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IMO-1959-1
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For every integer \(n\) prove that the fraction \(\frac{21n + 4}{14n + 3}\) cannot be reduced any further.
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The desired result \((14n + 3,21n + 4) = 1\) follows from
\[3(14n + 3) - 2(21n + 4) = 1.\]
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IMO-1959-2
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For which real numbers \(x\) do the following equations hold:
\[(a)\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = \sqrt{2},\] \[(b)\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = 1,\] \[(c)\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = 2?\]
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For the square roots to be real we must have \(2x - 1\geq 0\Rightarrow x\geq 1 / 2\) and \(x\geq\) \(\sqrt{2x - 1}\Rightarrow x^{2}\geq 2x - 1\Rightarrow (x - 1)^{2}\geq 0\) , which always holds. Then we have \(\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = c\iff\)
\[c^{2} = 2x + 2\sqrt{x^{2} - \sqrt{2x - 1}^{2}} = 2x + 2|x - 1| = \left\{ \begin{array}{ll}2, & 1 / 2\leq x\leq 1,\\ 4x - 2, & x\geq 1. \end{array} \right.\]
(a) \(c^{2} = 2\) . The equation holds for \(1 / 2\leq x\leq 1\)
(b) \(c^{2} = 1\) . The equation has no solution.
(c) \(c^{2} = 4\) . The equation holds for \(4x - 2 = 4\Rightarrow x = 3 / 2\)
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IMO-1959-3
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Let \(x\) be an angle and let the real numbers \(a\) , \(b\) , \(c\) , \(\cos x\) satisfy the following equation:
\[a\cos^2 x + b\cos x + c = 0.\]
Write the analogous quadratic equation for \(a\) , \(b\) , \(c\) , \(\cos 2x\) . Compare the given and the obtained equality for \(a = 4\) , \(b = 2\) , \(c = - 1\) .
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Multiplying the equality by \(4(a\cos^{2}x - b\cos x + c)\) , we obtain \(4a^{2}\cos^{4}x+\) \(2(4a c - 2b^{2})\cos^{2}x + 4c^{2} = 0\) . Plugging in \(2\cos^{2}x = 1 + \cos 2x\) we obtain (after quite a bit of manipulation):
\[a^{2}\cos^{2}2x + (2a^{2} + 4ac - 2b^{2})\cos 2x + (a^{2} + 4ac - 2b^{2} + 4c^{2}) = 0.\]
For \(a = 4\) \(b = 2\) , and \(c = - 1\) we get \(4\cos^{2}x + 2\cos x - 1 = 0\) and \(16\cos^{2}2x+\) \(8\cos 2x - 4 = 0\Rightarrow 4\cos^{2}2x + 2\cos 2x - 1 = 0\)
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IMO-1959-4
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Construct a right-angled triangle whose hypotenuse \(c\) is given if it is known that the median from the right angle equals the geometric mean of the remaining two sides of the triangle.
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Analysis. Let \(a\) and \(b\) be the other two sides of the triangle. From the conditions of the problem we have \(c^{2} = a^{2} + b^{2}\) and \(c / 2 = \sqrt{ab}\Leftrightarrow 3 / 2c^{2} = a^{2} + b^{2} + 2ab =\) \((a + b)^{2}\Leftrightarrow \sqrt{3 / 2} c = a + b\) . Given a desired \(\triangle ABC\) let \(D\) be a point on (AC such that \(CD = CB\) . In that case, \(AD = a + b = \sqrt{3 / 2} c\) , and also, since \(BC = CD\) , it follows that \(\angle ADB = 45^{\circ}\) .
Construction. From a segment of length \(c\) we elementarily construct a segment \(AD\) of length \(\sqrt{3 / 2} c\) . We then construct a ray ( \(DX\) such that \(\angle ADX = 45^{\circ}\)
and a circle \(k(A,c)\) that intersects the ray at point \(B\) . Finally, we construct the perpendicular from \(B\) to \(AD\) ; point \(C\) is the foot of that perpendicular.
Proof. It holds that \(AB = c\) , and, since \(CB = CD\) , it also holds that \(AC + CB = AC + CD = AD = \sqrt{3 / 2} c\) . From this it follows that \(\sqrt{AC \cdot CB} = c / 2\) . Since \(BC\) is perpendicular to \(AD\) , it follows that \(\angle BCA = 90^\circ\) . Thus \(ABC\) is the desired triangle.
Discussion. Since \(AB\sqrt{2} = \sqrt{2} c > \sqrt{3 / 2} c = AD > AB\) , the circle \(k\) intersects the ray \(DX\) in exactly two points, which correspond to two symmetric solutions.
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IMO-1959-5
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A segment \(AB\) is given and on it a point \(M\) . On the same side of \(AB\) squares \(AMKD\) and \(BMFE\) are constructed. The circumcircles of the two squares, whose centers are \(P\) and \(Q\) , intersect in \(M\) and another point \(N\) .
(a) Prove that lines \(FA\) and \(BC\) intersect at \(N\) .
(b) Prove that all such constructed lines \(MN\) pass through the same point \(S\), regardless of the selection of \(M\).
(c) Find the locus of the midpoints of all segments \(PQ\), as \(M\) varies along the segment \(AB\).
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(a) It suffices to prove that \(AF \perp BC\) , since then for the intersection point \(X\) we have \(\angle AXC = \angle BXF = 90^\circ\) , implying that \(X\) belongs to the circumcircles of both squares and thus that \(X = N\) . The relation \(AF \perp BC\) holds because from \(MA = MC\) , \(MF = MB\) , and \(\angle AMC = \angle FMB\) it follows that \(\triangle AMF\) is obtained by rotating \(\triangle BMC\) by \(90^\circ\) around \(M\) .(b) Since \(N\) is on the circumcircle of \(BMFE\) , it follows that \(\angle ANM = \angle MNB = 45^\circ\) . Hence \(MN\) is the bisector of \(\angle ANB\) . It follows that \(MN\) passes through the midpoint of the arc \(\widehat{AB}\) of the circle with diameter \(AB\) (i.e., the circumcircle of \(\triangle ABN\) ) not containing \(N\) .(c) Let us introduce a coordinate system such that \(A = (0,0)\) , \(B = (b,0)\) , and \(M = (m,0)\) . Setting in general \(W = (x_W, y_W)\) for an arbitrary point \(W\) and denoting by \(R\) the midpoint of \(PQ\) , we have \(y_R = (y_P + y_Q) / 2 = (m + b - m) / 4 = b / 4\) and \(x_R = (x_P + x_Q) / 2 = (m + m + b) / 4 = (2m + b) / 4\) , the parameter \(m\) varying from 0 to \(b\) . Thus the locus of all points \(R\) is the closed segment \(R_1R_2\) where \(R_1 = (b / 4, b / 4)\) and \(R_2 = (b / 4, 3b / 4)\) .
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IMO-1959-6
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Let \(\alpha\) and \(\beta\) be two planes intersecting at a line \(p\). In \(\alpha\) a point \(A\) is given and in \(\beta\) a point \(C\) is given, neither of which lies on \(p\). Construct \(B\) in \(\alpha\) and \(D\) in \(\beta\) such that \(ABCD\) is an equilateral trapezoid, \(AB \parallel CD\), in which a circle can be inscribed.
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Analysis. For \(AB \parallel CD\) to hold evidently neither must intersect \(p\) and hence constructing lines \(r\) in \(\alpha\) through \(A\) and \(s\) in \(\beta\) through \(C\) , both being parallel to \(p\) , we get that \(B \in r\) and \(D \in s\) . Hence the problem reduces to a planar problem in \(\gamma\) , determined by \(r\) and \(s\) . Denote by \(A'\) the foot of the perpendicular from \(A\) to \(s\) . Since \(ABCD\) is isosceles and has an incircle, it follows \(AD = BC = (AB + CD) / 2 = A'C\) . The remaining parts of the problem are now obvious.
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IMO-1960-1
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Find all the three-digit numbers for which one obtains, when dividing the number by 11, the sum of the squares of the digits of the initial number.
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Given the number \(\overline{acb}\) , since \(11 \mid \overline{acb}\) , it follows that \(c = a + b\) or \(c = a + b - 11\) . In the first case, \(a^{2} + b^{2} + (a + b)^{2} = 10a + b\) , and in the second case, \(a^{2} + b^{2} + (a + b - 11)^{2} = 10(a - 1) + b\) . In the first case the LHS is even, and hence \(b \in \{0, 2, 4, 6, 8\}\) , while in the second case it is odd, and hence \(b \in \{1, 3, 5, 7, 9\}\) . Analyzing the 10 quadratic equations for \(a\) we obtain that the only valid solutions are 550 and 803.
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IMO-1960-2
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For which real numbers \(x\) does the following inequality hold:
\[\frac{4x^{2}}{(1 - \sqrt{1 + 2x})^{2}} < 2x + 9?\]
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The LHS term is well-defined for \(x \geq -1 / 2\) and \(x \neq 0\) . Furthermore, \(4x^{2} / (1 - \sqrt{1 + 2x})^{2} = (1 + \sqrt{1 + 2x})^{2}\) . Since
\[f(x) = \left(1 + \sqrt{1 + 2x}\right)^{2} - 2x - 9 = 2\sqrt{1 + 2x} -7\]
is increasing and since \(f(45 / 8) = 0\) , it follows that the inequality holds precisely for \(- 1 / 2 \leq x < 45 / 8\) and \(x \neq 0\) .
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IMO-1960-3
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A right-angled triangle \(ABC\) is given for which the hypotenuse \(BC\) has length \(a\) and is divided into \(n\) equal segments, where \(n\) is odd. Let \(\alpha\) be the angle with which the point \(A\) sees the segment containing the middle of the hypotenuse. Prove that
\[\tan \alpha = \frac{4nh}{(n^2 - 1)a},\]
where \(h\) is the height of the triangle.
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Let \(B^{\prime}C^{\prime}\) be the middle of the \(n = 2k + 1\) segments and let \(D\) be the foot of the perpendicular from \(A\) to the hypotenuse. Let us assume \(\mathcal{B}(C,D,C^{\prime},B^{\prime},B)\) . Then from \(C D< B D,C D + B D = a\) , and \(C D\cdot B D = h^{2}\) we have \(C D^{2} - a\cdot C D + h^{2} =\) \(0\Rightarrow C D = (a - \sqrt{a^{2} - 4h^{2}})\big / 2\) . Let us define \(\angle D A C^{\prime} = \gamma\) and \(\angle D A B^{\prime} = \beta\) then \(\tan \beta = D B^{\prime} / h\) and \(\tan \gamma = D C^{\prime} / h\) . Since \(D B^{\prime} = C B^{\prime} - C D = (k + 1)a / (2k+\) \(1) - (c - \sqrt{c^{2} - 4h^{2}}) / 2\) and \(D C^{\prime} = k a / (2k + 1) - (c - \sqrt{c^{2} - 4h^{2}}) / 2\) , we have
\[\tan \alpha = \tan (\beta -\gamma) = \frac{\tan\beta - \tan\gamma}{1 + \tan\beta\cdot\tan\gamma} = \frac{\frac{a}{(2k + 1)h}}{1 + \frac{a^{2} - 4h^{2}}{4h^{2}} - \frac{a^{2}}{4h^{2}(2k + 1)^{2}}}\] \[= \frac{4h(2k + 1)}{4ak(k + 1)} = \frac{4nh}{(n^{2} - 1)a}.\]
The case \(\mathcal{B}(C,C^{\prime},D,B^{\prime},B)\) is similar.
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IMO-1960-4
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Construct a triangle \(ABC\) whose lengths of heights \(h_a\) and \(h_b\) (from \(A\) and \(B\) , respectively) and length of median \(m_a\) (from \(A\) ) are given.
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Analysis. Let \(A^{\prime}\) and \(B^{\prime}\) be the feet of the perpendiculars from \(A\) and \(B\) , respectively, to the opposite sides, \(A_{1}\) the midpoint of \(BC\) , and let \(D^{\prime}\) be the foot of the perpendicular from \(A_{1}\) to \(AC\) . We then have \(AA_{1} = m_{a}\) , \(AA^{\prime} = h_{a}\) , \(\angle AA^{\prime}A_{1} = 90^{\circ}\) , \(A_{1}D^{\prime} = h_{b} / 2\) , and \(\angle AD^{\prime}A_{1} = 90^{\circ}\) .
Construction. We construct the quadrilateral \(AD^{\prime}A_{1}A^{\prime}\) (starting from the circle with diameter \(AA_{1}\) ). Then \(C\) is the intersection of \(A^{\prime}A_{1}\) and \(AD^{\prime}\) , and \(B\) is on the line \(A_{1}C\) such that \(CA_{1} = A_{1}B\) and \(\mathcal{B}(B,A_{1},C)\) .
Discussion. We must have \(m_{a} \geq h_{a}\) and \(m_{a} \geq h_{b} / 2\) . The number of solutions is 0 if \(m_{a} = h_{a} = h_{b} / 2\) , 1 if two of \(m_{a}, h_{a}, h_{b} / 2\) are equal, and 2 otherwise.
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IMO-1960-5
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A cube \(ABCDA'B'C'D'\) is given.
(a) Find the locus of all midpoints of segments \(XY\) , where \(X\) is any point on segment \(AC\) and \(Y\) any point on segment \(B'D'\).
(b) Find the locus of all points \(Z\) on segments \(XY\) such that \(\overline{ZY} = 2\overline{XZ}\) .
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(a) The locus of the points is the square \(EFGH\) where these four points are the centers of the faces \(ABB'A', BCC'B', CDD'C'\) and \(DAA'D'\). (b) The locus of the points is the rectangle \(IJKL\) where these points are on \(AB'\), \(CB', CD'\), and \(AD'\) at a distance of \(AA' / 3\) with respect to the plane \(ABCD\).
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IMO-1960-6
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An isosceles trapezoid with bases \(a\) and \(b\) and height \(h\) is given.
(a) On the line of symmetry construct the point \(P\) such that both (nonbase) sides are seen from \(P\) with an angle of \(90^{\circ}\) .
(b) Find the distance of \(P\) from one of the bases of the trapezoid.
(c) Under what conditions for \(a\) , \(b\) , and \(h\) can the point \(P\) be constructed (analyze all possible cases)?
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Let \(E, F\) respectively be the midpoints of the bases \(AB, CD\) of the isosceles trapezoid \(ABCD\).
(a) The point \(P\) is on the intersection of \(EF\) and the circle with diameter \(BC\).
(b) Let \(x = EP\) . Since \(\triangle BEP \sim \triangle PFC\) , we have \(x(h - x) = ab / 4 \Rightarrow x_{1,2} = (h \pm \sqrt{h^2 - ab}) / 2\) .
(c) If \(h^2 > ab\) there are two solutions, if \(h^2 = ab\) there is only one solution, and if \(h^2 < ab\) there are no solutions.
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IMO-1960-7
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A sphere is inscribed in a regular cone. Around the sphere a cylinder is circumscribed so that its base is in the same plane as the base of the cone. Let \(V_1\) be the volume of the cone and \(V_2\) the volume of the cylinder.
(a) Prove that \(V_1 = V_2\) is impossible.
(b) Find the smallest \(k\) for which \(V_1 = kV_2\) , and in this case construct the angle at the vertex of the cone.
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Let \(A\) be the vertex of the cone, \(O\) the center of the sphere, \(S\) the center of the base of the cone, \(B\) a point on the base circle, and \(r\) the radius of the sphere. Let \(\angle SAB = \alpha\) . We easily obtain \(AS = r(1 + \sin \alpha) / \sin \alpha\) and \(SB = r(1 + \sin \alpha) \tan \alpha / \sin \alpha\) and hence \(V_1 = \pi SB^2 \cdot SA / 3 = \pi r^3 (1 + \sin \alpha)^2 / [3 \sin \alpha (1 - \sin \alpha)]\) . We also have \(V_2 = 2\pi r^3\) and hence
\[k = \frac{(1 + \sin\alpha)^2}{6\sin\alpha(1 - \sin\alpha)} \Rightarrow (1 + 6k)\sin^2\alpha +2(1 - 3k)\sin\alpha + 1 = 0.\]
The discriminant of this quadratic must be nonnegative: \((1 - 3k)^2 - (1 + 6k) \geq 0 \Rightarrow k \geq 4 / 3\) . Hence we cannot have \(k = 1\) . For \(k = 4 / 3\) we have \(\sin \alpha = 1 / 3\) , whose construction is elementary.
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IMO-1961-1
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Solve the following system of equations:
\[x + y + z = a,\] \[x^{2} + y^{2} + z^{2} = b^{2},\] \[xy = z^{2},\]
where \(a\) and \(b\) are given real numbers. What conditions must hold on \(a\) and \(b\) for the solutions to be positive and distinct?
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This is a problem solvable using elementary manipulations, so we shall state only the final solutions. For \(a = 0\) we get \((x,y,z) = (0,0,0)\) . For \(a\neq 0\) we get \((x,y,z)\in \{(t_{1},t_{2},z_{0}),(t_{2},t_{1},z_{0})\}\) , where
\[z_{0} = \frac{a^{2} - b^{2}}{2a}\quad \mathrm{and}\quad t_{1,2} = \frac{a^{2} + b^{2}\pm\sqrt{(3a^{2} - b^{2})(3b^{2} - a^{2})}}{4a}.\]
For the solutions to be positive and distinct the following conditions are necessary and sufficient: \(3b^{2} > a^{2} > b^{2}\) and \(a > 0\) .
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IMO-1961-2
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Let \(a, b,\) and \(c\) be the lengths of a triangle whose area is \(S\) . Prove that
\[a^{2} + b^{2} + c^{2}\geq 4S\sqrt{3}.\]
In what case does equality hold?
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Using \(S = b c\sin \alpha /2\) , \(a^{2} = b^{2} + c^{2} - 2b c\cos \alpha\) and \((\sqrt{3}\sin \alpha +\cos \alpha) / 2 = \cos (\alpha -60^{\circ})\) we have
\[a^{2} + b^{2} + c^{2}\geq 4S\sqrt{3}\Leftrightarrow b^{2} + c^{2}\geq b c(\sqrt{3}\sin \alpha +\cos \alpha)\Leftrightarrow\] \[\Leftrightarrow (b - c)^{2} + 2b c(1 - \cos (\alpha -60^{\circ}))\geq 0,\]
where equality holds if and only if \(b = c\) and \(\alpha = 60^{\circ}\) , i.e., if the triangle is equilateral.
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IMO-1961-3
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Solve the equation \(\cos^{n}x - \sin^{n}x = 1\) , where \(n\) is a given positive integer.
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For \(n\geq 2\) we have
\[1 = \cos^{n}x - \sin^{n}x\leq |\cos^{n}x - \sin^{n}x|\] \[\leq |\cos^{n}x| + |\sin^{n}x|\leq \cos^{2}x + \sin^{2}x = 1.\]
Hence \(\sin^{2}x = |\sin^{n}x|\) and \(\cos^{2}x = |\cos^{n}x|\) , from which it follows that \(\sin x\) , \(\cos x\in \{1,0, - 1\} \Rightarrow x\in \pi \mathbb{Z} / 2\) . By inspection one obtains the set of solutions
\[\{m\pi |m\in \mathbb{Z}\} \mathrm{for}evenn\mathrm{and}\{2m\pi ,2m\pi -\pi /2|m\in \mathbb{Z}\} \mathrm{for}oddn.\]
For \(n = 1\) we have \(1 = \cos x - \sin x = - \sqrt{2}\sin (x - \pi /4)\) , which yields the set of solutions
\[\{2m\pi ,2m\pi -\pi /2|m\in \mathbb{Z}\} .\]
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IMO-1961-4
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In the interior of \(\triangle P_{1}P_{2}P_{3}\) a point \(P\) is given. Let \(Q_{1}\) , \(Q_{2}\) , and \(Q_{3}\) respectively be the intersections of \(PP_{1}\) , \(PP_{2}\) , and \(PP_{3}\) with the opposing edges of \(\triangle P_{1}P_{2}P_{3}\) . Prove that among the ratios \(PP_{1} / PQ_{1}\) , \(PP_{2} / PQ_{2}\) , and \(PP_{3} / PQ_{3}\) there exists at least one not larger than 2 and at least one not smaller than 2.
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Let \(x_{i} = P P_{i} / P Q_{i}\) for \(i = 1,2,3\) . For all \(i\) we have
\[\frac{1}{x_{i} + 1} = \frac{P Q_{i}}{P_{i}Q_{i}} = \frac{S_{P P_{i}P_{k}}}{S_{P_{1}P_{2}P_{3}}},\]
where the indices \(j\) and \(k\) are distinct and different from \(i\) . Hence we have
\[f(x_{1},x_{2},x_{3}) = \frac{1}{x_{1} + 1} +\frac{1}{x_{2} + 1} +\frac{1}{x_{3} + 1}\] \[= \frac{S(P P_{2}P_{3}) + S(P P_{1}P_{3}) + S(P P_{2}P_{3})}{S(P_{1}P_{2}P_{3})} = 1.\]
It follows that \(1 / (x_{i} + 1)\geq 1 / 3\) for some \(i\) and \(1 / (x_{j} + 1)\leq 1 / 3\) for some \(j\) . Consequently, \(x_{i}\leq 2\) and \(x_{j}\geq 2\) .
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IMO-1961-5
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Construct a triangle \(ABC\) if the following elements are given: \(AC = b\) , \(AB = c\) , and \(\triangle AMB = \omega (\omega < 90^{\circ})\) , where \(M\) is the midpoint of \(BC\) . Prove that the construction has a solution if and only if
\[b\tan {\frac{\omega}{2}}\leq c< b.\]
In what case does equality hold?
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Analysis. Let \(C_1\) be the midpoint of \(AB\) . In \(\triangle AMB\) we have \(MC_1 = b / 2\) , \(AB = c\) , and \(\angle AMB = \omega\) . Thus, given \(AB = c\) , the point \(M\) is at the intersection of the circle \(k(C', b / 2)\) and the set of points \(e\) that view \(AB\) at an angle of \(\omega\) . The construction of \(ABC\) is now obvious.
Discussion. It suffices to establish the conditions for which \(k\) and \(e\) intersect. Let \(E\) be the midpoint of one of the arcs that make up \(e\) . A necessary and sufficient condition for \(k\) to intersect \(e\) is
\[\frac{c}{2} = C'A \leq \frac{b}{2} \leq C'E = \frac{c}{2} \cot \frac{\omega}{2} \Leftrightarrow b \tan \frac{\omega}{2} \leq c < b.\]
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IMO-1961-6
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A plane \(\epsilon\) is given and on one side of the plane three noncollinear points \(A\) , \(B\) , and \(C\) such that the plane determined by them is not parallel to \(\epsilon\) . Three arbitrary points \(A'\) , \(B'\) , and \(C'\) in \(\epsilon\) are selected. Let \(L\) , \(M\) , and \(N\) be the midpoints of \(AA'\) , \(BB'\) , and \(CC'\) , and \(G\) the centroid of \(\triangle LMN\) . Find the locus of all points obtained for \(G\) as \(A'\) , \(B'\) , \(C'\) are varied (independently of each other) across \(\epsilon\) .
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Let \(h(X)\) denote the distance of a point \(X\) from \(\epsilon\) , \(X\) restricted to being on the same side of \(\epsilon\) as \(A\) , \(B\) , and \(C\) . Let \(G_1\) be the (fixed) centroid of \(\triangle ABC\) and \(G_1'\) the centroid of \(\triangle A'B'C'\) . It is trivial to prove that \(G\) is the midpoint of \(G_1G_1'\) . Hence varying \(G_1'\) across \(\epsilon\) , we get that the locus of \(G\) is the plane \(\alpha\) parallel to \(\epsilon\) such that
\[X \in \alpha \Leftrightarrow h(X) = \frac{h(G_1)}{2} = \frac{h(A) + h(B) + h(C)}{6}.\]
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IMO-1962-1
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Find the smallest natural number \(n\) with the following properties:
(a) In decimal representation it ends with 6.
(b) If we move this digit to the front of the number, we get a number 4 times larger.
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From the conditions of the problem we have \(n = 10x + 6\) and \(4n = 6\cdot 10^{m} + x\) for some integer \(x\) . Eliminating \(x\) from these two equations, we get \(40n = 6\cdot\) \(10^{m + 1} + n - 6\Rightarrow n = 2(10^{m + 1} - 1) / 13\) . Hence we must find the smallest \(m\) such that this fraction is an integer. By inspection, this happens for \(m = 6\) , and for this \(m\) we obtain \(n = 153846\) , which indeed satisfies the conditions of the problem.
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IMO-1962-2
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Find all real numbers \(x\) for which
\[\sqrt{3 - x} -\sqrt{x + 1} >\frac{1}{2}.\]
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We note that \(f(x) = \sqrt{3 - x} - \sqrt{x + 1}\) is well-defined only for \(-1\leq x\leq 3\) and is decreasing (and obviously continuous) on this interval. We also note that \(f(-1) = 2 > 1 / 2\) and
\[f\left(1 - \frac{\sqrt{31}}{8}\right) = \sqrt{\left(\frac{1}{4} + \frac{\sqrt{31}}{4}\right)^2} -\sqrt{\left(\frac{1}{4} - \frac{\sqrt{31}}{4}\right)^2} = \frac{1}{2}.\]
Hence the inequality is satisfied for \(- 1\leq x< 1 - \sqrt{31} /8\)
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IMO-1962-3
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A cube \(A B C D A^{\prime}B^{\prime}C^{\prime}D^{\prime}\) is given. The point \(X\) is moving at a constant speed along the square \(A B C D\) in the direction from \(A\) to \(B\) . The point \(Y\) is moving with the same constant speed along the square \(B C C^{\prime}B^{\prime}\) in the direction from \(B^{\prime}\) to \(C^{\prime}\) . Initially, \(X\) and \(Y\) start out from \(A\) and \(B^{\prime}\) respectively. Find the locus of all the midpoints of \(X Y\) .
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By inspecting the four different stages of this periodic motion we easily obtain that the locus of the midpoints of \(XY\) is the edges of \(MNCQ\) , where \(M\) , \(N\) , and \(Q\) are the centers of \(ABB'A'\) , \(BCC'B'\) , and \(ABCD\) , respectively.
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IMO-1962-4
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Solve the equation
\[\cos^{2}x + \cos^{2}2x + \cos^{2}3x = 1.\]
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Since \(\cos 2x = 1 + \cos^2 x\) and \(\cos \alpha +\cos \beta = 2\cos \left(\frac{\alpha + \beta}{2}\right)\cos \left(\frac{\alpha - \beta}{2}\right)\) , we have \(\cos^2 x + \cos^2 2x + \cos^2 3x = 1\Leftrightarrow \cos 2x + \cos 4x + 2\cos^2 3x = 2\cos 3x(\cos x + \cos 3x) = 0\Leftrightarrow 4\cos 3x\cos 2x\cos x = 0\) . Hence the solutions are \(x\in \{\pi /2 + m\pi ,\pi /4 + m\pi /2,\pi /6 + m\pi /3\mid m\in \mathbb{Z}\}\) .
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IMO-1962-5
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On the circle \(k\) three points \(A\) , \(B\) , and \(C\) are given. Construct the fourth point on the circle \(D\) such that one can inscribe a circle in \(A B C D\) .
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Analysis. Let \(ABCD\) be the desired quadrilateral. Let us assume w.l.o.g. that \(AB > BC\) (for \(AB = BC\) the construction is trivial). For a tangent quadrilateral we have \(AD - DC = AB - BC\) . Let \(X\) be a point on \(AD\) such that \(DX = DC\) . We then have \(AX = AB - BC\) and \(\angle AXC = \angle ADC + \angle CDX = 180^\circ - \angle ABC / 2\) . Constructing \(X\) and hence \(D\) is now obvious.
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IMO-1962-6
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Let \(A B C\) be an isosceles triangle with circumradius \(r\) and inradius \(\rho\) . Prove that the distance \(d\) between the circumcenter and incenter is given by
\[d = \sqrt{r(r - 2\rho)}.\]
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This problem is a special case, when the triangle is isosceles, of Euler's formula, which holds for all triangles.
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IMO-1962-7
|
Prove that a tetrahedron \(S A B C\) has five different spheres that touch all six lines determined by its edges if and only if it is regular.
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The spheres are arranged in a similar manner as in the planar case where we have one incircle and three excircles. Here we have one "insphere" and four "exspheres" corresponding to each of the four sides. Each vertex of the tetrahedron effectively has three tangent lines drawn from it to each of the five spheres. Repeatedly using the equality of the three tangent segments from a vertex (in the same vein as for tangent planar quadrilaterals) we obtain \(SA + BC = SB + CA = SC + AB\) from the insphere. From the exsphere opposite of \(S\) we obtain \(SA - BC = SB - CA = SC - AB\) , hence \(SA = SB = SC\) and \(AB = BC = CA\) . By symmetry, we also have \(AB = AC = AS\) . Hence indeed, all the edges of the tetrahedron are equal in length and thus we have shown that the tetrahedron is regular.
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IMO-1963-1
|
Determine all real solutions of the equation \(\sqrt{x^{2} - p} + 2\sqrt{x^{2} - 1} = x\) where \(p\) is a real number.
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Obviously, \(x\geq 0\) and \(p\geq 0\) ; hence squaring the given equation yields an equivalent equation \(5x^{2} - p - 4 + 4\sqrt{(x^{2} - 1)(x^{2} - p)} = x^{2}\) , i.e., \(4\sqrt{(x^{2} - 1)(x^{2} - p)} = (p + 4) - 4x^{2}\) . If \(4x^{2}\leq (p + 4)\) , we may square the equation once again to get \(-16(p + 1)x^{2} + 16p = -8(p + 4)x^{2} + (p + 4)^{2}\) , which is equivalent to \(x^{2} = (4 - p)^{2} / [4(4 - 2p)]\) . We immediately get \(p< 2\) hence
\[x = \frac{4 - p}{2\sqrt{4 - 2p}}.\]
Placing this \(x\) in the original equation gives \(|3p - 4| = 4 - 3p\) hence \(0\leq p\leq \frac{4}{3}\) Thus for \(0\leq p\leq \frac{4}{3}\) we have \(x = (4 - p)^{2} / [4(4 - 2p)]\) , otherwise the solution doesn't exist.
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IMO-1963-2
|
Find the locus of points in space that are vertices of right angles of which one ray passes through a given point and the other intersects a given segment.
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Let \(A\) be the given point, \(BC\) the given segment, and \(\mathcal{B}_1,\mathcal{B}_2\) the closed balls with the diameters \(AB\) and \(AC\) respectively. Consider one right angle \(\angle AOK\) with \(K\in [BC]\) . If \(B',C'\) are the feet of the perpendiculars from \(B,C\) to \(AO\) respectively, then \(O\) lies on the segment \(B'C'\) , which implies that it lies on exactly one of the segments \(AB',AC'\) . Hence \(O\) belongs to exactly one of the balls \(\mathcal{B}_1,\mathcal{B}_2\) ; i.e., \(O\in \mathcal{B}_1\Delta \mathcal{B}_2\) . This is obviously the required locus.
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IMO-1963-3
|
Prove that if all the angles of a convex \(n\) -gon are equal and the lengths of consecutive edges \(a_{1},\ldots ,a_{n}\) satisfy \(a_{1}\geq a_{2}\geq \dots \geq a_{n}\) , then \(a_{1} = a_{2} = \dots = a_{n}\) .
|
Let \(\overrightarrow{OA_1},\overrightarrow{OA_2},\ldots ,\overrightarrow{OA_n}\) be the vectors corresponding respectively to the edges \(a_1,a_2,\ldots ,a_n\) of the polygon. By the conditions of the problem, these vectors satisfy \(\overrightarrow{OA_1} +\dots +\overrightarrow{OA_n} = \overrightarrow{0}\) , \(\angle A_1OA_2 = \angle A_2OA_3 = \dots = \angle A_naOA_1 = 2\pi /n\) and \(OA_1\geq OA_2\geq \dots \geq OA_n\) . Our task is to prove that \(OA_1 = \dots = OA_n\) . Let \(l\) be the line through \(O\) perpendicular to \(OA_n\) , and \(B_1,\ldots ,B_{n - 1}\) the projections of \(A_1,\ldots ,A_{n - 1}\) onto \(l\) respectively. By the assumptions, the sum of the \(\overrightarrow{OB_i}\) 's is \(\overrightarrow{0}\) . On the other hand, since \(OB_i\leq OB_{n - i}\) for all \(i\leq n / 2\) , all the sums \(\overrightarrow{OB_i} +\overrightarrow{OB_{n - i}}\) lie on the same side of the point \(O\) . Hence all these sums must be equal to \(\overrightarrow{0}\) . Consequently, \(OA_i = OA_{n - i}\) , from which the result immediately follows.
|
IMO-1963-4
|
Find all solutions \(x_{1},\ldots ,x_{5}\) to the system of equations
\[
\begin{cases}
x_5 + x_2 = y x_1,\\
x_1 + x_3 = y x_2,\\
x_2 + x_4 = y x_3,\\
x_3 + x_5 = y x_4,\\
x_4 + x_1 = y x_5.
\end{cases}
\]
where \(y\) is a real parameter.
|
Summing up all the equations yields \(2(x_{1} + x_{2} + x_{3} + x_{4} + x_{5}) = y(x_{1} + x_{2} + x_{3} + x_{4} - x_{5})\) . If \(y = 2\) , then the given equations imply \(x_{1} - x_{2} = x_{2} - x_{3} = \dots = x_{5} - x_{1}\) ; hence \(x_{1} = x_{2} = \dots = x_{5}\) , which is clearly a solution. If \(y\neq 2\) , then \(x_{1} + \dots +x_{5} = 0\) , and summing the first three equalities gives \(x_{2} = y(x_{1} + x_{2} + x_{3})\) . Using that \(x_{1} + x_{3} = yx_{2}\) we obtain \(x_{2} = (y^{2} + y)x_{2}\) , i.e., \((y^{2} + y - 1)x_{2} = 0\) . If \(y^{2} + y - 1\neq 0\) , then \(x_{2} = 0\) , and similarly \(x_{1} = \dots = x_{5} = 0\) . If \(y^{2} + y - 1 = 0\) , it is easy to prove that the last two equations are the consequence of the first three. Thus choosing any values for \(x_{1}\) and \(x_{5}\) will give exactly one solution for \(x_{2},x_{3},x_{4}\) .
|
IMO-1963-5
|
Prove that \(\cos \frac{\pi}{7} -\cos \frac{2\pi}{7} +\cos \frac{3\pi}{7} = \frac{1}{2}\) .
|
The LHS of the desired identity equals \(S = \cos (\pi /7) + \cos (3\pi /7) + \cos (5\pi /7)\) . Now
\[S\sin {\frac{\pi}{7}} = \frac{\sin{\frac{2\pi}{7}}}{2} +\frac{\sin{\frac{4\pi}{7}} - \sin{\frac{2\pi}{7}}}{2} +\frac{\sin{\frac{6\pi}{7}} - \sin{\frac{4\pi}{7}}}{2} = \frac{\sin{\frac{6\pi}{7}}}{2}\Rightarrow S = \frac{1}{2}.\]
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IMO-1963-6
|
Five students \(A,B,C,D,\) and \(E\) have taken part in a certain competition. Before the competition, two persons \(X\) and \(Y\) tried to guess the rankings. \(X\) thought that the ranking would be \(A,B,C,D,E\) ; and \(Y\) thought that the ranking would be \(D,A,E,C,B\) . At the end, it was revealed that \(X\) didn't guess correctly any rankings of the participants, and moreover, didn't guess any of the orderings of pairs of consecutive participants. On the other hand, \(Y\) guessed the correct rankings of two participants and the correct ordering of two pairs of consecutive participants. Determine the rankings of the competition.
|
The result is EDACB.
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IMO-1964-1
|
(a) Find all natural numbers \(n\) such that the number \(2^{n} - 1\) is divisible by 7. (b) Prove that for all natural numbers \(n\) the number \(2^{n} + 1\) is not divisible by 7.
|
Let \(n = 3k + r\) , where \(0\leq r< 2\) . Then \(2^{n} = 2^{3k + r} = 8^{k}\cdot 2^{r}\equiv 2^{r}\) (mod 7). Thus the remainder of \(2^{n}\) modulo 7 is 1, 2, 4 if \(n\equiv 0,1,2\) (mod 3). Hence \(2^{n} - 1\) is divisible by 7 if and only if 3 \(|n\) , while \(2^{n} + 1\) is never divisible by 7.
|
IMO-1964-2
|
Denote by \(a, b, c\) the lengths of the sides of a triangle. Prove that
\[a^{2}(b + c - a) + b^{2}(c + a - b) + c^{2}(a + b - c)\leq 3abc.\]
|
By substituting \(a = x + y\) , \(b = y + z\) , and \(c = z + x\) \((x,y,z > 0)\) the given inequality becomes
\[6xy z\leq x^{2}y + xy^{2} + y^{2}z + yz^{2} + z^{2}x + zx^{2},\]
which follows immediately by the AM- GM inequality applied to \(x^{2}y\) , \(xy^{2}\) , \(x^{2}z\) , \(xz^{2}\) , \(y^{2}z\) , \(yz^{2}\) .
|
IMO-1964-3
|
The incircle is inscribed in a triangle \(ABC\) with sides \(a, b, c\) . Three tangents to the incircle are drawn, each of which is parallel to one side of the triangle \(ABC\) . These tangents form three smaller triangles (internal to \(\triangle ABC\) ) with the sides of \(\triangle ABC\) . In each of these triangles an incircle is inscribed. Determine the sum of areas of all four incircles.
|
Let \(r\) be the radius of the incircle of \(\triangle ABC\) , \(r_{a},r_{b},r_{c}\) the radii of the smaller circles corresponding to \(A,B,C\) , and \(h_{a},h_{b},h_{c}\) the altitudes from \(A,B,C\) respectively. The coefficient of similarity between the smaller triangle at \(A\) and the triangle \(ABC\) is \(1 - 2r / h_{a}\) , from which we easily obtain \(r_{a} = (h_{a} - 2r)r / h_{a} = (s - a)r / s\) . Similarly, \(r_{b} = (s - b)r / s\) and \(r_{c} = (s - c)r / s\) . Now a straightforward computation gives that the sum of areas of the four circles is given by
\[\Sigma = \frac{(b + c - a)(c + a - b)(a + b - c)(a^{2} + b^{2} + c^{2})\pi}{(a + b + c)^{3}}.\]
|
IMO-1964-4
|
Each of 17 students talked with every other student. They all talked about three different topics. Each pair of students talked about one topic. Prove that there are three students that talked about the same topic among themselves.
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Let us call the topics \(T_{1},T_{2},T_{3}\) . Consider an arbitrary student \(A\) . By the pigeonhole principle there is a topic, say \(T_{3}\) , he discussed with at least 6 other students. If two of these 6 students discussed \(T_{3}\) , then we are done.
Suppose now that the 6 students discussed only \(T_{1}\) and \(T_{2}\) and choose one of them, say \(B\) . By the pigeonhole principle he discussed one of the topics, say \(T_{2}\) , with three of these students. If two of these three students also discussed \(T_{2}\) , then we are done. Otherwise, all the three students discussed only \(T_{1}\) , which completes the task.
|
IMO-1964-5
|
Five points are given in the plane. Among the lines that connect these five points, no two coincide and no two are parallel or perpendicular. Through each point we construct an altitude to each of the other lines. What is the maximal number of intersection points of these altitudes (excluding the initial five points)?
|
Let us first compute the number of intersection points of the perpendiculars passing through two distinct points \(B\) and \(C\) . The perpendiculars from \(B\) to the lines through \(C\) other than \(BC\) meet all perpendiculars from \(C\) , which counts to \(3\cdot 6 = 18\) intersection points. Each perpendicular from \(B\) to the 3 lines not containing \(C\) can intersect at most 5 of the perpendiculars passing through \(C\) , which counts to another \(3\cdot 5 = 15\) intersection points. Thus there are \(18 + 15 = 33\) intersection points corresponding to \(B,C\) .
It follows that the required total number is at most \(10\cdot 33 = 330\) . But some of these points, namely the orthocenters of the triangles with vertices at the given points, are counted thrice. There are 10 such points. Hence the maximal number of intersection points is \(330 - 2\cdot 10 = 310\) .
Remark. The jury considered only the combinatorial part of the problem and didn't require an example in which 310 points appear. However, it is "easily" verified that, for instance, the set of points \(A(1,1)\) , \(B(e,\pi)\) , \(C(e^{2},\pi^{2})\) , \(D(e^{3},\pi^{3})\) , \(E(e^{4},\pi^{4})\) works.
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IMO-1964-6
|
Given a tetrahedron \(ABCD\) , let \(D_1\) be the centroid of the triangle \(ABC\) and let \(A_1, B_1, C_1\) be the intersection points of the lines parallel to \(DD_1\) and passing through the points \(A, B, C\) with the opposite faces of the tetrahedron. Prove that the volume of the tetrahedron \(ABCD\) is one-third the volume of the tetrahedron \(A_1B_1C_1D_1\) . Does the result remain true if the point \(D_1\) is replaced with any point inside the triangle \(ABC\) ?
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We shall prove that the statement is valid in the general case, for an arbitrary point \(D_{1}\) inside \(\triangle ABC\) . Since \(D_{1}\) belongs to the plane \(ABC\) , there are real numbers \(a, b, c\) such that \((a + b + c)\overline{DD_1} = a\overline{DA} + b\overline{DB} + c\overline{DC}\) . Since \(AA_1 \parallel DD_1\) , it holds that \(\overline{AA_1} = k\overline{DD_1}\) for some \(k \in \mathbb{R}\) . Now it is easy to get \(\overline{DA_1} = -(b\overline{DB} + c\overline{DC}) / a\) , \(\overline{DB_1} = -(a\overline{DA} + c\overline{DC}) / b\) , and \(\overline{DC_1} = -(a\overline{DA} + b\overline{DB}) / c\) . This implies
\[\begin{array}{l}{\overrightarrow{D_1A_1} = -\frac{a^2\overrightarrow{DA} + b(a + 2b + c)\overrightarrow{DB} + c(a + b + 2c)\overrightarrow{DC}}{a(a + b + c)},}\\ {\overrightarrow{D_1B_1} = -\frac{a(2a + b + c)\overrightarrow{DA} + b^2\overrightarrow{DB} + c(a + b + 2c)\overrightarrow{DC}}{b(a + b + c)},}\\ {\overrightarrow{D_1C_1} = -\frac{a(2a + b + c)\overrightarrow{DA} + b(a + 2b + c)\overrightarrow{DB} + c^2\overrightarrow{DC}}{c(a + b + c)}} \end{array} \quad (D)\]
By using
\[6V_{D_1A_1B_1C_1} = \left|\left[\overrightarrow{D_1A_1},\overrightarrow{D_1B_1},\overrightarrow{D_1C_1}\right]\right| \quad \text{and} \quad 6V_{DABC} = \left|\left[\overrightarrow{DA},\overrightarrow{DB},\overrightarrow{DC}\right]\right|\]
we get
\[V_{D_1A_1B_1C_1} = \frac{\left|\left|\begin{array}{c c}{{a^{2}}}&{{b(a+2b+c)}}\\ {{a(2a+b+c)}}&{{b^{2}}}\end{array}\right|}\right|}{\left|\begin{array}{c c}{{a(2a+b+c)}}&{{b(2b+c)}}\\ {{a(2a+b+c)}}&{{b(2b+c)}}\end{array}\right|}\] \[\qquad \cdot \left|\left[\overrightarrow{DA},\overrightarrow{DB},\overrightarrow{DC}\right]\right|\] \[\qquad = 3V_{DABC}.\]
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IMO-1965-1
|
Find all real numbers \(x \in [0, 2\pi ]\) such that
\[2\cos x\leq |\sqrt{1 + \sin2x} -\sqrt{1 - \sin2x}|\leq \sqrt{2}.\]
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Let us set \(S = \left|\sqrt{1 + \sin2x} -\sqrt{1 - \sin2x}\right|\) . Notice that \(S^{2} = 2 - 2\sqrt{1 - \sin^{2}2x} =\) \(2 - 2|\cos 2x|\leq 2\) , implying \(S\leq \sqrt{2}\) . Thus the righthand inequality holds for all \(x\) .
It remains to investigate the left- hand inequality. If \(\pi /2\leq x\leq 3\pi /2\) , then \(\cos x\leq 0\) and the inequality trivially holds. Assume now that \(\cos x > 0\) . Then the inequality is equivalent to \(2 + 2\cos 2x = 4\cos^{2}x\leq S^{2} = 2 - 2|\cos 2x|\) , which is equivalent to \(\cos 2x\leq 0\) , i.e., to \(x\in [\pi /4,\pi /2]\cup [3\pi /2,7\pi /4]\) . Hence the solution set is \(\pi /4\leq x\leq 7\pi /4\) .
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IMO-1965-2
|
Consider the system of equations
\[
\begin{cases}
a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0,\\
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0,\\
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0.
\end{cases}
\]
whose coefficients satisfy the following conditions:
(a) \(a_{11}, a_{22}, a_{33}\) are positive real numbers;
(b) all other coefficients are negative;
(c) in each of the equations the sum of the coefficients is positive. Prove that \(x_1 = x_2 = x_3 = 0\) is the only solution to the system.
|
Suppose that \((x_{1},x_{2},x_{3})\) is a solution. We may assume w.l.o.g. that \(|x_{1}|\geq |x_{2}|\geq\) \(|x_{3}|\) . Suppose that \(|x_{1}| > 0\) . From the first equation we obtain that
\[0 = |x_{1}|\cdot \left|a_{11} + a_{12}\frac{x_{2}}{x_{1}} +a_{13}\frac{x_{3}}{x_{1}}\right|\geq |x_{1}|\cdot (a_{11} - |a_{12}| - |a_{13}|) > 0,\]
which is a contradiction. Hence \(|x_{1}| = 0\) and consequently \(x_{1} = x_{2} = x_{3} = 0\) .
|
IMO-1965-3
|
A tetrahedron \(ABCD\) is given. The lengths of the edges \(AB\) and \(CD\) are \(a\) and \(b\) , respectively, the distance between the lines \(AB\) and \(CD\) is \(d\) , and the angle between them is equal to \(\omega\) . The tetrahedron is divided into two parts by the plane \(\pi\) parallel to the lines \(AB\) and \(CD\) . Calculate the ratio of the volumes of the parts if the ratio between the distances of the plane \(\pi\) from \(AB\) and \(CD\) is equal to \(k\) .
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Let \(d\) denote the distance between the lines \(AB\) and \(CD\) . Being parallel to \(AB\) and \(CD\) , the plane \(\pi\) intersects the faces of the tetrahedron in a parallelogram \(EFGH\) . Let \(X\in AB\) be a point such that \(HX\parallel DB\) .
Clearly \(V_{AEHBFG} = V_{AXEH} + V_{XEHBFG}\) . Let \(MN\) be the common perpendicular to lines \(AB\) and \(CD\) ( \(M\in AB\) , \(N\in CD\) ) and let \(MN,BN\) meet the plane \(\pi\) at \(Q\) and \(R\) respectively. Then it holds that \(BR / RN = MQ / QN = k\) and consequently \(AX / XB = AE / EC = AH / HD = BF / FC = BG / GD = k\) . Now we have \(V_{AXEH} / V_{ABCD} = k^{3} / (k + 1)^{3}\) . Furthermore, if \(h = 3V_{ABCD} / S_{ABC}\) is the height of \(ABCD\) from \(D\) , then
\[V_{XEHBFG} = \frac{1}{2} S_{XBFE}\frac{k}{k + 1} h\mathrm{~and~}\]
\[S_{XBFE} = S_{ABC} - S_{AXE} - S_{EFC} = \frac{(k + 1)^{2} - 1 - k^{2}}{(k + 1)^{2}} = \frac{2k}{(1 + k)^{2}}.\]
These relations give us \(V_{XEHBFG} / V_{ABCD} = 3k^{2} / (1 + k)^{3}\) . Finally,
\[\frac{V_{AEHBFG}}{V_{ABCD}} = \frac{k^{3} + 3k^{2}}{(k + 1)^{3}}.\]
Similarly, \(V_{CEFDHG} / V_{ABCD} = (3k + 1) / (k + 1)^{3}\) , and hence the required ratio is \((k^{3} + 3k^{2}) / (3k + 1)\) .
|
IMO-1965-4
|
Find all sets of four real numbers \(x_1, x_2, x_3, x_4\) such that the sum of any of the numbers and the product of the other three is equal to 2.
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It is easy to see that all \(x_{i}\) are nonzero. Let \(x_{1}x_{2}x_{3}x_{4} = p\) . The given system of equations can be rewritten as \(x_{i} + p / x_{i} = 2\) , \(i = 1,2,3,4\) . The equation \(x + p / x = 2\) has at most two real solutions, say \(y\) and \(z\) . Then each \(x_{i}\) is equal either to \(y\) or to \(z\) . There are three cases:
(i) \(x_{1} = x_{2} = x_{3} = x_{4} = y\) . Then \(y + y^{3} = 2\) and hence \(y = 1\) . (ii) \(x_{1} = x_{2} = x_{3} = y\) , \(x_{4} = z\) . Then \(z + y^{3} = y + y^{2}z = 2\) . It is easy to obtain that the only possibilities for \((y,z)\) are \((-1,3)\) and \((1,1)\) . (iii) \(x_{1} = x_{2} = y\) , \(x_{3} = x_{4}\) . In this case the only possibility is \(y = z = 1\) . Hence the solutions for \((x_{1},x_{2},x_{3},x_{4})\) are \((1,1,1,1)\) , \((-1, - 1, - 1,3)\) , and the cyclic permutations.
|
IMO-1965-5
|
Given a triangle \(OAB\) such that \(\angle AOB = \alpha < 90^\circ\) , let \(M\) be an arbitrary point of the triangle different from \(O\) . Denote by \(P\) and \(Q\) the feet of the perpendiculars from \(M\) to \(OA\) and \(OB\) , respectively. Let \(H\) be the orthocenter of the triangle \(OPQ\) . Find the locus of points \(H\) when:
(a) \(M\) belongs to the segment \(AB\) ;
(b) \(M\) belongs to the interior of \(\triangle OAB\) .
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(a) Let \(A^{\prime}\) and \(B^{\prime}\) denote the feet of the perpendiculars from \(A\) and \(B\) to \(O B\) and \(O A\) respectively. We claim that \(H\in A^{\prime}B^{\prime}\) . Indeed, since \(M P H Q\) is a parallelogram, we have \(B^{\prime}P / B^{\prime}A = B M / B A = M Q / A A^{\prime} = P H / A A^{\prime}\) , which implies by Thales's theorem that \(H\in A^{\prime}B^{\prime}\) . It is easy to see that the locus of \(H\) is the whole segment \(A^{\prime}B^{\prime}\) .
(b) In this case the locus of points \(H\) is obviously the interior of the triangle \(O A^{\prime}B^{\prime}\) .
|
IMO-1965-6
|
We are given \(n \geq 3\) points in the plane. Let \(d\) be the maximal distance between two of the given points. Prove that the number of pairs of points whose distance is equal to \(d\) is less than or equal to \(n\) .
|
We recall the simple statement that every two diameters of a set must have a common point.
Consider any point \(B\) that is an endpoint of \(k\geq 2\) diameters \(B C_{1},B C_{2},\ldots ,B C_{k}\) We may assume w.l.o.g. that all the points \(C_{1},\ldots ,C_{k}\) lie on the arc \(C_{1}C_{k}\) , whose center is \(B\) and measure does not exceed \(60^{\circ}\) . We observe that for \(1< i< k\) any diameter with the endpoint \(C_{i}\) has to intersect both the diameters \(C_{1}B\) and \(C_{k}B\) Hence \(C_{i}B\) is the only diameter with an endpoint at \(C_{i}\) if \(i = 2,\ldots ,k - 1\) . In other words, with each point that is an endpoint of \(k\geq 2\) we can associate \(k - 2\) points that are endpoints of exactly one diameter.
We now assume that each \(A_{i}\) is an endpoint of exactly \(k_{i}\geq 0\) diameters, and that \(k_{1},\ldots ,k_{s}\geq 2\) , while \(k_{s + 1},\ldots ,k_{n}\leq 1\) . The total number \(D\) of diameters satisfies the inequality \(2D\leq k_{1} + k_{2} + \dots +k_{s} + (n - s)\) . On the other hand, by the above consideration we have \((k_{1} - 2) + \dots +(k_{s} - 2)\leq n - s\) , i.e., \(k_{1} + \dots +k_{s}\leq n + s\) . Hence \(2D\leq (n + s) + (n - s) = 2n\) , which proves the result.
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IMO-1966-1
|
Three problems \(A\) , \(B\) , and \(C\) were given on a mathematics olympiad. All 25 students solved at least one of these problems. The number of students who solved \(B\) and not \(A\) is twice the number of students who solved \(C\) and not \(A\) . The number of students who solved only \(A\) is greater by 1 than the number of students who along with \(A\) solved at least one other problem. Among the students who solved only one problem, half solved \(A\) . How many students solved only \(B\) ?
|
Let \(N_{a},N_{b},N_{c},N_{ab},N_{ac},N_{bc},N_{abc}\) denote the number of students who solved exactly the problems whose letters are stated in the index of the variable. From the conditions of the problem we have
\[N_{a} + N_{b} + N_{c} + N_{ab} + N_{bc} + N_{ac} + N_{abc} = 25,\]
\[N_{b} + N_{bc} = 2(N_{c} + N_{bc}),\qquad N_{a} - 1 = N_{ac} + N_{abc} + N_{ab},\qquad N_{a} = N_{b} + N_{c}.\]
From the first and third equations we get \(2N_{a} + N_{b} + N_{c} + N_{bc} = 26\) , and from the second and fourth we get \(4N_{b} + N_{c} = 26\) and thus \(N_{b}\leq 6\) . On the other hand, we have from the second equation \(N_{b} = 2N_{c} + N_{bc}\Rightarrow N_{c}\leq N_{b} / 2\Rightarrow 26\leq 9N_{b} / 2\Rightarrow\) \(N_{b}\geq 6\) ; hence \(N_{b} = 6\)
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IMO-1966-2
|
If \(a\) , \(b\) , and \(c\) are the sides and \(\alpha\) , \(\beta\) , and \(\gamma\) the respective angles of the triangle for which \(a + b = \tan \frac{\gamma}{2} (a\tan \alpha + b\tan \beta)\) , prove that the triangle is isosceles.
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Angles \(\alpha\) and \(\beta\) are less than \(90^{\circ}\) , otherwise if w.l.o.g. \(\alpha \geq 90^{\circ}\) we have \(\tan (\gamma /2)\cdot (a\tan \alpha +b\tan \beta)< b\tan (\gamma /2)\tan \beta \leq b\tan (\gamma /2)\cot (\gamma /2) = b< a+\) \(b\) . Since \(a\geq b\Leftrightarrow \tan a\geq \tan b\) , Chebyshev's inequality gives \(a\tan \alpha +b\tan \beta \geq\) \((a + b)(\tan \alpha +\tan \beta) / 2\) . Due to the convexity of the tan function we also have \((\tan \alpha +\tan \beta) / 2\geq \tan [(\alpha +\beta) / 2] = \cot (\gamma /2)\) . Hence we have
\[\tan {\frac{\gamma}{2}} (a\tan \alpha +b\tan \beta)\geq {\frac{1}{2}}\tan {\frac{\gamma}{2}} (a + b)(\tan \alpha +\tan \beta)\] \[\geq \tan {\frac{\gamma}{2}} (a + b)\cot {\frac{\gamma}{2}} = a + b.\]
The equalities can hold only if \(a = b\) . Thus the triangle is isosceles.
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IMO-1966-3
|
Prove that the sum of distances from the center of the circumsphere of the regular tetrahedron to its four vertices is less than the sum of distances from any other point to the four vertices.
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Consider a coordinate system in which the points of the regular tetrahedron are placed at \(A(-a, -a, -a)\) , \(B(-a, a, a)\) , \(C(a, -a, a)\) and \(D(a, a, -a)\) . Then the center of the tetrahedron is at \(O(0,0,0)\) . For a point \(X(x,y,z)\) we see that the sum \(XA + XB + XC + XD\) by the QM-AM inequality does not exceed \(2\sqrt{XA^2 + XB^2 + XC^2 + XD^2}\) . Now, since \(XA^2 = (x + a)^2 + (y + a)^2 + (z + a)^2\) etc., we easily obtain
\[X A^{2} + X B^{2} + X C^{2} + X D^{2} = 4(x^{2} + y^{2} + z^{2}) + 12a^{2}\] \[\geq 12a^{2} = O A^{2} + O B^{2} + O C^{2} + O D^{2}.\]
Hence \(XA + XB + XC + XD\geq 2\sqrt{OA^2 + OB^2 + OC^2 + OD^2} = OA + OB + OC+\) \(OD\)
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IMO-1966-4
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Prove the following equality:
\[\frac{1}{\sin 2x} +\frac{1}{\sin 4x} +\frac{1}{\sin 8x} +\dots +\frac{1}{\sin 2^n x} = \cot x - \cot 2^n x,\]
where \(n\in \mathbb{N}\) and \(x\notin \frac{\pi}{2k}\mathbb{Z}\) for every \(k\in \mathbb{N}\)
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It suffices to prove \(1 / \sin 2^{k}x = \cot 2^{k - 1}x - \cot 2^{k}x\) for any integer \(k\) and real \(x\) , i.e., \(1 / \sin 2x = \cot x - \cot 2x\) for all real \(x\) . We indeed have
\[\cot x - \cot 2x = \cot x - \frac{\cot^2x - 1}{2\cot x} = \frac{\left(\frac{\cos x}{\sin x}\right)^2 + 1}{2\frac{\cos x}{\sin x}} = \frac{1}{2\sin x\cos x} = \frac{1}{\sin 2x}.\]
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IMO-1966-5
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Solve the following 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 \(a_1\) , \(a_2\) , \(a_3\) , and \(a_4\) are mutually distinct real numbers.
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We define \(L_{1} = |a_{1} - a_{2}|x_{2} + |a_{1} - a_{3}|x_{3} + |a_{1} - a_{4}|x_{4}\) and analogously \(L_{2}, L_{3}\) , and \(L_{4}\) . Let us assume w.l.o.g. that \(a_{1} < a_{2} < a_{3} < a_{4}\) . In that case,
\[2|a_{1} - a_{2}||a_{2} - a_{3}|x_{2} = |a_{3} - a_{2}|L_{1} - |a_{1} - a_{3}|L_{2} + |a_{1} - a_{2}|L_{3}\] \[= |a_{3} - a_{2}| - |a_{1} - a_{3}| + |a_{1} - a_{2}| = 0,\] \[2|a_{2} - a_{3}||a_{3} - a_{4}|x_{3} = |a_{4} - a_{3}|L_{2} - |a_{2} - a_{4}|L_{3} + |a_{2} - a_{3}|L_{4}\] \[= |a_{4} - a_{3}| - |a_{2} - a_{4}| + |a_{2} - a_{3}| = 0.\]
Hence it follows that \(x_{2} = x_{3} = 0\) and consequently \(x_{1} = x_{4} = 1 / |a_{1} - a_{4}|\) . This solution set indeed satisfies the starting equations. It is easy to generalize this result to any ordering of \(a_{1}, a_{2}, a_{3}, a_{4}\) .
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IMO-1966-6
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Let \(M\) , \(K\) , and \(L\) be points on \((AB)\) , \((BC)\) , and \((CA)\) , respectively. Prove that the area of at least one of the three triangles \(\triangle MAL\) , \(\triangle KBM\) , and \(\triangle LCK\) is less than or equal to one-fourth the area of \(\triangle ABC\) .
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Let \(S\) denote the area of \(\triangle ABC\) . Let \(A_{1}, B_{1}, C_{1}\) be the midpoints of \(BC, AC, AB\) respectively. We note that \(S_{A_{1}B_{1}C} = S_{A_{1}B_{1}C_{1}} = S_{A_{1}B_{1}C_{1}} = S / 4\) . Let us assume w.l.o.g. that \(M \in [AC_{1}]\) . We then must have \(K \in [BA_{1}]\) and \(L \in [CB_{1}]\) . However, we then have \(S(KLM) > S(KLC_{1}) > S(KB_{1}C_{1}) = S(A_{1}B_{1}C_{1}) = S / 4\) . Hence, by the pigeonhole principle one of the remaining three triangles \(\triangle MAL, \triangle KBM\) , and \(\triangle LCK\) must have an area less than or equal to \(S / 4\) . This completes the proof.
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IMOLL-1967-1
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Prove that all numbers in the sequence
\[{\frac{107811}{3}},\quad{\frac{110778111}{3}},\quad{\frac{111077781111}{3}},\quad\dots\]
are perfect cubes.
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Let us denote the \(n\) th term of the given sequence by \(a_{n}\) . Then
\[a_{n} = \frac{1}{3}\left(\frac{10^{3n + 3} - 10^{2n + 3}}{9} +\frac{7}{9}\frac{10^{2n + 2} - 10^{n + 1}}{9} +\frac{10^{n + 2} - 1}{9}\right)\] \[\qquad = \frac{1}{27} (10^{3n + 3} - 3\cdot 10^{2n + 2} + 3\cdot 10^{n + 1} - 1) = \left(\frac{10^{n + 1} - 1}{3}\right)^{3}.\]
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IMOLL-1967-2
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Prove that \(\frac {1}{3}n^{2}+\frac {1}{2}n+\frac {1}{6}\geq (n!)^{2/n}\) ( \(n\) is a positive integer) and that equality is possible only in the case \(n=1.\)
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\((n!)^{2 / n} = ((1\cdot 2\dots n)^{1 / n})^{2}\leq (\frac{1 + 2 + \dots + n}{n})^{2} = (\frac{n + 1}{2})^{2}\leq \frac{1}{3} n^{2} + \frac{1}{2} n + \frac{1}{6}.\)
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IMOLL-1967-3
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Prove the trigonometric inequality \(\cos x<1-\frac {x^{2}}{2}+\frac {x^{4}}{16}\) , where \(x\in (0,\pi /2).\)
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Consider the function \(f:[0,\pi /2]\to \mathbb{R}\) defined by \(f(x) = 1 - x^{2} / 2 + x^{4} / 16-\) cosx.
It is easy to calculate that \(f^{\prime}(0) = f^{\prime \prime}(0) = f^{\prime \prime \prime}(0) = 0\) and \(f^{\prime \prime \prime \prime}(x) = 3 / 2 - \cos x\) Since \(f^{\prime \prime \prime \prime}(x) > 0\) \(f^{\prime \prime \prime}(x)\) is increasing. Together with \(f^{\prime \prime \prime}(0) = 0\) , this gives \(f^{\prime \prime \prime}(x) > 0\) for \(x > 0\) ; hence \(f^{\prime \prime}(x)\) is increasing, etc. Continuing in the same way we easily conclude that \(f(x) > 0\)
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IMOLL-1967-4
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Suppose medians \(m_{a}\) and \(m_{b}\) of a triangle are orthogonal. Prove that:
(a) The medians of that triangle correspond to the sides of a right-angled triangle.
(b) The inequality
\[5(a^{2}+b^{2}-c^{2})\geq 8ab\]
is valid, where \(a\), \(b\), and \(c\) are side lengths of the given triangle.
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(a) Let ABCD be a parallelogram, and \(K,L\) the midpoints of segments \(BC\) and \(CD\) respectively. The sides of \(\triangle AKL\) are equal and parallel to the medians of \(\triangle ABC\) .
(b) Using the formulas \(4m_{a}^{2} = 2b^{2} + 2c^{2} - a^{2}\) etc., it is easy to obtain that \(m_{a}^{2} + m_{b}^{2} = m_{c}^{2}\) is equivalent to \(a^{2} + b^{2} = 5c^{2}\) . Then
\[5(a^{2} + b^{2} - c^{2}) = 4(a^{2} + b^{2})\geq 8ab.\]
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IMOLL-1967-5
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Solve the system
\[\begin{array}{l}x^{2}+x-1=y,\\ y^{2}+y-1=z,\\ z^{2}+z-1=x. \end{array}\]
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If one of \(x,y,z\) is equal to 1 or \(-1\) , then we obtain solutions \((-1, - 1, - 1)\) and \((1,1,1)\) . We claim that these are the only solutions to the system.
Let \(f(t) = t^{2} + t - 1\) . If among \(x,y,z\) one is greater than 1, say \(x > 1\) , we have \(x< f(x) = y< f(y) = z< f(z) = x\) , which is impossible. It follows that \(x,y,z\leq 1\) .
Suppose now that one of \(x,y,z\) , say \(x\) , is less than \(- 1\) . Since \(\min_{t}f(t) = - 5 / 4\) , we have \(x = f(z)\in [- 5 / 4, - 1)\) . Also, since \(f([- 5 / 4, - 1)) = (- 1, - 11 / 16)\subseteq\) \((- 1,0)\) and \(f((- 1,0)) = [- 5 / 4, - 1)\) , it follows that \(y = f(x)\in (- 1,0)\) , \(z =\) \(f(y)\in [- 5 / 4, - 1)\) , and \(x = f(z)\in (- 1,0)\) , which is a contradiction. Therefore \(- 1\leq x,y,z\leq 1\) .
If \(- 1< x,y,z< 1\) , then \(x > f(x) = y > f(y) = z > f(z) = x\) , a contradiction. This proves our claim.
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IMOLL-1967-6
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Solve the system
\[\begin{array}{l}|x+y|+|1-x|=6,\\ |x+y+1|+|1-y|=4. \end{array}\]
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The given system has two solutions: \((-2, - 1)\) and \((-14 / 3,13 / 3)\) .
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IMOLL-1967-7
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Find all real solutions of the system of equations
\[\begin{array}{l}x_{1}+x_{2}+\cdots +x_{n}=a,\\ x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}=a^{2},\\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \end{array}\]
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Let \(S_{k} = x_{1}^{k} + x_{2}^{k} + \dots +x_{n}^{k}\) and let \(\sigma_{k}\) , \(k = 1,2,\dots ,n\) denote the \(k\) th elementary symmetric polynomial in \(x_{1},\ldots ,x_{n}\) . The given system can be written as \(S_{k} = a^{k}\) , \(k = 1,\ldots ,n\) . Using Newton's formulas
\[k\sigma_{k} = S_{1}\sigma_{k - 1} - S_{2}\sigma_{k - 2} + \dots +(-1)^{k}S_{k - 1}\sigma_{1} + (-1)^{k - 1}S_{k},\quad k = 1,2,\dots ,n,\]
the system easily leads to \(\sigma_{1} = a\) and \(\sigma_{k} = 0\) for \(k = 2,\ldots ,n\) . By Vieta's formulas, \(x_{1},x_{2},\ldots ,x_{n}\) are the roots of the polynomial \(x^{n} - ax^{n - 1}\) , i.e., \(a,0,0,\ldots ,0\) in some order.
Remark. This solution does not use the assumption that the \(x_{j}\) 's are real.
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IMOLL-1967-8
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\(ABCD\) is a parallelogram; \(AB=a\), \(AD=1\), \(\alpha\) is the size of \(\angle DAB\), and the three angles of the triangle \(ABD\) are acute. Prove that the four circles \(K_{A}\), \(K_{B}\), \(K_{C}\), \(K_{D}\), each of radius 1, whose centers are the vertices \(A\), \(B\), \(C\), \(D\), cover the parallelogram if and only if \(a\leq \cos \alpha +\sqrt {3}\sin \alpha\).
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The circles \(K_{A},K_{B},K_{C},K_{D}\) cover the parallelogram if and only if for every point \(X\) inside the parallelogram, the length of one of the segments \(XA,XB,XC,XD\) does not exceed 1.
Let \(O\) and \(r\) be the center and radius of the circumcircle of \(\triangle ABD\) . For every point \(X\) inside \(\triangle ABD\) , it holds that \(XA \leq r\) or \(XB \leq r\) or \(XD \leq r\) . Similarly, for \(X\) inside \(\triangle BCD\) , \(XB \leq r\) or \(XC \leq r\) or \(XD \leq r\) . Hence \(K_{A},K_{B},K_{C},K_{D}\) cover the parallelogram if and only if \(r \leq 1\) , which is equivalent to \(\angle ABD \geq 30^{\circ}\) . However, this last is exactly equivalent to \(a = AB = 2r\sin \angle ADB \leq 2\sin (\alpha + 30^{\circ}) = \sqrt{3}\sin \alpha +\cos \alpha\) .
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IMOLL-1967-9
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The circle \(k\) and its diameter \(AB\) are given. Find the locus of the centers of circles inscribed in the triangles having one vertex on \(AB\) and two other vertices on \(k\).
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The incenter of any such triangle lies inside the circle \(k\) . We shall show that every point \(S\) interior to the circle \(S\) is the incenter of one such triangle. If \(S\) lies on the segment \(AB\) , then it is obviously the incenter of an isosceles triangle inscribed in \(k\) that has \(AB\) as an axis of symmetry. Let us now suppose \(S\) does not lie on \(AB\) . Let \(X\) and \(Y\) be the intersection points of lines \(AS\) and \(BS\) with \(k\) , and let \(Z\) be the foot of the perpendicular from \(S\) to \(AB\) . Since the quadrilateral \(BZSY\) is cyclic, we have \(\angle ZXS = \angle ABS = \angle SXY\) and analogously \(\angle ZYS = \angle SYX\) , which implies that \(S\) is the incenter of \(\triangle XYZ\) .
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IMOLL-1967-10
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The square \(ABCD\) is to be decomposed into \(n\) triangles (nonoverlapping) all of whose angles are acute. Find the smallest integer \(n\) for which there exists a solution to this problem and construct at least one decomposition for this \(n\). Answer whether it is possible to ask additionally that (at least) one of these triangles has a perimeter less than an arbitrarily given positive number.
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Let \(n\) be the number of triangles and let \(b\) and \(i\) be the numbers of vertices on the boundary and in the interior of the square, respectively.
Since all the triangles are acute, each of the vertices of the square belongs to at least two triangles. Additionally, every vertex on the boundary belongs to at least three, and every vertex in the interior belongs to at least five triangles. Therefore
\[3n\geq 8 + 3b + 5i. \quad (1)\]
Moreover, the sum of angles at any vertex that lies in the interior, on the boundary, or at a vertex of the square is equal to \(2\pi ,\pi ,\pi /2\) respectively. The sum of all angles of the triangles equals \(n\pi\) , which gives us \(n\pi = 4\cdot \pi /2 + b\pi +\) \(2i\pi\) , i.e., \(n = 2 + b + 2i\) . This relation together with (1) easily yields that \(i\geq\) 2. Since each of the vertices inside the square belongs to at least five triangles, and at most two contain both, it follows that \(n\geq 8\)
It is shown in the figure that the square can be decomposed into eight acute triangles. Obviously one of them can have an arbitrarily small perimeter.
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IMOLL-1967-11
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Let \(n\) be a positive integer. Find the maximal number of noncongruent triangles whose side lengths are integers less than or equal to \(n\) .
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We have to find the number \(p_n\) of triples of positive integers \((a,b,c)\) satisfying \(a \leq b \leq c \leq n\) and \(a + b > c\) . Let us denote by \(p_n(k)\) the number of such triples with \(c = k\) , \(k = 1,2,\ldots ,n\) . For \(k\) even, \(p_n(k) = k + (k - 2) + (k - 4) + \dots + 2 = (k^2 + 2k) / 4\) , and for \(k\) odd, \(p_n(k) = (k^2 + 2k + 1) / 4\) . Hence
\[p_{n} = p_{n}(1) + p_{n}(2) + \dots +p_{n}(n) = \left\{ \begin{array}{ll}n(n + 2)(2n + 5) / 24, & \mathrm{for} 2\mid n,\] \[(n + 1)(n + 3)(2n + 1) / 24, & \mathrm{for} 2\nmid n. \end{array} \right.\]
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IMOLL-1967-12
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Given a segment \(AB\) of the length 1, define the set \(M\) of points in the following way: it contains the two points \(A, B\) , and also all points obtained from \(A, B\) by iterating the following rule: for every pair of points \(X, Y\) in \(M\) , the set \(M\) also contains the point \(Z\) of the segment \(XY\) for which \(YZ = 3XZ\) .
(a) Prove that the set \(M\) consists of points \(X\) from the segment \(AB\) for which the distance from the point \(A\) is either
\[AX = \frac{3k}{4^n}\quad \mathrm{or}\quad AX = \frac{3k - 2}{4^n},\]
where \(n, k\) are nonnegative integers.
(b) Prove that the point \(X_0\) for which \(AX_0 = 1 / 2 = X_0B\) does not belong to the set \(M\) .
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Let us denote by \(M_n\) the set of points of the segment \(AB\) obtained from \(A\) and \(B\) by not more than \(n\) iterations of \((\ast)\) . It can be proved by induction that
\[M_{n} = \left\{X\in AB\mid AX = \frac{3k}{4^{n}}\mathrm{or}\frac{3k - 2}{4^{n}}\mathrm{for~some~}k\in \mathbb{N}\right\} .\]
Thus (a) immediately follows from \(M = \bigcup M_n\) . It also follows that if \(a,b\in \mathbb{N}\) and \(a / b\in M\) , then 3 \(|a(b - a)\) . Therefore \(1 / 2\notin M\) .
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IMOLL-1967-13
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Find whether among all quadrilaterals whose interiors lie inside a semicircle of radius \(r\) there exists one (or more) with maximal area. If so, determine their shape and area.
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The maximum area is \(3\sqrt{3} r^2 /4\) (where \(r\) is the radius of the semicircle) and is attained in the case of a trapezoid with two vertices at the endpoints of the diameter of the semicircle and the other two vertices dividing the semicircle into three equal arcs.
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IMOLL-1967-14
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Which fraction \(p / q\) , where \(p, q\) are positive integers less than 100, is closest to \(\sqrt{2}\) ? Find all digits after the decimal point in the decimal representation of this fraction that coincide with digits in the decimal representation of \(\sqrt{2}\) (without using any tables).
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We have that
\[\left|\frac{p}{q} -\sqrt{2}\right| = \frac{|p - q\sqrt{2}|}{q} = \frac{|p^2 - 2q^2|}{q(p + q\sqrt{2})}\geq \frac{1}{q(p + q\sqrt{2})}, \quad (1)\]
because \(|p^2 - 2q^2 | \geq 1\) .
The greatest \(p,q\leq 100\) satisfying the equation \(|p^2 - 2q^2 | = 1\) are \((p,q) =\) (99,70). It is easy to verify using (1) that \(\frac{99}{70}\) best approximates \(\sqrt{2}\) among the fractions \(p / q\) with \(p,q\leq 100\) . The numbers \(\frac{99}{70} = 1.41428\ldots\) and \(\sqrt{2}\) coincide up to the fourth decimal digit: indeed, (1) gives \(7\cdot 10^{- 5}< \frac{p}{q} -\sqrt{2} < 8\cdot 10^{- 5}\)
Second solution. By using some basic facts about Farey sequences, one can find that \(\frac{41}{29} < \sqrt{2} < \frac{99}{70}\) and that \(\frac{41}{29} < \frac{p}{q} < \frac{99}{70}\) implies \(p\geq 41 + 99 > 100\) because \(99\cdot 29 - 41\cdot 70 = 1\) . Of the two fractions \(\frac{41}{29}\) and \(\frac{99}{70}\) , the latter is closer to \(\sqrt{2}\) .
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IMOLL-1967-15
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Suppose \(\tan \alpha = p / q\) , where \(p\) and \(q\) are integers and \(q \neq 0\) . Prove that the number \(\tan \beta\) for which \(\tan 2\beta = \tan 3\alpha\) is rational only when \(p^2 + q^2\) is the square of an integer.
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Given that \(\tan \alpha \in \mathbb{Q}\) , we have that \(\tan \beta\) is rational if and only if \(\tan \gamma\) is rational, where \(\gamma = \beta - \alpha\) and \(2\gamma = \alpha\) . Putting \(t = \tan \gamma\) we obtain \(\frac{p}{q} = \tan 2\gamma = \frac{2t}{1 - t^2}\) , which leads to the quadratic equation \(pt^2 + 2qt - p = 0\) . This equation has rational solutions if and only if its discriminant \(4(p^2 + q^2)\) is a perfect square, and the result follows.
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IMOLL-1967-16
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Prove the following statement: If \(r_1\) and \(r_2\) are real numbers whose quotient is irrational, then any real number \(x\) can be approximated arbitrarily well by numbers of the form \(z_{k_1, k_2} = k_1 r_1 + k_2 r_2\) , \(k_1, k_2\) integers; i.e., for every real number \(x\) and every positive real number \(p\) two integers \(k_1\) and \(k_2\) can be found such that \(|x - (k_1 r_1 + k_2 r_2)| < p\) .
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First let us notice that all the numbers \(z_{m_1,m_2} = m_1r_1 + m_2r_2\) \((m_1,m_2\in \mathbb{Z})\) are distinct, since \(r_1 / r_2\) is irrational. Thus for any \(n\in \mathbb{N}\) the interval \([- n(|r_1| + |r_2|),n(|r_1| + |r_2|)]\) contains \((2n + 1)^2\) numbers \(z_{m_1,m_2}\) , where \(|m_1|,|m_2|\leq n\) . Therefore some two of these \((2n + 1)^2\) numbers, say \(z_{m_1,m_2},z_{n_1,n_2}\) , differ by at most \(\frac{2n(|r_1| + |r_2|)}{(2n + 1)^2 - 1} = \frac{(|r_1| + |r_2|)}{2(n + 1)}\) . By taking \(n\) large enough we can achieve that \(z_{q_1,q_2} = |z_{m_1,m_2} - z_{n_1,n_2}|\leq p\) . If now \(k\) is the integer such that \(kz_{q_1,q_2}\leq x< (k + 1)z_{q_1,q_2}\) , then \(z_{kq_1,kq_2} = kz_{q_1,q_2}\) differs from \(x\) by at most \(p\) , as desired.
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IMOLL-1967-17
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Let \(k, m\) , and \(n\) be positive integers such that \(m + k + 1\) is a prime number greater than \(n + 1\) . Write \(c_s\) for \(s(s + 1)\) . Prove that the product \((c_{m+1} - c_k)(c_{m+2} - c_k) \dots (c_{m+n} - c_k)\) is divisible by the product \(c_1 c_2 \dots c_n\) .
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Using \(c_{r} - c_{s} = (r - s)(r + s + 1)\) we can easily get
\[\frac{(c_{m + 1} - c_{k})\cdots(c_{m + n} - c_{k})}{c_{1}c_{2}\cdots c_{n}} = \frac{(m - k + n)!}{(m - k)!n!}\cdot \frac{(m + k + n + 1)!}{(m + k + 1)!(n + 1)!}.\]
The first factor \(\frac{(m - k + n)!}{(m - k)!n!} = \binom{m - k + n}{n}\) is clearly an integer. The second factor is also an integer because by the assumption, \(m + k + 1\) and \((m + k)!(n + 1)!\) are coprime, and \((m + k + n + 1)!\) is divisible by both; hence it is also divisible by their product.
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IMOLL-1967-18
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If \(x\) is a positive rational number, show that \(x\) can be uniquely expressed in the form
\[x = a_1 + \frac{a_2}{2!} + \frac{a_3}{3!} + \dots ,\]
where \(a_1, a_2, \dots\) are integers, \(0 \leq a_n \leq n - 1\) for \(n > 1\) , and the series terminates. Show also that \(x\) can be expressed as the sum of reciprocals of different integers, each of which is greater than \(10^6\) .
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In the first part, it is sufficient to show that each rational number of the form \(m / n!\) , \(m, n \in \mathbb{N}\) , can be written uniquely in the required form. We prove this by induction on \(n\) .
The statement is trivial for \(n = 1\) . Let us assume it holds for \(n - 1\) , and let there be given a rational number \(m / n!\) . Let us take \(a_{n} \in \{0, \ldots , n - 1\}\) such that \(m - a_{n} = m n_{1}\) for some \(m_{1} \in \mathbb{N}\) . By the inductive hypothesis, there are unique \(a_{1} \in \mathbb{N}_{0}\) , \(a_{i} \in \{0, \ldots , i - 1\} (i = 1, \ldots , n - 1)\) such that \(m_{1} / (n - 1)! = \sum_{i = 1}^{n - 1} a_{i} / i!\) , and then
\[\frac{m}{n!} = \frac{m_{1}}{(n - 1)!} +\frac{a_{n}}{n!} = \sum_{i = 1}^{n}\frac{a_{i}}{i!},\]
as desired. On the other hand, if \(m / n! = \sum_{i = 1}^{n} a_{i} / i!\) , multiplying by \(n!\) we see that \(m - a_{n}\) must be a multiple of \(n\) , so the choice of \(a_{n}\) was unique and therefore the representation itself. This completes the induction.
In particular, since \(a_{i} \mid i!\) and \(i! / a_{i} > (i - 1)! \geq (i - 1)! / a_{i - 1}\) , we conclude that each rational \(q\) , \(0 < q < 1\) , can be written as the sum of different reciprocals.
Now we prove the second part. Let \(x > 0\) be a rational number. For any integer \(m > 10^{6}\) , let \(n > m\) be the greatest integer such that \(y = x - \frac{1}{m} - \frac{1}{m + 1} - \dots - \frac{1}{n} > 0\) . Then \(y\) can be written as the sum of reciprocals of different positive integers, which all must be greater than \(n\) . The result follows immediately.
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IMOLL-1967-19
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The \(n\) points \(P_1, P_2, \dots, P_n\) are placed inside or on the boundary of a disk of radius 1 in such a way that the minimum distance \(d_n\) between any two
of these points has its largest possible value \(D_{n}\) . Calculate \(D_{n}\) for \(n = 2\) to 7 and justify your answer.
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Suppose \(n \leq 6\) . Let us decompose the disk by its radii into \(n\) congruent regions, so that one of the points \(P_{j}\) lies on the boundaries of two of these regions. Then one of these regions contains two of the \(n\) given points. Since the diameter of each of these regions is \(2 \sin \frac{\pi}{n}\) , we have \(d_{n} \leq 2 \sin \frac{\pi}{n}\) . This value is attained if \(P_{i}\) are the vertices of a regular \(n\) -gon inscribed in the boundary circle. Hence \(D_{n} = 2 \sin \frac{\pi}{n}\) .
For \(n = 7\) we have \(D_{7} \leq D_{6} = 1\) . This value is attained if six of the seven points form a regular hexagon inscribed in the boundary circle and the seventh is at the center. Hence \(D_{7} = 1\) .
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IMOLL-1967-20
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In space, \(n\) points \((n \geq 3)\) are given. Every pair of points determines some distance. Suppose all distances are different. Connect every point with the nearest point. Prove that it is impossible to obtain a polygonal line in such a way. 1
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The statement so formulated is false. It would be true under the additional assumption that the polygonal line is closed. However, from the offered solution, which is not clear, it does not seem that the proposer had this in mind.
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IMOLL-1967-21
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Without using any tables, find the exact value of the product
\[P = \cos {\frac{\pi}{15}}\cos {\frac{2\pi}{15}}\cos {\frac{3\pi}{15}}\cos {\frac{4\pi}{15}}\cos {\frac{5\pi}{15}}\cos {\frac{6\pi}{15}}\cos {\frac{7\pi}{15}}.\]
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Using the formula \(\cos x \cos 2x \cos 4x \dots \cos 2^{n-1}x = \frac{\sin 2^{n}x}{2^{n}\sin x}\) , which is shown by simple induction, we obtain
\[\cos \frac{\pi}{15}\cos \frac{2\pi}{15}\cos \frac{4\pi}{15}\cos \frac{7\pi}{15} = -\cos \frac{\pi}{15}\cos \frac{2\pi}{15}\cos \frac{4\pi}{15}\cos \frac{8\pi}{15} = \frac{1}{16},\]
\[\cos {\frac{3\pi}{15}}\cos {\frac{6\pi}{15}} = \frac{1}{4},\qquad \cos {\frac{5\pi}{15}} = \frac{1}{2}.\]
Multiplying these equalities, we get that the required product \(P\) equals \(1 / 128\) .
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IMOLL-1967-22
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The distance between the centers of the circles \(k_{1}\) and \(k_{2}\) with radii \(r\) is equal to \(r\) . Points \(A\) and \(B\) are on the circle \(k_{1}\) , symmetric with respect to the line connecting the centers of the circles. Point \(P\) is an arbitrary point on \(k_{2}\) . Prove that
\[PA^{2} + PB^{2} \geq 2r^{2}.\]
When does equality hold?
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Let \(O_{1}\) and \(O_{2}\) be the centers of circles \(k_{1}\) and \(k_{2}\) and let \(C\) be the midpoint of the segment \(AB\) . Using the well-known relation for elements of a triangle, we obtain
\[PA^{2} + PB^{2} = 2PC^{2} + 2CA^{2}\geq 2O_{1}C^{2} + 2CA^{2} = 2O_{1}A^{2} = 2r^{2}.\]
Equality holds if \(P\) coincides with \(O_{1}\) or if \(A\) and \(B\) coincide with \(O_{2}\) .
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IMOLL-1967-23
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Prove that for an arbitrary pair of vectors \(f\) and \(g\) in the plane, the inequality
\[af^{2} + bfg + cg^{2} \geq 0\]
holds if and only if the following conditions are fulfilled: \(a \geq 0\) , \(c \geq 0\) , \(4ac \geq b^{2}\) .
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Suppose that \(a\geq 0\) , \(c\geq 0\) , \(4ac\geq b^{2}\) . If \(a = 0\) , then \(b = 0\) , and the inequality reduces to the obvious \(cg^{2}\geq 0\) . Also, if \(a > 0\) , then
\[a f^{2} + b f g + c g^{2} = a\left(f + \frac{b}{2a} g\right)^{2} + \frac{4a c - b^{2}}{4a} g^{2}\geq 0.\]
Suppose now that \(a f^{2} + b f g + c g^{2}\geq 0\) holds for an arbitrary pair of vectors \(f,g\) . Substituting \(f\) by \(t g\) ( \(t\in \mathbb{R}\) ) we get that \((a t^{2} + b t + c)g^{2}\geq 0\) holds for any real number \(t\) . Therefore \(a\geq 0\) , \(c\geq 0\) , \(4ac\geq b^{2}\) .
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IMOLL-1967-24
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Father has left to his children several identical gold coins. According to his will, the oldest child receives one coin and one-seventh of the remaining coins, the next child receives two coins and one-seventh of the remaining coins, the third child receives three coins and one-seventh of the remaining coins, and so on through the youngest child. If every child inherits an integer number of coins, find the number of children and the number of coins.
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Let \(m\) be the total number of coins and suppose that the \(k\) th child receive \(x_{k}\) coins. By the condition of the problem, the number of coins that remain after him was \(6(x_{k} - k)\) . This gives us a recurrence relation
\[x_{k + 1} = k + 1 + \frac{6(x_{k} - k) - k - 1}{7} = \frac{6}{7} x_{k} + \frac{6}{7},\]
which, together with the condition \(x_{1} = 1 + (m - 1) / 7\) , yields
\[x_{k} = \frac{6^{k - 1}}{7^{k}} (m - 36) + 6 \mathrm{for} 1\leq k\leq n.\]
Since we are given \(x_{n} = n\) , we obtain \(6^{n - 1}(m - 36) = 7^{n}(n - 6)\) . It follows that \(6^{n - 1} \mid n - 6\) , which is possible only for \(n = 6\) . Hence, \(n = 6\) and \(m = 36\) .
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IMOLL-1967-25
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Three disks of diameter \(d\) are touching a sphere at their centers. Moreover, each disk touches the other two disks. How do we choose the radius \(R\) of the sphere so that the axis of the whole figure makes an angle of \(60^{\circ}\) with the line connecting the center of the sphere with the point on the disks that is at the largest distance from the axis? (The axis of the figure is the line having the property that rotation of the figure through \(120^{\circ}\) about that line brings the figure to its initial position. The disks are all on one side of the plane, pass through the center of the sphere, and are orthogonal to the axes.)
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The answer is \(R = (4 + \sqrt{3})d / 6\) .
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IMOLL-1967-26
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Let \(ABCD\) be a regular tetrahedron. To an arbitrary point \(M\) on one edge, say \(CD\) , corresponds the point \(P = P(M)\) , which is the intersection of two lines \(AH\) and \(BK\) , drawn from \(A\) orthogonally to \(BM\) and from \(B\) orthogonally to \(AM\) . What is the locus of \(P\) as \(M\) varies?
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Let \(L\) be the midpoint of the edge \(AB\) . Since \(P\) is the orthocenter of \(\triangle ABM\) and \(ML\) is its altitude, \(P\) lies on \(ML\) and therefore belongs to the triangular area \(LCD\) . Moreover, from the similarity of triangles \(ALP\) and \(MLB\) we have
\[LP\cdot LM = LA\cdot LB = a^{2} / 4,\]
where \(a\) is the side length of tetrahedron \(ABCD\) . It easily follows that the locus of \(P\) is the image of the segment \(CD\) under the inversion of the plane \(LCD\) with center \(L\) and radius \(a / 2\) . This locus is the arc of a circle with center \(L\) and endpoints at the orthocenters of triangles \(ABC\) and \(ABD\) .
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IMOLL-1967-27
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Which regular polygons can be obtained (and how) by cutting a cube with a plane?
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Regular polygons with 3, 4, and 6 sides can be obtained by cutting a cube with a plane, as shown in the figure. A polygon with more than 6 sides cannot be obtained in such a way, for a cube has 6 faces. Also, if a pentagon is obtained by cutting a cube with a plane, then its sides lying on opposite faces are parallel; hence it cannot be regular.
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IMOLL-1967-28
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Find values of the parameter \(u\) for which the expression
\[y = \frac{\tan(x - u) + \tan x + \tan(x + u)}{\tan(x - u)\tan x\tan(x + u)}\]
does not depend on \(x\) .
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The given expression can be transformed into
\[y = \frac{4\cos2u + 2}{\cos2u - \cos2x} -3.\]
It does not depend on \(x\) if and only if \(\cos 2u = - 1 / 2\) , i.e., \(u = \pm \pi /3 + k\pi\) for some \(k\in \mathbb{Z}\)
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IMOLL-1967-29
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The triangles \(A_{0}B_{0}C_{0}\) and \(A^{\prime}B^{\prime}C^{\prime}\) have all their angles acute. Describe how to construct one of the triangles \(ABC\) , similar to \(A^{\prime}B^{\prime}C^{\prime}\) and circumscribing \(A_{0}B_{0}C_{0}\) (so that \(A\) , \(B\) , \(C\) correspond to \(A^{\prime}\) , \(B^{\prime}\) , \(C^{\prime}\) , and \(AB\) passes through \(C_{0}\) , \(BC\) through \(A_{0}\) , and \(CA\) through \(B_{0}\) ). Among these triangles \(ABC\) , describe, and prove, how to construct the triangle with the maximum area.
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Let arc \(l_{a}\) be the locus of points \(A\) lying on the opposite side from \(A_{0}\) with respect to the line \(B_{0}C_{0}\) such that \(\angle B_{0}A C_{0} = \angle A^{\prime}\) . Let \(k_{a}\) be the circle containing \(l_{a}\) , and let \(S_{a}\) be the center of \(k_{a}\) . We similarly define \(l_{b}\) , \(l_{c}\) , \(k_{b}\) , \(k_{c}\) , \(S_{b}\) , \(S_{c}\) . It is easy to show that circles \(k_{a}\) , \(k_{b}\) , \(k_{c}\) have a common point \(S\) inside \(\triangle ABC\) . Let \(A_{1}\) , \(B_{1}\) , \(C_{1}\) be the points on the arcs \(l_{a}\) , \(l_{b}\) , \(l_{c}\) diametrically opposite to \(S\) with respect to \(S_{a}\) , \(S_{b}\) , \(S_{c}\) respectively. Then \(A_{0}\in B_{1}C_{1}\) because \(\angle B_{1}A_{0}S = \angle C_{1}A_{0}S = 90^{\circ}\) ; similarly, \(B_{0}\in A_{1}C_{1}\) and \(C_{0}\in A_{1}B_{1}\) . Hence the triangle \(A_{1}B_{1}C_{1}\) is circumscribed about \(\triangle A_{0}B_{0}C_{0}\) and similar to \(\triangle A^{\prime}B^{\prime}C^{\prime}\) .
Moreover, we claim that \(\triangle A_{1}B_{1}C_{1}\) is the triangle \(ABC\) with the desired properties having the maximum side \(BC\) and hence the maximum area. Indeed, if \(ABC\) is any other such triangle and \(S_{b}^{\prime}\) , \(S_{c}^{\prime}\) are the projections of \(S_{b}\) and \(S_{c}\) onto the line \(BC\) , it holds that
\[BC = 2S_{b}^{\prime}S_{c}^{\prime}\leq 2S_{b}S_{c} = B_{1}C_{1},\]
which proves the maximality of \(B_{1}C_{1}\)
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IMOLL-1967-30
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Given \(m + n\) numbers \(a_{i}\) \((i = 1,2,\ldots ,m)\) , \(b_{j}\) \((j = 1,2,\ldots ,n)\) , determine the number of pairs \((a_{i},b_{j})\) for which \(|i - j|\geq k\) , where \(k\) is a nonnegative integer.
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We assume without loss of generality that \(m\leq n\) . Let \(r\) and \(s\) be the numbers of pairs for which \(i - j\geq k\) and of those for which \(j - i\geq k\) . The desired number is \(r + s\) . We easily find that
\[r = \left\{ \begin{array}{ll}(m - k)(m - k + 1) / 2, k< m,\] \[0, k\geq m, \end{array} \right.\] \[s = \left\{ \begin{array}{ll}m(2n - 2k - m + 1) / 2, k< n - m,\] \[(n - k)(n - k + 1) / 2, n - m\leq k< n,\] \[0, k\geq n. \end{array} \right.\]
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IMOLL-1967-31
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An urn contains balls of \(k\) different colors; there are \(n_{i}\) balls of the \(i\) th color. Balls are drawn at random from the urn, one by one, without replacement. Find the smallest number of draws necessary for getting \(m\) balls of the same color.
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Suppose that \(n_{1}\leq n_{2}\leq \dots \leq n_{k}\) . If \(n_{k}< m\) , there is no solution. Otherwise, the solution is
\[1 + (m - 1)(k - s + 1) + \sum_{i< s}n_{i},\]
where \(s\) is the smallest \(i\) for which \(m\leq n_{i}\) holds.
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IMOLL-1967-32
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Determine the volume of the body obtained by cutting the ball of radius \(R\) by the trihedron with vertex in the center of that ball if its dihedral angles are \(\alpha ,\beta ,\gamma\) .
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Let us denote by \(V\) the volume of the given body, and by \(V_{a}\) , \(V_{b}\) , \(V_{c}\) the
volumes of the parts of the given ball that lie inside the dihedra of the given trihedron. It holds that \(V_{a} = 2R^{3}\alpha /3\) , \(V_{b} = 2R^{3}\beta /3\) , \(V_{c} = 2R^{3}\gamma /3\) . It is easy to see that \(2(V_{a} + V_{b} + V_{c}) = 4V + 4\pi R^{3} / 3\) , from which it follows that
\[V = \frac{1}{3} R^{3}(\alpha +\beta +\gamma -\pi).\]
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IMOLL-1967-33
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In what case does the system
\[x + y + mz = a,\] \[x + my + z = b,\] \[mx + y + z = c,\]
have a solution? Find the conditions under which the unique solution of the above system is an arithmetic progression.
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If \(m\not\in \{-2,1\}\) , the system has the unique solution
\[x = \frac{b + a - (1 + m)c}{(2 + m)(1 - m)}, y = \frac{a + c - (1 + m)b}{(2 + m)(1 - m)}, z = \frac{b + c - (1 + m)a}{(2 + m)(1 - m)}.\]
The numbers \(x,y,z\) form an arithmetic progression if and only if \(a,b,c\) do so.
For \(m = 1\) the system has a solution if and only if \(a = b = c\) , while for \(m = - 2\) it has a solution if and only if \(a + b + c = 0\) . In both these cases it has infinitely many solutions.
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IMOLL-1967-34
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The faces of a convex polyhedron are six squares and eight equilateral triangles, and each edge is a common side for one triangle and one square. All dihedral angles obtained from the triangle and square with a common edge are equal. Prove that it is possible to circumscribe a sphere around this polyhedron and compute the ratio of the squares of the volumes of the polyhedron and of the ball whose boundary is the circumscribed sphere.
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Each vertex of the polyhedron is a vertex of exactly two squares and triangles (more than two is not possible; otherwise, the sum of angles at a vertex exceeds \(360^{\circ}\) ). By using the condition that the trihedral angles are equal it is easy to see that such a polyhedron is uniquely determined by its side length. The polyhedron obtained from a cube by "cutting" its vertices, as shown in the figure, satisfies the conditions.
hedron obtained from a cube by "cutting" its vertices, as shown in the figure, satisfies the conditions.
Now it is easy to calculate that the ratio of the squares of volumes of that polyhedron and of the ball whose boundary is the circumscribed sphere is equal to \(25 / (8\pi^{2})\) .
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IMOLL-1967-35
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Prove the identity
\[\sum_{k = 0}^{n}{\binom{n}{k}}\left(\tan {\frac{x}{2}}\right)^{2k}\left[1 + 2^{k}{\frac{1}{\left(1 - \tan^{2}(x / 2)\right)^{k}}}\right] = \sec^{2n}{\frac{x}{2}} + \sec^{n}x.\]
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The given sum can be rewritten as
\[\sum_{k = 0}^{n}\binom{n}{k}\left(\tan^{2}\frac{x}{2}\right)^{k} + \sum_{k = 0}^{n}\binom{n}{k}\left(\frac{2\tan^{2}\frac{x}{2}}{1 - \tan^{2}\frac{x}{2}}\right)^{k}.\]
Since \(\frac{2\tan^{2}(x / 2)}{1 - \tan^{2}(x / 2)} = \frac{1 - \cos x}{\cos x}\) , the above sum is transformed using the binomial formula into
\[\left(1 + \tan^{2}\frac{x}{2}\right)^{n} + \left(1 + \frac{1 - \cos x}{\cos x}\right)^{n} = \sec^{2n}\frac{x}{2} +\sec^{n}x.\]
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IMOLL-1967-36
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Prove that the center of the sphere circumscribed around a tetrahedron \(ABCD\) coincides with the center of a sphere inscribed in that tetrahedron if and only if \(AB = CD\) , \(AC = BD\) , and \(AD = BC\) .
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Suppose that the skew edges of the tetrahedron \(ABCD\) are equal. Let \(K, L, M, P, Q, R\) be the midpoints of edges \(AB, AC, AD, CD, DB, BC\) respectively. Segments \(KP, LQ, MR\) have the common midpoint \(T\) .
We claim that the lines \(KP\) , \(LQ\) and \(MR\) are axes of symmetry of the tetrahedron \(ABCD\) . From \(LM \parallel CD \parallel RQ\) and similarly \(LR \parallel MQ\) and \(LM = CD / 2 = AB / 2 = LR\) it follows that \(LMQR\) is a rhombus and therefore \(LQ \perp MR\) . We similarly show that \(KP\) is perpendicular to \(LQ\) and \(MR\) , and thus it is perpendicular to the plane \(LMQR\) . Since the lines \(AB\) and \(CD\) are parallel to the plane \(LMQR\) , they are perpendicular to \(KP\) . Hence the points \(A\) and \(C\) are symmetric to \(B\) and \(D\) with respect to the line \(KP\) , which means that \(KP\) is an axis of symmetry of the tetrahedron \(ABCD\) . Similarly, so are the lines \(LQ\) and \(MR\) .
\(LMQR\) . Since the lines \(AB\) and \(CD\) are parallel to the plane \(LMQR\) , they are perpendicular to \(KP\) . Hence the points \(A\) and \(C\) are symmetric to \(B\) and \(D\) with respect to the line \(KP\) , which means that \(KP\) is an axis of symmetry of the tetrahedron \(ABCD\) . Similarly, so are the lines \(LQ\) and \(MR\) .
The centers of circumscribed and inscribed spheres of tetrahedron \(ABCD\) must lie on every axis of symmetry of the tetrahedron, and hence both must coincide with \(T\) .
Conversely, suppose that the centers of circumscribed and inscribed spheres of the tetrahedron \(ABCD\) coincide with some point \(T\) . Then the orthogonal projections of \(T\) onto the faces \(ABC\) and \(ABD\) are the circumcenters \(O_1\) and \(O_2\) of these two triangles, and moreover, \(TO_1 = TO_2\) . Pythagoras's theorem gives \(AO_1 = AO_2\) , which by the law of sines implies \(\angle ACB = \angle ADB\) . Now it easily follows that the sum of the angles at one vertex of the tetrahedron is equal to \(180^\circ\) . Let \(D'\) , \(D''\) , and \(D'''\) be the points in the plane \(ABC\) lying outside \(\triangle ABC\) such that \(\triangle D'BC \cong \triangle DBC\) , \(\triangle D'CA \cong \triangle DCA\) , and \(\triangle D''AB \cong \triangle DAB\) . The angle \(D''AD'''\) is then straight, and hence \(A, B, C\) are midpoints of the segments \(D''D''', D''D', D'D''\) respectively. Hence \(AD = D''D''/2 = BC\) , and analogously \(AB = CD\) and \(AC = BD\) .
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IMOLL-1967-37
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Prove that for arbitrary positive numbers the following inequality holds:
\[\frac{1}{a} +\frac{1}{b} +\frac{1}{c}\leq \frac{a^{8} + b^{8} + c^{8}}{a^{3}b^{3}c^{3}}.\]
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Using the A-G mean inequality we obtain
\[8a^2 b^3 c^3 \leq 2a^8 + 3b^8 + 3c^8,\] \[8a^3 b^2 c^3 \leq 3a^8 + 2b^8 + 3c^8,\] \[8a^3 b^3 c^2 \leq 3a^8 + 3b^8 + 2c^8.\]
By adding these inequalities and dividing by \(3a^3 b^3 c^3\) we obtain the desired one.
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IMOLL-1967-38
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Does there exist an integer such that its cube is equal to \(3n^{2} + 3n + 7\) , where \(n\) is integer?
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Suppose that there exist integers \(n\) and \(m\) such that \(m^3 = 3n^2 + 3n + 7\) . Then from \(m^3 \equiv 1 \pmod{3}\) it follows that \(m = 3k + 1\) for some \(k \in \mathbb{Z}\) . Substituting into the initial equation we obtain \(3k(3k^2 + 3k + 1) = n^2 + n + 2\) . It is easy to check that \(n^2 + n + 2\) cannot be divisible by 3, and so this equality cannot be true. Therefore our equation has no solutions in integers.
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IMOLL-1967-39
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Show that the triangle whose angles satisfy the equality
\[\frac{\sin^{2}A + \sin^{2}B + \sin^{2}C}{\cos^{2}A + \cos^{2}B + \cos^{2}C} = 2\]
is a right- angled triangle.
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Since \(\sin^2 A + \sin^2 B + \sin^2 C + \cos^2 A + \cos^2 B + \cos^2 C = 3\) , the given equality is equivalent to \(\cos^2 A + \cos^2 B + \cos^2 C = 1\) , which by multiplying by 2 is transformed into
\[0 = \cos 2A + \cos 2B + 2\cos^2 C = 2\cos (A + B)\cos (A - B) + 2\cos^2 C\] \[= 2\cos C(\cos (A - B) - \cos C).\]
It follows that either \(\cos C = 0\) or \(\cos (A - B) = \cos C\) . In both cases the triangle is right- angled.
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IMOLL-1967-40
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Exactly one side of a tetrahedron is of length greater than 1. Show that its volume is less than or equal to \(1 / 8\) .
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Suppose \(CD\) is the longest edge of the tetrahedron \(ABCD\) , \(AB = a\) , \(CK\) and \(DL\) are the altitudes of the triangles \(ABC\) and \(ABD\) respectively, and \(DM\) is the altitude of the tetrahedron \(ABCD\) . Then \(CK^2 \leq 1 - a^2 /4\) , since \(CK\) is a leg of the right triangle whose other leg has length not less than \(a / 2\) and whose hypotenuse has length not greater than 1 ( \(AKC\) or \(BKC\) ). In the similar way we can show that \(DL^2 \leq 1 - a^2 /4\) . Since \(DM \leq DL\) , then \(DM^2 \leq 1 - a^2 /4\) . It follows that
\[V = \frac{1}{3}\left(\frac{a}{2} CK\right)DM\leq \frac{1}{6} a\left(1 - \frac{a^2}{4}\right) = \frac{1}{24} a(2 - a)(2 + a)\] \[= \frac{1}{24} [1 - (a - 1)^2 ](2 + a)\leq \frac{1}{24}\cdot 1\cdot 3 = \frac{1}{8}.\]
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IMOLL-1967-41
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A line \(l\) is drawn through the intersection point \(H\) of the altitudes of an acute-angled triangle. Prove that the symmetric images \(l_{a}, l_{b}, l_{c}\) of \(l\) with respect to sides \(BC, CA, AB\) have one point in common, which lies on the circumcircle of \(ABC\) .
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It is well known that the points \(K\) , \(L\) , \(M\) , symmetric to \(H\) with respect to \(BC, CA, AB\) respectively, lie on the circumcircle \(k\) of the triangle \(ABC\) . For \(K\) , this follows from an elementary calculation of angles of triangles \(HBC\) and noting that \(\angle KBC = \angle HBC = \angle KAC\) . For other points the proof is analogous.
Since the lines \(l_{a}, l_{b}\) pass through \(K\) and \(L\) and \(l_{b}\) is obtained from \(l_{a}\) by rotation about \(C\) for an angle \(2\gamma = \angle LCK\) , it follows that the intersection point \(P\) of \(l_{a}\) and \(l_{b}\) is at the circumcircle of \(KLC\) , that is, \(k\) . Similarly, \(l_{b}\) and \(l_{c}\) meet at a point on \(k\) ; hence they must pass through the same point \(P\) .
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IMOLL-1967-42
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Decompose into real factors the expression \(1 - \sin^{5}x - \cos^{5}x\) .
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\(E = (1 - \sin x)(1 - \cos x)[3 + 2(\sin x + \cos x) + 2\sin x\cos x + \sin x\cos x(\sin x + \cos x)]\) .
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IMOLL-1967-43
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The equation
\[x^{5} + 5\lambda x^{4} - x^{3} + (\lambda \alpha -4)x^{2} - (8\lambda +3)x + \lambda \alpha -2 = 0\]
is given.
(a) Determine \(\alpha\) such that the given equation has exactly one root independent of \(\lambda\) .
(b) Determine \(\alpha\) such that the given equation has exactly two roots independent of \(\lambda\) .
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We can write the given equation in the form
\[x^{5} - x^{3} - 4x^{2} - 3x - 2 + \lambda (5x^{4} + \alpha x^{2} - 8x + \alpha) = 0.\]
A root of this equation is independent of \(\lambda\) if and only if it is a common root of the equations
\[x^{5} - x^{3} - 4x^{2} - 3x - 2 = 0\quad \mathrm{and}\quad 5x^{4} + \alpha x^{2} - 8x + \alpha = 0.\]
The first of these two equations is equivalent to \((x - 2)(x^{2} + x + 1)^{2} = 0\) and has three different roots: \(x_{1} = 2\) , \(x_{2,3} = (- 1 \pm i\sqrt{3}) / 2\) .
(a) For \(\alpha = -64 / 5\) , \(x_{1} = 2\) is the unique root independent of \(\lambda\) .
(b) For \(\alpha = -3\) there are two roots independent of \(\lambda\) : \(x_{1} = \omega\) and \(x_{2} = \omega^{2}\) .
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IMOLL-1967-44
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Suppose \(p\) and \(q\) are two different positive integers and \(x\) is a real number. Form the product \((x + p)(x + q)\) .
(a) Find the sum \(S(x,n) = \sum (x + p)(x + q)\) , where \(p\) and \(q\) take values from 1 to \(n\) .
(b) Do there exist integer values of \(x\) for which \(S(x,n) = 0\) ?
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(a) \(S(x, n) = n(n - 1) \left[x^{2} + (n + 1)x + (n + 1)(3n + 2) / 12\right]\) .
(b) It is easy to see that the equation \(S(x,n) = 0\) has two roots
\[x_{1,2} = \left(-(n + 1)\pm \sqrt{(n + 1) / 3}\right) / 2.\]
They are integers if and only if \(n = 3k^{2} - 1\) for some \(k\in \mathbb{N}\)
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IMOLL-1967-45
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(a) Solve the equation
\[\sin^{3}x + \sin^{3}\left(\frac{2\pi}{3} +x\right) + \sin^{3}\left(\frac{4\pi}{3} +x\right) + \frac{3}{4}\cos 2x = 0.\]
(b) Suppose the solutions are in the form of arcs \(AB\) of the trigonometric circle (where \(A\) is the beginning of arcs of the trigonometric circle), and \(P\) is a regular \(n\) -gon inscribed in the circle with one vertex at \(A\) .
(1) Find the subset of arcs with the endpoint \(B\) at a vertex of the regular dodecagon.
(2) Prove that the endpoint \(B\) cannot be at a vertex of \(P\) if \(2,3 \nmid n\) or \(n\) is prime.
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(a) Using the formula \(4\sin^{3}x = 3\sin x - \sin 3x\) one can easily reduce the given equation to \(\sin 3x = \cos 2x\) . Its solutions are given by \(x = (4k + 1)\pi /10\) , \(k\in \mathbb{Z}\) .
(b) (1) The point \(B\) corresponding to the solution \(x = (4k + 1)\pi /10\) is a vertex of the regular dodecagon if and only if \((4k + 1)\pi /10 = 2m\pi /12\) , i.e., \(3(4k + 1) = 5m\) for some \(m\in \mathbb{Z}\) . This is possible if and only if \(5\mid 4k + 1\) , i.e., \(k\equiv 1\) (mod 5).
(2) Similarly, if the point \(B\) corresponding to \(x = (4k + 1)\pi /10\) is a vertex of a polygon \(P\) , then \((4k + 1)n = 20m\) for some \(m\in \mathbb{N}\) , which implies that \(4\mid n\) .
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IMOLL-1967-46
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If \(x, y, z\) are real numbers satisfying the relations \(x + y + z = 1\) and \(\arctan x + \arctan y + \arctan z = \pi /4\) , prove that
\[x^{2n + 1} + y^{2n + 1} + z^{2n + 1} = 1\]
for all positive integers \(n\) .
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Let us set \(\arctan x = a\) , \(\arctan y = b\) , \(\arctan z = c\) . Then \(\tan (a + b) = \frac{x + y}{1 - y}\) and \(\tan (a + b + c) = \frac{x + y + z - xyz}{1 - yz - zx - xy} = 1\) , which implies that
\[(x - 1)(y - 1)(z - 1) = xyz - xy - yz - zx + x + y + z - 1 = 0.\]
One of \(x,y,z\) is equal to 1, say \(z = 1\) , and consequently \(x + y = 0\) . Therefore
\[x^{2n + 1} + y^{2n + 1} + z^{2n + 1} = x^{2n + 1} + (-x)^{2n + 1} + 1^{2n + 1} = 1.\]
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IMOLL-1967-47
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Prove the inequality
\[x_{1}x_{2}\cdot \cdot \cdot x_{k}\left(x_{1}^{n - 1} + x_{2}^{n - 1} + \cdot \cdot \cdot +x_{k}^{n - 1}\right)\leq x_{1}^{n + k - 1} + x_{2}^{n + k - 1} + \cdot \cdot \cdot +x_{k}^{n + k - 1},\]
where \(x_{i} > 0 (i = 1,2,\ldots ,k), k \in N, n \in N\) .
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Using the A-G mean inequality we get
\[(n + k - 1)x_{1}^{n}x_{2}\cdot \cdot \cdot x_{k}\leq nx_{1}^{n + k - 1} + x_{2}^{n + k - 1} + \cdot \cdot \cdot +x_{k}^{n + k - 1},\] \[(n + k - 1)x_{1}x_{2}^{n}\cdot \cdot \cdot x_{k}\leq x_{1}^{n + k - 1} + nx_{2}^{n + k - 1} + \cdot \cdot \cdot +x_{k}^{n + k - 1},\] \[\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot\] \[(n + k - 1)x_{1}x_{2}\cdot \cdot \cdot x_{k}^{n}\leq x_{1}^{n + k - 1} + x_{2}^{n + k - 1} + \cdot \cdot \cdot +nx_{k}^{n + k - 1}.\]
By adding these inequalities and dividing by \(n + k - 1\) we obtain the desired one. Remark. This is also an immediate consequence of Muirhead's inequality.
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IMOLL-1967-48
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Determine all positive roots of the equation \(x^{z} = 1 / \sqrt{2}\) .
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Put \(f(x) = x\ln x\) . The given equation is equivalent to \(f(x) = f(1 / 2)\) , which has the solutions \(x_{1} = 1 / 2\) and \(x_{2} = 1 / 4\) . Since the function \(f\) is decreasing on \((0,1 / e)\) , and increasing on \((1 / e, + \infty)\) , this equation has no other solutions.
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IMOLL-1967-49
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Let \(n\) and \(k\) be positive integers such that \(1 \leq n \leq N + 1, 1 \leq k \leq N + 1\) . Show that
\[\min_{n \neq k} | \sin n - \sin k | < \frac{2}{N}.\]
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Since \(\sin 1,\sin 2,\ldots ,\sin (N + 1)\in (-1,1)\) , two of these \(N + 1\) numbers have distance less than \(2 / N\) . Therefore \(|\sin n - \sin k|< 2 / N\) for some integers \(1\leq k,n\leq\) \(N + 1,n\neq k\)
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IMOLL-1967-50
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The function \(\phi (x, y, z)\) , defined for all triples \((x, y, z)\) of real numbers, is such that there are two functions \(f\) and \(g\) defined for all pairs of real numbers such that
\[\phi (x, y, z) = f(x + y, z) = g(x, y + z)\]
for all real \(x, y\) , and \(z\) . Show that there is a function \(h\) of one real variable such that
\[\phi (x, y, z) = h(x + y + z)\]
for all real \(x, y\) , and \(z\) .
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Since \(\phi (x,y,z) = f(x + y,z) = \phi (0,x + y,z) = g(0,x + y + z)\) , it is enough to put \(h(t) = g(0,t)\) .
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