2024 AIME II Problems/Problem 10
Contents
[hide]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 .
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 ,
Thus,
Taking , we get \[ 4 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B-C}{2} . \]
We have
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
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
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.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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