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