Difference between revisions of "1994 AHSME Problems/Problem 29"
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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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+ | ==See Also== | ||
+ | |||
+ | {{AHSME box|year=1994|num-b=28|num-a=30}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:28, 9 January 2021
Problem
Points and on a circle of radius are situated so that , and the length of minor arc is . If angles are measured in radians, then
Solution
First note that arc length equals , where is the central angle in radians. Call the center of the circle . Then radian because the minor arc has length . Since is isosceles, . We use the Law of Cosines to find that Using half-angle formulas, we have that this ratio simplifies to
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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