HMMT 二月 2004 · 几何 · 第 7 题

HMMT February 2004 — Geometry — Problem 7

Yet another trapezoid ABCD has AD parallel to BC . AC and BD intersect at P . If [ ADP ] / [ BCP ] = 1 / 2, find [ ADP ] / [ ABCD ]. (Here the notation [ P · · · P ] denotes 1 n the area of the polygon P · · · P .) 1 n

解析

Yet another trapezoid ABCD has AD parallel to BC . AC and BD intersect at P . If [ ADP ] / [ BCP ] = 1 / 2, find [ ADP ] / [ ABCD ]. (Here the notation [ P · · · P ] denotes 1 n the area of the polygon P · · · P .) 1 n √ Solution: 3 − 2 2 A homothety (scaling) about P takes triangle ADP into BCP , since AD, BC are √ parallel and A, P, C ; B, P, D are collinear. The ratio of homothety is thus 2. It √ follows that, if we rescale to put [ ADP ] = 1, then [ ABP ] = [ CDP ] = 2, just by the √ √ ratios of lengths of bases. So [ ABCD ] = 3 + 2 2, so [ ADP ] / [ ABCD ] = 1 / (3 + 2 2). √ Simplifying this, we get 3 − 2 2.