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 (Less technical bary)
We are going to use barycentric coordinates on . Let
,
,
, and
,
,
. We have
and
so
and
. Since
, it follows that
Solving this gives
so
The equation for
is
Plugging in
and
gives
. Plugging in
gives
so
Now let
where
so
. It follows that
. It suffices to prove that
. Setting
, we get
. Furthermore we have
so it suffices to prove that
which is valid.
~KevinYang2.71
Solution 4 (Less bashy bary)
We employ barycentric coordinates. Set as the reference triangle with
,
, and
. We immediately have,
Since it passes through
, for some
, the equation of circle
is,
Plugging in
,
Plugging in
,
In conclusion the circle has formula,
is the second intersection of circle
with
. We let
for some
. Plugging this in,
We claim that
is the other solution.
Factoring out the
, this is clearly true.
We also check that, these are not the same value.
The triangle is acute, so this is impossible.
Since we had a quadratic in with at most two solutions, the second intersection
is indeed,
Therefore,
See Also
2023 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.