Label $O_1,O_2,O_3$ the centers of the referred circles. Internal common tangent of $\mathcal{K}_1,\mathcal{K}_2$ cuts $\mathcal{K}_3$ at $M,N$ such that $T,N$ are on the same side of the line $PQ.$ Inversion WRT $\odot(M,MT)$ fixes $\mathcal{K}_1,\mathcal{K}_2$ and carries $\mathcal{K}_3$ into the other external common tangent $\tau$ of $\mathcal{K}_1,\mathcal{K}_2$ $\Longrightarrow$ $MO_3 \perp \tau.$ Hence, if $H \equiv O_1O_2 \cap PQ \cap \tau,$ we have $\angle O_3NM=\angle O_3MN=\angle (\tau,O_1O_2)=\angle PHT$ $\Longrightarrow$ $NO_3 \perp PQ$ $\Longrightarrow$ $NO_3$ is perpendicular bisector of $\overline{PQ},$ i.e. $N$ is midpoint of the arc $PQ$ of $\mathcal{K}_3.$

http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A607&l=en

My solution: Let $ M $ be the midpoint of arc $ PQ $ . Let $ R, S $ be the tangent point of $ \mathcal{K}_3 $ with $ \mathcal{K}_1, \mathcal{K}_2 $, respectively . Let $ X, Y $ be the tangent point of $ PQ $ with $ \mathcal{K}_1, \mathcal{K}_2 $, respectively . From homothety with center $ R $ send $ \mathcal{K}_1 $ to $ \mathcal{K}_3 $ we get $ M, R, X $ are collinear . From homothety with center $ S $ send $ \mathcal{K}_2 $ to $ \mathcal{K}_3 $ we get $ M, S, Y $ are collinear . Since $ \angle YXM=\angle MSR $ , so we get $ R, S, X, Y $ are concyclic, hence we get $ M $ is the radical center of $ \{ \mathcal{K}_1, \mathcal{K}_2, (RSXY) \} $ , so $ M $ lie on the radical axis of $ \{ \mathcal{K}_1, \mathcal{K}_2 \} $ which is the internal common tangent of $ \mathcal{K}_1, \mathcal{K}_2 $ . Q.E.D

See the drawıng for notatıons; $VW\cap K_1K_2=U$ which is, from Menelaos in $\triangle K_1K_2K_3$, the ex-similicenter of the circles $\mathcal{K}_1,\mathcal{K}_2$, meaning that $PQ$ passes through $U$ and $\angle QSW=\angle RVY$, or $RVWS$ is cyclic and $X=RV\cap SW$ belongs to the radical axis of the 2 circles, from $XV\cdot XR=XW\cdot XS$. On the other way $V,W$ are the insimilicenters of $\mathcal{K}_1$ and $\mathcal{K}_3$, respectively $\mathcal{K}_2$ and $\mathcal{K}_3$, so $RV$ and $SW$ intersect the circle $\mathcal{K}_3$ the second time at the same point, and this is $X$, with $XK_3\parallel RK_1$, or $XK_3\bot PQ$, done. Best regards, sunken rock

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Really cool problem from Komal. Let the centers of $k_1,k_2,k_3$ be $A,B,C$ respectively. Relabel $T$ to $D$ and define $E,F$ as the intersections of $k_1,k_3$ and $k_2,k_3$. Define $M$ as the intersection of the tangent from $D$ and $k_3$; so we want to prove that $M$ is the circumcenter of triangle $PQD$. Due to the ratios of the circles, $CF, BE, AD$ concur. Hence, letting $R$ be the intersection of $MN$ with $BC$ we have that $F,E,R$ colinear (notice R is the harmonic conjugate of $D$ w.r.t $B,C$ due to the concurrency and that $R,D$ are harmonic conjugates(they are exsimilar and insimilar center of the circles)). So, we have $RE \cdot RF = RP \cdot RQ$ and notice that the circle $EFD$ is the incircle of triangle $ABC$ (due to the lengths of $BD,CD,BF,AF,AE,CE$); hence we have $RE \cdot RF = RD^2 = RP \cdot RQ$ which implies that the circumcircle of triangle $PQD$ (which I'm calling $\Gamma$ is tangent to $BC$. Now, let $G$ be the foot of the perpendicular from $D$ to $EF$. One notice that $\angle{QTG} = \angle{PDM}$ due to the tangency and the angle $\angle{MDC} = \angle{MDB} = 90$. Hence, if we extend $DM$ to hit $\Gamma$ at $H$, we have $TH$ a diameter. For the finish, we need to prove that $M$ is the midpoint of segment $TH$. Now, invert the problem around $D$ with arbitrary radius. With the inverted diagram, just notice that $k'_3$ and $\Gamma'$ are reflected w.r.t line $P'Q'$, which kills the problem. (We havve $H'$ midpoint of $TM'$.) $\blacksquare$.

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