44th Eötvös Competition Problems 1940

1.  Each button in a box is big or small, and white or black. There is at least one big button, at least one small button, at least one white button, and at least one black button. Show that there are two buttons with different size and color.

2.  m and n are different positive integers. Show that 2^2m + 1 and 2²ⁿ + 1 are coprime.

3.  T is a triangle. Show that there is a triangle T' whose sides are equal to the medians of T, and that T'' is similar to T. 

Solutions

Problem 1

Each button in a box is big or small, and white or black. There is at least one big button, at least one small button, at least one white button, and at least one black button. Show that there are two buttons with different size and color.

Solution

Suppose not. There is a big button. Suppose it is white. Then the small button must be white also (or it would be have different size and color from the big white). Now if the black button is big it is different size and color from the small white. If it is small it is different size and color from the big white. Contradiction. We obtain an exactly similar contradiction if the big button is black. 

Problem 2

m and n are different positive integers. Show that 2^2m + 1 and 2²ⁿ + 1 are coprime.

Solution

wlog m < n. Suppose p divides both 2^2m + 1 and 2²ⁿ + 1. Then since they are both odd, p must be odd. It must also divide their difference 2^2m(2^2n-m - 1) and hence 2^2n-m - 1. But (2^2n-m - 1)(2^2n-m + 1) = (2^2n-m+1 - 1), so p also divides (2^2n-m+1 - 1) and so by a trivial induction (2²ⁿ - 1). But it divides (2²ⁿ + 1), so it must divide their difference 2. But p is odd, so they have no common factors. 

Problem 3

T is a triangle. Show that there is a triangle T' whose sides are equal to the medians of T, and that T'' is similar to T.

Solution

Let L, M be the midpoints of BC, AC respectively. Take B' as the reflection of B in M, so that AB'CB is a parallelogram. Let N be the midpoint of B'C. Then MN is parallel to AB' and half its length. Hence it is also equal and parallel to BL. So MNLB is a parallelogram and hence LN = BM. Finally, AN has the same length as the median from C. So ALN has sides equal to the medians of ABC. Thus the triangle T certainly exists.

Note that MNCL is a parallelogram, so X is the midpoint of LN. Hence AX is a median of T. But X is the midpoint of MC, so AX = (3/4) AC. Now LN is the median from B (in length), and we have shown that the median to it is (3/4) the length of the side opposite to B. Similarly for the other sides. So T'' has sides equal to those of T times (3/4). In particular, it is similar to T.