APMO 01/3 solution

13th APMO 2001

Problem 3

Two equal-sized regular n-gons intersect to form a 2n-gon C. Prove that the sum of the sides of C which form part of one n-gon equals half the perimeter of C.

Solution

Let one regular n-gon have vertices P₁, P₂, ... , P_n and the other have vertices Q₁, Q₂, ... , Q_n. Each side of the 2n-gon forms a triangle with one of the P_i or Q_i. Note that P_i and Q_j must alternate as we go around the 2n-gon. For convenience assume that the order is P₁, Q₁, P₂, Q₂, ... , P_n, Q_n. Let the length of the side which forms a triangle with P_i be p_i, and the length of the side which forms a triangle with Q_i be q_i. Each of these triangles has one angle (180^o - 360^o/n). Adjacent triangles have one of their other angles equal (alternate angles), so all the triangles are similar. If the sides of the triangle vertex P_i have lengths a_i, b_i, p_i, then the side P_iP_i+1 is b_i + q_i + a_i+1, and a_i/a_i+1 = b_i/b_i+1 = p_i/p_i+1. Similarly, if the sides of the triangle vertex Q_i have lengths c_i, d_i, q_i, then the side Q_iQ_i+1 is d_i + p_i+1 + c_i+1 and c_i/c_i+1 = d_i/d_i+1 = q_i/q_i+1. But a_i/b_i = d_i/c_i (not c_i/d_i), because the triangles alternate in orientation.

Put a_i/p_i = h, b_i/p_i = k. Note that a_i + b_i > p_i, so h + k - 1 > 0. We have also c_i/q_i = k, d_i/q_i = h. Adding the expressions for P_iP_i+1 we get perimeter P_i = ∑(b_i + q_i + a_i+1) = k ∑ p_i + ∑ q_i + h ∑ p_i. Similarly, perimeter Q_i = (h + k) ∑ q_i + ∑ p_i. The two n-gons are equal, so (h + k - 1) ∑ p_i = (h + k - 1) ∑ q_i. Hence ∑ p_i = ∑ q_i, which is the required result.

13th APMO 2001