26th Eötvös Competition Problems 1922

1.  Show that given four non-coplanar points A, B, P, Q there is a plane with A, B on one side and P, Q on the other, and with all the points the same distance from the plane.

2.  Show that we cannot factorise x⁴ + 2x² + 2x + 2 as the product of two quadratics with integer coefficients.

3.  Let S be any finite set of distinct positive integers which are not divisible by any prime greater than 3. Prove that the sum of their reciprocals is less than 3. 

Solutions

Problem 1

Show that given four non-coplanar points A, B, P, Q there is a plane with A, B on one side and P, Q on the other, and with all the points the same distance from the plane.

Solution

Let U be the midpoint of AQ, V the midpoint of AP, W the midpoint of BP, and X the midpoint of BQ. Then UV and WX are both parallel to PQ and hence to each other. Similarly, XU and VW. So UVWX is a parallelogram and hence lies in a plane. This plane is the one required. AB is parallel to the plane, so A and B are equidistant from it. W is the midpoint of PB, so P and B are equidistant from the plane. Similarly, X is the midpoint of BQ, so B and Q are equidistant from the plane. So all four points are equidistant. 

Problem 2

Show that we cannot factorise x⁴ + 2x² + 2x + 2 as the product of two quadratics with integer coefficients.

Solution

Suppose we can. Both coefficients of x² must be 1. So suppose we have (x² + ax + b)(x² + cx + d). Now bd = 2. wlog |b| ≤ |d|, so either b = 1, d = 2 or b = -1, d = -2. The coefficient of x³ is 0, so a = -c. Now the coefficient of x² is 2 = b + d + ac = b + d - a². So a² = 5 or -1, neither of which is possible. Note that we did not need to know that the coefficient of x in the quartic was 2. 

Problem 3

Let S be any finite set of distinct positive integers which are not divisible by any prime greater than 3. Prove that the sum of their reciprocals is less than 3.

Solution

S is a subset of X = {2^m3ⁿ | m, n = 0, 1, 2, ... }. The sum for X is ∑ 1/(2^m3ⁿ) = (∑ 1/2^m)(∑ 1/3ⁿ) = (1/(1 - 1/2)) (1/(1 - 1/3)) = 2·3/2 = 3. The sum for any finite subset of X is smaller.