2024 AIME II Problems/Problem 10

Problem

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

Solution 1 (Similar Triangles and PoP)

Start off by (of course) drawing a diagram! Let and be the incenter and circumcenters of triangle , respectively. Furthermore, extend to meet at and the circumcircle of triangle at .

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 is the incenter, . Furthermore, and are both subtended by the same arc , so Therefore by AA similarity, . From this we can say that

Since is a chord of the circle and is a perpendicular from the center to that chord, must bisect . This can be seen by drawing and recognizing that this creates two congruent right triangles. Therefore,

We have successfully represented in terms of and . 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

by vertical angles and because both are subtended by arc . Thus .

Thus

Symmetrically, we get , so

Substituting, we get

Lemma 1: BD = CD = ID

Proof:

We commence angle chasing: we know . Therefore . Looking at triangle , we see that , and . Therefore because the sum of the angles must be , . Now is a straight line, so . Since , triangle is isosceles and thus .

A similar argument should suffice to show by symmetry, so thus .

Now we regroup and get

Now note that and are part of the same chord in the circle, so we can use Power of a point to express their product differently.

Solution 1 (Continued)

Now we have some sort of expression for in terms of and . Let's try to find first.

Drop an altitude from to , to , and to :

Since and , .

Furthermore, we know and , so . Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that since is the inradius.

Now notice that because of equal vertical angles and right angles. Furthermore, is the inradius so it's length is , which equals the length of . Therefore these two triangles are congruent, so .

Since , . Furthermore, .

We can now plug back into our initial equations for :

From ,

Alternatively, from ,

Now all we need to do is find .

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 , where is the circumradius and is the inradius. We will prove this formula first, but if you already know the proof, skip this part.

Theorem: in any triangle, let be the distance from the circumcenter to the incenter of the triangle. Then , where is the circumradius of the triangle and is the inradius of the triangle.

Proof:

Construct the following diagram:

Let , , . By the Power of a Point, . and , so

Now consider . Since all three points lie on the circumcircle of , the two triangles have the same circumcircle. Thus we can apply law of sines and we get . This implies

Also, , and is right. Therefore

Plugging in, we have

Thus

Now we can finish up our solution. We know that . Since , . Since is right, we can apply the pythagorean theorem: .

Plugging in from Euler's formula, .

Thus .

Finally .

~KingRavi

Solution 2 (Excenters)

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 3

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 with Ptolemy's Theorem, .

Thus, we have . Now, we have

~Bluesoul

Solution 4 (Trig)

Denote by and the circumradius and inradius, respectively.

First, we have

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)

Solution 5 (Trig)

Firstly, we can construct the triangle by drawing the circumcirlce (centered at with radius ) and incircle (centered at with radius ). Next, from , construct tangent lines to the incircle meeting the circumcirlce at point and , say, as shown in the diagram. By Euler's theorem (relating the distance between and to the circumradius and inradius), we have This leads to Let be the point of tangency where the incircle meets the side . Now we denote Notice that . Finally, the crux move is to recognize since 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 . Looking at , we learn that which means . Hence we have This completes the solution

-- VensL.

Solution 6 (Close to Solution 3)

Denote is inradius. It is known that

• Points on bisectors

vladimir.shelomovskii@gmail.com, vvsss

Solution 7

Call side , and similarly label the other sides. Note that . Also note that , so by the right angle, . However, we can double Angle Bisector theorem. The length of the angle bisector from A is . As a direct result, the length AI simplifies down to .

Draw the incircle and call the tangent to side AB F. Then, . But this length, by Pythagorean, is , so .

Also note that the area of the triangle is , by . By the incircle, we know that , and similarly, . By double-angle, . But the area of the triangle is simply , which is also . But we know this is from above, so . As a direct result, .

Apply this to the formula listed above to get , so . We're done. - sepehr2010

Solution 8

Let the intersection of the -angle bisector and the circumcircle be , and denote the -excenter as . Denote the tangent to the incircle from as and the tangent to the excircle from as .

Notice that our perpendicular condition implies , and Incenter-Excenter gives . Thus we have . From similar triangles we get . This implies .

Using areas we have that . Substituting gives 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

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