2000 AMC 10 Problems/Problem 19

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

$\mathrm{(A)}\ \frac{1}{2m+1} \qquad\mathrm{(B)}\ m \qquad\mathrm{(C)}\ 1-m \qquad\mathrm{(D)}\ \frac{1}{4m} \qquad\mathrm{(E)}\ \frac{1}{8m^2}$

Solution

$[asy] unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W); [/asy]$

Let the square have area $1$ , then it follows that the altitude of one of the triangles is $2m$ . The area of the other triangle is $\frac{x}{2}$ .

By similar triangles, we have $\frac{x}{1}=\frac{1}{2m}\Rightarrow \frac{x}{2}=\frac{1}{4m}$

This is choice $\boxed{\text{D}}$

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2000 AMC 10 Problems/Problem 19

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