Sketch for a): $AI_{A}'$ passes through $U-$ antigonal conjugate of $I,$ but since isogonal conjugate of $U$ (say $V$) is nothing but inverse of $I$ WRT $\odot(ABC) \implies AV \equiv l_a$ passes through inverse of $I$ hence the conclusion.

Lemma: Let $\odot(A), \odot(B), \odot(C)$ be circles in the plane such that $\odot(C)$ is the inverse of $\odot(B)$ with respect to $\odot(A)$. Then for every point $P$, $B(A(P))=A(C(P))$, where $X(P)$ denotes inversion about $X$. Notice that $X_A, I_A'$ are $\sqrt{bc}$ inverses since the circumcircle of triangle $AI_A'I_A$ passes through the reflection of $A$ over $BC$. Then by the above lemma with $\odot(A)$ $\sqrt{bc}$ inversion, $\odot(B)$ inversion in $\odot(ABC)$, and $P\equiv I_A$ (notice here that inversion in $\odot(C)$ degenerates to reflection in $BC$), we know that $X_A$ is the inverse of $I$ with respect to $\odot(ABC)$, solving part (a). Hence it suffices to show for part (b) that $\angle XOY=\tfrac{2\pi}{3}$ since quadrilateral $XYOI$ is cyclic. Compute $$d(O,XY)=r\cdot \frac{PO}{PI}=r\cdot \frac{R^2}{R^2-OI^2}=\frac{R}{2}$$as desired.

The following ingenious elementary solution is due to Wanlin Li. We will rephrase part (a) as follows: in triangle $ABC$ with orthic triangle $DEF$, if we let $A'$ be the reflection of $A$ across $EF$ then we wish to show that the $D$-isogonal of $\overline{DA'}$ passes through a common point on the Euler line with the analogous isogonals to $\overline{EB'}$, $\overline{FC'}$. Let $O$, $H$ be as usual. Actually, it is a well-known result that the circumcircles of $AOD$, $BOE$, $COF$ meet at a common point $P$ on line $OH$. So we will show that $\overline{DA}$ bisects $\angle PDA'$. Indeed, we note that $DHOA'$ is cyclic (by power of a point from $A$) and then observe \[ \measuredangle ADA'= \measuredangle HDA' = \measuredangle HOA = \measuredangle POA = \measuredangle PDA. \] [asy][asy] size(10cm); pair A = dir(115); pair B = dir(208); pair C = dir(332); pair D = foot(A, B, C); pair E = foot(B, C, A); pair F = foot(C, A, B); pair H = orthocenter(A, B, C); draw(unitcircle, lightblue); pair O = circumcenter(A, B, C); pair Z = foot(A, E, F); pair Ap = 2*Z-A; draw(A--B--C--cycle, lightblue); draw(D--E--F--cycle, lightblue); pair N = midpoint(H--O); pair M = midpoint(A--H); draw(A--D, lightblue); draw(circumcircle(D, E, F), heavycyan); pair P = -O+2*foot(circumcenter(A, O, D), O, H); draw(circumcircle(A, O, D), dashed+red); draw(circumcircle(Ap, O, D), dashed+red); draw(O--P, heavygreen); draw(P--D--Ap, heavygreen); draw(A--Ap, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(250)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$H$", H, dir(315)); dot("$O$", O, dir(350)); dot("$A'$", Ap, dir(Ap)); dot("$N$", N, dir(270)); dot("$M$", M, dir(M)); dot("$P$", P, dir(P)); /* TSQ Source: A = dir 115 B = dir 208 C = dir 332 D = foot A B C R250 E = foot B C A F = foot C A B H = orthocenter A B C R315 unitcircle 0.1 lightcyan / lightblue O = circumcenter A B C R350 Z := foot A E F A' = 2*Z-A A--B--C--cycle 0.1 lightcyan / lightblue D--E--F--cycle lightblue R220 N = midpoint H--O R270 M = midpoint A--H A--D lightblue circumcircle D E F 0.1 lightcyan / heavycyan P = -O+2*foot circumcenter A O D O H circumcircle A O D 0.1 lightred / dashed red circumcircle Ap O D 0.1 lightred / dashed red O--P heavygreen P--D--Ap heavygreen A--Ap lightblue */ [/asy][/asy] As for (b), by the Poncelet Porism from earlier we just need to show that $H$ is the inverse of $P$ with respect to $(DEF)$ (since then $PH \cdot PN$ is equal to the power). But $\overline{MN} \parallel \overline{AO}$, so $PDNM$ is cyclic; so inversion swaps $\overline{MD}$ and $(PMND)$ as needed.

Here is also a solution with barycentric coordinates. First, we calculate the barycentric coordinates of $P$. We have $I_A = (-a:b:c)$ and the foot from $I_A$ to $\overline{BC}$ has coordinates $(0:s-b:s-c)$ and so we deduce \begin{align*} I_A' &= (a^2 : 4(s-a)(s-b) - ab : 4(s-a)(s-c) - ac) \\ &= \left( a^2 : c^2-a^2-b^2+ab : b^2-a^2-c^2+ac \right). \end{align*}By now it is evident that $\overline{AI_A'}$, $\overline{BI_B'}$, $\overline{CI_C'}$ concur, and taking the isogonal conjugate we obtain \[ P = \left( - : - : c^2(2S_C-ab) \right). \]Since $O = (- : - : c^2S_C)$ and $I = (a:b:c)$ it follows already that $P \in \overline{OI}$. It remains to show $\angle XIY = 120^{\circ}$. We will actually show the following lemma holds for any chord $XY$ tangent to the incircle: Lemma: $\angle XIY = 120^{\circ}$ if and only if $\angle XOY = 120^{\circ}$ if and only if $XOIY$ is cyclic. Proof. By Poncelet Porism we can draw a point $Z$ on the circumcircle such that $\triangle XYZ$ has the same incircle as $\triangle ABC$. Then $\angle XIY = 90^{\circ} + \frac{1}{2} \angle Z$ and $\angle XOY = 2 \angle Z$, and all parts are equivalent to $\angle Z = 60^{\circ}$. $\blacksquare$ So, we would be done upon showing that \[ PO \cdot PI = PX \cdot PY. \]One can finish by computing lengths, but a trick due to Anant Mudgal is to instead show that $P$ lies on the radical axis of $(AIO)$ and $(ABC)$. If $(AIO)$ has equation $-a^2yz-b^2zx-c^2xy + (x+y+z)(vy+wz) = 0$, we get \begin{align*} abc &= bv + cw \\ (abc)^2 \cdot (8K^2) &= (16K^2)\left( b^2S_B v + c^2S_C w \right) \end{align*}whence Cramer's rule gives \begin{align*} -v \div w &= \det \begin{bmatrix} c & abc \\ c^2 S_C & \frac{1}{2} (abc)^2 \end{bmatrix} \div \det \begin{bmatrix} b & abc \\ b^2 S_B & \frac{1}{2} (abc)^2 \end{bmatrix} \\ &= \det \begin{bmatrix} c & 2 \\ c^2 S_C & abc \end{bmatrix} \div \det \begin{bmatrix} b & 2 \\ b^2 S_B & abc \end{bmatrix} \\ &= \left( c^2(2S_C-ab) \right) \div \left( b^2(2S_B-ac) \right) \end{align*}Hence the radical axis $vy+wz$ passes through $P$.

cjquines0 wrote: Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_A$ be the $A$-excentre, $I'_A$ be the reflection of $I_A$ in $BC$, and $l_A$ be the reflection of line $AI'_A$ in $AI$. Define points $I_B$, $I'_B$ and line $l_B$ analogously. Let $P$ be the intersection point of $l_A$ and $l_B$. Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$. Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\angle XIY = 120^{\circ}$. Let $I_A$ be the $A$-excenter, $P_A$ be the reflection of $I_A$ in $BC$, $A'$ be the touching point of the $A$-excircle on $BC$, $T$ the exsimilicenter of $\odot(I)$ and $\odot(O)$. Note that $AA', BB', CC'$ concur at the Nagell Point $N$ of triangle $ABC$ and $T, N$ are isogonal conjugates. Project $-1=(I_A, P_A, A', \infty)$ from $A$ onto $OI$ after reflection in $AI$. We conclude that if $S=AP_A \cap OI$ then $(O, T, I, S)=-1$. Evidently, $S$ is the common point of lines $OI, AP_A, BP_B, CP_C$ proving part a.). Note that $$\frac{r}{R-r}=\frac{VI}{IO}=\frac{VS}{SO} \implies OS=\frac{R}{R-2r} \cdot OI=\frac{R^2}{OI},$$since $OI^2=R(R-2Rr)$ where $R, r$ are the circumradius and inradius respectively. Hence, $\{S, I\}$ are inverses in $\odot(O)$, consequently, $SX \cdot SY=\text{Pow}(S, \odot(O))=SI \cdot SO,$ hence, $X, I, Y, O$ are concyclic. It suffices to show $\angle XOY=120^{\circ}$. To see this, let $d$ be the distance between $O$ and $XY$; then, $$\frac{d}{r}=\frac{SO}{SI}=\frac{\frac{R^2}{OI}}{\frac{R^2}{OI}-OI}=\frac{R}{2r} \implies d=\frac{R}{2},$$hence, $\angle XOY=120^{\circ}$ as desired. $\blacksquare$

My solution $(a)$ : Let $P_a$ be the inversion image of $I_a'$ from the inversion $I^A_{AB.AC} . R_{AI_a}$ . Let $A'$ reflects with $A$ about $BC$ Because $AA'I_aI_a'$ is an isoceles trapezoid so it's cyclic. Then , through the inversion, we get $P_a$ lies on $OI$. Similary we have $P_b,P_c$ also lie on $OI$ . Then, we need to prove $P_a$ is also the inversion image of $I_b'$ through the inversion centre $B$ It's equivalent to prove $\frac{BC}{BI_b'} = \frac{BP}{BA}$. From above we almost have $ \frac{BP}{BA} = \frac{CI_a'}{AI_a'}$. So, need to prove $\frac{BC}{I_aC} = \frac{BI_b'}{AI_a'}$ We have $\frac{AI_a'}{IO} = \frac{A'I_a}{IO} = \frac{AI_a}{R} \implies AI_a' = \frac{OI.AI_a}{R} \implies \frac{AI_a'}{BI_b'} = \frac{AI_a}{BI_b} = \frac{I_cI_b}{I_cI_a} = \frac{CB}{CI_a}$ as desire So , $P_a \equiv P_b \equiv P_c \equiv P$ lies on $OI$ $(b)$ From part $(a)$ we have $AI_a.AP = AB.AC$ so $\triangle I_a'AI_a \sim \triangle I_aAP$ Let $AP$ cuts $(O)$ at $G$, $AI_a'$ cuts $(O)$ at $L$. Then $\angle OGA = 90 - \angle ALG = \angle AI_a'I_a \implies \triangle OGP \sim \triangle AI_a'I_a \implies \triangle OGP \sim \triangle AIP$. Then , we have $AI.AO = AG.AP = AX.AY$ So , $XYOI$ is cyclic . Then it's surfice to prove that $\angle XIY = 120$. Let $M$ be the midpoint of $XY$ We have $\frac{OM}{r} = \frac{OP}{IP} = \frac{OP}{AI_a}.\frac{AI_a}{I_aI_a'}.\frac{I_a'I_a}{IP} = \frac{R}{AI_a'}.\frac{AI_a}{I_aI_a'}.\frac{AI_a'}{AI}= \frac{R}{2r_a}.\frac{r_a}{r} = \frac{R}{2r} \implies OM = \frac{R}{2}$. Then $\angle XOY = 120 = \angle XIY$

(a) Let $Q$ be the exsimilicenter between the incircle and circumcircle and $D,E$ be the foot from $A,I_A$ to $BC$. Note that $A(D,E;I_A,I_A')=-1$. Reflecting over $AI$ gives $A(O,Q;I,P)=-1$ and similarly we obtain $B(O,Q;I,P)=-1$, so $P$ lies on line $OIQ$. (b) We claim that $I$ and $P$ are inverses about the circumcircle. Let $I'$ be the inverse of $I$ with respect to the circumcircle, so it suffices to show that $(O,Q;I,I')=-1$. We prove it with the following length chase: Recall from Euler's formula that $OI=\sqrt{R(R-2r)}$. Also, as $Q$ is the exsimilicenter, we have that $\frac{IQ}{OQ}=\frac rR$, so $\frac{IO}{OQ}=\frac{R-r}R\implies OQ=\frac{R\sqrt{R(R-2r)}}{R-r}$ so if $Q'$ is the inverse of $Q$ with respect to the circumcircle, $OQ'=\frac{R(R-r)}{\sqrt{R(R-2r)}}$. We have that $OI'=\frac{R^2}{\sqrt{R(R-2r)}}$ and also that $OI=\frac{R(R-2r)}{\sqrt{R(R-2r)}}$, so $Q'$ is the midpoint of $II'$. Inverting about the circumcircle gives $(O,Q;I,I')=-1$ as desired. Now, by Poncelet's Porism, the tangents to the incircle at $X$ and $Y$ different from $XY$ meet at a point $Z$ on the circumcircle. Note that $O$ and $I$ are the circumcenter and incenter of $XYZ$. Inverting $XYP$ about the circumcircle gives $XYOI$ cyclic, so $2\angle XZY=\angle XOY=\angle XIY=90^\circ+\frac{\angle XZY}2\implies \angle XZY=60^\circ\implies \angle XIY=120^\circ$, as deisred.

a) Let $OI\cap I_aI_a'=X_{40}$ be the Bevan point of $\triangle ABC$, $M$ be the midpoint of $II_a$ which is also the midpoint of arc $BC$, $N$ the midpoint of major arc $BAC$, and $Q$ the inverse point of $I$ with respect to $(ABC)$. We will prove that $AQ$ and $AI_a'$ are isogonal, which will prove $Q\in \ell_A$ and likewise $\ell_B$, which will prove $P=Q\in OI$. It is well known that $p(I_a,(ABC))=2Rr_a$. Let us quicky prove this fact again. Let incircle $(I)$ touch $BC$ at $D$ and $DD'$ is diameter in $(I)$, then if $A$-excircle touch $BC$ at $E$ then $A, D', E$ are colinear. Let $K$ be a point on $AE$ so that $IK\parallel BC$, then consider triangle $\triangle D'DE$, $I$ is midpoint of $DD'$, hence $K$ must be the midpoint of $D'E$, so $KE=KD$. But we also have $BD=CE$, then $KB=KC$. In particular, $M, K, N$ colinear. If $X=AM\cap BC, Y=MN\cap BC$, by Shooting Lemma we have $ANXY$ cyclic, yet $IK\parallel XY$ hence $ANKI$ is cyclic. From here we have $$\frac{MI_a}{2R}=\frac{MI}{MN}=\frac{MP}{MA}=\frac{I_aE}{I_aA}=\frac{r_a}{I_aA}\Rightarrow p(I_a,(ABC))=I_aM\cdot I_aA=2Rr_a$$Now $X_{40}I_a=2OM$ and $I_aI_a'=2I_aE=2r_a$, so $$I_aI\cdot I_aA=2I_aM\cdot I_aA=2p(I_a,(ABC))=(2R)(2r_a)=I_aX_{40}\cdot I_aI_a'$$So $AIX_{40}I_a'$ is cyclic. Note that $\angle IQA=\angle OAI=\angle OMA=\angle X_{40}I_aA$, so $QAX_{40}I_a$ is also cyclic. Then we have $\angle QAI=\angle IX_{40}I_a=\angle IAI_a'$, so $AQ$ and $AI_a'$ are indeed isogonal. b) By Poncelet's Theorem, tangents to $(I)$ at $X$ and $Y$ other than line $PXY$, must intersect at a point on $(ABC)$, call it $Z$. Note that $$QI\cdot QO=OQ^2-OI\cdot OQ=OQ^2-R^2=p(Q,(ABC))=QX\cdot QY$$Hence $XYOI$ is cyclic. Note that $O, I$ are the circumcenter and incenter for $\triangle XYZ$ as well, so we get $\displaystyle 90+\frac{\angle XZY}{2}=\angle XIY=\angle XOY=2\angle XZY\Rightarrow \angle XZY=60^{\circ}\Rightarrow \angle XIY=120^{\circ}$. Q.E.D

Another approach: We start off with the following lemma. LEMMA Let $I'$ be the inverse of $I$ about $\odot (ABC)$. Then $I'$ lies on $l'_A$. PROOF Let $AI \cap \odot (ABC) = T, I_AI'_A \cap BC = D$. Also let $H$ be the orthocenter of $\triangle I_ABC$, and $M$ be the midpoint of $BC$. By Fact 5 (also known as Incenter Excenter Lemma), $T$ is the center of $\odot (IBI_AC)$. From now on, WLOG assume $AB \geq AC$. By using homotheties and the fact that the $A$-intouch point and $D$ are isotomic points on $BC$, it can easily be shown that $AD \parallel IM$. Also, $BI \parallel CH$ and $CI \parallel BH$ $\Rightarrow BHCI$ is a parallelogram $\Rightarrow I, M, H$ are collinear $\Rightarrow IH \parallel AD \Rightarrow \frac{I_AH}{I_AD} = \frac{I_AI}{I_AA} \Rightarrow \frac{I_AH}{\frac{I_AI_A'}{2}} = \frac{2I_AT}{I_AA} \Rightarrow \frac{I_AH}{I_AI_A'} = \frac{I_AT}{I_AA} \Rightarrow TH \parallel AI'_A$ ($*$) And, $I'$ is the inverse of $I$ about $\odot (ABC) \Rightarrow OI \cdot OI' = OA^2 \Rightarrow OA$ is tangent to $\odot (AII')$ $\Rightarrow \angle OI'A = \angle OAI = \angle OTA \Rightarrow OAI'T$ is cyclic $\Rightarrow \angle I'AI = \angle I'OT$ ($**$) Now, Let $TO \cap \odot (ABC) = T_1 \Rightarrow T_1$ is the inverse of $M$ w.r.t. $\odot (IBI_AC) \Rightarrow TT_1 \cdot TM = TI^2$ It is well known that in any triangle the distance of any vertex from the orthocenter is twice the distance of the circumcenter from the opposite side. $\Rightarrow I_AH = 2TM \Rightarrow 2TO \cdot \frac{I_AH}{2} = TI \cdot TI_A \Rightarrow \frac{TO}{TI} = \frac{I_AT}{I_AH}$ But, $TO \parallel I_AH \Rightarrow \angle OTI = \angle TI_AH \Rightarrow$ Using the above equality, we get $\triangle OTI \sim \triangle TI_AH$ $\Rightarrow \angle IOT = \angle I_ATH \Rightarrow$ Using ($*$) and ($**$), we get $\angle I'AI = \angle I_AAI'_A \Rightarrow I'$ lies on $l'_A$ $\Box$ Return to the problem at hand. From the symmetry of our LEMMA, we get that $I'$ lies on $l'_B$ also $\Rightarrow P = I'$ PART a) This follows from the fact that $P$ is the inverse of $I$ in $\odot (ABC)$ PART b) $PX \cdot PY = Pow_{\odot (ABC)}P = OP^2 - OA^2 = OP^2 - OP \cdot OI = OP(OP-OI) = PO \cdot PI \Rightarrow XYOI$ is cyclic. Now, Let the other tangents from $X$ and $Y$ to the incircle meet at a point $K$. Then by Poncelet's Porism, $K$ lies on $\odot (ABC)$. $\Rightarrow I$ is the incenter of $\triangle KXY$, and $O$ is the circumcenter of $\triangle KXY$. $\Rightarrow \angle XIY = 90^{\circ}+\frac{\angle XKY}{2}$ and $\angle XOY = 2\angle XKY$ But, As $XYOI$ is cylic, so $\angle XIY = \angle XOY \Rightarrow 90^{\circ}+\frac{\angle XKY}{2} = 2\angle XKY \Rightarrow \angle XKY = 60^{\circ} \Rightarrow \angle XIY = 120^{\circ}$ $\blacksquare$ REMARK 1: I believe that without part b) the problem is much more difficult, as the condition in part b) triggers the fact that $P$ is the inverse of $I$ w.r.t. $\odot (ABC)$ after which everything goes on in a flow (this happened for me atleast, obviously after I noticed the application of Poncelet's Porism). REMARK 2: The point $P$ is actually the isogonal conjugate of the Bevan Point of $\triangle ABC$ (the circumcenter of the excentral triangle). This easily follows from our Lemma. Also, My first G7, so yay

(a) Let $I'$ denote the inverse of $I$ in $(ABC)$. Let $\phi_{bc}$ denote the $\sqrt{bc}$ inversion map, and note that $\phi_{bc}$ is a Mobius transform. Therefore, $\phi_{bc}(I_A')$ is the inverse of $\phi_{bc}(I_A)=I$ in $\phi_{bc}(BC)=(ABC)$, so $\phi_{bc}(I_A')=I'$. Therefore, $AI'$ and $AI_A'$ are isogonal in $\angle A$, and similarly $BI'$ and $BI_A'$ are isogonal in $\angle B$. Therefore, $P=I'$. (b) Let $Z$ denote the intersection of the tangent to the incircle from $X$ and $(ABC)$. Poncelet Porisim says that the incircle of $XYZ$ is the incircle of $ABC$. Now, we have $X,Y,P$ collinear, so inverting about $(ABC)$, we have $OXYI$ concyclic. Thus, $\angle XOY=\angle XIY$, so \[2Z = \pi-(X/2-Y/2)=\pi/2+Z/2,\]so $Z=\pi/3$, so $\angle XIY=2\pi/3$. $\blacksquare$

Taking the advantage of the Bump, here is something (different?) cjquines0 wrote: Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_A$ be the $A$-excentre, $I'_A$ be the reflection of $I_A$ in $BC$, and $l_A$ be the reflection of line $AI'_A$ in $AI$. Define points $I_B$, $I'_B$ and line $l_B$ analogously. Let $P$ be the intersection point of $l_A$ and $l_B$. Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$. Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\angle XIY = 120^{\circ}$. Since \(I_AG= I_A'G \Rightarrow IE= EF=r\) Let \(H\) be the reflection of \(F\) in \(AI\) then \(IH = 2\times r.\) So: \(\dfrac{IH}{AO} = \dfrac{2\times r}{R} = \dfrac{\text{dist.(I, XY)}}{OM} = \dfrac{PI}{PO}\) From here we draw two conclusions; a. \(\dfrac{PI}{PO}\) is independent of the choice of vertex. This proves the first part. b. \(OM= R/2\) \(\Rightarrow \angle XOY = 120^{\circ}\). Then by Poncelet's Porism let \(\Delta XYZ\) be circumscribed upon \((I)\), we have \(\angle XZY = 60^{\circ} \Rightarrow \angle XIY = 90^{\circ}+ 60^{\circ}/2 = 120^{\circ}.\) \(\blacksquare\)

Homothety solution. (a) Let $DEF$ be the intouch triangle of $\triangle ABC$. Note that the homothety at $A$ which maps $A$-excircle to the incircle also maps the extouch point to the antipode of $D$. Thus it maps $I_A'$ to point $X$ such that $IX = 2ID$. Let $\gamma = \odot(I,2r)$. Let $X'$ be the reflection of $X$ across $AI$. Clearly $X,X'\in\gamma$. Moreover, it's clear by reflection that the tangent to $\gamma$ at $X$ is parallel to the tangent to $\odot(ABC)$ at $A$. Thus $AX'$ pass through exsimilicenter of $\gamma$ and $\odot(ABC)$. So $P$ is that exsimilicenter. (b) By Poncelet's porism, the other tangents from $X,Y$ to $\odot(I)$ meet at $Z\in\odot(ABC)$. Let the homothety at $P$ which maps $\odot(ABC)\to\gamma$ sends $X\to X'$ and $Y\to Y'$. Let $U,V$ be the reflections of $X',Y'$ across $XI$ and $YI$. Then by similar argument above, $IU\perp YZ$ and $IV\perp YZ$. Thus if we let $\triangle X_1Y_1Z_1$ be the intouch triangle. Then $3r = UX_1 = AX_1\cos\angle Z$. But $AX_1 = r\cot\tfrac{\angle Z}{2}$. Hence $\cos\angle Z\cot\tfrac{\angle Z}{2}=3$. By monotonicity, the only answer is $\angle Z=60^{\circ}$ so $\angle XIY=120^{\circ}$, done.

Here is a solution found with amar_04,mueller.25and Aryan_23 [asy][asy] size(10cm); pair A = dir(115); pair B = dir(208); pair C = dir(332); pair D=foot(A,B,C); pair F=foot(C,A,B); pair E=foot(B,A,C); pair K=foot(A,E,F); pair H=orthocenter(A,B,C); pair O=circumcenter(A,B,C); pair A1=2*K-A; pair P = -O+2*foot(circumcenter(A, O, D), O, H); draw(circumcircle(A,O,D),dashed); pair N9=(O+H)/2; draw(A--B--C--A); draw(D--E--F--D); draw(A--O--A1); draw(A--H--D); draw(O--H--P); draw(D--P); draw(circumcircle(O,H,D),dashed); dot("$A$", A, dir(80)); dot("$B$", B, dir(150)); dot("$C$", C, dir(330)); dot("$D$", D, dir(300)); dot("$E$", E, dir(390)); dot("$F$", F, dir(110)); dot("$A'$", A1, dir(60)); dot("$H$", H, dir(120)); dot("$O$", O, dir(40)); dot("$P$", P, dir(150)); dot("$N_9$", N9, dir(110)); [/asy][/asy] Proof: (for part a) Firstly embed the problem w.r.t to the excentral triangle. We obtain the following problem. Embeded Problem wrote: Let $\triangle{DEF}$ bethe orthic triangle of $\triangle{ABC}$ with $D,E,F$ on $\overline{BC},\overline{CA},\overline{AB}$ respectively. Now let $A'$ be the reflection of $A$ in $\overline{EF}$ and let $\ell$ be the line isogonal to $\overline{DA'}$ w.r.t $\angle FDE$. Then prove that if $\ell \cap \overline{OH}=P$ then $P \in \odot(AOD)$. Observe that to prove $(AODP)$ cyclic we just need to show $(HODA')$ cyclic since $\angle PDA=\angle HDA'$. But this is trivial after an inversion with radius $\sqrt{AH \cdot AD}$ centred at $A$ since under this inversion $O \mapsto A'$ and $H \mapsto D$. $\square$. Now for the second part. Notice that by poncelets porism we just need to show that $(H,P)$ are inverses w.r.t $\odot(DEF)$. To prove this notice that it is equivalent to showing $N_9H \cdot N_9P={N_9D}^2$. Now notice that it's well known that $$N_9H^2=\frac{9R^2-(a^2+b^2+c^2)}{4}$$where $a,b,c$ are the sides of the triangle and $R$ is the circumradius. Also we have that $\overline{N_9A}=\frac{R}{2}$ . Now we are all set to length chase. Notice that $$\overline{N_9H} \cdot \overline{N_9P}={N_9H}^2+N_9H\cdot HP=\frac{9R^2-(a^2+b^2+c^2)}{4}+\frac{AH \cdot HD}{2}$$. We have to show that this expression equals $N_9D^2=\frac{R^2}{4}$ so we are left to show that $8R^2=a^2+b^2+c^2-2\cdot AH\cdot HD$.Now we'll evaluate the left hand side.Let $A,B,C$ be the angles at vertex $A,B,C$. Notice that $a^2+b^2+c^2-2\cdot AH\cdot HD=4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C})-2\cdot(2R \cos{A})(2R\cos{B}\cos{C})=4R^2(\sin^2{A}+\sin^2{B}+\sin^2{C}-2\cos{A}\cos{B}\cos{C})=8R^2$ where the last part follows from the well knwon identity that $\sin^2A+\sin^2{B}+\sin^2{C}=2+2\cos{A} \cos{B} \cos{C}$. Hence we are done $\blacksquare$.

SidVicious wrote: Sketch for a): $AI_{A}'$ passes through $U-$ antigonal conjugate of $I,$ but since isogonal conjugate of $U$ (say $V$) is nothing but inverse of $I$ WRT $\odot(ABC) \implies AV \equiv l_a$ passes through inverse of $I$ hence the conclusion. How can we prove that $AI_{A}'$ passes through the antigonal conjugate of $I$ ?

Anyone...

Anyone...

a) Note that $I_A = (-a:b:c) = (-2a^2:2ab:2ac)$. Let $D = (0:s-b:s-c)$ be the foot of the altitude from $I_A$ to $BC$. Note that $D = (0: c^2-(a-b)^2:b^2-(a-c)^2)$. Then, \[ I_A'= 2D - I = (2a^2: 2c^2-2a^2-2b^2+2ab: 2b^2-2a^2-2c^2+2ac) \]\[ I_A' = (-a^2: 2S_C- ab: 2S_B - ac) .\]Note that clearly $AI_A'$ and $BI_B'$ will meet at \[ P'= \left( \frac{1}{2S_A-bc}, \frac{1}{2S_B-ac}, \frac{1}{2S_C-ab} \right)\]The isogonal conjugate of this point has coordinates \[ P = (a^2(S_A-bc), b^2(S_B-ac), c^2(S_C - ab) ).\]Since $O = (a^2S_A: b^2S_B: c^2S_C)$ and $I = (a:b:c)$, we clearly see that $P$ is a linear combination of $O$ and $I$, so $P$ is on $OI$. b) By Poncelet porism, we can find a point $Z$ such that $XZ$ and $YZ$ are tangent to the incircle. Note that $\angle XOY = 2 \angle XYZ$ and $\angle XIY = 90^\circ + \frac12 \angle XYZ$. When $\angle XOY = \angle XIY$, we have $\angle XYZ = 60^{\circ}$ and $\angle XIY = 120^{\circ}$. Therefore, it suffices to show $XOIY$ is cyclic, or that $PO \cdot PI = PX \cdot PY$. In other words, we just have to show that $P$ is on the radical axis of $(AOI)$ and $(ABC)$. To find the radical axis, we just have to find the equation of the $(AOI)$. This is \[ -a^2 yz - b^2xz -c^2xy+(x+y+z)(vy+wz)=0 \]To find $v$ and $w$, we plug in $O = (a^2S_A: b^2S_B: c^2S_C)$ and $I = (a:b:c)$. The equations we obtain are \[ -abc(a+b+c) + (a+b+c)(bv+cw) = 0, \]\[ -a^2b^2c^2 (S_AS_B+S_BS_C + S_CS_A) + (a^2S_A+b^2S_B+c^2S_C) (b^2S_Bv+c^2S_Cw) = 0.\]Note that $S_B + S_C = a^2$, so we can find that $(a^2S_A+b^2S_B+c^2S_C) = 2(S_AS_B+S_BS_C + S_CS_A) $. Thus, we can simplify the equations to \[ bv+cw = abc, b^2S_Bv + c^2S_Cw = \frac{a^2b^2c^2}{2} .\]The solution to this system is \[ w = \frac{ \frac{ab^2}{2} (2S_B - ac) }{bS_B - cS_C}, \qquad v = - \frac{ \frac{ac^2}{2} (2S_C - ab)}{bS_B -cS_C} .\]The equation of the radical axis is $vy+wz = 0$, and we see that $P$ is clearly on this line, so we are done.

Here's a relatively simple elementary solution (a little length computation at the end, but it's really easy): (1) Set $D$ = foot of angle bisector from $A$ to $BC$. (2) Let $X_0$ = second intersection of circle with center $I$ and radius $IL$ and $(ABC)$; then $X_0$ is reflection of $L$ over $IO$ (3) Thus $\angle IAO = \angle ILO = \angle IX_0O$ and hence $IAX_0O$ is cyclic. (4) Let $LO$ intersect circle w/ center $D$ and radius $DI_A$ again at $S$. Then $\angle ISO=\angle IAO=\angle ILO$ hence $ISL$ is isosceles. (5) Let reflection of $I_A$ in $BC$ be $T$, then $ISL$ and $DTI_A$ are homothetic with center $A$; hence $A,S,T$ are collinear. (6) Since $ASIX_0$ is cyclic and $IS=IX_0$, we see that $\angle SAI=\angle X_0AI$ and $\angle TAI=\angle XAI$, $X_0,X,A$ are collinear. (7) Hence the desired concurrency point $P$ is the radical axis of the circumcircles of $AOI$,$BOI$, and $COI$. Furthermore, remark $PI \cdot PO = PX_0 \cdot PA = PO^2-R^2$ and so $PO \cdot OI = R^2$ and $OI = \sqrt{R^2-2Rr}$, so the remainder of the problem is a (very) easy computation. [EDIT: reading above solutions, for the computation part I forgot to write the important step that $XYOI$ is cyclic but this is clear from PoP with respect to point $P$].

Here is a hybrid solution. For part (a), we use barycentrics wrt $\triangle ABC$. Recall that $I_A = (-a:b:c)$ and the $A$-extouch point is $(0 : a+c-b : a+b-c)$, so \begin{align*} I_A' &= 2\left(0, \frac{a+c-b}{2a}, \frac{a+b-c}{2a}\right) - \left(-\frac{a}{b+c-a}, \frac{b}{b+c-a}, \frac{c}{b+c-a}\right) \\ &= (a^2 : c^2-a^2-b^2+ab : b^2-a^2-c^2+ac).\end{align*}Therefore, by isogonal conjugation, $P$ is of the form $\left(\bullet : \frac{b^2}{c^2-a^2-b^2+ab} : \frac{c^2}{b^2-a^2-c^2+ac}\right)$ where $\bullet$ stands for some real number. Similarly, $P$ is also of the form $\left(\frac{a^2}{c^2-a^2-b^2+ab} : \bullet : \frac{c^2}{a^2-b^2-c^2+bc}\right)$, so we conclude that \[P = (a^2(a^2-b^2-c^2+bc) : - : -) = (a^2(bc-2S_A) : b^2(ac-2S_B) : c^2(ab-2S_C)). \](Notice that this also implies that $l_a, l_b, l_c$ concur.) Hence, \begin{align*} \left|\begin{matrix} a^2(bc-2S_A) & b^2(ac-2S_B) & c^2(ab-2S_C) \\ a & b & c \\ a^2S_A & b^2S_B & c^2S_C \end{matrix}\right| &= \left|\begin{matrix} a^2bc & b^2ac & c^2ab \\ a & b & c \\ a^2S_A & b^2S_B & c^2S_C \end{matrix}\right| \\ &= abc \left|\begin{matrix} a & b & c \\ a & b & c \\ a^2S_A & b^2S_B & c^2S_C \end{matrix}\right| = 0,\end{align*}which implies that $P,I,O$ collinear. For part (b), we first prove the following claim: Claim. Let $H$ be the orthocenter of $ABC$. Then $\angle I_A' A H = \angle AIO$. Proof. Let $IO$ cut $I_AI_A'$ at $T$. It suffices to prove that $A,I,T,I_A'$ are concyclic. Let $M$ be the midpoint of $II_A$ (which is also the midpoint of arc $BC$), and let $D$ be the $A$-extouch point, then it suffices to prove $A,M,D,T$ concyclic. Denote $R$ by the radius of $(ABC)$, then \[ A,M,D,T \text{ concyclic} \iff I_AM \cdot I_AA = I_AD \cdot I_AT \iff I_AO^2 - R^2 = I_AD\cdot 2R \]which is Euler's theorem. $\square$ Back to the original problem; the claim means that $\angle OIA = \angle I_A'AH = \angle PAO$. Combining this with (a), we see that $OI \cdot OP = R^2 \iff PI \cdot PO = PO^2 - R^2$. Therefore, points $X,Y,I,O$ are concyclic. Draw the other tangents to the incircle from $X,Y$, and they intersect at a point $Z\in(ABC)$ by Poncelet's Porism. But \[\frac{\pi}{2} + \frac{1}{2}\angle XZY = \angle XIY = \angle XOY = 2\angle XZY \iff \angle XZY = 60^{\circ}\]as desired.

EulerMacaroni wrote: Lemma: Let $\odot(A), \odot(B), \odot(C)$ be circles in the plane such that $\odot(C)$ is the inverse of $\odot(B)$ with respect to $\odot(A)$. Then for every point $P$, $B(A(P))=A(C(P))$, where $X(P)$ denotes inversion about $X$. Awesome Lemmma!!! @above you probably made a typo in the Claim section.

(a) Let $I'$ be the inverse of $I$ w.r.t. the circumcircle of $\triangle ABC$. We claim that $I'$ is the point $P$. Let $\triangle I_AI_BI_C$ be the excentral triangle of $\triangle ABC$, then $\triangle ABC$ is its orthic triangle. Let $Q$, $T$, $M$ be the midpoints of $\overline{I_AI}$, $\overline{I_BI_C}$ and $\overline{I_AI_B}$, $N_9$ and $O$ be the circumcenters of $\triangle ABC$ and $\triangle I_AI_BI_C$, repsectively, $S$ be the foot of the altitude from $I_A$ to $\overline{BC}$, $r$ be the circumradius of $\triangle ABC$. We have $$r^2=IN_9 \cdot I'N_9=I'N_9 \cdot N_9O=QN_9\cdot TN_9 \implies AI'I_AO \; \text{is cyclic} \implies \angle I'AI_A=\angle I'OI_A$$Now we want to show that $\angle I'OI_A=\angle I_A'AI_A$. To prove it suffices to show that $AIOI_A'$ is cyclic. Notice that $$\angle CMO=\angle CSO=90^{\circ} \implies CMSO \; \text{is cyclic} \implies I_AM \cdot I_AC= I_AS \cdot I_AO$$But at the same time from the NPC theorem $CMQA$ is cyclic $\implies I_AM \cdot I_AC=I_AQ \cdot I_AA$, thus $I_AS \cdot I_AO=I_AQ \cdot I_AA \implies I_AI_A' \cdot I_AO=2 \cdot I_AS \cdot I_AO=I_AQ \cdot I_AA = 2 \cdot I_AQ \cdot I_AA=I_AI \cdot I_AA$ which implies the desired concyclicity and we're done.

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Let $I'$ be the reflection of $I$ in $O$. CLAIM 1. $A,I',I_A',I$ are concyclic. Proof. Notice that $Q,M$ are the midpoints of $II'$ and $II_A$ respectively, therefore, $$I_AI'\times I_AI_A'=2OM\times 2I_AD=2R\times 2I_AB\cos\frac{C}{2}=4R\sin\frac{A}{2}\times I_AA\frac{\cos\frac{C}{2}}{\sin\frac{A}{2}}=I_AM\times I_AA$$as desired. $\blacksquare$ Therefore, if $H$ is the orthocenter of $\triangle ABC$, using the fact that $AH,AO$ are isogonal, $$\angle PAO=\angle I'AH=180^{\circ}-\angle AI_A'I_A=\angle AIO$$as desired. This proves $(a)$ since this implies $P$ lies on $l_B$ by symmetry. (b) Suppose the second tangent from $X$ and $Y$ to the incircle of $(ABC)$ intersect at $Z$, then $Z$ lies on $(ABC)$ by Poncelet Porism. Moreover, since $P$ is the image of $I$ under inverson, $$PI\times PO=PX\times PY$$hence $X,Y,I,O$ are concyclic. Therefore, $$90^{\circ}+\frac{\angle XZY}{2}=\angle XIY=\angle XOY=2\angle XZY$$which implies $\angle XZY=60^{\circ}$, and hence $\angle XIY=120^{\circ}$ as desired.

let $P'=IO \cap \ell_a$ let $N,M$ be the midpoints of ars $BC,AC$ (i) claim: $P'$ is the image of $I_a'$ under $\sqrt{AB.AC}$ inversion proof: let $A'$ be the reflection of $A$ wrt $BC$ note that $ O^*=A'$ and $I^*=I_a$ since $AI_a'I_aA'$ is cyclic so $I_a'^* \in IO$ $\blacksquare$ note that $I_a'-A'-AI \cap BC $ are collinear so $AONP'$ is cyclic so $BOMP'$ is also cyclic as above we will have that $P'$ is the image of $I_b'$ under $\sqrt{BC.BA}$ inversion so $P=P'$ (ii) note that $IA.IN=IO.IP$ so $P$ is the image of $O$ under the inversion centered at $I$ with radius $\sqrt{IA.IN}$ so $IOXY$ is cyclic now let $Z$ on $(ABC)$ such that $ZX,ZY$ are both tangent to the incircle (Poncelet Porism) since $O,I$ are the circumcenter,incenter of $\triangle ZXY $ and $XYOI$ is cylic so $\angle XZY=60$ so $\angle XOY=\angle XIY=120$ and we win

Flip99 wrote: Anyone... You can use feuerbach hyperbola

This is a direct conclusion from Komal A779

hsiangshen wrote: Flip99 wrote: Anyone... You can use feuerbach hyperbola How?

OK this problem is kinda easy under some dark and poisonous conic geometry... (a) Let $Na$=Nagel point, tangents of incircle=$D,E,F$ as usual, $D'$ be the antipode of $D$ wrt $(I)$. Notice that $ANa$ bisects $I_AI_A'\implies AI_A'$ passes through the reflection of $I$ over $D'$. Then by kariya theorem, if we construct $BI_B',CI_C'$ similarly, they must concur on the Feuerbach hyperbola, call it $X$. Now we prove that point is actually the reflection of $I$ over Feuerbach point=$X_{80}$. Notice that $A(I_A',I;D',H)=-1=(X,I;Na,H)$ on Feuerbach hyperbola. Therefore projecting from $I$ on $NaH:(IX,II,INa,IH)=-1=(IX\cap NaH,OI\cap NaH;Na,H)$, so $IX$ bisects $NaH$. On the other hand,$-1=(XX,IX;XNa,XH)\implies XX||NaH||OI$. So $(X,I)$ is a pair of antipode wrt Feuerbach hyperbola, $X=X_{80}$, as required. Finally, the problem makes the intersection isogonal conjugate so it clearly lies on the ispgonal conjugate of Feuerbach hyperbola=$OI$. (b) We know that the isogonal conjugate of $P$ is the antigonal conjugate of $I$. One of the equivalent definition of antigonal is the isogonal, inverse, isogonal of a point.(well-known) So $I,P$ are inverses wrt the circumcurcle. Then the rest can be done by poncelet or something like that but I haven't finished it.

Let $V$ be the reflection of $I$ wrt $O$ and $M$ be the midpoint of arc $BC$. Suppose that the A-excircle of $\triangle ABC$ touch the side $BC$ at $K$. (a) lemma. $AIVI'_A$ is cyclic. proof. From Euler's formula, we have $I_AO^2=OM^2+2OM\cdot I_AK.$ Note that $I_AV=2OM, I_AI'_A=2I_AK$ and $I_AI=2I_AM$ then $$I_AM\cdot I_AA=(OM^2+2OM\cdot I_AK)-OM^2=2OM\cdot I_AK\Longrightarrow I_AI\cdot I_AA=4OM\cdot I_AK=I_AV\cdot I_AI'_A,$$implying that $AIVI'_A$ is cyclic. We get $\measuredangle I'_AAM=\measuredangle I_AVI=\measuredangle MOI$ from lemma. Let $l_A\cap (ABC)=S$ and $l_B\cap (ABC)=T,$ then $\measuredangle MAS=\measuredangle MOI$. By simple angle chasing, it follows that $\measuredangle IOS=\measuredangle MOI$ so $ASIO$ is cyclic. Similarly, $BTIO$ is cyclic thus $P$ be the radical center of $(ABC), (ASIO), (BTIO)$ and we have the desired result. (b) Poncelet porism means that there exists point $Z$ on $(ABC)$ such that the incenter of $\triangle XYZ$ is $I$. Because of $P$ lies on radical axis of $(ABC)$ and $(ASOI),$ then $XYOI$ is also cyclic. Hence $$\angle XIY=90^\circ+\displaystyle\frac{1}{2}\angle XZY=\angle XOI=2\angle XZY\Longrightarrow \angle XZY=60^\circ.$$Therefore we conclude $\angle XIY=120^\circ.$ I used the same lemma at Taiwan TST 2019 6.

Projective, projective, projective. Give me a break. a) Let the medial triangle of the excentral triangle be $Q_A, Q_B, Q_C$. Let the projections of $I_A, I_B, I_C$ onto $\triangle ABC$ be $R_A, R_B, R_C$ and let $S_A, S_B, S_C$ be the midpoints of $I_A, I_B, I_C$ with $I$. Let $O$ be the circumcenter of $(ABC)$ and $O'$ be the circumcenter of $(I_AI_BI_C)$. Then since $(ABCQ_AQ_BQ_C)$ is the nine-point circle of the excentral triangle, $AI_A \cdot S_AI_A = CI_A \cdot Q_CI_A = OI_A \cdot R_AI_A$ so $\triangle I_AR_AA \sim \triangle I_A S_AO'$. Hence by similarity, $\angle I_A'AI_A = \angle O'IA = \angle OIA - \angle OAI$ so if $A'$ is $\ell_A \cap IO$ then $\angle OAA' = OIA$ so $A'$ is the inverse of $I$, $I'$ with respect to $(ABC)$ so we are done by symmetry. b) By Brokard's theorem, $X' = YI \cap (ABC)$ is the reflection of $X$ over $IO$, so by Poncelet's porism we have that $\frac{\angle XIY}{2} = \angle XX'I = \angle IX'Y = 2(\angle XIY - \frac{\pi}{2})$ so $\angle XIY = 120^{\circ}$.

Redefine $P$ as the inverse of $I$; it's clear via Poncelet spam that this point satisfies the second part. For the first part we assert more strongly that: Claim: $\triangle AI_aI_a'\overset+\sim \triangle API$. Proof: One of the few uses of SAS similarity? By angle chasing, $\angle I_a=\angle P$ follows easily. To finish, we show $I_aI_a'/I_aA=IP/AP$; indeed, the first ratio equals $2\cos\angle BI_aC=2\sin\frac A2$ because of similar triangles; thus, we're left to length chase $IP/AP$; this becomes \[\frac{OP}{AP}-\frac{OI}{OA}\frac{OA}{AP}=\frac{OI}{AI}-\frac{OI^2}{OA\cdot AI}=\frac{R}{AI}-\frac{R^2-2rR}{R\cdot AI}\]\[=\frac{R-(R-2r)}{AI}=2\sin\frac A2,\]so the ratios are equal, as needed. $\qquad\qquad\square$. The claim clearly implies the isogonality.

Writeup from an OTIS walkthrough. Let $V$ be the circumcenter of triangle $I_AI_BI_C$. Part (a) Consider the circumcircles of $AVI_A$ and cyclic permutations. As $I$ is their radical center and they are coaxial, their second intersection point $P$ must lie on $\overline{IV}$. On the other hand, $$I_AI_A' \cdot AV = 2\cos I_A R \cdot I_AA = I_AI \cdot I_AA$$with reference triangle $I_AI_BI_C$, thus $AIVI_A'$ is cyclic. To finish, set $P' = \overline{IV} \cap (AVI_A)$. Then $$\angle P'AI_A = \angle IVI_A = \angle I_AAI_A'$$thus $P$ lies on $\ell_A$. This finishes the first part. Part (b) Let $Z$ be the point on $(ABC)$ such that $XYZ$ and $ABC$ have the same incircle which exists by Poncelet porism. Claim. $XIOY$ is cyclic. Proof. It suffices to show that $P$ and $I$ are inverses with respect to $(ABC)$. To see this, set $M$ to be the midpoint of $\overline{II_A}$; then $M = (POA) \cap (ABC)$ by Reim's theorem. Then as $OM=OA$, $\angle MPO = \angle APO$ and $OI \cdot OP = OM^2$. Hence $$PI \cdot PO = PO^2 - R_{ABC}^2 = OX \cdot OY.$$ Finally, $\angle XIY = 90^\circ +\frac Z2 = 2Z = \angle XOY$, thus $\angle XIY = 120^\circ$.

500th post Also this took less than a hour with GGB (would have done it freehand but I've already wasted so much time in life ) 2016 ISL G7 wrote: Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_A$ be the $A$-excentre, $I'_A$ be the reflection of $I_A$ in $BC$, and $l_A$ be the reflection of line $AI'_A$ in $AI$. Define points $I_B$, $I'_B$ and line $l_B$ analogously. Let $P$ be the intersection point of $l_A$ and $l_B$. Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$. Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\angle XIY = 120^{\circ}$. Part (a) Let $Be$ be the Bevan point (circumcentre of the excentral triangle) of $\triangle ABC$.Clearly as $I_AI'_A \perp BC,I_BI'_B \perp CA$ , the lines $I_AI'_A,I_BI'_B$ are isogonal to $I_AI,I_BI$ respectively.Since $I$ is the orthocentre of the excentral triangle it's isogonal conjugate is $Be$.So $Be \equiv I_AI'_A \cap I_BI'_B$ $\square$ Now follow up with a $\sqrt{I_AI.I_AA}$ inversion centered at $I_A$.If $I_AI'_A \cap BC=F$ and $J$ be the reflection of $I_A$ over the Bevan point, then $F$ and $J$ gets swapped under this inversion. Now, $$I_AI.I_AA=I_AF.I_AJ=2I_AF.\frac{I_AJ}{2}=I_AI'_A.I_ABe $$Which gives that $BeIAI'_A$ and similarly $BeIBI'_B$ are cyclic .Now if $BeI \cap \ell_A=P'$, then $$\angle P'AI_A=\angle I_AAI'_A=\angle IAI'_A=\angle I_ABeI=\angle I_ABeP'$$Thus $I_ABeAP'$ is cyclic,this gives the following $$P'I.IBe=AI.I_A=BI.II_B $$which implies that $I_BBeBP'$ is cyclic as well.Now it is not hard to prove that $P' \in \ell_B$ by angle chase,thus $P \equiv P'$.Since $Be,I,O$ are collinear, we have that $P \in OI$ finishing the first part $\blacksquare$ Part (b)We use the well known property that $O$ is the midpoint of $IBe$ and the midpoint of $II_A$(say $S$) lie on the circumcircle of $\triangle ABC$. From the cyclic quadrilaterals we obtained in Part (a) we have , $$PI.2IO=PI.IBe=IA.II_A=2IA.IS\implies PI.IO=IA.IS=\text{Pow}_{\odot(\triangle ABC)}(I)$$Now we present a claim ; Claim : $P$ is the inverse of $I$ w.r.t $\odot(\triangle ABC)$. Proof : We invert around $I$ with radius $\sqrt{\text{Pow}_{\odot(\triangle ABC)}(I)}$.Clearly $P,O$ get swapped and $I$ and $P_{\infty,PI}$ get swapped.Now we let $PI \cap \odot(\triangle ABC)\equiv M,N$.Clearly $M,N$ get swapped under the inversion .Since inversion preserves cross ratios, we have $$(M,N;P,I)=(M,N;O,P_{\infty,PI})=-1$$This proves that $P$ and $I$ are inverses w.r.t $\odot(\triangle ABC)$ $\blacksquare$ Now note that $$OI.OP=R^2 \implies OP=\frac{R^2}{OI} \implies \frac{PI}{PO}=1-\frac{OI}{OP}=1-\frac{R^2}{OI^2}=1-\frac{R^2-2Rr}{R^2}=\frac{2r}{R}$$Where $r,R$ are circumradius and inradius of $\triangle ABC$ respectively.Now let $XY$ touch the incircle at $H$ and the midpoint of $XY$ be $L$.We have $\triangle PIH \sim \triangle POL$.Thus , $$OL=IH.\frac{PO}{PI}=\frac{R}{2r}.r=\frac{R}{2} \implies \angle XOY=120^{\circ}$$One final observation, since $I,P$ are inverses we have $PX.PY=PO.PI$ which gives $XYOI$ is cyclic and thus we get that $$\angle XIY=\angle XOY=120^{\circ}$$This took more time to write than to solve.So much skill issue $\blacksquare$

I spent almost over 4 hours trying to complecks bash part (a) but in vain. Then I noticed the use of inversion distance formula and bruh moment. Why is this $C$ centered considering the hardness of the problem. It makes it even more harder (at least for me ;-; ). So of course, we $A$-center the problem. The statement thus reads. A-centered problem statement wrote: Let $I$ be the incenter of a non-equilateral triangle $ABC$, $I_B$ be the $B$-excenter, $I'_B$ be the reflection of $I_B$ in $CA$, and $l_B$ be the reflection of line $BI'_B$ in $BI$. Define points $I_C$, $I'_C$ and line $l_C$ analogously. Let $P$ be the intersection point of $l_B$ and $l_C$. Prove that $P$ lies on line $OI$ where $O$ is the circumcenter of triangle $ABC$. Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\angle XIY = 120^{\circ}$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-28.83788,16.19840); pair B = (-55.62356,-38.63888); pair C = (40,-40); pair I = (-21.42183,-17.61800); pair O = (-7.65223,-28.11036); pair I_B = (68.23612,37.48706); pair I_C = (-82.14257,4.50850); pair I_A = (4.71935,-136.81900); pair P = (-118.47836,56.33847); pair I_Ap = (7.48934,57.78395); pair Ap = (-30.40953,-94.21634); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle(O, 49.11312), linewidth(0.5)); draw(B--I_B, linewidth(0.5)); draw(C--I_C, linewidth(0.5)); draw(P--O, linewidth(0.5) + linetype("4 4") + red); draw(I_A--A, linewidth(0.5)); draw(circle(I, 21.50552), linewidth(0.5)); draw(Ap--I_A, linewidth(0.5) + blue); draw(A--I_Ap, linewidth(0.5) + blue); draw(P--A, linewidth(0.5)); dot("$A$", A, dir(90)); dot("$B$", B, SW); dot("$C$", C, SE); dot("$I$", I, NE); dot("$O$", O, SE); dot("$I_B$", I_B, NE); dot("$I_C$", I_C, NW); dot("$I_A$", I_A, SE); dot("$P$", P, NW); dot("$I_A'$", I_Ap, NE); dot("$A'$", Ap, NW); [/asy][/asy] We first solve part (a). I claim that all $\left\{l_A,l_B,l_C\right\}$ pass through the inverse of $I$ w.r.t. $\odot(ABC)$. Note that this actually finishes. So we prove that $l_A$ passes through $I^*$. Redefine $P$ as the image of $I_A'$ under $\sqrt{bc}$-inversion. Note that we need to prove that $I^*$ lies on the isogonal of the line $AI_A$. Thus if we show that $P$ is the image of $I$ under $\mathbf{I}(\odot(O,OA))$, then we are done. Let $A'$ be the reflection of $A$ over $BC$. It is well known that under $\sqrt{bc}$-inversion, $O\leftrightarrow A$, $I\leftrightarrow I_A$. Note that $AA'I_AI_A'$ is an isosceles trapezium and so, it is cyclic. Thus after inverting, we get that $\overline{P-I-O}$ are collinear. Now using inversion distance formula, we get that, \[ OP = \dfrac{AB\cdot AC}{AA'\cdot AI_A'}\cdot A'I_A' = \dfrac{AB\cdot AC}{AA'\cdot AI_A'}\cdot AI_A. \]We also get, \[ OI = \dfrac{AB\cdot AC}{AA'\cdot AI_A}\cdot A'I_A. \] Multiplying the above identities, we get, \[ OP\cdot OI = \left(\dfrac{AB\cdot AC}{AA'}\right)^2 = \left(\dfrac{AB\cdot AC}{\dfrac{AB\cdot AC}{AO}}\right)^2 = AO^2.\] This concludes that $P$ is indeed the image of $I$ under $\mathbf{I}(\odot(O,OA))$ and we are done. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-39.92648,31.75707); pair B = (-55.62356,-38.63888); pair C = (40,-40); pair I = (-25.10426,-14.53006); pair O = (-7.42914,-12.43734); pair P = (-175.32738,-32.31636); pair X = (-60.67652,0.75126); pair Y = (30.61622,27.08194); pair L = (-27.31701,-63.56166); pair M = (-22.63116,40.27054); import graph; size(15cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle(O, 54.85638), linewidth(0.5)); draw(P--O, linewidth(0.5) + linetype("4 4") + red); draw(circle(I, 24.54075), linewidth(0.5)); draw(P--A, linewidth(0.5)); draw(P--Y, linewidth(0.5)); draw(X--L, linewidth(0.5)); draw(Y--L, linewidth(0.5)); draw(circle(M, 54.85638), linewidth(0.5) + blue); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$I$", I, dir(90)); dot("$O$", O, dir(90)); dot("$P$", P, NW); dot("$X$", X, dir(180)); dot("$Y$", Y, NE); dot("$L$", L, dir(270)); dot("$M$", M, dir(90)); [/asy][/asy] Since $\overline{P-X-Y}$ are collinear, thus after inverting $\mathbf{I}(\odot(O,OA))$, we get that $OIXY$ is cyclic. Now let the tangent to the incircle from $X$ intersect $\odot(ABC)$ at $L\;(\neq Y)$. Thus by Poncelet Porism, we know that $\triangle LYX$ has the same incircle. Now let $M$ be the midpoint of the arc $\widehat{XY}$ not containing $L$. Thus $I$ is the incenter of $\triangle LYX$. Thus by Fact 5, we get that $MX = MY = MI$ which means that $M$ is the center of $\odot(OIXY)$. This means $MX=MO$. But on the other hand, $OX = OM$. This forces that $\triangle OMX$ is an equilateral triangle. This implies that $\angle MOX = 60^\circ$. Similarly we also get that $\angle YOM = 60^\circ$. Adding these two values, we get that $\angle YOX = 120^\circ$. Now since $OIXY$ is cyclic, we finally get $\angle XIY = \angle XOY = 120^\circ$ and we are done.

Sol:- Part (a):- It suffices to show that $AI_A',BI_B',CI_C'$ concur at a point on feuerbach hyperbola $h$. Let $H,I,N_a$ be orthocenter,incenter, nagel point of $\Delta ABC$. We know that $H,I,N_a \in h$. Let $Q$ be the unique point on $h$ such that $-1=(N_a,H;I,Q)_h$. $D$ be the tangency point of $A$ excircle with $BC$.$(N_a,H;I,Q)_h=-1=(D,\infty_{I_AI_A'};I_A,I_A') \stackrel{A}{=}(N_a,H;I,AI_A' \cap h)_h \implies A-Q-I_A'$ are collinear. In similar way $Q \in BI_B'$ and $Q \in CI_C'$. Hence $P$ ,the isogonal conjugate of $Q$ lies on $OI$. Part(b):- Let $I_C$ be $C$ excenter,$M$ be the midpoint of arc $BC$ not containing $A$. $I'$ be the bevan point (circumcenter of $\Delta I_AI_BI_C$) ,we know $O$ is midpoint of $II'$ and $I' \in I_AD$. Note that $\Delta I_ABCMD \stackrel{-}{\sim} \Delta I_AI_BI_CI'A \implies MDI'A$ cyclic ,since $MD \parallel II'$ by reims we obtain $II_A'I'A$ cyclic.Now we have $\measuredangle PAI=\measuredangle I_AAI_A'$ and $\measuredangle AIP=\measuredangle AI_A'I'=\measuredangle AI_A'I_A \implies \Delta PAI \stackrel{+}{\sim} \Delta I_AAI_A' \implies API_AI'$ is cyclic. Since $OM \parallel I_AI'$ by reims theorem we obtain $APMO$ cyclic.By shooting lemma $OI \cdot OP=OA^2 $, thus by converse of shooting lemma we obtain $OIXY$ concyclic.Let the second tangents from $X,Y$ to incircle meet $(ABC)$ at $Z$ (due to poncelet porism). Let $\angle XZY= x$. Then $90^\circ +\frac{x}{2}=\angle XIY=\angle XOY=2x \implies x=60^\circ \implies \angle XIY=120^\circ.$

(Note: in solving this, part (b) was used to identify the point $P$)