2000 AMC 12 Problems/Problem 17
Contents
Problem
A circle centered at has radius and contains the point . The segment is tangent to the circle at and . If point lies on and bisects , then
Solution 1
Since is tangent to the circle, is a right triangle. This means that , and . By the Angle Bisector Theorem, We multiply both sides by to simplify the trigonometric functions, Since , . Therefore, the answer is .
Solution 2
Alternatively, one could notice that OC approaches the value 1/2 as theta gets close to 90 degrees. The only choice that is consistent with this is (D).
Solution 3 (with minimal trig)
Let's assign a value to so we don't have to use trig functions to solve. is a good value for , because then we have a -- because is tangent to Circle .
Using our special right triangle, since , , and .
Let . Then . since bisects , we can use the angle bisector theorem:
.
Now, we only have to use a bit of trig to guess and check: the only trig facts we need to know to finish the problem is:
.
With a bit of guess and check, we get that the answer is .
Solution 4
Let = x, = h, and = y. = - .
Because = x, and = 1 (given in the problem), = 1-x.
Using the Angle Bisector Theorem, = h(1-x) = xy. Solving for x gives us x = .
. Solving for y gives us y = .
Substituting this for y in our initial equation yields x = .
Using the distributive property, x = and finally or
Solution 5
Since is tangent to the circle, and thus we can use trig ratios directly.
By the angle bisector theorem, we have
Seeing the resemblance of the ratio on the left-hand side to we turn the ratio around to allow us to plug in Another source of motivation for this also lies in the idea of somehow adding 1 to the right-hand side so that we can substitute for a given value, i.e. , and flipping the fraction will preserve the , whilst adding one right now would make the equation remain in direct terms of
Solution 6 (tangent half angle)
. By sine law,
Let . . Because and ,
Video Solution
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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