Nice problem. I coordinate-bashed setting $P$ as the origin and $(x_1,y_1) ~ ,(x_2,y_2) ~ ,(x_3,y_3) $ as the centres. I got the desired concurrence at $(x_1+x_2+x_3 , y_1+ , y_2+ y_3)$

2019 IGO Intermediate P3 wrote: Three circles $\omega_1$, $\omega_2$ and $\omega_3$ pass through one common point, say $P$. The tangent line to $\omega_1$ at $P$ intersects $\omega_2$ and $\omega_3$ for the second time at points $P_{1,2}$ and $P_{1,3}$, respectively. Points $P_{2,1}$, $P_{2,3}$, $P_{3,1}$ and $P_{3,2}$ are similarly defined. Prove that the perpendicular bisector of segments $P_{1,2}P_{1,3}$, $P_{2,1}P_{2,3}$ and $P_{3,1}P_{3,2}$ are concurrent. Proposed by Mahdi Etesamifard Let $\measuredangle$ denote directed angles modulo $\pi$. We will represent $P_{a,b}$ as $P_{ab}$. [asy][asy] size(10cm); defaultpen(fontsize(9pt)); pair A=(0.72,-3.15); pair B=(-0.29,3.1); pair C=(-2.94,1.04); pair D=(2.28,1.12); pair K=(-0.65,-1.31); pair L=(2.35,-2.89); pair M12=(-1.1,-1.07); pair M23=(-0.68,1.08); pair M31=(0.39,-1.14); pair O1=(0.1,-3.25); pair O2=(2.28,1.45); pair O3=(-2.96,2.67); pair P=(0.03,1.09); pair P12=(4.54,1.16); pair P13=(-5.9,1); pair P21=(1.4,-7.39); pair P23=(-0.61,5.11); pair P31=(-3.52,-5.64); pair P32=(1.31,3.51); pair X=(-0.32,-3.32); pair Y=(2.32,-1.26); dot("$A$", A,SE); dot("$B$", B,E); dot("$C$", C,S); dot("$D$", D,SE); dot("$L$", L,SE); dot("$K$", K,SW); dot("$M_{12}$",M12,NW); dot("$M_{23}$",M23,NW); dot("$M_{31}$",M31,NE); dot("$O_1$",O1,S); dot("$O_2$",O2,NE); dot("$O_3$",O3,NW); dot("$P$",P,SE); dot("$P_{12}$",P12,E); dot("$P_{13}$",P13,W); dot("$P_{21}$",P21,S); dot("$P_{23}$",P23,N); dot("$P_{31}$",P31,SW); dot("$P_{32}$",P32,NW); dot("$X$",X,W); dot("$Y$", Y, E); draw(circumcircle(P,P32,P12)); draw(circumcircle(P,P23,P13)); draw(circumcircle(P,P31,P21)); draw(P13--P12); draw(P23--P21); draw(P32--P31); draw(O1--P31,dashed); draw(O2--P32,dashed); draw(O1--X); draw(X--P); draw(M23--K); draw(M31--K); draw(M12--K,dotted); draw(K--L,dotted+red); draw(O3--P,dashed+red); draw(O3--C,dashed+brown); draw(O3--B,dashed+brown); draw(K--X,dashed+brown); draw (K--Y,dashed+brown); draw(O1--L, dashed+orange); draw(O1--P,dashed+orange); draw(O2--P,dashed+orange); draw(O2--L,dashed+orange); [/asy][/asy] Let $O_1,O_2,O_3$ be the centers of $\omega_1,\omega_2,\omega_3$. Let $M_{12},M_{23},M_{31},A,B,C,D$ be the midpoint of $P_{31}P_{32},P_{13}P_{12},P_{23}P_{21},P_{21}P,P_{23}P,P_{13}P,P_{12}P$ respectively. Let $K$ be a point such that $KM_{31}\perp P_{23}P_{21}$ and $KM_{23}\perp P_{13}P_{12}$. "So, it suffice to show $KM_{12}\perp P_{31}P_{32}$". Let $O_2D\cap O_1A=L$. Let $X\in O_1L$ such that $KX\perp O_1L$ and $Y\in O_2L$ such that $KY\perp O_2L$. Claim 1: $PO_2LO_1$ is a parallelogram. Proof: As $O_1P\perp P_{13}P_{12}\perp O_2D\equiv O_2L\implies PO_1\|O_2L$. Now, $$\measuredangle LO_2P=\measuredangle DO_2P=\measuredangle P_{12}PP_{21}=\measuredangle PO_1A=\measuredangle PO_1L.$$Thus, $PO_2LO_1$ is a parallelogram. $\square$ Claim 2: $L$ lies on the perpendicular bisector of $P_{31}P_{32}$. Proof: From Claim 1 $O_2L=O_1P=O_1P_{31}$ and $O_2P_{32}=O_2P=LO_1$. It suffice to show $\measuredangle LO_2P_{32}=\measuredangle LO_1P_{31}$ as then $\triangle LO_2P_{32}\cong \triangle LO_1P_{31}\implies LP_{32}=LP_{31}$ from which the claim follows. Indeed, \begin{align*} \measuredangle LO_2P_{32} &=\measuredangle LO_2P+\measuredangle PO_2P_{32} \\&=\measuredangle PO_1L-2\measuredangle O_2PP_{32}\\&=\measuredangle PO_1L+2\measuredangle P_{23}PP_{32}\\&=\measuredangle PO_1L+2\measuredangle P_{21}PP_{31}\\&=\measuredangle PO_1L +2\measuredangle O_1PP_{31}+2\measuredangle APO_1 \\&=\measuredangle PO_1L-\measuredangle P_{31}O_1P-2\measuredangle PO_1L\\&=\measuredangle LO_1P_{31}\end{align*}$\square$ Claim 3: $KL\|O_3P$. Proof: We have $$\measuredangle BO_3C=-\measuredangle BPC=-\measuredangle APD=-\measuredangle PO_2L=\measuredangle O_1LY\qquad $$which gives $O_3CPB$ and $LYKO_1$ are similar. So, $\measuredangle KLY=\measuredangle PO_3C\implies O_3P\|KL$ since, $O_3C\|O_1P\|O_2L\equiv LY$. $\square$. From Claim 2 $O_3P\perp P_{31}P_{32}\perp LM_{12}$ and so by Claim 3 $LM_{12}\|O_3P\|KL$ which gives $K,L,M_{12}$ are collinear which in turn gives $KM_{12}\perp P_{31}P_{32}$ as desired. $\blacksquare$.

Looks Projective. . Ig this solution is much more easier and natural? The solution looks lengthy but it's actually a 2 or 3 liner. Dadgarnia wrote: Three circles $\omega_1$, $\omega_2$ and $\omega_3$ pass through one common point, say $P$. The tangent line to $\omega_1$ at $P$ intersects $\omega_2$ and $\omega_3$ for the second time at points $P_{1,2}$ and $P_{1,3}$, respectively. Points $P_{2,1}$, $P_{2,3}$, $P_{3,1}$ and $P_{3,2}$ are similarly defined. Prove that the perpendicular bisector of segments $P_{1,2}P_{1,3}$, $P_{2,1}P_{2,3}$ and $P_{3,1}P_{3,2}$ are concurrent. Proposed by Mahdi Etesamifard We just write the same problem with different notations. Notation Changed Problem wrote: Let $\omega_1,\omega_2,\omega_3$ be three circles intersecting at a point $O$. Let the tangent to $\omega_1$ at $O$ intersect $\{\omega_2,\omega_3\}$ at $\{P,Q\}$ respectively. Let the tangent to $\omega_2$ at $O$ intersect $\{\omega_1,\omega_3\}$ at $\{U,V\}$ and let the tangent to $\omega_3$ at $O$ intersect $\{\omega_1,\omega_2\}$ at $\{M,N\}$ respectively. Then the Perpendicular Bisectors of $PQ,UV,MN$ are concurrent. Let $X,Y,Z$ be the midpoints of $UV,MN,PQ$ respectively. We prove that $O\in\odot(XYZ)$. Now Invert around $O$ with an arbitary radius to get the following equivalent problem. Inverted Problem wrote: $ABC$ be a triangle and $O$ is an arbitary point. Let the parallel lines $\{\ell_1,\ell_2,\ell_3\}$ from $O$ to $\{BC,CA,AB\}$ respectively . $\ell_1\cap\{AB,AC\}=\{M,N\},\ell_2\cap\{BA,BC\}=\{U,V\},\ell_3\cap\{CA,CB\}=\{P,Q\}$ respectively. Let $\{X,Y,Z\}$ be the Harmonic Conjugates of $O$ WRT Segments $PQ,UV,MN$ respectively. Then $\overline{X-Y-Z}$. To prove the Inverted Problem ,we first prove the following generalization $(\bigstar)$. Generalization wrote: Let $\mathcal C$ be a conic with $(A_1^*,A_2^*),(B_1^*,B_2^*),(C_1^*,C_2^*)$ as pairs of an Involution. (Or in other words $A_1^*A_2^*,B_1^*B_2^*,C_1^*C_2^*$ are concurrent at a point $O$). If $\{X,Y,Z\}$ are the Harmonic Conjugates of $O$ WRT segments $A_1^*A_2^*,B_1^*B_2^*,C_1^*C_2^*$ respectively then $\overline{X-Y-Z}$. Note that $(A_1^*,A_2^*;X,O)=(B_1^*,B_2^*;Y,O)=(C_1^*,C_2^*;Z,O)=-1$. So, $X,Y,Z$ lies on the Polar of $O$ WRT $\mathcal C$.So $\overline{X-Y-Z}$. Coming back to the problem. Notice that $UM\cap PQ=MN\cap QV=NP\cap VU=\infty$. So, $\{U,M,Q,V,N,P\}$ lies on a Conic $\mathcal C$ by Converse of Pascal's Theorem . So from $(\bigstar)$ we get $\overline{X-Y-Z}$. Now Invert back to get that $O\in\odot(XYZ)$. So, the Perpendiular bisectors of $UV,PQ,MN$ are concurrent at the $O$ antipode of $\odot(XYZ)$. $\blacksquare$

Consider the vector sum of $\overrightarrow{PO_i}$ where $O_i$ is the center of $\omega_i$

By some angle chasing we get: $\angle{P_{2.1}PP_{3.1}}$=$\angle{P_{3.2}PP_{2.3}}$=$\angle{PP_{1.2}P_{3.2}}$=$\angle{P_{2.3}P_{1.3}P}$ $\angle{P_{3.1}P_{2.1}P}$=$\angle{P_{3.1}PP_{1.3}}$=$\angle{P_{3.2}PP_{1.2}}$=$\angle{P_{1.3}P_{2.3}P}$ $\angle{P_{2.1}P_{3.1}P}$=$\angle{P_{2.1}PP_{1.2}}$=$\angle{P_{1.3}PP_{2.3}}$=$\angle{P_{1.2}P_{3.2}P}$ Let lines $P_{2.1}P_{3.1} $$\cap$ $ P_{1.3}P_{2.3}$=$A$; $P_{2.1}P_{3.1} $$\cap$ $ P_{1.2}P_{3.2}$=$B$ ; $ P_{1.2}P_{3.2}$$\cap$ $ P_{1.3}P_{2.3}$=$C$, which these three points form triangle $\triangle{ABC}$. Since the $\triangle{AP_{2.3}P_{2.1}}$ is isosceles, then perpendicular bisector of $P_{2.3}P_{2.1}$ and angle bisector of $\angle{P_{2.3}AP_{2.1}}$ are the same line, similarly with $\angle{P_{3.1}BP_{3.2}}$ and $\angle{P_{1.3}CP_{1.2}}$. We know that the bisectors of triangle are concurrent, so the perpendicular bisector of segments $P_{1,2}P_{1,3}$, $P_{2,1}P_{2,3}$ and $P_{3,1}P_{3,2}$ are concurrent.

bingbing666 wrote: Consider the vector sum of $\overrightarrow{PO_i}$ where $O_i$ is the center of $\omega_i$ Can you elaborate?