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Difference between revisions of "2010 AIME II Problems/Problem 14"
(→‎Solution 3)
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==Solution 3==
 
==Solution 3==
  
−
Let <math>\angle{ACP}</math> be equal to <math>x</math>. Then by Law of Sines, <math>PB = -\frac{\cos{x}}{\cos{3x}}</math> and <math>AP = \frac{\sin{x}}{\sin{3x}}</math>. We then obtain <math>\cos{3x} = 4\cos^3{x} - 3\cos{x}</math> and <math>\sin{3x} = 3\sin{x} - 4\sin^3{x}</math>. Solving, we determine that <math>\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}</math>. Plugging this in gives that <math>\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}</math>. The answer is <math>7</math>.
+
Let <math>\angle{ACP}</math> be equal to <math>x</math>. Then by Law of Sines, <math>PB = -\frac{\cos{x}}{\cos{3x}}</math> and <math>AP = \frac{\sin{x}}{\sin{3x}}</math>. We then obtain <math>\cos{3x} = 4\cos^3{x} - 3\cos{x}</math> and <math>\sin{3x} = 3\sin{x} - 4\sin^3{x}</math>. Solving, we determine that <math>\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}</math>. Plugging this in gives that <math>\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}</math>. The answer is <math>\boxed{007}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:19, 25 February 2017

Problem

Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$.

Solution

Let $O$ be the circumcenter of $ABC$ and let the intersection of $CP$ with the circumcircle be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.

Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the pythagorean theorem, $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$. Now note that $EP = \frac {1}{2}$. By the pythagorean theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$. Hence it now follows that,

\[\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}\]

This gives that the answer is $\boxed{007}$.

An alternate finish for this problem would be to use Power of a Point on $BA$ and $CD$. By Power of a Point Theorem, $CP\cdot PD=1\cdot 2=BP\cdot PA$. Since $BP+PA=4$, we can solve for $BP$ and $PA$, giving the same values and answers as above.

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Solution 2

Let $AC=b$, $BC=a$ by convention. Also, Let $AP=x$ and $BP=y$. Finally, let $\angle ACP=\theta$ and $\angle APC=2\theta$.

We are then looking for $\frac{AP}{BP}=\frac{x}{y}$

Now, by arc interceptions and angle chasing we find that $\triangle BPD \sim \triangle CPA$, and that therefore $BD=yb.$ Then, since $\angle ABD=\theta$ (it intercepts the same arc as $\angle ACD$) and $ADB$ is right,

$\cos\theta=\frac{DB}{AB}=\frac{by}{4}$.


Using law of sines on $APC$, we additionally find that $\frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.$ Simplification by the double angle formula $\sin 2\theta=2\sin \theta\cos\theta$ yields

$\cos \theta=\frac{b}{2x}$.


We equate these expressions for $\cos\theta$ to find that $xy=2$. Since $x+y=AB=4$, we have enough information to solve for $x$ and $y$. We obtain $x,y=2 \pm \sqrt{2}$

Since we know $x>y$, we use $\frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$

Solution 3

Let $\angle{ACP}$ be equal to $x$. Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$. We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$. Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$. Plugging this in gives that $\frac{AP}{PB} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = 3 + 2\sqrt{2}$. The answer is $\boxed{007}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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