Track / Stage 3 / 111 of 260 #111 of 2000

Problem 111

AMC 10/12, early questions

Geometry Difficulty 3.4 Find the answer

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 $AB/BC=$

$\textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}}$

Your answer

Official solution

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

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.