MathDB

5

Part of 1967 IMO Shortlist

Problems(8)

IMO LongList 1967, Bulgaria 5

Source: IMO LongList 1967, Bulgaria 5

12/16/2004
Solve the system of equations: x2+x−1=yy2+y−1=zz2+z−1=x. \begin{matrix} x^2 + x - 1 = y \\ y^2 + y - 1 = z \\ z^2 + z - 1 = x. \end{matrix}
linear algebramatrixalgebrapolynomialsystem of equationsIMO ShortlistIMO Longlist
IMO LongList 1967, Hungary 5

Source: IMO LongList 1967, Hungary 5

12/16/2004
Prove that for an arbitrary pair of vectors ff and gg in the space the inequality af2+bfg+cg2≥0af^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.
algebravectorInequalitygeometric inequality3D geometryIMO ShortlistIMO Longlist
IMO LongList 1967, Mongolia 5

Source: IMO LongList 1967, Mongolia 5

12/16/2004
Faces of a convex polyhedron are six squares and 8 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 the polyhedron, and compute the ratio of the squares of volumes of that polyhedron and of the ball whose boundary is the circumscribed sphere.
geometry3D geometryspherepolyhedronIMO ShortlistIMO Longlist
IMO LongList 1967, Poland 5

Source: IMO LongList 1967, Poland 5

12/16/2004
Show that a triangle whose angles AA, BB, CC satisfy the equality sin⁡2A+sin⁡2B+sin⁡2Ccos⁡2A+cos⁡2B+cos⁡2C=2 \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} = 2 is a rectangular triangle.
trigonometryalgebraTriangleTrigonometric IdentitiesIMO ShortlistIMO Longlist
IMO LongList 1967, Romania 5

Source: IMO LongList 1967, Romania 5

12/16/2004
If x,y,zx,y,z are real numbers satisfying relations x+y+z = 1   \textrm{and}   \arctan x + \arctan y + \arctan z = \frac{\pi}{4}, prove that x2n+1+y2n+1+z2n+1=1x^{2n+1} + y^{2n+1} + z^{2n+1} = 1 holds for all positive integers nn.
trigonometryalgebrasystem of equationsTrigonometric EquationsIMO ShortlistIMO Longlist
IMO LongList 1967, Socialists Republic Of Czechoslovakia 5

Source: IMO LongList 1967, Socialists Republic Of Czechoslovakia 5

12/16/2004
Let nn be a positive integer. Find the maximal number of non-congruent triangles whose sides lengths are integers ≤n.\leq n.
countingtriangle inequalitycombinatoricsTriangleIMO ShortlistIMO Longlist
IMO LongList 1967, Sweden 5

Source: IMO LongList 1967, Sweden 5

12/16/2004
In the plane a point OO is and a sequence of points P1,P2,P3,…P_1, P_2, P_3, \ldots are given. The distances OP1,OP2,OP3,…OP_1, OP_2, OP_3, \ldots are r1,r2,r3,…r_1, r_2, r_3, \ldots Let α\alpha satisfies 0<α<1.0 < \alpha < 1. Suppose that for every nn the distance from the point PnP_n to any other point of the sequence is ≥rnα.\geq r^{\alpha}_n. Determine the exponent β\beta, as large as possible such that for some CC independent of nn rn≥Cnβ,n=1,2,…r_n \geq Cn^{\beta}, n = 1,2, \ldots
geometrypoint seteuclidean distanceIMO ShortlistIMO Longlist
IMO LongList 1967, Soviet Union 5

Source: IMO LongList 1967, Soviet Union 5

12/16/2004
A linear binomial l(z)=Az+Bl(z) = Az + B with complex coefficients AA and BB is given. It is known that the maximal value of ∣l(z)∣|l(z)| on the segment −1≤x≤1-1 \leq x \leq 1 (y=0)(y = 0) of the real line in the complex plane z=x+iyz = x + iy is equal to M.M. Prove that for every zz ∣l(z)∣≤Mρ,|l(z)| \leq M \rho, where ρ\rho is the sum of distances from the point P=zP=z to the points Q1:z=1Q_1: z = 1 and Q3:z=−1.Q_3: z = -1.
geometryfunctionalgebraInequalityLinear FunctionIMO ShortlistIMO Longlist