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 $H$ be the orthocenter of acute triangle $ABC$ , let $F$ be the foot of the altitude from $C$ to $AB$ , and let $P$ be the reflection of $H$ across $BC$ . Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$ . Prove that $C$ is the midpoint of $XY$ .

Solution

Let AP intersects BC at D. Extend FC to the point E on the circumcircle $\omega$ of $AFP$ . Since $H$ is the orthocenter of $\Delta ABC$ , we know that $HD = DP$ or $HP = 2HD$ , and $AH \cdot HD = CH \cdot HF$ . Next we use the power of H in $\omega$ : $AH \cdot HP = CH \cdot HE$ . These relations imply that $HE = 2HF$ .

Hence $C, D$ are midpoints of $HE, HP$ respectively. By midline theorem, $CD // EP$ . Since $AD \perp CD$ , we have $AD \perp EP$ . This implies that $\angle APE = 90^{\circ}$ . Consequently, $AE$ is the diameter of $\omega$ . Let $G$ be the midpoint of $AE$ which is also the center of $\omega$ . $G,C$ are midpoints of $AE, EH$ respectively. By the midline theorem again, we have $GC//AH$ , consequently, $GC \perp BC$ . This implies that $GC$ is the perpendicular bisector of the chord $XY$ hence $C$ is the midpoint of $XY$ . ~ Dr. Shi davincimath.com

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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2025 USAMO Problems/Problem 4

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Problem

Solution

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