# 1994 AHSME Problems/Problem 29

## Problem

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 $AB/BC=$ $[asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); label("A", (-13,0), W); label("B", (12,5), NE); label("C", (12,-5), SE); [/asy]$ $\textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}}$

## Solution 1

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 $$\frac{\cos\frac{1}{4}}{\sin{\frac{1}{2}}} = \frac{\cos\frac{1}{4}}{\sqrt{1 - \cos^2{\frac{1}{2}}}} = \frac{\cos\frac{1}{4}}{\sqrt{(1 + \cos{\frac{1}{2}})(1 - \cos{\frac{1}{2}})}} = \frac{\cos{\frac{1}{4}}}{2\cos{\frac{1}{4}}\sin{\frac{1}{4}}}$$ $$= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}$$

## Solution 2

$[asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); dot((12,0)); dot((0,0)); draw((-13,0)--(12,0)--cycle); label("A", (-13,0), W); label("B", (12,5), NE); label("C", (12,-5), SE); label("D", (12,0), NW); label("O", (0,0), NE); [/asy]$

Let the center of this circle be $O$, $\angle BOC = \theta$, the radius of $\odot O$ be $r$.

By the definition of radian, $\theta =$ $\overarc{BC}$ $/r=1$

$\angle BAC=\frac{\angle BOC}{2} = \frac12$

$\sin \angle BAD = \frac{BD}{AB}$, $\sin \frac{\angle BAC}{2} = \frac{\frac{BC}{2}}{AB}$

$\frac{BC}{AB}=2 \sin \frac{\frac12}{2} = 2\sin \frac14$

$\frac{AB}{BC}= \frac{1}{2\sin \frac14} = \boxed{\textbf{(A) } \frac{1}{2} \csc \frac{1}{4} }$