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Difference between revisions of "1997 AJHSME Problems/Problem 24"

(Created page with "==Problem== Diameter <math>ACE</math> is divided at <math>C</math> in the ratio <math>2:3</math>. The two semicircles, <math>ABC</math> and <math>CDE</math>, divide the circula...")
(No difference)

Revision as of 22:36, 31 July 2011

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

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 $\stackrel\frown{AE}$ on top, plus the area of the semicircle $\stackrel\frown{CDE}$ on the bottom, minus the area of semicircle $\stackrel\frown{ABC}$ 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 $25\pi - 15\pi = 10\pi$.

So the ratio of shaded:unshaded is $15\pi : 10\pi = 3:2$, and teh answer is $\boxed{C}$


See Also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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
All AJHSME/AMC 8 Problems and Solutions
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