Question

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 $C$ being the largest. Then

$A + B = C$

$A + B + 210 = C$

$A^2 + B^2 = C^2$

$20A + 21B = 29C$

$\dfrac{1}{A^2} + \dfrac{1}{B^2} = \dfrac{1}{C^2}$

Solution:

Since $20^2 + 21^2 = 841 = 29^2,$ the triangle is right-angled, and its hypotenuse of length $29$ is a diameter of the circle. Thus the largest region $C$ is the semicircle on one side of that diameter.

The other semicircle consists of the triangle together with regions $A$ and $B.$ Since the two semicircles are congruent and the triangle has area $\tfrac12 \cdot 20 \cdot 21 = 210,$ we get $A + B + 210 = C.$

Thus, the correct answer is B.

1999 AMC 12 Problem 21

Solution:

Problem 21 in Other Years