AMC12 2000 A

AMC12 2000 A · Q21

AMC12 2000 A · Q21. 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

通过直角三角形的斜边上的一点，作两条分别平行于该三角形两条直角边的直线，使三角形被分成一个正方形和两个更小的直角三角形。其中一个小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

WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the height of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.

不失一般性，设正方形的边长为 $1$。简单的角度追踪可知这两个直角三角形相似。因此三角形对应边的比相同。由于 $A = \frac{1}{2}bh = \frac{h}{2}$，面积为 $m$ 的那个三角形的底边为 $2m$。因此 $\frac{2m}{1} = \frac{1}{x}$，其中 $x$ 为另一个三角形的高。于是 $x = \frac{1}{2m}$，该三角形的面积为 $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.

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