2025 USAMO Problems/Problem 4
- The following problem is from both the 2025 USAMO #4 and 2025 USAJMO #5, so both problems redirect to this page.
Problem
Let be the orthocenter of acute triangle
, let
be the foot of the altitude from
to
, and let
be the reflection of
across
. Suppose that the circumcircle of triangle
intersects line
at two distinct points
and
. Prove that
is the midpoint of
.
Solution 1
Let AP intersects BC at D. Extend FC to the point E on the circumcircle of
. Since
is the orthocenter of
, we know that
or
, and
. Next we use the power of H in
:
. These relations imply that
.
Hence are midpoints of
respectively. By midline theorem,
. Since
, we have
. This implies that
. Consequently,
is the diameter of
. Let
be the midpoint of
which is also the center of
.
are midpoints of
respectively. By the midline theorem again, we have
, consequently,
. This implies that
is the perpendicular bisector of the chord
hence
is the midpoint of
. ~ Dr. Shi davincimath.com
Solution 2
Denote as the center of
,
as the center of
,
as the midpoint of
,
as the midpoint of
, and
as the midpoint of
. It suffices to show that
.
Claim: is cyclic.
Proof: Since and
, KM is a midline of
and
.
as well since
, so
lies on
.
Next, note that
lies on
, so the perpendicular bisector of
through
passes through
. In other words,
, and
are collinear. Since
and
are both perpendicular to
, it follows that they are parallel.
Since
and
, then
.
Finally, we have that
and thus
is cyclic. It follows that
, so
, as desired.
-mop
Solution 3
Connect and have
intersect
at
. Also extend
past point
and have it intersect with the circle at point
.
Since is the reflection of
over
, we know that
. Since
is the orthocenter, we can draw the altitude and tell that
,
, and
are collinear.
We know
and
, so
by AA, so
.
and
. From this, we can tell that
. Therefore,
and
.
If we connect , we can tell that that
due to
being the reflection of
and
being perpendicular to
, so
. In addition,
. Also,
because they are vertical angles.
So, because of SAA. From this we can conclude that
, so
is the midpoint of
.
Solution 4
Let be the foot of the altitude from
to
By Power of a Point, we have
and
Adding, we get
It is well known that
Then, let
We have
Thus,
and since
we have
Hence,
is the midpoint of
~TThB0501
See Also
2025 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
2025 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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