Let $O$, $I$, and $\omega$ be the circumcenter, the incenter, and the incircle of nonequilateral $\triangle ABC$. Let $\omega_A$ be the unique circle tangent to $AB$ and $AC$, such that the common chord of $\omega_A$ and $\omega$ passes through the center of $\omega_A$ . Let $O_A$ be the center of $\omega_A$. Define $\omega_B, O_B, \omega_C, O_C$ similarly. If $\omega$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively, prove that the perpendiculars from $D$, $E$, $F$ to $O_BO_C , O_CO_A , O_AO_B$ are concurrent on the line $OI$. Pitchayut Saengrungkongka
Problem
Source: IMEO Problem 6
Tags: IMEO, geometry, concurrency
MarkBcc168
16.07.2020 02:22
My problem! Didn't expect this to appear in the contest since it's quite technical for an Olympiad problem. I will post the solutions I submitted few hours later.
Physicsknight
16.07.2020 09:07
Lemma- Given circle $(O,R)$ and $P$ outside $(O). $ Draw $2$ tangents $PA,PB$ to $(O)$ at $A$ and $B.PO\cap AB=M.$ $A$ is (unique) circle $(I)$ tangent to $PA,PB$ such that common chord of $(I) $ and $(O)$ passes through centre $I.$ We have $P,I,M,O$ are harmonic $((PM,IO)=-1)$ Proof: Let $XY$ be the diametre of $(O)$ such that $XY\parallel AB,$ it is clear that if $(I)$ cuts $(O)$ at $X_O$ and $Y_O. $ $P,X,X_O $ are colinear and $P,Y,Y_O$ are colinear. Hence $(XX_O,AB)=-1,$ with $Y\in (O)$ then $Y(XX_O,AB)=-1. $ Note that, $YX\parallel AB, $ so that $YXO $ passes through midpoint of $AB,$ or $XO,Y,M$ are colinear. Similarly to $X,Y_O,M. $ $M$ is the orthocentre of $PXY,$ with $X_OY_O $ cuts $PM$ at $I. $ $(PM,IO)=-1,$ as desired Back to main problem Let $IA$ cuts $EF$ at $A_O,$ $IB$ cuts $FD$ at $B_O $ and $IC$ cuts $DE$ at $C_O. $ Let $A_1,B_1,C_1$ are the midpoints of $AA_O,BB_O,CC_O.$ From the lemma, we have $(AA_O,O_AI)=-1.$ Hence with $r$ is the radius of $(I)$ then $IA_1\cdot IO_A=IA_O\cdot IA=r^2.$ Similarly, $r^2=IA_1\cdot IO_A=IB_1\cdot IO_B=IC1.IO_C\,(\star).$ Let $I_a,I_b,I_c$ are the excenters of $A,B,C$ of $ABC. AI_a,BI_b,CI_c$ are the altitudes of $I_aI_bI_c.$ It is clear that if $A_2,B_2,C_2$ are the midpoints of $DI_a,EI_b,FI_c$ then $(A_2)=(BB_OC_OC),(B_2)=(CC_OA_OA),(C_2)=(AA_OB_OB),$ and from $(\star). $ $IA_2\perp O_BO_C.$ We also have $(O)$ is the Euler circle of $I_aI_bI_c,$ and $I_bI_c\parallel EF,I_cI_a\parallel FD,I_aI_b\parallel DE. $ $OI $passes through center of $\odot(I_aI_bI_c),$ then $I_aD,I_bE,I_cF,OI$ are concurrent at $S.$ Therefore there exists a dilatation with center $S,$ call $\mathcal {H}: A_2\to D,B_2\to E,C_2\to F,I\to R. $ $3$ given perpendiculars are concurrent at $R$ with $R$ lies on $OI. $ This is easier than, I solved your's most difficult problem.
MarkBcc168
16.07.2020 17:08
As promised, here are my proposed solutions. Let $M_A, M_B, M_C$ be the midpoints of $\overline{EF}$, $\overline{FD}$, $\overline{DE}$ respectively. Let $X, Y, Z$ denote the midpoints of $\overline{AM_A}$, $\overline{BM_B}$, $\overline{CM_C}$ respectively. We divide the solution into three steps. Step 1: $O_A$ and $X$ are inverses with respect to $\omega$. We present three proofs for this step.
Let $\omega_A$ touches $AB$ and $AQ$ at $P$ and $Q$ respectively. Let $P'$ and $Q'$ be the reflection of $P$ and $Q$ across $O_A$. Since the radical axis of $\omega$ and $\omega_A$ is the segment joining the midpoints of $EQ$ and $FP$, we deduce that $P', Q'$ lies on $EF$.
Notice that the tangents of $\omega_A$ at $P'$ and the tangent of $\omega$ at $F$ are parallel. Thus the insimilicenter of $\omega$ and $\omega_A$ lies on $P'F$, hence it must be the midpoint of $EF$. Since $A$ is the exsimilicenter of these circles, we deduce that $(O_A, I; A, M_A)=-1$. Therefore $IO_A\cdot IX = IA\cdot IM_A = r^2$, hence we are done.
Let $\omega\cap\omega_A = \{U,V\}$. Consider an inversion at $A$ that sends $\omega_A\to\omega$. This inversion must fix $U,V$ while mapping $O_A\to M_A$. Thus $AO_A\cdot AM_A = AU^2$ or $\angle AUM_A = 90^{\circ}$.
Since $IM_A\cdot IA = IE^2 = IU^2$, we get that $IU$ is tangent to the circle $\gamma$ with diameter $AM_A$. However, $X$ is the center of $\gamma$, hence $\angle IUX = 90^{\circ}$ or $IX\cdot IO_A = IU^2$ as desired.
Let $\alpha = \tfrac{\angle A}{2}$, $r$ be the inradius, and $r_a$ be the radius of $\omega$. Therefore
$$r_a^2 = -\mathrm{Pow}(O_A,\omega) = r^2 - IO_A^2 \implies (r-r_a)(r+r_a) = IO_A^2$$However, $AI = \tfrac{r}{\sin\alpha}$ and $AO_A = \tfrac{r_a}{\sin\alpha}$. Thus
\begin{align*}
(r-r_a)(r+r_a) = \left(\frac{r-r_a}{\sin\alpha}\right)^2
&\implies (\sin\alpha)^2(r+r_a) = r-r_a \\
&\implies r_a = r\cdot\frac{1-(\sin\alpha)^2}{1+(\sin\alpha)^2} \\
&\implies IO_A = \frac{1}{\sin\alpha}\cdot \frac{2r(\sin\alpha)^2}{1+(\sin\alpha)^2} = \frac{2r\sin\alpha}{1+(\sin\alpha)^2}
\end{align*}Thus
$$\frac{r^2}{IO_A} = \frac 12\left(r\sin\alpha + \frac{r}{\sin\alpha}\right) = \frac 12(IA+IM_A) = IX$$as desired.
Before the second step, we define a few points. Let $I_A, I_B, I_C$ be the $A,B,C$-excenters. Let $I_A', I_B', I_C'$ be the reflections of $I_A, I_B, I_C$ across $I$. Step 2: $O_BO_C\perp DI_A'$ First, we note that $B,C,M_B, M_C$ are concyclic due to power of point from $I$. Let $S$ denote the center of this circle. Notice that $SY\parallel DF\parallel BI_A$ and $SZ\parallel DE\parallel CI_A$. Thus $S$ is actually the midpoint of $DI_A$ due to homothety from $D$. Now we note that from Step 1, we get that $Y,Z,O_b, O_c$ are concyclic. Thus \begin{align*} \measuredangle(IS, O_BO_C) &= \measuredangle SIO_B + \measuredangle IO_BO_C \\ &=\measuredangle SIY + \measuredangle YO_BO_C \\ &= \measuredangle SZY + \measuredangle YZO_C \\ &= \measuredangle SZO_C \\ &= 90^{\circ} \end{align*}so $IS\perp O_BO_C$. Finally, we remark that $IS\parallel I_A'D$ due to $IS$ being the mid-segment of $\triangle I_AI_A'D$. Step 3: Conclusion From Step 2, the concurrency point is the homothetic center of $\triangle DEF$ and $\triangle I_A'I_B'I_C'$. It's well known (as inversion around $\omega$ sends $\odot(ABC)\to\odot(M_AM_BM_C)$), that $OI$ is the Euler Line of $\triangle DEF$. Moreover, $OI$ is also the Euler Line of $\triangle I_AI_BI_C$, hence it's the Euler Line of $\triangle I_A'I_B'I_C'$. Thus $OI$ passes through the center of homothety, yielding the result.
It's easy to show that the perpendiculars from $D,E,F$ to $\overline{O_BO_C}$, $\overline{O_CO_A}$, $\overline{O_AO_B}$ are concurrent: just recognize that $\triangle DEF$ and $\triangle O_AO_BO_C$ are orthologic. This is the reason why the second part is necessary.
Finding the claim in Step 1 seems necessary, although there are probably many proofs. However, Step 2 and 3 can be complete without introduction of $\triangle I_A'I_B'I_C'$. After we get that $IS\perp O_BO_C$, we can proceed as follows.
By homothety, $DI_A, EI_B, FI_C$ are concurrent at $T\in OI$. Let $H$ be the orthocenter of $\triangle DEF$. Then upon considering the harmonic pencil $I(S,{\infty}_{DI_A}; D, I_A)=-1$, we translate the pencil to $D$ and project onto $OI$, we see that the concurrency point is the harmonic conjugate of $T$ w.r.t. $HI$.
In terms of Kimberling Center, the concurrency is $X_{3340}$ of $\triangle ABC$ and has the barycentric coordinate
$$\left(\frac{a(3b+3c-a)}{b+c-a} : \frac{b(3a+3c-b)}{a+c-b} : \frac{c(3a+3b-c)}{a+b-c}\right).$$However, due to the circles $\omega_A, \omega_B, \omega_C$, knowing those points does not offer any help in solving this problem.
pinetree1
24.07.2020 06:00
Here's a solution with Daniel Hong. Similar in flavor to the official solution, but with a slightly different finish. Let $A', B', C'$ be the midpoints of $\overline{EF}, \overline{FD}, \overline{DE}$. The key observation is that: Claim: We have $(A, A'; O_A, I) = -1$. Proof. Let $\omega_A$ and $\omega$ intersect at $X$ and $Y$, and let $\omega_A$ hit $\overline{AC}$ and $\overline{AB}$ at $R$ and $S$. Consider the inversion $\Gamma$ centered at $A$ swapping $\omega$ and $\omega_A$. Note that $\Gamma$ fixes $X$ and $Y$, while swapping $\{R, E\}$, $\{S, F\}$, and $\{O_A, A'\}$. Since $\omega_A$ has diameter $\overline{XY}$, we know that $\angle XRS = 90^\circ$; hence \begin{align*} 90^\circ &= \angle XRY = \angle YRA - \angle XRA = \angle EYA - \angle EXA \\ &= \angle XAY + \angle XEY = \angle XAY + \tfrac{1}{2}\angle XIY. \end{align*}Since $\triangle IXY$ is isosceles, we have $\angle IXY = 90^\circ - \tfrac{1}{2}\angle XIY = \angle XAY$. In particular, this implies $\overline{IX}$ and $\overline{IY}$ are tangent to $(AXY)$. Now $\overline{XY}$ is the polar of $I$ with respect to $(AXY)$, which proves the claim. $\blacksquare$ [asy][asy] size(300); defaultpen(fontsize(10pt)); pair A, B, C, D, E, F, I, AA, BB, CC, EE, FF, X, Y, OA, OB, OC, P, KA, KB, KC; A = dir(120); B = dir(210); C = dir(330); I = incenter(A, B, C); D = foot(I, B, C); E = foot(I, C, A); F = foot(I, A, B); AA = midpoint(E--F); BB = midpoint(F--D); CC = midpoint(D--E); P = extension(BB, CC, B, C); OB = extension(P, extension(C, BB, B, CC), B, BB); OC = extension(P, extension(C, BB, B, CC), C, CC); X = IP(CP(midpoint(A--AA), A), incircle(A, B, C), 0); Y = IP(CP(midpoint(A--AA), A), incircle(A, B, C), 1); OA = midpoint(X--Y); EE = foot(OA, A, C); FF = foot(OA, A, B); KA = circumcenter(B, BB, C); KB = circumcenter(C, CC, A); KC = circumcenter(A, AA, B); draw(A--B--C--cycle, orange); draw(incircle(A, B, C), red); draw(D--E--F--cycle, heavyred); draw(CP(OA, EE), heavygreen+dotted); draw(CP(OB, foot(OB, B, C)), heavygreen+dotted); draw(CP(OC, foot(OC, B, C)), heavygreen+dotted); draw(KA--KB--KC--cycle, magenta+linewidth(0.9)); draw(circumcircle(A, B, C), red); draw(A--I^^B--I^^C--I, lightblue); draw(arc(KA, abs(KA-B), 20, 160), heavycyan+dashed); draw(circumcircle(A, X, Y), lightblue+dashed); draw(OB--OC, heavygreen+linewidth(0.8)); draw(KA--I, purple+linewidth(0.8)); dot("$A$", A, dir(A)); dot("$B$", B, dir(190)); dot("$C$", C, dir(350)); dot("$D$", D, dir(270)); dot("$E$", E, dir(45)); dot("$F$", F, dir(150)); dot("$I$", I, dir(150)); dot("$R$", EE, dir(25)); dot("$S$", FF, dir(170)); dot("$O_A$", OA, dir(20)); dot("$O_B$", OB, dir(100)); dot("$O_C$", OC, dir(270)); dot("$A'$", AA, dir(130)); dot("$B'$", BB, dir(90)); dot("$C'$", CC, dir(100)); dot("$K_A$", KA, dir(270)); dot("$K_B$", KB, dir(45)); dot("$K_C$", KC, dir(150)); dot("$X$", X, dir(180)); dot("$Y$", Y, dir(20)); [/asy][/asy] By Power of a Point at $I$, we have $BB'C'C$ cyclic, and let $K_A$ be the center of its circumcircle. Define $K_B$ and $K_C$ similarly. Note that $\triangle K_AK_BK_C$ and $\triangle DEF$ are homothetic, as $\overline{K_BK_C}$ and $\overline{EF}$ are both perpendicular to $\overline{AI}$. Claim: Let $T$ be the center of the homothety $\mathcal H$ sending $\triangle DEF$ to $\triangle K_AK_BK_C$. Then $T$ lies on $\overline{OI}$. Proof. Note that the altitude from $K_A$ to $\overline{K_BK_C}$ is the perpendicular bisector of $\overline{B'C'}$, and so the orthocenter of $\triangle K_AK_BK_C$ is the circumcenter of $\triangle A'B'C'$, and also the nine-point center of $\triangle DEF$. Since $T$ lies on the line connecting the orthocenters of $\triangle DEF$ and $\triangle K_AK_BK_C$, we see that $T$ must lie on the Euler line of $\triangle DEF$. But this is precisely line $OI$, which proves the claim. $\blacksquare$ From the first claim, we see that $\overline{O_BO_C}$ is the polar of $I$ with respect to $(BB'C'C)$, and so $\overline{O_BO_C}\perp \overline{IK_A}$. Thus the desired concurrency point is $J = \mathcal H^{-1}(I)$, which lies on $\overline{OI}$ by the second claim. This completes the proof.
amar_04
09.09.2020 18:48
Beautiful Problem! There is a cute Generalization of the Problem which I have given below which is far more easier to solve than the Original Problem! IMEO 2020 P6 wrote: Let $O$, $I$, and $\omega$ be the circumcenter, the incenter, and the incircle of nonequilateral $\triangle ABC$. Let $\omega_A$ be the unique circle tangent to $AB$ and $AC$, such that the common chord of $\omega_A$ and $\omega$ passes through the center of $\omega_A$ . Let $O_A$ be the center of $\omega_A$. Define $\omega_B, O_B, \omega_C, O_C$ similarly. If $\omega$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively, prove that the perpendiculars from $D$, $E$, $F$ to $O_BO_C , O_CO_A , O_AO_B$ are concurrent on the line $OI$. Pitchayut Saengrungkongka $\textbf{LEMMA 1:-}$ Let $\omega$ be a circle and $P$ be an arbitary point. Let $\{A,B\}\in\omega$ such that $\overline{PA},\overline{PB}$ are tangents to $\omega$. Let $\gamma$ be a circle with center $M$ and tangent to $\overline{PA},\overline{PB}$ such that $M\in\ell=\omega\cap\gamma$. Then $(O,M;Q,P)$ forms a harmonic bundle. Proof:- Let $\omega\cap\gamma=\{X,Y\}$. Consider the Inversion $\Psi$ around $P$ which swaps $\{\omega,\gamma\}$. So, $\Psi$ swpas $\{M,Q\}$ and fixes $\{X,Y\}$ where $Q=\overline{AB}\cap\overline{OP}$. So, $\measuredangle QXP=\measuredangle PXQ=\frac{\pi}{2}$ also $OX^2=OY^2=OP\cdot OQ\implies\overline{XY}$ is the Polar of $O$ WRT $\odot(AQ)$. So, $(O,M;Q,P)=-1$. ____________________________________________________________________________________________ $\textbf{LEMMA 2:-}$ Let $\Phi:\ell_1\mapsto\ell_2$ be a Homography between two pencil of lines $\ell_1$ and $\ell_2$. Then $\Phi(\ell_1)=\ell_2$ if and only if there $\exists$ a Line $\mathcal T$ such that $\ell_1\cap\ell_2\in\mathcal T$ Proof:- See Theorem 2.1 (Steiner Conic) 1 here ____________________________________________________________________________________________ Coming back to the Problem. Let $\Delta XYZ$ denote the Anticomplementary Triangle of $\Delta ABC$. By the so called Prism Lemma we have $\overline{YZ},\overline{BC},\overline{O_BO_C}$ are concurrent at a point from $\textbf{LEMMA 1}$ and similarly others. Now we are ready for a Generalization. ____________________________________________________________________________________________ $\textbf{GENERALIZATION:-}$ $ABC$ be a triangle with Contanct triangle $DEF$. Let $\Delta XYZ$ be the Anticomplementary triangle of $\Delta DEF$. Let $\tau$ be the Perspectix of $\Delta ABC$ and $\Delta XYZ$ cutting $\overline{BC},\overline{CA},\overline{AB}$ at points $P,Q,R$ respectively. Let $U,V,W\in\overline{AI},\overline{BI},\overline{CI}$ respectively such that $\{P,V,W\},\{Q,U,W\},\{R,V,U\}$ are collinear. Then the Line Joining the Orthology Centers of $\Delta UVW$ and $\Delta XYZ$ passes through Circumcenter $O$ of $\Delta ABC$. Proof:- Let $Z$ be the Intersection of the Perpendiculars from $D,E,F$ on $\overline{VW},\overline{UW},\overline{UV}$ respectively. The Concurrency is obvious by Sondats Theorem as $\Delta XYZ$ and $\Delta UVW$ are Orthologic. Let $\mathcal{T}_1,\mathcal{T}_2$ denote the Perpendicular from $D$ to $\overline{VW}$ and the Perpendicular from $E$ to $\overline{UW}$ respectively. Now Animate $V$ on $\overline{BI}$. Then $V\mapsto\overline{PV}\mapsto\overline{WQ}\mapsto\mathcal{T}_1\mapsto\mathcal{T}_2$ is a Homography. When $V$ is a point on $\overline{BI}$ such that $\overline{PV}\|\overline{CI}$ then $W=\infty_{\overline{CI}}$. So, $\mathcal{T}_1\equiv\mathcal{T}_2$. Hence by $\textbf{LEMMA 2}$ we get that $Z$ lies on a fixed line $\mathcal{L}$. Now to determine the Line it suffices to check two positions of $V$ on $\overline{BI}$.When $V\equiv B$ then $Z\equiv I$ and when $V\equiv X$ then $Z$ is the De- Long Champ Point of $\Delta XYZ$. Hence, the Fixed Line is the Euler Line $(\mathcal{E})$ of $\Delta DEF$. Hence, $Z\in\overline{OI}$. $\qquad\blacksquare$
bruckner
06.02.2022 17:28
There is a solution for this problem using complex numbers. As usual, let $\omega$ be the unit circle and $D=d$, etc., $A=\frac{2ef}{e+f}$, etc. To calculate $O_A$, we look at the homothety $\omega \rightarrow \omega_A$ to consider a diameter $X_1X_2$ of $\omega$ such that $X_1X_2$ and $EF$ are parallel and points $Y_1, Y_2$ on $\omega$ different from $X_1, X_2$ such that $A,X_1,Y_1$ (as well as $A,X_2,Y_2$) are collinear. We would have then $O_A=\frac{1}{2}(y_1+y_2)$. Calculating it: $$ x+y=\frac{2ef}{e+f}+\frac{2}{e+f}xy \Leftrightarrow y\frac{e+f-2x}{e+f}=\frac{2ef-x(e+f)}{e+f} \Leftrightarrow y=\frac{2ef-x(e+f)}{e+f-2x}$$$$\Rightarrow \frac{1}{2}(y_1+y_2)=\frac{1}{2}\left( \frac{2ef-x_1(e+f)}{e+f-2x_1} + \frac{2ef-x_2(e+f)}{e+f-2x_2} \right)=\frac{1}{2}\left( \frac{4ef(e+f)+4ef(e+f)}{(e+f)^2+4ef} \right)$$$$\Rightarrow O_A=\frac{4ef(e+f)}{e^2+6ef+f^2}$$(We used $x_1+x_2=0$ and $x_1x_2=ef$) Now, we calculate the line passing through $D$ perpencidular to $O_BO_C$: $$\frac{d-x}{\frac{4df(d+f)}{d^2+6df+f^2}-\frac{4de(d+e)}{d^2+6de+e^2}} \in i\mathbb{R} \Leftrightarrow \frac{d-x}{4d\left( \frac{(df+f^2)(d^2+6de+e^2)-(de+e^2)(d^2+6df+f^2)}{(d^2+6df+f^2)(d^2+6de+e^2)} \right)} \in i \mathbb{R}$$$$\Leftrightarrow \frac{(d^2+6df+f^2)(d^2+6de+e^2)}{4d} \frac{d-x}{d^3(f-e)+d^2(f^2-e^2)+d(5ef(f-e))} \in i \mathbb{R}$$$$\Leftrightarrow \frac{(d^2+6df+f^2)(d^2+6de+e^2)}{4d^2(f-e)} \frac{d-x}{d^2+d(e+f)+5ef} \in i \mathbb{R}$$$$\Leftrightarrow \frac{d-x}{d(d^2+d(e+f)+5ef)}=\frac{1-d\bar{x}}{5d^2+d(e+f)+ef}$$
Let $L_A$ be the midpoint of minor arc $BC$ in $(ABC)$ (define $L_B,L_C$ similarly) and consider the homothety that sends $L_AL_BL_C$ to $DEF$ (exists since corresponding sides are parallel).
The Euler Line of $L_AL_BL_C$ is $OI$ and the Euler LIne of $DEF$ passes through $I$, but since they are parallel they must be the same.
) and substitute $\bar{x}=x\frac{de+df+ef}{def(d+e+f)}$ in the previous relation (in order to prove $x$ is symmetric in $d,e,f$, which finishes the problem). $$ \frac{d-x}{d(d^2+d(e+f)+5ef)}=\frac{1 - x \frac{de+df+ef}{ef(d+e+f)}}{5d^2+d(e+f)+ef}$$$$\Leftrightarrow d(5d^2+d(e+f)+ef-d^2-d(e+f)-5ef) = x \left( 5d^2+d(e+f)+ef - \frac{d(de+df+ef)}{ef(d+e+f)} (d^2+d(e+f)+5ef) \right)$$$$\Leftrightarrow 4def(d^2-ef)(d+e+f)=x((5d^2+d(e+f)+ef)ef(d+e+f)-d(de+df+ef)(d^2+d(e+f)+5ef))$$$$\Leftrightarrow 4def(d^2-ef)(d+e+f)=x(-d^4(e+f)-d^3(e-f)^2+def(e-f)^2+e^2f^2(e+f))$$$$\Leftrightarrow 4def(d^2-ef)(d+e+f)=x(d^2-ef)(-d^2(e+f)-ef(e+f)-d(e-f)^2)$$$$\Leftrightarrow 4def(d+e+f)=x(-d^2e-d^2f-e^2f-ef^2-de^2-df^2+2def)$$ And we are done.
math_comb01
30.03.2024 15:58
Nice problem! Let $A',B',C'$ be the midpoints of $EF,DF,DE$. Key Claim: $A'$ is the insimilscenter of $\omega_a,\omega$
It is clear that $\omega_a$ must be unique, so we give a construction of $\omega_a$, draw a line paralllel to $EF$ through $I$ intersecting the incircle again at $G,H$, extend $GA',HA'$ to meet $\omega$ at $Y,X$ then let the midpoint of $XY$ be $O_A$ define $\omega_a$ as the circle centered at $O_A$ passing through $X,Y$, note that we clearly have $A,G,X$ collinear, now it follows by brocard that $A$ is the exsimiliscenter of $\omega_a,\omega$
Note that due to incircle inversion we have $BCB'C'$ and friends cyclic. Let $P_A,P_B,P_C$ denote the centers of these circles. Key Claim 2: The perspector of $\triangle P_AP_BP_C ,\triangle DEF \in \overline{OI}$
The key is that, the perpendicular from $P_A$ to $P_BP_C$ is actually the perendicular bisector of $B'C'$, and therefore the perspector lies on euler line of contact triangle which is well known to be $OI$.
Notice that by the first claim we have $O_A$ as the pole of $B'C'$ in $BB'CC'$, and therefore $O_BO_C \perp IP_A$ implying the desired concurrence point lies on $OI$, as desired. $\blacksquare$