HMMT 二月 2010 · GEN1 赛 · 第 3 题

英文解析

3. [ 4 ] A rectangular piece of paper is folded along its diagonal (as depicted below) to form a non-convexpentagon that has an area of of the area of the original rectangle. Find the ratio of the longer side 7
   of the rectangle to the shorter side of the rectangle.10
   √
   Answer: 5
   ′
   B D
   BCAE
   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
   4 AEC must have area , since 3
   General Test, Part 110
   [ ABCD ] = [ ABC ] + [ ACD ]
   ′
   = [ AB C ] + [ ACD ]
   ′
   = [ AB E ] + 2[ AEC ] + [ EDC ]
   ′
   = [ ACDEB ] + [ AEC ] = [ ABCD ] + [ AEC ] ,7
   3 310
   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 ,
   and 4 CDA are all congruent, and all have area . Since 4 AEC and 4 EDC share altitude DC ,15
   [ DEC ]2
   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