20th Eötvös Competition Problems 1913

1.  Prove that n! n! > nⁿ for n > 2.

2.  Let A and B be diagonally opposite vertices of a cube. Prove that the midpoints of the 6 edges not containing A or B form a regular (planar) hexagon.

3.  If d is the greatest common divisor of a and b, and D is the greatest common divisor of A and B, show that dD is the greatest common divisor of aA, aB, bA and bB. 

Solutions

Problem 1

Prove that n! n! > nⁿ for n > 2.

Solution

We may write the lhs as ∏_0≤k<n (k+1)(n-k). For k > 0 and n-k > 1, we have (k+1)(n-k) = k(n-k) + (n-k) > k + (n-k) = n. In other words, all terms except 1·n and n·1 are > n. There are n terms, so for n > 2 the product is > nⁿ. 

Problem 2

Let A and B be diagonally opposite vertices of a cube. Prove that the midpoints of the 6 edges not containing A or B form a regular (planar) hexagon.

Solution

Let the side of the cube be 2k. Let X be any of the midpoints. Then AX = BX = √(4k² + k²). So X lies on the sphere center A radius k√5 and on the sphere center B radius k√5. Hence X lies on the intersection of these two spheres which is a circle. In particular, the midpoints are planar.

Let XY be any two adjacent midpoints. Then XY = √(k² + k²), so the sides have equal length. Since the hexagon is cyclic, it follows that it must be regular. 

Problem 3

If d is the greatest common divisor of a and b, and D is the greatest common divisor of A and B, show that dD is the greatest common divisor of aA, aB, bA and bB.

Solution

Obviously dD divides all four numbers. Put a = da', b = db', A = DA', B = DB'. Then a' and b' are relatively prime, and A' and B' are relatively prime. We have to show that the gcd of a'A', a'B', b'A', b'B' is 1. Suppose a prime p > 1 divides them all. Suppose p does not divide a'. Then since it divides a'A' and a'B' it must divide A' and B', contradicting the fact that A' and B' are relatively prime. So p must divide a'. Similarly it must divide b'. Contradiction. So no prime p can divide them all, so their gcd is 1, as required.