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Difference between revisions of "1994 AHSME Problems/Problem 29"

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</asy>
 
</asy>
 
<math> \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}} </math>
 
<math> \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}} </math>
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==Solution==
+
 
 +
==Solution 1==
 
First note that arc length equals <math>r\theta</math>, where <math>\theta</math> is the central angle in radians. Call the center of the circle <math>O</math>. Then <math>\angle{BOC} = 1</math> radian because the minor arc <math>BC</math> has length <math>r</math>. Since <math>ABC</math> is isosceles, <math>\angle{AOB} = \pi - \tfrac{1}{2}</math>. We use the Law of Cosines to find that <cmath>\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}}.</cmath>  
 
First note that arc length equals <math>r\theta</math>, where <math>\theta</math> is the central angle in radians. Call the center of the circle <math>O</math>. Then <math>\angle{BOC} = 1</math> radian because the minor arc <math>BC</math> has length <math>r</math>. Since <math>ABC</math> is isosceles, <math>\angle{AOB} = \pi - \tfrac{1}{2}</math>. We use the Law of Cosines to find that <cmath>\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}}.</cmath>  
 
Using half-angle formulas, we have that this ratio simplifies to <cmath>\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}}}</cmath> <cmath>= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}</cmath>
 
Using half-angle formulas, we have that this ratio simplifies to <cmath>\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}}}</cmath> <cmath>= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}</cmath>
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 +
==Solution 2==
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 +
<asy>
 +
draw(Circle((0,0), 13));
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draw((-13,0)--(12,5)--(12,-5)--cycle);
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dot((-13,0));
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dot((12,5));
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dot((12,-5));
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dot((12,0));
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dot((0,0));
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draw((-13,0)--(12,0)--cycle);
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label("A", (-13,0), W);
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label("B", (12,5), NE);
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label("C", (12,-5), SE);
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label("D", (12,0), NW);
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label("O", (0,0), NE);
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</asy>
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Let the center of this circle be <math>O</math>, <math>\angle BOC = \theta</math>, the radius of <math>\odot O</math> be <math>r</math>.
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 +
Because the length of minor arc <math>BC</math> is <math>r</math>, <math>\frac{\theta}{2 \pi} \cdot 2 \pi r = r</math>, <math>\theta = 1</math>
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<math>\angle BAC=\frac{\angle BOC}{2} = \frac12</math>
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<math>\sin \angle BAD = \frac{BD}{AB}</math>, <math>\sin \frac{\angle BAC}{2} = \frac{\frac{BC}{2}}{AB}</math>
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<math>\frac{BC}{AB}=2 \sin \frac{\frac12}{2} = \sin \frac14</math>
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<math>\frac{AB}{BC}= \boxed{\textbf{(A) } \frac{1}{2} \csc \frac{1}{4} }</math>
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 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==See Also==
 
==See Also==

Revision as of 08:03, 28 September 2023

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 $r$.

Because the length of minor arc $BC$ is $r$, $\frac{\theta}{2 \pi} \cdot 2 \pi r = r$, $\theta = 1$

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

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

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

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

~isabelchen

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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
All AHSME Problems and Solutions

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