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

Latest revision as of 20:32, 7 August 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 $FAP$ 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 .

Video Solution - A 2-minute proof

https://youtu.be/dnALS0aNWNQ

Solution 6

Quick angle chasing gives $\angle FAP = \angle BCF = \angle BCP = \alpha$ . Let $O$ be the circumcenter of $\triangle AFP$ .

Thus $\angle FOP= 2 \angle FAP = 2\alpha$ (because O and A lie on the same side of segment $AF$ ).

As $\angle FOP = \angle FCP = 2 \alpha$ , the quadrilateral FOCP is cyclic.

Observe that $2\angle FPO = 180-2\alpha$ , so $\angle FPO = 90^{\circ} - \alpha$ .

From the properties of cyclic quadrilaterals, $\angle FCO = \angle FPO$

Thus $\angle BCO = \angle BCF + \angle FCO = \alpha + (90^{\circ} - \alpha) =90^{\circ}$ . Therefore $OC$ is perpendicular to chord $XY$ , $XC=CY$ .

Solution 7

Let the line perpendicular to $BC$ and going through $C$ be line $l$ . Let the midpoint of $AF$ be $M$ , and let the line perpendicular to $AF$ and going through $M$ be line $n$ . Let $O$ be the intersection of $l$ and $n$ , and let $D$ be the intersections of $n$ and $AH$ .

Because $AH\parallel l$ and $CF\parallel n$ , $CODH$ is a parallelogram. From this, we get $OC=DH$ . Looking at triangle $\triangle{FAP}$ , because $n\parallel HF$ , and $M$ is the midpoint of $AF$ , it is clear that $D$ is the midpoint of $AH$ .

Let the intersection of $HP$ and $BC$ be $Z$ . $PZ=ZH$ because $P$ is a reflection of $H$ across $BC$ . $AP=AH+HP=2DH+2PZ$

Substituting the equation from earlier, we get $AP=2OC+2PZ$

Draw a line perpendicular to $OC$ starting from $O$ . Let the intersection of that line and $AP$ be $E$ . Because $BC$ is perpendicular to $OC$ which is perpendicular to $OE$ , $BC\parallel OE$ . By the same logic, $AP\parallel OC$ . This means $OCZE$ is a parallelogram (specifically a rectangle). Therefore $EZ=OC$ . $EP=EZ+PZ$ and by substituting, we get $EP=OC+PZ$ , and because $AP=2OC+2PZ$ , that means $2EP=AP$ , so $E$ is a midpoint of $AP$ . This means $OE$ is the perpendicular bisector of $AP$ , and because $AP$ is a chord of the circumcircle of $\triangle{FAP}$ , $OE$ goes through the center of the circle. By the same logic, $n$ also goes through the center of the circle, since it is the perpendicular bisector of $AF$ . This means the center of the circle is $O$ , because it is the only point on both $OE$ and $n$ . Because $O$ is on line $OC$ and line $OC$ is perpendicular to line $BC$ , line $OC$ must perpendicularly bisect the chord of circle $O$ containing $C$ that lies on line $BC$ (basically $OC$ must perpendicularly bisect $XY$ ). This means $C$ is the midpoint of $XY$ .

-Nolan Lin

@@ Line 79: / Line 79: @@
 Therefore <math>OC</math> is perpendicular to chord <math>XY</math>, <math>XC=CY</math>.
 [[File:Usamo_2025_problem4.png|thumb|center|600px]]
+== Solution 7 ==
+Let the line perpendicular to <math>BC</math> and going through <math>C</math> be line <math>l</math>. Let the midpoint of <math>AF</math> be <math>M</math>, and let the line perpendicular to <math>AF</math> and going through <math>M</math> be line <math>n</math>. Let <math>O</math> be the intersection of <math>l</math> and <math>n</math>, and let <math>D</math> be the intersections of <math>n</math> and <math>AH</math>.
+Because <math>AH\parallel l</math> and <math>CF\parallel n</math>, <math>CODH</math> is a parallelogram. From this, we get <math>OC=DH</math>. Looking at triangle <math>\triangle{FAP}</math>, because <math>n\parallel HF</math>, and <math>M</math> is the midpoint of <math>AF</math>, it is clear that <math>D</math> is the midpoint of <math>AH</math>.
+Let the intersection of <math>HP</math> and <math>BC</math> be <math>Z</math>. <math>PZ=ZH</math> because <math>P</math> is a reflection of <math>H</math> across <math>BC</math>.
+<math>AP=AH+HP=2DH+2PZ</math>
+Substituting the equation from earlier, we get
+<math>AP=2OC+2PZ</math>
+Draw a line perpendicular to <math>OC</math> starting from <math>O</math>. Let the intersection of that line and <math>AP</math> be <math>E</math>. Because <math>BC</math> is perpendicular to <math>OC</math> which is perpendicular to <math>OE</math>, <math>BC\parallel OE</math>. By the same logic, <math>AP\parallel OC</math>. This means <math>OCZE</math> is a parallelogram (specifically a rectangle). Therefore <math>EZ=OC</math>. <math>EP=EZ+PZ</math> and by substituting, we get <math>EP=OC+PZ</math>, and because <math>AP=2OC+2PZ</math>, that means <math>2EP=AP</math>, so <math>E</math> is a midpoint of <math>AP</math>. This means <math>OE</math> is the perpendicular bisector of <math>AP</math>, and because <math>AP</math> is a chord of the circumcircle of <math>\triangle{FAP}</math>, <math>OE</math> goes through the center of the circle. By the same logic, <math>n</math> also goes through the center of the circle, since it is the perpendicular bisector of <math>AF</math>. This means the center of the circle is <math>O</math>, because it is the only point on both <math>OE</math> and <math>n</math>. Because <math>O</math> is on line <math>OC</math> and line <math>OC</math> is perpendicular to line <math>BC</math>, line <math>OC</math> must perpendicularly bisect the chord of circle <math>O</math> containing <math>C</math> that lies on line <math>BC</math> (basically <math>OC</math> must perpendicularly bisect <math>XY</math>). This means <math>C</math> is the midpoint of <math>XY</math>.
+-Nolan Lin
+[[File:USAMO4diagram.png|thumb|center|600px]]

Page

Toolbox

Search