1979 AHSME Problems/Problem 21

Problem 21

The length of the hypotenuse of a right triangle is $h$ , and the radius of the inscribed circle is $r$ . The ratio of the area of the circle to the area of the triangle is

$\textbf{(A) }\frac{\pi r}{h+2r}\qquad \textbf{(B) }\frac{\pi r}{h+r}\qquad \textbf{(C) }\frac{\pi}{2h+r}\qquad \textbf{(D) }\frac{\pi r^2}{r^2+h^2}\qquad \textbf{(E) }\text{none of these}$

Solution

Solution by e_power_pi_times_i

Since $rs = A$ , where $r$ is the inradius, $s$ is the semiperimeter, and $A$ is the area, we have that the ratio of the area of the circle to the area of the triangle is $\frac{\pi r^2}{rs} = \frac{\pi r}{s}$ . Now we try to express $s$ as $h$ and $r$ . Denote the points where the incircle meets the triangle as $X,Y,Z$ , where $O$ is the incenter, and denote $AX = AY = z, BX = BZ = y, CY = CZ = x$ . Since $XOZB$ is a square (tangents are perpendicular to radius), $r = BX = BZ = y$ . The perimeter can be expressed as $2(x+y+z)$ , so the semiperimeter is $x+y+z$ . The hypotenuse is $AY+CY = z+x$ . Thus we have $s = x+y+z = (z+x)+y = h+r$ . The answer is $\boxed{\textbf{(B) } \frac{\pi r}{h+r}}$ .

1979 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1979 AHSME Problems/Problem 21

Problem 21

Solution

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