1979 AHSME Problems/Problem 21
Problem 21
The length of the hypotenuse of a right triangle is , and the radius of the inscribed circle is . The ratio of the area of the circle to the area of the triangle is
Solution
Solution by e_power_pi_times_i
Since , where is the inradius, is the semiperimeter, and is the area, we have that the ratio of the area of the circle to the area of the triangle is . Now we try to express as and . Denote the points where the incircle meets the triangle as , where is the incenter, and denote . Since is a square (tangents are perpendicular to radius), . The perimeter can be expressed as , so the semiperimeter is . The hypotenuse is . Thus we have . The answer is .
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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