8th International Mathematical Olympiad 1966 Problems & Solutions

A1.  Problems A, B and C were posed in a mathematical contest. 25 competitors solved at least one of the three. Amongst those who did not solve A, twice as many solved B as C. The number solving only A was one more than the number solving A and at least one other. The number solving just A equalled the number solving just B plus the number solving just C. How many solved just B?

A2.  Prove that if BC + AC = tan C/2 (BC tan A + AC tan B), then the triangle ABC is isosceles.

A3.  Prove that a point in space has the smallest sum of the distances to the vertices of a regular tetrahedron iff it is the center of the tetrahedron.

B1.  Prove that 1/sin 2x + 1/sin 4x + ... + 1/sin 2ⁿx = cot x - cot 2ⁿx for any natural number n and any real x (with sin 2ⁿx non-zero).

B2.  Solve the equations:     |a_i - a₁| x₁ + |a_i - a₂| x₂ + |a_i - a₃| x₃ + |a_i - a₄| x₄ = 1, i = 1, 2, 3, 4, where a_i are distinct reals.

B3.  Take any points K, L, M on the sides BC, CA, AB of the triangle ABC. Prove that at least one of the triangles AML, BKM, CLK has area ≤ 1/4 area ABC. 

Solutions

Problem A1

Problems A, B and C were posed in a mathematical contest. 25 competitors solved at least one of the three. Amongst those who did not solve A, twice as many solved B as C. The number solving only A was one more than the number solving A and at least one other. The number solving just A equalled the number solving just B plus the number solving just C. How many solved just B?

Answer

6.

Solution

Let a solve just A, b solve just B, c solve just C, and d solve B and C but not A. Then 25 - a - b - c - d solve A and at least one of B or C. The conditions give:

b + d = 2(c + d); a = 1 + 25 - a - b - c - d; a = b + c.

Eliminating a and d, we get: 4b + c = 26. But d = b - 2c ≥ 0, so b = 6, c = 2. 

Problem A2

Prove that if BC + AC = tan C/2 (BC tan A + AC tan B), then the triangle ABC is isosceles.

Solution

A straight slog works. Multiply up to get (a + b) cos A cos B cos C/2 = a sin A cos B sin C/2 + b cos A sin B sin C/2 (where a = BC, b = AC, as usual). Now use cos(A + C/2) = cos A cos C/2 - sin A sin C/2 and similar relation for cos (B + C/2) to get: a cos B cos(A + C/2) + b cos A cos (B + C/2) = 0. Using C/2 = 90^o - A/2 - B/2, we find that cos(A + C/2) = - cos(B + C/2) (and = 0 only if A = B). Result follows. 

Problem A3

Prove that a point in space has the smallest sum of the distances to the vertices of a regular tetrahedron iff it is the center of the tetrahedron.

Solution

Let the tetrahedron be ABCD and let P be a general point. Let X be the midpoint of CD. Let P' be the foot of the perpendicular from P to the plane ABX. We show that if P does not coincide with P', then PA + PB + PC + PD > P'A + P'B + P'C + P'D.

PA > P'A (because angle PP'A = 90^o) and PB > P'B. P'CD is isosceles and PCD is not but P is the same perpendicular distance from the line CD as P'. It follows that PC + PD > P'C + P'D. The easiest way to see this is to reflect C and D in the line PP' to give C' and D'. Then PC = PC', and PC' + PD > C'D = P'C' + P'D = P'C + P'D.

So if P has the smallest sum, it must lie in the plane ABX and similarly in the plane CDY, where Y is the midpoint of AB, and hence on the line XY. Similarly, it must lie on the line joining the midpoints of another pair of opposite sides and hence must be the center. 

Problem B1

Prove that 1/sin 2x + 1/sin 4x + ... + 1/sin 2ⁿx = cot x - cot 2ⁿx for any natural number n and any real x (with sin 2ⁿx non-zero).

Solution

cot y - cot 2y = cos y/sin y - (2 cos²y - 1)/(2 sin y cos y) = 1/(2 sin y cos y) = 1/sin 2y. The result is now easy. Use induction. True for n = 1 (just take y = x). Suppose true for n, then taking y = 2ⁿx, we have 1/sin 2ⁿ⁺¹x = cot 2ⁿx - cot 2ⁿ⁺¹x and result follows for n + 1. 

Problem B2

Solve the equations:

    |a_i - a₁| x₁ + |a_i - a₂| x₂ + |a_i - a₃| x₃ + |a_i - a₄| x₄ = 1, i = 1, 2, 3, 4, where a_i are distinct reals.

Answer

x₁ = 1/(a₁ - a₄), x₂ = x₃ = 0, x₄ = 1/(a₁ - a₄).

Solution

Take a₁ > a₂ > a₃ > a₄. Subtracting the equation for i=2 from that for i=1 and dividing by (a₁ - a₂) we get:

      - x₁ + x₂ + x₃ + x₄ = 0.

Subtracting the equation for i=4 from that for i=3 and dividing by (a₃ - a₄) we get:

      - x₁ - x₂ - x₃ + x₄ = 0.

Hence x₁ = x₄. Subtracting the equation for i=3 from that for i=2 and dividing by (a₂ - a₃) we get:

      - x₁ - x₂ + x₃ + x₄ = 0.

Hence x₂ = x₃ = 0, and x₁ = x₄ = 1/(a₁ - a₄). 

Problem B3

Take any points K, L, M on the sides BC, CA, AB of the triangle ABC. Prove that at least one of the triangles AML, BKM, CLK has area ≤ 1/4 area ABC.

Solution

If not, then considering ALM we have 4·AL·AM·sin A > AB·AC·sin A, so 4·AL·AM > AB·AC = (AM + BM)(AL + CL), so 3·AL·AM > AM·CL + BM·AL + BM·CL. Set k = BK/CK, l = CL/AL, m = AM/BM, and this inequality becomes:

      3 > l + 1/m + l/m.

Similarly, considering the other two triangles we get: 3 > k + 1/l + k/l, and 3 > m + 1/k + m/k.

Adding gives: 9 > k + l + m + 1/k + 1/l + 1/m + k/l + l/m + m/k, which is false by the arithmetic/geometric mean inequality.

Solutions are also available in:   Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in   István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.