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Difference between revisions of "1963 AHSME Problems/Problem 22"

m (Was OBE not BOE.)
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Line 3: Line 3:
 
Acute-angled <math>\triangle ABC</math> is inscribed in a circle with center at <math>O</math>; <math>\stackrel \frown {AB} = 120^\circ</math> and <math>\stackrel \frown {BC} = 72^\circ</math>.
 
Acute-angled <math>\triangle ABC</math> is inscribed in a circle with center at <math>O</math>; <math>\stackrel \frown {AB} = 120^\circ</math> and <math>\stackrel \frown {BC} = 72^\circ</math>.
  
−
A point <math>E</math> is taken in minor arc <math>AC</math> such that <math>OE</math> is perpendicular to <math>AC</math>. Then the ratio of the magnitudes of <math>\angle BOE</math> and <math>\angle BAC</math> is:
+
A point <math>E</math> is taken in minor arc <math>AC</math> such that <math>OE</math> is perpendicular to <math>AC</math>. Then the ratio of the magnitudes of <math>\angle OBE</math> and <math>\angle BAC</math> is:
  
 
<math>\textbf{(A)}\ \frac{5}{18}\qquad
 
<math>\textbf{(A)}\ \frac{5}{18}\qquad

Latest revision as of 20:00, 29 April 2023

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 $\boxed{\textbf{(D)}}$.

See Also

1963 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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