13th Eötvös Competition Problems 1906

1.  Let α be a real number, not an odd multiple of π. Prove that tan α/2 is rational iff cos α and sin α are rational.

2.  Show that the centers of the squares on the outside of the four sides of a rhombus form a square.

3.  (a₁, a₂, ... , a_n) is a permutation of (1, 2, ... , n). Show that ∏ (a_i - i) is even if n is odd. 

Solutions

Problem 1

Let α be a real number, not an odd multiple of π. Prove that tan α/2 is rational iff cos α and sin α are rational.

Solution

Put t = tan α/2. Then cos α = (1-t²)/(1+t²), sin α = 2t/(1+t²). So certainly cos α and sin &alphap are rational if t is.

Put k = cos α. Then since α is not an odd multiple of π, k is not -1. Hence t² = (1-k)/(1+k), which is rational. Hence 2t = (1+t²) sin α is rational, and so also t. 

Problem 2

Show that the centers of the squares on the outside of the four sides of a rhombus form a square.

Solution

Take the rhombus as ABCD and the centers as PQRS as shown. Let the center of the rhombus be O. By symmetry about AC, we have OP = OS. Now ∠AOD = ∠ ASD = 90^o, so AODS is cyclic. Hence ∠AOS = ∠ADS = 45^o. Similarly for all the other angles. So POR and QOS are straight and perpendicular. Hence PQRS is a square.

Problem 3

(a₁, a₂, ... , a_n) is a permutation of (1, 2, ... , n). Show that ∏ (a_i - i) is even if n is odd.

Solution

Put n = 2m+1. Suppose there are k values of i for which both a_i and i are odd. Then since there are m+1 odd values in total, there must be m+1-k values for which a_i is odd, but i is not and m+1-k values for which a_i is even and i is odd. The remaining k-1 values have both a_i and i even. Since a_i - i is odd iff a_i and i have opposite parity, there are 2(m+1-k) values for which that is true. Since that is even, there must be at least one value for which a_i and i have the same parity. Hence the product is even.