Difference between revisions of "2011 AMC 12B Problems/Problem 6"
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<math>\angle BOC</math> and <math>\angle BAC</math> must add up to <math>180^{\circ}</math> since quadrilateral <math>ABOC</math> is cyclic, so <math>\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}</math>. | <math>\angle BOC</math> and <math>\angle BAC</math> must add up to <math>180^{\circ}</math> since quadrilateral <math>ABOC</math> is cyclic, so <math>\angle BAC = 180 - 144 = \boxed{\textbf{(C) } 36}</math>. | ||
+ | |||
+ | -Solution by Joeya | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}} | {{AMC12 box|year=2011|num-b=5|num-a=7|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:35, 26 January 2021
Problem
Two tangents to a circle are drawn from a point . The points of contact and divide the circle into arcs with lengths in the ratio . What is the degree measure of ?
Solution
In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection between two tangents dividing a circle into arc length A and arc length B = 1/2 (Arc A° - Arc B°).
In order to utilize this theorem, the degree measures of the arcs must be found. First, set A (Arc length A) equal to 3d, and B (Arc length B) equal to 2d.
Setting 3d+2d = 360° will find d = 72°, and so therefore Arc length A in degrees will equal 216° and arc length B will equal 144°.
Finally, simply plug the two arc lengths into the tangent-tangent intersection theorem, and the answer:
1/2 (216°-144°) = 1/2 (72°)
Solution 2
Let arc A be 3x and arc B be 2x. Then the angle formed by the tangents is by the arc length formula. Also note that , which simplifies to Hence the angle formed by the tangents is equal to .
Solution 3
Let the center of the circle be . The radii extending from to the points of tangency and are both perpendicular to lines and . Thus , and quadrilateral is cyclic, since the opposite angles add to .
Since the ratio of the arc lengths is , the shorter arc (the arc cut off by ) is .
and must add up to since quadrilateral is cyclic, so .
-Solution by Joeya
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 12 Problems and Solutions |
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