HMMT 二月 2010 · GEN1 赛 · 第 3 题

HMMT February 2010 — GEN1 Round — Problem 3

[ 4 ] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex 7 pentagon that has an area of of the area of the original rectangle. Find the ratio of the longer side 10 of the rectangle to the shorter side of the rectangle.

解析

[ 4 ] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convex 7 pentagon that has an area of of the area of the original rectangle. Find the ratio of the longer side 10 of the rectangle to the shorter side of the rectangle. √ Answer: 5 ′ B D E A C B Given a polygon P P · · · P , let [ P P · · · P ] denote its area. Let ABCD be the rectangle. Suppose 1 2 k 1 2 k ′ we fold B across AC , and let E be the intersection of AD and B C . Then we end up with the pentagon ′ ACDEB , depicted above. Let’s suppose, without loss of generality, that ABCD has area 1. Then 3 4 AEC must have area , since 10 General Test, Part 1 [ ABCD ] = [ ABC ] + [ ACD ] ′ = [ AB C ] + [ ACD ] ′ = [ AB E ] + 2[ AEC ] + [ EDC ] ′ = [ ACDEB ] + [ AEC ] 7 = [ ABCD ] + [ AEC ] , 10 3 3 That is, [ AEC ] = [ ABCD ] = . 10 10 ′ 1 ′ Since 4 ECD is congruent to 4 EAB , both triangles have area . Note that 4 AB C , 4 ABC , 5 1 and 4 CDA are all congruent, and all have area . Since 4 AEC and 4 EDC share altitude DC , 2 [ DEC ] DE 2 = = . Because 4 CAE is isosceles, CE = EA . Let AE = 3 x . The CE = 3 x , DE = 2 x , EA [ AEC ] 3 √ √ √ AD AE + ED 3+2 √ and CD = x 9 − 4 = x 5. Then = = = 5. DC DC 5