8th Iberoamerican Mathematical Olympiad Problems 1993

A1.  A palindrome is a positive integers which is unchanged if you reverse the order of its digits. For example, 23432. If all palindromes are written in increasing order, what possible prime values can the difference between successive palindromes take?

A2.  Show that any convex polygon of area 1 is contained in some parallelogram of area 2.

A3.  Find all functions f on the positive integers with positive integer values such that (1) if x < y, then f(x) < f(y), and (2) f(y f(x)) = x²f(xy).

B1.  ABC is an equilateral triangle. D is on the side AB and E is on the side AC such that DE touches the incircle. Show that AD/DB + AE/EC = 1.

B2.  If P and Q are two points in the plane, let m(PQ) be the perpendicular bisector of PQ. S is a finite set of n > 1 points such that: (1) if P and Q belong to S, then some point of m(PQ) belongs to S, (2) if PQ, P'Q', P"Q" are three distinct segments, whose endpoints are all in S, then if there is a point in all of m(PQ), m(P'Q'), m(P"Q") it does not belong to S. What are the possible values of n?

B3.  We say that two non-negative integers are related if their sum uses only the digits 0 and 1. For example 22 and 79 are related. Let A and B be two infinite sets of non-negative integers such that: (1) if a ∈ A and b ∈ B, then a and b are related, (2) if c is related to every member of A, then it belongs to B, (3) if c is related to every member of B, then it belongs to A. Show that in one of the sets A, B we can find an infinite number of pairs of consecutive numbers. 

Solutions

Problem 1

A palindrome is a positive integers which is unchanged if you reverse the order of its digits. For example, 23432. If all palindromes are written in increasing order, what possible prime values can the difference between successive palindromes take?

Solution

Answer: 2, 11.

Let x be a palindrome and x' the next highest palindrome. If x < 101, then it is easy to see by inspection that x' - x = 1, 2 or 11, so the only prime differences are 2 and 11.

So assume x > 100. If x and x' have the same final digit, then their difference is divisible by 10 and hence not prime. So they must have different digits. Thus either x = d9...9d and x' = d'0...0d', where d < 9 and d' = d+1, or x' has one more digit than x and d = 9, d' = 1. In the first case x' - x = 11. In the second case x' - x = 2. So again the only prime differences are 2 and 11. 

Problem 2

Show that any convex polygon of area 1 is contained in some parallelogram of area 2.

Solution

Let the vertices X, Y of the polygon be the two which are furthest apart. The polygon must lie between the lines through X and Y perpendicular to XY (for if a vertex Z lay outside the line through Y, then ZY > XY). Take two sides of a rectangle along these lines and the other two sides as close together as possible. There must be a vertices U and V on each of the other two sides. But now the area of the rectangle is twice the area of XUYV, which is at most the area of the polygon. [In the case of a triangle one side of the rectangle will be XY.] 

Problem A3

Find all functions f on the positive integers with positive integer values such that (1) if x < y, then f(x) < f(y), and (2) f(y f(x)) = x²f(xy).

Solution

Answer: f(x) = x².

Note that (1) implies f is (1, 1).

Put y = 1. Then f( f(x) ) = x² f(x).

Put y = f(z), then f( f(z) f(x) ) = x² f(x f(z) ) = x²z² f(xz) = f( f(xz) ). But f is (1, 1) so f(xz) = f(x) f(z).

Now suppose f(m) > m² for some m. Then by (1), f( f(m) ) > f(m² = f(m.m) = f(m)². But f( f(m) ) = m² f(m), so m² > f(m). Contradiction.

Similarly, suppose f(m) < m². Then m² f(m) = f( f(m) ) < f(m²) = f(m)², so m² < f(m). Contradiction. So we must have f(m) = m². 

Problem B1

ABC is an equilateral triangle. D is on the side AB and E is on the side AC such that DE touches the incircle. Show that AD/DB + AE/EC = 1.

Solution

Put BD = x, CE = y, BC = a. Then since the two tangents from B to the incircle are of equal length, and similarly the two tangents from D and E, we have ED + BC = BD + CE, or ED = x + y - a. By the cosine law, ED² = AE² + AD² - AE.AD. Substituting and simplifying, we get a = 3xy/(x + y). Hence AD/DB = (2y - x)/(x + y) and AE/EC = (2x - y)/(x + y) with sum 1. 

Problem B2

If P and Q are two points in the plane, let m(PQ) be the perpendicular bisector of PQ. S is a finite set of n > 1 points such that: (1) if P and Q belong to S, then some point of m(PQ) belongs to S, (2) if PQ, P'Q', P"Q" are three distinct segments, whose endpoints are all in S, then if there is a point in all of m(PQ), m(P'Q'), m(P"Q") it does not belong to S. What are the possible values of n?

Answer

n = 3 (equilateral triangle), 5 (regular pentagon).

Solution

There are n(n-1)/2 pairs of points. Each has a point of S on its bisector. But each point of S is on at most two bisectors, so 2n ≥ n(n-1)/2. Hence n ≤ 5.

The equilateral triangle and regular pentagon show that n = 3, 5 are possible.

Consider n = 4. There are 6 pairs of points, so at least one point of S must be on two bisectors. wlog A is on the bisectors of BC and BD. But then it is also on the bisector of CD. Contradiction.

Many thanks to Tsimerman for this.

Problem B3

We say that two non-negative integers are related if their sum uses only the digits 0 and 1. For example 22 and 79 are related. Let A and B be two infinite sets of non-negative integers such that: (1) if a ∈ A and b ∈ B, then a and b are related, (2) if c is related to every member of A, then it belongs to B, (3) if c is related to every member of B, then it belongs to A. Show that in one of the sets A, B we can find an infinite number of pairs of consecutive numbers.

Solution

Suppose there is a member of A with last digit d. Then every member of B must have one of two possible last digits. Suppose there are members of B with both possibilities. Then every member of A must have last digit d. So either every member of A has the same last digit or every member of B has the same last digit (or both). Suppose every member of A has the same last digit d.

But now if n belongs to B and n + d has last digit 0, then n+1 + d has last digit 1. Moreover, if m is any member of A, then m+n has last digit 0 and other digits all 0 or 1. Hence m+n+1 last last digit 1 and other digits all 0 or 1, so n+1 must also belong to B. Similarly, if n is in B and n+d has last digit 1, then n-1 must also belong to B. So in either case there are infinitely many pairs of consecutive numbers in B.