2000 AMC 12 Problem 21

Problem 21 of 25HarderGeometry

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. What is the ratio of the area of the other small right triangle to the area of the square?

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Solution

Let the square have side

1.

The small triangle sharing one side of the square has a perpendicular leg

r,

so its area is

\tfrac12 \cdot 1 \cdot r = m,

giving

r = 2m.

The two small triangles are similar, so the other triangle’s leg along the square is

\dfrac1r,

and its area is

\frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}.

Thus, the correct answer is D.

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