Difference between revisions of "2024 AIME II Problems/Problem 10"

Latest revision as of 16:09, 13 October 2024

Problem

Let $\triangle ABC$ have circumcenter $O$ and incenter $I$ with $\overline{IA}\perp\overline{OI}$ , circumradius $13$ , and inradius $6$ . Find $AB\cdot AC$ .

Solution 1 (Similar Triangles and PoP)

Start off by (of course) drawing a diagram! Let $I$ and $O$ be the incenter and circumcenters of triangle $ABC$ , respectively. Furthermore, extend $AI$ to meet $BC$ at $L$ and the circumcircle of triangle $ABC$ at $D$ .

$[asy] size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26); pair A = (c/3,8.65*c/10); draw(circumcircle(A,B,C)); pair I=incenter(A,B,C); pair O=circumcenter(A,B,C); pair L=extension(A,I,C,B); dot(I^^O^^A^^B^^C^^D^^L); draw(A--L); draw(A--D); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(C--B--D--cycle); draw(A--C--B); draw(A--B); draw(B--I--C^^A--I); draw(incircle(A,B,C)); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); label("$D$",D,S); label("$I$",I,NW); label("$L$",L,SW); label("$O$",O,E); label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); draw(I--O); draw(A--O); draw(rightanglemark(A,I,O)); [/asy]$

We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations.

Solution 1.1

Since $I$ is the incenter, $\angle BAL \cong \angle DAC$ . Furthermore, $\angle ABC$ and $\angle ADC$ are both subtended by the same arc $AC$ , so $\angle ABC \cong \angle ADC.$ Therefore by AA similarity, $\triangle ABL \sim \triangle ADC$ . From this we can say that $\frac{AB}{AD} = \frac{AL}{AC} \implies AB \cdot AC = AL \cdot AD$

Since $AD$ is a chord of the circle and $OI$ is a perpendicular from the center to that chord, $OI$ must bisect $AD$ . This can be seen by drawing $OD$ and recognizing that this creates two congruent right triangles. Therefore, $AD = 2 \cdot ID \implies AB \cdot AC = 2 \cdot AL \cdot ID$

We have successfully represented $AB \cdot AC$ in terms of $AL$ and $ID$ . Solution 1.2 will explain an alternate method to get a similar relationship, and then we'll rejoin and finish off the solution.

Solution 1.2

$\angle ALB \cong \angle DLC$ by vertical angles and $\angle LBA \cong \angle CDA$ because both are subtended by arc $AC$ . Thus $\triangle ABL \sim \triangle CDL$ .

Thus $\frac{AB}{CD} = \frac{AL}{CL} \implies AB = CD \cdot \frac{AL}{CL}$

Symmetrically, we get $\triangle ALC \sim \triangle BLD$ , so $\frac{AC}{BD} = \frac{AL}{BL} \implies AC = BD \cdot \frac{AL}{BL}$

Substituting, we get $AB \cdot AC = CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL}$

Lemma 1: BD = CD = ID

Proof:

We commence angle chasing: we know $\angle DBC \cong DAC = \gamma$ . Therefore $\angle IBD = \alpha + \gamma$ . Looking at triangle $ABI$ , we see that $\angle IBA = \alpha$ , and $\angle BAI = \gamma$ . Therefore because the sum of the angles must be $180$ , $\angle BIA = 180-\alpha - \gamma$ . Now $AD$ is a straight line, so $\angle BID = 180-\angle BIA = \alpha+\gamma$ . Since $\angle IBD = \angle BID$ , triangle $IBD$ is isosceles and thus $ID = BD$ .

A similar argument should suffice to show $CD = ID$ by symmetry, so thus $ID = BD = CD$ .

Now we regroup and get $CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL} = ID^2 \cdot \frac{AL^2}{BL \cdot CL}$

Now note that $BL$ and $CL$ are part of the same chord in the circle, so we can use Power of a point to express their product differently. $BL \cdot CL = AL \cdot LD \implies AB \cdot AC = ID^2 \cdot \frac{AL}{LD}$

Solution 1 (Continued)

Now we have some sort of expression for $AB \cdot AC$ in terms of $ID$ and $AL$ . Let's try to find $AL$ first.

Drop an altitude from $D$ to $BC$ , $I$ to $AC$ , and $I$ to $BC$ :

$[asy] size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); pair A = (c/3,8.65*c/10); pair F = (2*c/3-0.14, 4-0.29); pair G = (c/2-0.68,0); draw(circumcircle(A,B,C)); pair I=incenter(A,B,C); pair O=circumcenter(A,B,C); pair L=extension(A,I,C,B); dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G); draw(A--L); draw(A--D); draw(D--E); draw(I--F); draw(I--G); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(C--B--D--cycle); draw(A--C--B); draw(A--B); draw(B--I--C^^A--I); draw(incircle(A,B,C)); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); label("$D$",D,S); label("$I$",I,NW); label("$L$",L,SW); label("$O$",O,E); label("$E$",E,N); label("$F$",F,NE); label("$G$",G,SW); label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); draw(I--O); draw(A--O); draw(rightanglemark(A,I,O)); draw(rightanglemark(B,E,D)); draw(rightanglemark(I,F,A)); draw(rightanglemark(I,G,L)); [/asy]$

Since $\angle DBE \cong \angle IAF$ and $\angle BED \cong \angle IFA$ , $\triangle BDE \sim \triangle AIF$ .

Furthermore, we know $BD = ID$ and $AI = ID$ , so $BD = AI$ . Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that $DE = IF = 6$ since $IF$ is the inradius.

Now notice that $\triangle IGL \sim \triangle DEL$ because of equal vertical angles and right angles. Furthermore, $IG$ is the inradius so it's length is $6$ , which equals the length of $DE$ . Therefore these two triangles are congruent, so $IL = DL$ .

Since $IL+DL = ID$ , $ID = 2 \cdot IL$ . Furthermore, $AL = AI + IL = ID + IL = 3 \cdot IL$ .

We can now plug back into our initial equations for $AB \cdot AC$ :

From $1.1$ , $AB \cdot AC = 2 \cdot AL \cdot ID = 2 \cdot 3 \cdot IL \cdot 2 \cdot IL$

$\implies AB \cdot AC = 3 \cdot (2 \cdot IL) \cdot (2 \cdot IL) = 3 \cdot ID^2$

Alternatively, from $1.2$ , $AB \cdot AC = ID^2 \cdot \frac{AL}{DL}$ $\implies AB \cdot AC = ID^2 \cdot \frac{3 \cdot IL}{IL} = 3 \cdot ID^2$

Now all we need to do is find $ID$ .

The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that $OI^2 = R(R-2r)$ , where $R$ is the circumradius and $r$ is the inradius. We will prove this formula first, but if you already know the proof, skip this part.

Theorem: in any triangle, let $d$ be the distance from the circumcenter to the incenter of the triangle. Then $d^2 = R \cdot (R-2r)$ , where $R$ is the circumradius of the triangle and $r$ is the inradius of the triangle.

Proof:

Construct the following diagram:

$[asy] size(300); import olympiad; real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); pair A = (c/3,8.65*c/10); pair F = (2*c/3-0.14, 4-0.29); pair G = (c/2-0.68,0); draw(circumcircle(A,B,C)); pair I=incenter(A,B,C); pair O=circumcenter(A,B,C); pair L=extension(A,I,C,B); dot(I^^O^^A^^B^^C^^D^^L^^F); draw(A--L); draw(A--D); draw(I--F); path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} draw(C--B--D--cycle); draw(A--C--B); draw(A--B); draw(A--I); draw(incircle(A,B,C)); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); label("$D$",D,S); label("$I$",I,NW); label("$L$",L,SW); label("$O$",O,S); label("$F$",F,NE); label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); pair H = (10*c/8-1.46,2*c/3-1.85), J = (-0.55,1.4); dot(H^^J); label("$H$", H, E); label("$J$", J, W); draw(I--O); draw(I--H); draw(I--J); draw(rightanglemark(I,F,A)); [/asy]$

Let $OI = d$ , $OH = R$ , $IF = r$ . By the Power of a Point, $IH \cdot IJ = AI \cdot ID$ . $IH = R+d$ and $IJ = R-d$ , so $(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD$

Now consider $\triangle ACD$ . Since all three points lie on the circumcircle of $\triangle ABC$ , the two triangles have the same circumcircle. Thus we can apply law of sines and we get $\frac{CD}{\sin(\angle DAC)} = 2R$ . This implies

$(R+d)\cdot (R-d) = AI \cdot 2R \cdot \sin(\angle DAC)$

Also, $\sin(\angle DAC)) = \sin(\angle IAF))$ , and $\triangle IAF$ is right. Therefore $\sin(\angle IAF) = \frac{IF}{AI} = \frac{r}{AI}$

Plugging in, we have

$(R+d)\cdot (R-d) = AI \cdot 2R \cdot \frac{r}{AI} = 2R \cdot r$

Thus $R^2-d^2 = 2R \cdot r \implies d^2 = R \cdot (R-2r)$

Now we can finish up our solution. We know that $AB \cdot AC = 3 \cdot ID^2$ . Since $ID = AI$ , $AB \cdot AC = 3 \cdot AI^2$ . Since $\triangle AOI$ is right, we can apply the pythagorean theorem: $AI^2 = AO^2-OI^2 = 13^2-OI^2$ .

Plugging in from Euler's formula, $OI^2 = 13 \cdot (13 - 2 \cdot 6) = 13$ .

Thus $AI^2 = 169-13 = 156$ .

Finally $AB \cdot AC = 3 \cdot AI^2 = 3 \cdot 156 = \textbf{468}$ .

~KingRavi

Solution 2 (Excenters)

By Euler's formula $OI^{2}=R(R-2r)$ , we have $OI^{2}=13(13-12)=13$ . Thus, by the Pythagorean theorem, $AI^{2}=13^{2}-13=156$ . Let $AI\cap(ABC)=M$ ; notice $\triangle AOM$ is isosceles and $\overline{OI}\perp\overline{AM}$ which is enough to imply that $I$ is the midpoint of $\overline{AM}$ , and $M$ itself is the midpoint of $II_{a}$ where $I_{a}$ is the $A$ -excenter of $\triangle ABC$ . Therefore, $AI=IM=MI_{a}=\sqrt{156}$ and $AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.$

Note that this problem is extremely similar to 2019 CIME I/14.

Solution 3

Denote $AB=a, AC=b, BC=c$ . By the given condition, $\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6$ , where $A$ is the area of $\triangle{ABC}$ .

Moreover, since $OI\bot AI$ , the second intersection of the line $AI$ and $(ABC)$ is the reflection of $A$ about $I$ , denote that as $D$ . By the incenter-excenter lemma with Ptolemy's Theorem, $DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c$ .

Thus, we have $\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c$ . Now, we have $\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}$

~Bluesoul

Solution 4 (Trig)

Denote by $R$ and $r$ the circumradius and inradius, respectively.

First, we have $r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1)$

Second, because $AI \perp IO$ , \begin{align*} AI & = AO \cos \angle IAO \\ & = AO \cos \left( 90^\circ - C - \frac{A}{2} \right) \\ & = AO \sin \left( C + \frac{A}{2} \right) \\ & = R \sin \left( C + \frac{180^\circ - B - C}{2} \right) \\ & = R \cos \frac{B - C}{2} . \end{align*}

Thus, \begin{align*} r & = AI \sin \frac{A}{2} \\ & = R \sin \frac{A}{2} \cos \frac{B-C}{2} \hspace{1cm} (2) \end{align*}

Taking $(1) - (2)$ , we get \[ 4 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B-C}{2} . \]

We have \begin{align*} 2 \sin \frac{B}{2} \sin \frac{C}{2} & = - \cos \frac{B+C}{2} + \cos \frac{B-C}{2} . \end{align*}

Plugging this into the above equation, we get \[ \cos \frac{B-C}{2} = 2 \cos \frac{B+C}{2} . \hspace{1cm} (3) \]

Now, we analyze Equation (2). We have \begin{align*} \frac{r}{R} & = \sin \frac{A}{2} \cos \frac{B-C}{2} \\ & = \sin \frac{180^\circ - B - C}{2} \cos \frac{B-C}{2} \\ & = \cos \frac{B+C}{2} \cos \frac{B-C}{2} \hspace{1cm} (4) \end{align*}

Solving Equations (3) and (4), we get \[ \cos \frac{B+C}{2} = \sqrt{\frac{r}{2R}}, \hspace{1cm} \cos \frac{B-C}{2} = \sqrt{\frac{2r}{R}} . \hspace{1cm} (5) \]

Now, we compute $AB \cdot AC$ . We have \begin{align*} AB \cdot AC & = 2R \sin C \cdot 2R \sin B \\ & = 2 R^2 \left( - \cos \left( B + C \right) + \cos \left( B - C \right) \right) \\ & = 2 R^2 \left( - \left( 2 \left( \cos \frac{B+C}{2} \right)^2 - 1 \right) + \left( 2 \left( \cos \frac{B-C}{2} \right)^2 - 1 \right) \right) \\ & = 6 R r \\ & = \boxed{\textbf{(468) }} \end{align*} where the first equality follows from the law of sines, the fourth equality follows from (5).

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 5 (Trig)

Firstly, we can construct the triangle $\triangle ABC$ by drawing the circumcirlce (centered at $O$ with radius $R = OA = 13$ ) and incircle (centered at $I$ with radius $r = 6$ ). Next, from $A$ , construct tangent lines to the incircle meeting the circumcirlce at point $B$ and $C$ , say, as shown in the diagram. By Euler's theorem (relating the distance between $O$ and $I$ to the circumradius and inradius), we have $OI = \sqrt{R^2 - 2rR} = \sqrt{13}.$ This leads to $AI = \sqrt{R^2 - OI^2} = \sqrt{156}.$ Let $P$ be the point of tangency where the incircle meets the side $\overline{AC}$ . Now we denote $\theta \coloneqq \angle BAI = \angle IAP \qquad \text{and} \qquad \phi \coloneqq \angle OAI.$ Notice that $\angle BAO = \angle BAI - \angle OAI = \theta - \phi$ . Finally, the crux move is to recognize $AB = 2R \cos(\theta - \phi) \qquad \text{and} \qquad AC = 2R \cos(\theta + \phi)$ since $O$ is the circumcenter. Then multiply these two expressions and apply the compound-angle formula to get \begin{aligned} AB \cdot AC &= 4R^2 \cos(\theta - \phi) \cos(\theta + \phi) \\[0.3em] &= 4R^2\left( \cos^2\theta \cos^2\phi - \sin^2\theta \sin^2\phi \right) \\[0.3em] &= 4\cos^2\theta(\underbrace{R\cos\phi}_{AI \, = \, \sqrt{156}})^2 - 4\sin^2\theta(\underbrace{R\sin\phi}_{OI \, = \, \sqrt{13}})^2 \\[0.3em] &= 52 (12\cos^2\theta - \sin^2 \theta) \\[0.3em] AB \cdot AC &= 52 (12 - 13\sin^2\theta), \end{aligned} where in the last equality, we make use of the substitution $\cos^2\theta = 1 - \sin^2\theta$ . Looking at $\triangle AIP$ , we learn that $\sin \theta = \frac{r}{AI} = \frac{6}{\sqrt{156}}$ which means $\sin^2 \theta = \frac{3}{13}$ . Hence we have $AB \cdot AC = 52\left( 12 - 13 \cdot \tfrac{3}{13} \right) = 52 \cdot 9 = \boxed{468}.$ This completes the solution

-- VensL.

Solution 6 (Close to Solution 3)

Denote $E = \odot ABC \cap AI, AB = c, AC = b, BC=a, r$ is inradius. $AO = EO = R \implies AI = EI.$ It is known that $\frac {AI}{EI} = \frac {b+c}{a} – 1 = 1 \implies b + c = 2a.$

Points on bisectors

$[ABC] =\frac{ (a+b+c) r}{2} = \frac {3ar}{2} = \frac {abc}{4R} \implies bc = 6Rr = \boxed{468}.$ vladimir.shelomovskii@gmail.com, vvsss

Solution 7

Call side $BC = a$ , and similarly label the other sides. Note that ${OI}^2 = R^2 - 2Rr$ . Also note that $AO = R$ , so by the right angle, $AI = \sqrt{2Rr}$ . However, we can double Angle Bisector theorem. The length of the angle bisector from A is $\sqrt{(bc)(1 - \frac{a^2}{(b+c)^2})}$ . As a direct result, the length AI simplifies down to $\frac{\sqrt{(bc)(b+c-a)}}{\sqrt{{a+b+c}}}$ .

Draw the incircle and call the tangent to side AB F. Then, $AF = \frac{b+c-a}{2}$ . But this length, by Pythagorean, is $\sqrt{120}$ , so $b+c-a = 2\sqrt{120}$ .

Also note that the area of the triangle is $[ABC] = \frac{abc}{52}$ , by $\frac{abc}{4S} = R$ . By the incircle, we know that $\sin{\frac{A}{2}} = \frac{6}{\sqrt{156}}$ , and similarly, $\cos{\frac{A}{2}} = \frac{\sqrt{120}}{\sqrt{156}}$ . By double-angle, $\sin A = \frac{\sqrt{120}}{13}$ . But the area of the triangle $[ABC]$ is simply $\frac{1}{2}bc \sin A$ , which is also $2\sqrt{120}bc$ . But we know this is $abc$ from above, so $a = 2\sqrt{120}$ . As a direct result, $a+b+c = 6\sqrt{120}$ .

Apply this to the formula $\frac{\sqrt{(bc)(b+c-a)}}{\sqrt{a+b+c}}$ listed above to get $2Rr = 156 = \frac{bc}{3}$ , so $bc = 468$ . We're done. - sepehr2010

Solution 8

Let the intersection of the $A$ -angle bisector and the circumcircle be $M$ , and denote the $A$ -excenter as $I_A$ . Denote the tangent to the incircle from $AC$ as $E$ and the tangent to the excircle from $AC$ as $E_A$ .

Notice that our perpendicular condition implies $AI = IM$ , and Incenter-Excenter gives $IM = MI_A$ . Thus we have $AI_A = 3AI$ . From similar triangles we get $3(s-a) = 3AE = AE_A = s$ . This implies $a = \frac23 S$ .

Using areas we have that $\frac{abc}{4R} = rs$ . Substituting gives $\frac{sbc}{6R} = rs \implies bc = 6Rr = \boxed{468}$ and we're done. - thoom

Video Solution

https://youtu.be/_zxBvojcAQ4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://www.youtube.com/watch?v=pPBPfpo12j4

~MathProblemSolvingSkills.com

2024 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2024 AIME II Problems/Problem 10"

Latest revision as of 16:09, 13 October 2024

Contents

Problem

Solution 1 (Similar Triangles and PoP)

Solution 1.1

Solution 1.2

Solution 1 (Continued)

Solution 2 (Excenters)

Solution 3

Solution 4 (Trig)

Solution 5 (Trig)

Solution 6 (Close to Solution 3)

Solution 7

Solution 8

Video Solution

Video Solution

See also

@@ Line 1: / Line 1: @@
 ==Problem==
 Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math> with <math>\overline{IA}\perp\overline{OI}</math>, circumradius <math>13</math>, and inradius <math>6</math>. Find <math>AB\cdot AC</math>.
+==Solution 1 (Similar Triangles and PoP)==
+Start off by (of course) drawing a diagram! Let <math>I</math> and <math>O</math> be the incenter and circumcenters of triangle <math>ABC</math>, respectively. Furthermore, extend <math>AI</math> to meet <math>BC</math> at <math>L</math> and the circumcircle of triangle <math>ABC</math> at <math>D</math>.
 <asy>
@@ Line 7: / Line 11: @@
 real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
 pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26);
-pair A = (c/3,8.7*c/10);
+pair A = (c/3,8.65*c/10);
 draw(circumcircle(A,B,C));
 pair I=incenter(A,B,C);
@@ Line 25: / Line 29: @@
 label("$A$",A,N);
 label("$D$",D,S);
-label("$I$",I,NE);
+label("$I$",I,NW);
 label("$L$",L,SW);
 label("$O$",O,E);
@@ Line 36: / Line 40: @@
 draw(I--O);
+draw(A--O);
 draw(rightanglemark(A,I,O));
 </asy>
-Solution in Progress
+We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations.
+==Solution 1.1==
+Since <math>I</math> is the incenter, <math>\angle BAL \cong \angle DAC</math>. Furthermore, <math>\angle ABC</math> and <math>\angle ADC</math> are both subtended by the same arc <math>AC</math>, so <math>\angle ABC \cong \angle ADC.</math> Therefore by AA similarity, <math>\triangle ABL \sim \triangle ADC</math>.
+From this we can say that <cmath>\frac{AB}{AD} = \frac{AL}{AC} \implies AB \cdot AC = AL \cdot AD </cmath>
+Since <math>AD</math> is a chord of the circle and <math>OI</math> is a perpendicular from the center to that chord, <math>OI</math> must bisect <math>AD</math>. This can be seen by drawing <math>OD</math> and recognizing that this creates two congruent right triangles. Therefore, <cmath>AD = 2 \cdot ID \implies AB \cdot AC = 2 \cdot AL \cdot ID</cmath>
+We have successfully represented <math>AB \cdot AC</math> in terms of <math>AL</math> and <math>ID</math>. Solution 1.2 will explain an alternate method to get a similar relationship, and then we'll rejoin and finish off the solution.
+==Solution 1.2==
+<math>\angle ALB \cong \angle DLC</math> by vertical angles and <math>\angle LBA \cong \angle CDA</math> because both are subtended by arc <math>AC</math>. Thus <math>\triangle ABL \sim \triangle CDL</math>.
+Thus <cmath>\frac{AB}{CD} = \frac{AL}{CL} \implies AB = CD \cdot \frac{AL}{CL}</cmath>
+Symmetrically, we get <math>\triangle ALC \sim \triangle BLD</math>, so
+<cmath>\frac{AC}{BD} = \frac{AL}{BL} \implies AC = BD \cdot \frac{AL}{BL}</cmath>
+Substituting, we  get <cmath>AB \cdot AC = CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL}</cmath>
+Lemma 1: BD = CD = ID
+Proof:
+We commence angle chasing: we know <math>\angle DBC \cong DAC = \gamma</math>. Therefore <cmath>\angle IBD = \alpha + \gamma</cmath>.
+Looking at triangle <math>ABI</math>, we see that <math>\angle IBA = \alpha</math>, and <math>\angle BAI = \gamma</math>. Therefore because the sum of the angles must be <math>180</math>, <math>\angle BIA = 180-\alpha - \gamma</math>. Now <math>AD</math> is a straight line, so <cmath>\angle BID = 180-\angle BIA = \alpha+\gamma</cmath>.
+Since <math>\angle IBD = \angle BID</math>, triangle <math>IBD</math> is isosceles and thus <math>ID = BD</math>.
+A similar argument should suffice to show <math>CD = ID</math> by symmetry, so thus <math>ID = BD = CD</math>.
+Now we regroup and get <cmath>CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL} = ID^2 \cdot \frac{AL^2}{BL \cdot CL}</cmath>
+Now note that <math>BL</math> and <math>CL</math> are part of the same chord in the circle, so we can use Power of a point to express their product differently. <cmath>BL \cdot CL = AL \cdot LD \implies AB \cdot AC = ID^2 \cdot \frac{AL}{LD}</cmath>
+==Solution 1 (Continued)==
+Now we have some sort of expression for <math>AB \cdot AC</math> in terms of <math>ID</math> and <math>AL</math>. Let's try to find <math>AL</math> first.
+Drop an altitude from <math>D</math> to <math>BC</math>, <math>I</math> to <math>AC</math>, and <math>I</math> to <math>BC</math>:
+<asy>
+size(300);
+import olympiad;
+real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
+pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0);
+pair A = (c/3,8.65*c/10);
+pair F = (2*c/3-0.14, 4-0.29);
+pair G = (c/2-0.68,0);
+draw(circumcircle(A,B,C));
+pair I=incenter(A,B,C);
+pair O=circumcenter(A,B,C);
+pair L=extension(A,I,C,B);
+dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G);
+draw(A--L);
+draw(A--D);
+draw(D--E);
+draw(I--F);
+draw(I--G);
+path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
+draw(C--B--D--cycle);
+draw(A--C--B);
+draw(A--B);
+draw(B--I--C^^A--I);
+draw(incircle(A,B,C));
+label("$B$",B,SW);
+label("$C$",C,SE);
+label("$A$",A,N);
+label("$D$",D,S);
+label("$I$",I,NW);
+label("$L$",L,SW);
+label("$O$",O,E);
+label("$E$",E,N);
+label("$F$",F,NE);
+label("$G$",G,SW);
+label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8));
+label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8));
+label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8));
+label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8));
+label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8));
+label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8));
+draw(I--O);
+draw(A--O);
+draw(rightanglemark(A,I,O));
+draw(rightanglemark(B,E,D));
+draw(rightanglemark(I,F,A));
+draw(rightanglemark(I,G,L));
+</asy>
+Since <math>\angle DBE \cong \angle IAF</math> and <math>\angle BED \cong \angle IFA</math>, <math>\triangle BDE \sim \triangle AIF</math>.
+Furthermore, we know <math>BD = ID</math> and <math>AI = ID</math>, so <math>BD = AI</math>. Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that <math>DE = IF = 6</math> since <math>IF</math> is the inradius.
+Now notice that <math>\triangle IGL \sim \triangle DEL</math> because of equal vertical angles and right angles. Furthermore, <math>IG</math> is the inradius so it's length is <math>6</math>, which equals the length of <math>DE</math>. Therefore these two triangles are congruent, so <math>IL = DL</math>.
+Since <math>IL+DL = ID</math>, <math>ID = 2 \cdot IL</math>. Furthermore, <math>AL = AI + IL = ID + IL = 3 \cdot IL</math>.
+We can now plug back into our initial equations for <math>AB \cdot AC</math>:
+From <math>1.1</math>, <math>AB \cdot AC = 2 \cdot AL \cdot ID = 2 \cdot 3 \cdot IL \cdot 2 \cdot IL</math>
+<cmath>\implies AB \cdot AC = 3 \cdot (2 \cdot IL) \cdot (2 \cdot IL) = 3 \cdot ID^2</cmath>
+Alternatively, from <math>1.2</math>, <math>AB \cdot AC = ID^2 \cdot \frac{AL}{DL}</math>
+<cmath>\implies AB \cdot AC = ID^2  \cdot \frac{3 \cdot IL}{IL} = 3 \cdot ID^2</cmath>
+Now all we need to do is find <math>ID</math>.
+The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that <math>OI^2 = R(R-2r)</math>, where <math>R</math> is the circumradius and <math>r</math> is the inradius. We will prove this formula first, but if you already know the proof, skip this part.
+Theorem: in any triangle, let <math>d</math> be the distance from the circumcenter to the incenter of the triangle. Then <math>d^2 = R \cdot (R-2r)</math>, where <math>R</math> is the circumradius of the triangle and <math>r</math> is the inradius of the triangle.
+Proof:
+Construct the following diagram:
+<asy>
+size(300);
+import olympiad;
+real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
+pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0);
+pair A = (c/3,8.65*c/10);
+pair F = (2*c/3-0.14, 4-0.29);
+pair G = (c/2-0.68,0);
+draw(circumcircle(A,B,C));
+pair I=incenter(A,B,C);
+pair O=circumcenter(A,B,C);
+pair L=extension(A,I,C,B);
+dot(I^^O^^A^^B^^C^^D^^L^^F);
+draw(A--L);
+draw(A--D);
+draw(I--F);
+path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
+draw(C--B--D--cycle);
+draw(A--C--B);
+draw(A--B);
+draw(A--I);
+draw(incircle(A,B,C));
+label("$B$",B,SW);
+label("$C$",C,SE);
+label("$A$",A,N);
+label("$D$",D,S);
+label("$I$",I,NW);
+label("$L$",L,SW);
+label("$O$",O,S);
+label("$F$",F,NE);
+label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8));
+label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8));
+pair H = (10*c/8-1.46,2*c/3-1.85), J = (-0.55,1.4);
+dot(H^^J);
+label("$H$", H, E);
+label("$J$", J, W);
+draw(I--O);
+draw(I--H);
+draw(I--J);
+draw(rightanglemark(I,F,A));
+</asy>
+Let <math>OI = d</math>, <math>OH = R</math>, <math>IF = r</math>. By the Power of a Point, <math>IH \cdot IJ = AI \cdot ID</math>.
+<math>IH = R+d</math> and <math>IJ = R-d</math>, so <cmath>(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD</cmath>
+Now consider <math>\triangle ACD</math>. Since all three points lie on the circumcircle of <math>\triangle ABC</math>, the two triangles have the same circumcircle. Thus we can apply law of sines and we get <math>\frac{CD}{\sin(\angle DAC)} = 2R</math>. This implies
+<cmath>(R+d)\cdot (R-d) = AI \cdot 2R \cdot \sin(\angle DAC)</cmath>
+Also, <math>\sin(\angle DAC)) = \sin(\angle IAF))</math>, and <math>\triangle IAF</math> is right. Therefore <cmath>\sin(\angle IAF) = \frac{IF}{AI} = \frac{r}{AI}</cmath>
+Plugging in, we have
+<cmath>(R+d)\cdot (R-d) = AI \cdot 2R \cdot \frac{r}{AI} = 2R \cdot r</cmath>
+Thus <cmath>R^2-d^2 = 2R \cdot r \implies d^2 = R \cdot (R-2r)</cmath>
+Now we can finish up our solution. We know that <math>AB \cdot AC = 3 \cdot ID^2</math>. Since <math>ID = AI</math>, <math>AB \cdot AC = 3 \cdot AI^2</math>. Since <math>\triangle AOI</math> is right, we can apply the pythagorean theorem: <math>AI^2 = AO^2-OI^2 = 13^2-OI^2</math>.
+Plugging in from Euler's formula, <math>OI^2 = 13 \cdot (13 - 2 \cdot 6) = 13</math>.
+Thus <math>AI^2 = 169-13 = 156</math>.
+Finally <math>AB \cdot AC = 3 \cdot AI^2 = 3 \cdot 156 = \textbf{468}</math>.
 ~KingRavi
-==Solution==
+==Solution 2 (Excenters)==
 By Euler's formula <math>OI^{2}=R(R-2r)</math>, we have <math>OI^{2}=13(13-12)=13</math>. Thus, by the Pythagorean theorem, <math>AI^{2}=13^{2}-13=156</math>. Let <math>AI\cap(ABC)=M</math>; notice <math>\triangle AOM</math> is isosceles and <math>\overline{OI}\perp\overline{AM}</math> which is enough to imply that <math>I</math> is the midpoint of <math>\overline{AM}</math>, and <math>M</math> itself is the midpoint of <math>II_{a}</math> where <math>I_{a}</math> is the <math>A</math>-excenter of <math>\triangle ABC</math>. Therefore, <math>AI=IM=MI_{a}=\sqrt{156}</math> and <cmath>AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.</cmath>
@@ Line 48: / Line 243: @@
-==Solution 2==
+==Solution 3==
 Denote <math>AB=a, AC=b, BC=c</math>. By the given condition, <math>\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6</math>, where <math>A</math> is the area of <math>\triangle{ABC}</math>.
-Moreover, since <math>OI\bot AI</math>, the second intersection of the line <math>AI</math> and <math>(ABC)</math> is the reflection of <math>A</math> about <math>I</math>, denote that as <math>D</math>. By the incenter-excenter lemma, <math>DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c</math>.
+Moreover, since <math>OI\bot AI</math>, the second intersection of the line <math>AI</math> and <math>(ABC)</math> is the reflection of <math>A</math> about <math>I</math>, denote that as <math>D</math>. By the incenter-excenter lemma with Ptolemy's Theorem, <math>DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c</math>.
 Thus, we have <math>\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c</math>. Now, we have <math>\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}</math>
@@ Line 58: / Line 253: @@
 ~Bluesoul
-==Solution 3==
+==Solution 4 (Trig)==
 Denote by <math>R</math> and <math>r</math> the circumradius and inradius, respectively.
 First, we have
-\[
+<cmath>\[
 r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1)
-\]
+\]</cmath>
 Second, because <math>AI \perp IO</math>,
@@ Line 124: / Line 319: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Solution 5 (Trig)==
+[[File:2024AIMEIIProblem10.png|450px|center]]
+Firstly, we can construct the triangle <math>\triangle ABC</math> by drawing the circumcirlce (centered at <math>O</math> with radius <math>R = OA = 13</math>) and incircle (centered at <math>I</math> with radius <math>r = 6</math>). Next, from <math>A</math>, construct tangent lines to the incircle meeting the circumcirlce at point <math>B</math> and <math>C</math>, say, as shown in the diagram. By Euler's theorem (relating the distance between <math>O</math> and <math>I</math> to the circumradius and inradius), we have
+<cmath>
+	OI = \sqrt{R^2 - 2rR} = \sqrt{13}.
+</cmath>
+This leads to
+<cmath>
+	AI = \sqrt{R^2 - OI^2} = \sqrt{156}.
+</cmath>
+Let <math>P</math> be the point of tangency where the incircle meets the side <math>\overline{AC}</math>. Now we denote
+<cmath>
+	\theta \coloneqq \angle BAI = \angle IAP \qquad \text{and} \qquad \phi \coloneqq \angle OAI.
+</cmath>
+Notice that <math>\angle BAO = \angle BAI - \angle OAI = \theta - \phi</math>. Finally, the crux move is to recognize
+<cmath>
+	AB = 2R \cos(\theta - \phi) \qquad \text{and} \qquad AC = 2R \cos(\theta + \phi)
+</cmath>
+since <math>O</math> is the circumcenter. Then multiply these two expressions and apply the compound-angle formula to get
+\begin{aligned}
+	AB \cdot AC
+	&= 4R^2 \cos(\theta - \phi) \cos(\theta + \phi) \\[0.3em]
+	&= 4R^2\left(
+	\cos^2\theta \cos^2\phi - \sin^2\theta \sin^2\phi
+	\right) \\[0.3em]
+	&= 4\cos^2\theta(\underbrace{R\cos\phi}_{AI \, = \, \sqrt{156}})^2 - 4\sin^2\theta(\underbrace{R\sin\phi}_{OI \, = \, \sqrt{13}})^2 \\[0.3em]
+	&= 52 (12\cos^2\theta - \sin^2 \theta) \\[0.3em]
+	AB \cdot AC
+	&= 52 (12 - 13\sin^2\theta),
+\end{aligned}
+where in the last equality, we make use of the substitution <math>\cos^2\theta = 1 - \sin^2\theta</math>. Looking at <math>\triangle AIP</math>, we learn that <math>\sin \theta = \frac{r}{AI} = \frac{6}{\sqrt{156}}</math> which means <math>\sin^2 \theta = \frac{3}{13}</math>. Hence we have
+<cmath>
+	AB \cdot AC = 52\left( 12 - 13 \cdot \tfrac{3}{13} \right) = 52 \cdot 9 = \boxed{468}.
+</cmath>
+This completes the solution
+-- VensL.
+==Solution 6 (Close to Solution 3)==
+[[File:2024 AIME II 10.png|230px|right]]
+Denote <math>E = \odot ABC \cap AI, AB = c, AC = b, BC=a, r</math> is inradius.
+<cmath>AO = EO = R \implies AI = EI.</cmath>
+It is known that <math>\frac {AI}{EI} = \frac {b+c}{a} – 1 = 1 \implies b + c = 2a.</math>
+*[[Barycentric_coordinates | Points on bisectors]]
+<cmath>[ABC] =\frac{ (a+b+c) r}{2} = \frac {3ar}{2} = \frac {abc}{4R} \implies bc = 6Rr = \boxed{468}.</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 7==
+Call side <math>BC = a</math>, and similarly label the other sides. Note that <math>{OI}^2 = R^2 - 2Rr</math>. Also note that <math>AO = R</math>, so by the right angle, <math>AI = \sqrt{2Rr}</math>. However, we can double Angle Bisector theorem. The length of the angle bisector from A is <math>\sqrt{(bc)(1 - \frac{a^2}{(b+c)^2})}</math>. As a direct result, the length AI simplifies down to <math>\frac{\sqrt{(bc)(b+c-a)}}{\sqrt{{a+b+c}}}</math>.
+Draw the incircle and call the tangent to side AB F. Then, <math>AF = \frac{b+c-a}{2}</math>. But this length, by Pythagorean, is <math>\sqrt{120}</math>, so <math>b+c-a = 2\sqrt{120}</math>.
+Also note that the area of the triangle is <math>[ABC] = \frac{abc}{52}</math>, by <math>\frac{abc}{4S} = R</math>. By the incircle, we know that <math>\sin{\frac{A}{2}} = \frac{6}{\sqrt{156}}</math>, and similarly, <math>\cos{\frac{A}{2}} = \frac{\sqrt{120}}{\sqrt{156}}</math>. By double-angle, <math>\sin A = \frac{\sqrt{120}}{13}</math>. But the area of the triangle <math>[ABC]</math> is simply <math>\frac{1}{2}bc \sin A</math>, which is also <math>2\sqrt{120}bc</math>. But we know this is <math>abc</math> from above, so <math>a = 2\sqrt{120}</math>. As a direct result, <math>a+b+c =
+\sqrt{120}</math>.
+Apply this to the formula <math>\frac{\sqrt{(bc)(b+c-a)}}{\sqrt{a+b+c}}</math> listed above to get <math>2Rr = 156 = \frac{bc}{3}</math>, so <math>bc = 468</math>. We're done. - sepehr2010
+==Solution 8==
+Let the intersection of the <math>A</math>-angle bisector and the circumcircle be <math>M</math>, and denote the <math>A</math>-excenter as <math>I_A</math>. Denote the tangent to the incircle from <math>AC</math> as <math>E</math> and the tangent to the excircle from <math>AC</math> as <math>E_A</math>.
+Notice that our perpendicular condition implies <math>AI = IM</math>, and Incenter-Excenter gives <math>IM = MI_A</math>. Thus we have <math>AI_A = 3AI</math>. From similar triangles we get <math>3(s-a) = 3AE = AE_A = s</math>. This implies <math>a = \frac23 S</math>.
+Using areas we have that <math>\frac{abc}{4R} = rs</math>. Substituting gives <math>\frac{sbc}{6R} = rs \implies bc = 6Rr = \boxed{468}</math> and we're done. - thoom
 ==Video Solution==
@@ Line 131: / Line 399: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Video Solution==
+https://www.youtube.com/watch?v=pPBPfpo12j4
+~MathProblemSolvingSkills.com
 ==See also==