Maths Olympiad Prep

Track / Stage 5 / 332 of 400 #932 of 1964

Problem 932

AIME late
Geometry Difficulty 5.8 Prove it

Russian problems Problem 45 The convex hexagon ABCDEF has all angles equal. Prove that AB - DE = EF - BC = CD - FA. (b) Given six lengths a 1 , ... , a 6 satisfying a 1 - a 4 = a 5 - a 2 = a 3 - a 6 , show that you can construct a hexagon with sides a 1 , ... , a 6 and equal angles.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

(a) Extend AB, CD, EF. We get an equilateral triangle with sides AF + AB + BC, BC + CD + DE, ED + EF + FA. Hence AB - DE = CD - FA = EF - BC, as required. (b) Take an equilateral triangle with sides s, t, u lengths a 2 + a 3 + a 4 , a 4 + a 5 + a 6 , and a 6 + a 1 + a 2 respectively. Construct BC length a 2 parallel to t with B on u and C on s. Construct DE length a 4 parallel to u with D on s and E on t. Construct FA length a 6 parallel to s with F on t and A on u. Then ABCDEF is the required hexagon, with AB = a 1 , BC = a 2 etc. Russian 41-50 (C) John Scholes [email protected] 23 Sep 1998

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.