1974 AHSME Problems/Problem 16

Problem

A circle of radius $r$ is inscribed in a right isosceles triangle, and a circle of radius $R$ is circumscribed about the triangle. Then $R/r$ equals

$\mathrm{(A)\ } 1+\sqrt{2} \qquad \mathrm{(B) \ }\frac{2+\sqrt{2}}{2} \qquad \mathrm{(C) \ } \frac{\sqrt{2}-1}{2} \qquad \mathrm{(D) \ } \frac{1+\sqrt{2}}{2} \qquad \mathrm{(E) \ }2(2-\sqrt{2})$

Solution

$[asy] draw((0,0)--(1,0)--(0,1)--cycle); draw(circle((0.5,0.5),sqrt(2)/2)); draw(circle((1-sqrt(2)/2,1-sqrt(2)/2),1-sqrt(2)/2)); label("$A$",(0,0),SW); label("$B$",(0,1),NNW); label("$C$",(1,0),SSE); label("$D$",(0.5,0.5),NE); draw((0,0)--(0.5,0.5)); draw((1-sqrt(2)/2,1-sqrt(2)/2)--(0,1-sqrt(2)/2)); draw((1-sqrt(2)/2,1-sqrt(2)/2)--(1-sqrt(2)/2,0)); label("$E$",(1-sqrt(2)/2,0),S); label("$F$",(0,1-sqrt(2)/2),W); label("$I$",(1-sqrt(2)/2,1-sqrt(2)/2),NNW); [/asy]$

Label the points as in the figure above. Let the side length $AB=AC=s$ . Therefore, $AC=s\sqrt{2}$ . Since the circumradius of a right triangle is equal to half of the length of the hypotenuse, we have $R=\frac{s\sqrt{2}}{2}$ .

Now to find the inradius. Notice that $IFAE$ is a square with side length $r$ , and also $AD=AC=s$ . Therefore, $s=AD=AI+ID=r\sqrt{2}+r$ , and so $r=\frac{s}{\sqrt{2}+1}$ .

Finally, $\frac{R}{r}=\frac{\frac{s\sqrt{2}}{2}}{\frac{s}{\sqrt{2}+1}}=\frac{\sqrt{2}+1}{\sqrt{2}}=\frac{2+\sqrt{2}}{2}, \boxed{\text{B}}$ .

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1974 AHSME Problems/Problem 16

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