Difference between revisions of "2024 AIME II Problems/Problem 10"
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Let <math>OI = d</math>, <math>OH = R</math>, <math>IF = r</math>. By the Power of a Point, <math>IH \cdot IJ = AI \cdot ID</math>. | Let <math>OI = d</math>, <math>OH = R</math>, <math>IF = r</math>. By the Power of a Point, <math>IH \cdot IJ = AI \cdot ID</math>. | ||
<math>IH = R+d</math> and <math>IJ = R-d</math>, so <cmath>(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD</cmath> | <math>IH = R+d</math> and <math>IJ = R-d</math>, so <cmath>(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD</cmath> | ||
+ | |||
+ | Now consider <math>\triangle ACD</math>. Since all three points lie on the circumcircle of <math>\triangle ABC</math>, the two triangles have the same circumcircle. Thus we can apply law of sines and we get <math>\frac{CD}{\sin(\angle DAC)} = 2</math>R$. This implies | ||
+ | |||
+ | <cmath>(R+d)\cdot (R-d) = AI \cdot 2R \cdot \sin(\angle DAC)</cmath> | ||
Revision as of 18:50, 13 February 2024
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 .
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 R$. This implies
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.