1st Iberoamerican Mathematical Olympiad Problems 1985

A1.  Find all integer solutions to: a + b + c = 24, a² + b² + c² = 210, abc = 440.

A2.  P is a point inside the equilateral triangle ABC such that PA = 5, PB = 7, PC = 8. Find AB.

A3.  Find the roots r₁, r₂, r₃, r₄ of the equation 4x⁴ - ax³ + bx² - cx + 5 = 0, given that they are positive reals satisfying r₁/2 + r₂/4 + r₃/5 + r₄/8 = 1.

B1.  The reals x, y, z satisfy x ≠ 1, y ≠ 1, x ≠ y, and (yz - x²)/(1 - x) = (xz - y²)/(1 - y). Show that (yx - x²)/(1 - x) = x + y + z.

B2.  The function f(n) is defined on the positive integers and takes non-negative integer values. It satisfies (1) f(mn) = f(m) + f(n), (2) f(n) = 0 if the last digit of n is 3, (3) f(10) = 0. Find f(1985).

B3.  O is the circumcenter of the triangle ABC. The lines AO, BO, CO meet the opposite sides at D, E, F respectively. Show that 1/AD + 1/BE + 1/CF = 2/AO. 

Solutions

Problem A1

Find all integer solutions to: a + b + c = 24, a² + b² + c² = 210, abc = 440.

Solution

ab + bc + ca = ( (a + b + c)² - (a² + b² + c²) )/2 = 183, so a, b, c are roots of the cubic x³ - 24x² + 183x - 440 = 0. But it easily factorises as (x - 5)(x - 8)(x - 11) = 0, so the only solutions are permutations of (5, 8, 11). 

Problem A2

P is a point inside the equilateral triangle ABC such that PA = 5, PB = 7, PC = 8. Find AB.

Solution

Answer: √129.

Let the side length be x. Using the cosine formula, we have cos APB = (74 - x²)/70, cos APC = (89 - x²)/80, cos BPC = (113 - x²)/112. But cos BPC = cos APC cos BPC - sin APC sin BPC, so (113 - x²)/112 = (74 - x²)/79 (89 - x²)/80 - √( (1 - (74 - x²)²/70²) (1 - (89 - x²)²/80²) ).

We isolate the square root term, then square. We multiply through by 25.256.49 and, after some simplification, we get x⁶ - 138x⁴ + 1161x² = 0. Hence x = 0, ±3, ±√129. We discard the zero and negative solutions. x = 3 corresponds to a point P outside the triangle. So the unique solution for a point P inside the triangle is x = √129.

Alternative solution by Johannes Tang

Rotate the triangle about C through 60^o. Let P go to P'. We have AP' = 7, CP' = 8 and angle PCP' = 60^o, so PP'C is equilateral. Hence angle CPP' = 60^o. Also PP' = 8. Using the cosine formula on triangle APP' we find angle APP' = 60^o. Hence angle APC = 120^o. Now applying cosine formula to triangle APC, we get result. 

Problem A3

Find the roots r₁, r₂, r₃, r₄ of the equation 4x⁴ - ax³ + bx² - cx + 5 = 0, given that they are positive reals satisfying r₁/2 + r₂/4 + r₃/5 + r₄/8 = 1.

Solution

We have r₁r₂r₃r₄ = 5/4 and hence (r₁/2) (r₂/4) (r₃/5) (r₄/8) = 1/4⁴. But AM/GM gives that (r₁/2) (r₂/4) (r₃/5) (r₄/8) ≤ ( (r₁/2 + r₂/4 + r₃/5 + r₄/8)/4 )⁴ = 1/4⁴ with equality iff r₁/2 = r₂/4 = r₃/5 = r₄/8. Hence we must have r₁ = 1/2, r₂ = 1, r₃ = 5/4, r₄ = 2. 

Problem B1

The reals x, y, z satisfy x ≠ 1, y ≠ 1, x ≠ y, and (yz - x²)/(1 - x) = (xz - y²)/(1 - y). Show that (yx - x²)/(1 - x) = x + y + z.

Solution

We have yz - x² - y²z + yx² = xz - y² - x²z + xy². Hence z(y - x - y² + x²) = -y² + xy² - x²y + x². Hence z = (x + y - xy)/(x + y - 1).

So yz = x + y + z - xy - xz, so yz - x² = x + y + z - x² - xy - xz = (x + y + z)(1 - x), so (yz - x²)/(1 - x) = (x + y + z). 

Problem B2

The function f(n) is defined on the positive integers and takes non-negative integer values. It satisfies (1) f(mn) = f(m) + f(n), (2) f(n) = 0 if the last digit of n is 3, (3) f(10) = 0. Find f(1985).

Solution

If f(mn) = 0, then f(m) + f(n) = 0 (by (1)). But f(m) and f(n) are non-negative, so f(m) = f(n) = 0. Thus f(10) = 0 implies f(5) = 0. Similarly f(3573) = 0 by (2), so f(397) = 0. Hence f(1985) = f(5) + f(397) = 0.

Problem B3

O is the circumcenter of the triangle ABC. The lines AO, BO, CO meet the opposite sides at D, E, F respectively. Show that 1/AD + 1/BE + 1/CF = 2/AO.

Solution

Projecting onto the altitude from A, we have AD cos(C - B) = AC sin C = 2R sin B sin C, so 2R/AD = cos(C - B)/(sin B sin C).

Hence 2R/AD + 2R/BE + 2R/CF =cos(C - B)/(sin B sin C) + cos(A - C)/(sin C sin A) + cos(B - A)/(sin A sin B). So 2R sin A sin B sin C (1/AD + 1/BE + 1/CF) = sin A cos(B - C) + sin B cos(C - A) + sin C cos(A - B) = 3 sin A sin B sin C + sin A cos B cos C + sin B cos A cos C + sin C cos A cos B = 3 sin A sin B sin C + sin(A + B) cos C + sin C cos A cos B = 3 sin A sin B sin C + sin C (cos C + cos A cos B) = 3 sin A sin B sin C + sin C (-cos(A + B) + cos A cos B) = 4 sin A sin B sin C. Hence 1/AD + 1/BE + 1/CF = 2/R.