Difference between revisions of "2025 USAMO Problems/Problem 4"
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− | { | + | Let AP intersects BC at D. Extend FC to the point E on the circumcircle <math>\omega</math> of <math>AFP</math>. Since <math>H</math> is the orthocenter of <math>\Delta ABC</math>, we know that <math>HD = DP</math> or <math>HP = 2HD</math>, and <math>AH \cdot HD = CH \cdot HF</math>. Next we use the power of H in <math>\omega</math>: <math>AH \cdot HP = CH \cdot HE</math>. These relations imply that <math>HE = 2HF</math>. |
+ | |||
+ | Hence <math>C, D</math> are midpoints of <math>HE, HP</math> respectively. By midline theorem, <math>CD // EP</math>. Since <math>AD \perp CD</math>, we have <math>AD \perp EP</math>. This implies that <math>\angle APE = 90^{\circ}</math>. Consequently, <math>AE</math> is the diameter of <math>\omega</math>. Let <math>G</math> be the midpoint of <math>AE</math> which is also the center of <math>\omega</math>. <math>G,C</math> are midpoints of <math>AE, EH</math> respectively. By the midline theorem again, we have <math>GC//AH</math>, consequently, <math>GC \perp BC</math>. This implies that <math>GC</math> is the perpendicular bisector of the chord <math>XY</math> hence <math>C</math> is the midpoint of <math>XY</math>. ~ Dr. Shi davincimath.com | ||
==See Also== | ==See Also== |
Revision as of 11:20, 22 March 2025
- The following problem is from both the 2025 USAMO #4 and 2025 USAJMO #5, so both problems redirect to this page.
Contents
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
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
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.