9th Mexican Mathematical Olympiad Problems 1995

9th Mexican Mathematical Olympiad Problems 1995

A1.  N students are seated at desks in an m x n array, where m, n ≥ 3. Each student shakes hands with the students who are adjacent horizontally, vertically or diagonally. If there are 1020 handshakes, what is N?

A2.  6 points in the plane have the property that 8 of the distances between them are 1. Show that three of the points form an equilateral triangle with side 1.

A3.  A, B, C, D are consecutive vertices of a regular 7-gon. AL and AM are tangents to the circle center C radius CB. N is the point of intersection of AC and BD. Show that L, M, N are collinear.

B1.  Find 26 elements of {1, 2, 3, ... , 40} such that the product of two of them is never a square. Show that one cannot find 27 such elements.

B2.  ABCDE is a convex pentagon such that the triangles ABC, BCD, CDE, DEA and EAB have equal area. Show that (1/4) area ABCDE < area ABC < (1/3) area ABCDE.

B3.  A 1 or 0 is placed on each square of a 4 x 4 board. One is allowed to change each symbol in a row, or change each symbol in a column, or change each symbol in a diagonal (there are 14 diagonals of lengths 1 to 4). For which arrangements can one make changes which end up with all 0s? 

Solutions

Problem A1

N students are seated at desks in an m x n array, where m, n ≥ 3. Each student shakes hands with the students who are adjacent horizontally, vertically or diagonally. If there are 1020 handshakes, what is N?

Answer

280

Solution

Students in the corner shake hands 3 times, those on the sides 5 times and those in the middle 8 times. So the total number of handshakes is (4·3 + (2m-4+2n-4)5 + (m-2)(n-2)8)/2 = (12 + 10m + 10n - 40 + 8mn - 16m - 16n + 32)/2 = (16mn - 12m - 12n + 8)/4 = (4m - 3)(4n - 3)/4 - 1/4 = 1020, so (4m-3)(4n-3) = 4081 = 7·11·53. Now 7 = 3 mod 4, 11 = 3 mod 4 and 53 = 1 mod 4, so we must have 4m-3 = 77, 4n-3 = 53 (or vice versa) and hence N = mn = 20·14 = 280. 

Problem A2

6 points in the plane have the property that 8 of the distances between them are 1. Show that three of the points form an equilateral triangle with side 1.

Solution

Suppose not. Call two points related if they are a distance 1 apart. If any point is related to 5 others, then none of the others can be related, so we only have 5 distances of 1. Contradiction.

Now if X and Y are any two points, then at most two points can be related to X and Y (for there are two isosceles triangles with base XY and equal sides 1, one on each side of XY). So if a point X is related to 4 others, then none of the 4 can be related to each other and the last point Y (not related to X) can be related to at most 2 of the points related to X. So we have at most 6 distances of 1. Contradiction.

Now if we have at most one point related to 3 others, then we have at most (1·3 + 5·2)/2 < 8 distances of 1. So there must be two points X and Y, each related to 3 others. Since there are only 6 points in all, X must be related to Y. So we have A and B related to X, and C and D related to Y (giving a total of 5 distances of 1 so far). Now A and B cannot be related (or ABX is equilateral with side 1), and similarly C and D cannot be related. X is related to A and Y, and C and D are related to Y, so at most one of C, D is related to A (otherwise we would have three points related to A and Y). Similarly, at most one of C, D is related to B. So we have at most 7 distances of 1. Contradiction. 

Problem B1

Find 26 elements of {1, 2, 3, ... , 40} such that the product of two of them is never a square. Show that one cannot find 27 such elements.

Solution

We can include at most one of 1,2²,3²,4²,5²,6², at most one of 2,2·2², 2·3², 2·4², at most one of 3, 3·2², 3·3², and at most one of each of the pairs (5,20), (6,24), (7,28), (10,40). Thus we must exclude at least 5+3+2+4 = 14 numbers, so we cannot find 27 numbers.

On the other hand the set of 26 numbers 1, 11, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40 works. 

Problem B2

ABCDE is a convex pentagon such that the triangles ABC, BCD, CDE, DEA and EAB have equal area. Show that (1/4) area ABCDE < area ABC < (1/3) area ABCDE.

Solution

This is a watered down version of --> 

Problem B3

A 1 or 0 is placed on each square of a 4 x 4 board. One is allowed to change each symbol in a row, or change each symbol in a column, or change each symbol in a diagonal (there are 14 diagonals of lengths 1 to 4). For which arrangements can one make changes which end up with all 0s?

Answer

any arrangement which has an even number of Xs equal to 0

Solution

Any change switches an even number of the 8 Xs. So if we start with an odd number of Xs as 0, we can never get them all 0. But suppose we start with an even number 0. For each of the 4 central squares, there is a diagonal that contains that central square and no others. So we can always get all the central squares 0. By a combination of border rows/columns and diagonals of length 2, we can then change each X to a 0 in turn except one (without affecting the four central squares). But the last X will not need changing, since there must always be an even number equal to 0. Finally, we can use diagonals length one to change the four corners to 0.