1963 AHSME Problems/Problem 22

Problem

Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$ ; $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$ .

A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$ . Then the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is:

$\textbf{(A)}\ \frac{5}{18}\qquad \textbf{(B)}\ \frac{2}{9}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{1}{3}\qquad \textbf{(E)}\ \frac{4}{9}$

Solution

$[asy] draw(circle((0,0),1)); dot((-1,0)); pair A=(-1,0),B=(0.5,0.866),C=(0.978,-0.208),O=(0,0),E=(-0.105,-0.995); label("A",(-1,0),W); dot((0.5,0.866)); label("B",(0.5,0.866),NE); dot((0.978,-0.208)); label("C",(0.978,-0.208),SE); dot((0,0)); label("O",(0,0),NE); dot(E); label("E",E,S); draw(A--B--C--A); draw(E--O); [/asy]$

Because $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$ , $\stackrel \frown {AC} = 168^\circ$ . Also, $OA = OC$ and $OE \perp AC$ , so $\angle AOE = \angle COE = 84^\circ$ . Since $\angle BOC = 72^\circ$ , $\angle BOE = 156^\circ$ . Finally, $\triangle BOE$ is an isosceles triangle, so $\angle OBE = 12^\circ$ . Because $\angle BAC = \frac{1}{2} \cdot 72 = 36^\circ$ , the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is $\frac{12}{36} = \frac{1}{3}$ , which is answer choice $\boxed{\textbf{(D)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
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1963 AHSME Problems/Problem 22

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