Difference between revisions of "2009 AMC 10B Problems/Problem 16"
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− | As <math>\triangle ABC</math> is equilateral, we have <math>\angle BAC = \angle BCA = 60^\circ</math>, hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\ | + | As <math>\triangle ABC</math> is equilateral, we have <math>\angle BAC = \angle BCA = 60^\circ</math>, hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\angle AOC = 120^\circ</math>, and from symmetry we have <math>\angle AOB = \angle COB = 60^\circ</math>. Finally this gives us <math>\angle ABO = \angle CBO = 30^\circ</math>. |
We know that <math>DO = AO</math>, as <math>D</math> lies on the circle. From <math>\triangle ABO</math> we also have <math>AO = BO \sin 30^\circ = \frac{BO}2</math>, Hence <math>DO = \frac{BO}2</math>, therefore <math>BD = BO - DO = \frac{BO}2</math>, and <math>\frac{BD}{BO} = \boxed{\frac 12}</math>. | We know that <math>DO = AO</math>, as <math>D</math> lies on the circle. From <math>\triangle ABO</math> we also have <math>AO = BO \sin 30^\circ = \frac{BO}2</math>, Hence <math>DO = \frac{BO}2</math>, therefore <math>BD = BO - DO = \frac{BO}2</math>, and <math>\frac{BD}{BO} = \boxed{\frac 12}</math>. |
Revision as of 17:43, 18 June 2015
Problem
Points and lie on a circle centered at , each of and are tangent to the circle, and is equilateral. The circle intersects at . What is ?
Solution
Solution 1
As is equilateral, we have , hence . Then , and from symmetry we have . Finally this gives us .
We know that , as lies on the circle. From we also have , Hence , therefore , and .
Solution 2
As in the previous solution, we find out that . Hence and are both equilateral.
We then have , hence is the incenter of , and as is equilateral, is also its centroid. Hence , and as , we have , therefore , and as before we conclude that .
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AMC 10 Problems and Solutions |
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