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 $FAP$ 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 $FAP$ . 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 $FAP$ , $K$ 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 $XY$ .

Solution 4

Let $D$ be the foot of the altitude from $A$ to $BC.$ By Power of a Point, we have $BF\cdot BA = BX \cdot BY = (BC-CX)(BC+CY) = BC^2 + BC(CY-CX) - CX\cdot CY$ and $DB\cdot DC = DX\cdot DY = (CX-CD)(CD+CY) = CX\cdot CY - CD(CY-CX) - CD^2.$ Adding, we get $BF \cdot BA + DB\cdot DC = BC^2-DC^2 + (BC-CD)(CY-CX).$ It is well known that the reflection of $H$ over $AB,$ which we denote by $P,$ lies on $(ABC).$ Then, let $\angle{BAP} = \angle{BCP} = \angle{BCH} = \theta.$ We have $BF\cdot BA + DB\cdot DC = (BC\sin\theta)\cdot BA + DB\cdot DC = BC\cdot (BA \sin\theta) + DB\cdot DC = DB\cdot (BC+DC) = (BC-DC)(BC+DC) = BC^2-DC^2.$ Thus, $(BC-CD)(CY-CX)=0,$ and since $BC\neq CD$ we have $CY=CX.$ Hence, $C$ is the midpoint of $XY.$ ~TThB0501

Solution 5

Let Q be the antipode of B. Claim — AHQC is a parallelogram, and AP CQ is an isosceles trapezoid.

Proof. As AH ⊥ BC ⊥ CQ and CF ⊥ AB ⊥ AQ. Let M be the midpoint of QC.

Claim — Point M is the circumcenter of triangle AFP.

Proof. It’s clear that MA = MP from the isosceles trapezoid.

As for MA = MF, let N denote the midpoint of AF; then MN is a midline of the parallelogram, so MN ⊥ AF. Since CM ⊥ BC and M is the center of (AF P), it follows CX = CY .