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1994 AHSME Problems/Problem 29

Revision as of 21:39, 16 February 2016 by 0x5f3759df (talk | contribs)


Points $A, B$ and $C$ on a circle of radius $r$ are situated so that $AB=AC, AB>r$, and the length of minor arc $BC$ is $r$. If angles are measured in radians, then $AB/BC=$ [asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); label("A", (-13,0), W); label("B", (12,5), NE); label("C", (12,-5), SE); [/asy] $\textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}}$


First note that arc length equals $r\theta$, where $\theta$ is the central angle in radians. Call the center of the circle $O$. Then $\angle{BOC} = 1$ radian because the minor arc $BC$ has length $r$. Since $ABC$ is isosceles, $\angle{AOB} = \pi - \tfrac{1}{2}$. We use the Law of Cosines to find that \[\frac{AB}{BC} = \frac{\sqrt{2r^2 - 2r^2\cos{(\pi - \frac{1}{2})}}}{\sqrt{2r^2 - 2r^2\cos1}} = \frac{\sqrt{1 + \cos{(\frac{1}{2})}}}{\sqrt{1 - \cos1}}.\] Using half-angle formulas, we have that this ratio simplifies to \[\frac{\cos\frac{1}{4}}{\sin{\frac{1}{2}}} = \frac{\cos\frac{1}{4}}{\sqrt{1 - \cos^2{\frac{1}{2}}}} = \frac{\cos\frac{1}{4}}{\sqrt{(1 + \cos{\frac{1}{2}})(1 - \cos{\frac{1}{2}})}} = \frac{\cos{\frac{1}{4}}}{2\cos{\frac{1}{4}}\sin{\frac{1}{4}}}\] \[= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}\]

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