2023 USAJMO Problems/Problem 2
Contents
Problem
(Holden Mui) In an acute triangle 
, let 
be the midpoint of 
. Let 
be the foot of the perpendicular from 
to 
. Suppose that the circumcircle of triangle 
intersects line 
at two distinct points 
and 
. Let 
be the midpoint of 
. Prove that 
.
Solution 1
The condition is solved only if 
is isosceles, which in turn only happens if 
is perpendicular to .
Now, draw the altitude from 
to , and call that point 
. Because of the Midline Theorem, the only way that this condition is met is if 
, or if 
.
By 
similarity, 
. Using similarity ratios, we get that 
. Rearranging, we get that 
. This implies that 
is cyclic.
Now we start using Power of a Point. We get that 
, and from before. This leads us to get that 
.
Now we assign variables to the values of the segments. Let 
and 
. The equation from above gets us that 
. As 
from the problem statements, this gets us that 
and 
, and we are done.
-dragoon and rhydon516 (:
Solution 2
Let 
be the foot of the altitude from onto . We want to show that 
for obvious reasons.
Notice that 
is cyclic and that lies on the radical axis of 
and 
. By Power of a Point, 
. As 
, we have , as desired.
- Leo.Euler
Solution 3
We are going to use barycentric coordinates on 
. Let 
, 
, 
, and 
, 
, 
. We have 
and 
so 
and 
. Since 
, it follows that
\begin{align*}
a^2\left(\frac{1}{2}\cdot\frac{1}{x+2}+\frac{1}{2}\left(\frac{1}{x+2}-1\right)\right)+b^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\left(\frac{1}{x+2}-1\right)\right)\\
+c^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\frac{1}{x+2}\right)=0.
\end{align*}Solving this gives
\[
x=\frac{2b^2-2c^2}{a^2-3b^2-c^2}
\]so
\[
P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right).
\]The equation for 
is
\[
-a^2yz-b^2zx-c^2xy+ux+vy+wz=0.
\]
Plugging in and gives 
. Plugging in gives
\begin{align*}
-a^2\left(\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right)^2-b^2\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\\
-c^2\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}+w\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}=0
\end{align*}
so
\[
w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}.
\]
Now let 
where
\begin{align*}
-a^2t(1-t)+w(1-t)&=0\\
\implies t&=\frac{w}{a^2}
\end{align*}
so 
. It follows that 
. It suffices to prove that 
. Setting 
, we get 
. Furthermore we have 
so it suffices to prove that
\begin{align*}
a^2\left(-\frac{w}{2a^2}+\frac{1}{2}-\frac{u}{2a^2}\right)+b^2\left(-\frac{1}{2}\right)+c^2\left(\frac{1}{2}\right)=0\\
\implies w=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}
\end{align*}
which is valid. 
~KevinYang2.71
