PUMaC 2010 · 几何（A 组） · 第 4 题

PUMaC 2010 — Geometry (Division A) — Problem 4

In regular hexagon ABCDEF , AC , CE are two diagonals. Points M , N are on AC , CE 2 respectively and satisfy AC : AM = CE : CN = r . Suppose B, M, N are collinear, find 100 r .

解析

In regular hexagon ABCDEF , AC , CE are two diagonals. Points M , N are on AC , CE 2 respectively and satisfy AC : AM = CE : CN = r . Suppose B, M, N are collinear, find 100 r . 1 [Answer] 300 [Solution] Let the side length of hexagon be 1. Extend N C and AB to intersect at some √ point G . Then AB = 1, BG = 2, GC = 3, let CN = x . We use [ XY Z ] to denote area of triangle XY Z . Then √ [ BCG ] : [ BCN ] = CG : CN = 3 : x [ BAN ] : [ BGN ] = BA : BG = 1 : 2 Consequently, √ √ 3 + x 3 + x AM : CM = [ BCN ] : [ BAN ] = : x = . 2 2 x By condition, AM : AC = CN : CE , which translates to √ 3 + x x √ = √ . 3 + 3 x 3 √ √ Solve for x , the only positive solution is 1. Hence r = CE/CN = 3 / 1 = 3.