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> | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME box|year=1994|num-b=28|num-a=30}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:28, 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 
![[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]](http://latex.artofproblemsolving.com/c/a/d/cad0445f8a3e2981969ac6b261c962573da90c76.png)
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 ![\[\frac{AB}{BC} = \frac{\sqrt{2r^2 - 2r^2\cos{(\pi - \frac{1}{2})}}}{\sqrt{2r^2 - 2r^2\cos1}} = \frac{\sqrt{1 + \cos{(\frac{1}{2})}}}{\sqrt{1 - \cos1}}.\]](http://latex.artofproblemsolving.com/b/b/c/bbcb4203f52f3c00815b2294805a86befd30722f.png)
Using half-angle formulas, we have that this ratio simplifies to ![\[\frac{\cos\frac{1}{4}}{\sin{\frac{1}{2}}} = \frac{\cos\frac{1}{4}}{\sqrt{1 - \cos^2{\frac{1}{2}}}} = \frac{\cos\frac{1}{4}}{\sqrt{(1 + \cos{\frac{1}{2}})(1 - \cos{\frac{1}{2}})}} = \frac{\cos{\frac{1}{4}}}{2\cos{\frac{1}{4}}\sin{\frac{1}{4}}}\]](http://latex.artofproblemsolving.com/3/6/1/361770ab7ba9346bf073a184266fa6b6de91c77b.png)
![\[= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}\]](http://latex.artofproblemsolving.com/3/9/0/390164fc95f13b9a5e2dfa40658e0c71b95a5ce3.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
