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Difference between revisions of "2000 AMC 12 Problems/Problem 17"

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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:
  
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<math>sin\theta =\frac{Opposite}{Hypotenuse}</math>
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<math>\sin\theta =\frac{\text{Opposite}}{\text{Hypotenuse}}</math>
  
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<math>cos\theta =\frac{Adjacent}{Hypotenuse}</math>
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<math>\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}</math>
  
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<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>.
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== Solution 4 ==
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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.
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 +
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>.
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<math>\sin\theta = \frac{opposite}{hypotenuse} = \frac{y}{h}</math>. Solving for y gives us y = <math>h \sin\theta</math>.
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 +
Substituting this for y in our initial equation yields x = <math>\dfrac{h}{h+h\sin \theta}</math>.
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 +
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>
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== 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.
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 +
<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
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<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>
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 +
== 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>
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==Solution 7 (if you forgot angle bisector but remember LoS)==
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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>
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 +
and
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<cmath>\dfrac{\sin\alpha}{1-x}=\dfrac{\sin90^\circ}{BC}=\dfrac1{BC}.</cmath>
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Combining the two give <math>\dfrac{\sin\alpha}x=\sin\theta\cdot\dfrac{\sin\alpha}{1-x}</math>.
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 +
Solving, this gives <math>\boxed{x=\frac{1}{\sin{\theta}+1}}</math>.
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 +
~Technodoggo
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 +
== Video Solution ==
 +
https://youtu.be/zGnRV6eNiV0
  
 
== See also ==
 
== See also ==

Latest revision as of 00:38, 20 August 2024

Problem

A circle centered at $O$ has radius $1$ and contains the point $A$. The segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle ABO$, then $OC =$

[asy] import olympiad; size(6cm); unitsize(1cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); labelmargin=0.2; dotfactor=3; pair O=(0,0); pair A=(1,0); pair B=(1,1.5); pair D=bisectorpoint(A,B,O); pair C=extension(B,D,O,A); draw(Circle(O,1)); draw(O--A--B--cycle); draw(B--C); label("$O$",O,SW); dot(O); label("$\theta$",(0.1,0.05),ENE); dot(C); label("$C$",C,S); dot(A); label("$A$",A,E); dot(B); label("$B$",B,E);[/asy]

$\text {(A)}\ \sec^2 \theta - \tan \theta \qquad \text {(B)}\ \frac 12 \qquad \text {(C)}\ \frac{\cos^2 \theta}{1 + \sin \theta}\qquad \text {(D)}\ \frac{1}{1+\sin\theta} \qquad \text {(E)}\ \frac{\sin \theta}{\cos^2 \theta}$

Solution 1

Since $\overline{AB}$ is tangent to the circle, $\triangle OAB$ is a right triangle. This means that $OA = 1$, $AB = \tan \theta$ and $OB = \sec \theta$. By the Angle Bisector Theorem, \[\frac{OB}{OC} = \frac{AB}{AC} \Longrightarrow AC \sec \theta = OC \tan \theta\] We multiply both sides by $\cos \theta$ to simplify the trigonometric functions, \[AC=OC \sin \theta\] Since $AC + OC = 1$, $1 - OC = OC \sin \theta \Longrightarrow$ $OC = \dfrac{1}{1+\sin \theta}$. Therefore, the answer is $\boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}$.

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 $\theta$ so we don't have to use trig functions to solve. $60$ is a good value for $\theta$, because then we have a $30-60-90 \triangle$ -- $\angle BAC=90$ because $AB$ is tangent to Circle $O$.

Using our special right triangle, since $AO=1$, $OB=2$, and $AB=\sqrt{3}$.

Let $OC=x$. Then $CA=1-x$. since $BC$ bisects $\angle ABO$, we can use the angle bisector theorem:

$\frac{2}{x}=\frac{\sqrt{3}}{1-x}$

$2-2x=\sqrt{3}x$

$2=(\sqrt{3}+2)x$

$x=\frac{2}{\sqrt{3}+2}$.

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:

$\sin\theta =\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\cos\theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}$

$\tan\theta =\frac{\text{Opposite}}{\text{Adjacent}}$.

With a bit of guess and check, we get that the answer is $\boxed{D}$.

Solution 4

Let $OC$ = x, $OB$ = h, and $AB$ = y. $AC$ = $OA$ - $OC$.

Because $OC$ = x, and $OA$ = 1 (given in the problem), $AC$ = 1-x.

Using the Angle Bisector Theorem, $\frac{h}{y}$ = $\frac{x}{1-x}$ $\Longrightarrow$ h(1-x) = xy. Solving for x gives us x = $\frac{h}{h+y}$.

$\sin\theta = \frac{opposite}{hypotenuse} = \frac{y}{h}$. Solving for y gives us y = $h \sin\theta$.

Substituting this for y in our initial equation yields x = $\dfrac{h}{h+h\sin \theta}$.

Using the distributive property, x = $\dfrac{h}{h(1+\sin \theta)}$ and finally $\dfrac{1}{1+\sin \theta}$ or $\boxed{\textbf{(D)}}$

Solution 5

Since $\overline{AB}$ is tangent to the circle, $\angle OAB=90^{\circ}$ and thus we can use trig ratios directly.

\[\sin{\theta}=\frac{\overline{AB}}{\overline{BO}}, \cos{\theta}=\frac{1}{\overline{BO}}, \tan{\theta}=\overline{AB}\]

By the angle bisector theorem, we have

\[\frac{\overline{OB}}{\overline{AB}}=\frac{\overline{OC}}{\overline{CA}}\]

Seeing the resemblance of the ratio on the left-hand side to $\sin{\theta},$ we turn the ratio around to allow us to plug in $\sin{\theta}.$ 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. $\overline{OA}=1$, and flipping the fraction will preserve the $\overline{OC}$, whilst adding one right now would make the equation remain in direct terms of $\overline{CA}.$

\[\frac{\overline{AB}}{\overline{OB}}=\sin{\theta}=\frac{\overline{CA}}{\overline{OC}}\Rightarrow \sin{\theta}+1=\frac{\overline{CA}+\overline{OC}}{\overline{OC}}=\frac{1}{\overline{OC}}\]

\[\sin{\theta}+1=\frac{1}{\overline{OC}} \Rightarrow \boxed{\overline{OC}=\frac{1}{\sin{\theta}+1}}\]

Solution 6 (tangent half angle)

$\angle CBO = 45^{\circ} - \frac{\theta}{2}, \angle ACB = 45^{\circ} + \frac{\theta}{2}, OB = \frac{1}{\cos(\theta)}$. By sine law, $\frac{OC}{\sin(\angle CBO)} = \frac{OB}{\sin(\angle OCB)} = \frac{OB}{\sin(\angle ACB)}$

\[OC = \frac{\sin(45^{\circ} - \frac{\theta}{2})}{\sin(45^{\circ} + \frac{\theta}{2})}OB = \frac{\sin(45^{\circ} - \frac{\theta}{2})}{\cos(45^{\circ} - \frac{\theta}{2})}OB = \tan(45^{\circ} - \frac{\theta}{2})OB = \frac{1-\tan(\theta/2)}{1+\tan(\theta/2)}OB\]

Let $t = \tan(\theta/2)$. $OC = \frac{1-t}{1+t}OB = \frac{1-t^2}{1+2t+t^2}$. Because $\sin(\theta) = \frac{2t}{1+t^2}$ and $\cos(\theta) = \frac{1-t^2}{1+t^2}$, \[OC = \frac{\cos(\theta)}{1+\sin(\theta)}OB = \boxed{\textbf{(D)} \dfrac{1}{1+\sin \theta}}\]

Solution 7 (if you forgot angle bisector but remember LoS)

Let $x=\overline{OC}$, and let $\angle OBC=\angle ABC=\alpha$. We know that $\overline{AC}=\overline{OA}-\overline{OC}=1-x$. By the Law of Sines, \[\dfrac{\sin\alpha}x=\dfrac{\sin\theta}{BC}\]

and

\[\dfrac{\sin\alpha}{1-x}=\dfrac{\sin90^\circ}{BC}=\dfrac1{BC}.\]

Combining the two give $\dfrac{\sin\alpha}x=\sin\theta\cdot\dfrac{\sin\alpha}{1-x}$.

Solving, this gives $\boxed{x=\frac{1}{\sin{\theta}+1}}$.

~Technodoggo

Video Solution

https://youtu.be/zGnRV6eNiV0

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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