HMMT 二月 2001 · 几何 · 第 10 题
HMMT February 2001 — Geometry — Problem 10
题目详情
英文原题
- A is the center of a semicircle, with radius AD lying on the base. B lies on the basebetween A and D , and E is on the circular portion of the semicircle such that EBA is aright 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
英文解析
- A is the center of a semicircle, with radius AD lying on the base. B lies on the basebetween A and D , and E is on the circular portion of the semicircle such that EBA is aright 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 .