PUMaC 2010 · 几何（A 组） · 第 7 题

PUMaC 2010 — Geometry (Division A) — Problem 7

Square ABCD is divided into four rectangles by EF and GH . EF is parallel to AB and GH ◦ parallel to BC . ∠ BAF = 18 . EF and GH meet at point P . The area of rectangle P F CH is twice that of rectangle AGP E . Given that the value of ∠ F AH in degrees is x , find the nearest integer to x .

解析

Square ABCD is divided into four rectangles by EF and GH . EF is parallel to AB and GH parallel to BC . EF and GH meet at point P . The area of rectangle P F CH is twice that of rectangle AGP E . If maximal value of ∠ F AH in degrees is x , find the nearest integer to x . [Answer] 45 [Solution] Let side length of square be 1. Let AG = x , AE = y . Then (1 − x )(1 − y ) = 2 xy ⇒ x + y = 1 − xy . There are two ways to proceed from here. 3 (Analytic Method) We have tan ∠ HAD = x , tan ∠ F AB = y , by compound angle formula x + y tan( ∠ HAD + ∠ F AB ) = = 1 , 1 − xy ◦ ◦ which implies that ∠ HAD + ∠ F AB = 45 , i.e. ∠ HAF = 45 . (Geometric Solution) Rotate 4 ABF 90 degrees counterclockwise to 4 ADM . We inteprete the LHS and RHS of equation x + y = 1 − xy as follow: LHS is twice the sum of areas of 4 ABF and 4 ADH , which is the area of 4 AHM ; RHS is area of pentagon ABF HD , which is equal to area of quadrilateral AF HM . Therefore, 4 AF H and 4 AHM have equal area. Also since ◦ AF = AM , AH = AH , the two triangles must be congruent, which implies ∠ F AH = 45 .