AMC10 2004 A

AMC10 2004 A · Q20

AMC10 2004 A · Q20. It mainly tests Triangles (properties), Pythagorean theorem.

Points E and F are located on square ABCD so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$?

点 E 和 F 位于正方形 ABCD 上，使得 $\triangle BEF$ 为等边三角形。$\triangle DEF$ 的面积与 $\triangle ABE$ 的面积之比是多少？

(A) 4/3 $\frac{4}{3}$

(B) 3/2 $\frac{3}{2}$

(D) 2 2

(E) $1 + \sqrt{3}$ $1 + \sqrt{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) First, assume that $AB = 1$, and let $ED = DF = x$. By the Pythagorean Theorem $x^2 + x^2 = EF^2 = EB^2 = 1^2 + (1 - x)^2$, so $x^2 = 2(1 - x)$. Hence the desired ratio of the areas is $$ \frac{\mathrm{Area}(\triangle DEF)}{\mathrm{Area}(\triangle ABE)} = \frac{x^2}{1-x} = 2. $$

（D）首先，假设 $AB = 1$，并令 $ED = DF = x$。由勾股定理，$x^2 + x^2 = EF^2 = EB^2 = 1^2 + (1 - x)^2$，因此 $x^2 = 2(1 - x)$。所以所求的面积比为 $$ \frac{\mathrm{Area}(\triangle DEF)}{\mathrm{Area}(\triangle ABE)} = \frac{x^2}{1-x} = 2. $$

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