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 $0<A<C$, $\frac 1{C^2} < \frac1{A^2} < \frac 1{A^2} + \frac 1{B^2}$, thus $\mathrm{(E)}$ is false as well.

See also

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS