Difference between revisions of "2000 AMC 12 Problems/Problem 17"
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<math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math> | <math>\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math> | ||
− | == Solution == | + | == Solution 1 == |
− | Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a right triangle. This means that <math>OA = 1</math>, <math> | + | Since <math>\overline{AB}</math> is tangent to the circle, <math>\triangle OAB</math> is a right triangle. This means that <math>OA = 1</math>, <math>AB = \tan \theta</math> and <math>OB = \sec \theta</math>. By the [[Angle Bisector Theorem]], <cmath> \frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta </cmath> We multiply both sides by <math>\cos \theta</math> to simplify the trigonometric functions, <cmath> AC=OC \sin \theta </cmath> Since <math>AC + OC = 1</math>, <math>1 - OC = OC \sin \theta \Longrightarrow</math> <math>OC = \dfrac{1}{1+\sin \theta}</math>. Therefore, the answer is <math>\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}</math>. |
+ | |||
+ | == 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 <math>\theta</math> so we don't have to use trig functions to solve. <math>60</math> is a good value for <math>\theta</math>, because then we have a <math>30-60-90 \triangle</math> -- <math>\angle BAC=90</math> because <math>AB</math> is tangent to Circle <math>O</math>. | ||
+ | |||
+ | Using our special right triangle, since <math>AO=1</math>, <math>OB=2</math>, and <math>AB=\sqrt{3}</math>. | ||
+ | |||
+ | Let <math>OC=x</math>. Then <math>CA=1-x</math>. since <math>BC</math> bisects <math>\angle ABO</math>, we can use the angle bisector theorem: | ||
+ | |||
+ | <math>\frac{2}{x}=\frac{\sqrt{3}}{1-x}</math> | ||
+ | |||
+ | <math>2-2x=\sqrt{3}x</math> | ||
+ | |||
+ | <math>2=(\sqrt{3}+2)x</math> | ||
+ | |||
+ | <math>x=\frac{2}{\sqrt{3}+2}</math>. | ||
+ | |||
+ | 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{\text{Opposite}}{\text{Hypotenuse}}</math> | ||
+ | |||
+ | <math>\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}</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>. | ||
+ | |||
+ | == 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> <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 = <math>h \sin\theta</math>. | ||
+ | |||
+ | Substituting this for y in our initial equation yields x = <math>\dfrac{h}{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> | ||
+ | |||
+ | == 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. | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Seeing the resemblance of the ratio on the left-hand side to <math>\sin{\theta},</math> we turn the ratio around to allow us to plug in <math>\sin{\theta}.</math> 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. <math>\overline{OA}=1</math>, and flipping the fraction will preserve the <math>\overline{OC}</math>, whilst adding one right now would make the equation remain in direct terms of <math>\overline{CA}.</math> | ||
+ | |||
+ | <cmath>\frac{\overline{AB}}{\overline{OB}}=\sin{\theta}=\frac{\overline{CA}}{\overline{OC}}\Rightarrow \sin{\theta}+1=\frac{\overline{CA}+\overline{OC}}{\overline{OC}}=\frac{1}{\overline{OC}}</cmath> | ||
+ | |||
+ | <cmath>\sin{\theta}+1=\frac{1}{\overline{OC}} \Rightarrow \boxed{\overline{OC}=\frac{1}{\sin{\theta}+1}}</cmath> | ||
+ | |||
+ | == Solution 6 (tangent half angle) == | ||
+ | |||
+ | <math>\angle CBO = 45^{\circ} - \frac{\theta}{2}, \angle ACB = 45^{\circ} + \frac{\theta}{2}, OB = \frac{1}{\cos(\theta)}</math>. | ||
+ | By sine law, <math>\frac{OC}{\sin(\angle CBO)} = \frac{OB}{\sin(\angle OCB)} = \frac{OB}{\sin(\angle ACB)}</math> | ||
+ | |||
+ | <cmath>OC = \frac{\sin(45^{\circ} - \frac{\theta}{2})}{\sin(45^{\circ} + \frac{\theta}{2})}OB = \frac{\sin(45^{\circ} - \frac{\theta}{2})}{\cos(45^{\circ} - \frac{\theta}{2})}OB = \tan(45^{\circ} - \frac{\theta}{2})OB = \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}OB</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> | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zGnRV6eNiV0 | ||
== See also == | == See also == | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
[[Category:Introductory Trigonometry Problems]] | [[Category:Introductory Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:50, 17 July 2022
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
,
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 |
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