11th Mexican Mathematical Olympiad Problems 1997

11th Mexican Mathematical Olympiad Problems 1997

A1.  Find all primes p such that 8p⁴ - 3003 is a (positive) prime.

A2.  ABC is a triangle with centroid G. P, P' are points on the side BC, Q is a point on the side AC, R is a point on the side AB, such that AR/RB = BP/PC = CQ/QA = CP'/P'B. The lines AP' and QR meet at K. Show that P, G and K are collinear.

A3.  Show that it is possible to place the numbers 1, 2, ... , 16 on the squares of a 4 x 4 board (one per square), so that the numbers on two squares which share a side differ by at most 4. Show that it is not possible to place them so that the difference is at most 3.

B1.  3 non-collinear points in space determine a unique plane, which contains the points. What is the smallest number of planes determined by 6 points in space if no three points are collinear and the points do not all lie in the same plane?

B2.  ABC is a triangle. P, Q, R are points on the sides BC, CA, AB such that BQ, CR meet at A', CR, AP meet at B', AP, BQ meet at C' and we have AB' = B'C', BC' = C'A', CA' = A'B'. Find area PQR/area ABC.

B3.  Show that we can represent 1 as 1/5 + 1/a₁ + 1/a₂ + ... + 1/a_n (for positive integers a_i) in infinitely many different ways.

Solutions

Problem A1

Find all primes p such that 8p⁴ - 3003 is a (positive) prime.

Solution

p = 3 fails, p = 5 gives the prime 1997. If p > 5, then p⁴ = 1 mod 5, so 8p⁴ - 3003 = 0 mod 5 and is therefore composite.

Thanks to Suat Namli

Problem A3

Show that it is possible to place the numbers 1, 2, ... , 16 on the squares of a 4 x 4 board (one per square), so that the numbers on two squares which share a side differ by at most 4. Show that it is not possible to place them so that the difference is at most 3.

Solution

1  2  3  4
 5  6  7  8
 9 10 11 12
13 14 15 16

shows that a difference of ≤ 4 is possible. Now suppose a difference of ≤ 3 is possible. Let k be the smallest number of moves (each 1 square N, S, E or W) to get from 1 to 16. Evidently k must be at least 5 (because 1 + 4·3 < 16). But if k = 5, then every difference along the path from 1 to 16 must be exactly 3 and so the numbers along the path must be 1,4,7,10,13,16. However, there are at least two paths length k (we can go either way around the rectangle). Contradiction. So we must have k = 6 and hence 1 and 16 must be at opposite corners. Hence 2 cannot be 6 moves from 16. It must be at least 5 moves (2 + 4·3 < 16), so it must be adjacent to 1. Similarly, 3 must be adjacent to 1. Similarly 14 and 15 must be adjacent to 16. But now 2 and 15 are only 4 moves apart. Contradiction (2 + 4·3 < 15).

Problem B3

Show that we can represent 1 as 1/5 + 1/a₁ + 1/a₂ + ... + 1/a_n (for distinct positive integers a_i) in infinitely many different ways.

Solution

We use 1/n by 1/n+1 + 1/(n²+n) repeatedly. We start with 1 = 1/2 + 1/3 + 1/6. Then 1/3 = 1/4 + 1/12 and 1/4 = 1/5 + 1/20, so we have 1 = 1/5 + 1/2 + 1/6 + 1/12 + 1/20. So we have found one representation of the required type. Now having found k, take the largest denominator N in any of the representations and replace 1/N by 1/(N+1) + 1/(N²+N). That gives another representation which must be different from any of the previous ones. Hence we can find infinitely many.

Thanks to Suat Namli