Difference between revisions of "2000 AMC 12 Problems/Problem 17"
m (→Solution (with minimal trig)) |
m (→Solution 3 (with minimal trig)) |
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Line 52: | Line 52: | ||
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: | 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: | ||
− | <math>sin\theta =\frac{Opposite}{Hypotenuse}</math> | + | <math>\sin\theta =\frac{\text{Opposite}}{\text{Hypotenuse}}</math> |
− | <math>cos\theta =\frac{Adjacent}{Hypotenuse}</math> | + | <math>\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}</math> |
− | <math>tan\theta =\frac{Opposite}{Adjacent}</math>. | + | <math>\tan\theta =\frac{\text{Opposite}}{\text{Adjacent}}</math>. |
With a bit of guess and check, we get that the answer is <math>\boxed{D}</math>. | With a bit of guess and check, we get that the answer is <math>\boxed{D}</math>. |
Revision as of 10:59, 26 December 2019
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 .
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.