2nd Iberoamerican Mathematical Olympiad Problems 1987

A1.  Find f(x) such that f(x)²f( (1-x)/(1+x) ) = 64x for x not 0, ±1.

A2.  In the triangle ABC, the midpoints of AC and AB are M and N respectively. BM and CN meet at P. Show that if it is possible to inscribe a circle in the quadrilateral AMPN (touching every side), then ABC is isosceles.

A3.  Show that if (2 + √3)^k = 1 + m + n√3, for positive integers m, n, k with k odd, then m is a perfect square.

B1.  Define the sequence p₁, p₂, p₃, ... as follows. p₁ = 2, and p_n is the largest prime divisor of p₁p₂ ... p_n-1 + 1. Prove that 5 does not occur in the sequence.

B2.  Show that the roots r, s, t of the equation x(x - 2)(3x - 7) = 2 are real and positive. Find tan^-1r + tan^-1s + tan^-1t.

B3.  ABCD is a convex quadrilateral. P, Q are points on the sides AD, BC respectively such that AP/PD = BQ/QC = AB/CD. Show that the angle between the lines PQ and AB equals the angle between the lines PQ and CD.

Solutions

Problem A1

Find f(x) such that f(x)²f( (1-x)/(1+x) ) = 64x for x not 0, ±1.
Solution

Put x = (1-y)/(1+y), then (1--x)/(1+x) = y, so f( (1-y)/(1+y) )² f(y) = 64(1-y)/(1+y). Hence f( (1-x)/(1+x) )² f(x) = 64(1-x)/(1+x). But f(x)⁴ f( (1-x)/(1+x) )² = 64²x², so f(x)³ = 64 x²(1 + x)/(1 - x). Hence f(x) = 4 ( x²(1 + x)/(1 - x) )^1/3.

Problem A2

In the triangle ABC, the midpoints of AC and AB are M and N respectively. BM and CN meet at P. Show that if it is possible to inscribe a circle in the quadrilateral AMPN (touching every side), then ABC is isosceles.
Solution

If the quadrilateral has an inscribed circle then AM + PN = AN + PM (consider the tangents to the circle from A, M, P, N). But if AB > AC, then BM > CN (see below). We have AN = AB/2, PM = BM/3, AM = AC/2, PN = CN/3, so it follows that AM + PN < AN + PM. Similarly, AB < AC implies AM + PN > AN + PM, so the triangle must be isosceles.
To prove the result about the medians, note that BM² = BC² + CM² - 2 BC.CM cos C = (BC - CM cos C)² + (CM sin C)². Similarly, CN² = (BC - BN cos B)² + (BN sin B)². But MN is parallel to BC, so CM sin C = BN sin B. But AB > AC, so BN > CM and B < C, so cos B > cos C, hence BN cos B > CM cos C and BC - CM cos C > BC - BN cos B. So BM > CN.

Problem A3

Show that if (2 + √3)^k = 1 + m + n√3, for positive integers m, n, k with k odd, then m is a perfect square.
Solution

We have (2 + √3)⁴ = 97 + 56√3 = 14 (7 + 4√3) - 1 = 14 (2 + √3)² - 1. Hence (2 + √3)^k+2 = 14 (2 + √3)^k - (2 + √3)^k-2. Thus if (2 + √3)^k = a_k + b_k√3, then a_k+2 = 14 a_k - a_k-2.
Now suppose the sequence c_k satisfies c₁ = 1, c₂ = 5, c_k+1 = 4 c_k - c_k-1. We claim that c_k² - c_k-1c_k+1 = 6. Induction on k. We have c₃ = 19, so c₂² - c₁c₃ = 25 - 19 = 6. Thus the result is true for k = 2. Suppose it is true for k. Then c_k+1 = 4 c_k - c_k-1, so c_k+1² = 4 c_kc_k+1 - c_k-1c_k+1 = 4 c_kc_k+1 - c_k² + 6 = c_k(4 c_k+1 - c_k) + 6 = c_kc_k+2 + 6, so the result is true for k+1.
Now put d_k = c_k² + 1. We show that d_k+2 = 14 d_k+1 - d_k. Induction on k. We have d₁ = 2, d₂ = 26, d₃ = 362 = 14 d₂ - d₁, so the result is true for k = 1. Suppose it is true for k. We have c_k+3 - 4 c_k+2 + c_k+1 = 0. Hence 12 + 2 c_k+3c_k+1 - 8 c_k+2c_k+1 + 2 c_k+1² = 12. Hence 2 c_k+2² - 8 c_k+2c_k+1 + 2 c_k+1² = 12. Hence 16 c_k+2² - 8 c_k+2c_k+1 + c_k+1² + 1 = 14 c_k+2² + 14 - c_k+1² - 1, or (4 c_k+2 - c_k+1)² + 1 = 14 (c_k+2² + 1) - (c_k+1² + 1), or c_k+3² + 1 = 14 (c_k+2² + 1) - (c_k+1² + 1), or d_k+3 = 14 d_k+2 - d_k+1. So the result is true for all k.
But a₁ = 2, a₃ = 26 and a_2k+3 = 14 a_2k+1 - a_2k-1, and d₁ = 2, d₂ = 26 and d_k+1 = 14 d_k - d_k-1. Hence a_2k-1 = d_k = c_k² + 1.

Problem B1

Define the sequence p₁, p₂, p₃, ... as follows. p₁ = 2, and p_n is the largest prime divisor of p₁p₂ ... p_n-1 + 1. Prove that 5 does not occur in the sequence.
Solution

See Australian 82/5

Problem B2

Show that the roots r, s, t of the equation x(x - 2)(3x - 7) = 2 are real and positive. Find tan^-1r + tan^-1s + tan^-1t.
Solution

Put f(x) = x(x - 2)(3x - 7) - 2 = 3x³ - 13x² + 14x - 2. Then f(0) = -2, f(1) = 2, so there is a root between 0 and 1. f(2) = -2, so there is another root between 1 and 2. f(3) = 4, so the third root is between 2 and 3. f(x) = 0 has three roots, so they are all real and positive.
We have tan(a + b + c) = (tan a + tan b + tan c - tan a tan b tan c)/(1 - (tan a tan b + tan b tan c + tan c tan a)). So putting a = tan^-1r, b = tan^-1s, c = tan^-1t, we have, tan(a + b + c) = ( (r + s + t) - rst)/(1 - (rs + st + tr) ) = (13/3 - 2/3)/(1 - 14/3) = -1. So a + b + c = -π/4 + kπ. But we know that each of r, s, t is real and positive, so a + b + c lies in the range 0 to 3π/2. Hence a + b + c = 3π/4.

Problem B3

ABCD is a convex quadrilateral. P, Q are points on the sides AD, BC respectively such that AP/PD = BQ/QC = AB/CD. Show that the angle between the lines PQ and AB equals the angle between the lines PQ and CD.
Solution

If AB is parallel to CD, then it is obvious that PQ is parallel to both. So assume AB and CD meet at O. Take O as the origin for vectors. Let e be a unit vector in the direction OA and f a unit vector in the direction OC. Take the vector OA to be ae, OB to be be, OC to be cf, and OD to be df. Then OP is ( (d - c)ae + (a - b)df)/(d - c + a - b) and OQ is ( (d - c)be + (a - b)cf)/(d - c + a - b). Hence PQ is (c - d)(a - b)(e + f)/(d - c + a - b). But e and f are unit vectors, so e+ f makes the same angle with each of them and hence PQ makes the same angle with AB and CD.