1st USA Mathematical Olympiad 1972 Problems

1st USA Mathematical Olympiad 1972 Problems

1.  Let (a, b, ... , k) denote the greatest common divisor of the integers a, b, ... k and [a, b, ... , k] denote their least common multiple. Show that for any positive integers a, b, c we have (a, b, c)² [a, b] [b, c] [c, a] = [a, b, c]² (a, b) (b, c) (c, a).

2.  A tetrahedron has opposite sides equal. Show that all faces are acute-angled.

3.  n digits, none of them 0, are randomly (and independently) generated, find the probability that their product is divisible by 10. 

4.  Let k be the real cube root of 2. Find integers A, B, C, a, b, c such that | (Ax² + Bx + C)/(ax² + bx + c) - k | < | x - k | for all non-negative rational x.

5.  A pentagon is such that each triangle formed by three adjacent vertices has area 1. Find its area, but show that there are infinitely many incongruent pentagons with this property. 

Solution

1st USA Maths Olympiad 1972

Problem 1
Let (a, b, ... , k) denote the greatest common divisor of the integers a, b, ... k and [a, b, ... , k] denote their least common multiple. Show that for any positive integers a, b, c we have (a, b, c)² [a, b] [b, c] [c, a] = [a, b, c]² (a, b) (b, c) (c, a).
Solution

If we express a, b, c as a product of primes then the gcd has each prime to the smallest power and the lcm has each prime to the largest power. So the equation given is equivalent to showing that 2 min(r, s, t) + max(r, s) + max(s, t) + max(t, r) = 2 max(r, s, t) + min(r, s) + min(s, t) + min(t, r) for non-negative integers r, s, t. Assume r ≤ s ≤ t. Then each side is 2r + s + 2t. Problem 2

A tetrahedron has opposite sides equal. Show that all faces are acute-angled.
Solution

Let the tetrahedron be ABCD. Let M be the midpoint of BC. We have AM + MD > AD (*). Now the triangles ABC and DCB are congruent because AB = DC, BC = CB and AC = DB. Hence AM = DM. Also AD = BC = 2MC. So (*) implies that AM > MC. But that implies that angle BAC is acute. Similarly for all the other angles.

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Problem 3n digits, none of them 0, are randomly (and independently) generated, find the probability that their product is divisible by 10.
Solution

Answer: 1 - (8/9)ⁿ - (5/9)ⁿ + (4/9)ⁿ.
A number is divisible by 10 iff it has an even number and a 5 amongst its digits. The probability of no 5 is (8/9)ⁿ. The probability of no even number is (5/9)ⁿ. The probability of no 5 and no even number is (4/9)ⁿ. Hence result.

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Problem 4Let k be the real cube root of 2. Find integers A, B, C, a, b, c such that | (Ax² + Bx + C)/(ax² + bx + c) - k | < | x - k | for all non-negative rational x.
Solution

Taking the limit, we must have (Ak²+ Bk + C) = k(ak² + bk + c), so A = b, B = c, C = 2a. Now notice that (bx² + cx + 2a) - k(ax² + bx + c) = (b - ak)x² + (c - bk)x + (2a - ck) = (x - k)( (b - ak)x + c - ak²). So we require | (b - ak)x + c - ak²| < | ax² + bx + c | for all x ≥ 0.

There are many ways to satisfy this. For example, take a = 1, b = c = 2. Then (b - ak)x is always positive and less than bx for positive x, and c - ak² is positive and less than c.

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Problem 5A pentagon is such that each triangle formed by three adjacent vertices has area 1. Find its area, but show that there are infinitely many incongruent pentagons with this property.
Solution

Let the pentagon be ABCDE. Triangles BCD and ECD have the same area, so B and E are the same perpendicular distance from CD, so BE is parallel to CD. The same applies to the other diagonals (each is parallel to the side with which it has no endpoints in common). Let BD and CE meet at X. Then ABXE is a parallelogram, so area BXE = area EAB = 1. Also area CDX + area EDX = area CDX + area BCX = 1. Put area EDX = x. Then DX/XB = area EDX/area BXE = x/1 and also = area CDX/area BCX = (1-x)/x. So x² + x - 1 = 0, x = √5 - 1)/2 (we know x < 0, so it cannot be the other root). Hence area ABCDE = 3 + x = (√5 + 5)/2.

Take any triangle XCD of area (3 - √5)/2 and extend DX to B, so that BCD has area 1, and extend CX to E so that CDE has area 1. Then take BA parallel to CE and EA parallel to BD. It is easy to check that the pentagon has the required property.