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

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

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Solution

Let the square have side

1.

One small triangle has legs

1

and

r,

with area

\tfrac12 r = m,

r = 2m.

The two small triangles are similar, so the other has legs

1

and

\tfrac1r,

with area

\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.

Since the square has area

1,

the desired ratio is

\dfrac{1}{4m}.

Thus, the correct answer is D.

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