id stringlengths 10 15 | question stringlengths 63 2.3k | solutions stringlengths 20 28.5k |
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IMO-1959-1 | For every integer \(n\) prove that the fraction \(\frac{21n + 4}{14n + 3}\) cannot be reduced any further. | The desired result \((14n + 3,21n + 4) = 1\) follows from
\[3(14n + 3) - 2(21n + 4) = 1.\] |
IMO-1959-2 | For which real numbers \(x\) do the following equations hold:
\[(a)\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = \sqrt{2},\] \[(b)\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = 1,\] \[(c)\sqrt{x + \sqrt{2x - 1}} +\sqrt{x - \sqrt{2x - 1}} = 2?\] | 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}^{... |
IMO-1959-3 | 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 \(a = 4\) , \(b = 2\) , \(c = - 1\) . | 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):
\[a^{2}\cos^{2}2x + (2a^{2} + 4ac - 2b^{2})\cos 2x + (a^{2} + 4ac - 2b^{2} + 4c^{2}) = 0.\]... |
IMO-1959-4 | 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. | 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 ... |
IMO-1959-5 | 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 th... | (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... |
IMO-1959-6 | 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 circ... | 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 \(\gamma\) , de... |
IMO-1960-1 | 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. | 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 \i... |
IMO-1960-2 | For which real numbers \(x\) does the following inequality hold:
\[\frac{4x^{2}}{(1 - \sqrt{1 + 2x})^{2}} < 2x + 9?\] | 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 prec... |
IMO-1960-3 | 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
\[\tan \alpha = \frac{4nh}{(n^2 - 1)a},\]... | 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} =... |
IMO-1960-4 | 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. | 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 \(AA_{1} = m_{a}\) , \(AA^{\prime} = h_{a}\) , ... |
IMO-1960-5 | 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 \(\overline{ZY} = 2\overline{XZ}\) . | (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 \(A... |
IMO-1960-6 | 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 fo... | 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... |
IMO-1960-7 | 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 ... | 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 \... |
IMO-1961-1 | 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? | 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}... |
IMO-1961-2 | 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? | 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}... |
IMO-1961-3 | Solve the equation \(\cos^{n}x - \sin^{n}x = 1\) , where \(n\) is a given positive integer. | 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} /... |
IMO-1961-4 | 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}\... | 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} +... |
IMO-1961-5 | 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 hol... | 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 constru... |
IMO-1961-6 | 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... | 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 o... |
IMO-1962-1 | 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. | 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 intege... |
IMO-1962-2 | Find all real numbers \(x\) for which
\[\sqrt{3 - x} -\sqrt{x + 1} >\frac{1}{2}.\] | 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(\fr... |
IMO-1962-3 | 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}\... | 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. |
IMO-1962-4 | Solve the equation
\[\cos^{2}x + \cos^{2}2x + \cos^{2}3x = 1.\] | 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 = ... |
IMO-1962-5 | 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 \(A B C D\) . | 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 + \an... |
IMO-1962-6 | 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
\[d = \sqrt{r(r - 2\rho)}.\] | This problem is a special case, when the triangle is isosceles, of Euler's formula, which holds for all triangles. |
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. | 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. Repea... |
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. | 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} ... |
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. | 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 \(O\) lie... |
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... |
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 \(y\neq 2\) , ... |
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 = \... |
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'... | The result is EDACB. |
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... | 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... |
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. | 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 \(T_{1}\) a... |
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 fro... |
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 ... | 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 \parall... |
IMO-1965-1 | Find all real numbers \(x \in [0, 2\pi ]\) such that
\[2\cos x\leq |\sqrt{1 + \sin2x} -\sqrt{1 - \sin2x}|\leq \sqrt{2}.\] | 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 \(\pi /2\leq x\leq 3\pi /2\)... |
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... | 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_{... |
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 ... | 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 th... |
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. | 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... |
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... | (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}\... |
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 \(B\)... |
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 t... | 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... |
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. | 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... |
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. | 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 no... |
IMO-1966-4 | 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 \(k\in \mathbb{N}\) | 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 ... |
IMO-1966-5 | 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 \... | 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}... |
IMO-1966-6 | 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 \(\triangle ABC\) . | 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}]\... |
IMOLL-1967-1 | Prove that all numbers in the sequence
\[{\frac{107811}{3}},\quad{\frac{110778111}{3}},\quad{\frac{111077781111}{3}},\quad\dots\]
are perfect cubes. | 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{1... |
IMOLL-1967-2 | 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 \(n=1.\) | \((n!)^{2 / n} = ((1\cdot 2\dots n)^{1 / n})^{2}\leq (\frac{1 + 2 + \dots + n}{n})^{2} = (\frac{n + 1}{2})^{2}\leq \frac{1}{3} n^{2} + \frac{1}{2} n + \frac{1}{6}.\) |
IMOLL-1967-3 | Prove the trigonometric inequality \(\cos x<1-\frac {x^{2}}{2}+\frac {x^{4}}{16}\) , where \(x\in (0,\pi /2).\) | 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) >... |
IMOLL-1967-4 | 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. | (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} = ... |
IMOLL-1967-5 | Solve the system
\[\begin{array}{l}x^{2}+x-1=y,\\ y^{2}+y-1=z,\\ z^{2}+z-1=x. \end{array}\] | 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 impossi... |
IMOLL-1967-6 | Solve the system
\[\begin{array}{l}|x+y|+|1-x|=6,\\ |x+y+1|+|1-y|=4. \end{array}\] | The given system has two solutions: \((-2, - 1)\) and \((-14 / 3,13 / 3)\) . |
IMOLL-1967-7 | Find all real solutions of the system of equations
\[\begin{array}{l}x_{1}+x_{2}+\cdots +x_{n}=a,\\ x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}=a^{2},\\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \quad \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \end{array}\] | 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 -... |
IMOLL-1967-8 | \(ABCD\) 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 ... | 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 \(\t... |
IMOLL-1967-9 | 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 \(k\). | 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 symme... |
IMOLL-1967-10 | 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... | 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 le... |
IMOLL-1967-11 | Let \(n\) be a positive integer. Find the maximal number of noncongruent triangles whose side lengths are integers less than or equal to \(n\) . | 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 f... |
IMOLL-1967-12 | 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 \(XY\) fo... | 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... |
IMOLL-1967-13 | 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. | 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. |
IMOLL-1967-14 | 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). | 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 ... |
IMOLL-1967-15 | 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. | 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 \... |
IMOLL-1967-16 | 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 \(p\)... | 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 \(|m_1|,|m_2|\leq n\) . ... |
IMOLL-1967-17 | 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 \(c_1 c_2 \dots c_n\) . | 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 clearl... |
IMOLL-1967-18 | 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 rec... | 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 rationa... |
IMOLL-1967-19 | 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. | 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 h... |
IMOLL-1967-20 | 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 | 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. |
IMOLL-1967-21 | Without using any tables, find the exact value of the product
\[P = \cos {\frac{\pi}{15}}\cos {\frac{2\pi}{15}}\cos {\frac{3\pi}{15}}\cos {\frac{4\pi}{15}}\cos {\frac{5\pi}{15}}\cos {\frac{6\pi}{15}}\cos {\frac{7\pi}{15}}.\] | 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} = \fr... |
IMOLL-1967-22 | 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} \ge... | 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 \(P\) ... |
IMOLL-1967-23 | 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: \(a \geq 0\) , \(c \geq 0\) , \(4ac \geq b^{2}\) . | 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}... |
IMOLL-1967-24 | 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... | 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}... |
IMOLL-1967-25 | 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 ... | The answer is \(R = (4 + \sqrt{3})d / 6\) . |
IMOLL-1967-26 | 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? | 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
\[LP\cdot LM = LA\cdot LB = a^{2} / 4,\]
... |
IMOLL-1967-27 | Which regular polygons can be obtained (and how) by cutting a cube with a plane? | 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; ... |
IMOLL-1967-28 | 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 \(x\) . | 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 \(k\in \mathbb{Z}\) |
IMOLL-1967-29 | 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 \(A^{\prime}\) , \(B^{\prime}\)... | 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_{... |
IMOLL-1967-30 | 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. | 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.... |
IMOLL-1967-31 | 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. | 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. |
IMOLL-1967-32 | 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 \(\alpha ,\beta ,\gamma\) . | 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 \(... |
IMOLL-1967-33 | 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. | 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 ... |
IMOLL-1967-34 | 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 ... | 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... |
IMOLL-1967-35 | Prove the identity
\[\sum_{k = 0}^{n}{\binom{n}{k}}\left(\tan {\frac{x}{2}}\right)^{2k}\left[1 + 2^{k}{\frac{1}{\left(1 - \tan^{2}(x / 2)\right)^{k}}}\right] = \sec^{2n}{\frac{x}{2}} + \sec^{n}x.\] | 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 transfor... |
IMOLL-1967-36 | 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 \(AD = BC\) . | 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 \(ABCD\) . F... |
IMOLL-1967-37 | Prove that for arbitrary positive numbers the following inequality holds:
\[\frac{1}{a} +\frac{1}{b} +\frac{1}{c}\leq \frac{a^{8} + b^{8} + c^{8}}{a^{3}b^{3}c^{3}}.\] | 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. |
IMOLL-1967-38 | Does there exist an integer such that its cube is equal to \(3n^{2} + 3n + 7\) , where \(n\) is integer? | 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 ... |
IMOLL-1967-39 | 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. | 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).\... |
IMOLL-1967-40 | Exactly one side of a tetrahedron is of length greater than 1. Show that its volume is less than or equal to \(1 / 8\) . | 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 le... |
IMOLL-1967-41 | 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 \(ABC\) . | 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 \(\angle KBC = \angle HBC = \angle KAC\)... |
IMOLL-1967-42 | Decompose into real factors the expression \(1 - \sin^{5}x - \cos^{5}x\) . | \(E = (1 - \sin x)(1 - \cos x)[3 + 2(\sin x + \cos x) + 2\sin x\cos x + \sin x\cos x(\sin x + \cos x)]\) . |
IMOLL-1967-43 | 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 ind... | 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} + \a... |
IMOLL-1967-44 | 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 \(S(x,n) = 0\) ? | (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 \(k\in \mathbb{N}\) |
IMOLL-1967-45 | (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 r... | (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... |
IMOLL-1967-46 | 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 \(n\) . | 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 \(... |
IMOLL-1967-47 | 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 \(x_{i} > 0 (i = 1,2,\ldots ,k), k \in N, n \in N\) . | 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 ... |
IMOLL-1967-48 | Determine all positive roots of the equation \(x^{z} = 1 / \sqrt{2}\) . | 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. |
IMOLL-1967-49 | Let \(n\) and \(k\) be positive integers such that \(1 \leq n \leq N + 1, 1 \leq k \leq N + 1\) . Show that
\[\min_{n \neq k} | \sin n - \sin k | < \frac{2}{N}.\] | 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 \(1\leq k,n\leq\) \(N + 1,n\neq k\) |
IMOLL-1967-50 | 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 on... | 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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