Difference between revisions of "2024 AIME II Problems/Problem 10"
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~Bluesoul | ~Bluesoul | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Denote by <math>R</math> and <math>r</math> 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 <math>AI \perp IO</math>, | ||
+ | \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 <math>(1) - (2)</math>, 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 <math>AB \cdot AC</math>. 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== | ||
+ | |||
+ | https://youtu.be/_zxBvojcAQ4 | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
==See also== | ==See also== |
Revision as of 16:18, 9 February 2024
Problem
Let have circumcenter
and incenter
with
, circumradius
, and inradius
. Find
.
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 Fact 5,
.
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.