id stringlengths 10 10 | source stringlengths 74 74 | problem stringlengths 57 1.29k | solutions listlengths 0 9 |
|---|---|---|---|
IMO-1959-1 | https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_1 | Prove that the fraction \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\). | [
"Denoting the greatest common divisor of \\(a, b\\) as \\((a,b)\\), we use the Euclidean algorithm:\n\n\\[\n(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1\n\\]\n\nTheir greatest common divisor is 1, so \\(\\frac{21n+4}{14n+3}\\) is irreducible. Q.E.D.",
"Proof by contradiction:\n\nAssume that \\(\\dfrac{14n+3}{21... |
IMO-1959-2 | https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_2 | For what real values of \(x\) is
\[
\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,
\]
given (a) \(A=\sqrt{2}\), (b) \(A=1\), (c) \(A=2\), where only non-negative real numbers are admitted for square roots? | [
"The square roots imply that \\(x\\ge \\frac{1}{2}\\).\n\nSquare both sides of the given equation:\n\n\\[\nA^2 = \\Big( x + \\sqrt{2x - 1}\\Big) + 2 \\sqrt{x + \\sqrt{2x - 1}} \\sqrt{x - \\sqrt{2x - 1}} + \\Big( x - \\sqrt{2x - 1}\\Big)\n\\]\n\nAdd the first and the last terms to get:\n\n\\[\nA^2 = 2x + 2 \\sqrt{... |
IMO-1959-3 | https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_3 | Let \(a,b,c\) be real numbers. Consider the quadratic equation in \(\cos{x}\) :
\[
a\cos ^{2}x + b\cos{x} + c = 0.
\]
Using the numbers \(a,b,c\), form a quadratic equation in \(\cos{2x}\), whose roots are the same as those of the original equation. Compare the equations in \(\cos{x}\) and \(\cos{2x}\) for \(a=4, b=2, c=-1\). | [
"Let the original equation be satisfied only for \\(\\cos{x}=m, \\cos{x}=n\\). Then we wish to construct a quadratic with roots \\(2m^2 -1, 2n^2 -1\\).\n\nClearly, the sum of the roots of this quadratic must be\n\n\\[\n2 (m^2 + n^2) - 2 = 2 \\left(\\frac{b^2-2ac}{a^2}\\right) - 2 = \\frac{2b^2 - 4ac - 2a^2}{a^2},\n... |
IMO-1959-4 | https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_4 | Construct a right triangle with a given hypotenuse \(c\) such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle. | [
"We denote the catheti of the triangle as \\(a\\) and \\(b\\). We also observe the well-known fact that in a right triangle, the median to the hypotenuse is of half the length of the hypotenuse. (This is true because if we inscribe the triangle in a circle, the hypotenuse is the diameter, so a segment from any poin... |
IMO-1959-5 | https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_5 | An arbitrary point \(M\) is selected in the interior of the segment \(AB\). The squares \(AMCD\) and \(MBEF\) are constructed on the same side of \(AB\), with the segments \(AM\) and \(MB\) as their respective bases. The circles about these squares, with respective centers \(P\) and \(Q\), intersect at \(M\) and also at another point \(N\). Let \(N'\) denote the point of intersection of the straight lines \(AF\) and \(BC\).
(a) Prove that the points \(N\) and \(N'\) coincide.
(b) Prove that the straight lines \(MN\) pass through a fixed point \(S\) independent of the choice of \(M\).
(c) Find the locus of the midpoints of the segments \(PQ\) as \(M\) varies between \(A\) and \(B\). | [
"## Part A\n\nSince the triangles \\(AFM, CBM\\) are congruent, the angles \\(AFM, CBM\\) are congruent; hence \\(AN'B\\) is a right angle. Therefore \\(N'\\) must lie on the circumcircles of both quadrilaterals; hence it is the same point as \\(N\\).\n\n\\[\n1IMO5A.JPG\n\\]\n\n## Part B\n\nWe observe that \\(\\fra... |
IMO-1959-6 | https://artofproblemsolving.com/wiki/index.php/1959_IMO_Problems/Problem_6 | Two planes, \(P\) and \(Q\), intersect along the line \(p\). The point \(A\) is in the plane \(P\), and the point \({C}\) is in the plane \(Q\); neither of these points lies on the straight line \(p\). Construct an isosceles trapezoid \(ABCD\) (with \(AB\) parallel to \(DC\)) in which a circle can be inscribed, and with vertices \(B\) and \(D\) lying in the planes \(P\) and \(Q\), respectively. | [
"We first observe that we must have both lines \\(AB\\) (which we shall denote \\(a\\)) and \\(DC\\) (which we shall denote \\(c\\)) parallel to \\(p\\), since if one of them is not, then neither can be and they must both intersect \\(p\\) (since they are both coplanar with \\(p\\)), making them skew.\n\nNow we not... |
IMO-1960-1 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_1 | Determine all three-digit numbers \(N\) having the property that \(N\) is divisible by 11, and \(\dfrac{N}{11}\) is equal to the sum of the squares of the digits of \(N\). | [
"Let \\(N = 100a + 10b+c\\) for some digits \\(a,b,\\) and \\(c\\). Then\n\n\\[\n100a + 10b+c = 11m\n\\]\n\nfor some \\(m\\). We also have \\(m=a^2+b^2+c^2\\). Substituting this into the first equation and simplification, we get\n\n\\[\n100a+10b+c = 11a^2 +11b^2 +11c^2\n\\]\n\nFor an integer divisible by \\(11\\), ... |
IMO-1960-2 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_2 | For what values of the variable \(x\) does the following inequality hold:
\[
\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?
\] | [
"Set \\(x = -\\frac{1}{2} + \\frac{a^2}{2}\\), where \\(a\\ge0\\). \\(\\frac{4\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)^2}{\\left(1-\\sqrt{1+2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)}\\right)^2}<2\\left(-\\frac{1}{2}+\\frac{a^2}{2}\\right)+9\\)\n\nAfter simplifying, we get \\((a+1)^2<a^2+8\\)\n\nSo \\(a^2+2a+1... |
IMO-1960-3 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_3 | In a given right triangle \(ABC\), the hypotenuse \(BC\), of length \(a\), is divided into \(n\) equal parts (\(n\) an odd integer). Let \(\alpha\) be the acute angle subtending, from \(A\), that segment which contains the midpoint of the hypotenuse. Let \(h\) be the length of the altitude to the hypotenuse of the triangle. Prove that:
\[
\tan{\alpha}=\frac{4nh}{(n^2-1)a}.
\] | [
"Using coordinates, let \\(A=(0,0)\\), \\(B=(b,0)\\), and \\(C=(0,c)\\). Also, let \\(PQ\\) be the segment that contains the midpoint of the hypotenuse with \\(P\\) closer to \\(B\\).\n\n\\[\n[asy] size(8cm); pair A,B,C,P,Q; A=(0,0); B=(4,0); C=(0,3); P=(2.08,1.44); Q=(1.92,1.56); dot(A); dot(B); dot(C); dot(P); do... |
IMO-1960-4 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_4 | Construct triangle \(ABC\), given \(h_a\), \(h_b\) (the altitudes from \(A\) and \(B\)), and \(m_a\), the median from vertex \(A\). | [
"Let \\(M_a\\), \\(M_b\\), and \\(M_c\\) be the midpoints of sides \\(\\overline{BC}\\), \\(\\overline{CA}\\), and \\(\\overline{AB}\\), respectively. Let \\(H_a\\), \\(H_b\\), and \\(H_c\\) be the feet of the altitudes from \\(A\\), \\(B\\), and \\(C\\) to their opposite sides, respectively. Since \\(\\triangle AB... |
IMO-1960-5 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_5 | Consider the cube \(ABCDA'B'C'D'\) (with face \(ABCD\) directly above face \(A'B'C'D'\)).
a) Find the locus of the midpoints of the segments \(XY\), where \(X\) is any point of \(AC\) and \(Y\) is any point of \(B'D'\);
b) Find the locus of points \(Z\) which lie on the segment \(XY\) of part a) with \(ZY = 2XZ\). | [
"Let \\(A=(0,0,2)\\), \\(B=(2,0,2)\\), \\(C=(2,0,0)\\), \\(D=(0,0,0)\\), \\(A'=(0,2,2)\\), \\(B'=(2,2,2)\\), \\(C'=(2,2,0)\\), and \\(D'=(0,2,0)\\). Then there exist real \\(x\\) and \\(y\\) in the closed interval \\([0,2]\\) such that \\(X=(x,0,2-x)\\) and \\(Y=(y,2,y)\\).\n\nThe midpoint of \\(XY\\) has coordinat... |
IMO-1960-6 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_6 | Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let \(V_1\) be the volume of the cone and \(V_2\) be the volume of the cylinder.
a) Prove that \(V_1 \neq V_2\);
b) Find the smallest number \(k\) for which \(V_1 = kV_2\); for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone. | [
"Part (a):\n\nLet \\(R\\) denote the radius of the cone, and let \\(r\\) denote the radius of the cylinder and sphere. Let \\(l\\) denote the slant height of the cone, and let \\(h\\) denote the height of the cone.\n\nConsider a plane that contains the axis of the cone. This plane will slice the cone and sphere int... |
IMO-1960-7 | https://artofproblemsolving.com/wiki/index.php/1960_IMO_Problems/Problem_7 | An isosceles trapezoid with bases \(a\) and \(c\) and altitude \(h\) is given.
a) On the axis of symmetry of this trapezoid, find all points \(P\) such that both legs of the trapezoid subtend right angles at \(P\);
b) Calculate the distance of \(P\) from either base;
c) Determine under what conditions such points \(P\) actually exist. Discuss various cases that might arise. | [
"(a) The intersection of the circle with diameter one of the legs with the axis of symmetry.\n\n(b) Let \\(x\\) be the distance from \\(P\\) to one of the bases; then \\(h - x\\) must be the distance from \\(P\\) to the other base. Similar triangles give \\(\\frac{x}{\\frac{a}{2}} = \\frac{\\frac{c}{2}}{h - x}\\), ... |
IMO-1961-1 | https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_1 | Solve the system of equations:
\[
\begin{matrix} \quad x + y + z \!\!\! &= a \; \, \\ x^2 +y^2+z^2 \!\!\! &=b^2 \\ \qquad \qquad xy \!\!\! &= z^2 \end{matrix}
\]
where \(a\) and \(b\) are constants. Give the conditions that \(a\) and \(b\) must satisfy so that \(x, y, z\) (the solutions of the system) are distinct positive numbers. | [
"Note that \\(x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2\\), so the first two equations become\n\n\\[\n\\begin{matrix} \\quad (x + y) + z \\!\\!\\! &= a \\; \\; (*) \\\\ (x+y)^2 - z^2 \\!\\!\\! &=b^2 (**) \\end{matrix}\n\\]\n\n.\n\nWe note that \\((x+y)^2 - z^2 = \\Big[ (x+y)+z \\Big]\\Big[ (x+y)-z\\Big]\\), so if ... |
IMO-1961-2 | https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_2 | Let \(a\), \(b\), and \(c\) be the lengths of a triangle whose area is S. Prove that
\[
a^2 + b^2 + c^2 \ge 4S\sqrt{3}
\]
In what case does equality hold? | [
"By Heron's formula, we have\n\n\\[\nS = \\sqrt{s(s-a)(s-b)(s-c)}.\n\\]\n\nThis can be simplified to\n\n\\[\nS = \\sqrt{\\left(\\frac{a+b+c}{2}\\right)\\left(\\frac{-a+b+c}{2}\\right)\\left(\\frac{a-b+c}{2}\\right)\\left(\\frac{a+b-c}{2}\\right)}.\n\\]\n\nNext, we can factor out all of the \\(2\\)s and use a clever... |
IMO-1961-3 | https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_3 | Solve the equation
\[
\cos^n{x} - \sin^n{x} = 1
\]
where \(n\) is a given positive integer. | [
"Since \\(cos^2x + sin^2x = 1\\), we cannot have solutions with \\(n\\ne2\\) and \\(0<|cos(x)|,|sin(x)|<1\\). Nor can we have solutions with \\(n=2\\), because the sign is wrong. So the only solutions have \\(sin (x) = 0\\) or \\(cos (x) = 0\\), and these are: \\(x =\\) multiple of \\(\\pi\\), and \\(n\\) even; \\(... |
IMO-1961-4 | https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_4 | In the interior of triangle \(P_1P_2P_3\) a point \(P\) is given. Let \(Q_1,Q_2,Q_3\) be the intersections of \(PP_1, PP_2,PP_3\) with the opposing edges of triangle \(P_1P_2P_3\). Prove that among the ratios \(\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}\) there exists one not larger than \(2\) and one not smaller than \(2\). | [
"Let \\([ABC]\\) denote the area of triangle \\(ABC\\).\n\nSince triangles \\(P_1P_2P_3\\) and \\(PP_2P_3\\) share the base \\(P_2P_3\\), we have \\(\\frac{[PP_2P_3]}{[P_1P_2P_3]}=\\frac{PQ_1}{P_1Q_1}\\).\n\nSimilarly, \\(\\frac{[PP_1P_3]}{[P_1P_2P_3]}=\\frac{PQ_2}{P_2Q_2}, \\frac{[PP_1P_2]}{[P_1P_2P_3]}=\\frac{PQ_... |
IMO-1961-5 | https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_5 | Construct a triangle ABC if the following elements are given: \(AC = b, AB = c\), and \(\angle AMB = \omega \left(\omega < 90^{\circ}\right)\) where M is the midpoint of BC. Prove that the construction has a solution if and only if
\[
b \tan{\frac{\omega}{2}} \le c < b
\]
In what case does equality hold? | [
"Prolong BA to a point D such that \\(BD = 2AB\\). Take circle through B and D such that the minor arc BD is equal to \\(2*\\omega\\) so that for points P on the major arc BD we have \\(\\angle BPD = \\omega\\). Draw a circle with center A and radius AC, and the point of intersection of this circle and the major ar... |
IMO-1961-6 | https://artofproblemsolving.com/wiki/index.php/1961_IMO_Problems/Problem_6 | Consider a plane \(\epsilon\) and three non-collinear points \(A,B,C\) on the same side of \(\epsilon\); suppose the plane determined by these three points is not parallel to \(\epsilon\). In plane \(\epsilon\) take three arbitrary points \(A',B',C'\). Let \(L,M,N\) be the midpoints of segments \(AA', BB', CC'\); Let \(G\) be the centroid of the triangle \(LMN\). (We will not consider positions of the points \(A', B', C'\) such that the points \(L,M,N\) do not form a triangle.) What is the locus of point \(G\) as \(A', B', C'\) range independently over the plane \(\epsilon\)? | [
"We will consider the various points in terms of their coordinates in space. We have \\(L=\\frac{A+A^\\prime}{2},M=\\frac{B+B^\\prime}{2},N=\\frac{C+C^\\prime}{2}\\). Since the centroid of a triangle is the average of the triangle's vertices, we have \\(G=\\frac{1}{3}\\left(L+M+N\\right)=\\frac{1}{6}\\left(A+B+C+A^... |
IMO-1962-1 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_1 | Find the smallest natural number \(n\) which has the following properties:
(a) Its decimal representation has 6 as the last digit.
(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number \(n\). | [
"As the new number starts with a \\(6\\) and the old number is \\(1/4\\) of the new number, the old number must start with a \\(1\\).\n\nAs the new number now starts with \\(61\\), the old number must start with \\(\\lfloor 61/4\\rfloor = 15\\).\n\nWe continue in this way until the process terminates with the new n... |
IMO-1962-2 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_2 | Determine all real numbers \(x\) which satisfy the inequality:
\[
\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}
\]
(Note that the problem as written in the official PDF https://www.imo-official.org/year_info.aspx?year=1962 does not have the outer \(\sqrt{\ }\)) | [
"Obviously we need \\(\\sqrt{3-x} \\geq \\sqrt{x+1}\\) for the outer square root to be defined, \\(x\\leq 3\\) for the first inner square root to be defined, and \\(x\\geq -1\\) for the second inner square root to be defined. Solving these we get that the left hand side is defined for \\(x\\in \\left[ -1,1 \\right]... |
IMO-1962-3 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_3 | Consider the cube \(ABCDA'B'C'D'\)(\(ABCD\) and \(A'B'C'D'\) are the upper and lower bases, respectively, and edges \(AA'\), \(BB'\), \(CC'\), \(DD'\) are parallel). The point \(X\) moves at constant speed along the perimeter of the square \(ABCD\) in the direction \(ABCDA\), and the point \(Y\) moves at the same rate along the perimeter of the square \(B'C'CB\) in the direction \(B'C'CBB'\). Points \(X\) and \(Y\) begin their motion at the same instant from the starting positions \(A\) and \(B'\), respectively. Determine and draw the locus of the midpoints of the segments \(XY\). | [
"First we prove a small lemma: If the particles \\(X\\) and \\(Y\\) move along straight lines at constant velocities, then the locus of the midpoint of \\(XY\\) is also a line. This is rather trivial, since all lines in 3D space may take the parametric form \\((at+n, bt+m, ct+p)\\), with \\(t\\) being time, and the... |
IMO-1962-4 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_4 | Solve the equation \(\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1\). | [
"First, note that we can write the left hand side as a cubic function of \\(\\cos^2 x\\). So there are at most \\(3\\) distinct values of \\(\\cos^2 x\\) that satisfy this equation. Therefore, if we find three values of \\(x\\) that satisfy the equation and produce three different \\(\\cos^2 x\\), then we found all... |
IMO-1962-5 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_5 | On the circle \(K\) there are given three distinct points \(A,B,C\). Construct (using only straightedge and compass) a fourth point \(D\) on \(K\) such that a circle can be inscribed in the quadrilateral thus obtained. | [
"Aviso: La siguiente solución está en español\n\ngeneralidad ya que es simétrico para todas las combinaciones posibles)\n\nCircunferencia K en 2 puntos, uno de ello se nombrará con E (va a estar del mismo lado que B con respecto a AC) y el otro se nombrará F\n\niguales, abrir con el compás la distancia BE y se traz... |
IMO-1962-6 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_6 | Consider an isosceles triangle. Let \(r\) be the radius of its circumscribed circle and \(\rho\) the radius of its inscribed circle. Prove that the distance \(d\) between the centers of these two circles is
\[
d=\sqrt{r(r-2\rho)}
\]
. | [
"<geogebra>d1f93636341cbe0bc2f98c788171d8a55d94f8c8</geogebra> Instead of an isosceles triangle, let us consider an arbitrary triangle \\(ABC\\). Let \\(ABC\\) have circumcenter \\(O\\) and incenter \\(I\\). Extend \\(AI\\) to meet the circumcircle again at \\(L\\). Then extend \\(LO\\) so it meets the circumcircle... |
IMO-1962-7 | https://artofproblemsolving.com/wiki/index.php/1962_IMO_Problems/Problem_7 | The tetrahedron \(SABC\) has the following property: there exist five spheres, each tangent to the edges \(SA, SB, SC, BC, CA, AB\), or to their extensions.
(a) Prove that the tetrahedron \(SABC\) is regular.
(b) Prove conversely that for every regular tetrahedron five such spheres exist. | [
"\\[\nIMO 1962 P7 01.png\n\\]\n\nPart (a)\n\nLet points \\(P_{SA}, P_{SB}, P_{SC}, P_{BC}, P_{CA}, P_{AB}\\) be the points where the smallest sphere is tangent to the edges \\(SA, SB, SC, BC, CA, AB\\) respectively.\n\nFor each face of the tetrahedron there is a circular cross section of the smallest sphere. Since ... |
IMO-1963-1 | https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_1 | Find all real roots of the equation
\[
\sqrt{x^2-p}+2\sqrt{x^2-1}=x
\]
,
where \(p\) is a real parameter. | [
"Assuming \\(x \\geq 0\\), square the equation, obtaining\n\n\\[\n4\\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2\n\\]\n\n. If we have \\(p + 4 \\geq 4x^2\\), we can square again, obtaining\n\n\\[\nx^2 = \\frac {(p - 4)^2}{4(4 - 2p)} \\implies x = \\pm\\frac {p - 4}{2\\sqrt {4 - 2p}}\n\\]\n\nWe must have \\(4 - 2p > 0 ... |
IMO-1963-2 | https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_2 | Point \(A\) and segment \(BC\) are given. Determine the locus of points in space which are the vertices of right angles with one side passing through \(A\), and the other side intersecting the segment \(BC\). | [
"Let \\(\\omega_1\\) be the circle with diameter \\(AB\\), and let \\(\\omega_2\\) be the circle with diameter \\(AC\\). Then the locus is simply the set of points inside either \\(\\omega_1\\) or \\(\\omega_2\\), but not both.\n\nTo see this, suppose the right angle's ray that does not pass through \\(A\\) interse... |
IMO-1963-3 | https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_3 | In an \(n\)-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
\[
a_1\ge a_2\ge \cdots \ge a_n.
\]
Prove that \(a_1=a_2=\cdots = a_n\). | [
"Let \\(a_1 = p_1p_2\\), \\(a_2 = p_2p_3\\), etc.\n\nPlot the \\(n\\)-gon on the cartesian plane such that \\(p_1p_2\\) is on the \\(x\\)-axis and the entire shape is above the \\(x\\)-axis. There are two cases: the number of sides is even, and the number of sides is odd:\n\n\\[\n\\textbf{Case 1: Even}\n\\]\n\nIn t... |
IMO-1963-4 | https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_4 | Find all solutions \(x_1,x_2,x_3,x_4,x_5\) of the system
\[
\begin{eqnarray*} x_5+x_2&=&yx_1\\ x_1+x_3&=&yx_2\\ x_2+x_4&=&yx_3\\ x_3+x_5&=&yx_4\\ x_4+x_1&=&yx_5,\end{eqnarray*}
\]
where \(y\) is a parameter. | [
"Notice: The following words are Chinese.\n\n首先,我们可以将以上5个方程相加,得到:\n\n\\[\n2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)\n\\]\n\n当\\(x_1+x_2+x_3+x_4+x_5=0\\)时,因为\\(x_1,x_2,x_3,x_4,x_5\\)关于原方程组轮换对称,所以\n\n\\[\nx_1=x_2=x_3=x_4=x_5=0\n\\]\n\n若反之,则方程两边同除以\\((x_1+x_2+x_3+x_4+x_5)\\),得到\\(y=2\\),显然解为\n\n\\[\nx_1=x_2=x_3=x_... |
IMO-1963-5 | https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_5 | Prove that \(\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}\). | [
"Because the sum of the \\(x\\)-coordinates of the seventh roots of unity is \\(0\\), we have\n\n\\[\n\\cos{\\frac{2\\pi}{7}} + \\cos{\\frac{4\\pi}{7}} + \\cos{\\frac{6\\pi}{7}} + \\cos{\\frac{8\\pi}{7}} + \\cos{\\frac{10\\pi}{7}} + \\cos{\\frac{12\\pi}{7}} + \\cos{\\frac{14\\pi}{7}} = 0\n\\]\n\n\\[\n\\cos{\\frac{2... |
IMO-1963-6 | https://artofproblemsolving.com/wiki/index.php/1963_IMO_Problems/Problem_6 | Five students, \(A,B,C,D,E\), took part in a contest. One prediction was that the contestants would finish in the order \(ABCDE\). This prediction was very poor. In fact no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order \(DAECB\). This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished. | [
"We are given that no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so in order \\(ABCDE\\). None of them finished in that order. Also only two of them had their actual positions in \\(DAECB\\). After imposing these two conditions the list of po... |
IMO-1964-1 | https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_1 | (a) Find all positive integers \(n\) for which \(2^n-1\) is divisible by \(7\).
(b) Prove that there is no positive integer \(n\) for which \(2^n+1\) is divisible by \(7\). | [
"We claim \\(2^n\\) is equivalent to \\(2, 4,\\) and \\(1\\) \\(\\pmod{7}\\) for \\(n\\) congruent to \\(1\\), \\(2\\), and \\(0\\) \\(\\pmod{3}\\), respectively.\n\n(a) From the statement above, only \\(n\\) divisible by \\(3\\) will work.\n\n(b) Again from the statement above, \\(2^n\\) can never be congruent to ... |
IMO-1964-2 | https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_2 | Suppose \(a, b, c\) are the sides of a triangle. Prove that
\[
a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.
\] | [
"Let \\(b+c-a = x\\), \\(c+a-b = y\\), and \\(a+b-c = z\\). Then, \\(a = \\frac{y+z}{2}\\), \\(b = \\frac{x+z}{2}\\), and \\(c = \\frac{x+y}{2}\\). By AM-GM,\n\n\\[\n\\frac{x+y}{2} \\geq \\sqrt{xy},\n\\]\n\n\\[\n\\frac{y+z}{2} \\geq \\sqrt{yz},\n\\]\n\n\\[\n\\textrm{and }\\frac{x+z}{2} \\geq \\sqrt{xz}.\n\\]\n\nMul... |
IMO-1964-3 | https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_3 | A circle is inscribed in a triangle \(ABC\) with sides \(a,b,c\). Tangents to the circle parallel to the sides of the triangle are contructed. Each of these tangents cuts off a triangle from \(\triangle ABC\). In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of \(a,b,c\)). | [
"Let the tangent to the in circle parallel to BC cut AB,AC at D & E respectively. Similarly let the tangent to the same parallel to AB cut AC,BC at F & G respectively and the tangent to the same parallel to AC cuts BC,AB at H,M respectively. Let the incircle touch the sides BC,CA,AB at P,Q,R respectively and let th... |
IMO-1964-4 | https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_4 | Seventeen people correspond by mail with one another - each one with all the rest. In their letters only three different topics are discussed. Each pair of correspondents deals with only one of these topics. Prove that there are at least three people who write to each other about the same topic. | [
"Lemma: Consider a complete graph with 6 vertices colored with 2 colors. There exists a monochromatic triangle.\n\nProof: Consider one vertex and all connections leading out from it. Call it \\(V_1\\). It has 5 edges coming out from it. By the Pigeonhole Principle, there are at least 3 of the same color. Call this ... |
IMO-1964-5 | https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_5 | Suppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maximum number of intersections that these perpendiculars can have. | [
"Suppose, those five points are \\((A, B, C, D, E)\\). Now, we want to create some special structure. Let, we take the line \\(BC\\) and draw a perpendicular from \\(A\\) on \\(BC\\), andd call it \\(P_1\\). We can do this set up in \\(\\binom{5}{1}\\binom{4}{2}=30\\) ways. There will \\(30\\) such \\(P_i\\) s.\n\n... |
IMO-1964-6 | https://artofproblemsolving.com/wiki/index.php/1964_IMO_Problems/Problem_6 | In tetrahedron \(ABCD\), vertex \(D\) is connected with \(D_0\), the centroid of \(\triangle ABC\). Lines parallel to \(DD_0\) are drawn through \(A,B\) and \(C\). These lines intersect the planes \(BCD, CAD\) and \(ABD\) in points \(A_1, B_1,\) and \(C_1\), respectively. Prove that the volume of \(ABCD\) is one third the volume of \(A_1B_1C_1D_0\). Is the result true if point \(D_o\) is selected anywhere within \(\triangle ABC\)? | [
"\\[\nIMO 1964 P6 01.png\n\\]\n\nLet \\(A_{2}\\) be the point where line \\(AD_{0}\\) intersects line \\(BC\\)\n\nLet \\(B_{2}\\) be the point where line \\(BD_{0}\\) intersects line \\(AC\\)\n\nLet \\(C_{2}\\) be the point where line \\(CD_{0}\\) intersects line \\(AB\\)\n\nFrom centroid properties we have:\n\n\\[... |
IMO-1965-1 | https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_1 | Determine all values \(x\) in the interval \(0\leq x\leq 2\pi\) which satisfy the inequality
\[
2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right| \leq \sqrt{2}.
\] | [
"We shall deal with the left side of the inequality first (\\(2\\cos x \\leq \\left| \\sqrt{1+\\sin 2x} - \\sqrt{1-\\sin 2x } \\right|\\)) and the right side after that.\n\nIt is clear that the left inequality is true when \\(\\cos x\\) is non-positive, and that is when \\(x\\) is in the interval \\([\\pi/2, 3\\pi/... |
IMO-1965-2 | https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_2 | Consider the system of equations
\[
a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0
\]
\[
a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0
\]
\[
a_{31}x_1 + a_{32}x_2 + a_{33}x_3 = 0
\]
with unknowns \(x_1\), \(x_2\), \(x_3\). The coefficients satisfy the conditions:
(a) \(a_{11}\), \(a_{22}\), \(a_{33}\) are positive numbers;
(b) the remaining coefficients are negative numbers;
(c) in each equation, the sum of the coefficients is positive.
Prove that the given system has only the solution \(x_1 = x_2 = x_3 = 0\). | [
"Clearly if the \\(x_i\\) are all equal, then they are equal to 0. Now let's assume WLOG that \\(x_1=0\\). If \\(x_2\\) or \\(x_3\\) is 0, then the other is clearly zero, so let's consider the case where neither are 0. \\(a_{12}\\) and \\(a_{21}\\) are negative, so exactly one of \\(x_2\\) or \\(x_3\\) is positive.... |
IMO-1965-3 | https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_3 | Given the tetrahedron \(ABCD\) whose edges \(AB\) and \(CD\) have lengths \(a\) and \(b\) respectively. The distance between the skew lines \(AB\) and \(CD\) is \(d\), and the angle between them is \(\omega\). Tetrahedron \(ABCD\) is divided into two solids by plane \(\varepsilon\), parallel to lines \(AB\) and \(CD\). The ratio of the distances of \(\varepsilon\) from \(AB\) and \(CD\) is equal to \(k\). Compute the ratio of the volumes of the two solids obtained. | [
"Let the plane meet \\(AD\\) at \\(X\\), \\(BD\\) at \\(Y\\), \\(BC\\) at \\(Z\\) and \\(AC\\) at \\(W\\). Take a plane parallel to \\(BCD\\) through \\(WX\\) and let it meet \\(AB\\) in \\(P\\).\n\n\\[\nProb 1965 3.png\n\\]\n\nSince the distance of \\(AB\\) from \\(WXYZ\\) is \\(k\\) times the distance of \\(CD\\)... |
IMO-1965-4 | https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_4 | Find all sets of four real numbers \(x_1\), \(x_2\), \(x_3\), \(x_4\) such that the sum of any one and the product of the other three is equal to \(2\). | [
"Let \\(P = x_1x_2x_3x_4\\) be the product of the four real numbers.\n\nThen, for \\(i = 1,2,3,4\\) we have: \\(x_i + \\prod_{j \\neq i}x_j = 2\\).\n\nMultiplying by \\(x_i\\) yields:\n\n\\(x^2_i + P = 2x_i \\Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \\Longleftrightarrow x_i = 1 \\pm t\\) where \\(t = \\pm ... |
IMO-1965-5 | https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_5 | Consider \(\triangle OAB\) with acute angle \(AOB\). Through a point \(M \neq O\) perpendiculars are drawn to \(OA\) and \(OB\), the feet of which are \(P\) and \(Q\) respectively. The point of intersection of the altitudes of \(\triangle OPQ\) is \(H\). What is the locus of \(H\) if \(M\) is permitted to range over (a) the side \(AB\), (b) the interior of \(\triangle OAB\)? | [
"Let \\(O(0,0),A(a,0),B(b,c)\\). Equation of the line \\(AB: y=\\frac{c}{b-a}(x-a)\\). Point \\(M \\in AB : M(\\lambda,\\frac{c}{b-a}(\\lambda-a))\\). Easy, point \\(P(\\lambda,0)\\). Point \\(Q = OB \\cap MQ\\), \\(MQ \\bot OB\\). Equation of \\(OB : y=\\frac{c}{b}x\\), equation of \\(MQ : y=-\\frac{b}{c}(x-\\lamb... |
IMO-1965-6 | https://artofproblemsolving.com/wiki/index.php/1965_IMO_Problems/Problem_6 | In a plane a set of \(n\) points (\(n\geq 3\)) is given. Each pair of points is connected by a segment. Let \(d\) be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length \(d\). Prove that the number of diameters of the given set is at most \(n\). | [
"Image of problem Solution. Credits to user awe-sum.\n\n## Remarks (added by pf02, October 2024)\n\n1. As a public service, I will upload the \"Image of problem Solution\" mentioned above to this web page. That way, a reader can see the \"Solution\" immediately, without having to go to another web site, and we are ... |
IMO-1966-1 | https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_1 | In a mathematical contest, three problems, \(A\), \(B\), and \(C\) were posed. Among the participants there were \(25\) students who solved at least one problem each. Of all the contestants who did not solve problem \(A\), the number who solved \(B\) was twice the number who solved \(C\). The number of students who solved only problem \(A\) was one more than the number of students who solved \(A\) and at least one other problem. Of all students who solved just one problem, half did not solve problem \(A\). How many students solved only problem \(B\)? | [
"Let us draw a Venn Diagram.\n\n\\[\nIMO1966.png\n\\]\n\nLet \\(a\\) be the number of students solving both B and C. Then for some positive integer \\(x\\), \\(2x - a\\) students solved B only, and \\(x - a\\) students solved C only. Let \\(2y - 1\\) be the number of students solving A; then \\(y\\) is the number o... |
IMO-1966-2 | https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_2 | Let \(a\), \(b\), and \(c\) be the lengths of the sides of a triangle, and \(\alpha,\beta,\gamma\) respectively, the angles opposite these sides. Prove that if
\[
a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}),
\]
the triangle is isosceles. | [
"We'll prove that the triangle is isosceles with \\(a=b\\). We'll prove that \\(a=b\\). Assume by way of contradiction WLOG that \\(a>b\\). First notice that as \\(\\gamma = \\pi -\\alpha-\\beta\\) then and the identity \\(\\tan\\left(\\frac \\pi 2 - x \\right)=\\cot x\\) our equation becomes:\n\n\\[\na+b=\\cot \\f... |
IMO-1966-3 | https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_3 | Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space. | [
"We will need the following lemma to solve this problem:\n\n\\(\\emph{Lemma:}\\) Let \\(MNOP\\) be a regular tetrahedron, and \\(T\\) a point inside it. Let \\(x_1, x_2, x_3, x_4\\) be the distances from \\(T\\) to the faces \\(MNO, MNP, MOP\\), and \\(NOP\\). Then, \\(x_1 + x_2 + x_3 + x_4\\) is constant, independ... |
IMO-1966-4 | https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_4 | Prove that for every natural number \(n\), and for every real number \(x \neq \frac{k\pi}{2^t}\) (\(t=0,1, \dots, n\); \(k\) any integer)
\[
\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}
\] | [
"First, we prove \\(\\cot \\theta - \\cot 2\\theta = \\frac {1}{\\sin 2\\theta}\\).\n\nLHS\\(\\ =\\ \\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}\\)\n\n\\[\n= \\frac{2\\cos^2 \\theta}{2\\cos \\theta \\sin \\theta}-\\frac{2\\cos^2 \\theta -1}{\\sin 2\\theta}\n\\]\n\n\\[\n=\\frac{2\\cos^... |
IMO-1966-5 | https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_5 | Solve the system of equations
\[
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_1 - a_2| x_2 + |a_1 - a_3| x_3 + |a_1 - a_4| x_4 = 1 \\ |a_2 - a_1| x_1 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_2 - a_3| x_3 + |a_2 - a_4| x_4 = 1 \\ |a_3 - a_1| x_1 + |a_3 - a_2| x_2 + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |a_3 - a_4| x_4 = 1 \\ |a_4 - a_1| x_1 + |a_4 - a_2| x_2 + |a_4 - a_3| x_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1
\]
where \(a_1, a_2, a_3, a_4\) are four different real numbers. | [
"Take a1 > a2 > a3 > a4. Subtracting the equation for i=2 from that for i=1 and dividing by (a1 - a2) we get:\n\n\\[\n- x1 + x2 + x3 + x4 = 0.\n\\]\n\nSubtracting the equation for i=4 from that for i=3 and dividing by (a3 - a4) we get:\n\n\\[\n- x1 - x2 - x3 + x4 = 0.\n\\]\n\nHence x1 = x4. Subtracting the equation... |
IMO-1966-6 | https://artofproblemsolving.com/wiki/index.php/1966_IMO_Problems/Problem_6 | In the interior of sides \(BC, CA, AB\) of triangle \(ABC\), any points \(K, L,M\), respectively, are selected. Prove that the area of at least one of the triangles \(AML, BKM, CLK\) is less than or equal to one quarter of the area of triangle \(ABC\). | [
"Let the lengths of sides \\(BC\\), \\(CA\\), and \\(AB\\) be \\(a\\), \\(b\\), and \\(c\\), respectively. Let \\(BK=d\\), \\(CL=e\\), and \\(AM=f\\).\n\nNow assume for the sake of contradiction that the areas of \\(\\Delta AML\\), \\(\\Delta BKM\\), and \\(\\Delta CLK\\) are all at greater than one fourth of that ... |
IMO-1967-1 | https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_1 | Let \(ABCD\) be a parallelogram with side lengths \(AB = a\), \(AD = 1\), and with \(\angle BAD = \alpha\). If \(\Delta ABD\) is acute, prove that the four circles of radius \(1\) with centers \(A\), \(B\), \(C\), \(D\) cover the parallelogram if and only if
\(a\leq \cos \alpha+\sqrt{3}\sin \alpha\) \(\ \ \ \ \ \ \ \ \ \ (1)\) | [
"To start our proof we draw a parallelogram with the requested sides. We notice that by drawing the circles with centers A, B, C, D that the length of \\(a\\) must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.\n\nTo prove our conjecture we draw a parall... |
IMO-1967-2 | https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_2 | Prove that if one and only one edge of a tetrahedron is greater than \(1\), then its volume is \(\le \frac{1}{8}\). | [
"Assume \\(CD>1\\) and let \\(AB=x\\). Let \\(P,Q,R\\) be the feet of perpendicular from \\(C\\) to \\(AB\\) and \\(\\triangle ABD\\) and from \\(D\\) to \\(AB\\), respectively.\n\nSuppose \\(BP>PA\\). We have that \\(CP=\\sqrt{CB^2-BT^2}\\le\\sqrt{1-\\frac{x^2}4}\\), \\(CQ\\le CP\\le\\sqrt{1-\\frac{x^2}4}\\). We a... |
IMO-1967-3 | https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_3 | Let \(k, m, n\) be natural numbers such that \(m+k+1\) is a prime greater than \(n+1.\) Let \(c_s=s(s+1).\) Prove that the product
\[
(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)
\]
is divisible by the product \(c_1c_2\cdots c_n\). | [
"For any \\(m, n\\), \\(c_m - c_n = m(m+1) - n(n+1) = m^2 + m - n^2 - n = (m+n)(m-n) + m - n = (m+n+1)(m-n)\\).\n\nWe can therefore write the product in the problem as follows:\n\n\\[\n(c_{m+1}-c_k)(c_{m+2}-c_k)\\cdots (c_{m+n}-c_k)\n\\]\n\n\\[\n= \\left(\\prod_{a = 1}^n m+a+k+1\\right)\\left(\\prod{a = 1}^n m+a-k\... |
IMO-1967-4 | https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_4 | Let \(A_0B_0C_0\) and \(A_1B_1C_1\) be any two acute-angled triangles. Consider all triangles \(ABC\) that are similar to \(\triangle A_1B_1C_1\) (so that vertices \(A_1\), \(B_1\), \(C_1\) correspond to vertices \(A\), \(B\), \(C\), respectively) and circumscribed about triangle \(A_0B_0C_0\) (where \(A_0\) lies on \(BC\), \(B_0\) on \(CA\), and \(AC_0\) on \(AB\)). Of all such possible triangles, determine the one with maximum area, and construct it. | [
"We construct a point \\(P\\) inside \\(A_0B_0C_0\\) s.t. \\(\\angle X_0PY_0=\\pi-\\angle X_1Z_1Y_1\\), where \\(X,Y,Z\\) are a permutation of \\(A,B,C\\). Now construct the three circles \\(\\mathcal C_A=(B_0PC_0),\\mathcal C_B=(C_0PA_0),\\mathcal C_C=(A_0PB_0)\\). We obtain any of the triangles \\(ABC\\) circumsc... |
IMO-1967-5 | https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_5 | Let \(a_1,\ldots,a_8\) be reals, not all equal to zero. Let
\[
c_n = \sum^8_{k=1} a^n_k
\]
for \(n=1,2,3,\ldots\). Given that among the numbers of the sequence \((c_n)\), there are infinitely many equal to zero, determine all the values of \(n\) for which \(c_n = 0.\) | [
"\\(c_n\\) must be zero for all odd \\(n\\).\n\nProof: WLOG suppose that \\(a_1 \\geq a_2 \\geq ... \\geq a_8\\). If \\(a_1+a_8 > 0\\) then for sufficiently high odd \\(n\\), \\(c_n\\) will be dominated by \\(a_1\\) alone i.e. it will always be positive. Similarly if \\(a_1+a_8 < 0\\); hence \\(a_1=-a_8\\). Now for... |
IMO-1967-6 | https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems/Problem_6 | In a sports contest, there were \(m\) medals awarded on \(n\) successive days \((n > 1)\). On the first day, one medal and \(\frac{1}{7}\) of the remaining \(m - 1\) medals were awarded. On the second day, two medals and \(\frac{1}{7}\) of the now remaining medals were awarded; and so on. On the n-th and last day, the remaining \(n\) medals were awarded. How many days did the contest last, and how many medals were awarded altogether? | [
"This is not a particularly elegant solution, but if you start from 1 and go all the way in a clever method, by only guessing those that are 1 more than a multiple of 7, you arrive at the answer of 36.\n\n## Comment (added by pf02, August 2024)\n\nIndeed, as the author says, the above is not an elegant solution. Al... |
IMO-1968-1 | https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_1 | Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another. | [
"In triangle \\(ABC\\), let \\(BC=a\\), \\(AC=b\\), \\(AB=c\\), \\(\\angle ABC=\\alpha\\), and \\(\\angle BAC=2\\alpha\\). Using the Law of Sines gives that\n\n\\[\n\\frac{b}{\\sin{\\alpha}}=\\frac{a}{\\sin{2\\alpha}}\\Rightarrow \\frac{\\sin{2\\alpha}}{\\sin{\\alpha}}=2\\cos{\\alpha}=\\frac{a}{b}\n\\]\n\nTherefore... |
IMO-1968-2 | https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_2 | Find all natural numbers \(x\) such that the product of their digits (in decimal notation) is equal to \(x^2 - 10x - 22\). | [
"Let the decimal expansion of \\(x\\) be \\(\\overline{d_1d_2d_3\\dots d_n}\\), where \\(d_i\\) are base-10 digits. We then have that \\(x\\geq d_1\\cdot 10^{n-1}\\). However, the product of the digits of \\(x\\) is \\(d_1d_2d_3\\dots d_n\\leq d_1\\cdot 10\\cdot 10\\dots 10=d_1\\cdot 10^{n-1}\\), with equality only... |
IMO-1968-3 | https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_3 | Consider the system of equations
\[
ax_1^2 + bx_1 + c = x_2
\]
\[
ax_2^2 + bx_2 + c = x_3
\]
\[
\cdots
\]
\[
ax_{n-1}^2 + bx_{n-1} + c = x_n
\]
\[
ax_n^2 + bx_n + c = x_1
\]
with unknowns \(x_1, x_2, \cdots, x_n\) where \(a, b, c\) are real and \(a \neq 0\). Let \(\Delta = (b - 1)^2 - 4ac\). Prove that for this system
(a) if \(\Delta < 0\), there is no solution,
(b) if \(\Delta = 0\), there is exactly one solution,
(c) if \(\Delta > 0\), there is more than one solution. | [
"Adding the \\(n\\) equations together yields\n\n\\[\n\\sum_{i=1}^{n} (ax_i^2+(b-1)x_i+c)=0\n\\]\n\nLet \\(s_i=ax_i^2+(b-1)x_i+c\\).\n\n(a) If \\(\\Delta<0\\), then there is no solution to the quadratic equation \\(ax^2+(b-1)x+c\\), as the determinant is negative. This implies that either \\(s_i>0\\) for all \\(i\\... |
IMO-1968-4 | https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_4 | Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which are the sides of a triangle. | [
"Let the edges of one of the faces of the tetrahedron have lengths \\(a\\), \\(b\\), and \\(c\\). Let \\(d\\), \\(e\\), and \\(f\\) be the lengths of the sides that are not adjacent to the sides with lengths \\(a\\), \\(b\\), and \\(c\\), respectively.\n\nWithout loss of generality, assume that \\(\\max(a,b,c,d,e,f... |
IMO-1968-5 | https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_5 | Let \(f\) be a real-valued function defined for all real numbers \(x\) such that, for some positive constant \(a\), the equation
\[
f(x + a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2}
\]
holds for all \(x\).
(a) Prove that the function \(f\) is periodic (i.e., there exists a positive number \(b\) such that \(f(x + b) = f(x)\) for all \(x\)).
(b) For \(a = 1\), give an example of a non-constant function with the required properties. | [
"(a) Since\n\n\\[\nf(x+a) \\ge \\frac{1}{2}\n\\]\n\nis true for any \\(x\\), and\n\n\\[\nf(x+a)(1-f(x+a)) = \\frac{1}{4} - (f(x)-(f(x))^2) = (\\frac{1}{2}-f(x))^2\n\\]\n\nWe have:\n\n\\[\nf(x+2a) = \\frac{1}{2} + \\sqrt{(\\frac{1}{2}-f(x))^2} = \\frac{1}{2} + (f(x) - \\frac{1}{2}) = f(x)\n\\]\n\nTherefore \\(f\\) i... |
IMO-1968-6 | https://artofproblemsolving.com/wiki/index.php/1968_IMO_Problems/Problem_6 | For every natural number \(n\), evaluate the sum
\[
\sum_{k = 0}^\infty\bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] = \Big[\frac{n + 1}{2}\Big] + \Big[\frac{n + 2}{4}\Big] + \cdots + \bigg[\frac{n + 2^k}{2^{k + 1}}\bigg] + \cdots
\]
(The symbol \([x]\) denotes the greatest integer not exceeding \(x\).) | [
"I shall prove that the summation is equal to \\(n\\).\n\nLet the binary representation of \\(n\\) be \\(\\overline{d_xd_{x-1}\\dots d_1d_0}_2\\), where \\(d_i\\in\\{0,1\\}\\) for all \\(i\\), and \\(x=\\lfloor \\log_2 n\\rfloor\\). Note that if \\(d_i=0\\), then \\(\\Big[\\frac{n + 2^i}{2^{i+1}}\\Big]=\\Big[\\frac... |
IMO-1969-1 | https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_1 | Prove that there are infinitely many natural numbers \(a\) with the following property: the number \(z = n^4 + a\) is not prime for any natural number \(n\). | [
"Suppose that \\(a = 4k^4\\) for some \\(k\\). We will prove that \\(a\\) satisfies the property outlined above.\n\nThe polynomial \\(n^4 + 4k^4\\) can be factored as follows:\n\n\\[\nn^4 + 4k^4\n\\]\n\n\\[\n= n^4 + 4n^2k^2 + 4k^4 - 4n^2k^2\n\\]\n\n\\[\n= (n^2 + 2k^2)^2 - (2nk)^2\n\\]\n\n\\[\n= (n^2 + 2k^2 - 2nk)(n... |
IMO-1969-2 | https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_2 | Let \(a_1, a_2,\cdots, a_n\) be real constants, \(x\) a real variable, and
\[
f(x)=\cos(a_1+x)+\frac{1}{2}\cos(a_2+x)+\frac{1}{4}\cos(a_3+x)+\cdots+\frac{1}{2^{n-1}}\cos(a_n+x).
\]
Given that \(f(x_1)=f(x_2)=0,\) prove that \(x_2-x_1=m\pi\) for some integer \(m.\) | [
"Because the period of \\(\\cos(x)\\) is \\(2\\pi\\), the period of \\(f(x)\\) is also \\(2\\pi\\).\n\n\\[\nf(x_1)=f(x_2)=f(x_1+x_2-x_1)\n\\]\n\nWe can get \\(x_2-x_1 = 2k\\pi\\) for \\(k\\in N^*\\). Thus, \\(x_2-x_1=m\\pi\\) for some integer \\(m.\\)",
"By the cosine addition formula,\n\n\\[\nf(x)=(\\cos{a_1}+\\... |
IMO-1969-3 | https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_3 | For each of \(k = 1\), \(2\), \(3\), \(4\), \(5\) find necessary and sufficient conditions on \(a > 0\) such that there exists a tetrahedron with \(k\) edges length \(a\) and the remainder length \(1\). | [
"A plodding question. Take the tetrahedron to be ABCD.\n\nTake k = 1 and AB to have length a, the other edges length 1. Then we can hinge triangles ACD and BCD about CD to vary AB. The extreme values evidently occur with A, B, C, D coplanar. The least value, 0, when A coincides with B, and the greatest value √3, wh... |
IMO-1969-4 | https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_4 | A semicircular arc \(\gamma\) is drawn with \(AB\) as diameter. \(C\) is a point on \(\gamma\) other than \(A\) and \(B\), and \(D\) is the foot of the perpendicular from \(C\) to \(AB\). We consider three circles, \(\gamma_1, \gamma_2, \gamma_3\), all tangent to the line \(AB\). Of these, \(\gamma_1\) is inscribed in \(\triangle ABC\), while \(\gamma_2\) and \(\gamma_3\) are both tangent to \(CD\) and \(\gamma\), one on each side of \(CD\). Prove that \(\gamma_1, \gamma_2\), and \(\gamma_3\) have a second tangent in common. | [
"Denote the triangle sides \\(a = BC, b = CA, c = AB\\). Let \\(\\omega\\) be the circumcircle of the right angle triangle \\(\\triangle ABC\\) centered at the midpoint \\(O\\) of its hypotenuse \\(c = AB\\). Let \\(R, S, T\\) be the tangency points of the circles \\(K_1, K_2, K_3\\) with the line AB. In an inversi... |
IMO-1969-5 | https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_5 | Given \(n > 4\) points in the plane such that no three are collinear, prove that there are at least \(C(n-3, 2) = {n-3 \choose 2} = \frac{(n-3)(n-4)}{2}\) convex quadrilaterals whose vertices are four of the given points. | [
"Orient the points so that one of them corresponds to the origin (A), another lies on the y-axis (B), and all the others are in either quandrant I or IV. (In other words, you are creating a boundary line.) Select one of the points (C) which minimizes the angle BAC. You cannot have two of them because then they woul... |
IMO-1969-6 | https://artofproblemsolving.com/wiki/index.php/1969_IMO_Problems/Problem_6 | Prove that for all real numbers \(x_1, x_2, y_1, y_2, z_1, z_2\), with \(x_1 > 0, x_2 > 0, x_1y_1 - z_1^2 > 0, x_2y_2 - z_2^2 > 0\), the inequality
\[
\frac{8}{(x_1 + x_2)(y_1 + y_2) - (z_1 + z_2)^2} \leq \frac{1}{x_1y_1 - z_1^2} + \frac{1}{x_2y_2 - z_2^2}
\]
is satisfied. Give necessary and sufficient conditions for equality. | [
"Let \\(A=x_1y_1 - z_1^2>0\\) and \\(B=x_2y_2 - z_2^2>0\\)\n\nFrom AM-GM:\n\n\\(\\sqrt{AB} \\le \\frac{A+B}{2}\\) with equality at \\(A=B\\)\n\n\\[\n4AB \\le (A+B)^2\n\\]\n\n\\[\n\\frac{4}{A+B} \\le \\frac{A+B}{AB}\n\\]\n\n\\[\n\\frac{8}{2(A+B)} \\le \\frac{A+B}{AB}\n\\]\n\n\\(\\frac{8}{2(A+B)} \\le \\frac{1}{A}+\\... |
IMO-1970-1 | https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_1 | Let \(M\) be a point on the side \(AB\) of \(\triangle ABC\). Let \(r_1, r_2\), and \(r\) be the inscribed circles of triangles \(AMC, BMC\), and \(ABC\). Let \(q_1, q_2\), and \(q\) be the radii of the escribed circles of the same triangles that lie in the angle \(ACB\). Prove that
\[
\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}
\]
. | [
"We use the conventional triangle notations.\n\nLet \\(I\\) be the incenter of \\(ABC\\), and let \\(I_{c}\\) be its excenter to side \\(c\\). We observe that\n\n\\[\nr \\left[ \\cot\\left(\\frac{A}{2}\\right) + \\cot\\left(\\frac{B}{2}\\right) \\right] = c\n\\]\n\n,\n\nand likewise,\n\n\\[\n\\begin{matrix} c & = &... |
IMO-1970-2 | https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_2 | Let \(a, b\), and \(n\) be integers greater than 1, and let \(a\) and \(b\) be the bases of two number systems. \(A_{n-1}\) and \(A_{n}\) are numbers in the system with base \(a\) and \(B_{n-1}\) and \(B_{n}\) are numbers in the system with base \(b\); these are related as follows:
\[
A_{n} = x_{n}x_{n-1}\cdots x_{0}, A_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}
\]
,
\[
B_{n} = x_{n}x_{n-1}\cdots x_{0}, B_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}
\]
,
\[
x_{n} \neq 0, x_{n-1} \neq 0.
\]
Prove:
\[
\frac{A_{n-1}}{A_{n}} < \frac{B_{n-1}}{B_{n}}
\]
if and only if
\[
a > b
\]
. | [
"Suppose \\(a>b\\). Then for all integers \\(0 \\le k \\le n\\), \\(x_n x_k a^n b^k \\ge x_n x_k b^n a^k\\), with equality only when \\(k=n\\) or \\(x_k = 0\\). (In particular, we have strict inequality for \\(k=n-1\\).) In summation, this becomes\n\n\\[\nx_n a^n \\sum_{k=0}^n x_k b^k > x_n b^n \\sum_{k=0}^n x_k a^... |
IMO-1970-3 | https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_3 | The real numbers \(a_0, a_1, \ldots, a_n, \ldots\) satisfy the condition:
\[
1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots.
\]
The numbers \(b_{1}, b_{2}, \ldots, b_n, \ldots\) are defined by
\[
b_n = \sum_{k=1}^{n} \left( 1 - \frac{a_{k-1}}{a_{k}} \right) \frac{1}{\sqrt{a_k}}
\]
(a) Prove that \(0 \leq b_n < 2\) for all \(n\).
(b) given \(c\) with \(0 \leq c < 2\), prove that there exist numbers \(a_0, a_1, \ldots\) with the above properties such that \(b_n > c\) for large enough \(n\). | [
"\\[\nb_n = \\sum_{k=1}^{n} \\frac{1 - \\frac{a_{k-1}}{a_{k}} }{\\sqrt{a_k}} = \\sum_{k=1}^{n} \\frac{a_k - a_{k-1}}{a_k\\sqrt{a_k}} = \\sum_{k=1}^{n} (a_k - a_{k-1})\\left(a_k^{-\\dfrac{3}{2}}\\right)\n\\]\n\nLet \\(X_k\\) be the rectangle with the verticies: \\((a_{k-1},0)\\); \\((a_{k},0)\\); \\((a_{k},a_k^{-\\d... |
IMO-1970-4 | https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_4 | Find the set of all positive integers \(n\) with the property that the set \(\{ n, n+1, n+2, n+3, n+4, n+5 \}\) can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set. | [
"The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The oth... |
IMO-1970-5 | https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_5 | In the tetrahedron \(ABCD\), angle \(BDC\) is a right angle. Suppose that the foot \(H\) of the perpendicular from \(D\) to the plane \(ABC\) in the tetrahedron is the intersection of the altitudes of \(\triangle ABC\). Prove that
\[
( AB+BC+CA )^2 \leq 6( AD^2 + BD^2 + CD^2 ).
\]
For what tetrahedra does equality hold? | [
"Let us show first that angles \\(ADB\\) and \\(ADC\\) are also right. Let \\(H\\) be the intersection of the altitudes of \\(ABC\\) and let \\(CH\\) meet \\(AB\\) at \\(E\\). Planes \\(CED\\) and \\(ABC\\) are perpendicular and \\(AB\\) is perpendicular to the line of intersection \\(CE\\). Hence \\(AB\\) is perpe... |
IMO-1970-6 | https://artofproblemsolving.com/wiki/index.php/1970_IMO_Problems/Problem_6 | In a plane there are \(100\) points, no three of which are collinear. Consider all possible triangles having these point as vertices. Prove that no more than \(70 \%\) of these triangles are acute-angled. | [
"At most \\(3\\) of the triangles formed by \\(4\\) points can be acute. It follows that at most \\(7\\) out of the \\(10\\) triangles formed by any \\(5\\) points can be acute. For given \\(10\\) points, the maximum number of acute triangles is: the number of subsets of \\(4\\) points times \\(\\frac{3}{\\text{the... |
IMO-1971-1 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_1 | Prove that the following assertion is true for \(n=3\) and \(n=5\), and that it is false for every other natural number \(n>2:\)
If \(a_1, a_2,\cdots, a_n\) are arbitrary real numbers, then \((a_1-a_2)(a_1-a_3)\cdots (a_1-a_n)+(a_2-a_1)(a_2-a_3)\cdots (a_2-a_n)+\cdots+(a_n-a_1)(a_n-a_2)\cdots (a_n-a_{n-1})\ge 0.\) | [
"Denote \\(E_n\\) the expression in the problem, and denote \\(S_n\\) the statement that \\(E_n \\ge 0\\).\n\nTake \\(a_1 < 0\\), and the remaining \\(a_i = 0\\). Then \\(E_n = a_1^{n-1} < 0\\) for \\(n\\) even. So the proposition is false for even \\(n\\).\n\nSuppose \\(n \\ge 7\\) and odd. Take any \\(c > a > b\\... |
IMO-1971-2 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_2 | Consider a convex polyhedron \(P_1\) with nine vertices \(A_1, A_2, \cdots, A_9;\) let \(P_i\) be the polyhedron obtained from \(P_1\) by a translation that moves vertex \(A_1\) to \(A_i(i=2,3,\cdots, 9).\) Prove that at least two of the polyhedra \(P_1, P_2,\cdots, P_9\) have an interior point in common. | [
"WLOG let \\(A_1\\) be the origin \\(0\\). Take any point \\(A_i\\), then \\(P_i=A_i+P_1\\), lies in \\(2 P_1\\), the polyhedron \\(P_1\\) stretched by the factor \\(2\\) on \\(P_1=0\\). More general: take any \\(p,q\\) in any convex shape \\(S\\). Then \\(p+q \\in 2S\\). Prove: since \\(S\\) is convex, \\(\\frac{p... |
IMO-1971-3 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_3 | Prove that the set of integers of the form \(2^k - 3(k = 2; 3; \cdots)\) contains an infinite subset in which every two members are relatively prime. | [
"Wlog, assume \\(a_{n}\\ge 3\\). Then say \\(p_{1}, p_{2}, \\ldots, p_{k}\\in \\mathbb{N}\\) are all the (pairwise distinct) primes dividing \\(2^{a_{n}}-3\\) and let \\(a_{n+1}= a_{n}\\cdot \\prod_{i=1}^{k}(p_{i}-1)\\). Obviously \\(p_{i}\\) is odd, for any \\(i \\in \\overline{1,k}\\). So \\(\\prod_{i=1}^{k}p_{i}... |
IMO-1971-4 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_4 | All the faces of tetrahedron \(ABCD\) are acute-angled triangles. We consider all closed polygonal paths of the form \(XYZTX\) defined as follows: \(X\) is a point on edge \(AB\) distinct from \(A\) and \(B\); similarly, \(Y, Z, T\) are interior points of edges \(BC, CD, DA\), respectively. Prove:
(a) If \(\angle DAB + \angle BCD \neq \angle CDA + \angle ABC\), then among the polygonal paths, there is none of minimal length.
(b) If \(\angle DAB + \angle BCD = \angle CDA + \angle ABC\), then there are infinitely many shortest polygonal paths, their common length being \(2AC \sin(\alpha / 2)\), where \(\alpha = \angle BAC + \angle CAD + \angle DAB\). | [
"Rotate the triangle \\(BCD\\) around the edge \\(BC\\) until \\(ABCD\\) are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting \\(X\\) and \\(Z\\). Therefore, \\(XYB=ZYC\\). Summing the four equations like this, we get exactly \\(\\angle ABC+\\angle ADC=\\angle BCD+\\angle B... |
IMO-1971-5 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_5 | Prove that for every natural number \(m\), there exists a finite set \(S\) of points in a plane with the following property: For every point \(A\) in \(S\), there are exactly \(m\) points in \(S\) which are at unit distance from \(A\). | [
"I shall prove a more general statement about the unit distance graph(\\(V=\\mathbb{R}^2\\), adjacency iff the Euclidean distance between the points is \\(1\\)) and this will follow as a consequence. if \\(G,H\\) occur as unit distance graphs, then so does \\(G\\times H\\)( here \\(G\\times H\\) is described as \\(... |
IMO-1971-6 | https://artofproblemsolving.com/wiki/index.php/1971_IMO_Problems/Problem_6 | Let \(A = (a_{ij})(i, j = 1, 2, \cdots, n)\) be a square matrix whose elements are non-negative integers. Suppose that whenever an element \(a_{ij} = 0\), the sum of the elements in the \(i\)th row and the \(j\)th column is \(\geq n\). Prove that the sum of all the elements of the matrix is \(\geq n^2 / 2\). | [
"Take the the row or column (without loss of generality, a row) with the minimum possible sum \\(S\\) of its elements. Let \\(z\\) be the number of zeros in this row. We will assume \\(S < \\frac{n}{2}\\) once the other case is obvious. Clearly \\(S \\ge n - z \\Rightarrow z \\ge n - S\\). The total sum \\(T\\) of ... |
IMO-1972-1 | https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_1 | Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum. | [
"There are \\(2^{10}-2=1022\\) distinct subsets of our set of 10 two-digit numbers. The sum of the elements of any subset of our set of 10 two-digit numbers must be between \\(10\\) and \\(90+91+92+93+94+95+96+97+98+99 < 10 \\cdot 100 = 1000\\). (There are fewer attainable sums.) As \\(1000 < 1022\\), the Pigeonhol... |
IMO-1972-2 | https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_2 | Prove that if \(n \geq 4\), every quadrilateral that can be inscribed in a circle can be dissected into \(n\) quadrilaterals each of which is inscribable in a circle. | [
"Our initial quadrilateral will be \\(ABCD\\).\n\nFor \\(n=4\\), we do this:\n\nTake \\(E\\in AB,F\\in CD,\\ EF\\|AD\\) with \\(E,F\\) sufficiently close to \\(A,D\\) respectively. Take \\(U\\in AD,V\\in EF\\) such that \\(AEVU\\) is an isosceles trapezoid, with \\(V\\) close enough to \\(F\\) (or \\(U\\) close eno... |
IMO-1972-3 | https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_3 | Let \(m\) and \(n\) be arbitrary non-negative integers. Prove that
\[
\frac{(2m)!(2n)!}{m!n!(m+n)!}
\]
is an integer. (\(0! = 1\).) | [
"Denote the given expression as \\(f(m,n)\\). We intend to show that \\(f(m,n)\\) is integral for all \\(m,n \\geq 0\\). To start, we would like to find a recurrence relation for \\(f(m,n)\\). First, let's look at \\(f(m,n-1)\\):\n\n\\[\n\\begin{align*} f(m,n-1) &=\\frac{(2m)!(2n-2)!}{m!(n-1)!(m+n-1)!}\\\\ ... |
IMO-1972-4 | https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_4 | Find all solutions \((x_1, x_2, x_3, x_4, x_5)\) of the system of inequalities
\[
\begin{align*} (x_1^2 - x_3x_5)(x_2^2 - x_3x_5) \leq 0 \\ (x_2^2 - x_4x_1)(x_3^2 - x_4x_1) \leq 0 \\ (x_3^2 - x_5x_2)(x_4^2 - x_5x_2) \leq 0 \\ (x_4^2 - x_1x_3)(x_5^2 - x_1x_3) \leq 0 \\ (x_5^2 - x_2x_4)(x_1^2 - x_2x_4) \leq 0 \end{align*}
\]
where \(x_1, x_2, x_3, x_4, x_5\) are positive real numbers. | [
"Add the five inequalities together to get\n\n\\[\n(x_1^2 - x_3 x_5)(x_2^2 - x_3 x_5) + (x_2^2 - x_4 x_1)(x_3^2 - x_4 x_1) + (x_3^2 - x_5 x_2)(x_4^2 - x_5 x_2) +\n\\]\n\n\\[\n(x_4^2 - x_1 x_3)(x_5^2 - x_1 x_3) + (x_5^2 - x_2 x_4)(x_1^2 - x_2 x_4) \\leq 0\n\\]\n\nExpanding, multiplying by \\(2\\), and re-combining t... |
IMO-1972-5 | https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_5 | Let \(f\) and \(g\) be real-valued functions defined for all real values of \(x\) and \(y\), and satisfying the equation
\[
f(x + y) + f(x - y) = 2f(x)g(y)
\]
for all \(x, y\). Prove that if \(f(x)\) is not identically zero, and if \(|f(x)| \leq 1\) for all \(x\), then \(|g(y)| \leq 1\) for all \(y\). | [
"Let \\(u>0\\) be the least upper bound for \\(|f(x)|\\) for all \\(x\\). So, \\(|f(x)| \\leq u\\) for all \\(x\\). Then, for all \\(x,y\\),\n\n\\[\n2u \\geq |f(x+y)+f(x-y)| = |2f(x)g(y)|=2|f(x)||g(y)|\n\\]\n\nTherefore, \\(u \\geq |f(x)||g(y)|\\), so \\(|f(x)| \\leq u/|g(y)|\\).\n\nSince \\(u\\) is the least upper... |
IMO-1972-6 | https://artofproblemsolving.com/wiki/index.php/1972_IMO_Problems/Problem_6 | Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane. | [
"Let our planes be \\(\\pi_1,\\pi_2,\\pi_3,\\pi_4\\), which we assume to be parallel to the \\(xy\\)-plane, listed in the increasing order of their \\(z\\)-coordinates. First take a plane \\(\\pi\\) orthogonal to \\(\\pi_i\\), which cuts \\(\\pi_1,\\pi_2,\\pi_3\\) along three lines \\(d_1,d_2,d_3\\). On these three... |
IMO-1973-1 | https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_1 | Point \(O\) lies on line \(g;\) \(\overrightarrow{OP_1}, \overrightarrow{OP_2},\cdots, \overrightarrow{OP_n}\) are unit vectors such that points \(P_1, P_2, \cdots, P_n\) all lie in a plane containing \(g\) and on one side of \(g.\) Prove that if \(n\) is odd,
\[
\left|\overrightarrow{OP_1}+\overrightarrow{OP_2}+\cdots+ \overrightarrow{OP_n}\right|\ge1.
\]
Here \(\left|\overrightarrow{OM}\right|\) denotes the length of vector \(\overrightarrow{OM}.\) | [
"We prove it by induction on the number \\(2n+1\\) of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the \\(2n-1\\) vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm \\(\\ge 1\\) betwen two with norm \\(1\\). T... |
IMO-1973-2 | https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_2 | Determine whether or not there exists a finite set \(M\) of points in space not lying in the same plane such that, for any two points \(A\) and \(B\) of \(M\); one can select two other points \(C\) and \(D\) of \(M\) so that lines \(AB\) and \(CD\) are parallel and not coincident. | [
"In order to solve this problem we can start by finding at least one finite set \\(M\\) that satisfies the condition.\n\nWe start by defining our first set \\(M_{8}\\) with the vertices of a cube of side \\(k\\) as follows: \\(M_{8} = \\{ (0,0,0), (k,0,0), (k,k,0), (0,k,0), (0,0,k), (k,0,k), (k,k,k), (0,k,k) \\}\\)... |
IMO-1973-3 | https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_3 | Let \(a\) and \(b\) be real numbers for which the equation \(x^4 + ax^3 + bx^2 + ax + 1 = 0\) has at least one real solution. For all such pairs \((a, b)\), find the minimum value of \(a^2 + b^2\). | [
"Substitute \\(z=x+1/x\\) to change the original equation into \\(z^2+az+b-2=0\\). This equation has solutions \\(z=\\frac{-a \\pm \\sqrt{a^2+8-4b}}{2}\\). We also know that\n\n\\[\n|z|=|x+1/x| \\geq 2.\\ \\ \\ \\ \\ \\ \\ \\ (1)\n\\]\n\nSo,\n\n\\[\n\\left | \\frac{-a \\pm \\sqrt{a^2+8-4b}}{2} \\right | \\geq 2\\ \... |
IMO-1973-4 | https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_4 | A soldier needs to check on the presence of mines in a region having the shape of an equilateral triangle. The radius of action of his detector is equal to half the altitude of the triangle. The soldier leaves from one vertex of the triangle. What path shouid he follow in order to travel the least possible distance and still accomplish his mission? | [
"Let our triangle be \\(\\triangle ABC\\), let the midpoint of \\(AB\\) be \\(D\\), and let the midpoint of \\(CD\\) be \\(E\\). Let the height of the triangle be \\(h\\). Draw circles around points \\(B\\) and \\(C\\) with radius \\(\\frac{h}{2}\\), and label them \\(c_1\\) and \\(c_2\\). Let the intersection of \... |
IMO-1973-5 | https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_5 | \(G\) is a set of non-constant functions of the real variable \(x\) of the form
\[
f(x) = ax + b, a \text{ and } b \text{ are real numbers,}
\]
and \(G\) has the following properties:
(a) If \(f\) and \(g\) are in \(G\), then \(g \circ f\) is in \(G\); here \((g \circ f)(x) = g[f(x)]\).
(b) If \(f\) is in \(G\), then its inverse \(f^{-1}\) is in \(G\); here the inverse of \(f(x) = ax + b\) is \(f^{-1}(x) = (x - b) / a\).
(c) For every \(f\) in \(G\), there exists a real number \(x_f\) such that \(f(x_f) = x_f\).
Prove that there exists a real number \(k\) such that \(f(k) = k\) for all \(f\) in \(G\). | [
"First, observe that for each function \\(f(x)=ax+b\\) in \\(G\\), if \\(a=1\\) then \\(b=0\\). This is a result of (c); for example, \\(f(x)=x+1\\) could not be in \\(G\\) because it does not have a fixed point. Or if \\(f(x)=x\\), then every point is a fixed point.\n\nAlso, for each function \\(f(x)=ax+b\\) in \\... |
IMO-1973-6 | https://artofproblemsolving.com/wiki/index.php/1973_IMO_Problems/Problem_6 | Let \(a_1, a_2,\cdots, a_n\) be \(n\) positive numbers, and let \(q\) be a given real number such that \(0<q<1.\) Find \(n\) numbers \(b_1, b_2, \cdots, b_n\) for which
(a) \(a_k<b_k\) for \(k=1,2,\cdots, n,\)
(b) \(q<\dfrac{b_{k+1}}{b_k}<\dfrac{1}{q}\) for \(k=1,2,\cdots,n-1,\)
(c) \(b_1+b_2+\cdots+b_n<\dfrac{1+q}{1-q}(a_1+a_2+\cdots+a_n).\) | [
"We notice that the constraints are linear, in the sense that if bi is a solution for ai, q, and bi' is a solution for ai', q, then for any k, k' > 0 a solution for kai + k'ai', q is kbi + k'bi'. Also a \"near\" solution for ah = 1, other ai = 0 is b1 = qh-1, b2 = qh-2, ... , bh-1 = q, bh = 1, bh+1 = q, ... , bn = ... |
IMO-1974-1 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_1 | Three players \(A, B\) and \(C\) play the following game: On each of three cards an integer is written. These three numbers \(p, q, r\) satisfy \(0 < p < q < r\). The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled again; the counters remain with the players.
This process (shuffling, dealing, giving out counters) takes place for at least two rounds. After the last round, \(A\) has 20 counters in all, \(B\) has 10 and \(C\) has 9. At the last round \(B\) received \(r\) counters. Who received \(q\) counters on the first round? | [
"Answer: player \\(C\\).\n\nLet \\(n\\) be the number of rounds played, then obviously \\(n (p + q + r) = 20 + 10 + 9 = 39\\). So \\(n\\) must be a divisor of 39, i. e. \\(n \\in \\{ 1, 3, 13, 39 \\}\\). But \\(p \\geq 1,\\) \\(q \\geq p + 1 \\geq 2,\\) \\(r \\geq q + 1 \\geq 3,\\) so \\(p + q + r \\geq 1 + 2 + 3 =... |
IMO-1974-2 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_2 | In the triangle \(ABC\), prove that there is a point \(D\) on side \(AB\) such that \(CD\) is the geometric mean of \(AD\) and \(DB\) if and only if \(\sin{A}\sin{B} \leq \sin^2 (\frac{C}{2})\). | [
"Let a point \\(D\\) on the side \\(AB\\). Let \\(CF\\) the altitude of the triangle \\(\\triangle ABC\\), and \\(C'\\) the symmetric point of \\(C\\) through \\(F\\). We bring a parallel line \\(L\\) from \\(C'\\) to \\(AB\\). This line intersects the ray \\(CD\\) at the point \\(E\\), and we know that \\(DE=DC\\)... |
IMO-1974-3 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_3 | Prove that the number \(\sum^n_{k=0}\binom{2n+1}{2k+1}2^{3k}\) is not divisible by \(5\) for any integer \(n\ge0.\) | [
"Everything that follows takes place in \\(\\mathbb F_5(\\sqrt 2)\\), i.e. the field we get by adjoining a root of \\(x^2-2=0\\) to \\(\\mathbb F_5\\), the field with \\(5\\) elements.\n\nWe have \\(\\sum_{k=0}^n\\binom{2n+1}{2k+1}2^{3k}=\\sum_{k=0}^n\\binom{2n+1}{2n-2k}3^k=\\sum_{k=0}^n\\binom{2n+1}{2(n-k)}2^{-k}\... |
IMO-1974-4 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_4 | Consider decompositions of an \(8\times8\) chessboard into \(p\) non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If \(a_i\) is the number of white squares in the \(i\)-th rectangle, then \(a_1<a_2<\cdots<a_p.\)
Find the maximum value of \(p\) for which such a decomposition is possible. For this value of \(p,\) determine all possible sequences \(a_1, a_2, \cdots, a_p.\) | [
"Since each rectangle has the same number of black squares as white squares, \\(a_1+a_2+\\ldots +a_p=\\frac{64}{2}=32\\). Clearly \\(a_i\\ge i\\) for \\(i=1\\) to \\(i=p\\) so \\(32=a_1+a_2+\\ldots + a_p\\ge 1+2+\\ldots +p=\\frac{p(p+1)}{2}\\) so this forces \\(p\\le 7\\). It is possible to decompose the board into... |
IMO-1974-5 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_5 | Determine all possible values of
\[
S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}
\]
where \(a, b, c, d,\) are arbitrary positive numbers. | [
"Note that\n\n\\[\n2 = \\frac{a}{a+b}+\\frac{b}{a+b}+\\frac{c}{c+d}+\\frac{d}{c+d} > S > \\frac{a}{a+b+c+d}+\\frac{b}{a+b+c+d}+\\frac{c}{a+b+c+d}+\\frac{d}{a+b+c+d} = 1.\n\\]\n\nWe will now prove that \\(S\\) can reach any range in between \\(1\\) and \\(2\\).\n\nChoose any positive number \\(a\\). For some variabl... |
IMO-1974-6 | https://artofproblemsolving.com/wiki/index.php/1974_IMO_Problems/Problem_6 | Let \(P\) be a non-constant polynomial with integer coefficients. If \(n(P)\) is the number of distinct integers \(k\) such that \((P(k))^2=1,\) prove that \(n(P)-\deg(P)\le2,\) where \(\deg(P)\) denotes the degree of the polynomial \(P.\) | [
"Lemma: Let \\(P(x)\\) be a polynomial with integer coefficients which is not constant. Then if \\(P(x)\\) obtains \\(1\\) (or \\(-1\\)) as its values for at least four times then \\(P(x)\\neq -1\\) ( or \\(P(x)\\neq 1\\)) for all \\(x\\). Proof. Assume that \\(P(a)=P(b)=P(c)=P(d)=1\\) for \\(a,b,c,d\\) distince. T... |
IMO-1975-1 | https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_1 | Let \(x_i, y_i\) \((i=1,2,\cdots,n)\) be real numbers such that
\[
x_1\ge x_2\ge\cdots\ge x_n \text{ and } y_1\ge y_2\ge\cdots\ge y_n.
\]
Prove that, if \(z_1, z_2,\cdots, z_n\) is any permutation of \(y_1, y_2, \cdots, y_n,\) then
\[
\sum^n_{i=1}(x_i-y_i)^2\le\sum^n_{i=1}(x_i-z_i)^2.
\] | [
"We can expand and simplify the inequality a bit, and using the fact that \\(z\\) is a permutation of \\(y\\), we can cancel some terms.\n\n\\[\n\\sum^n_{i=1}x_i^2 + \\sum^n_{i=1}y_i^2 - 2\\sum^n_{i=1}x_iy_i \\leq \\sum^n_{i=1}x_i^2 + \\sum^n_{i=1}z_i^2 - 2\\sum^n_{i=1}x_iz_i\n\\]\n\n\\[\n\\sum^n_{i=1}x_iy_i \\geq ... |
IMO-1975-2 | https://artofproblemsolving.com/wiki/index.php/1975_IMO_Problems/Problem_2 | Let \(a_1, a_2, a_3, \cdots\) be an infinite increasing sequence of positive integers. Prove that for every \(p \geq 1\) there are infinitely many \(a_m\) which can be written in the form
\[
a_m = xa_p + ya_q
\]
with \(x, y\) positive integers and \(q > p\). | [
"If we can find \\(p\\ne q\\) such that \\((a_p,a_q)=1\\), we're done: every sufficiently large positive integer \\(n\\) can be written in the form \\(xa_p+ya_q,\\ x,y\\in\\mathbb N\\). We can thus assume there are no two such \\(p\\ne q\\). We now prove the assertion by induction on the first term of the sequence,... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.