Difference between revisions of "2024 AIME II Problems/Problem 10"
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First, we have | First, we have | ||
− | \[ | + | <cmath>\[ |
r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1) | r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1) | ||
− | \] | + | \]</cmath> |
Second, because <math>AI \perp IO</math>, | Second, because <math>AI \perp IO</math>, |
Revision as of 21:47, 21 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
. 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,
.
Thus, we have . Now, we have
~Bluesoul
Solution 4 (Trig)
Denote by and
the circumradius and inradius, respectively.
First, we have
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.