# 1999 AHSME Problems/Problem 21

## Problem

A circle is circumscribed about a triangle with sides $20,21,$ and $29,$ thus dividing the interior of the circle into four regions. Let $A,B,$ and $C$ be the areas of the non-triangular regions, with $C$ be the largest. Then $\mathrm{(A) \ }A+B=C \qquad \mathrm{(B) \ }A+B+210=C \qquad \mathrm{(C) \ }A^2+B^2=C^2 \qquad \mathrm{(D) \ }20A+21B=29C \qquad \mathrm{(E) \ } \frac 1{A^2}+\frac 1{B^2}= \frac 1{C^2}$

## Solution $20^2 + 21^2 = 841 = 29^2$. Therefore the triangle is a right triangle. But then its hypotenuse is a diameter of the circumcircle, and thus $C$ is exactly one half of the circle. Moreover, the area of the triangle is $\frac{20\cdot 21}{2} = 210$. Therefore the area of the other half of the circumcircle can be expressed as $A+B+210$. Thus the answer is $\boxed{\mathrm{(B)}}$.

To complete the solution, note that $\mathrm{(A)}$ is clearly false. As $A+B < C$, we have $A^2 + B^2 < (A+B)^2 < C^2$ and thus $\mathrm{(C)}$ is false. Similarly $20A + 21B < 21(A+B) < 21C < 29C$, thus $\mathrm{(D)}$ is false. And finally, since $0, $\frac 1{C^2} < \frac1{A^2} < \frac 1{A^2} + \frac 1{B^2}$, thus $\mathrm{(E)}$ is false as well.

## Solution 2 (Alternative to realize that the triangle is Right)

Copy the link into the browser to see a diagram (I don't know how to link it and I tried using the brackets, but it didn't work)https://geogebra.org/classic/psg3ugzm

Let $\circ{O}$ be the circumcircle of $\triangle{XYZ}$ in the problem, and let the circle have a radius of $r$. Let $XY=20, YZ=21, XZ=29$.

Using the law of cosines: $29^2=20^2+21^2-2*20*21*\cos{XYZ}$.

Thus $\cos{XYZ}=0$, thus the triangle is right. Thus follow as above: "Moreover, the area of the triangle is $\frac{20\cdot 21}{2} = 210$. Therefore the area of the other half of the circumcircle can be expressed as $A+B+210$. Thus the answer is $\boxed{\mathrm{(B)}}$" (I quoted the solution above to show you where to continue).

~hastapasta

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