$XP,XA$ meets $\odot ABC$ again at $E,D$,respectively.Easy to know $\angle ADI=\angle PEI=\frac{\pi}{2}$. Then if we let $K$ be a point on the ray $IO$ such that $IK=2IO$,we will have $\angle XYK=\angle XAK=\angle XPK=\frac{\pi}{2}$.$\Box$

Attachments:

We proceed by complex bashing. Identify the incircle of $\triangle ABC$ with the unit circle, and let it touch sides $\overline{BC},\overline{CA},\overline{AB}$ at $D,E,F$ respectively. Rename $I$ to $J$. Then $$|d|=|e|=|f|=1$$$$a = \frac{2ef}{e+f}$$$$b = \frac{2df}{d+f}$$$$c = \frac{2de}{d+e}$$$$j = 0$$$$o = \frac{2def(d+e+f)}{(d+e)(d+f)(e+f)}$$$$p = \frac{2def}{(d+e)(d+f)}$$Now the coordinate of $X$ satisfies $$\overline{x} = -\frac{x}{ef} = \frac{2d-x}{d^2} \implies x = \frac{2def}{ef-d^2}$$and then we find the coordinate of $Y$ as $$y = \frac12\left(x+\frac{o\overline{x}}{\overline{o}}\right) = \frac{def}{de+df+ef}$$We have the vectors \begin{align*} a-p &= \frac{2ef(d-e)(d-f)}{(d+e)(d+f)(e+f)} \\ y-x &= \frac{def(d^2+2de+2df+ef)}{(d^2-ef)(de+df+ef)} \\ a-y &= \frac{ef(de+df+2ef)}{(e+f)(de+df+ef)} \\ p-x &= \frac{2d^2ef(2d+e+f)}{(d+e)(d+f)(d^2-ef)} \end{align*}Then $$\frac{(a-p)(y-x)}{(a-y)(p-x)} = \frac{(d-e)(d-f)(d^2+2de+2df+ef)}{d(2d+e+f)(de+df+2ef)}$$which is real. $\blacksquare$

Point Circles! Let $Z$ be on $AP$ s.t. $ZI^2 = ZA \cdot ZP$, now clearly $Z,X$ lie on the radical axis of of point circle $I$ and $(ABC)$ and therefore $OI \perp XZ$, so now let $OI \cap XZ = Y’$ then we have $Y’$ on $OI$ and also $\measuredangle XY’I = 90^{\circ}$ therefore $Y \equiv Y’$ and we’re done by power of point on $Z$

egxa wrote: In triangle $ABC$, the incenter is $I$ and the circumcenter is $O$. Let $AI$ intersects $(ABC)$ second time at $P$ . The line passes through $I$ and perpendicular to $AI$ intersects $BC$ at $X$. The feet of the perpendicular from $X$ to $IO$ is $Y$. Prove that $A,P,X,Y$ cyclic.

Attachments:

Let $K=I \perp AX$ and $M=I \perp BX$. So $IMYXK$ is cyclic,now we can prove two things: 1.$AKCB$ is cyclic. Proof:$\angle BIM=\angle PIC \implies \angle IBC=\angle CIX$ so $IX^2=XC\cdot XB$,at the same time $IX^2=XK\cdot XA \implies AKCB$ is cyclic. 2.$P,M,K$ are collinear. Proof:It's well known that $PA \cap BC ,A,M,K$ are concyclic(if $P,M,K$ are collinear),rest is easy angle chasing. If we prove that $\angle KYO=\angle OYP$ we're done,but $OP \parallel IM \implies OPYK$ cyclic,$OP=OK$.

Any solution with isogonal lines ?

math_comb01 wrote: Point Circles! Let $Z$ be on $AP$ s.t. $ZI^2 = ZA \cdot ZP$, now clearly $Z,X$ lie on the radical axis of of point circle $I$ and $(ABC)$ and therefore $OI \perp XZ$, so now let $OI \cap XZ = Y’$ then we have $Y’$ on $OI$ and also $\measuredangle XY’I = 90^{\circ}$ therefore $Y \equiv Y’$ and we’re done by power of point on $Z$ Nice use of point circles... I couldn't prove that Z was on XY. Could you explain why X is on the radical axis of circle I and (ABC)?

Lemma: In triangle $ABC$, $D,E,F$ are altitudes from $A,B,C$ respectively and $H$ is the orthocenter. $M$ is the midpoint of $BC$ and $O$ is the circumcenter of $(ABC)$. $OM$ intersects $(DEFM)$ at $M,S$. If $K$ is the perpendicular from $O$ to $AD$, then $SMKH$ is an isosceles trapezoid. Proof: Let $N_9$ be the circumcenter of the nine point circle. $N_9$ is the midpoint of $OH$. $N_9$ lies on the perpendicular bisector of $SM$ since $S,M\in (N_9)$. Also $HK\perp KO$ so $N_9$ is the circumcenter of $(HKO)$. Hence $N_9$ also lies on the perpendicular bisector of $HK$. And $HK\perp BC\perp OM$ gives that $HK\parallel SM$ so $SMKH$ is an isosceles trapezoid. Let $D,E,F$ be the perpendiculars from $I$ to $BC,CA,AB$ respectively. Let $S$ be the Sharky-Devil Point where $X,S,A$ are collinear. Let's invert from $I$ with radius $ID$. It's well known that the perpendicular from $D$ to $EF$, $I$ and $S$ are collinear. $(DISYX)\rightarrow DX^*Y^*S^*$ which is the perpendicular from $D$ to $EF$. $(ABCP)\rightarrow (A^*B^*C^*P^*)$ so $P^*$ lies on the nine point circle of $DEF$. $IO$ is the euler line of $DEF$ and $Y^*$ lies on the perpendicular from $D$ to $EF$ thus $Y^*$ is the orthocenter of $DEF$. We should prove that $X^*Y^*A^*P^*$ is cyclic. It's true from lemma.

SBYT wrote: we will have $\angle XYK=\angle XAK=\angle XPK=\frac{\pi}{2}$.$\Box$ Can you please explain how?

$XI$ is a tangent line on $\odot BIC$.$XI^2=XB\cdot XC=XD\cdot XA=XE\cdot XP$.So $\angle XEI=\angle XDI=\frac{\pi}{2}$. Then by $ADIK$ and $IEPK$ are both right angled trapezoid.So $\angle XAK=\angle XPK=\frac{\pi}{2}$.

Thanks a lot

Solved with GrantStar: Let $R$ be the second intersection of $\overline{XA}$ with $(ABC)$. By radical axis on $(ARI)$, $(BIC)$ and $(ABC)$ we have that $\overline{XI}$ is the radical axis of $(ARI)$ and $(BIC)$, hence $\angle IRA = 90^{\circ}$. Similarly, let $T$ be the second intersection of $\overline{XP}$ with $(ABC)$. By radical axis on $(ITP)$, $(IBC)$ and $(ABC)$ it follows that $IX$ is the radical axis of$(IBC)$ and $(ITP)$, hence $\angle ITP = 90^{\circ}$. Let $I'$ be the reflection of $I$ over $O$. Since the perpendicular bisector of $\overline{AR}$ passes through $O$, the perpendicular to $\overline{AR}$ through $A$ passes through $I'$. Similarly, the perpendicular to $\overline{TP}$ passes through $I'$. So, $APXY$ is cyclic with diameter $XI'$.

Let $S$ and $D$ be the foot of the perpendiculars from $I$ to lines $AX$ and $BC$ resperctively. $Claim 1$: $S$ is the $A-SharkyDevil$ of triangle ABC. Since $AX$ is perpendicular to $AP$ and $P$ is the center of circle $(BIC)$ $XI^2=XB.XC$ Similarly $XI$ is tangent to the circle with diameter $AI$ $XI^2=XS.XA$ Thus $XB.XC=XS.XA$ so $S$ lies on $(ABC)$ hence $S$ is the $A-SharkyDevil$ of triangle ABC. $Claim 2$: It is known that points $S,D,P$ are collinear(properties of the $SharkyDevil$ point ) $Claim 3$: $S,I,D,Y,X$ are concyclic $\angle$$ISX$ $+$ $\angle$$IDX$$=90+90=180$ thus $S,I,D,X$ are concyclic $\angle$$XDI$$=90=$$\angle$$XYI$ thus $X,Y,D,I$ are concyclic $Claim4$: $S,O,P,Y$ are concyclic Since $ID$ is parallel to $OP$ $\angle$$YOP$$=$$\angle$$YID$$=$$\angle$$YSD$$=$$\angle$$YSP$ $Claim5$: $A,X,Y,P$ are concyclic Since $S,O,P,Y$ are concyclic and $OS=OP$ $\angle$$SXI$$=$$\angle$$SYI$$=$$\angle$$SYO$$=$$\angle$$OYP$$=$$\angle$$IYP$ and $\angle$$SXI$$=$$90-$$\angle$$XAP$ and $\angle$$XYP$$=$$\angle$$IYP$$+90$ hence $\angle$$XYP$$=180-$$\angle$$XAP$ $Q.E.D$

Nice geo : Let $AI$ and $XY$ meet at $Z$ and let $w$ be the circumcircle of $ABC$. Claim: $XI^2=pow(X,w)$ Consider the circle centered at $P$ passing through $B$, $I$, and $C$. Notice $XI\perp AP$ so $XI$ is tangent to this circle. Then $XI^2=XB\cdot XC=pow(X,w)$. Claim: Line $XY$ is the radical axes of $I$ and $w$ Since $pow(X,I)=XI^2=pow(X,w)$ and $XY\perp IO$ the result follows. Claim: The points $A$, $P$, $X$, and $Y$ lie on a circle It is sufficient to show that $ZP\cdot ZA=ZX\cdot ZY$. Since $Z$ lies on the radical axes of $I$ and $w$ we get that $ZI^2=ZP\cdot ZA$. But triangle $ZIY$ and $ZXI$ are similar so $ZI^2=ZX\cdot ZY$. The result follows.

Attachments:

utterly stupid solution: Let $XY, AP$ concur at $Z$. Then $\triangle XZI \sim \triangle IZY$, giving $ZX \cdot ZY = ZI^2$, so we want to prove $ZI^2 = ZA \cdot ZP$, equivalent to showing $pow (Z, (O)) - pow (Z, (I)) = r^2 - R^2$. By linearity of power of a point, and noting that $XY$ is parallel to the radical axis of the incenter and circumcenter, it suffices to show that $X$ also satisfies this power relation. However, it is trivial to note that $XI^2 = XB \cdot XC$ since $XI$ is tangent to $(BIC)$ by angle chasing, this finishes.

Let $Q$ be the intersection of lines $XY$ and $AP$. Since $\angle XYI = \angle XIA = 90^{\circ}$, it follows that $QY \cdot QX = QI^2$. Since line $XY$ is the radical axis of $(ABC)$ and $I$, it then follows that $QX \cdot QY = QA \cdot QP$, implying $APXY$ cyclic.

I will use #2's diagram. Let midpoint of major arc $BAC$ is $N_a$. Radical axis on $(DYXE),(ABC),(APX)$ we should show that $XY,DE,AP$ concurrent. Miquel of complate quadrilateral $ADEP$. We should show that $O,I,AE \cap DP$ collinear and this is Pascal on $PN_aEAA'D$ where $A'$ is a antipode.

I will provide two different solutions from similarity and linearity of PoP + Coaxality Lemma. Let $DEF$ be the orthic triangle of $ABC$. Let $A'$ be the reflection of $A$ across $IO$ and $I'$ be the reflection of $I$ across $BC$. It is known that $P, A',I'$ are collinear and $AIO \sim AI'P$. $Q$ is the antipode of $P$ in $(ABC)$. $G$ is the midpoint of $BC$, $S = QI \cap (ABC)$. $A'_1$ is the reflection of $A$ across $QP$. $M$ is the midpoint of $OX$. Solution 1 Claim 1: $IYXD \sim QA'PA'_1$ Proof. $\angle YIX = \angle A'AP = \angle A'QP \implies PA'Q \sim XYI$, $\angle QPA'_1 = \angle QPA = \angle DIP = \angle IXD \implies QA'_1P \sim IDX$. Combining gives the desired result.$\square$ Claim 2: $AA'Q \sim IA'O$ Proof. $\angle IOA' = \angle APA' = \angle AQA'$ and $\angle AA'Q = \angle IAO = \angle IA'O$.$\square$ Claim 3: $S \in PX$ Proof. There are various ways to do this. $SBI \sim SIC \implies \frac{SB}{SC} = \frac{IB^2}{IC^2} = \frac{XB}{XC}$ which gives the desired result due to ex-angle bisector theorem.$\square$ Main Claim: $AIY \sim I'PG$ Proof. $\angle AIY = 180 - \angle AIO = 180 - \angle II'P = \angle GPI'$. Note that $\frac{ID}{IY} \overset{1}= \frac{QA'_1}{QA'} = \frac{QA}{QA'} \overset{2} = \frac{OI}{R}$ $$\frac{PI'}{PG} = \frac{AI}{IY} \iff \frac{PI'}{IO} \cdot \frac{IO}{PG} = \frac{AI \cdot OI}{ID \cdot R} \iff \frac{PI}{AO} \cdot \frac{IO}{PG} = \frac{AI \cdot OI}{ID \cdot R} \iff \frac{AI}{IF} = \frac{PC}{PG}$$which is true because $AIF \sim CFG$.$\square$ Now, $\angle AYX = 90 + \angle AYI = 90 + \angle I'GP = 90 + \angle DIG = \angle AD'C = 180 - \angle APS$ as desired. Solution 2 Because $IYXD$ and $OYXG$ are cyclic, it is sufficient to prove that $\frac{AI^2}{PI^2} = \frac{AM^2-MO^2}{PG \cdot PO}$. $AM^2 - MO^2 = \frac{1}{2} (AO^2+AX^2 - XO^2) = \frac{1}{2} (AO^2+AI^2 + IX^2 - XO^2) = \frac{1}{2} (AO^2+AI^2 + IX^2 - GO^2 - GX^2) =$ $ \frac{1}{2} (AO^2+AI^2 + PG^2 - GO^2 - PG \cdot PQ) = \frac{1}{2} (AO^2+AI^2 - GO^2 - GB^2) = \frac{1}{2} AI^2$. On the other hand, $\frac{AI^2}{PI^2} \cdot PG \cdot PO = \frac{AI^2}{PI^2} \cdot \frac{1}{2}PI^2 = \frac{1}{2} AI^2$ as desired.

Let $XY$ meet $AP$ at $K$. We are going to show $KP.KA = KY.KX$ to finish the problem. Let $R$ be the radius of $(ABC)$. By power of a point: $KP.KA = KO^2 - R^2$. Also we know that $XI$ is tangent to $(BIC)$ thus: $XI^2 = XB.XC$. Also $XB.XC = XO^2 - R^2$. Substract these two equations: $KP.KA - XI^2 = KO^2 - XO^2$. Now, we know that $OI \perp XK$. Hence: $KO^2 - XO^2 = KI^2 - XI^2$. So we conclude: $KP.KA - XI^2 = KI^2 - XI^2 \implies KP.KA = KI^2$. By euclid, $KI^2 = KY.KX$ hence: $KP.KA = KY.KX$ and we are done. $\blacksquare$

Similar to some solutions... From angle chasing we got $XI^2=XB \cdot XC$. That means $X$ lies on radical axis of circles $(I)$ and $(ABC)$. $OI$ perpendicular to $XY$ means $XY$ is radical axis line of circles $(I)$ and $(ABC)$. Let $XY \cap AP$ at $K$. It is obvious that $KA \cdot KP=KI^2$.From euclid theorem for right angles $KI^2=KY \cdot KX$.Which means $APXY$ is cyclic.