2000 AMC 12 Problems/Problem 21

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Problem

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

$\text {(A)}\ \frac{1}{2m+1} \qquad \text {(B)}\ m \qquad \text {(C)}\ 1-m \qquad \text {(D)}\ \frac{1}{4m} \qquad \text {(E)}...$

Solution

Without loss of generality 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 height of the triangle with area $m$ is $2m$ . Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the base of the other triangle. $x = \frac{1}{2m}$ , and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{(D)}$ .

2000 AMC 12 (Problems • Resources)
Preceded by Problem 20	Followed by Problem 22
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2000 AMC 12 Problems/Problem 21

From AoPSWiki

Problem

Solution

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