16th Iberoamerican Mathematical Olympiad Problems 2001

A1.  Show that there are arbitrarily large numbers n such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides n.

A2.  ABC is a triangle. The incircle has center I and touches the sides BC, CA, AB at D, E, F respectively. The rays BI and CI meet the line EF at P and Q respectively. Show that if DPQ is isosceles, then ABC is isosceles. 

A3.  Let X be a set with n elements. Given k > 2 subsets of X, each with at least r elements, show that we can always find two of them whose intersection has at least r - nk/(4k - 4) elements.

B1.  Call a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest number of trios that can be subsets of a set of n distinct real numbers?

B2.  Two players play a game on a 2000 x 2001 board. Each has one piece and the players move their pieces alternately. A short move is one square in any direction (including diagonally) or no move at all. On his first turn each player makes a short move. On subsequent turns a player must make the same move as on his previous turn followed by a short move. This is treated as a single move. The board is assumed to wrap in both directions so a player on the edge of the board can move to the opposite edge. The first player wins if he can move his piece onto the same square as his opponent's piece. For example, suppose we label the squares from (0, 0) to (1999, 2000), and the first player's piece is initially at (0, 0) and the second player's at (1996, 3). The first player could move to (1999, 2000), then the second player to (1996, 2). Then the first player could move to (1998, 1998), then the second player to (1995, 1). Can the first player always win irrespective of the initial positions of the two pieces?

B3.  Show that a square with side 1 cannot be covered by five squares with side less than 1/2. 

Solutions

Problem A1

Show that there are arbitrarily large numbers n such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides n.

Solution

3232 = 16 x 202 and 10000 = 16 x 625. So any number with 3232 as its last 4 digits is divisible by 16. So consider N = 22223232. Its sum of digits is 18, so it is divisible by 9. Hence it is divisible by 9.16 = 144. But any four digits have at most four 2s and at most two 3s, so the product of any four digits divides 144 and hence N. But now we can extend N by inserting an additional 9m 2s at the front. Its digit sum is increased by 18m, so it remains divisible by 144 and it is still divisible by the product of any four digits.

Alternative solution

The number 111111111 with nine 1s is divisible by 9. Hence the number with twenty-seven 1s which equals 111111111 x 1000000001000000001 is divisible by 27. So N, the number with twenty-seven 3s, is divisible by 3⁴. Now the number with 27n 3s is divisible by N and hence by 3⁴. 

Problem A2

ABC is a triangle. The incircle has center I and touches the sides BC, CA, AB at D, E, F respectively. The rays BI and CI meet the line EF at P and Q respectively. Show that if DPQ is isosceles, then ABC is isosceles.

Solution

AF = AE, so ∠AFE = 90^o - A/2. Hence ∠BFP = 90^o + A/2. But ∠FBP = B/2, so ∠FPB = C/2. But BFP and BDP are congruent (BF = BD, BP common, ∠FBP = ∠FDP), so ∠DPB = C/2 and ∠DPQ = C. Similarly, ∠DQP = B. Hence ∠PDQ = A. So DQP and ABC are similar. So if one is isosceles, so is the other.

Thanks to Johann Peter Gustav Lejeune Dirichlet

Problem B1

Call a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest number of trios that can be subsets of a set of n distinct real numbers?

Answer

(m-1)m for n = 2m
m² for n = 2m+1

Solution

Let X be one of the elements. What is the largest number of trios that can have X as middle element? Obviously, at most max(b,a), where b is the number of elements smaller than X and a is the number larger. Thus if n = 2m, the no. of trios is at most 0 + 1 + 2 + ... + m-1 + m-1 + m-2 + ... + 1 + 0 = (m-1)m. If n = 2m+1, then the no. is at most 0 + 1 + 2 + ... + m-1 + m + m-1 + ... + 1 + 0 = m².

These maxima can be achieved by taking the numbers 1, 2, 3, ... , n.