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

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-- VensL.
 
-- VensL.
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==Solution 6 (Close to Solution 3)==
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[[File:2024 AIME II 10.png|230px|right]]
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Denote <math>E = \odot ABC \cap AI, AB = c, AC = b, BC=a, r</math> is inradius.
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<cmath>AO = EO = R \implies AI = EI.</cmath>
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It is known that <math>\frac {AI}{EI} = \frac {b+c}{a} – 1 = 1 \implies b + c = 2a.</math>
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*[[Barycentric_coordinates | Points on bisectors]]
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<cmath>[ABC] =\frac{ (a+b+c) r}{2} = \frac {3ar}{2} = \frac {abc}{4R} \implies bc = 6Rr = \boxed{468}.</cmath>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Solution 7==
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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>.
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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>.
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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 =
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6\sqrt{120}</math>.
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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
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==Solution 8==
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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>.
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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>.
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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==
 
==Video Solution==

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)

2024AIMEIIProblem10.png


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)

2024 AIME II 10.png

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 $\frac {AI}{EI} = \frac {b+c}{a} – 1 = 1 \implies b + c = 2a.$

\[[ABC] =\frac{ (a+b+c) r}{2} = \frac {3ar}{2} = \frac {abc}{4R} \implies bc = 6Rr = \boxed{468}.\] 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

See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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