2000 AMC 10 Problems/Problem 19

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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

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

Solution

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 $\boxed{\text{D}}$

(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle $k$ times changes each of the areas $k^2$ times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)

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

From AoPSWiki

Problem

Solution

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