Difference between revisions of "2000 AMC 12 Problems/Problem 17"
(→Solution 4) |
(→Solution 5) |
||
Line 74: | Line 74: | ||
== Solution 5 == | == Solution 5 == | ||
− | Since <math>\overline{AB}</math> is tangent to the circle, <math>\angle OAB=90^{\circ}</math> and thus we can use trig ratios directly. | + | Since <math>\overline{AB}</math> is tangent to the circle, <math>\angle OAB=90^{\circ}</math> and thus we can use trig ratios directly. |
<cmath>\sin{\theta}=\frac{\overline{AB}}{\overline{BO}}, \cos{\theta}=\frac{1}{\overline{BO}}, \tan{\theta}=\overline{AB}</cmath> | <cmath>\sin{\theta}=\frac{\overline{AB}}{\overline{BO}}, \cos{\theta}=\frac{1}{\overline{BO}}, \tan{\theta}=\overline{AB}</cmath> | ||
+ | |||
+ | By the angle bisector theorem, we have | ||
<cmath>\frac{\overline{OB}}{\overline{AB}}=\frac{\overline{OC}}{\overline{CA}}</cmath> | <cmath>\frac{\overline{OB}}{\overline{AB}}=\frac{\overline{OC}}{\overline{CA}}</cmath> |
Revision as of 17:34, 18 November 2020
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 = h .
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
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.