Difference between revisions of "2025 USAMO Problems/Problem 4"

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== Solution ==
 
== Solution ==
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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>.
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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.

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

See Also

2025 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAMO Problems and Solutions
2025 USAJMO (ProblemsResources)
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. AMC Logo.png