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

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2000 AMC 10 Problems/Problem 19

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