PUMaC 2015 · 团队赛 · 第 7 题

PUMaC 2015 — Team Round — Problem 7

[ 5 ] Charlie noticed his golden ticket was golden in two ways! In addition to being gold, it was √ 1+ 5 a rectangle whose side lengths had ratio the golden ratio φ = . He then folds the ticket 2 so that two opposite corners (vertices connected by a diagonal) coincide and makes a sharp crease (the ticket folds just as any regular piece of paper would). The area of the resulting b shape can be expressed as a + bφ . What is ? a

解析

[ 5 ] Charlie noticed his golden ticket was golden in two ways! In addition to being gold, it was √ 1+ 5 a rectangle whose side lengths had ratio the golden ratio φ = . He then folds the ticket 2 so that two opposite corners (vertices connected by a diagonal) coincide and makes a sharp crease (the ticket folds just as any regular piece of paper would). The area of the resulting b shape can be expressed as a + bφ . What is ? a Solution: 2 Let the side lengths of the rectangle be s, sφ . Then looking at figure 1 a , the area of the pentagon is the area of ABCO plus the area of ODEA . But referring back to the uncreased ′ rectangle, this is same as the area of the rectangle minus the area of AOD and A OB . By ′ symmetry, these have the same area. Then we know that AA ⊥ CD (convince yourself of this) so AOD is a right triangle. Now: ( √ ) 2 2 2 2 3 ′ 2 AO · OD AE · A E AO OD s φ 1 + φ s φ + s φ [ AOD ] = = · · = · = 2 ′ 2 2 2 2 φ 8 φ AE A E ′ 2 Where we used the fact that 4 AOD ∼ 4 AEA . Now using the fact that φ − φ − 1 = 0, we have that: 2 2 2 2 2 2 2 s φ s 3 s φ s φ s s s φ 2 2 [ ABCDE ] = s φ − 2[ AOD ] = s φ − − = − + = + 4 4 φ 4 4 4 4 2 So b/a = 2 . Author: Nathan Mytelka, Roy Zhao