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Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 4"
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==Problem==
 
==Problem==
 
[[Point]]s <math>A</math>, <math>B</math>, and <math>C</math> are on the [[circumference]] of a unit [[circle]] so that the measure of <math>\widehat{AB}</math> is <math>72^{\circ}</math>, the measure of <math>\widehat{BC}</math> is <math>36^{\circ}</math>, and the measure of <math>\widehat{AC}</math> is <math>108^\circ</math>. The area of the triangular shape bounded by <math>\widehat{BC}</math> and [[line segment]]s <math>\overline{AB}</math> and <math>\overline{AC}</math> can be written in the form <math>\frac{m}{n} \cdot \pi</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m + n</math>.
 
[[Point]]s <math>A</math>, <math>B</math>, and <math>C</math> are on the [[circumference]] of a unit [[circle]] so that the measure of <math>\widehat{AB}</math> is <math>72^{\circ}</math>, the measure of <math>\widehat{BC}</math> is <math>36^{\circ}</math>, and the measure of <math>\widehat{AC}</math> is <math>108^\circ</math>. The area of the triangular shape bounded by <math>\widehat{BC}</math> and [[line segment]]s <math>\overline{AB}</math> and <math>\overline{AC}</math> can be written in the form <math>\frac{m}{n} \cdot \pi</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m + n</math>.
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==Solution==
 
==Solution==
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{{image}}
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[[Image:2007_4_MockAIME-4.png]]
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Let the center of the circle be <math>O</math>.  The area of the desired region is easily seen to be that of sector <math>BOC</math> plus the area of triangle <math>AOB</math> minus the area of triangle <math>AOC</math>.  Using the area formula <math>K_{\triangle XYZ} = \frac{1}{2} XY \cdot YZ \cdot \sin Y</math> to compute the areas of the two triangles, this is <math>\pi \cdot \frac{36}{360} + \frac{1}{2}\sin 72^\circ - \frac{1}{2}\sin108^{\circ} = \frac{1}{10}\cdot \pi</math>, so the answer is <math>1 + 10 = 011</math>.
 
Let the center of the circle be <math>O</math>.  The area of the desired region is easily seen to be that of sector <math>BOC</math> plus the area of triangle <math>AOB</math> minus the area of triangle <math>AOC</math>.  Using the area formula <math>K_{\triangle XYZ} = \frac{1}{2} XY \cdot YZ \cdot \sin Y</math> to compute the areas of the two triangles, this is <math>\pi \cdot \frac{36}{360} + \frac{1}{2}\sin 72^\circ - \frac{1}{2}\sin108^{\circ} = \frac{1}{10}\cdot \pi</math>, so the answer is <math>1 + 10 = 011</math>.
  
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==See also==
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{{Mock AIME box|year=2006-2007|n=4|num-b=3|num-a=5|source=125025}}
  
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*[[Mock AIME 4 2006-2007 Problems/Problem 5| Next Problem]]
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[[Category:Intermediate Geometry Problems]]
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*[[Mock AIME 4 2006-2007 Problems/Problem 3| Previous Problem]]
 
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*[[Mock AIME 4 2006-2007 Problems]]
 

Latest revision as of 15:52, 8 October 2007

Problem

Points $A$, $B$, and $C$ are on the circumference of a unit circle so that the measure of $\widehat{AB}$ is $72^{\circ}$, the measure of $\widehat{BC}$ is $36^{\circ}$, and the measure of $\widehat{AC}$ is $108^\circ$. The area of the triangular shape bounded by $\widehat{BC}$ and line segments $\overline{AB}$ and $\overline{AC}$ can be written in the form $\frac{m}{n} \cdot \pi$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

2007 4 MockAIME-4.png

Let the center of the circle be $O$. The area of the desired region is easily seen to be that of sector $BOC$ plus the area of triangle $AOB$ minus the area of triangle $AOC$. Using the area formula $K_{\triangle XYZ} = \frac{1}{2} XY \cdot YZ \cdot \sin Y$ to compute the areas of the two triangles, this is $\pi \cdot \frac{36}{360} + \frac{1}{2}\sin 72^\circ - \frac{1}{2}\sin108^{\circ} = \frac{1}{10}\cdot \pi$, so the answer is $1 + 10 = 011$.

See also

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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