Difference between revisions of "2024 AIME II Problems/Problem 10"
(→Solution 1 (Continued)) |
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Alternatively, from <math>1.2</math>, <math>AB \cdot AC = ID^2 \cdot \frac{AL}{DL}</math> | Alternatively, from <math>1.2</math>, <math>AB \cdot AC = ID^2 \cdot \frac{AL}{DL}</math> | ||
− | <cmath>\implies AB \cdot = ID^2 \frac{3 \cdot IL}{IL} = 3 \cdot ID^2</cmath> | + | <cmath>\implies AB \cdot AC = ID^2 \cdot \frac{3 \cdot IL}{IL} = 3 \cdot ID^2</cmath> |
Now all we need to do is find <math>ID</math>. | Now all we need to do is find <math>ID</math>. | ||
+ | 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 <math>OI^2 = R(R-2r)</math>, where <math>R</math> is the circumradius and <math>r</math> is the inradius. We will prove this formula first, but if you already know the proof, skip this part. | ||
+ | Theorem: in any triangle, let <math>d</math> be the distance from the circumcenter to the incenter of the triangle. Then <math>d^2 = R \cdot (R-2r)</math>, where <math>R</math> is the circumradius of the triangle and <math>r</math> is the inradius of the triangle. | ||
+ | Proof: | ||
+ | |||
+ | Construct the following diagram: | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | 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 B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); | ||
+ | pair A = (c/3,8.65*c/10); | ||
+ | pair F = (2*c/3-0.14, 4-0.29); | ||
+ | pair G = (c/2-0.68,0); | ||
+ | draw(circumcircle(A,B,C)); | ||
+ | pair I=incenter(A,B,C); | ||
+ | pair O=circumcenter(A,B,C); | ||
+ | pair L=extension(A,I,C,B); | ||
+ | dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G); | ||
+ | draw(A--L); | ||
+ | draw(A--D); | ||
+ | draw(D--E); | ||
+ | draw(I--F); | ||
+ | draw(I--G); | ||
+ | path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | ||
+ | draw(C--B--D--cycle); | ||
+ | draw(A--C--B); | ||
+ | draw(A--B); | ||
+ | draw(B--I--C^^A--I); | ||
+ | draw(incircle(A,B,C)); | ||
+ | label("$B$",B,SW); | ||
+ | label("$C$",C,SE); | ||
+ | label("$A$",A,N); | ||
+ | label("$D$",D,S); | ||
+ | label("$I$",I,NW); | ||
+ | label("$L$",L,SW); | ||
+ | label("$O$",O,E); | ||
+ | label("$E$",E,N); | ||
+ | label("$F$",F,NE); | ||
+ | label("$G$",G,SW); | ||
+ | label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); | ||
+ | label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); | ||
+ | label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); | ||
+ | label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); | ||
+ | |||
+ | |||
+ | draw(I--O); | ||
+ | draw(A--O); | ||
+ | draw(rightanglemark(A,I,O)); | ||
+ | draw(rightanglemark(B,E,D)); | ||
+ | draw(rightanglemark(I,F,A)); | ||
+ | draw(rightanglemark(I,G,L)); | ||
+ | </asy> | ||
Revision as of 15:41, 13 February 2024
Contents
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:
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
~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.