# 1997 AJHSME Problems/Problem 24

## Problem

Diameter $ACE$ is divided at $C$ in the ratio $2:3$. The two semicircles, $ABC$ and $CDE$, divide the circular region into an upper (shaded) region and a lower region. The ratio of the area of the upper region to that of the lower region is

$[asy] pair A,B,C,D,EE; A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray); draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)); draw(circle((5,0),5)); dot(A); dot(B); dot(C); dot(D); dot(EE); label("A",A,W); label("B",B,N); label("C",C,E); label("D",D,N); label("E",EE,W); [/asy]$

$\text{(A)}\ 2:3 \qquad \text{(B)}\ 1:1 \qquad \text{(C)}\ 3:2 \qquad \text{(D)}\ 9:4 \qquad \text{(E)}\ 5:2$

## Solution

$[asy] pair A,B,C,D,EE; A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray); draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)); draw(circle((5,0),5)); draw(A--EE); dot(A); dot(B); dot(C); dot(D); dot(EE); label("A",A,W); label("B",B,N); label("C",C,dir(40)); label("D",D,N); label("E",EE,W);[/asy]$

Draw $\overline{AE}$ to divide the big circle in half. Assign $AC = 4$ and $CE = 6$ so that the radii work out to be integers. ($4x$ and $6x$ can be used instead, but the $x$ will cancel in the ratio.)

The shaded region is equal to the area of semicircle $\overarc{AE}$ on top, plus the area of the semicircle $CDE$ on the bottom, minus the area of semicircle $\overarc{ABC}$ on top.

The radii of those three semicircles are $5, 3,$ and $2$, respectively.Thus, the area of the shaded region is $\frac{1}{2}\pi\cdot5^2 + \frac{1}{2}\pi\cdot3^2 - \frac{1}{2}\pi\cdot2^2 =\frac{\pi}{2}(5^2 + 3^2 - 2^2)=15\pi$

The total area of the circle is $\pi \cdot 5^2 = 25\pi$.Thus, the unshaded area is $25\pi - 15\pi = 10\pi$.Therefore the ratio of shaded:unshaded is $15\pi : 10\pi =\boxed{ \text{(C)}\ 3:2}$.