2024 AIME II Problems/Problem 10

Problem

Let have circumcenter and incenter with , circumradius , and inradius . Find .

Solution 1 (Similar Triangles)

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 A=(0,0),B=(c,0), C = (c-sqrt(c^2-64))/6;
draw(circumcircle(A,B,C));
pair D=(c/2-1,-sqrt(25-(c/2)^2));
pair I=incenter(A,B,C);
pair O = circumcenter(A,B,C);
pair L=extension(C,D,A,B);
dot(I^^A^^B^^C^^D);
draw(C--D);
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;}
draw(A--B--D--cycle);
draw(circumcircle(A,B,D));
draw(A--C--B);
draw(A--I--B^^C--I);
draw(incircle(A,B,C));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,S);
label("$I$",I,NE);
label("$O$", O, N);
label("$L$",L,SW);
label("$\alpha$",A,5*dir(midangle(C,A,I)),fontsize(8));
label("$\alpha$",A,5*dir(midangle(I,A,B)),fontsize(8));
label("$\beta$",B,12*dir(midangle(A,B,I)),fontsize(8));
label("$\beta$",B,12*dir(midangle(I,B,C)),fontsize(8));
label("$\gamma$",C,5*dir(midangle(A,C,I)),fontsize(8));
label("$\gamma$",C,5*dir(midangle(I,C,B)),fontsize(8));
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Solution in Progress ~KingRavi

Solution

By Euler's formula , we have . Thus, by the Pythagorean theorem, . Let ; notice is isosceles and which is enough to imply that is the midpoint of , and itself is the midpoint of where is the -excenter of . Therefore, and

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

Solution 2

Denote . By the given condition, , where is the area of .

Moreover, since , the second intersection of the line and is the reflection of about , denote that as . By the incenter-excenter lemma, .

Thus, we have . Now, we have

~Bluesoul

Solution 3

Denote by and 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 , \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 , 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 . 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)

Video Solution

https://youtu.be/_zxBvojcAQ4

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

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.