1999 AMC 12 Problem 21

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}

Check whether

20, 21, 29

form a right triangle

The hypotenuse is a diameter, so the largest region

C

is a semicircle

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.

Problem 21 in other years