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

Revision as of 20:34, 10 February 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

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 2

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, $DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c$ .

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 3

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 $AI \perp IO$ , $\begin{aligned} A I & = A O \cos ∠ I A O \\ = A O \cos (90^{\circ} - C - \frac{A}{2}) \\ = A O \sin (C + \frac{A}{2}) \\ = R \sin (C + \frac{180^{\circ} - B - C}{2}) \\ = R \cos \frac{B - C}{2} . \end{aligned}$

Thus, $\begin{aligned} r & = A I \sin \frac{A}{2} \\ = R \sin \frac{A}{2} \cos \frac{B - C}{2} (2) \end{aligned}$

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

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

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 $\begin{aligned} \frac{r}{R} & = \sin \frac{A}{2} \cos \frac{B - C}{2} \\ = \sin \frac{180^{\circ} - B - C}{2} \cos \frac{B - C}{2} \\ = \cos \frac{B + C}{2} \cos \frac{B - C}{2} (4) \end{aligned}$

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 $\begin{aligned} A B \cdot A C & = 2 R \sin C \cdot 2 R \sin B \\ = 2 R^{2} (- \cos (B + C) + \cos (B - C)) \\ = 2 R^{2} (- (2 {(\cos \frac{B + C}{2})}^{2} - 1) + (2 {(\cos \frac{B - C}{2})}^{2} - 1)) \\ = 6 R r \\ = (468) \end{aligned}$ 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)

Video Solution

https://youtu.be/_zxBvojcAQ4

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

@@ Line 12: / Line 12: @@
 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 Fact 5, <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, <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>

2024 AIME II (Problems • Answer Key • Resources)
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