2000 AMC 12 Problems/Problem 17
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
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 .
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 .
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 = h .
Substituting this for y in our initial equation yields x = .
Using the distributive property, x = and finally or
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 ,
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