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

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.

$[asy] /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */ pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2); /* segments and figures */ draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35)); /* points and labels */ label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94); } draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){ return (0.28*cos(t)+0.59,0.28*sin(t)+0); } draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){ return (0.28*cos(t)+2,0.28*sin(t)+0); } draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle); [/asy]$

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

@@ Line 1: / Line 1: @@
-== Problem 14 ==
+== Problem ==
-Triangle <math>ABC</math> with right angle at <math>C</math>, <math>\angle BAC < 45^\circ</math> and <math>AB = 4</math>. Point <math>P</math> on <math>\overbar{AB}</math> is chosen such that <math>\angle APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> can be represented in the form <math>p + q\sqrt{r}</math>, where <math>p</math>, <math>q</math>, <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>.
+[[Triangle]] <math>ABC</math> with [[right angle]] at <math>C</math>, <math>\angle BAC < 45^\circ</math> and <math>AB = 4</math>. Point <math>P</math> on <math>\overline{AB}</math> is chosen such that <math>\angle APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> can be represented in the form <math>p + q\sqrt{r}</math>, where <math>p</math>, <math>q</math>, <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>.
-== Solution ==
+== Solution 1==
+Let <math>O</math> be the [[circumcenter]] of <math>ABC</math> and let the intersection of <math>CP</math> with the [[circumcircle]] be <math>D</math>. It now follows that <math>\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>.
-Label the center of the circumcircle of <math>ABC</math> as <math>O</math> and the intersection of <math>CP</math> with the circumcircle as <math>D</math>. It now follows that <math>\angle{DOA} = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>.
+Denote <math>E</math> the projection of <math>O</math> onto <math>CD</math>. Now <math>CD = CP + DP = 3</math>. By the [[Pythagorean Theorem]], <math>OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}</math>. Now note that <math>EP = \frac {1}{2}</math>. By the Pythagorean Theorem, <math>OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}</math>. Hence it now follows that,
-Denote <math>E</math> the projection of <math>O</math> onto <math>CD</math>. Now <math>CD = CP + DP = 3</math>. By the pythagorean theorem, <math>OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}</math>. Now note that <math>EP = \frac {1}{2}</math>. By the pythagorean theorem, <math>OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}</math>. Hence it now follows that,
+<cmath>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</cmath>
-<math>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</math>
+This gives that the answer is <math>\boxed{007}</math>.
+An alternate finish for this problem would be to use Power of a Point on <math>BA</math> and <math>CD</math>. By Power of a Point Theorem, <math>CP\cdot PD=1\cdot 2=BP\cdot PA</math>. Since <math>BP+PA=4</math>, we can solve for <math>BP</math> and <math>PA</math>, giving the same values and answers as above.
+<center><asy>   /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */
+import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */
+pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2);
+  /* segments and figures */
+draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35));
+  /* points and labels */
+label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){
+ return (0.28*cos(t)+0.23,0.28*sin(t)+0.94);
+}
+draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){
+ return (0.28*cos(t)+0.59,0.28*sin(t)+0);
+}
+draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){
+ return (0.28*cos(t)+2,0.28*sin(t)+0);
+}
+draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle);
+</asy></center>
+==Solution 2==
+Let <math>AC=b</math>, <math>BC=a</math> by convention. Also, Let <math>AP=x</math> and <math>BP=y</math>.  Finally, let <math> \angle ACP=\theta</math> and <math> \angle APC=2\theta</math>.
+We are then looking for <math> \frac{AP}{BP}=\frac{x}{y}</math>
+Now, by arc interceptions and angle chasing we find that <math> \triangle BPD \sim \triangle CPA</math>, and that therefore <math> BD=yb.</math>  Then, since <math> \angle ABD=\theta</math> (it intercepts the same arc as <math> \angle ACD</math>) and <math> ADB</math> is right,
+<math> \cos\theta=\frac{DB}{AB}=\frac{by}{4}</math>.
+Using law of sines on <math>APC</math>, we additionally find that <math> \frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.</math> Simplification by the double angle formula <math> \sin 2\theta=2\sin \theta\cos\theta</math> yields
+<math> \cos \theta=\frac{b}{2x}</math>.
+We equate these expressions for <math> \cos\theta</math> to find that <math> xy=2</math>. Since <math> x+y=AB=4</math>, we have enough information to solve for <math>x</math> and <math>y</math>.  We obtain <math> x,y=2 \pm \sqrt{2}</math>
+Since we know <math>x>y</math>, we use <math> \frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}</math>
+==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>\boxed{007}</math>.
+(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)==
+Let <math>\alpha=\angle{ACP}</math>, <math>\beta=\angle{ABC}</math>, and <math>x=BP</math>. By Law of Sines,
+<math>\frac{1}{\sin(\beta)}=\frac{x}{\sin(90-\alpha)}\implies \sin(\beta)=\frac{\cos(\alpha)}{x}</math> (1), and
+<math>\frac{4-x}{\sin(\alpha)}=\frac{4\sin(\beta)}{\sin(2\alpha)} \implies 4-x=\frac{2\sin(\beta)}{\cos(\alpha)}</math>. (2)
+Then, substituting (1) into (2), we get
+<math>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}</math>
-This gives that the answer is <math>\boxed{007}</math>.
+The answer is <math>\boxed{007}</math>.
+~Rowechen
+==Solution 5==
+Let <math>\angle{ACP}=x</math>. Then, <math>\angle{APC}=2x</math> and <math>\angle{A}=180-3x</math>. Let the foot of the angle bisector of <math>\angle{APC}</math> on side <math>AC</math> be <math>D</math>. Then,
+<math>CD=DP</math> and <math>\triangle{DAP}\sim{\triangle{APC}}</math> due to the angles of these triangles.
+Let <math>CD=a</math>. By the Angle Bisector Theorem, <math>\frac{1}{a}=\frac{AP}{AD}</math>, so <math>AD=a\cdot{AP}</math>. Moreover, since <math>CD=DP=a</math>, by similar triangle ratios, <math>\frac{AP}{a+a\cdot{AP}}=a</math>. Therefore, <math>AP = \frac{a^2}{1-a^2}</math>.
+Construct the perpendicular from <math>D</math> to <math>AP</math> and denote it as <math>F</math>. Denote the midpoint of <math>CP</math> as <math>M</math>. Since <math>PD</math> is an angle bisector, <math>PF</math> is congruent to <math>PM</math>, so <math>PF=\frac{1}{2}</math>.
+Also, <math>\triangle{DFA}\sim{\triangle{BCA}}</math>. Thus, <math>\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}</math>. After some major cancellation, we have <math>7a^4-8a^2+2=0</math>, which is a quadratic in <math>a^2</math>. Thus, <math>a^2 = \frac{4\pm\sqrt{2}}{7}</math>.
+Taking the negative root implies <math>AP<BP</math>, contradiction. Thus, we take the positive root to find that <math>AP=2+\sqrt{2}</math>. Thus, <math>BP=2-\sqrt{2}</math>, and our desired ratio is <math>\frac{2+\sqrt{2}}{2-\sqrt{2}}\implies{3+2\sqrt{2}}</math>.
+The answer is <math>\boxed{007}</math>.
+==Solution 6==
+Let <math>O</math> be the circumcenter of <math>\triangle ABC</math>. Since <math>\triangle ABC</math> is a right triangle, <math>O</math> will be on <math>\overline{AB}</math> and <math>\overline{AO} \cong \overline{OB} \cong \overline{OC} = 2</math>. Let <math>\overline{OP} = x</math>.
+Next, let's do some angle chasing. Label <math>\angle ACP = \theta^{\circ}</math>, and <math>\angle APC = 2\theta^{\circ}</math>. Thus, <math>\angle PAC = (180-3\theta)^{\circ}</math>, and by isosceles triangles,  <math>\angle ACO = (180-3\theta)^{\circ}</math>. Then, by angle subtraction, <math>\angle OCP = (\theta - (180-3\theta))^{\circ} = (4\theta - 180)^{\circ}</math>.
+Using the Law of Sines: <cmath>\frac{x}{\sin (4\theta-180)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}</cmath>Using trigonometric identies, <math>\sin (4\theta-180)^{\circ}=-\sin (4\theta)^{\circ}=-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}</math>. Plugging this back into the Law of Sines formula gives us: <cmath>\frac{x}{-2\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}}=\frac{2}{\sin (2\theta)^{\circ}}</cmath>
+<cmath>-4\sin (2\theta)^{\circ}\cos (2\theta)^{\circ}=x\sin (2\theta)^{\circ}</cmath>
+<cmath>-4\cos (2\theta)^{\circ}=x</cmath>
+<cmath>\cos(2\theta)^{\circ}=\frac{-x}4</cmath>
+Next, using the Law of Cosines: <cmath>2^2=1^2+x^2-2\cdot 1\cdot x\cdot \cos (2\theta)^{\circ}</cmath>
+Substituting <math>\cos(2\theta)^{\circ}=\frac{-x}4</math> gives us:
+<cmath>2^2=1^2+x^2-2\cdot 1\cdot x\cdot \frac{-x}4</cmath>
+<cmath>4=1+x^2+\frac{x^2}{2}</cmath>
+Solving for x gives <math>x=\sqrt 2</math>
+Finally: <math>\frac{\overline{AP}}{\overline{BP}}=\frac{\overline{AO}+\overline{OP}}{\overline{BO}-\overline{OP}}=\frac{2+\sqrt 2}{2-\sqrt 2}=3+2\sqrt2</math>, which gives us an answer of <math>3+2+2=\boxed{007}</math>. ~adyj
 == See also ==
-{{AIME box|year=2010|num-b=14|num-a=15|n=I}}
+{{AIME box|year=2010|num-b=13|num-a=15|n=II}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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