3\\cdot\\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.\n\\]\n\nThis value is constant, so the proof of the lemma is complete.\n\n\\[\n\\emph{Proof of problem statement:}\n\\]\n\nLet our tetrahedron be \\(ABCD\\), and the center of its circumscribed sphere be \\(O\\). Construct a new regular tetrahedron, \\(WXYZ\\), such that the centers of the faces of this tetrahedron are at \\(A\\), \\(B\\), \\(C\\), and \\(D\\).\n\nFor any point \\(P\\) in \\(ABCD\\),\n\n\\[\nOA + OB + OC + OD = \\sum \\textrm{Distances from }O\\textrm{ to faces of }WXYZ\n\\]\n\n\\[\n= \\sum \\textrm{Distances from }P\\textrm{ to faces of }WXYZ \\leq PA + PB + PC + PD,\n\\]\n\nwith equality only occurring when \\(AP\\), \\(BP\\), \\(CP\\), and \\(DP\\) are perpendicular to the faces of \\(WXYZ\\), meaning that \\(P = O\\). This completes the proof. \\(\\square\\)\n\n~mathboy100\n\n## Remarks (added by pf02, September 2024)\n\n1. The text of the Lemma needed a little improvement, which I did.\n\n2. The Solution above is not complete. It considered only points \\(P\\) inside the tetrahedron, but the problem specifically said \"any other point in space\".\n\n3. I will give another solution below, in which I will also fill in the gap of the solution above, mentioned in the preceding paragraph.",
"We will first prove the problem in the 2-dimensional case. We do this to convey the idea of the proof, and because we will use this in one spot in proving the 3-dimensional case. So let us prove that:\n\nThe sum of the distances of the vertices of an equilateral triangle \\(\\triangle ABC\\) from the center \\(O\\) of its circumscribed circle is less than the sum of the distances of these vertices from any other point \\(P\\) in the plane.\n\nWe will do the proof in three steps:\n\n\\(\\mathbf{1.}\\) We will show that if \\(P\\) is in one of the exterior regions, then there is a point \\(P_1\\) on the boundary of the triangle (a vertex, or on a side), such that \\(PA + PB + PC > P_1A + P_1B + P_1C\\).\n\n\\(\\mathbf{2.}\\) Then we will show that if \\(P\\) is on the boundary, then \\(PA + PB + PC > OA + OB + OC\\).\n\n\\(\\mathbf{3.}\\) For the final step, we will show that if \\(P\\) is a point of minimum for \\(PA + PB + PC\\) inside the triangle, then the extensions of \\(PA, PB, PC\\) are perpendicular to the opposite sides \\(BC, AC, AB\\). This implies that \\(P = O\\).\n\n\\(\\mathbf{Proof\\ of\\ 1:}\\) If the point \\(P\\) is outside the triangle, it can be in one of six regions as seen in the pictures below.\n\n\\[\nProb 1966 3 fig1.png\n\\]\n\nIf \\(P\\) is in a region delimited by extensions of two sides of the triangle, as in the picture on the left, we notice that by taking \\(P_1 = A\\), \\(PA + PB + PC > P_1A + P_1B + P_1C\\) (because \\(P_1A = 0\\) and \\(P_1B < PB\\) as sides in an obtuse triangles, and similarly \\(P_1C < PC\\)).\n\nIf \\(P\\) is in a region delimited by a segment which is a side of the triangle and by the extensions of two sides, as in the picture on the right, take \\(P_1 =\\) the foot of the perpendicular from \\(P\\) to \\(AB\\). Then \\(PA + PB + PC > P_1A + P_1B + P_1C\\) (because the triangle \\(\\triangle PP_1C\\) is obtuse, and because the triangles \\(\\triangle PP_1A, \\triangle PP_1B\\) are right triangles).\n\n\\(\\mathbf{Proof\\ of\\ 2:}\\) Now assume that \\(P_1 = A\\). A direct, simple computation shows that \\(P_1A + P_1B + P_1C > OA + OB + OC\\) (indeed, if we take the side of the triangle to be \\(1\\), then \\(P_1A + P_1B + P_1C = 2\\), and \\(OA + OB + OC = 3 \\cdot \\frac{\\sqrt{3}}{3} = \\sqrt{3}\\)).\n\nNow assume that \\(P_1\\) is on \\(AB\\). If \\(P_1\\) is not the midpoint of \\(AB\\), let \\(P_2\\) be the midpoint. Then \\(P_1A + P_1B + P_1C > P_2A + P_2B + P_2C\\) (because \\(P_1A + P_1B = P_2A + P_2B = AB\\) and \\(P_1C > P_2C\\)). A direct, simple computation shows that \\(P_2A + P_2B + P_2C > OA + OB + OC\\) (indeed, if we take the side of the triangle to be \\(1\\), \\(P_2A + P_2B + P_2C = 1 + \\frac{\\sqrt{3}}{2}\\) and \\(OA + OB + OC = \\sqrt{3}\\)).\n\n\\(\\mathbf{Proof\\ of\\ 3:}\\) Assume that \\(P\\) is inside the triangle \\(\\triangle ABC\\). In this case, we make a proof by contradiction. We will show that if \\(P\\) is a point where \\(PA + PB + PC\\) is minimum, then the extensions of \\(PA, PB, PC\\) are perpendicular to the opposite sides \\(BC, AC, AB\\). (This statement implies that \\(P = O\\).) If this were not true, at least one of \\(PA \\perp BC, PB \\perp AC, PC \\perp AB\\) would be false. We can assume that \\(PC\\) is not perpendicular to \\(AB\\). Then draw the ellipse with focal points \\(A, B\\) which goes through \\(P\\).\n\n\\[\nProb 1966 3 fig2.png\n\\]\n\nNow consider the point \\(P_1\\) on the ellipse such that \\(CP_1 \\perp AB\\). Because of the properties of the ellipse, \\(CP_1 < CP\\), and because of the definition of the ellipse \\(PA + PB = P_1A + P_1B\\). We conclude that \\(PA + PB + PC > P_1A + P_1B + P_1C\\), which contradicts the assumption that \\(P\\) was such that \\(PA + PB + PC\\) was minimum.\n\nThis proves the 2-dimensional case.\n\nNOTE: a very picky reader might object that the proof used that a minimum of \\(PA + PB + PC\\) exists, and is achieved at a point \\(P\\) inside the triangle. This can be justified simply by noting that \\(PA + PB + PC > 0\\) and quoting the theorem from calculus (or is it topology?) which says that a continuous function on a closed, bounded set has a minimum, and there is a point where the minimum is achieved. Because of the arguments in the proof, this point can not be on the boundary of the triangle, so it is inside.\n\nNow we will give the proof in the 3-dimensional case. We will do the proof in three steps. It is extremely similar to the proof in the 2-dimensional case, we just need to go from 2D to 3D, so I will skip some details.\n\n\\(\\mathbf{1.}\\) We will show that if \\(P\\) is in one of the exterior regions, then there is a point \\(P_!\\) on the boundary of the tetrahedron (a vertex, or on a edge, or on a side, such that \\(PA + PB + PC + PD > P_1A + P_1B + P_1C + P_1D\\).\n\n\\(\\mathbf{2.}\\) Then we will show that if \\(P\\) is on the boundary, then \\(PA + PB + PC + PD > OA + OB + OC + OD\\).\n\n\\(\\mathbf{3.}\\) For the final step, consider the plane going through the edge \\(CD\\) perpendicular to the edge \\(AB\\), the plane going through \\(AB\\) perpendicular to \\(CD\\), the plane going through \\(CA\\) perpendicular to \\(BD\\), etc. There are six such planes, and they all contain \\(O\\), the center of the circumscribed sphere. We will show that if \\(P\\) is a point of minimum for \\(PA + PB + PC + PD\\) inside the tetrahedron, then \\(P\\) is in each of the six planes described above. This implies that \\(P = O\\).\n\n\\(\\mathbf{Proof\\ of\\ 1:}\\) Let \\(P\\) be in one of the exterior regions. Assume \\(P\\) is in a prism shaped region delimited by extensions of three sides meeting in a vertex (there are 4 of them). Assume it is at vertex \\(A\\), the sides being the extensions of planes \\(ABC, ABD, ACD\\). Then take \\(P_1 = A\\). We have \\(PA + PB + PC + PD > P_1A + P_1B + P_1C + P_1D\\) because of obtuse triangles formed with \\(PP_1\\).\n\nNow assume \\(P\\) is in one of the wedge shaped regions, formed by an edge and the extensions of two sides going through them. (there are six such regions.) Assume this is the line \\(AB\\) and the extensions of \\(ABC, ABD\\). Then take \\(P_1\\) to be the foot of the perpendicular from \\(P\\) to \\(AB\\). Again, we have the desired inequality because \\(PP_1\\) formed some right and obtuse triangles.\n\nNow assume \\(P\\) is in the truncated prism region delimited by a side and the extensions of the faces going through the edges of this side. (There are four such regions.) Assume this is the side \\(ABC\\), and extensions of the sides \\(DAB, DBC, DCA\\). Then take \\(P_1\\) to be the foot of the perpendicular from \\(P\\) to the plane \\(ABC\\). Again, we have the desired inequality because of right and obtuse triangles formed by \\(PP_1\\).\n\n\\(\\mathbf{Proof\\ of\\ 2:}\\) Assume \\(P_1 = A\\). If we take the edge of the tetrahedron to be \\(1\\), a direct computation gives us that \\(P_1A + P_1B + P_1C + P_1D = 3\\), and \\(OA + OB + OC + OD = 4 \\cdot \\frac{\\sqrt{6}}{4} = \\sqrt{6}\\).\n\nAssume \\(P_1\\) is on \\(AB\\). If \\(P_1\\) is not the midpoint of \\(AB\\), take \\(P_2\\) to be the midpoint of \\(AB\\). Then \\(P_1A + P_1B + P_1C + P_1D > P_2A + P_2B + P_2C + P_2D\\) because of right triangles formed by \\(P_2C, P_2D\\). And, if we take the edge of the tetrahedron to be \\(1\\), a direct computation yields that \\(P_2A + P_2B + P_2C + P_2D = 1 + 2 \\cdot \\frac{\\sqrt{3}}{2} = 1 + \\sqrt{3}\\), which is bigger than \\(OA + OB + OC + OD = \\sqrt{6}\\).\n\nAssume \\(P_1\\) is on \\(ABC\\). If \\(P_1\\) is not the circumcenter of \\(\\triangle ABC\\) then take \\(P_2\\) to be the circumcenter. We have \\(P_1D > P_2D\\) because \\(P_2D \\perp ABC\\). We also have \\(P_1A + P_1B + P_1C > P_2A + P_2B + P_2C\\) because we proved the 2-dimensional analogue of the problem. And, if we take the edge of the tetrahedron to be \\(1\\), we have \\(P_2A + P_2B + P_2C + P_2D = \\sqrt{3} + \\frac{\\sqrt{6}}{3}\\), which is bigger than \\(OA + OB + OC + OD = \\sqrt{6}\\).\n\nNOTE: In the above paragraph, we used that the similar result is true in the 2-dimensional case, with an equilateral triangle instead of a regular tetrahedron.\n\nNOTE: This part of the proof concludes filling in the gap in the first \"Solution\", written above. (A reader may complain that the proof in Solution 2 is very long (compared to the first \"Solution\"), but the first \"Solution\" should have done this too, one way or another.)\n\n\\(\\mathbf{Proof\\ of\\ 3:}\\) Now consider the six planes going through one edge, perpendicular to the opposite edge. They intersect at the circumcenter of the tetrahedron. Assume \\(P\\) is a point in the interior of the tetrahedron where \\(PA + PB + PC + PD\\) achieves its minimum value. Then \\(P\\) is in each of the six plane.\n\nProve this statement by contradiction. Assume that there is a plane among the six, so that \\(P\\) is not on it. Assume the plane is the one going through \\(CD\\), perpendicular to \\(AB\\). To make it more explicit, this is the plane going through \\(C, D, E\\), where \\(E\\) is the midpoint of \\(AB\\).\n\nConsider the ellipsoid with focal points \\(A, B\\) going through \\(P\\). This can be obtained as the set of points \\(Q\\) in space so that \\(QA + QB = PA + PB\\). It can also be obtained as the surface obtained when we form the ellipse with focal points \\(A, B\\) in the plane \\(ABC\\) (as the set of points \\(Q\\) so that \\(QA + QB = PA + PB\\)), and we rotate this ellipse from the plane \\(ABC\\) around its axis \\(AB\\). Let \\(P_1\\) be the foot of the perpendicular from \\(P\\) to the plane \\(CDE\\). We have \\(PC > P_1C, PD > P_1D\\) because \\(PP_1 \\perp CDE\\). We also have \\(PA + PB > P_1A + P_1B\\) because \\(P_1\\) is in the interior of the ellipsoid. (Indeed, the intersection of the plane \\(CDE\\) and the ellipsoid is the circle generated by rotating the ends of the small axis of the ellipse in the plane \\(ABC\\). Since the point \\(P\\) is not on the plane CDE, it must be on a smaller circle, so its projection to the plane \\(CDE\\) will be inside.)\n\nThis concludes the proof of the problem.\n\n[Solution by pf02, September 2024]"
] |
IMO-1966-4
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https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_4
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Prove that for every natural number \(n\), and for every real number \(x \neq \frac{k\pi}{2^t}\) (\(t=0,1, \dots, n\); \(k\) any integer)
\[
\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}
\]
|
[
"First, we prove \\(\\cot \\theta - \\cot 2\\theta = \\frac {1}{\\sin 2\\theta}\\).\n\nLHS\\(\\ =\\ \\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}\\)\n\n\\[\n= \\frac{2\\cos^2 \\theta}{2\\cos \\theta \\sin \\theta}-\\frac{2\\cos^2 \\theta -1}{\\sin 2\\theta}\n\\]\n\n\\[\n=\\frac{2\\cos^2 \\theta}{\\sin 2\\theta}-\\frac{2\\cos^2 \\theta -1}{\\sin 2\\theta}\n\\]\n\n\\[\n=\\frac {1}{\\sin 2\\theta}\n\\]\n\nUsing the above formula, we can rewrite the original series as\n\n\\(\\cot x - \\cot 2x + \\cot 2x - \\cot 4x + \\cot 4x + \\dots + \\cot 2^{n-1} x - \\cot 2^n x\\).\n\nWhich gives us the desired answer of \\(\\cot x - \\cot 2^n x\\)."
] |
IMO-1966-5
|
https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_5
|
Solve the system of equations
\[
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\ |a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\ |a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3 - a_4| x_4 = 1 \\ |a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1
\]
where \(a_1, a_2, a_3, a_4\) are four different real numbers.
|
[
"Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:\n\n\\[\n- x1 + x2 + x3 + x4 = 0.\n\\]\n\nSubtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:\n\n\\[\n- x1 - x2 - x3 + x4 = 0.\n\\]\n\nHence x1 = x4. Subtracting the equation for i=3 from that for i=2 and dividing by (a2 - a3) we get:\n\n\\[\n- x1 - x2 + x3 + x4 = 0.\n\\]\n\nHence \\(x2 = x3 = 0\\), and \\(x1 = x4 = 1/(a1 - a4)\\).\n\n## Remarks (added by pf02, September 2024)\n\nThe solution above is in the realm of flawed and incorrect. It is flawed because you can not claim to have solved a system of equations by having solved a particular case. It is correct in solving the particular case, but it is incorrect in stating that the result obtained is a solution to the system in general.\n\nBelow I will give a complete solution to the problem. The first few lines will be a repetition of the \"solution\" above, and I will repeat them for the sake of completeness and of a more tidy writing.",
"There are 24 possibilities when we count the ordering of \\(a_1, a_2, a_3, a_4\\), and each ordering gives a different system of equations. Let us consider one of them, like in the \"solution\" above.\n\nAssume \\(a_1 > a_2 > a_3 > a_4\\). In this case, the system is\n\n\\[\n\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (a_1 - a_2) x_2 + (a_1 - a_3) x_3 + (a_1 - a_4) x_4 = 1 \\\\ (a_1 - a_2) x_1 + \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (a_2 - a_3) x_3 + (a_2 - a_4) x_4 = 1 \\\\ (a_1 - a_3) x_1 + (a_2 - a_3) x_2 + \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (a_3 - a_4) x_4 = 1 \\\\ (a_1 - a_4) x_1 + (a_2 - a_4) x_2 + (a_3 - a_4) x_3 \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ = 1\n\\]\n\nSubtract the second equation from the first, and divide by \\((a_1 - a_2)\\). Also, subtract the fourth equation from the third, and divide by \\((a_3 - a_4)\\). We obtain\n\n\\[\n-x_1 + x_2 + x_3 + x_4 = 0 \\\\ -x_1 - x_2 - x_3 + x_4 = 0\n\\]\n\nIt follows that \\(x_1 - x_4 = 0\\) and \\(x_2 + x_3 = 0\\).\n\nSubtract the third equation from the second, and divide by \\((a_2 - a_3)\\). We obtain\n\n\\[\n-x_1 - x_2 + x_3 + x_4 = 0\n\\]\n\nSince \\(x_1 - x_4 = 0\\), it follows that \\(x_2 - x_3 = 0\\). Combining with \\(x_2 + x_3 = 0\\), we get \\(x_2 = x_3 = 0\\). Replacing these in the first equation of the system, we get \\(x_4 = \\frac{1}{a_1 - a_4}\\), so we also have \\(x_1 = \\frac{1}{a_1 - a_4}\\).\n\nNow we have two ways of proceeding. We could consider each of the other 23 cases, and solve it by a similar method. The task is made easy if we notice that each case is obtained from the first case by a permutation of indices, so it can be viewed as a change of notation. With some care, we can just write the solution in each case. For example, in the case \\(a_2 > a_1 > a_3 > a_4\\), we will obtain \\(x_1 = x_3 = 0\\) and \\(x_2 = x_4 = \\frac{1}{a_2 - a_4}\\).\n\nWe will proceed differently, but we will use the same idea. Let \\(m, n, p, q\\) be the indices such that \\(a_m > a_n > a_p > a_q\\). Written in a compact way, our system becomes\n\n\\(\\sum_{\\substack{i = 1 \\\\ i \\ne j}}^4 |a_j - a_i| x_i = 1, \\ \\ \\ \\ j = 1, 2, 3, 4\\).\n\nMake the following change of notation: \\(a_m \\rightarrow b_1 \\\\ a_n \\rightarrow b_2 \\\\ a_p \\rightarrow b_3 \\\\ a_q \\rightarrow b_4\\)\n\nand\n\n\\[\nx_m \\rightarrow y_1 \\\\ x_n \\rightarrow y_2 \\\\ x_p \\rightarrow y_3 \\\\ x_q \\rightarrow y_4\n\\]\n\nIn the new notation we have \\(b_1 > b_2 > b_3 > b_4\\) and the system becomes\n\n\\(\\sum_{\\substack{k = 1 \\\\ k \\ne l}}^4 |b_l - b_k| y_k = 1, \\ \\ \\ \\ l = 1, 2, 3, 4\\).\n\nThis is exactly the system we solved above, just with a new notation (\\(b, y\\) instead of \\(a, x\\)). So the solutions are \\(y_2 = y_3 = 0, y_1 = y_4 = \\frac{1}{b_1 - b_4}\\).\n\nReturning to our original notation, we have \\(x_n = x_p = 0, x_m, x_q = \\frac{1}{a_m - a_q}\\).\n\nIn conclusion, here is a compact way of giving the solution to the system: let \\(m\\) be the index of the largest of the \\(a_i\\)'s, and q = the index of the smallest of the \\(a_i\\)'s, and let \\(n, p\\) be the other two indices. Then \\(x_n = x_p = 0, x_m, x_q = \\frac{1}{a_m - a_q}\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1966-6
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https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_6
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In the interior of sides \(BC, CA, AB\) of triangle \(ABC\), any points \(K, L,M\), respectively, are selected. Prove that the area of at least one of the triangles \(AML, BKM, CLK\) is less than or equal to one quarter of the area of triangle \(ABC\).
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[
"Let the lengths of sides \\(BC\\), \\(CA\\), and \\(AB\\) be \\(a\\), \\(b\\), and \\(c\\), respectively. Let \\(BK=d\\), \\(CL=e\\), and \\(AM=f\\).\n\nNow assume for the sake of contradiction that the areas of \\(\\Delta AML\\), \\(\\Delta BKM\\), and \\(\\Delta CLK\\) are all at greater than one fourth of that of \\(\\Delta ABC\\). Therefore\n\n\\[\n\\frac{AM\\cdot AL\\sin{\\angle BAC}}{2}>\\frac{AB\\cdot AC\\sin{\\angle BAC}}{8}\n\\]\n\nIn other words, \\(AM\\cdot AL>\\frac{1}{4}AB\\cdot AC\\), or \\(f(b-e)>\\frac{bc}{4}\\). Similarly, \\(d(c-f)>\\frac{ac}{4}\\) and \\(e(a-d)>\\frac{ab}{4}\\). Multiplying these three inequalities together yields\n\n\\[\ndef(a-d)(b-e)(c-f)>\\frac{a^2b^2c^2}{64}\n\\]\n\nWe also have that \\(d(a-d)\\leq \\frac{a^2}{4}\\), \\(e(b-e)\\leq \\frac{b^2}{4}\\), and \\(f(c-f)\\leq \\frac{c^2}{4}\\) from the Arithmetic Mean-Geometric Mean Inequality. Multiplying these three inequalities together yields\n\n\\[\ndef(a-d)(b-e)(c-f)\\leq\\frac{a^2b^2c^2}{64}\n\\]\n\nThis is a contradiction, which shows that our assumption must have been false in the first place. This proves the desired result.",
"Let \\(AR : AB = x, BP : BC = y, CQ : CA = z\\). Then it is clear that the ratio of areas of \\(AQR, BPR, CPQ\\) to that of \\(ABC\\) equals \\(x(1-y), y(1-z), z(1-x)\\), respectively. Suppose all three quantities exceed \\(\\frac{1}{4}\\). Then their product also exceeds \\(\\frac{1}{64}\\). However, it is clear by AM-GM that \\(x(1-x) \\le \\frac{1}{4}\\), and so the product of all three quantities cannot exceed \\(\\frac{1}{64}\\) (by the associative property of multiplication), a contradiction. Hence, at least one area is less than or equal to \\(\\frac{1}{4} [ABC]\\).\n\n## Remarks (added by pf02, September 2024)\n\nSolution 2 is written in a very sloppy way. However, an interested reader can make sense of it. More importantly, the two solutions are identical. If it wasn't for the sloppy writing, Solution 2 could be obtained from the first Solution after applying a word by word translation which replaces line segments by ratios.\n\nBelow I will give another solution. It is formally different from the previous solutions, even if not at a deep level.",
"Let \\(\\triangle ABC\\) and \\(K, L, M\\) be as in the problem. Denote \\(x = \\frac{AM}{AB}, y = \\frac{BK}{BC}, z = \\frac{CL}{CA}\\) as in Solution 2. Note that \\(x, y, z, \\in (0, 1)\\) because \\(K, L, M\\) are in the interior of the respective sides.\n\n\\[\nProb 1966 6.png\n\\]\n\nUsing the fact that the area of a triangle is half of the product of two sides and \\(\\sin\\) of the angle between them (like in the first Solution), we have that \\(\\mathbf{area} AML = x(1 - z) \\mathbf{area} ABC, \\mathbf{area} BKM = y(1 - x) \\mathbf{area} ABC, \\mathbf{area} CLK = z(1 - y) \\mathbf{area} ABC\\).\n\nNow the problem has nothing to do with geometry anymore: we just have to show that given three numbers \\(x, y, z, \\in (0, 1)\\), at least one of \\(x(1 - z), y(1 - x), z(1 - y)\\) is \\(\\le \\frac{1}{4}\\).\n\nIf \\(y(1 - x) \\le \\frac{1}{4}\\), we are done. Otherwise, we have \\(y(1 - x) > \\frac{1}{4}\\). It follows that \\(y > \\frac{1}{4(1 - x)}\\) (recall that \\(0 < x, y, z < 1\\)). In particular, it follows that \\(\\frac{1}{4(1 - x)} < 1\\), which implies \\(3 - 4x > 0\\).\n\nIf \\(z(1 - y) \\le \\frac{1}{4}\\), we are done. Otherwise, we have \\(z(1 - y) > \\frac{1}{4}\\). Using the inequality on \\(y\\) from the previous paragraph, we have \\(z \\left( 1 - \\frac{1}{4(1 - x)} \\right) > \\frac{1}{4}\\), or after a few computations, \\(z \\cdot \\frac{3 - 4x}{1 - x} > 1\\). Using the observation about \\(3 - 4x\\) from the preceding paragraph, we get \\(z > \\frac{1 - x}{3 - 4x}\\).\n\nNow consider \\(x(1 - z)\\). Using the inequality on \\(z\\) from the previous paragraph, we have that \\(x(1 - z) < x \\left( 1 - \\frac{1 - x}{3 - 4x} \\right)\\). To finish the solution to the problem, it is enough to show that \\(x \\cdot \\left( 1 - \\frac{1 - x}{3 - 4x} \\right) \\le \\frac{1}{4}\\).\n\nAfter some easy computations (and using again that \\(3 - 4x > 0\\)), this becomes \\(3(4x^2 - 4x + 1) \\ge 0\\), which is true because. \\(4x^2 - 4x + 1 = (2x - 1)^2\\).\n\n[Solution by pf02, September 2024]"
] |
IMO-1967-1
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https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_1
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Let \(ABCD\) be a parallelogram with side lengths \(AB = a\), \(AD = 1\), and with \(\angle BAD = \alpha\). If \(\Delta ABD\) is acute, prove that the four circles of radius \(1\) with centers \(A\), \(B\), \(C\), \(D\) cover the parallelogram if and only if
\(a\leq \cos \alpha+\sqrt{3}\sin \alpha\) \(\ \ \ \ \ \ \ \ \ \ (1)\)
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[
"To start our proof we draw a parallelogram with the requested sides. We notice that by drawing the circles with centers A, B, C, D that the length of \\(a\\) must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.\n\nTo prove our conjecture we draw a parallelogram with \\(a=2\\) and draw a segment \\(DB\\) so that \\(\\angle ADB=90^{\\circ}\\)\n\nThis is the parallelogram which we claim has the maximum length on \\(a\\) and the highest value on any one angle.\n\nWe now have two triangles inside a parallelogram with lengths \\(1, 2\\) and \\(x\\), \\(x\\) being segment \\(DB\\). Using the Pythagorean theorem we conclude:\n\n\\[\n1^2+x^2=2^2\\\\x=\\sqrt{3}\n\\]\n\nUsing trigonometric functions we can compute:\n\n\\[\ncos\\alpha=\\frac{1}{2}\\\\sin\\alpha=\\frac{\\sqrt{3}}{2}\n\\]\n\nNotice that by applying the \\(arcsine\\) and \\(arccos\\) functions, we can conclude that our angle \\(\\alpha=60^{\\circ}\\)\n\nTo conclude our proof we make sure that our values match the required values for maximum length of \\(a\\)\n\n\\[\na\\leq\\cos\\alpha+\\sqrt{3}\\sin\\alpha\\\\\\\\a\\leq\\frac{1}{2}+\\sqrt{3}\\cdot \\frac{\\sqrt{3}}{2}\\\\\\\\a\\leq 2\n\\]\n\nNotice that as \\(\\angle\\alpha\\) decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as \\(\\angle\\alpha\\) increases, the value of (1) decreases below 2, confirming that (1) is only implied when \\(\\Delta ABD\\) is acute.\n\n--Bjarnidk 02:16, 17 May 2013 (EDT)\n\n## Remarks (added by pf02, September 2024)\n\n\\(\\mathbf{Remark\\ 1}\\). I am sorry to be so harshly critical, but the solution above is deeply flawed. Not only it has errors, but the logic is flawed.\n\nIt shows that when \\(a = 2, \\alpha = \\frac{\\pi}{3}\\) the parallelogram is covered by the circles of radius \\(1\\) centered at \\(A, B, C, D\\), and the inequality in the problem is true. (Even this is incomplete, while giving too many, unnecessary details.) (Note that this is not a case which satisfies the conditions of the problem since \\(\\triangle ABD\\) is right, not acute.)\n\nIn the last two lines it gives some reasoning about other values of \\(\\alpha\\) which is incomprehensible to this reader.\n\nIn one short sentence: this is not a solution.\n\n\\(\\mathbf{Remark\\ 2}\\). The problem itself is mildly flawed. To see this, denote \\(S1, S2\\) the following two statements:\n\nS1: The parallelogram \\(ABCD\\) is covered by the four circles of radius \\(1\\) centered at \\(A, B, C, D\\).\n\nS2: We have \\(a \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\\).\n\nThe problem says that if \\(\\triangle ABD\\) is acute, \\(S1\\) and \\(S2\\) are equivalent, i.e. they imply each other.\n\nNotice that \\(S2\\) can be rewritten as \\(a \\le 2 \\cos \\left( \\alpha - \\frac{\\sqrt{\\pi}}{3} \\right)\\).\n\nNow notice that if \\(a \\le 1\\) then S1 is obviously true. See the picture below:\n\n\\[\nProb 1967 1 fig1.png\n\\]\n\nAlso, notice that if \\(a \\le 1\\) and \\(\\alpha \\in \\left( 0, \\frac{\\sqrt{\\pi}}{2} \\right)\\) then \\(S2\\) is true as well. Indeed \\(\\left( \\alpha - \\frac{\\sqrt{\\pi}}{3} \\right) \\in \\left( -\\frac{\\sqrt{\\pi}}{3}, \\frac{\\sqrt{\\pi}}{6} \\right)\\), so \\(\\cos\\) is \\(> \\frac{1}{2}\\) on this interval, so the right hand side of \\(S2\\) is \\(> 1 \\ge a\\).\n\nWe see that if \\(a \\le 1\\) and \\(\\triangle ABD\\) is acute, both \\(S1\\) and \\(S2\\) are true. We can not say that one implies the other in the usual meaning of the word \"imply\": the two statements just happen to be both true.\n\nIf we take \\(a > 1\\) then the problem is a genuine problem, and there is something to prove.\n\n\\(\\mathbf{Remark\\ 3}\\). In the proofs I give below, we will see where we need that \\(\\triangle ABD\\) is acute. We will see that \\(\\alpha < \\frac{\\pi}{2}\\) is needed for the technicalities of the proof. The fact that \\(\\angle ADB\\) is acute will be needed at one crucial point in the proof.\n\nIn fact, it is possible to modify \\(S2\\) to a statement \\(S3\\) similar to \\(S2\\) so that \\(S1\\) and \\(S3\\) are equivalent without any assumption on \\(\\alpha\\). I will not go into this, I will just give a hint: Denote \\(\\beta = \\angle ABC\\). If \\(\\alpha\\) is acute, \\(\\beta\\) is obtuse, and we can easily reformulate \\(S2\\) in terms of \\(\\beta\\).\n\n\\(\\mathbf{Remark\\ 4}\\). Below, I will give two solutions. Solution 2 is one I carried out myself and relies on a straightforward computation. Solution 3 (it happens to be similar to Solution 2) is inspired by an idea by feliz shown on the web page https://artofproblemsolving.com/community/c6h21154p137323 The author calls their text a solution, but it is quite confused, so I would not call it a good solution. The idea though is good and nice, and it yields a nice solution.",
"We can assume \\(a > 1\\). Indeed, refer to Remark 2 above to see that if \\(a \\le 1\\) there is nothing to prove.\n\nNote that instead of the statement \\(S1\\) we can consider the following statement \\(S1'\\):\n\n\\(S1'\\): the circles of radius \\(1\\) centered at \\(A, B, D\\) cover \\(\\triangle ABD\\).\n\nThis is equivalent to \\(S1\\) because of the symmetry between \\(\\triangle ABD\\) and \\(\\triangle BCD\\).\n\nLet \\(F\\) be the intersection above \\(AB\\) of the circles of radius \\(1\\) centered at \\(A, B\\). The three circles cover \\(\\triangle ABD\\) if an only if \\(F\\) is inside the circle of radius 1 centered at \\(D\\).\n\n\\[\nProb 1967 1 fig2.png\n\\]\n\nThis needs an explanation: Let \\(H\\) be the midpoint of \\(BD\\), and consider \\(\\triangle FHD\\). All the vertices of this triangle are in the circle centered at \\(D\\), so the whole triangle is inside this circle. It is obvious that \\(\\triangle FHB\\) is inside the circle centered at \\(B\\), and that \\(\\triangle FAD, \\triangle FAB\\) are inside the circles centered at \\(A, B\\).\n\nWe will now show that \\(F\\) is inside the circle of radius 1 centered at \\(D\\) if an only if \\(DF \\le 1\\).\n\nThe plan is to calculate \\(DF\\) in terms of \\(a, \\alpha\\) and impose this condition. Let \\(FG \\perp AB\\), \\(DE \\perp AB\\) and \\(FF' \\parallel GE\\). From the right triangle \\(\\triangle AFG\\) we have \\(FG = \\sqrt{1 - \\left( \\frac{a}{2} \\right)^2} = \\frac{\\sqrt{4 - a^2}}{2}\\). From the right triangle \\(\\triangle DFF'\\) we have\n\n\\[\nDF = \\sqrt{(DF')^2 + (FF')^2} = \\sqrt{(DE - FG)^2 + (AG - AE)^2} = \\sqrt{\\left( \\sin \\alpha - \\frac{\\sqrt{4 - a^2}}{2} \\right)^2 + \\left( \\frac{a}{2} - \\cos \\alpha \\right)^2}\n\\]\n\n(Note that here we used the fact that \\(\\alpha\\) is acute. These equalities would look slightly differently otherwise.)\n\nNow look at the condition \\(DF \\le 1\\), or equivalently \\(DF^2 \\le 1\\). Making all the computations and simplifications, we have \\(\\sqrt{4 - a^2} \\sin \\alpha \\ge 1 - a \\cos \\alpha\\).\n\nNow I would like to square both sides. In order to get an equivalent inequality, we need to know that \\(1 - a \\cos \\alpha \\ge 0\\). This follows from the fact that \\(\\angle ADB\\) is acute. Indeed, denote \\(\\angle ADB = \\beta\\). From the law of sines in \\(\\triangle ADB\\) we have \\(\\frac{1}{\\sin \\angle ABD} = \\frac{a}{\\sin \\beta}\\). Successively this becomes \\(\\frac{1}{\\sin (\\pi - \\alpha - \\beta)} = \\frac{a}{\\sin \\beta}\\) or \\(\\frac{1}{\\sin (\\alpha + \\beta)} = \\frac{a}{\\sin \\beta}\\) or \\((1 - a \\cos \\alpha) \\sin \\beta = a \\sin \\alpha \\cos \\beta\\). From here we see that \\(\\beta < \\frac{\\pi}{2}\\) implies the right hand side is positive, so \\(1 - a \\cos \\alpha > 0\\).\n\nGoing back to our inequality, we can square both sides, and after rearranging terms we get that \\(DF \\le 1\\) if and only if\n\n\\(a^2 - 2a \\cos \\alpha + (1 - 4 \\sin^2 \\alpha) \\le 0\\).\n\nView this as an equation of degree \\(2\\) in \\(a\\). The value of the polynomial in \\(a\\) is \\(\\le 0\\) when \\(a\\) is between its solutions, that is\n\n\\(\\cos \\alpha - \\sqrt{3} \\sin \\alpha \\le a \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\\).\n\nNote that \\(\\cos \\alpha - \\sqrt{3} \\sin \\alpha = 2 \\cos \\left( \\alpha + \\frac{\\pi}{3} \\right)\\). If \\(\\alpha \\in \\left( 0, \\frac{\\pi}{2} \\right)\\) then \\(\\left( \\alpha + \\frac{\\pi}{3} \\right) \\in \\left( \\frac{\\pi}{3}, \\frac{5\\pi}{6} \\right)\\), and it follows that \\(2 \\cos \\left( \\alpha + \\frac{\\pi}{3} \\right) \\le 1\\).\n\nOn the other hand, remember that we are in the case \\(a > 1\\), so the left inequality is always true. It follows that\n\n\\(DF \\le 1\\) (i.e. the three circles of radius \\(1\\) centered at \\(A, B, D\\) cover \\(\\triangle ABD\\)) if an only if \\(a \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\\).\n\n[Solution by pf02, September 2024]",
"This solution is very similar to Solution 2, except that we choose another point instead of \\(F\\). This will in fact simplify the proof. Start like in Solution 2.\n\nWe can assume \\(a > 1\\). Indeed, refer to Remark 2 above to see that if \\(a \\le 1\\) there is nothing to prove.\n\nNote that instead of the statement \\(S1\\) we can consider the following statement \\(S1'\\):\n\n\\(S1'\\): the circles of radius \\(1\\) centered at \\(A, B, D\\) cover \\(\\triangle ABD\\).\n\nThis is equivalent to \\(S1\\) because of the symmetry between \\(\\triangle ABD\\) and \\(\\triangle BCD\\).\n\nLet \\(O\\) be the center of the circle circumscribed to \\(\\triangle ABD\\). Let \\(M, N, P\\), be the midpoints of \\(AB, AD, BD\\). The three circles cover \\(\\triangle ABD\\) if an only if \\(O\\) is inside the circle of radius 1 centered at \\(D\\).\n\n\\[\nProb 1967 1 fig3.png\n\\]\n\nThis needs an explanation. In fact, since \\(OA = OB = OD\\), the point \\(O\\) is inside or on the circle centered at \\(D\\) if and only if \\(OD \\le 1\\), if and only if \\(O\\) is in or inside the circles centered at \\(A, B, D\\). Since \\(OP \\perp BD\\), we have \\(DP < OD, PB < OB\\), so the triangles \\(\\triangle OPD, \\triangle OPB\\) are inside the circles centered at \\(D, B\\) respectively. By drawing \\(OA, OM\\) we can easily verify that the whole triangle is inside the circles centered at \\(A, D, B\\).\n\nNote that in the above argument we used that \\(O\\) is inside \\(\\triangle ABD\\), which is true because the triangle is acute.\n\nDenote \\(R\\) the radius of the circle circumscribed to \\(\\triangle ABD\\). From the law of sines, we have \\(\\frac{BD}{\\sin \\alpha} = 2R\\), and from the law of cosines we have \\(BD^2 = 1 + a^2 -2a \\cos \\alpha\\). So \\(R = OD \\le 1\\) translates to \\(\\frac{\\sqrt{1 + a^2 -2a \\cos \\alpha}}{2 \\sin \\alpha} \\le 1\\)\n\nSince \\(\\sin \\alpha > 0\\), we can simplify this inequality, and get\n\n\\(a^2 - 2a \\cos \\alpha + (1 - 4 \\sin^2 \\alpha) \\le 0\\).\n\nBut this is exactly the inequality we encountered in Solution 2, so proving that this is equivalent to\n\n\\[\na \\le \\cos \\alpha + \\sqrt{3} \\sin \\alpha\n\\]\n\nis identical to what has been done above.\n\n[Solution based on an idea by feliz; see link in Remark 4, or below.]\n\nA solution can also be found here [1]"
] |
IMO-1967-2
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https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_2
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Prove that if one and only one edge of a tetrahedron is greater than \(1\), then its volume is \(\le \frac{1}{8}\).
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[
"Assume \\(CD>1\\) and let \\(AB=x\\). Let \\(P,Q,R\\) be the feet of perpendicular from \\(C\\) to \\(AB\\) and \\(\\triangle ABD\\) and from \\(D\\) to \\(AB\\), respectively.\n\nSuppose \\(BP>PA\\). We have that \\(CP=\\sqrt{CB^2-BT^2}\\le\\sqrt{1-\\frac{x^2}4}\\), \\(CQ\\le CP\\le\\sqrt{1-\\frac{x^2}4}\\). We also have \\(DQ^2\\le\\sqrt{1-\\frac{x^2}4}\\). So the volume of the tetrahedron is \\(\\frac13\\left(\\frac12\\cdot AB\\cdot DR\\right)CQ\\le\\frac{x}6\\left(1-\\frac{x^2}4\\right)\\).\n\nWe want to prove that this value is at most \\(\\frac18\\), which is equivalent to \\((1-x)(3-x-x^2)\\ge0\\). This is true because \\(0<x\\le 1\\).\n\nThe above solution was posted and copyrighted by jgnr. The original thread can be found here: [1]\n\n## Remarks (added by pf02, September 2024)\n\nThe solution above is essentially correct, and it is nice, but it is so sloppily written that it borders the incomprehensible. Below I will give an edited version of it for the sake of completeness.\n\nThen, I will give a second solution to the problem.\n\nA few notes which may be of interest.\n\nThe condition that one side is greater than \\(1\\) is not really necessary. The statement is true even if all sides are \\(\\le 1\\). What we need is that no more than one side is \\(> 1\\).\n\nThe upper limit of \\(1/8\\) for the volume of the tetrahedron is actually reached. This will become clear from both solutions.",
"Assume that five of the edges are \\(\\le 1\\). Take them to be the edges other than \\(CD\\). Denote \\(AB = x\\). Let \\(P, Q, R\\) be the feet of perpendiculars from \\(C\\) to \\(AB\\), from \\(C\\) to the plane \\(ABD\\), and from \\(D\\) to \\(AB\\), respectively.\n\n\\[\nProb 1967 2 fig1.png\n\\]\n\nAt least one of the segments \\(AP, PB\\) has to be \\(\\ge \\frac{x}{2}\\). Suppose \\(PB \\ge \\frac{x}{2}\\). (If \\(AP\\) were bigger that \\(\\frac{x}{2}\\) the argument would be the same.) We have that \\(CP = \\sqrt{BC^2 - PB^2} \\le \\sqrt{1 - \\frac{x^2}{4}}\\). By the same argument in \\(\\triangle ABD\\) we have \\(DR \\le \\sqrt{1 - \\frac{x^2}{4}}\\). Since \\(CQ \\perp\\) plane \\(ABD\\), we have \\(CQ \\le CP\\), so \\(CQ \\le \\sqrt{1 - \\frac{x^2}{4}}\\).\n\nThe volume \\(V\\) of the tetrahedron is\n\n\\(V = \\frac{1}{3} \\cdot (\\)area of \\(\\triangle ABD) \\cdot\\)(height from \\(C) = \\frac{1}{3} \\cdot \\left( \\frac{1}{2} \\cdot AB \\cdot DR \\right) \\cdot CQ \\le \\left( \\frac{1}{6} \\cdot x \\cdot \\sqrt{1 - \\frac{x^2}{4}} \\cdot \\sqrt{1 - \\frac{x^2}{4}} \\right) = \\frac{x}{6} \\left( 1 - \\frac{x^2}{4} \\right)\\).\n\nWe need to prove that \\(\\frac{x}{6} \\left( 1 - \\frac{x^2}{4} \\right) \\le \\frac{1}{8}\\). Some simple computations show that this is the same as \\((1 - x)(3 - x - x^2) \\ge 0\\). This is true because \\(0 < x \\le 1\\), and \\(-x^2 - x + 3 > 0\\) on the interval \\((0, 1]\\).\n\n## Note\n\n\\(V = \\frac{1}{8}\\) is achieved when \\(x = 1\\) and all inequalities are equalities. This is the case when all sides except \\(CD\\) are \\(= 1\\), \\(P = R\\) are the midpoint of \\(AB\\), and \\(Q = P\\) (in which case the planes \\(ABC, ABD\\) are perpendicular). In this case, \\(CD = \\frac{\\sqrt{6}}{2}\\), and \\(V = \\frac{1}{8}\\) as can be seen from an easy computation.\n\n[This is an edited version of the solution by jgnr.]",
"Let \\(\\mathcal{T}\\) be the set of tetrahedrons with five edges \\(\\le 1\\). This proof will show that there is a \\(T \\in \\mathcal{T}\\) with one edge \\(> 1\\) and such that \\(\\mathbf{volume} (T) = \\frac{1}{8}\\), and that for any \\(U \\in \\mathcal{T}\\) either \\(U = T\\) or there is a finite sequence of tetrahedrons \\(T_1, \\dots, T_n\\) such that\n\n\\(\\mathbf{volume} (U) = \\mathbf{volume} (T_1) < \\dots < \\mathbf{volume} (T_n) = \\mathbf{volume} (T)\\).\n\nThe statement of the problem is a consequence of these facts.\n\nWe begin with two simple propositions.\n\n## Proposition\n\nLet \\(ABCD\\) be a tetrahedron, and consider the transformations which rotate \\(\\triangle ABC\\) around \\(AB\\) while keeping \\(\\triangle ABD\\) fixed. We get a set of tetrahedrons, two of which, \\(ABC_1D\\) and \\(ABC_2D\\) are shown in the picture below. The lengths of all sides except \\(CD\\) are constant through this transformation.\n\n\\[\nProb 1967 2 fig2.png\n\\]\n\n1. Assume that the angles between the planes \\(ABD\\) and \\(ABC\\), and \\(ABD\\) and \\(ABC_1\\) are both acute. If the perpendicular from \\(C_1\\) to the plane \\(ABD\\) is larger that the perpendicular from \\(C\\) to the plane \\(ABD\\) then the volume of \\(ABC_1D\\) is larger than the volume of \\(ABCD\\).\n\n2. Furthermore, the tetrahedron \\(ABC_2D\\) obtained when the position of \\(C_2\\) is such that the planes \\(ABD\\) and \\(ABC_2\\) are perpendicular has the maximum volume of all tetrahedrons obtained from rotating \\(\\triangle ABC\\) around \\(AB\\).\n\nThese statements are intuitively clear, since the volume \\(V\\) of the tetrahedron \\(ABCD\\) is given by\n\n\\(V = \\frac{1}{3} \\cdot (\\)area of \\(\\triangle ABD) \\cdot (\\)height from \\(C)\\).\n\nA formal proof is very easy, and I will skip it.\n\n## Corollary\n\nGiven a tetrahedron \\(T\\), and an edge \\(e_1\\) of it, we can find another tetrahedron \\(U\\) such that \\(\\mathbf{volume}(U) > \\mathbf{volume}(T)\\), with an edge \\(f_1 > e_1\\), and such that all the other edges of \\(U\\) are equal to the corresponding edges of \\(T\\), \\(\\mathbf{unless}\\) the edge \\(e_1\\) stretches between sides of \\(T\\) which are perpendicular. When we chose a bigger \\(f_1\\), if \\(e_1 < 1\\) we can choose \\(f_1 = 1\\). Or, we can choose \\(f_1\\) such that it stretches between sides which are perpendicular.\n\n(By \"stretches between two sides\" I mean that the end points of the edge are the vertices on the two sides which are not common to the two sides. In the picture above, \\(DC_2\\) stretches between the sides \\(ABD, AC_2B\\) of \\(ABC_2D\\).)\n\n## Lemma\n\nAssume we have a tetrahedron \\(T\\) with edges \\(e_1, \\dots, e_6\\), such that \\(e_2, \\dots, e_6 \\le 1\\). If there is an edge \\(e_m < 1\\) among \\(e_2, \\dots, e_6\\) then there is a tetrahedron \\(U\\) with volume bigger than the volume of \\(T\\), whose edges are equal to those of \\(T\\), except for \\(e_m\\), which is replaced by an edge of size \\(1\\).\n\n## Proof\n\nCase 1: If \\(T\\) does not have any sides which are perpendicular, then the existence of \\(U\\) follows from the corollary.\n\nCase 2: Assume \\(T\\) has exactly two sides which are perpendicular (like \\(ABC_2D\\) in the picture above). If \\(C_2D < 1\\) and it were the only edge \\(< 1\\), then all the other edges are \\(= 1\\) (because they were assumed to be \\(\\le 1\\)). In this case \\(\\triangle ABD, \\triangle AC_2B\\) are equilateral with sides \\(= 1\\), and the planes can not be perpendicular since \\(C_2D < 1\\). (Indeed, an easy computation shows that if we take two equilateral triangles \\(\\triangle ABD, \\triangle AC_2B\\) and place them perpendicular to each other, then \\(CD = \\frac{\\sqrt{6}}{2} > 1\\).) So \\(e_m\\) must be one of the other sides. Then again, the existence of \\(U\\) follows from the corollary.\n\nCase 3: Assume that three sides are perpendicular.\n\n\\[\nProb 1967 2 fig3.png\n\\]\n\nAssume the perpendicular sides are the ones meeting at \\(A\\), i.e. each pair of the planes meeting at \\(A\\) are perpendicular. Since at least two of the edges \\(BD, BC, CD \\le 1\\) it follows that \\(AB, AC, AD < 1\\) (the sides of the right angle in a right triangle are less than the hypotenuse). Just for the sake of notation, assume \\(e_m = AB < 1\\). We can apply the corollary, and find a tetrahedron \\(U\\) with volume bigger than the volume of \\(T\\), with edges equal to those of \\(T\\), except that \\(AB\\) is replaced by an edge \\(= 1\\).\n\nNow the problem is very easy to prove. Let \\(T\\) be a tetrahedron with edges \\(e_1, \\dots, e_6\\), such that \\(e_2, \\dots, e_6 \\le 1\\). Apply the lemma as many times as necessary (up to five times), successively replacing each edge \\(< 1\\) by an edge \\(= 1\\). We obtain a tetrahedron \\(U\\) with five edges \\(f_2, \\dots, f_6 = 1\\), and one edge \\(f_1 = e_1\\). If \\(f_1\\) stretches between two perpendicular sides, we are done. If not, apply the corollary one more time to obtain a bigger tetrahedron in which \\(f_1\\) is replaced by a larger edge which stretches between two perpendicular sides.\n\nWe obtain the same result as in the first solution: the largest tetrahedron is the one formed by two equilateral triangles with sides \\(= 1\\), having one side in common, with the two planes containing the triangles perpendicular. An easy calculation shows that the edge which is \\(> 1\\) is in fact of length \\(\\frac{\\sqrt{6}}{2}\\), and the volume of this tetrahedron is \\(\\frac{1}{8}\\).\n\n[Solution by pf02, September 2024]"
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