PUMaC 2011 · 几何（B 组） · 第 6 题
PUMaC 2011 — Geometry (Division B) — Problem 6
题目详情
- [ 6 ] A rectangular piece of paper has corners labeled A, B, C , and D , with BC = 80 and CD = 120. Let M be the midpoint of side AB . The corner labeled A is folded along line M D and the corner labeled B is folded along line M C until the segments AM and M B coincide. Let S denote the point in space where A and B meet. If H is the foot of the perpendicular from S to the original plane of the paper, find HM .
解析
- Pick P on DM and R on CM so the AP is perpendicular to DM and BR is perpendicular to CM . Because of the way the paper is being folded, the projection of A onto the plane of the paper is always along line AP , and the projection of B along line BR . Thus, the two lines will intersect in exactly the point H . Since △ HM B ∼ △ M BC , we have HM/M B = M B/BC , so HM = ( M B/BC ) · M B = (60 / 80) · 60 = (3 / 4) · 60 = 45 . ′