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Difference between revisions of "2010 AIME II Problems/Problem 14"

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(You can derive that <math>\cos{3x} = 4\cos^3{x} - 3\cos{x},</math> and similarly for <math>\sin{3x},</math> by considering the expansion of <math>(\text{cis}(x))^3,</math> equating real parts to <math>\cos{x}</math> and imaginary parts to <math>\sin{x},</math> then substituting with <math>1-\sin^2{x}</math> to finish.)
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(You can derive that <math>\cos{3x} = 4\cos^3{x} - 3\cos{x},</math> and similarly for <math>\sin{3x},</math> by considering the expansion of <math>(\text{cis}(x))^3,</math> equating real parts to <math>\cos{x}</math> and imaginary parts to <math>\sin{x},</math> then substituting with <math>1-\sin^2{x}</math> to finish. ~happypi31415)
  
 
==Solution 4 (The quickest and most elegant)==
 
==Solution 4 (The quickest and most elegant)==

Latest revision as of 23:31, 18 November 2023

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 1

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 $\boxed{007}$.


(You can derive that $\cos{3x} = 4\cos^3{x} - 3\cos{x},$ and similarly for $\sin{3x},$ by considering the expansion of $(\text{cis}(x))^3,$ equating real parts to $\cos{x}$ and imaginary parts to $\sin{x},$ then substituting with $1-\sin^2{x}$ to finish. ~happypi31415)

Solution 4 (The quickest and most elegant)

Let $\alpha=\angle{ACP}$, $\beta=\angle{ABC}$, and $x=BP$. By Law of Sines,

$\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}$ (1), and

$\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}$. (2)

Then, substituting (1) into (2), we get

$4-x=\frac{2}{x} \implies x^2-4x+2=0 \implies x=2-\sqrt{2} \implies \frac{4-x}{x}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}$

The answer is $\boxed{007}$. ~Rowechen

Solution 5

Let $\angle{ACP}=x$. Then, $\angle{APC}=2x$ and $\angle{A}=180-3x$. Let the foot of the angle bisector of $\angle{APC}$ on side $AC$ be $D$. Then,

$CD=DP$ and $\triangle{DAP}\sim{\triangle{APC}}$ due to the angles of these triangles.

Let $CD=a$. By the Angle Bisector Theorem, $\frac{1}{a}=\frac{AP}{AD}$, so $AD=a\cdot{AP}$. Moreover, since $CD=DP=a$, by similar triangle ratios, $\frac{AP}{a+a\cdot{AP}}=a$. Therefore, $AP = \frac{a^2}{1-a^2}$.

Construct the perpendicular from $D$ to $AP$ and denote it as $F$. Denote the midpoint of $CP$ as $M$. Since $PD$ is an angle bisector, $PF$ is congruent to $PM$, so $PF=\frac{1}{2}$.

Also, $\triangle{DFA}\sim{\triangle{BCA}}$. Thus, $\frac{FA}{AC}=\frac{AD}{AB}\Longrightarrow\frac{\frac{a^2}{1-a^2}-\frac{1}{2}}{a+\frac{a^3}{1-a^2}}=\frac{\frac{a^3}{1-a^3}}{4}$. After some major cancellation, we have $7a^4-8a^2+2=0$, which is a quadratic in $a^2$. Thus, $a^2 = \frac{4\pm\sqrt{2}}{7}$.

Taking the negative root implies $AP<BP$, contradiction. Thus, we take the positive root to find that $AP=2+\sqrt{2}$. Thus, $BP=2-\sqrt{2}$, and our desired ratio is $\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies{3+2\sqrt{2}}$.

The answer is $\boxed{007}$.

Solution 6

Let $O$ be the circumcenter of $\triangle ABC$. Since $\triangle ABC$ is a right triangle, $O$ will be on $\overline{AB}$ and $\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2$. Let $\overline{OP} = x$.

Next, let's do some angle chasing. Label $\angle ACP = \theta^{\circ}$, and $\angle APC = 2\theta^{\circ}$. Thus, $\angle PAC = (180-3\theta)^{\circ}$, and by isosceles triangles, $\angle ACO = (180-3\theta)^{\circ}$. Then, by angle subtraction, $\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}$.

Using the Law of Sines: \[\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\]Using trigonometric identies, $\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}$. Plugging this back into the Law of Sines formula gives us: \[\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}\]

\[-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}\] \[-4\cos (2\theta)^{\circ}=x\] \[\cos(2\theta)^{\circ}=\frac{-x}4\]

Next, using the Law of Cosines: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}\] Substituting $\cos(2\theta)^{\circ}=\frac{-x}4$ gives us: \[2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4\] \[4=1+x^2+\frac{x^2}{2}\]

Solving for x gives $x=\sqrt 2$

Finally: $\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2$, which gives us an answer of $3+2+2=\boxed{007}$. ~adyj

See also

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

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