8th Mexican Mathematical Olympiad Problems 1994

8th Mexican Mathematical Olympiad Problems 1994

A1.  The sequence 1, 2, 4, 5, 7, 9 ,10, 12, 14, 16, 17, ... is formed as follows. First we take one odd number, then two even numbers, then three odd numbers, then four even numbers, and so on. Find the number in the sequence which is closest to 1994.

A2.  The 12 numbers on a clock face are rearranged. Show that we can still find three adjacent numbers whose sum is 21 or more.

A3.  ABCD is a parallelogram. Take E on the line AB so that BE = BC and B lies between A and E. Let the line through C perpendicular to BD and the line through E perpendicular to AB meet at F. Show that ∠DAF = ∠BAF.

B1.  A capricious mathematician writes a book with pages numbered from 2 to 400. The pages are to be read in the following order. Take the last unread page (400), then read (in the usual order) all pages which are not relatively prime to it and which have not been read before. Repeat until all pages are read. So, the order would be 2, 4, 5, ... , 400, 3, 7, 9, ... , 399, ... . What is the last page to be read?

B2.  ABCD is a convex quadrilateral. Take the 12 points which are the feet of the altitudes in the triangles ABC, BCD, CDA, DAB. Show that at least one of these points must lie on the sides of ABCD.

B3.  Show that we cannot tile a 10 x 10 board with 25 pieces of type A, or with 25 pieces of type B, or with 25 pieces of type C.

Solutions

Problem A1

The sequence 1, 2, 4, 5, 7, 9 ,10, 12, 14, 16, 17, ... is formed as follows. First we take one odd number, then two even numbers, then three odd numbers, then four even numbers, and so on. Find the number in the sequence which is closest to 1994.

Answer
1993 and 1995

Solution

In the first 1 + 2 + ... + (n+1) numbers we have n gaps of 1, the other gaps are all 2. Now 1 + 2 + ... + 44 = 990, so in the first 990 numbers we have 43 gaps of 1, so the 990th number is 1 + 2·989 - 43 = 1936. The next number is 1937 and we then have 44 more odd numbers which takes us past 1994. So we get 1993 and 1995 but not 1994. 

Problem A2

The 12 numbers on a clock face are rearranged. Show that we can still find three adjacent numbers whose sum is 21 or more.

Solution

Suppose it is possible to arrange them so that each three adjacent numbers have sum ≤ 20. If ABCD are adjacent numbers and A + B + C = 20, then we cannot have B + C + D = 20, because A ≠ D. There are 12 sets of three adjacent numbers. The total of all 12 sets is just 3(1 + 2 + ... + 12). But at most 6 can have sum 20, and the other 6 can have sum at most 19. So the sum of all 12 sets is at most 6·39 = 3(1 + 2 +... + 12). Thus 6 sets must have sum 20 and 6 sets must have sum 19. So suppose we have ABCDEFG ... with A + B + C = 20. Then B + C + D = D + E + F = 19, and C + D + E = E + F + G = 20. So D = A - 1, and G = D + 1 = A. Contradiction. 

Problem B1

A capricious mathematician writes a book with pages numbered from 2 to 400. The pages are to be read in the following order. Take the last unread page (400), then read (in the usual order) all pages which are not relatively prime to it and which have not been read before. Repeat until all pages are read. So, the order would be 2, 4, 5, ... , 400, 3, 7, 9, ... , 399, ... . What is the last page to be read?

Answer

37

Solution

First read multiples of 2 and 5 (up to 400). Then since 399 = 3·7·19, read remaining multiples of 3, 7 and 19 up to 399. Then read 397 (a prime). Then read remaining multiples of 17 and 23 up to 391 (=17·23). Now all primes up to 23 have been covered except 11 and 13. Now the largest non-prime number not yet read must be divisible by 11 or 13 (since all multiples of smaller primes have already been read and 29² > 400), so the candidates are 377 = 13·29 and 341 = 11·31. So we keep reading primes until we get down to 377 and then read all remaining multiples of 13. We then read primes until we get down to 341, and then read all remaining multiples of 11. At this point the only remaining numbers are primes > 31. So we read them in reverse order, the last being 37. 

Problem B2

ABCD is a convex quadrilateral. Take the 12 points which are the feet of the altitudes in the triangles ABC, BCD, CDA, DAB. Show that at least one of these points must lie on the sides of ABCD.

Solution

One of the feet of the altitudes of ABC lies on AB or BC unless ∠B is obtuse. Similarly for the other triangles. But the angles add to 360^o, so they cannot all be obtuse. 

Problem B3

Show that we cannot tile a 10 x 10 board with 25 pieces of type A, or with 25 pieces of type B, or with 25 pieces of type C.

Solution

Consider type A. The yellow piece must go into a corner. But then the only way to cover A is with the blue piece. Now the only way to cover B is with a translate of the blue piece, and so on. We can never round off at the next corner.

Color the board alternately black and white like a chessboard. Then type B pieces cover either 1 or 3 black squares. Suppose there are k type B pieces covering 1 black square, then there are 25-k covering 3 black squares, so we must have k + 75 - 3k = 50, or k = 25/2 which does not work.

Color the squares with four colors A, B, C, D in the usual pattern:

A B C D A B C D A B
B C D A B C D A B C
C D A B C D A B C D
D A B C D A B C D A
...
A B C D A B C D A B
B C D A B C D A B C

Now any type C piece must cover equal numbers of A, B, C, D. But the 10 x 10 square has 25 A, 26 B, 25 C and 24 D, so the tiling is impossible.