AMC10 2000 A

AMC10 2000 A · Q19

AMC10 2000 A · Q19. It mainly tests Triangles (properties), Similarity.

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

在一个直角三角形的斜边上一点，作与两条直角边平行的直线，将三角形分割成一个正方形和两个更小的直角三角形。其中一个小的直角三角形的面积是正方形面积的$m$倍。另一个小直角三角形面积与正方形面积的比值为

(A) $\frac{1}{2m+1}$ $\frac{1}{2m+1}$

(B) $m$ $m$

(D) $\frac{1}{4m}$ $\frac{1}{4m}$

(E) $\frac{1}{8m^2}$ $\frac{1}{8m^2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Without loss of generality, let the side of the square have length $1$ unit and let the area of triangle $ADF$ be $m$. Let $AD=r$ and $EC=s$. Because triangles $ADF$ and $FEC$ are similar, $\frac{s}{1}=\frac{1}{r}$. Since $\frac{1}{2}r=m$, the area of triangle $FEC$ is $\frac{1}{2}s=\frac{1}{2r}=\frac{1}{4m}$.

答案（D）：不失一般性，设正方形的边长为 $1$ 个单位，且三角形 $ADF$ 的面积为 $m$。设 $AD=r$，$EC=s$。由于三角形 $ADF$ 与 $FEC$ 相似，所以 $\frac{s}{1}=\frac{1}{r}$。又因为 $\frac{1}{2}r=m$，所以三角形 $FEC$ 的面积为 $\frac{1}{2}s=\frac{1}{2r}=\frac{1}{4m}$。

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