2025 USAMO Problems/Problem 4: Difference between revisions

Revision as of 10:20, 22 March 2025

The following problem is from both the 2025 USAMO #4 and 2025 USAJMO #5, so both problems redirect to this page.

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

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 == Solution ==
-{{solution}}
+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==

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: Difference between revisions

Revision as of 10:20, 22 March 2025

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