2000 AMC 12 Problems/Problem 21
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 times the area of the square. The ratio of the area of the other small right triangle to the area of the square is
Solution
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Without loss of generality let a side of the square be . Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since
, the height of the triangle with area
is
. Therefore
where
is the base of the other triangle.
, and the area of that triangle is
.
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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