olymmath

IMO 1966

International Mathematical Olympiad 1966.

Day 1

Problem 1

  • Day 1

In a mathematical contest, three problems, A,B,CA,B,C were posed. Among the participants ther were 25 students who solved at least one problem each. Of all the contestants who did not solve problem AA , the number who solved BB was twice the number who solved CC . The number of students who solved only problem AA was one more than the number of students who solved AA and at least one other problem. Of all students who solved just one problem, half did not solve problem AA . How many students solved only problem BB ?

Answer, solutions and source

Problem 2

  • Day 1

Let a,b,ca,b,c be the lengths of the sides of a triangle, and α,β,γ\alpha, \beta, \gamma respectively, the angles opposite these sides. Prove that if a+b=tan⁡γ2(atan⁡α+btan⁡β)a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta}) the triangle is isosceles.

Answer, solutions and source

Problem 3

  • Day 1

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.

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Prove that for every natural number nn , and for every real number x≠kπ2tx \neq \frac{k\pi}{2^t} ( t=0,1,…,nt=0,1, \dots, n ; kk any integer) 1sin⁡2x+1sin⁡4x+⋯+1sin⁡2nx=cot⁡x−cot⁡2nx\frac{1}{\sin{2x}}+\frac{1}{\sin{4x}}+\dots+\frac{1}{\sin{2^nx}}=\cot{x}-\cot{2^nx}

Answer, solutions and source

Problem 5

  • Day 2

Solve the system of equations ∣a1−a2∣x2+∣a1−a3∣x3+∣a1−a4∣x4=1|a_1-a_2|x_2+|a_1-a_3|x_3+|a_1-a_4|x_4=1 ∣a2−a1∣x1+∣a2−a3∣x3+∣a2−a4∣x4=1|a_2-a_1|x_1+|a_2-a_3|x_3+|a_2-a_4|x_4=1 ∣a3−a1∣x1+∣a3−a2∣x2+∣a3−a4∣x4=1|a_3-a_1|x_1+|a_3-a_2|x_2+|a_3-a_4|x_4=1 ∣a4−a1∣x1+∣a4−a2∣x2+∣a4−a3∣x3=1|a_4-a_1|x_1+|a_4-a_2|x_2+|a_4-a_3|x_3=1 where a1,a2,a3,a4a_1, a_2, a_3, a_4 are four different real numbers.

Answer, solutions and source

Problem 6

  • Day 2

Let ABCABC be a triangle, and let PP , QQ , RR be three points in the interiors of the sides BCBC , CACA , ABAB of this triangle. Prove that the area of at least one of the three triangles AQRAQR , BRPBRP , CPQCPQ is less than or equal to one quarter of the area of triangle ABCABC .

Alternative formulation: Let ABCABC be a triangle, and let PP , QQ , RR be three points on the segments BCBC , CACA , ABAB , respectively. Prove that

min⁡{∣AQR∣,∣BRP∣,∣CPQ∣}≤14⋅∣ABC∣\min\left\{\left|AQR\right|,\left|BRP\right|,\left|CPQ\right|\right\}\leq\frac14\cdot\left|ABC\right| ,

where the abbreviation ∣P1P2P3∣\left|P_1P_2P_3\right| denotes the (non-directed) area of an arbitrary triangle P1P2P3P_1P_2P_3 .

Answer, solutions and source