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Difference between revisions of "2025 USAMO Problems/Problem 4"
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Connect <math>HP</math> and have <math>HP</math> intersect <math>XY</math> at <math>W</math>. Also extend <math>FC</math> past point <math>C</math> and have it intersect with the circle at point <math>D</math>.
 
Connect <math>HP</math> and have <math>HP</math> intersect <math>XY</math> at <math>W</math>. Also extend <math>FC</math> past point <math>C</math> and have it intersect with the circle at point <math>D</math>.
 +
 
Since <math>P</math> is the reflection of <math>H</math> over <math>BC</math>, we know that <math>HP\perp XY</math>. Since <math>H</math> is the orthocenter, we can draw the altitude and tell that <math>A</math>, <math>H</math>, and <math>P</math> are collinear.
 
Since <math>P</math> is the reflection of <math>H</math> over <math>BC</math>, we know that <math>HP\perp XY</math>. Since <math>H</math> is the orthocenter, we can draw the altitude and tell that <math>A</math>, <math>H</math>, and <math>P</math> are collinear.
 
We know <math>m\angle{AFH} = m\angle{CWH}=90^{\circ}</math> and <math>m\angle{FHA} = m\angle{WHC}</math>, so <math>\triangle{AHF} \sim \triangle{CHW}</math> by AA, so <math>m\angle{FAH} = m\angle{WCH}</math>.
 
We know <math>m\angle{AFH} = m\angle{CWH}=90^{\circ}</math> and <math>m\angle{FHA} = m\angle{WHC}</math>, so <math>\triangle{AHF} \sim \triangle{CHW}</math> by AA, so <math>m\angle{FAH} = m\angle{WCH}</math>.
 +
 
<math>m\angle{FAP} = \frac{1}{2}m\overarc{FP} = \frac{1}{2}(m\overarc{FX} + m\overarc{XP})</math> and <math>m\angle{FCX} = \frac{1}{2}(m\overarc{FX} + m\overarc{DY})</math>. From this, we can tell that <math>m\overarc{XP} = m\overarc{DY}</math>. Therefore, <math>m\overarc{XD} = m\overarc{PY}</math> and <math>XD = PY</math>.
 
<math>m\angle{FAP} = \frac{1}{2}m\overarc{FP} = \frac{1}{2}(m\overarc{FX} + m\overarc{XP})</math> and <math>m\angle{FCX} = \frac{1}{2}(m\overarc{FX} + m\overarc{DY})</math>. From this, we can tell that <math>m\overarc{XP} = m\overarc{DY}</math>. Therefore, <math>m\overarc{XD} = m\overarc{PY}</math> and <math>XD = PY</math>.
 +
 
If we connect <math>HY</math>, we can tell that that <math>PY = HY</math> due to <math>P</math> being the reflection of <math>H</math> and <math>WY</math> being perpendicular to <math>HP</math>, so <math>XD = HY</math>. In addition, <math>m\angle{HYW} = m\angle{PYW} = \frac{1}{2} m\overarc{XP} = \frac{1}{2} m\overarc{DY} = m\angle{YXD}</math>. Also, <math>m\angle{HCY} = m\angle{XCD}</math> because they are vertical angles.
 
If we connect <math>HY</math>, we can tell that that <math>PY = HY</math> due to <math>P</math> being the reflection of <math>H</math> and <math>WY</math> being perpendicular to <math>HP</math>, so <math>XD = HY</math>. In addition, <math>m\angle{HYW} = m\angle{PYW} = \frac{1}{2} m\overarc{XP} = \frac{1}{2} m\overarc{DY} = m\angle{YXD}</math>. Also, <math>m\angle{HCY} = m\angle{XCD}</math> because they are vertical angles.
 +
 
So, <math>\triangle{HCY} \cong \triangle{XCD}</math> because of SAA. From this we can conclude that <math>XC = CY</math>, so <math>C</math> is the midpoint of <math>XY</math>.
 
So, <math>\triangle{HCY} \cong \triangle{XCD}</math> because of SAA. From this we can conclude that <math>XC = CY</math>, so <math>C</math> is the midpoint of <math>XY</math>.
  

Revision as of 01:00, 23 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 1

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

Solution 2

Denote $O_1$ as the center of $(ABC)$, $O_2$ as the center of $AFP$, $K$ as the midpoint of $AF$, $M$ as the midpoint of $AC$, and $N$ as the midpoint of $AP$. It suffices to show that $\angle{O_2CB}=90$.

Claim: $O_1MO_2C$ is cyclic.

Proof: Since $AK=FK$ and $AM=MC$, KM is a midline of $\triangle{AFC}$ and $KM\parallel FC$. $KO_2\parallel FC$ as well since $\angle{AKO_2}=\angle{AFC}=90$, so $M$ lies on $KO_2$. Next, note that $P$ lies on $(ABC)$, so the perpendicular bisector of $AP$ through $N$ passes through $O_1$. In other words, $N, O_1$, and $O_2$ are collinear. Since $NO_2$ and $BC$ are both perpendicular to $AP$, it follows that they are parallel. Since $KO_2\parallel FC$ and $NO_2\parallel BC$, then $\angle KO_2N=\angle{FCB}$. Finally, we have that \[\angle{MO_2O_1}=\angle{KO_2N}=\angle{FCB}=90-B=\angle{MCO_1},\] and thus $O_1MO_2C$ is cyclic. It follows that $\angle O_1O_2C=\angle{O_1MC}=90$, so $\angle{O_2CB}=180-\angle{O_1MC}=90$, as desired.

-mop

Solution 3

Connect $HP$ and have $HP$ intersect $XY$ at $W$. Also extend $FC$ past point $C$ and have it intersect with the circle at point $D$.

Since $P$ is the reflection of $H$ over $BC$, we know that $HP\perp XY$. Since $H$ is the orthocenter, we can draw the altitude and tell that $A$, $H$, and $P$ are collinear. We know $m\angle{AFH} = m\angle{CWH}=90^{\circ}$ and $m\angle{FHA} = m\angle{WHC}$, so $\triangle{AHF} \sim \triangle{CHW}$ by AA, so $m\angle{FAH} = m\angle{WCH}$.

$m\angle{FAP} = \frac{1}{2}m\overarc{FP} = \frac{1}{2}(m\overarc{FX} + m\overarc{XP})$ and $m\angle{FCX} = \frac{1}{2}(m\overarc{FX} + m\overarc{DY})$. From this, we can tell that $m\overarc{XP} = m\overarc{DY}$. Therefore, $m\overarc{XD} = m\overarc{PY}$ and $XD = PY$.

If we connect $HY$, we can tell that that $PY = HY$ due to $P$ being the reflection of $H$ and $WY$ being perpendicular to $HP$, so $XD = HY$. In addition, $m\angle{HYW} = m\angle{PYW} = \frac{1}{2} m\overarc{XP} = \frac{1}{2} m\overarc{DY} = m\angle{YXD}$. Also, $m\angle{HCY} = m\angle{XCD}$ because they are vertical angles.

So, $\triangle{HCY} \cong \triangle{XCD}$ because of SAA. From this we can conclude that $XC = CY$, so $C$ is the midpoint of $XY$.

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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