# 1986 AHSME Problems/Problem 21

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## Problem

In the configuration below, $\theta$ is measured in radians, $C$ is the center of the circle, $BCD$ and $ACE$ are line segments and $AB$ is tangent to the circle at $A$. $[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(0,-1), E=(0,1), C=(0,0), D=dir(10), F=dir(190), B=(-1/sin(10*pi/180))*dir(10); fill(Arc((0,0),1,10,90)--C--D--cycle,mediumgray); fill(Arc((0,0),1,190,270)--B--F--cycle,mediumgray); draw(unitcircle); draw(A--B--D^^A--E); label("A",A,S); label("B",B,W); label("C",C,SE); label("\theta",C,SW); label("D",D,NE); label("E",E,N); [/asy]$

A necessary and sufficient condition for the equality of the two shaded areas, given $0 < \theta < \frac{\pi}{2}$, is $\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad \textbf{(C)}\ \tan\theta = 4\theta\qquad \textbf{(D)}\ \tan 2\theta =\theta\qquad\\ \textbf{(E)}\ \tan\frac{\theta}{2}=\theta$

## Solution

Well, the shaded sector's area is basically $\text{(ratio of } \theta \text{ to total angle of circle)} \times \text{(total area)} = \frac{\theta}{2\pi} \cdot (\pi r^2) = \frac{\theta}{2} \cdot (AC)^2$.

In addition, if you let $\angle{ACB} = \theta$, then $$\tan \theta = \frac{AB}{AC}$$ $$AB = AC\tan \theta = r\tan \theta$$ $$[ABC] = \frac{AB \cdot AC}{2} = \frac{r^2\tan \theta}{2}$$Then the area of that shaded thing on the left becomes $$\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2}$$We want this to be equal to the sector area so $$\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2} = \frac{\theta \cdot r^2}{2}$$ $$\frac{r^2\tan \theta}{2} = \theta \cdot r^2$$ $$\tan \theta = 2\theta$$

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