Difference between revisions of "2000 AMC 12 Problems/Problem 17"
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Let <math>t = \tan(\theta/2)</math>. <math>OC = \frac{1-t}{1+t}OB = \frac{1-t^2}{1+2t+t^2}</math>. Because <math>\sin(\theta) = \frac{2t}{1+t^2}</math> and <math>\cos(\theta) = \frac{1-t^2}{1+t^2}</math>, <cmath>OC = \frac{\cos(\theta)}{1+\sin(\theta)}OB = \boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</cmath> | Let <math>t = \tan(\theta/2)</math>. <math>OC = \frac{1-t}{1+t}OB = \frac{1-t^2}{1+2t+t^2}</math>. Because <math>\sin(\theta) = \frac{2t}{1+t^2}</math> and <math>\cos(\theta) = \frac{1-t^2}{1+t^2}</math>, <cmath>OC = \frac{\cos(\theta)}{1+\sin(\theta)}OB = \boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</cmath> | ||
+ | |||
+ | ==Solution 7 (if you forgot angle bisector but remember LoS)== | ||
+ | |||
+ | Let <math>x=\overline{OC}</math>, and let <math>\angle OBC=\angle ABC=\alpha</math>. We know that <math>\overline{AC}=\overline{OA}-\overline{OC}=1-x</math>. By the Law of Sines, | ||
+ | <cmath>\dfrac{\sin\alpha}x=\dfrac{\sin\theta}{BC}</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>\dfrac{\sin\alpha}{1-x}=\dfrac{\sin90^\circ}{BC}=\dfrac1{BC}.</cmath> | ||
+ | |||
+ | Combining the two give <math>\dfrac{\sin\alpha}x=\sin\theta\cdot\dfrac{\sin\alpha}{1-x}</math>. | ||
+ | |||
+ | Solving, this gives <math>\boxed{x=\frac{1}{\sin{\theta}+1}}</math>. | ||
+ | |||
+ | ~Technodoggo | ||
== Video Solution == | == Video Solution == |
Latest revision as of 23:38, 19 August 2024
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 ,
Solution 7 (if you forgot angle bisector but remember LoS)
Let , and let . We know that . By the Law of Sines,
and
Combining the two give .
Solving, this gives .
~Technodoggo
Video Solution
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.