HMMT 二月 2001 · 几何 · 第 10 题

HMMT February 2001 — Geometry — Problem 10

A is the center of a semicircle, with radius AD lying on the base. B lies on the base between A and D , and E is on the circular portion of the semicircle such that EBA is a right angle. Extend EA through A to C , and put F on line CD such that EBF is a line. √ √ √ √ √ √ 2 − 2 2 5+ 10 2 5 − 10 Now EA = 1, AC = 2, BF = , CF = , and DF = . Find DE . 4 4 4

解析

A is the center of a semicircle, with radius AD lying on the base. B lies on the base between A and D , and E is on the circular portion of the semicircle such that EBA is a right angle. Extend EA through A to C , and put F on line CD such that EBF is a line. √ √ √ √ √ √ 2 − 2 2 5+ 10 2 5 − 10 Now EA = 1, AC = 2, BF = , CF = , and DF = . Find DE . 4 4 4 Solution: Let θ = ∠ AED and x = DE . By the law of cosines on triangle ADE , we have 2 2 1 = 1+ x − 2 x cos θ ⇒ 2 x cos θ = x . Then by the law of cosines on triangle CDE (note that √ ( √ ) ( √ ) ( √ ) ( √ ) 2 2 2 2 2 CD = 5), we have 5 = 1 + 2 + x − 2 1 + 2 x cos θ = 1 + 2 + x − 1 + 2 x . √ √ Solving the quadratic equation gives x = 2 − 2 .