1973 AHSME Problems/Problem 15

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Problem

A sector with acute central angle $\theta$ is cut from a circle of radius 6. The radius of the circle circumscribed about the sector is

$\textbf{(A)}\ 3\cos\theta \qquad \textbf{(B)}\ 3\sec\theta \qquad \textbf{(C)}\ 3 \cos \frac12 \theta \qquad \textbf{(D)}\ 3 \sec \frac12 \theta \qquad \textbf{(E)}\ 3$

Solution

Let $O$ be the center of the circle and $A,B$ be two points on the circle such that $\angle AOB = \theta$. If the circle circumscribes the sector, then the circle must circumscribe $\triangle AOB$.

[asy] draw((-120,-160)--(0,0)--(120,-160)); draw((-60,-80)--(0,-125)--(60,-80),dotted); draw((0,0)--(0,-125)); draw(arc((0,0),200,233.13,306.87)); [/asy]

Draw the perpendicular bisectors of $OA$ and $OB$ and mark the intersection as point $C$, and draw a line from $C$ to $O$. By HL Congruency and CPCTC, $\angle AOC = \angle BOC = \theta /2$.

Let $R$ be the circumradius of the triangle. Using the definition of cosine for right triangles, \[\cos (\theta /2) = \frac{3}{R}\] \[R = \frac{3}{\cos (\theta /2)}\] \[R = 3 \sec (\theta /2)\] Answer choices A, C, and E are smaller, so they are eliminated. However, as $\theta$ aproaches $90^\circ$, the value $3\sec\theta$ would approach infinity while $3\sec \tfrac12 \theta$ would approach $\tfrac{3\sqrt{2}}{2}$. A super large circle would definitely not be a circumcircle if $\theta$ is close to $90^\circ$, so we can confirm that the answer is $\boxed{\textbf{(D)}\ 3 \sec \frac12 \theta}$.

See Also

1973 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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