40th Eötvös Competition Problems 1936

1.  Show that 1/(1·2) + 1/(3·4) + 1/(5·6) + ... + 1/( (2n-1)·2n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n).

2.  ABC is a triangle. Show that, if the point P inside the triangle is such that the triangles PAB, PBC, PCA have equal area, then P must be the centroid.

3.  Given any positive integer N, show that there is just one solution to m + ½(m + n - 1)(m + n - 2) = N in positive integers. 

Solutions

Problem 1

Show that 1/(1·2) + 1/(3·4) + 1/(5·6) + ... + 1/( (2n-1)·2n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n).

Solution

We have 1/( (2m-1)·2m) = 1/(2m-1) - 1/2m, so the sum is 1 - 1/2 + 1/3 - 1/4 + ... + 1/(2n-1) - 1/2n = h - k, where h = 1 + 1/3 + 1/5 + ... + 1/(2n-1), and k = 1/2 + 1/4 + ... + 1/2n. But h - k = (h+k) - 2k = (1 + 1/2 + 1/3 + ... + 1/2n) - (1 + 1/2 + 1/3 + ... + 1/n) = 1/(n+1) + 1/(n+2) + ... + 1/2n. 

Problem 2

ABC is a triangle. Show that, if the point P inside the triangle is such that the triangles PAB, PBC, PCA have equal area, then P must be the centroid.

Solution

Extend AP to meet BC at K. Then triangles PAB and PAC have the same base PA and ratio of heights BK/CK. Hence K is the midpoint of BC, so P lies on the median from A. Similarly, it lies on the other medians, so it must be the centroid. 

Problem 3

Given any positive integer N, show that the is just one solution to m + ½(m + n - 1)(m + n - 2) = N in positive integers.

Solution

Put T_r = ½r(r-1). So 1 = T₂ < T₃ < T₄ < ... . Also T_r+1 - T_r = r. Take r such that T_r < N ≤ T_r+1. Then 1 < N - T_r ≤ T_r+1 - T_r = r. Put m = N - T_r, then 1 ≤ m ≤ r. Put n = r+1-m, then 1 ≤ n, and r = m+n-1, so we have a solution as required. Now since m is positive, we cannot take r any larger. If we take it smaller, then we will have m > r and hence n = r+1-m ≤ 0, which is not allowed. So the solution is unique.