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2000 AMC 10 Problem 19

Problem 19 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. The ratio of the area of the other small right triangle to the area of the square is

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Solution

Let the square have side 1.1. One small triangle has legs 11 and r,r, with area 12r=m,\tfrac12 r = m, so r=2m.r = 2m. The two small triangles are similar, so the other has legs 11 and 1r,\tfrac1r, with area 12⋅1r=14m.\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}. Since the square has area 1,1, the desired ratio is 14m.\dfrac{1}{4m}. Thus, the correct answer is D.
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Tagged: similarity · area ratio · right triangle

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