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>. | |
==Solution== | ==Solution== | ||
− | + | {{image}} | |
− | {{ | + | 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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Revision as of 13:19, 20 January 2007
Problem
Points , , and are on the circumference of a unit circle so that the measure of is , the measure of is , and the measure of is . The area of the triangular shape bounded by and line segments and can be written in the form , where and are relatively prime positive integers. Find .
Solution
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Let the center of the circle be . The area of the desired region is easily seen to be that of sector plus the area of triangle minus the area of triangle . Using the area formula to compute the areas of the two triangles, this is , so the answer is .