http://rmm.lbi.ro/index.php?id=solutions_math

Congrats F.Ivlev ! Very very cute problem

Lemma:- Let $\ell$ be a line and $\Gamma$ be circle.Suppose $P$ is the pole of $\ell$ wrt $\Gamma$.Now let $\triangle ABC$ be inscribed in $\Gamma$.Let $\triangle A'B'C'$ be the circumcevian triangle of $\triangle ABC$ wrt $P$.Then $\ell$ is the perspective axis of $\triangle ABC$ and $\triangle A'B'C'$. Proof:- Let $A_1=BC\cap B'C', B_1=AC\cap A'C', C_1=AB\cap A'B'$. Now applying Pascal's theorem on the hexagon $B'C'ABCA'$ we get $A_1,C_1, C'A\cap CA'$ are collinear. Again $\triangle AC'B$ and $\triangle A'CB'$ are perspective.So $C_1, C'A\cap CA', BC'\cap B'C$ are collinear.Hence we conclude that $A_1,C_1, BC'\cap B'C$ are collinear.But this line is nothing but the polar of $P$ i.e $\ell$.Hence we conclude that $A_1B_1C_1\equiv \ell$. Back to the main proof Clearly $AA'$ is the radical axis of $\omega _b, \omega_c$, $BB'$ is the radical axis of $\omega_c, \omega_a$ and $CC'$ is the radical axis of $\omega_a, \omega_b$.So by radical axis theorem $AA', BB', CC'$ are concurrent. Let $(I)$ be the incircle of $\triangle ABC$ and $\triangle DEF$ is its intouch triangle.$D', E', F'$ are the touch points of $(I)$ with $\omega_a, \omega_b, \omega_c$ respectively.Now let tangents at $D', E', F'$ meets $BC, CA, AB$ at $X,Y,Z$ respectively.Then its easy to show that $XYZ$ is the radical axis of $(I)$ and $\odot ABC$.So $X$ is the pole of $DD'$ wrt $(I)$ and similar for others.So $DD', EE', FF'$ concur at the pole of $XYZ$ wrt $(I)$.Now consider the circles $(I), \omega_b, \omega_c$.Then by radical axis theorem the lines $AA', E'E', F'F'$ are concurrent ,say at $X_1$.Then $X_1$ is the pole of $E'F'$ wrt $(I)$.Since $A, X_1, A'$ are collinear, their polars i.e $EF,E'F',\text{Polar of A'}$ are concurrent.So $\triangle DEF$ and the triangle formed by the polars of $A', B', C'$ are perspective wrt the perspective axis of $\triangle DEF, \triangle D'E'F'$.But according to our lemma this perspective axis is the polar of the perspective point of $\triangle DEF, \triangle D'E'F'$, i.e the radical axis of $(I), \odot ABC$.So we conclude that $AA', BB', CC'$ concur at the pole of the radical axis of $(I), \odot ABC$ wrt $(I)$.Since $IO\perp \text{Radical axis of (I),circumcircle of ABC}$ we conclude that the pole of the radical axis of $(I), \odot ABC$ lies on $IO$.Hence $AA', BB', CC'$ concur on $IO$.

Daaaamn! That problem was so incredibly similar to my solution of this problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2280575&sid=06651024ba8c2a2d88a8d3bf555bf086#p2280575 (I don't know why, but it's written that this is from 2010 MO, but it's from 2011 MO). During the polish final I solved that using radical axis of incenter and vertices and homothety. But during Romanian I didn't see that sixth problem is so similar :/. If I only drew one line :<... 75% of first official solution is like copy+paste from my solution from polish final :/. Of course very nice and hard problem .

Here is my solution : By radical axis theorem , we can immediately conclude that the three lines intersect at the radical centre $ R $ of the three circles . To show that this radical centre is on $ IO $ , consider the inversion w.r.t to a circle centered at $ R $ and with radius $ \sqrt{ - \text{ power at R to the circles } } $ and then the reflection w.r.t $ R $ . ( It is actually the same as the case that the three circles are disjoint so that we can choose an inversion at $ R $ sending these circles to themselves . ) We find that the images of $ \omega_A ~,~ \omega_B ~,~ \omega_C $ are themselves and the incircle becomes the larger circle tangent to them ( internally ) . From the property of this transformation ( inversion + homothety ) , $ R ~,~ I $ and the centre of this circle are collinear . We now show that this centre is actually $ O $ ! Let $ A_1 $ be the mid-pt of arc $BC$ of $ \omega_A $ that does not intersect the incircle . $ B_1 ~,~ C_1 $ are defined in similar mannar . It is well-known that the power at $A_1$ to the incircle is equal to $ A_1B^2 $ and $ A_1C^2 $ , so $ A_1 $ is the radical centre of the incircle , 'circle' $ B$ and 'circle' $C$ . ( $ B_1 ~,~ C_1 $ have similar property ). Therefore , we conclude that $ A_1 ~,~ B_1 $ lie on the radical axis of the incircle and 'circle' $C$ , and for other pairs , we have similar conclusion . Let $ D ~,~ E ~,~ F $ be the mid-pts of arc $ BC ~,~ CA ~,~ AB $ ( not containing the third vertex of $ \Delta ABC $ . ) of $(ABC)$ respectively . Then , we deduce that $ \Delta A_1B_1C_1 $ is homothetic to $ \Delta DEF $ and the centre of homothety is the intersection of the perpendicular bisectors of $ AB ~,~ BC ~,~ CA $ , which is $ O $ . Since $ O $ is the circumcenter of $ \Delta DEF $ , so is $ \Delta A_1B_1C_1 $ . But $ A_1 $ is an intersection of $ \omega_A $ and $ (A_1B_1C_1) $ and the centres of these two circles lie on the perpendicular bisector of $ BC $ which passes through $ A_1 $ , we therefore deduce that $ (A_1B_1C_1) $ is tangent to $ \omega_A $ . Similarly , $ (A_1B_1C_1) $ is tangent to $ \omega_B ~,~ \omega_C $ . In other words , $ (A_1B_1C_1) $ is the larger circle we previously described , which has the centre $ O $ !

Bonus points for the one who generalizes this as much as possible.

General problem Let $ABC$ be a triangle and a point $P$. $A_1B_1C_1$ is pedal triangle of $P$. $A_2B_2C_2$ is antipedal triangle of $P$. $A_1A_2,B_1B_2,C_1C_2$ cut pedal circle $(A_1B_1C_1)$ again at $A_3,B_3,C_3$. Let circumcircle $(ABC_3),(CAB_3)$ intersect again at $A_4$. Similarly, we have $B_4,C_4$. Prove that $AA_4,BB_4,CC_4$ are concurrent.

buratinogigle wrote: General problem Let $ABC$ be a triangle and a point $P$. $A_1B_1C_1$ is pedal triangle of $P$. $A_2B_2C_2$ is antipedal triangle of $P$. $A_1A_2,B_1B_2,C_1C_2$ cut pedal circle $(A_1B_1C_1)$ again at $A_3,B_3,C_3$. Let circumcircle $(ABC_3),(CAB_3)$ intersect again at $A_4$. Similarly, we have $B_4,C_4$. Prove that $AA_4,BB_4,CC_4$ are concurrent. only concurrent ? but can't it be immediately proved by radical axis theorem ?

Sorry, you are right, it is trivial , I will try again!

Thank you for all who say that like this problem. It's my I was very wonderful that this problem was taken on this Olympiad. I didn't expecte that it's so nice, but.

If we call $P_a, P_b,P_c$ be the intersections of $\omega_A, \omega_B, \omega_C$ with $(I)$, respectively, then $AP_a, BP_b, CP_c$ are concurrent. I find it by sketchpad but I don't know how to prove it. Who can help me?

yumeidesu wrote: If we call $P_a, P_b,P_c$ be the intersections of $\omega_A, \omega_B, \omega_C$ with $(I)$, respectively, then $AP_a, BP_b, CP_c$ are concurrent. I find it by sketchpad but I don't know how to prove it. Who can help me? Keep the notations in my first post in mind , There is a celebrated theorem which asserts that , for three points $A' , B' , C'$ on $(I)$ $AA' , BB' , CC'$ are concurrent if and only if $A' T_a , B' T_b, C' T_c$ . It can easily proved by trigonometric Ceva theorem . According to what I proved in my post we know $P_a T_a , P_b T_b, P_c T_c$ are concurrent and above theorem implies that $AP_a, BP_b, CP_c$ are concurrent.

I think this problem is a combine problem from: IMO Shortlist 2002: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=118682&sid=ff0420e5b46b2284383a55c03e1bc918#p118682 Vietnam TST 2003: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=268390&sid=ff0420e5b46b2284383a55c03e1bc918#p268390

Let the point of tangency between $\omega_A$ and the incircle be $A''$, and so on, and the excenters be $I_A$, $I_B$, $I_C$. Let the incircle be tangent to $BC$, $AC$, and $AB$ at $T_A$, $T_B$, and $T_C$, respectively. Lemma 1: $A''T_A$ passes through $I_A$. Proof: This is 2002 G7. Lemma 2: $I_AT_A$, $I_BT_B$, $I_CT_C$ are concurrent. Sketch of proof: Trig Ceva wrt $\triangle I_AI_BI_C$. Lemma 3: $A''T_A$, $B''T_B$, and $AA'$ are concurrent. Proof: Let the tangents to the incircle at $A''$ and $B''$ intersect at $N$. By the Radical Axis Theorem, $N$ lies on $AA'$. Let $T_AT_B$ and $A''B''$ intersect at $M$, then it's easy to show that $NA$ is the polar of $M$ wrt to $(I)$. Thus, $A''T_A$ and $B''T_B$ intersect on $NA$. From Lemmas 2 and 3, we can conclude that $AA'$, $BB'$, $CC'$, $A''T_A$, $B''T_B$, and $C''T_C$ are all concurrent. Call this point $Q$. Let $A''T_A$ meet $\omega_A$ again at $P_A$, and define $P_B$ and $P_C$ the same way. Since $P_AP_BA''B''$ is cyclic, $P_AP_B \parallel T_AT_B$. The same goes for $P_BP_C$ and $P_AP_C$, so $\triangle P_AP_BP_C$ is homothetic to $\triangle T_AT_BT_C$ wrt $Q$. Note that the homothety from $A''$ takes $T_A$ to $P_A$; thus, the tangent $l_A$ to $\omega_A$ at $P_A$ is parallel to $BC$; furthermore, $A''P_A$ bisects arc $BC$ in $\omega_A$. Define $l_B$ and $l_C$ similarly; then note that from the homothety from $Q$, $(P_AP_BP_C)$ is the incircle of the triangle formed by the intersections of $l_A$, $l_B$, and $l_C$. Let $O'$ be the circumcenter of $(P_AP_BP_C)$. Then $O'P_A \perp l_A$, so $O'P_A$ is the perpendicular bisector of $BC$. In addition, $O'P_B$ and $O'P_C$ bisect $AC$ and $AB$, respectively. Thus, $O = O'$. But $O'I$ passes through $Q$ by homothety, so $OI$ passes through $Q$, as desired.

As above, let $A''$ and $T_A$ be the points of tangency of (I) and $w_A$ and $BC$ respectively, etc. For notation's sake, for any point Z on (I), define ZZ to be the tangent to (I) at Z. Again, as previously noted, by the radical axis theorem applied to $w_A$, (I) and (O), we have that $A''A''$ and ${T_AT_A}\equiv{BC}$ intersect on the radical axis of (O) and (I), which means that if M is the pole of this radical axis with respect to (I), then M (which clearly lies on $OI$) is on $A''T_A$. Now let $\ell$ be the line through M and ${T_AB''}\cap{T_BA''}$. Since $M = {T_AA''}\cap{T_BB''}'$, by applying Pascal's Theorem to hexagons $T_AT_AA''T_BT_BB''$ and $A''A''T_AB''B''T_B$, we have that $C={T_AT_A}\cap{T_BT_B}$ is on $\ell$, as is ${A''A''}\cap{B''B''}$. Now if we consider the circles $w_A$,$w_B$, and (I), and their radical axes, we see that ${A''A''}\cap{B''B''}$, $C$ and $C'$ are collinear, which means that $C'$ is on $\ell$ as well.

Very cool problem My solution is not as nice as some others, but I thought it was pretty cool still.

r1234 wrote: Lemma:- Let $\ell$ be a line and $\Gamma$ be circle.Suppose $P$ is the pole of $\ell$ wrt $\Gamma$.Now let $\triangle ABC$ be inscribed in $\Gamma$.Let $\triangle A'B'C'$ be the circumcevian triangle of $\triangle ABC$ wrt $P$.Then $\ell$ is the perspective axis of $\triangle ABC$ and $\triangle A'B'C'$. Proof:- Let $A_1=BC\cap B'C', B_1=AC\cap A'C', C_1=AB\cap A'B'$. Now applying Pascal's theorem on the hexagon $B'C'ABCA'$ we get $A_1,C_1, C'A\cap CA'$ are collinear. Again $\triangle AC'B$ and $\triangle A'CB'$ are perspective.So $C_1, C'A\cap CA', BC'\cap B'C$ are collinear.Hence we conclude that $A_1,C_1, BC'\cap B'C$ are collinear.But this line is nothing but the polar of $P$ i.e $\ell$.Hence we conclude that $A_1B_1C_1\equiv \ell$. Back to the main proof Clearly $AA'$ is the radical axis of $\omega _b, \omega_c$, $BB'$ is the radical axis of $\omega_c, \omega_a$ and $CC'$ is the radical axis of $\omega_a, \omega_b$.So by radical axis theorem $AA', BB', CC'$ are concurrent. Let $(I)$ be the incircle of $\triangle ABC$ and $\triangle DEF$ is its intouch triangle.$D', E', F'$ are the touch points of $(I)$ with $\omega_a, \omega_b, \omega_c$ respectively.Now let tangents at $D', E', F'$ meets $BC, CA, AB$ at $X,Y,Z$ respectively.Then its easy to show that $XYZ$ is the radical axis of $(I)$ and $\odot ABC$.So $X$ is the pole of $DD'$ wrt $(I)$ and similar for others.So $DD', EE', FF'$ concur at the pole of $XYZ$ wrt $(I)$.Now consider the circles $(I), \omega_b, \omega_c$.Then by radical axis theorem the lines $AA', E'E', F'F'$ are concurrent ,say at $X_1$.Then $X_1$ is the pole of $E'F'$ wrt $(I)$.Since $A, X_1, A'$ are collinear, their polars i.e $EF,E'F',\text{Polar of A'}$ are concurrent.So $\triangle DEF$ and the triangle formed by the polars of $A', B', C'$ are perspective wrt the perspective axis of $\triangle DEF, \triangle D'E'F'$.But according to our lemma this perspective axis is the polar of the perspective point of $\triangle DEF, \triangle D'E'F'$, i.e the radical axis of $(I), \odot ABC$.So we conclude that $AA', BB', CC'$ concur at the pole of the radical axis of $(I), \odot ABC$ wrt $(I)$.Since $IO\perp \text{Radical axis of (I),circumcircle of ABC}$ we conclude that the pole of the radical axis of $(I), \odot ABC$ lies on $IO$.Hence $AA', BB', CC'$ concur on $IO$. Nice Lemma It can be generalised for any conic: Lemma: (Generalised) Suppose line $\ell$ outside conic $\Gamma(ABC)$. Let the pole of $\ell$ wrt $\Gamma(ABC)$ be $P$ and let $A', B', C'$ be the (conic?)umcevian triangle of $P$ wrt $\triangle ABC$. Proof: Under a projective transformation, take the conic to a circle $\odot ABC$ and $\ell$ to infinity. Then, $P$ goes to the centre of $\odot \triangle ABC$, so $\triangle A'B'C'$ and $\triangle ABC$ are homothetic, so done. My proof goes very similar: $T_A, T_B, T_C$ be the tangency points of those circles with the incircle. Let the tangent $\equiv \ell_A$ through $T_A$ intersect $BC$ at $X$ and define $Y, Z$ similarly. (Note that $X$ is the midpoint of $DD'$ where $D'$ is the harmonic conjugate of $D$ wrt $BC$). Then, $XYZ$ is the radical axis of $\odot ABC$ and the incircle, hence $OI \perp XYZ$. Let $P = DT_A \cap ET_B$ wrt incircle. Then, by radical axis theorem, $AA'$ meets $\ell_B \cap \ell_C$. Then by pascals we get $T_CF \cap T_BE, \ell_B \cap \ell_C, P$ are collinear. But we get $A, P, T_CF \cap T_BE$ are also collinear by pascals, so $A, \ell_B \cap \ell_C, P$ are collinear. We conclude $P$ is the radical centre. Note that $P$ is the polar of $XYZ$ of incircle, so done. One does notice some interesting similarities. For instance, this sort of set-up was present in IMO 2002 shortlist, and 2009 Serbian MO (I think).

mahanmath wrote: Congrats F.Ivlev ! Very very cute problem

Nice solution, I have another idea to prove the concurrency of $P_aT_a, P_bT_b, P_cT_c, OI$. Let $T_bT_c \cap BC=S_a$. Similarly we define $S_bS_c$. It's easy to see that $P_aT_a$ is the angle bisector of $\angle BPC$. Hence $P_aT_a$ is the radical axis of $(I)$ and $(P_aT_aS_a)$, which is indeed the circle with diameter $T_aS_a$ (because $(BCT_aS_a)=-1$ ). Likewise, $P_bT_b$ is the radical axis of $(I)$ and $(P_bS_bT_b)$. Consider 3 circles $(I), (P_aT_aS_a), (P_bT_bS_b)$, their radical axises concur at a point $P$. Let $H$ be the orthocenter of $\triangle T_aT_bT_c$, it's easy to prove that $HI$ is the common radical axis of 3 circles $(P_aS_aT_a), (P_bT_bS_c); (P_cS_cT_c)$, and from a well-known fact, we know that $O,I,H$ lies on the Euler line of $\triangle T_aT_bT_c$. So $P_aT_a;P_bT_c$ and $OI$ are concurrent. Similarly, we shall have all 4 lines are concurrent at $P$.

WakeUp wrote: Let $ABC$ be a triangle and let $I$ and $O$ denote its incentre and circumcentre respectively. Let $\omega_A$ be the circle through $B$ and $C$ which is tangent to the incircle of the triangle $ABC$; the circles $\omega_B$ and $\omega_C$ are defined similarly. The circles $\omega_B$ and $\omega_C$ meet at a point $A'$ distinct from $A$; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA',BB'$ and $CC'$ are concurrent at a point on the line $IO$. (Russia) Fedor Ivlev Let $\Gamma$ be the incircle of $\triangle ABC$, which is tangent to $BC,CA,AB$ at $P,Q,R$ respectively. Let $P \equiv \Gamma \cap \omega_A, Q \equiv \Gamma \cap \omega_B, R \equiv \Gamma \cap \omega_C$, and finally, let $P_1 \equiv P'P \cap \omega_A$, with similar definitions for $Q_1, R_1$. $\Gamma$ and $\omega_A$ are homothetic center $P'$, so $P_1$ is the midpoint of the arc $\overarc{BC}$ not containing $P'$. Hence \begin{align*}\angle P_1BC = \angle P_1P'C = \angle P_1P'B \implies \triangle P_1BP \sim \triangle P_1PB \implies \frac{P_1B}{P_1P'} = \frac{P_1P}{P_1B} \implies P_1B^2 = P_1P \cdot P_1P'. \end{align*}$\therefore P_1$ is on the radical axis of $\Gamma$ and degenerate circle $B$. Similarly, $P_1$ lies on the radical axis of $\Gamma$ and $C$, with analogous results for $Q_1, R_1$. $\therefore P_1Q_1 \perp CI \perp PQ$, with analogous results for $Q_1R_1, R_1P_1$. It follows that there exists a homothecy $\Theta: \triangle PQR \to \triangle P_1Q_1R_1$ with center $X$, the point of concurrency of $PP_1, QQ_1, RR_1$. $PX \cdot XP' = QX \cdot XQ'$, since $PQP'Q'$ is cyclic. Noting $\frac{P_1X}{XP} = \frac{Q_1X}{XQ}$ gives $P_1X \cdot XP' = Q_1X \cdot XQ'$, so that $X$ lies on the radical axis $CC'$ of $\omega_A, \omega_B$. Similarly, $AA', BB'$ pass through $X$. $IP \parallel OP_1$ and $IQ \parallel OQ_1$, so it follows that $O$ is the image of $I$ in $\Theta$ . Hence $AA', BB', CC'$ concur at a point on $IO$, as required. Note: The point $X$ of the required concurrency is in fact the exsimilicenter of $\triangle ABC$, the homothetic center of the incircle and the circumcircle, and hence the homothetic center of the intouch and circummidarc triangles. It is $X_{56}$ in Kimberling's online encyclopedia of triangle centers.

Suppose $\omega_A,\omega_B,\omega_C$ are tangent to the incircle at $X,Y,Z$. By homothety and a well-known lemma, $XD$ intersects $\omega_A$ again at some point $M$, which is the radical center of the incircle, $B$, and $C$. Define $N,P$ similarly. Clearly $MNP$ is homothetic to the intouch triangle. But $DI\perp BC$, so $M$, the center of $\omega_A$, and the circumcenter of $MNP$ are collinear, and $(MNP)$ is tangent to $\omega_A$. Note that if $O$ is the circumcenter of $MNP$, $MO\perp BC$ and $M$ lies on the perpendicular bisector of $BC$, so $O$ is also the circumcenter of $ABC$. We now have two circles $\Gamma_1$, the incircle and $\Gamma_2$, the circumcircle of $MNP$, with three circles $\omega_A,\omega_B,\omega_C$ externally tangent to both. We aim to show that the radical center of the three circles lies on the line through the centers of $\Gamma_1$ and $\Gamma_2$. In fact, we claim that this radical center is the exsimilicenter of $\Gamma_1$ and $\Gamma_2$. This fact holds in general, but we will use the points already defined in the problem to prove it. Let $S$ be the exsimilicenter of the two circles. Then $S$ is the homothetic center of $DEF$ and $MNP$. By Monge's theorem or by our previous results, $S$ is collinear with $X,D,M$. So \[ \frac{SF}{SE}=\frac{SP}{SN} \] and \[ SY\cdot SF = SE\cdot SZ \] imply \[ SY\cdot SP = SZ\cdot SN \] Therefore $S$ has equal powers with respect to $\omega_B$ and $\omega_C$, so it is the radical axis of all $3$.

My solution : Let $ D\equiv \odot (I) \cap BC, E\equiv\odot (I) \cap CA, F\equiv\odot (I) \cap AB $ . Let $ X\equiv\omega_A \cap \odot (I), Y\equiv\omega_B \cap \odot (I), Z\equiv\omega_C \cap \odot (I) $ . Let $ \mathcal{P}(P, \odot) $ be the power of a point $ P $ with respect to a circle $ \odot $ From homothety with center $ X $ ( maps $ \odot (I) \mapsto \omega_A $ ) $ \Longrightarrow M_A\equiv XD \cap \omega_A $ is the midpoint of arc $ BC $ in $ \omega_A $ . Similarly, $ M_B\equiv YE \cap \omega_B, M_C\equiv ZF \cap \omega_C $ is the midpoint of arc $ CA $, arc $ AB $ in $ \omega_B, \omega_C $, respectively . Since $ \mathcal{P}(M_B,\odot(I))=\mathcal{P}(M_B,C)=\mathcal{P}(M_B,A), \mathcal{P}(M_C,\odot(I))=\mathcal{P}(M_C,B)=\mathcal{P}(M_C,A) $ , so $ M_BM_C $ is the radical axis of $ \{ \odot (I), A \} \Longrightarrow M_BM_C $ pass through the midpoint of $ AE $ and $ AF $ . Similarly, $ M_CM_A, M_AM_B $ is B-midline, C-midline of $ \triangle BFD, \triangle CDE $, resp $ \Longrightarrow \triangle DEF , \triangle M_AM_BM_C $ are homothetic . Let $ T \equiv M_AD \cap M_BE \cap M_CF $ be the homothety center of $ \triangle DEF $ and $ \triangle M_AM_BM_C $ . From Reim theorem and $ M_BM_C \parallel EF \Longrightarrow Y, Z, M_B, M_C $ are concyclic , so $ T $ is the radical center of $ \{ \odot (YZM_BM_C), \omega_B, \omega_C \} \Longrightarrow \mathcal{P}(T, \omega_B )=\mathcal{P}(T, \omega_C ) \Longrightarrow T \in AA' $ . Similarly, we can prove $ T \in BB' $ and $ T \in CC' \Longrightarrow AA', BB', CC' $ are concurrent at $ T $ . From $ ID \parallel OM_A, IE \parallel OM_B, IF \parallel OM_C \Longrightarrow \triangle DEF \cup I $ and $ \triangle M_AM_BM_C \cup O $ are homothetic $ \Longrightarrow T \in OI $ . Q.E.D ____________________________________________________________ Remark : It's well-known that $ TD, TE, TF $ pass through the A-excenter $ I_a $ , B-excenter $ I_b $ , C-excenter $ I_c $ of $ \triangle ABC $, respectively ( see here ) , so $ T $ is the homothety center of $ \triangle DEF $ and $ \triangle I_aI_bI_c \Longrightarrow T $ is $ X_{57} $ ( in ETC ) of $ \triangle ABC $ .

Incircle $(I)$ touches $BC,CA,AB$ at $D,E,F$ and $\omega_A,\omega_B,\omega_C$ touch $(I)$ at $T_A,T_B,T_C.$ Denote $\tau_A,\tau_B,\tau_c$ the tangents of $(I)$ at $T_A,T_B,T_C$ and $X \equiv \tau_b \cap \tau_C,$ $Y \equiv \tau_C \cap \tau_A,$ $Z \equiv \tau_A \cap \tau_B.$ Since $XT_B=XT_C,$ then $X$ is on radical axis $AA'$ of $\omega_B,\omega_C$ and similarly $Y \in BB'$ and $Z \in CC'$ $\Longrightarrow$ $\triangle ABC$ and $\triangle XYZ$ are perspective through the radical center $T$ of $\omega_A,\omega_B,\omega_C$ and their perspectrix is the radical axis $\tau$ of $(I),(O).$ Since $\tau$ does not intersect $(I),$ then there exists a homology sendind $\tau$ to infinity and taking $(I)$ into another circle $(J).$ Thus in this figure, $\triangle ABC$ and $\triangle XYZ$ become symmetric WRT $J$ $\Longrightarrow$ $AX,BY,CZ,DT_A,ET_B,FT_C$ concur at $J;$ their common midpoint. Hence, in the original figure $AA' \equiv AX,$ $BB' \equiv BY,$ $CC' \equiv CZ,$ $DT_A,$ $ET_B,$ $FT_C$ concur at $T \equiv X_{57}$ lying on $OI.$

A nice result!

Q.E.D. An additional challenge would be to prove that $AX,BY,CZ$ are concurrent, which is true!

Let $\triangle MNP$ be the intouch triangle of $\triangle ABC$, and let $\gamma$ be the incircle. Now $\omega_b$ and $\omega_c$ touch to $\gamma$ at $B_1$ and $C_1$ respectively, and the common tangents from this points cut at $A'', X \equiv B_1N \cap C_1P$, From Pascal's Theorem for fourth points in $B_1,C_1,P,N$ we get $A,X$ and $A''$ are collinear, also by radical axis on $\omega_b,\omega_c$ and $\gamma$ we get $A'' \in AA' \Rightarrow X \in AA'$ Moreover the common tangent between $\omega_b$ and the incircle touch $AC$ at $B_2$, let $\ell$ be the radical axis between $\gamma$ and $\odot (ABC)$, by radical axis theorem on $\gamma, \odot (ABC), \omega$ we get $B_2 \in \ell$, so the polar lines of $A_2,B_2,C_2$ (defined analogously) with respect to $\gamma$ are concurrent at $IO$,( the pole of $\ell)$, and this point is $X$ is a point fixed that belongs to $AA'$ and $IO$ as desired.

It's just a point that out-divides OI as R+r/2:r

Let the incircle be tangent to $BC$ and $\omega_A$ at $D$ and $D'$ respectively and similarly define $E$, $F$, $E'$, and $F'$. Let $D'D$ intersect $\omega_A$ again at $K_a$, and define $K_b$ and $K_c$ similarly. By homothety, we have that $K_a$ is the midpoint of arc $BC$ on $\omega_A$. Inversion about $K_a$ with radius $K_aB=K_aC$ swaps line $BC$ with $\omega_A$, so $K_aB^2=K_aC^2=K_aD\cdot K_aD'$. Hence, $K_a$ is the radical center of $B$, $C$, and the incircle. Note that $K_aK_bK_c$ is homothetic with $DEF$ because $K_b$ and $K_c$ both lie on the radical axis of $A$ and the incircle which is parallel to $EF$. Let $L$ be the center of homothety of $K_aK_bK_c$ and $DEF$. The perpendiculars from $D$ to $BC$, $E$ to $CA$, and $F$ to $AB$ concur at $I$, and the perpendiculars from $K_a$ to $BC$, $K_b$ to $CA$, and $K_c$ to $AB$ concur at $O$. Hence, $L$ lies on $IO$. Now, note that by Power of a Point $LD\cdot LD'=LE\cdot LE'=LF\cdot LF'$ and because it's the center of homothety $LK_a\cdot LD'=LK_b\cdot LE'=LK_c\cdot LF'$, so $L$ is the radical center of $\omega_A$, $\omega_B$, and $\omega_C$, or where $AA'$, $BB'$, and $CC'$ concur.

Define $I_A, I_B, I_C$ as the excenters of triangle $ABC$ corresponding to the vertices in the indices. Let $DEF$ be the contact triangle of $ABC$. It is well-known that triangles $DEF$ and $I_AI_BI_C$ are homothetic. Define $T$ as the center of this homothety and note that $T$ lies on the line $IO$ dividing it in the ratio $2r \, : \, 2R+r$. We will show that lines $AA',BB',CC'$ are concurrent at $T$. Let $\omega_A$ touch $\omega$ at $P$. Dilation at $P$ mapping $\omega_A$ to $\omega$ implies $PD$ bisects angle $BPC$. From IMO ShortList 2002 G7, we know $I_A$ lies on line $PD$. Let the $A$-excircle touch side $BC$ at $K$ and $M$ be the intersection of perpendicular bisector of $BC$ with $DI_A$. Obviously, $MD=MK \Longrightarrow MD=MI_A$ and from Fact 5, we know $M$ lies on $\omega_A$. Finally, we observe that $$\operatorname{Pow}(T, \omega_A)=TP\cdot TM=\frac{TD}{TM}\cdot \operatorname{Pow}(T, \omega)=\frac{2r}{2R+r}\cdot (TI^2-r^2),$$which is symmetric. It follows that $T$ is the radical center of $\omega_A, \omega_B, \omega_C$. Lines $AA', BB', CC'$ are concurrent at $T$, which lies on $IO$. $\, \square$

Let $(I),(O)$ denote the incircle, circumcircle of triangle $ABC$ respectively. Let $\ell$ be the radical axis of $(I)$ and $(O)$, and let $P_a, P_b, P_c$ be the intersections of $\ell$ with lines $BC, CA, AB$ respectively. Define $D,E,F$ as the intersections of $(I)$ with lines $BC, CA, AB$ respectively. Define $D', E', F'$ as the reflections of $D,E,F$ about lines $P_aI, P_bI, P_cI$ respectively. By the Radical Axis Theorem on $\omega_A, (I), (O)$, we have that $P_a$ is the radical center of these three circles, so $P_a$ lies on the radical axis of $\omega_A$ and $(I)$. This means that $D'$ lies on $\omega_A$, so $\omega_A = (BD'C)$. Similarly, $\omega_B = (CE'A)$ and $\omega_C = (AF'B)$. Now we will use the Radical Axis Theorem on $\omega_A, \omega_B, (I)$. Letting $C''$ be the intersection of $P_aD'$ and $P_bE'$, we see that $C,C',C''$ are collinear. Thus we only need to show that $AA'',BB'',CC'',IO$ are concurrent, where $A'',B''$ are defined similarly. Focus on $(I)$. Let $X$ be the intersection of lines $DD', EE'$. Note that $X$ is the pole of $\ell$, so $X$ also lies on $FF'$. Let $Q_c$ be the intersection of lines $DE, D'E'$. Define $Q_a, Q_b$ similarly. Note that $Q_c$ is the pole of line $CC''$. But we also know that $X$ lies on the polar of $Q_c$ (by Brokard's Theorem). Thus $CC''$ passes through $X$. We see immediately that $AA'', BB'',CC'', IO$ concur at $X$, as desired. $\blacksquare$

Isn't this a bit easy for a RMM P6 ? Or maybe because I was lucky to try (but fail to solve) 2002 ISL G7 today... Anways, here's a solution using geogebra Define $E_A$ to be the $A$-excircle, $T_A$ to be the touchpoint of the incircle with $\overline{BC}$, $X_A$ to be the touchpoint of $\omega_A$ with incircle $\omega$ and $F_A$ to be the midpoint of $\overline{E_AT_A}$. Define $E_B, E_C, X_B, X_C, F_B, F_C, T_B, T_C$ similarly. By v_enhance's solution to 2002 ISL G7, we have $E_A, F_A, T_A, X_A$ colinear $F_A \in \omega_A$ A very easy angle chasing gives $E_BE_C || T_BT_C$, and similarly for the other sides, so $E_AT_A, T_BE_B, T_CE_C$ concur at a point $Y$, the center of homothety mapping $\Delta T_AT_BT_C$ to $\Delta E_AE_BE_C$. I claim that $Y$ is the desired concurrency point. Indeed, letting $X$ be the circumcircle of $\Delta E_AE_BE_C$. Observe that by very easy angle chasing/very well known lemma, $O$ and $I$ are the nine-point center and orthocenter of $\Delta E_AE_BE_C$, so $X,O,I$ are collinear (they all lie on the Euler line of $\Delta E_AE_BE_C$). Since a homothety at $Y$ maps $\Delta T_AT_BT_C$ to $\Delta E_AE_BE_C$, the same homothety maps (Circumcenter of $\Delta T_AT_BT_C = I$) to (Circumcenter of $\Delta T_AT_BT_C = X$) too, so $I, X, Y$ are colinear. So together, we have $O, I, X, Y$ colinear, or $Y \in OI$ By radical center theorem, clearly $AA', BB', CC'$ are concurrent, and the concurrency point is the radical center of $\{ \omega_A, \omega_B, \omega_C \}$. So we need to show that $Pow_Y(\omega_A)$ is symmetric in $A,B,C$. But this is not hard. Note that $\frac{YE_A}{YT_A} = \frac{YX}{YI}$ for homothety reasons. So $$ Pow_Y(\omega_A) = YX_A \cdot YF_A = \frac{YX_A \cdot YE_A}{2} = \frac{1}{2} \cdot (YX_A \cdot YT_A) \cdot (\frac{YE_A}{YT_A}) = \frac{1}{2} (\frac{YX}{YI}) (Pow_Y(Incircle))$$, which is evidently symmetric in $A, B,C$

Surprisingly, I don't think anyone has this sol yet. It's actually rather short, but yeah this problem is really cute. I do think that it is kind of ruined by 2002 ISL G7 though because this makes it more of an "extension" than a standalone problem, but this is still a pretty cool problem.

Let $DEF$ be the contact triangle of $ABC$, and let $I_AI_BI_C$ be the excentral triangle. We claim the desired point is $P$, the exsimilicenter of $(DEF)$ and $(I_AI_BI_C)$. This lies on the line joining their centers, which is the Euler line of $I_AI_BI_C$. It's well known that $O$ is on this Euler line (by incircle inversion), so $P\in OI$. It suffices now to show that $P$ has equal power with respect to the circles $\omega_A$, $\omega_B$, $\omega_C$. Note that the triangles $DEF$ and $I_AI_BI_C$ are homothetic (USAMTS Round 2 anyone?), so $P=I_AD\cap I_BE\cap I_CF$. Let the scale factor be $k$, so $PI_A/PD=PI_B/PE=PI_C/PF=k$. The main idea to this problem is ISL 2002 G7, but we'll show the relevant pieces here without citation. We'll show that $I_AD$ passes through $X$, the tangency point of $\omega_A$ and the incircle, and that the other intersection $X'$ of $I_AD$ with $\omega_A$ is the midpoint of $I_AD$. This finishes, because then the power of $P$ with respect to $\omega_A$ is \[PX\cdot PX' = PX\cdot PD\cdot\frac{k+1}{2} = \frac{k+1}{2}\mathrm{Pow}_{(DEF)}(P),\]which is symmetric in $A$, $B$, $C$. Now we'll show the claims. It's well known that $M,D,I_A$ are collinear (here $M$ is the midpoint of the altitude $AU$), so to show $X,D,I_A$ collinear, it suffices to show that $X,M,D$ collinear, or that $M$ is on the polar of $T$ with respect to $\omega:=(DEF)$, or that $T$ is on the polar of $M$ with respect to $\omega$. Here $T$ is the intersection of the common tangent to $\omega$ and $\omega_A$ with $BC$. Define the projective map from line $AU$ to line $BC$ by sending $G\in AU$ to the intersection of the polar of $G$ with respect to $\omega$ with $BC$. This preserves cross ratio, so \[-1=(AU;M\infty) = (SD;T'\infty),\]where $S=EF\cap BC$, and $T'$ is the intersection of the polar of $M$ with $BC$. It suffices to show that $T'=T$, so it suffices to show $T$ is the midpoint of $DS$. Indeed, we have \[TD^2=TX^2=TB\cdot TC,\]so $B$ and $C$ are harmonic conjugate in $DS'$, where $S'=2D-T$. But we know $(BC;DS)=-1$, so we're done. This shows that $X,D,I_A$ collinear. Now we'll show that $X'$ is the midpoint of $DI_A$. Note that $TD=TS=TX$, so $X$ is on the semicircle with diameter $DS$, so $\angle DXS=\pi/2$. Furthermore, we have $(BC;DS)=-1$, so we must have that $XD$ and $XS$ are the internal and external angle bisectors of $\angle BXC$. Thus, $X'$ is the arc midpoint of $BC$ in $\omega_A$, so $X'L$ is the perpendicular bisector of $BC$, where $L$ is the arc midpoint of $BC$ in $(ABC)$. Thus, $X'L\parallel DI$, and since $L$ is the midpoint of $II_A$, we see that $X'$ is the midpoint of $DI_A$. This completes the proof as we showed how the above claims imply the problem.

khina wrote: Surprisingly, I don't think anyone has this sol yet. It's actually rather short, but yeah this problem is really cute. I do think that it is kind of ruined by 2002 ISL G7 though because this makes it more of an "extension" than a standalone problem, but this is still a pretty cool problem.

Why can you use brianchon ? would you pleas explain it more ?

Let $D,E,F$ the touching points of $\omega$ on $BC,CA,AB$ and $A_1$ be the tangency point of $\omega_A, \omega$. Define $B_1,C_1$ similarly. Furthermore, let $X=EF \cap BC$ and $M_A$ the midpoint of $XD$. $Y,Z,M_B,M_C$ are defined similarly. By ISL 2002 G7, $A_1,D,I_A$ are collinear, where $I_A$ is the $A$-excenter of $ABC$. Therefore, $A_2=A_1A_1 \cap BC$ satisfies $A_2A_1^2=A_2B.A_2C$, but since it's well known that $(B,C;D;X)=-1$, and the midpoint $M_A$ of $XD$ also satisfies $XD^2=XB.XC$, we have that $A_2=X$. Furthermore, $X,Y,Z$ are collinear, because $XYZ$ is the radical axis of $(ABC), \omega$ (all of them have same power of point WRT $(ABC),\omega$). Now, observe that $\Pi_{\omega}(X)=DA_1, \Pi_{\omega}(Y)=EB_1,\Pi_{\omega}(Z)=FC_1$, so since $X,Y,Z$ are collinear, $DA_1,EB_1,FC_1$ are concurrent. Let $P= DA_1 \cap EB_1 \cap FC_1$. By La Hire's Theorem, $\Pi_{\omega}(P)= \ell$, where $\ell$ is the line $XYZ$. Thus, $IP \perp \ell$. On the other hand, since $\ell$ is the radical axis of $(ABC), \omega$, we have that $OI \perp \ell$, so $O,I,P$ are collinear. By Pascal's Theorem on $EEB_1FFC_1$ and $B_1B_1EC_1C_1F$, we have that $Q_A=B_1B_1 \cap C_1C_1$ lies on $AP$. Also, by the Radical Axis Theorem on $\omega_B,\omega_C,\omega$, $Q_A$ lies on $AA'$, so $P$ lies on $AA'$. Similarly $P$ lies on $BB',CC'$, and since $P \in OI$, we are done. $\blacksquare$

Solution with help from khina :heart_eyes: Let $\omega$ denote the incircle. The first step is to identify the concurrence point. Let $\triangle DEF$ denote the intouch triangle and $\triangle I_AI_BI_C$ denote the excentral triangle of $\triangle ABC$. Since the two triangles are similar with parallel sides they have a center of homothety $T$ that lies on the line through the two circumcenters of the triangles, or $IO$. We spend the rest of the proof showing $T$ lies on $AA'$, which suffices by symmetry. By ISL 2002 G7, the circle through $B$ and $C$ tangent to $\omega$ passes through the second intersection of $DI_A$ and $\omega$, or $U$. Define $V,W$ similarly. It is enough to show that $EV,FW,AA'$ concur. Clearly $A_1=WW\cap VV$ lies on $AA'$, so it is enough to show $AA_1,EV,FW$ concur. Now the problem ought to be true no matter what configuration $E,V,F,W$ form: take a homography sending $EV\cap FW$ to the center of $\omega$, then the result follows by symmetry, since $A=EE\cap FF$ and $A_1=VV\cap WW$.

Let $\triangle I_AI_BI_C$ and $\triangle DEF$ be the excentral and incentral triangles of $\triangle ABC,$ respectively, and let $\omega_A,\omega_B,$ and $\omega_C$ touch $\omega$ at $P,Q,$ and $R,$ respectively. Also, let $S=\overline{QQ}\cap\overline{RR},$ $T=\overline{EQ}\cap\overline{FR}$ and $U=\overline{ER}\cap\overline{FQ},$ noticing $S$ lies on $\overline{AA'}$ by Radical Axis. By ISL 2002/G7, $P,Q,$ and $R$ lie on $\overline{DI_A},\overline{EI_B},$ and $\overline{FI_C},$ and by Vietnam TST 2003/2, $T$ lies on $\overline{IO}$ and $\overline{DI_A}.$ Also, $S$ and $A$ both lie on $\overline{TU}$ by Pascal on $QQFRRE$ and $EERFFQ,$ respectively. Similarly, $T$ also lies on $\overline{BB'}$ and $\overline{CC'}.$ $\square$

This is actually a nice config... let $K_a=\overline{EF}\cap\overline{BC}$, $\gamma_a=(K_aD)$, and cyclic variants. Let $H,\ell$ denote the orthocenter and Euler line of $\triangle DEF$, respectively. Observe that $\overline{OI}$ is just $\ell$, and that $\gamma_a$, etc are coaxial Apollonian circles. Define $X$ as the radical center of $\gamma_a,\gamma_b,\gamma_c,\omega$ (which exists since the former 3 circles are coaxial). We'll show this is the desired concurrency point. Also, define $J_a=\omega_a\cap\gamma_a\cap\omega$ and variants. Clearly, $\overline{DJ_a}$ is the raxis of $(\gamma_a,\omega)$, i.e. $X\in\overline{DJ_a}$. Lemma 1: $\ell$ is the raxis of $\gamma_a$ and variants.

To finish, let tangents to $\omega$ at $J_b,J_c$ meet @ $U$; then, $\overline{AU}$ is the raxis of $\omega_b,\omega_c$. Clearly this is the polar of $\overline{J_bJ_c}\cap\overline{EF}$. Recalling that $X=\overline{EJ_b}\cap\overline{FJ_c}$, follows by Brokard that $X\in\overline{AU}$, the end. Remark: This concurrency point also appears inm Brazil 2013/6...

Let $DEF$ be the intouch triangle and $\omega$ be the incircle. Let $T_A$ be the intersection of the midlines of $\triangle BDF$ and $\triangle CDE$, i.e. the radical center of $B$, $C$, and $\omega$; define $T_B$ and $T_C$ similarly. Then $\omega_A$ is the circle $(BCT_A)$ by a well-known lemma. Finally, let $M_A$ be the midpoint of arc $BC$ not containing $A$, and define $M_B$ and $M_C$ similarly. Now note that $AA'$, $BB'$, and $CC'$ concur by radical center. Invert about this point fixing $\omega_A$, $\omega_B$, and $\omega_C$. Then $\omega$ is sent to some circle internally tangent to $\omega_A,\omega_B,\omega_C$, and hence the radical center lies on the line joining $I$ with the center of this circle. Note that $\overline{T_BT_C}$, $\overline{EF}$, and $\overline{M_BM_C}$ are all parallel, and cyclic variants hold. Thus $\triangle T_AT_BT_C$ and $\triangle DEF$ are homothetic. Since $D$ is the "lowest" point in $\omega$, $T_A$ is thus the "lowest" point in $(T_AT_BT_C)$; since it's also clearly the "lowest" point in $(BCT_A)$ it follows that $(T_AT_BT_C)$ and $(BCT_A)$ are tangent. Cyclic variants hold as well, so it follows that $\omega$ gets sent to $(T_AT_BT_C)$. Further, note that $\triangle T_AT_BT_C$ and $\triangle M_AM_BM_C$ are homothetic. Since $\overline{T_AM_A}$ is the perpendicular bisector of $\overline{BC}$ and similar statements hold, it follows that $O$ is the center of homothety; hence $O$ is also the center of $(T_AT_BT_C)$, which finishes. $\blacksquare$

Let the tangency points of the circles be $T_A$, $T_B$, and $T_C$. Let $P$ be the radical center. Let $DEF$ by the intouch triangle of $ABC$. Then we know that the intersections of the tangents at $T_B$ and $T_C$ to the incircle are on AA' by radax. Also the intersection of the tangents at $E$ and $F$ are on the radax. Then in follows that $FT_C \cap ET_B$ is on $AA'$ since it is known that quadrilaterals with incircles have the same intersection of diagonals as their tangential quadrilateral. Now let $X$, $Y$, and $Z$ be the poles of $DT_A$, $ET_B$, $FT_C$. Observe that since $XT_A^2=XB \cdot XC$ that $X$ is on the radical axis of $(ABC)$ and $(DEF)$. So $XYZ$ is a line that is exactly the radax of $(ABC)$ and $(DEF)$. Then the pole of $XYZ$ must be on $T_AD$, $T_BE$, and $T_CF$. Thus the pole of $XYZ$ is $P$ from the concurrences from earlier. Since $XYZ \perp OI$, it follows that $P$ is on $OI$, so we are done.

Let $DEF$ be the intouch triangle, let $I_A$, $I_B$, and $I_C$ be the excenters. Let $A_1$ be the foot from $I$ to $DI_A$, let $A_2$ be the reflection of $D$ over $A_1$, and let $M_A$ be the midpoint of $DI_A$. Since $BCA_1II_A$ is cyclic, $DB\cdot DC=DA_1\cdot DI_A=DA_2\cdot DM_A$, which implies $BCA_2M_A$ is cyclic. Since $A_2$ lies on the incircle and $DM_A$, the homothety centered at $A_2$ mapping $D$ to $M_A$ must map the incircle to the circumcircle of $BCA_2M_A$, so $A_2$ lies on $\omega_A$. Therefore, since $DEF$ and $I_AI_BI_C$ are homothetic, the triangle formed by the midpoints of $DI_A$, $EI_B$, and $FI_C$, which is $M_AM_BM_C$ is homothetic to $DEF$. Let $X$ be the center of homothety of those triangles. $\operatorname{Pow}_{\omega_A}(X)=XA_2\cdot XM_A=\operatorname{Pow}_{(DEF)}(X)\frac{XM_A}{XD}$, so since $\frac{XM_A}{XD}$ is the ratio of homothety from $M_AM_BM_C$ to $DEF$, $X$ is the point of concurrency of $AA'$, $BB'$, and $CC'$, which lies on $IO$.

Let $DEF$ be the contact triangle of $ABC$, and let $I_AI_BI_C$ be the excentral triangle. We claim the desired point is $P$, the exsimilicenter of $(DEF)$ and $(I_AI_BI_C)$. We claim that the intersection of $\omega_A$ and the incircle lies on the line $I_AD$. Lets take base of altitude of $A$ as $D$ and the midpoint of $AD$ as $M$, intersection of $EF$ and $BC$ as $Y$, antipode of $D$ wrt incircle as $Z$, the intersection of incircle and $I_AD$ as $K$. We can easily see that by taking a pencil on $D$ that $LDKZ$ is a harmonic quadrilateral. Which implies that $Y$, $L$ and $Z$ are collinear. Because $(Y, D; B, C) = -1$ and take $X$ as midpoint of segment $YD$ it becomes clear that $XD^2 = XB \cdot XC$ it further implies that $XK$ is tangent to $\omega_A$ which proves the claim. Then we can further assume that midpoint of $I_AD$ lies on $\omega_A$. This comes from that $KI_A$ is angle bisector of $\angle BKC$ and midpoint of $KI_A$ lying on the perpendicular bisector of $BC$. Now finally we just have to prove that power of point of $P$ wrt to the three circles are equal. If we take the midpoints of $I_AD$, $I_BE$ and $I_CF$ as $V$, $U$ and $W$. Which is same as proving $T_AP \cdot PV=T_BP \cdot PU$ which comes from the fact that triangles $(DEF)$ and $(I_AI_BI_C)$ are homothetic. $T_AT_BDE$ is cyclic and $DE$ and $UV$ being parallel implies the result. Now the homothety form $P$ takes circumcenter of $(I_AI_BI_C)$ to $I$ so it means that $P$ lies on the Euler line of triangle $(I_AI_BI_C)$ which is indeed $OI$ proving the result.

Nvm this solution has been posted before. Why is this question so projective however. First I claim that if $\omega_a$ is tangent to incircle $\omega=(DEF)$ at $T_a$ and other cyclic definitions then $T_aD$, $T_bE$, $T_cF$ concur on $OI$. Let the tangents at $T_a$ and $D$ of $\omega$ concur at $S_a$. Takes the polar dual of this entire configuration. I claim that the poles of $T_aD$, $T_bE$ and $T_cF$, which are $S_a$, $S_b$ and $S_c$ respectively, are collinear. Note however $S_aT_a^2=S_aB\cdot S_aC$, and hence $S_a$’s power to both $\omega$ and $\Omega=(ABC)$ are the same. Hence $S_a$, $S_b$, $S_c$ lie on the radical axis of $\Omega$ and $\omega$. Hence we are done as then $\overline{S_aS_bS_c}$’s pole, $S$, is a point on $OI$. Now consider $A_1$ as the intersection of the tangent to $T_b$ and $T_c$ to $\omega$; I claim that $A_1$ lies on $AA_1$, which is just the radical axis of $\omega_b$ and $\omega_c$. However this follows from radical axis from $\omega$, $\omega_b$ and $\omega_c$. However $AA_1$ also passes through $S$ by Brianchon on the degenerate hexagon $AFXA’T_cYE$ and $AXT_bA’YE$ ($X=\overline{AF}\cap\overline{A’T_b}$ and $Y=\overline{AE}\cap\overline{A’T_c}$). Thus we are done (surprisingly quickly!)