Difference between revisions of "1999 AHSME Problems/Problem 21"

Revision as of 13:35, 5 July 2013

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.

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AHSME Problems and Solutions

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

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 == See also ==
 {{AHSME box|year=1999|num-b=20|num-a=22}}
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Difference between revisions of "1999 AHSME Problems/Problem 21"

Revision as of 13:35, 5 July 2013

Problem

Solution

See also