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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 mm 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 1.1. The small triangle sharing one side of the square has a perpendicular leg r,r, so its area is 12⋅1⋅r=m,\tfrac12 \cdot 1 \cdot r = m, giving r=2m.r = 2m. The two small triangles are similar, so the other triangle’s leg along the square is 1r,\dfrac1r, and its area is 12⋅1⋅1r=12r=14m. \frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}. Thus, the correct answer is D.
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