Difference between revisions of "2000 AMC 12 Problems/Problem 17"
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== Solution 4 == | == Solution 4 == | ||
− | Let <math>OC</math> = x, <math>OB</math> = h, and <math>AB</math> = y. <math>AC</math> = <math>OA</math> - <math>OC</math>. Because <math>OC</math> = x, and <math>OA</math> = 1 (given in the problem), <math>AC</math> = 1-x. Using the [[Angle Bisector Theorem]], <math>\frac{h}{y}</math> = <math>\frac{x}{1-x}</math> \Longrightarrow h(1-x) = xy. Solving for x gives us x = <math>\frac{h}{h+y}</math>. <math>\sin\theta = \frac{opposite}{hypotenuse} | + | Let <math>OC</math> = x, <math>OB</math> = h, and <math>AB</math> = y. <math>AC</math> = <math>OA</math> - <math>OC</math>. |
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+ | Because <math>OC</math> = x, and <math>OA</math> = 1 (given in the problem), <math>AC</math> = 1-x. | ||
+ | |||
+ | Using the [[Angle Bisector Theorem]], <math>\frac{h}{y}</math> = <math>\frac{x}{1-x}</math> <math>\Longrightarrow</math> h(1-x) = xy. Solving for x gives us x = <math>\frac{h}{h+y}</math>. | ||
+ | |||
+ | <math>\sin\theta = \frac{opposite}{hypotenuse} = \frac{y}{h}</math>. Solving for y gives us y = h <math>\sin\theta</math>. | ||
+ | |||
+ | Substituting this for y in our initial equation yields x = <math>\dfrac{h}{h+\sin \theta}</math>. | ||
+ | |||
+ | Using the distributive property, x = <math>\dfrac{h}{h(1+\sin \theta)}</math> and finally <math>\dfrac{1}{1+\sin \theta}</math> or <math>\boxed{\textbf{(D)}}</math> | ||
== See also == | == See also == |
Revision as of 11:49, 11 June 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
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.