I managed to reduce this to proving XY was perpendicular to BC, but i couldn't manage to prove that. Is this even true?

djmathman wrote: Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $AHC$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$. It should be: let $P$ be the second intersection of the circumcircle of triangle $ABC$ with the internal bisector of the angle $\angle BAC$ EDIT: oops this is not the original problem but this version also has $XY$ equaling the circumradius.

Reflecting the whole thing over line AC gives a nice solution

msinghal wrote: Reflecting the whole thing over line AC gives a nice solution It also gives a nice complex bash took me around 20 minutes, thankfully... I finished the test with five to spare so any more synthetic effort would have finished me.

Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly)

pi37 wrote: Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly) I was thinking about that but I really didn't want to take care of that. I wasn't sure whether to add at the end that directed angles/lengths could resolve this, but decided against it.

$Y$ is on circumcircle by angle chasing. Let $O_1$ be the reflection of $O$ through $AC$. Then look at the rotohomothety with center $O$ that sends $AOY$ to $O_1OX$ (these triangles are similar). It turns out $AO_1=R$ and so $XY=R$, so we're done.

Starting to worry that reducing to BC perpendicular to XY was not actually useful for finding a solution...

Well, it could be used to find the spiral similarity in my solution... I'm not sure.

pi37 wrote:

Why $(O_1)$ is the reflection of $(O)$ across $AC$?

pi37 wrote: Question: Do you think not addressing possible configuration issues will lose points? (i.e. if I said assume the configuration is as shown, other cases are handled similarly) Your solution was the same as the official solution, so it is safe to assume that you won't lose any points .

here my solution: Let $R$ be the radius of $(ABC)$. The radius of $AHC$ also equal to $R$ and \[ \frac{AX}{R}=\sin{\sin \angle ACP}{\sin \angle ABP}=\frac{\sin \angle PCB}{\sin \angle PBC}=\frac{BP}{CP} .\] We have that $\angle XAY=\angle XAP-\angle YAP=\angle PBC$ and \[ \frac{AX}{AY}=\frac{BP}{BC}. \] So $ \triangle AXY $ and $ \angle BPC$ are similar. And \[ \frac{XY}{CP}=\frac{AY}{CP}=\frac{R}{CP}. \]

Wait, darn, I should have tried this more during the USAMO. By radical axes, $OO_1$ is perpendicular to $AC$, and $OX$ to $AB$, and $XO_1$ to $AP$, so $OO_1=OX$ follows. Easy angle chasing gives $Y$ is on the circle. Then you're basically done, you can write out the coordinates for $X$ and $Y$ very easily. Just do $A=(cos(A),sin(A))$, $B=(cos(B),sin(B))$, and $X$, $Y$ are not hard to find, for some angles $A$ and $B$ (you need $A-B=2c$ or something...). Or you know $OX=2Rcos(b)$ and $OX$ is perpendicular to $AB$, and you now where $Y$ is on the circle because the angle $ACY$ is just $\frac {b+c} {2}$. So you can make $M$ the midpoint of the arc and prove $OX$ is parallel to and equal to $MY$ easily. But I would not notice $M$ because I am silly unlike the other posters.

This problem can be easily solved using Barycentric coordinates....coordinates of $H$ is well known.Now if one finds out the coordintes of $P$ everything is fine.(I mean what one requires is patience during the bash). USAMO Geometry problems often have good solutions using barycentric coordinates...

We also can see that $\triangle YPX= \triangle OXB \ ( \text{A.S.S})$.

A general problem Let $ABC$ be a triangle with circumcirlce $(O)$ and $(K)$ is a circle passing through $A,C$. Bisector of $\angle BAC$ cuts $(K)$ again at $D$. Let $L$ be circumcenter of triangle $DAB$. $N$ lies on $(O)$ such that $CN\perp AD$. Prove that $LN=AK$.

Define $\Omega:=(ABC)$. Define $\Psi:\bullet\mapsto\bullet'$ as reflection over $\overline{AC}$. Let $Q=\Psi(P)$; note that $Q\in \Omega$ by reflecting $P$, the orthocenter of $\triangle AYC$. First, we claim $Y\in \Omega$. Indeed, $\angle AYC=\pi-\angle APC=\pi-\angle AQC$. Use complex numbers with respect to $\Omega:=(ABC)$, and set $c=1$. So $Y'$ is the orthocenter of $\triangle AQC$, hence $Y'=a+q+1$. Now we want to find $X'$, the center of $(AQB')$. Since $\angle QAC=\tfrac12 \angle BAC$ but $Q\not = M_A$ the arc midpoint of $BC$, we have $b=1/q^2$. Hence $b'=a+c-ac/b=a+1-aq^2$. By the circumcenter formula, \begin{align*} X' &= \begin{vmatrix} a & 1 & 1 \\ q & 1 & 1 \\ a+1-aq^2 & (a+1-aq^2)(\tfrac{1}{a}+1-\tfrac{1}{aq^2}) & 1 \end{vmatrix} \div \begin{vmatrix} a & \tfrac{1}{a} & 1 \\ q & \tfrac{1}{q} & 1 \\ a+1-aq^2 & \tfrac{1}{a}+1-\tfrac{1}{aq^2} & 1 \end{vmatrix} \\ &= \begin{vmatrix} a & 0 & 1 \\ q & 0 & 1 \\ 1-aq^2 & (a+1-aq^2)(\tfrac{1}{a}+1-\tfrac{1}{aq^2})-1 & 0 \end{vmatrix} \div \frac{1}{aq^2}\begin{vmatrix} a & a+q^2 & 1 \\ q & q+aq & 1 \\ 1-aq^2 & 0 & 0 \end{vmatrix} \\ &= \frac{-[(a+1-aq^2)(\tfrac{1}{a}+1-\tfrac{1}{aq^2})-1](a-q)}{-\tfrac{1}{aq^2} \cdot (1-aq^2)(q+aq-a-q^2)} \\ &= \frac{(a+1-aq^2)\cdot \frac{1}{aq^2}(q^2+aq^2-1)-1}{\frac{1}{aq^2}(1-aq^2)(q-1)} \\ &= \frac{(a+1-aq^2)(q^2+aq^2-1)-aq^2}{(1-aq^2)(q-1)} \\ &= (a+1)(q+1). \end{align*}where the second equation follows by (1) subtracting the first row from the third in both dets, (2) subtracting third column from second in first det, (3) factoring out $\tfrac{1}{aq^2}$ of the second column of the second det, (4) adding the first column to the second column of the second det. Now \begin{align*} |X'-Y'| &= | a+q+1 - (a+1)(q+1)| = |aq| =1, \end{align*}as needed.

$O$ is the center of $(ABC)$, $Q$ is the center of $(AHC)$, $F$ is the midpoint of $BC$, $M$ is the midpoint of arc $BC$, $E = AB \cap MY$, $G = AM \cap CY$. Claim: $Y$ lies on $(ABC)$. Proof: By some basic orthocenter properties, $$\measuredangle ABC = -\measuredangle AHC = -\measuredangle APC = \measuredangle AYC$$so $ABYC$ is a cyclic quadrilateral. \(\blacksquare\) Claim: $YM || XO$. Proof: Because $XO \perp AB$, it suffices to prove $YM \perp AB$. We have $$\measuredangle BYE = \measuredangle EAM = \measuredangle BAM = \measuredangle MAC = \measuredangle MBC.$$We want to prove $EBY \sim FMB $, and because $\measuredangle MGC = \measuredangle MFC = 90^{\circ}$, $F, C, M, G$ are concyclic. Therefore, we have $$\implies \measuredangle OMB = \measuredangle EBY = \measuredangle AMY \implies \measuredangle FMA = \measuredangle BMY$$$$\implies \measuredangle BMY = \measuredangle BCY = \measuredangle FCG = \measuredangle FMG = \measuredangle FMA$$as desired. Therefore, by AA similarity, $\measuredangle YEB = \measuredangle BFM = 90^{\circ}$. \(\blacksquare\) $AP$ is the radical axis of $(APB)$ and $(APC)$, so $XQ \perp AP$, and by definition $YC \perp AP$ so $YC || XQ$. Also, $\measuredangle AOC = 2\measuredangle ABC = \measuredangle CQA$, so the circle centered at $Q$ has the same radius as $(ABC)$. Let $I$ be on $(AHC)$ such that $QI || OM$. Because $QI = OM$, $QOMI$ is a parallelogram and $OQ = IM$, $OQ || IM$. Now (notice we proved <BAY = <OMA earlier) $$\measuredangle YAM = \measuredangle BAM-\measuredangle BAY = \measuredangle BAM-\measuredangle OMA = \measuredangle BAM-\measuredangle MAO = \measuredangle BAM - \measuredangle MAC + \measuredangle OAC = \measuredangle OAC$$$$\implies \measuredangle YOM = \measuredangle OAQ$$which means $AOQ \cong OYM \cong CQO$ and $YM = OQ$. By properties of orthocenters, $\measuredangle GYP = \measuredangle PAC = \measuredangle MYG$, and because $PM \perp YG$, $PY = MY$. Finally, we have $PY = IM$ and $PY \perp AC, OQ \perp AC \implies PY || OQ || MI$ which means $PYMI$ is an isosceles parallelogram, so $YMI$ is isosceles. Because $YMI \cong XOQ$ by ASA, $XOQ$ is isosceles which means by SAS $XOY \cong MYO$, and $XY = OM$ as desired. \(\blacksquare\) Remarks: I accidentally put grid lines in my geogebra diagram but deal with it .

[asy][asy] size(250); pair A = (4.0103415548596,10.5702321209724), B = origin, C = (14, 0); pair O = circumcenter(A, B, C); pair H = orthocenter(A, B, C), D = extension(A, incenter(A, B, C), B, C); pair X1[] = intersectionpoints(A--D, circumcircle(A, H, C)); pair P = X1[1]; pair X = circumcenter(A, B, P), Y = orthocenter(A, C, P), HB = 2*foot(H, A, C)-H; pair O1 = circumcenter(A, C, P); draw(X--O--O1--cycle, heavygreen); draw(A--B--C--cycle); draw(B--HB, blue); draw(A--P, red); draw(B--P, red); draw(C--P, red); draw(C--Y--X, orange); draw(C--O1, orange); draw(O--Y, dashed); draw(circumcircle(A, B, C), magenta); draw(circumcircle(C, H, A), purple); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$P$", P, S); dot("$H$", H, S); dot("$H_B$", HB, NE); dot("$X$", X, NW); dot("$Y$", Y, SW); dot("$O_1$", O1, NE); dot("$O$", O, SE); [/asy][/asy] First, we claim that $Y$ lies on $(ABC)$. First, since $Y$ is the orthocenter of $(APC)$ by definition, the reflection of $Y$ over $\overline{AC}$ lies on $(APC)$. However it is well-known that $(APC) = (AHC)$ and $(ABC)$ are reflections over $\overline{AC}$. Hence, $Y$ must lie on $(ABC)$, proving the claim. $\blacksquare$ Now, the crux of the problem is the following Claim. Let $O_1$ be the circumcenter of $(AHC)$. Then $OXO_1$ is isosceles. Proof. Because $X$ lies on the perpendicular bisector of $\overline{AB}$, it follows that $\overline{OX} \perp \overline{AB}$. Notice also that $\overline{O_1X} \perp \overline{AP}$, because $\overline{AP}$ is the radical axis of $(APB)$ and $(APC)$, and $\overline{OO_1} \perp \overline{BC}$ because $(ABC)$ and $(ACH)$ are reflections over $\overline{AC}$. Now notice that $$\measuredangle OXO_1 = 90^{\circ} - \measuredangle(AP, XO) = \measuredangle BAP$$by our two perpendicularities, while $$\measuredangle XO_1O = 90^{\circ} - \measuredangle(OO_1, AC) = \measuredangle PAC.$$But $P$ lies on the angle bisector, so these angles are equal, proving the claim. $\blacksquare$ To finish, this implies that the perpendicular bisector of $\overline{O_1X}$ passes through $O$. But because $\overline{XO_1} \perp \overline{AP}, \overline{CY} \perp \overline{AP}$ by the orthocenter definition, while the perpendicular bisector of $\overline{CY}$ also passes through $O$, it follows that the perpendicular bisectors of these two lines concur. However, this means that $XYCO_1$ is an isosceles trapezoid, so $O_1C = XY$. But since $(AHC)$ and $(ABC)$ are congruent, $$OY=O_1C=XY,$$finishing the problem. $\square$

Here's a walkthrough (outline) to my solution. (a) Let $Y$ instead be a point on arc $ABC$ such that $YP\perp AC$ (b) Angle chase to confirm that $Y$ is actually the orthocenter of $\triangle APC$, this also follows from the fact that $(AHC)$ is the reflection of $(ABC)$ over $AC$ (c) Angle chase to notice that $\triangle YSM_A$ is isoceles (d) Let $X$ instead be the point such that $XOM_AY$ is a parallelogram, where $O$ is the center of $(ABC)$, $M_A$ the midpoint of minor arc $BC$ (e) Angle chase to notice that $YM_A\perp AB$, note that this also gives $X$ lies on the perpendicularbisector of $\overline{AB}$ (f) Drop perpendiculars from $X,Y$ to $AM_A$, do some ratio chasing by introducing $L_A$, the midpoint of arc $BAC$. (g) Conclude that $X$ also lies on the perpendicularbisector of $\overline{AP}$, thus we have $X$ is the center of $(ABP)$.

Something same to mathpirate Let $AP$ intersect $(ABC)$ at $M$. We claim that $Y$ lies on $(ABC)$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 10., ymin = -10.5, ymax = 2.; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); /* draw figures */ draw((0.8655617412366693,0.6216229568176963)--(-1.92,-4.78), linewidth(0.8)); draw((-1.92,-4.78)--(8.98,-4.76), linewidth(0.8)); draw((8.98,-4.76)--(0.8655617412366693,0.6216229568176963), linewidth(0.8)); draw(circle((3.5289007120376135,-4.17088806049879), 5.482840169518616), linewidth(0.8)); draw(circle((6.316661029199057,0.03251101731648583), 5.482840169518617), linewidth(0.8)); draw(circle((-0.9539487016641175,-1.8591284433689232), 3.076482693166482), linewidth(0.8)); draw((-0.9539487016641175,-1.8591284433689232)--(-0.9438884614073338,-7.341959383317089), linewidth(0.8) + ffxfqq); draw((0.8655617412366693,0.6216229568176963)--(4.710129680802359,1.1831977109320253), linewidth(0.8)); draw((4.710129680802359,1.1831977109320253)--(8.98,-4.76), linewidth(0.8)); draw((-0.9438884614073338,-7.341959383317089)--(8.98,-4.76), linewidth(0.8)); draw((-0.9438884614073338,-7.341959383317089)--(4.710129680802359,1.1831977109320253), linewidth(0.8)); draw((0.8655617412366693,0.6216229568176963)--(3.538960952294398,-9.653719000446957), linewidth(0.8)); draw(circle((-0.1282848169802912,-3.7524012016009203), 2.0654787792506077), linewidth(0.8)); draw(circle((2.3333146277043664,-6.489309204924442), 3.3863066036607337), linewidth(0.8)); draw((-0.9469953755773727,-5.648691160646276)--(1.8438718557541116,-3.1385603055018167), linewidth(0.8)); draw((-0.9539487016641175,-1.8591284433689232)--(-1.92,-4.78), linewidth(0.8)); draw((-1.92,-4.78)--(1.8438718557541116,-3.1385603055018167), linewidth(0.8)); draw((1.9702516157285461,1.0857410227694846)--(-1.92,-4.78), linewidth(0.8)); draw((-0.9539487016641175,-1.8591284433689232)--(3.5289007120376135,-4.17088806049879), linewidth(0.8) + ffxfqq); draw((3.5289007120376135,-4.17088806049879)--(3.538960952294398,-9.653719000446957), linewidth(0.8) + ffxfqq); draw((3.538960952294398,-9.653719000446957)--(-0.9438884614073338,-7.341959383317089), linewidth(0.8) + ffxfqq); /* dots and labels */ dot((0.8655617412366693,0.6216229568176963),dotstyle); label("$A$", (0.3729002823643257,1.0265932261440223), NE * labelscalefactor); dot((-1.92,-4.78),dotstyle); label("$B$", (-2.4152150566409243,-5.7012503092816855), NE * labelscalefactor); dot((8.98,-4.76),dotstyle); label("$C$", (9.383765798279844,-5.054730810381917), NE * labelscalefactor); dot((3.5289007120376135,-4.17088806049879),dotstyle); label("$O$", (3.6054977768631664,-3.9637291559885597), NE * labelscalefactor); dot((0.8677603171614444,-0.576600922184725),dotstyle); label("$H$", (0.5143264227486499,-1.3776511603894888), NE * labelscalefactor); dot((1.8438718557541116,-3.1385603055018167),dotstyle); label("$P$", (3.1004044183477224,-2.670690158189024), NE * labelscalefactor); dot((-0.9438884614073338,-7.341959383317089),dotstyle); label("$Y$", (-2.3141963849378357,-7.2569378535092515), NE * labelscalefactor); dot((-0.9539487016641175,-1.8591284433689232),dotstyle); label("$X$", (-1.122176058841388,-1.5796885037956663), NE * labelscalefactor); dot((3.538960952294398,-9.653719000446957),dotstyle); label("$M$", (3.5852940425225484,-10.186479332898823), NE * labelscalefactor); dot((4.710129680802359,1.1831977109320253),dotstyle); label("$P'$", (4.83792557164085,1.5518903190000835), NE * labelscalefactor); dot((-0.9469953755773727,-5.648691160646276),dotstyle); label("$R$", (-0.718101372029033,-6.246751136478364), NE * labelscalefactor); dot((-0.9485925747692824,-4.778217601054623),dotstyle); label("$S$", (-0.8595275124133572,-4.569841186207092), NE * labelscalefactor); dot((1.9702516157285461,1.0857410227694846),dotstyle); label("$H'$", (2.049810232635599,1.2892417725720529), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that \begin{eqnarray*} \measuredangle PAC+\measuredangle PCA=\measuredangle P'AB+\measuredangle CBP'=\measuredangle CBA \end{eqnarray*}Which means $Y$ lies on $(ABC)$. Clearly $OM\perp BC$ and $OX\perp AB$. Now $$\measuredangle MYC=\measuredangle MAC=\measuredangle PHA+90^\circ -\measuredangle CBA=\measuredangle HCP+\measuredangle PCY=\measuredangle ACY$$Which means $MY\perp AB$. Now we if we can prove that $XY\perp BC$ then we are basically done. Because then $MYXO$ will be a parallelogram and we get our desired result. Now suppose that $R=(OBX)\cap MYP$. Then an simple angle chase gives that $$\measuredangle PRY=\measuredangle PMY=\measuredangle AP'P=\measuredangle PBX=\measuredangle PRX$$Therefore $X,R,X$ collinear and similarly $B,R,M$ collinear. And another angle chasing gives that $$\measuredangle SRB+\measuredangle SBR=\measuredangle XPB+\measuredangle MBC=90^\circ$$And we are done. $\square$

Let $O_1$ be the center of $(AHC).$ Notice $(ABC)$ and $(AHC)$ are reflections over $\overline{AC},$ and since the reflection of $Y$ over $\overline{AC}$ lies on $(AHC),$ we see $Y$ lies on $(ABC).$ Notice $\overline{AB}\perp\overline{OX},\overline{AP}\perp\overline{O_1X},$ and $\overline{AB}\perp\overline{OO_1}.$ Hence, $$\measuredangle XO_1O=90-\measuredangle(\overline{XO_1},\overline{AC})=\measuredangle PAC=\measuredangle BAP=90-\measuredangle(\overline{AP},\overline{XO})=\measuredangle OXO_1.$$But $$\angle AOY=2\angle ADY=180-2\angle DYC=180-\angle A=\angle O_1OX$$so $\triangle OAY\sim\triangle OO_1X$ and $O$ is the center of the spiral similarity $\overline{XY}\mapsto\overline{O_1A}.$ Therefore, $\triangle AOO_1\cong\triangle YOX.$ $\square$

Let the midpoints of $AB$ and $AC$ be $M$ and $N$ respectively, the circumcenter of $ABC$ be $O$, the reflections of $O$ and $Y$ over $AC$ be $O_B$ and $Y_1$ respectively, $K = AP \cap OM$, and $T = AP \cap O_BX$. The Orthocenter Reflection Lemma implies $(ABC)$ and $(AHC)$ are symmetric about $AC$, so $O_B$ is the center of $(AHC)$. Now, because the aforementioned lemma yields $Y_1 \in (AHPC)$, we have $Y \in (ABC)$. Since $AP$ is the common chord of $(APB)$ and $(AHPC)$, we know $AP \perp O_BX$. Thus, because $XA = XB$ gives $X \in OM$, $$\angle XOO_B = \angle MON = 180^{\circ} - \angle A$$and $$\angle OXO_B = \angle KXT = 90^{\circ} - \angle XKT = \angle MAK = \frac{\angle A}{2}$$which means $OO_B = OX$. Let $l$ be the perpendicular bisector of $O_BX$. Now, since $OC = OY$ and $$CY \perp AP \perp O_BX$$follows from the orthocenter condition, we know $C$ and $Y$ are symmetric about $l$. Hence, reflection properties imply $$XY = O_BC = OC$$as desired. $\blacksquare$

Claim 1: $Y\in (ABC)$ Proof: Note that the circle $(AHC)$ is the reflection of $(ABC)$ over line $AC$. Thus, $P'$, the reflection of $P$ over $AC$, lies on $(ABC)$. Since $P$ is the orthocenter of $AYC$, we also have that $Y,A,C,P'$ is cycle, which finally shows that $Y\in (ACP') =(ABC)$ $\square$ Claim 2: $\triangle OXO_1$ is isosceles Proof: Note that \[\angle (OX,XO_1) = \angle (AB,AP) = \angle (AP,AC) = \angle (O_1X, OO_1)\]due to perpendicularity conditions $\square$ Claim 3: $YXO_1C$ is an isosceles trapezoid. Proof: Clearly, $XO_1\perp AP \perp YC\Longrightarrow XO_1\parallel YC$. Additionally, the perpendicular bisectors both pass through $O$ and are parallel, so they must be the same line, and therefore it is an isosceles trapezoid. $\square$ Thus, $XY = O_1C = R$ and we're done. $\blacksquare$.

this problem is so boring Let $M$ be the midpoint of arc $BC$ in $(ABC)$, and let $O$ be the circumcenter of $ABC$. We prove that $XYMO$ is a parallelogram, which finishes. Note that $ABYC$ is cyclic since $\angle ABC = 180 - \angle APC = \angle AYC$. Claim: $\overline{XO} \parallel \overline{MY}$. Proof. Note that $\overline{XO} \perp \overline{BC}$, so it is enough to show that $\overline{YM} \perp \overline{BC}$. Indeed, $\angle BAD = \frac{\angle A}{2}$, while $\angle YDA = \angle YCA = 90 - \frac{\angle A}{2}$ since $P$ is the orthocenter in $AYC$. Since $\angle BAD + \angle YDA = 90$, we are done. $\square$ Claim: $\overline{XY} \perp \overline{BC}$. Proof. Note that $\angle PYM = \angle PYC + \angle CYD = \angle A$, and $\angle PXB=2\angle PAB=\angle A$. Also note that $M$ is the reflection of $P$ over $YC$, hence $PYM$ is isosceles. Since $PXB$ is also isosceles, in fact we have $PXB \sim PYM$. Now by spiral similarity, $\overline{BM} \cap \overline{XY}$ is on $(PXB)$, $(PYC)$ and hence their angle between them is equal to $\angle XPB = 90 - \frac{\angle A}{2}$. Also, $\angle MBC = \frac{A}{2}$ and hence $\overline{XY} \perp \overline{BC}$. $\square$ To finish, note that $\overline{OM} \perp \overline{BC}$ and we are done since $XY$ is parallel to $OM$. $\blacksquare$

Due to orthocenter reflecting stuff, the circumcircle of $APHC$ is the reflection of the circumcircle of $\triangle ABC$ about $\overline{AC}$. Let $O_1$ and $O_2$ be the centers of $(ABC)$ and $(APC)$, respectively. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AP}$, respectively. Then, clearly, $X$ is the intersection of rays $O_1 M$ and $O_2 N$, since those are the perpendicular bisectors of $\overline{AB}$ and $\overline{AP}$. Since $\angle XMA = \angle XNA = 90^{\circ}$, it follows that $XMNA$ is cyclic. Letting $T$ be the intersection of $\overline{AC}$ and $\overline{O_1 O_2}$, we have $\angle ATO_2 = 90^{\circ}$ due to symmetry, so $\angle ANO_2 = \angle ATO_2 = 90^{\circ}$, so $ANTO_2$ is cyclic. Thus, \[ \angle O_1X O_2 = \angle MXN = \angle MAN = \angle CAN = \angle TAN = \angle TO_2 N = \angle O_1 O_2 X\]so $\triangle O_1 X O_2$ is isosceles with $O_1 X = O_1 O_2$ and $\angle O_1XO_2 = \angle O_1 O_2 X = \frac{\angle A}{2}$. Now, we claim that $\triangle O_1 YX \cong \triangle O_2 A O_1$.

gives us that $\angle YO_1 X = \angle AO_1 O_2$. Since $YO_1 = AO_2$, $O_1 X = O_2 O_1$ and $\angle YO_1 X = \angle AO_1 O_2$, we have $\triangle YO_1 X \cong \triangle AO_1 O_2$ by SAS congruence. Thus, $XY = AO_2 = AO_1 = R$.

this problem is middle of the line imo not too easy, not too hard, not too bashy, not too clean Let the center of $(ABC)$ be $O$ and the center of $(AHC)$ be $O'$. It is well known that $(ABC)$ and $(AHC)$ are reflections of each other, so $AO=AO'$. Note that \[\angle AYC = 180^\circ - \angle APC = 180^\circ - \angle AHC = \angle ABC,\] hence $Y$ lies on $(ABC)$. Also, $\overline{OX}$ and $\overline{OO'}$ are perpendicular to $\overline{AB}$ and $\overline{AC}$, respectively. Thus letting $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ give that $AMON$ is cyclic, which means $\angle MON = \angle XOO' = 180^\circ - \angle BAC$. Then, notice \[\angle AOY = 2 \angle ACY = 2 (90^\circ-\angle PAC) = 180^\circ - 2 \angle PAC = 180^\circ - \angle XOO'.\] This implies that rays $OO'$ and $OX$ can be rotated around $O$ to coincide with rays $OA$ and $OY$, respectively. Hence, we can rotate rays $OO'$ and $OA$ around $O$ to coincide with rays $OX$ and $OY$, giving $\angle AOO' = \angle XOY$. Finally, realize that \[\angle OXO' = \angle BAP = \angle CAP = \angle O'XO,\] so $\triangle OXO'$ is isosceles with $OX=OX'$. Combine this with the aforementioned angle condition and the fact that $OA=OY$, we get that $\triangle OXY \cong \triangle OO'A$; therefore, $XY=OA'=OA$. $\square$

Quite a fun problem! Trig bash ftw. First, notice that $\angle B = 180^{\circ} - \angle AHC = 180^{\circ}-\angle APC = \angle AYC$ so $(ABYC)$ cyclic. Now note that $X$ is simply the intersection of the perpendicular bisector of $AB$ with the perpendicular bisector of $AP$. In particular, $AP \perp YC$. Let $O'$ be the center of $(AHC)$ and the reflection of $O$ over $AC$, and let $M$ be the midpoint of $AC$. It follows that the projection in the direction of $CY$ from the three points $O, M, O'$ map to three points $O, X', X$ along the perpendicular bisector of $AB$, such that $X'$ is the midpoint of $OX$. It suffices to show that $XY = YO = R$, so we simply need to show that $YX' \perp OX \perp AB$, which is equivalent to $d(X, AB) = d(Y, AB)$. Now, we begin the trig bash. Let $M'$ be the midpoint of $AB$. Then by LoS in $\triangle MM'X$, we have $d(X, AB) = \sin \angle M'MX \cdot \frac{M'M}{\sin \angle M'XM} = \sin \left( \frac{B - C}{2} \right) \cdot R \cdot \frac{\sin A}{\sin \frac{A}{2}}$ Now by generalized LoS we have $d(Y, AB) = BY \cdot \sin \angle ABY = 2R \cdot \sin \left( \frac{B - C}{2} \right) \cdot \cos \frac{A}{2}$. Thus equating the two, it is equivalent to $\sin A = 2\sin \frac{A}{2} \cos \frac{A}{2}$, which follows from double angle.

Here's a complex solution very similar to Evan's (even down to the point names!) We will instead introduce the point $Q$, the reflection of $P$ over $\overline{AC}$, which lies on $(ABC)$. WLOG assume $(ABC)$ has unit radius. Observe that the bisector condition implies that $q^2 = \frac{c^3}b$ (say by considering the arc midpoint). Now, let $x$ and $y$ represent the complex numbers associated with the reflections of $X$ and $Y$ over $\overline{AC}$. We have $y = a+q+c$, and $$x = a+\frac{-(q-a)(b'-a)(\overline{q-a} - \overline{b'-a})}{(q-a)(\overline{b'-a}) - (\overline{q-a})(b'-a)}$$by the circumcenter formula, where $b'$ is the reflection of $B$ over $\overline{AC}$. It follows that \begin{align*} x-y &= q+c+\frac{-(q-a)\left(\frac 1c-\frac b{ac}\right)-\left(\frac 1q - \frac 1a\right)\left(c-\frac{ac}b\right)}{(q-a)\left(\frac 1c-\frac b{ac}\right) - \left(\frac 1q-\frac 1a\right)\left(c-\frac{ac}b\right)} \\ &= q+c+\frac{c(q-a)\left(\frac{ac}q-c-a+b\right)}{bq-ab+\frac{ac^2}q-c^2} \\ &= q+c + \frac{c(q-a)\left(\frac{ac}q-c-a+\frac{c^3}{q^2}\right)}{\frac{c^3}q-\frac{ac^3}{q^2}+\frac{ac^2}q - c^2} \\ &= q+c+\frac{c(q-a)(acq-cq^2-aq^2+c^3)}{c^2(c-q)(q-c)} \\ &= q+c-\frac{aq+c^2+cq}c \\ &= -\frac{aq}c. \end{align*}This clearly has magnitude $1$, as needed.

Let $N$ be the minor arc midpoint. First notice that $Y\in(ABC)$ since $\angle ABC=180-\angle AHC=180-\angle APC=\angle AYC$. Next $\angle CNP=\angle CBA=180-\angle CHA=180-\angle CPA=\angle CPN$, and $CY\perp AP$, so $CY$ perpendicularly bisects $PN$. Next if $Z=CY\cap AN$ then $\measuredangle ZYN=\measuredangle ZAB$ implies $(AZY)$ passes through $AB\cap YN$, so $AB\perp YN$ and thus $XO\parallel YN$, where $O$ is the circumcenter. Finally notice that the projections from $X,O$ to $AN$ have distance $\tfrac12PN=ZN$, and thus $XO\parallel YN$ implies $XO=YN$. Thus $XONY$ is a parallelogram and $XY=ON$ as desired.