Lemma 1: Two arcs $w_1$ and $w_2$ with centres $O_1$ and $O_2$ connect points $A$ and $B$, both constructed on the same half plane with respect to line $AB$. Circles $S_1$ and $S_2$ with centres $C_1$ and $C_2$ tangent to $w_1$ and $w_2$ in points $P_1, Q_1; P_2, Q_2$ respectively. Then common external tangents to $S_1$ and $S_2$ meet on $AB$.

Lemma 2: Three circles $S_1, S_2, S_3$ are given on the plane. Then three intersection points of common external tangents of pair of circles $(S_1, S_2), (S_2, S_3), (S_3, S_1)$ lie on a line (if such points exist).

Denote the circle that tangent to $\gamma_3, \gamma_2, h_1, h_2$ by $w_1$. Denote the circle that tangent to $\gamma_1, \gamma_2, h_1, h_2$ by $w_2$. Denote the circle that tangent to $\gamma_1, \gamma_2, h_2, h_3$ by $w_3$. Denote the circle that tangent to $\gamma_3, \gamma_2, h_2$ by $w_4$. Now it is enough to prove that $w_4$ tangents to $h_3$.

Fantastic solution! Another idea for lemma 1: By Monge-d'Alembert (lemma 2) the center of homothety between $S_1$ and $S_2$ lies on $Q_1Q_2$ (just consider the homotheties) and for the same reason also on $P_1P_2$. Call their intersection $X$ and it is enough to show, it lies on the radical axis of $w_1$ and $w_2$. Further, there exists an inversion with center $X$ and radius $r$ that takes $S_1$ to $S_2$. This inversion takes $Q_1$ to $Q_2$ and $P_1$ to $P_2$. Hence \[p(X, w_1) = XP_2\cdot XP_1 = r^2 = XQ_2\cdot XQ_1 = p(X,w_2).\] and lemma 1 is done!

In the first lemma I would start with the known fact that the points $P_1,P_2,Q_1,Q_2$ are concyclic. (From the isosceles triangles $O_1P_1P_2$, $O_2Q_1Q_2$, $C_1P_1Q_1$ and $C_2P_2Q_2$, it can be verified that $\angle P_1P_2Q_2+\angle Q_2Q_1P_1=\angle Q_1P_1P_2+\angle P_2Q_2Q_1$.) Then the lines $AB$, $P_1P_2$ and $Q_1Q_2$ are the radical axes of pairs of circles $k_1$, $k_2$ and $P_1P_2Q_1Q_2$ so they pass through the radical center $X$. Finally, as mentioned, $X=P_1P_2\cap Q_1Q_2$ is the external homothety center of $S_1$ and $S_2$.

Yes , it's not hard to prove that if figures V_11V_21_V12V_22 , V_31V_21V_22V_32 , V_22V_12V_13V23 are described then V_32V_22V_23V_33 is outscribed too , but intersecting how to prove that exist circle witch is inscribed in V_31V_11V_13V_33

Another solution (I heard this observation from Ali Khezeli). Place the picture in the Poincaré's half-plane model of the hyperbolic plane, then the three arcs and the three half lines will be equidistant curves/hypercycles or lines. In Poincaré's models the circles look like circles. The key to the solution is that if two equidistant curves have the same end-points, then all circles which are tangent to them have the same size; their common diameter is the distance between the two cycles. For example, in the picture below all blue circles have the same radius, all green circles have the same radius, all brown circles have the same radius and all magenta circles have the same radius. The magenta family of circles has a common element with the blue family, the blue family has a common element with the brown, and the brown has a common element with the green. Therefore, all magenta, blue, brown and green circles have the same size (in hyperbolic) and hence also the magenta and the green family have a common element. Concerning skytin's observation, the circles inscribed into the curved quadrilaterals have the same (hyperbolic) diameter $d$. The (hyperbolic) distance between the point $V_{22}$ and the cycles $\gamma_1$, $\gamma_3$, $h_1$ and $h_3$ are all $d$. So the quadrilateral $V_{11}V_{13}V_{33}V_{31}$ in indeed circumscribed and its incenter is $V_{22}$ in the model.

My quite lengthy solution with inversion, but if it's ok then I believe it's woth it! Complete the circular arcs $\gamma_1,\gamma_2,\gamma_3$ to full circles which we will now referr to as $\gamma_1,\gamma_2,\gamma_3$ respectively. Given a sphere $S$, we say that $C$ is a "great circle'' of $S$ if it is the intersection of $S$ with some plane passing though the center of $S$. Let $\pi$ be the plane where our problem's configuration is given. Consider spheres $\Gamma_1,\Gamma_2,\Gamma_3$ such that $\gamma_1,\gamma_2,\gamma_3$ are great circles of $\Gamma_1,\Gamma_2,\Gamma_3$ respectively. Then $\Gamma_1,\Gamma_2,\Gamma_3$ intersect on a circle $C$ with diameter $AC$ such that the projection of $C$ onto plane $\pi$ is segment $AC$. Consider half-planes $H_1,H_2,H_3$ which are perpendicular to $\pi$ and cut $\pi$ at $h_1,h_2,h_3$ respectively. Then $H_1,H_2,H_3$ concurr at a line $l$ passing through $B$ and perpendicular to $\pi$. Let $C$ and $l$ intersect at $P$ and $Q$. $P$ is our candidate for inversion. Supose $C_1,C_2,C_3$ are the circles "inscribed'' in the curved quadrilaterals $V_{11}V_{21}V_{22}V_{12}$, $V_{12}V_{22}V_{23}V_{13}$ and $V_{21}V_{31}V_{32}V_{22}$ respectively. Consider the spheres $S_1,S_2,S_3$ for which $C_1,C_2,C_3$ are great circles respectively. Note that for all possible $i,j$: $C_i$ and $\gamma_j$ are tangent iff $S_i$ and $\Gamma_j$ are tangent. In fact, if each pair is tangent, then they are tangent at the same point. The same is true for $C_i$ and $H_j$, which are tangent iff $S_i$ and $H_j$ are tangent, and tangency ocurrs at the same point. Invert at $P$. This inversion leaves $H_1,H_2,H_3$ fixed, and maps $\Gamma_1,\Gamma_2,\Gamma_3$ to planes $G_1,G_2,G_3$ respectively (check out the attachment below). Moreover, these planes concurr at line $t$, the inverse of circle $C$. The six planes $H_1,H_2,H_3,G_1,G_2,G_3$ concurr at point $Q'$, the inverse of $Q$, and determine four regions, each bounded by four of these planes, which we will call "tetrahedral regions''. These four regions are those determined by the set of planes: $\{H_1,H_2,G_1,G_2\}$, $\{H_1,H_2,G_2,G_3\}$, $\{H_2,H_3,G_1,G_2\}$ and $\{H_2,H_3,G_2,G_3\}$. The spheres $S_1,S_2,S_3$ are mapped to spheres $S_1',S_2',S_3'$ respectively, each interior to one of these tetrahedral regions and tangent to the corresponding four of the six given planes which determine this region. Before continuing note that if there is a sphere $S$ with center $O$ tangent to each plane of the tetrahedral region determined by the planes $H_i,H_j,G_k,G_l$, then for any point $N$ on the half-line $Q'O$, there is a sphere centered at $N$ and tangent to these four planes: just consider homothety with center $Q'$ maping $O$ to $N$; it maps $S$ onto the desired sphere. We will call this observation 1. If we prove that the tetrahedral region bounded by the planes $H_2,H_3,G_2,G_3$ is inscriptable we will be done because: let $C_4$ be the circle tangent to $h_2,h_3$ and $\gamma_2$. Let $S_4$ be the sphere whith great circle $C_4$. Then $S_4$ is tangent to $H_2,H_3$ and $\Gamma_2$. Inversion at $P$ maps $S_4$ to sphere $S_4'$ tangent to planes $H_2,H_3$ and to $G_2$. Because the tetrahedral region determined by the planes $H_2,H_3,G_2,G_3$ is inscriptible, $S_4'$ will also be tangent to $G_3$, and so the pre-images $S_4$ and $\Gamma_3$ of $S_4'$ and $G_3$ are tangent. Moreover, since the centers of $\Gamma_3$ and $S_4$ lie on plane $\pi$, this tangency point must also lie on plane $\pi$, and so $C_4$ will be tangent to $\gamma_3$ at this point. So let's prove the above claim. Recall that the planes $H_1,H_2,H_3$ intersect on line $l$ through $P$ and $G_1,G_2,G_3$ intersect on line $t$ through $Q'$. Let these two lines, which concurr at $Q'$, determine the plane $\pi_1$. Take any plane $\pi_2$ parallel to $\pi_1$ and intersecting the six planes $H_1,H_2,H_3,G_1,G_2,G_3$; call these intersections the lines $l_1,l_2,l_3,l_4,l_5,l_6$ respectively. The key here is that the lines $l_1,l_2,l_3$ are parallel, and so are $l_4,l_5,l_6$ because the planes $H_1,H_2,H_3$ concurr only on plane $\pi_1$ and so do the planes $G_1,G_2,G_3$, however $\pi_2$ was chosen to be parallel to $\pi_1$. Call this observation 2. Because of observation 1, since we know that the three tetrahedral regions bounded by the set of planes $\{H_1,H_2,G_1,G_2\}$, $\{H_1,H_2,G_2,G_3\}$ and $\{H_2,H_3,G_1,G_2\}$ are inscriptable, we can supose that there are spheres $\Omega_1,\Omega_2,\Omega_3$ with centers $O_1,O_2,O_3$ respectively inscribed in each of them and such that $O_1,O_2,O_3$ lie on plane $\pi_2$. Let us denote by $d(p,K)$ the distance from point $p$ to plane $K$. We claim that the distance from $O_1$ to the planes $H_1$ and $H_2$ is the same as the distance from $O_2$ to the planes $H_1$ and $H_2$, and similarly with points $O_1,O_3$ and planes $G_1,G_2$. This is true because: consider the internal angle bisector of planes $l_1$ and $l_2$, which is itself a plane $\pi_B$, and contains line $l$. This plane intersects plane $\pi_2$ on line $l_B$, and by an argument in the same spirit as that used in observation 2, $l_B$ will be parallel to the lines $l_1$ and $l_2$. We know $O_1$ and $O_2$ lie on line $l_B$. Consider translation of space by vector $O_1O_2$. Because $O_1O_2$ is paralel to $l_1$ and $l_2$ this translation leaves $H_1,H_2$ and $\pi_2$ fixed. It takes sphere $\Omega_1$ to a sphere $\Omega_1'$ centered at $O_2$ and tangent to the images of $H_1$ and $H_2$ under this translation, which are themsevles respectively. Since there is a unique sphere tangent to the planes $H_1$ and $H_2$ and centered at $O_2$, we conclude that $\Omega_1'=\Omega_2$, and in particular our claim that $d(O_1,H_1)=d(O_2,H_1)=d(O_1,H_2)=d(O_2,H_2)$ follows. Analogously $d(O_1,G_1)=d(O_3,G_1)=d(O_1,G_2)=d(O_3,G_2)$. Call this observation 3. Using this repeatedly finishes the proof: let the internal angle bisector of planes $H_2,H_3$ intersect the plane $\pi_2$ at line $l_H$ and the internal angle bisector of planes $G_2,G_3$ intersect the plane $\pi_2$ at line $l_G$. As before, $l_H$ is paralel to $l_2$ and $l_3$, and $l_G$ is paralel to $l_5$ and $l_6$. Let $l_H$ and $l_G$ intersect at point $O_4$. We claim that $O_4$ is equidistant from the planes $H_2,H_3,G_2,G_3$. Indeed, as before, consider translations, first by vector $O_3O_4$. As in observation 3, it takes $\Omega_3$ to sphere $\Omega_{H}$ such that it has center $O_4$ and is tangent to $H_2$ and $H_3$, therefore $d(O_4,H_2)=d(O_3,H_2)$. A second translation by vector $O_2O_4$ maps $\Omega_2$ to sphere $\Omega_{G}$ such that it has center $O_4$ and is tangent to $G_2$ and $G_3$, therefore $d(O_4,G_2)=d(O_2,G_2)$. Now, because of the other distances which we have proven to be the same, we have that \[d(O_4,H_2)=d(O_3,H_2)=d(O_3,G_2)=d(O_1,G_2)=d(O_1,H_2)=d(O_2,H_2)=d(O_2,G_2)=d(O_4,G_2).\] It follows that the spheres $\Omega_{H}$ and $\Omega_{G}$ have the same center and the same radius, so they are the same sphere. In particular, the tetrahedral region bounded by the planes $H_2,H_3,G_2,G_3$ contains an inscribed sphere and so is inscribable. This is what we wanted to prove, and what finishes our proof.

Never thought this day would come I admit seeking out a hint for solving this, apparently the lemma was known to me but it didn't strike in this setting. Anyway, posting my solution for this IMO ShortList 2010 G7 wrote: Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same half-plane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be three rays starting at $B,$ lying in the same half-plane, $h_2$ being between $h_1$ and $h_3.$ For $i, j = 1, 2, 3,$ denote by $V_{ij}$ the point of intersection of $h_i$ and $\gamma_j$ (see the Figure below). Denote by $\widehat{V_{ij}V_{kj}}\widehat{V_{kl}V_{il}}$ the curved quadrilateral, whose sides are the segments $V_{ij}V_{il},$ $V_{kj}V_{kl}$ and arcs $V_{ij}V_{kj}$ and $V_{il}V_{kl}.$ We say that this quadrilateral is $circumscribed$ if there exists a circle touching these two segments and two arcs. Prove that if the curved quadrilaterals $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are circumscribed, then the curved quadrilateral $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed, too. Proposed by Géza Kós, Hungary [asy][asy] pathpen=black; size(400); pair A=(0,0), B=(4,0), C=(10,0); draw(L(A,C,0.3)); MP("A",A); MP("B",B); MP("C",C); pair X=(5,-7); path G1=D(arc(X,C,A)); pair Y=(5,7), Z=(9,6); draw(Z--B--Y); struct T {pair C;real r;}; T f(pair X, pair B, pair Y, pair Z) { pair S=unit(Y-B)+unit(Z-B); real s=abs(sin(angle((Y-B)/(Z-B))/2)); real t=10, r=abs(X-A); pair Q; for(int k=0;k<30;++k) { Q=B+t*S; t-=(abs(X-Q)-r)/abs(S)-s*t; } T T=new T; T.C=Q; T.r=s*t*abs(S); return T; } void g(pair Q, real r) { real t=0; for(int k=0;k<30;++k) { X=(5,t); t+=(abs(X-Q)+r-abs(X-A)); } } pair Z1=(1.07,6); draw(B--Z1); T T=f(X,B,Y,Z1); draw(CR(T.C,T.r)); T T=f(X,B,Y,Z); draw(CR(T.C,T.r)); g(T.C,T.r); path G2=D(arc(X,C,A)); T T=f(X,B,Y,Z1); draw(CR(T.C,T.r)); T=f(X,B,Y,Z); draw(CR(T.C,T.r)); g(T.C,T.r); path G3=D(arc(X,C,A)); pen p=black+fontsize(8); MC("\gamma_1",G1,0.85,p); MC("\gamma_2",G2,0.85,NNW,p); MC("\gamma_3",G3,0.85,WNW,p); MC("h_1",B--Z1,0.95,E,p); MC("h_2",B--Y,0.95,E,p); MC("h_3",B--Z,0.95,E,p); path[] G={G1,G2,G3}; path[] H={B--Z1,B--Y,B--Z}; pair[][] al={{S+SSW,S+SSW,3*S},{SE,NE,NW},{2*SSE,2*SSE,2*E}}; for(int i=0;i<3;++i) for(int j=0;j<3;++j) MP("V_{"+string(i+1)+string(j+1)+"}",IP(H[i],G[j]),al[i][j],fontsize(8));[/asy][/asy] We start with the following crucial result. Claim: Consider two circles $\gamma_1, \gamma_2$ which intersect at $A, B$. Let $\omega_1, \omega_2$ be circles tangent to both $\gamma_1, \gamma_2$. Then, the exsimilcenter of $\omega_1, \omega_2$ lies on line $\overline{AB}$. (Proof) Let $\gamma_1$ touch $\omega_i$ at $S_i$ and $\gamma_2$ touch $\omega_i$ at $T_i$ for $i=1,2$. By Monge D'Alembert, it is clear that the exsimilicenter of $\omega_1, \omega_2$ is actually the intersection point of lines $\overline{S_1S_2}$ and $\overline{T_1T_2}$. Apply inversion at $A$; notice that $\gamma_1, \gamma_2$ are transformed to lines passing through a point $B$ and $\omega_1, \omega_2$ to circles tangent to them, say at $S_1, S_2$ and $T_1, T_2$ respectively. Let $\odot(AS_1S_2)$ meet $\overline{AB}$ at $S$ and $\odot(AT_1T_2)$ meet $\overline{AB}$ at $T$. It suffices to show $S \equiv T$. Note that by equating power of $B$, we get $$BS\cdot BA=BS_1\cdot BS_2=BT_1\cdot BT_2=BT\cdot BA,$$hence, the claim is established. $\blacksquare$ Consider a circle $\mathcal{C}_4$ tangent to $h_2, \gamma_2, \gamma_3$. We will see that it is also tangent to line $h_3$ establishing the result. Label the top-left, bottom-left, bottom-right circumscribed circles $\mathcal{C}_1, \mathcal{C}_2, \mathcal{C}_3$ respectively. By Monge D'Alembert on $\mathcal{C}_1, \mathcal{C}_2, \mathcal{C}_4$ we see that the exsimilicenter for $\mathcal{C}_2, \mathcal{C}_4$ lies on line $\overline{AC}$. Apply Monge D'Alembert now, on $\mathcal{C}_2, \mathcal{C}_3, \mathcal{C}_4$ to get that the exsimilicenter for $\mathcal{C}_3, \mathcal{C}_4$ lies on $\overline{AC}$. However, that point also lies on $h_2$ hence $B$ is the desired point. Evidently, $\mathcal{C}_4$ is tangent to line $h_3$ as desired. $\blacksquare$

I'd like to point out that Dr Ahnnos Renbuarg has published a 57 page long handout regarding the use of Cartesian Coordinates in solving this and couple of other G7's.

From what I can see, the above solutions find a circle tangent to $\gamma_2, \gamma_3, h_2$ and then prove that circle is also tangent to $h_3$. Here I'll instead find a circle tangent to $h_2,h_3,\gamma_2$ and show it's also tangent to $\gamma_3$, which is perhaps more roundabout but easier for me to think of. Let the circles inscribed in $V_{11}V_{21}V_{12}V_{22}, V_{21}V_{31}V_{32}V_{22}, V_{12}V_{22}V_{23}V_{13}$ be $\omega_1,\omega_2,\omega_3$, respectively, with centers $O_1,O_2,O_3$. Let $\omega_1$ meet $\gamma_1, \gamma_2$ at $P_1,Q_1$ and $\omega_2$ meet $\gamma_1, \gamma_2$ at $P_2,Q_2$. Then by Monge, $P_1P_2,Q_1Q_2$ meet at the exsimilicenter $X$ of $\omega_1, \omega_2$. Furthermore, the inversion centered at $X$ of power $XP_1\cdot XP_2$ fixes the common tangents to these two circles as well as $\gamma_1$, thus $\omega_1$ maps to a new circle tangent to $\gamma_1$ at $P_2$ and also tangent to the common tangents of $\gamma_1,\gamma_2$, meaning the image of $\gamma_1$ is $\gamma_2$ and vice versa. It clearly follows that $XP_1\cdot XP_2$ is the power which switches the two circles, as is $XQ_1\cdot XQ_2$ by a similar argument, hence the two quantities are equal, so $X\in AC$. Now let $\ell_1, \ell_2$ be the bisectors of the angles formed by $h_1,h_2$ and $h_2,h_3$, respectively. Since $h_2$ as a common internal tangent contains the insimilicenter of $\omega_1,\omega_2$ while $BA$ contains the exsimilicenter, clearly $(BA, h_2; \ell_1, \ell_2)=-1$. Therefore, if we consider a circle $\omega_4$ tangent to $h_2,h_3, \gamma_2$ with center $O_4$, then by the same reasoning in reverse, since $O_3,O_4\in \ell_1, \ell_2$, the exsimilicenter of $\omega_3, \omega_4$ is a point $Y$ lying on $AC$. Now let $\omega_3$ meet $\gamma_2, \gamma_3$ at $P_3, Q_3$, and let $\omega_4$ meet $\gamma_2$ at $P_4$, so by Monge it follows that $P_3,P_4,Y$ are collinear. By the same reasoning as earlier, the inversion centered at $Y$ with power $YA\cdot YC = YP_3\cdot YP_4$ switches $\omega_3, \omega_4$ but it fixes $\gamma_3$, so the image of $\omega_3$ must remain tangent to $\gamma_3$, meaning $\omega_4$ is tangent to $\gamma_3$ as desired.

So I was going through my old notebook and found this solution. Mainly for storage and providing motivation. Ok, so firstly we need to construct the diagram. Let the ``incenter" of $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}$ be $X_1$, and the center, radius of $\gamma_i$ be $C_i, R_i$. To construct such a curvilinear quadrilateral, we need to find $X_1$, which we can get by the tangency conditions by finding a locus. Note that we have $C_1X_1+C_2X_1=R_1+R_2$, so the locus of $X_1$ is an ellipse, say $\epsilon_{12}$. Also, $X_1$ is on the bisector of $h_1, h_2$. We need another constraint on $X_1$, so that the location of $h_2$ gets fixed by $h_1$, so that the only degree of freedom in the system is due to $h_1$. Let's look at $\epsilon_{12}$ in more detail. If the distance of $X_1$ from the directrix closer to $C_2$ is $d$, we have $R_2+r = ed$ where $e$ is the eccentricity of the ellipse. Continuing similarly, we have the distance of $A$ from the directrix as $R_2/e$. So the distance of $X_1$ from $AC$ is $r/e$. So the ratio of the radius of a circle to the distance of its center from $AC$ is fixed, and it equals $1/e$. But then this ratio is also fixed for a given angle bisector, due to similarity. Thus, we get a necessary and sufficient condition, i.e. equality of the ratios for opposite arcs and opposite sides for the existence of an incircle (easy to see that the condition is necessary and sufficient). Now applying this to $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}$, the ratio for $\gamma_1, \gamma_2$, say $r_{\gamma_1, \gamma_2}$ and that for $h_1, h_2$, say $r_{h_1, h_2}$ are the same. Similarly, applying this for $\widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}}$ and $\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$, we get $r_{\gamma_3, \gamma_2} = r_{h_1, h_2}$ and $r_{\gamma_3, \gamma_2} = r_{h_3, h_2}$. These equalities give $r_{\gamma_1, \gamma_2} = r_{h_3, h_2}$, so $\widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}$ is circumscribed as well.

$Lemma $= For every $1\leq i <j\leq 3$ consider the circles $\rho(P,r) $ in the half plane $H$ which is tangent externally and internally to circular arcs $\gamma_{i} $ and $\gamma_{j} $ respectively .Hence $O_{i}P=R_{i}+r$ and $O_{j}=R_{j}-r$ . Ellipse $\mathcal {E}_{ij}$ is the locus of the centres of the circles $\rho (P,r)$ , end points $A$ and $C$ and $r =v_{ij}\cdot d(P) $ constant $v_{ij}$ for all the circles . $O_{i}P+O_{j}P=R_{i}+R_{j} $ . Arcs of the ellipse $\mathcal {E}_{ij} $ lies between the arcs $\gamma_{i} $ , $\gamma_{j} $ . Some points of $P $ lies on ellipse $\mathcal {E}_{ij} $. Hence $O_{i}P>R_{i} $ and $O_{j}P <R_{j}$ . Considering $P$ doesn't lie in the line $O_j,O_j$ Menelaus theorem applied to the $\triangle O_{i} O_{j}P$ . Homothety centres are collinear .

Dear Mathlinkers, in order to have a possible proof... http://jl.ayme.pagesperso-orange.fr/Docs/III.Rubedo.pdf p. 5-7. Sincerely Jean-Louis

If $\gamma_i$ were concurrent lines, this would be the famous problem, we'll try to prove it the similar way: if two of $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{22}V_{32}}\widehat{V_{33}V_{23}}, \widehat{V_{11}V_{31}}\widehat{V_{33}V_{13}}$ are circumscribed, then so is third, the result follows immediately. Invert the $\mathbb{R}^3$ picture in $X$, which doesn't lie on a plane $\alpha$ with these arcs. Now $\alpha$ is a sphere $S$, $h_i$ are circles on $S$ passing through $X$ and $B$, $\gamma_i$ are circles on $S$ passing through $A$ and $C$. Note that $A$, $B$, $C$, $X$ are now coplanar, so let $Z=AC\cap BX$. Consider $\mathbb{R}P^3$ projective transformation which saves $S$ and sends $Z$ to its center. Circles on $S$ are still circles on $S$, because they are defined as intersection of planes with $S$. $(X,B)$ and $(A,C)$ are now antipode pairs. Considering this as a spherical geometry problem, $\gamma_i$ and $h_i$ are now "lines" on $S$. Then, $V_{11}V_{21}V_{22}V_{12}$ is circumscribed iff (arc lengths) $CV_{11}+V_{11}B=CV_{22}+V_{22}B$, $V_{22}V_{32}V_{33}V_{23}$ is circumscribed iff $CV_{22}+V_{22}B=CV_{33}+V_{33}B$ and $V_{11}V_{31}V_{33}V_{13}$ is circumscribed iff $CV_{11}+V_{11}B=CV_{33}+V_{33}B$. Obviously, if two pairs of sums $CV_{11}+V_{11}B, CV_{22}+V_{22}B, CV_{33}+V_{33}B$ are equal, then so is third, so we're done.

I really enjoyed this problem! btw here is my proof by anti-homologous points and monge's theorem. name the intersection of $\gamma_2$ with incircles of $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}$ and $\widehat{V_{23}V_{13}}\widehat{V_{21}V_{31}}$,$P$ and $Q$,also name the intersection of $\gamma_1$ with incircles of $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}$ and $\widehat{V_{23}V_{13}}\widehat{V_{21}V_{31}}$ as $R$ and $S$. now since the pairs $(P,Q)$ and $(R,S)$ are anti-homologous points we get that lines $PQ$,$RS$ and $AC$ are concurrent. then it is obvious that exsimilicenters of the set of incircles $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ are on the line $AC$. Now consider that the forth circle is tagent to $h_2$,$h_3$ and $\gamma_2$ now by applying monge's theorem on the incircles $\widehat{V_{11}V_{21}}\widehat{V_{22}V_{12}}, \widehat{V_{12}V_{22}}\widehat{V_{23}V_{33}},\widehat{V_{22}V_{32}}\widehat{V_{32}V_{22}}$ we get that the forth circle should be tangnet to the arc $\gamma_3$ as desired $\boxed{\mathbf{Q.E.D}}$

skytin wrote: Yes , it's not hard to prove that if figures V_11V_21_V12V_22 , V_31V_21V_22V_32 , V_22V_12V_13V23 are described then V_32V_22V_23V_33 is outscribed too , but intersecting how to prove that exist circle witch is inscribed in V_31V_11V_13V_33 My solution to skytin's Problem, based on SebP's amazing method: Method 1: H1,H2G1,G2; H2H3G2G3; H1H2G2G3; H2H3G2G3 => H1H3G1G3 Relabel denote line l1,l2,l3; m1,m2,m3 the intersetion of H1,2,3 and G1,2,3 with plane O1,2,3,4, li,mj intersect at Aij, denote S the intersection of Hi,Gj it's sufficient to prove that the parallelogram O1,2,3,4 and A11A31A33A13 are homothetic since we can then take S'A11A31A33A13 homothetic to SO1O2O3O4 and sphere centered S' with radius r is tangent to H1,H3,G1,G3 Then with some calculations, it's not hard to show O1,2,3,4 and A11A31A33A13 are homothetic with ratio 2(h-r)^2/(h*(h-2r)), where h is the distance of S to plane O1,2,3,4. done. Method 2: H1,H2G1,G2; H2H3G2G3 => H1H3G1G3 Relabel denote l_ij the intersection of Hi,Gj, S the intersection of Hi,Gj s Similarly with method 1, first take two spheres centered O1O2 tangent to H1,H2G1,G2; H2H3G2G3 with same radius r It's sufficient to show O1O2// plane(S,l11,l33) Then consider any sphere centered O3O4 tangent to H1,H2G1,G2; H2H3G2G3 it's sufficient to show the external center of homothety of spheres (O3) (O4) on plane(S,l11,l33) take any plane H tangent to sphere (O3) (O4) project everything into this plane it's sufficient to prove a 2d problem: two ellipses tangent inside ABCX, DEFX, such that line AB, XCF, ED intersect at P; line BC, AXD, EF intersect at Q then the intersection of external common tangents of the two ellipses lies on line BE This could be easily proved by projective transformation that maps PQ to infinity line done.

This problem used to be really scary. It isn't anymore. We will prove a lemma which readily solve the problem. $\textbf{Lemma.}$ Fix circular arcs $\gamma_1$, $\gamma_2$, $\gamma_3$ all meeting at $A$ and $C$, and let $\omega_1$ and $\omega_2$ be circles with external center of homothety on line $AC$, with $\omega_1$ tangent to $\gamma_1$, $\gamma_2$, and $\omega_2$ tangent to $\gamma_2$, $\gamma_3$. Let $\omega_1$ be closer than $\omega_2$ to line $AC$. If $P$ is a variable point on $AC$, the curved quadrilateral formed by the tangents at $P$ to $\omega_1$, $\gamma_2$, and $\gamma_3$ is circumscribed. $\textit{Proof.}$ Denote by $\Omega$ the circle tangent to $\gamma_2$ such that $\Omega$ and $\gamma_2$ have $A$ as their exsimilicenter. We will show that the other circular arc $\gamma_3'$ through $A$ and $C$ tangent to $\Omega$ is tangent to $\omega_2$. To do this, we will prove a subclaim: $\textbf{Subclaim.}$ For $X$ variable on $AC$, the quadrilateral formed by the tangents from $X$ to $\omega_1$, $\gamma_2$, $\gamma_3'$ is circumscribed, provided it actually forms a quadrilateral. $\textit{Proof.}$ Perform an inversion at $A$ with radius $AC$. Then $\Omega$ and $\omega_1$ become the incircle and $A$-excircle respectively of some triangle. We will denote the images by $\Omega'$ and $\omega_1'$. Let $Y$ be a variable point on $AC$, and consider the circle $\kappa'$ which is tangent to the tangents at $Y$ to $\omega_1'$, whose center is collinear with $C$ and the center of $\Omega'$. By Monge's theorem, the exsimilicenter of $\kappa'$ and $\omega_1$ lies on $AY$, but $C$ lies on $AY$ and the line through the centers and hence it must be the exsimilicenter. Let $\kappa$ be the image of $\kappa'$ under the inversion. Remark that the exsimilicenter of $\kappa$ and $\kappa'$ is $A$. We apply Monge again on $\kappa$, $\kappa'$, and $\omega_1'$ to get that the exsimilicenter of $\kappa$ and $\Omega'$ lies on $AC$. We apply Monge AGAIN on $\kappa$, $\omega_1'$, and $\omega_1$ to get that the exsimilicenter of $\kappa$ and $\omega_1$ lies on $AC$. Since $Y$ is completely variable, we can obtain any circle tangent to $\gamma_1$ and $\gamma_2$ on the same side of the arcs and line $AC$ as $\Omega$. This does the trick. $\square$ Now we choose $X$ to be the exsimilicenter of $\omega_1$ and $\omega_2$. It can then be ovserved that there is one unique circle that isn't $\omega_1$ that is tangent to the tangents at $X$ to $\omega_1$ and $\gamma_2$. It thus follows that $\omega_2$ must be tangent to $\gamma_3'$. $\square$. To finish the problem, simply move $B$ to be the exsimilicenter of the incircles of $\widehat{V_{12}V_{22}}\widehat{V_{23}V_{13}},\widehat{V_{21}V_{31}}\widehat{V_{32}V_{22}}$ (which is on $AC$ by the subclaim), but then we apply the lemma by moving $B$ back to the original position. Yay.

Popop gave hints but the cooking is burnt. MONGE SPAM FTW Claim: Let $\Gamma$ be a circle with chord $AB$. Let $\Omega$ be a circle outside $\Gamma$ tangent to $\Gamma$. Let $C$ be a point on $AB$. Then if $\Omega_C$ is $\Omega$ scaled at $C$ to be tangent at $\Gamma$, then the center of $\Omega_C$ lies on a fixed ellipse. Proof. Let $I$ be the exsimilicenter of $\Gamma$ and $\Omega$. By Monge's we have that the tangency point of $\Omega_C$ is $D = CI \cap \Omega$. Let $O_1$ be the center of $\Gamma$ and $O_2$ the center of $\Omega$. Let $O_3$ be the center of $\Omega_C$. Redefine everything in terms of $D$. Then $C = IC \cap AB$, $\Omega_3 = CO_2 \cap O_1D$. Now, define $Z = O_1D \cap AB$. It then follows that \[ z = (O_1, Z; D, O_3) \overset{C}= (O_1, AB \cap O_1O_2; I, O_2) \]is fixed. Let $\theta$ be the angle at $Z$ between $ZO_1DO_3$ and $ZACB$. Then we have that $x = C \cdot \cos(\theta)$ for some fixed $C$. Then if we let $O_1O_3 = y$, it follows that \[ \frac{r}{x+r} \big\slash \frac{y}{x+y} = z \]which rearranges as $y = \frac{rx}{r(z-1) + xz}$ so thus it is a conic section. This can only be a parabola or hyperbola if $O_3$ is sometimes at infinity, which since $z < 1$ can't happen. $\blacksquare$ Anyways, this implies that given a unique $\gamma_1, \gamma_2$, and a fixed quadrilateral circle, there must be a unique arc $\gamma_3'$ regardless of the positioning of $C$. As such, it remains to solve the problem when $C = A$. Inverting around $C$ gives the following claim. Claim: Let three lines $a_1, a_2, a_3$ through $A$ and three lines $b_1, b_2, b_3$ through $B$ intersect such that three of the four quadrilaterals formed by pairwise intersections of $\{a_i, a_{i+1}\} \times \{b_j, b_{j+1}\}$ are tangential. Then the fourth is as well. Proof. Let the three circles be $C_1, C_2, C_3$ such that $C_1, C_2$ have exsimilicenter $A$, $C_1, C_3$ have exsimilicenter $B$, $C_2, C_3$ have exsimilicenter $C$. Define $C_4'$ to have exsimilicenter $A$ with $C_3$ as well as being tangent to the common tangent of $C_1, C_2, C_3$. Then, by Monge's on $C_1, C_2, C_3$ it follows that $AC$ is collinear with $B$. Monge's on $C_1, C_2, C_4'$ gives that $AC$ is collinear with the exsimilicenter of $C_2, C_4$. This gives that $C_2, C_4$ have exsimilicenter $B$ which finishes. $\blacksquare$

Here's an approach with special relativity which solves the problem and the generalisations mentioned above.

Applying the same argument, we can prove the generalisation shown in the colourful diagram in Géza Kós' post. A more interesting problem is skytin's generalisation, that $\widehat{V_{11}V_{31}}\widehat{V_{33}V_{13}}$ also has an incircle.

A more detailed solution with diagrams can be found in my blog post here, along with a proof of Casey's theorem and generalisations of other famous theorems using similar methods.

Drawing the centers $O_1$, $O_2$, $P_1$, $P_2$ of the respective circles, note $W_1$, $O_1$, $P_1$. etc. Now an easy angle chase gives that $W_1W_2X_1X_2$ cyclic using some isosceles triangles. Now observe by radical axis that $W_1X_1$, $W_2X_2$, and $AB$ concur at $K$. Now note too that by insimilicenter monge on $(O_1)$, $(O_2)$, and $(P_1)$ that the exsimilicenter of $(O_1)$ and $(O_2)$ lies on $W_1X_1$, similarly on $W_2X_2$. Call it $K$. Also, the exsimilicenter of $(O_2)$ and $(T_1T_2Y_1Y_2)$ is on $AB$ by monge on those with $(O_1)$. We label the circles, starting from top-right and going anticlockwise: $C_1$, $C_2$, $C_3$, $C_4$, with $C_4$ defined as the circle tangent at $Z_1$, $Z_2$, and $U_1$. Note now $Y_1Y_2Z_1Z_2$ is cyclic and $Y_1Z_1$, $Y_2Z_2$ pass through $K$. Now monge on $C_2$, $C_3$, $C_4$ gives that the exsimilicenter of $C_2$ and $C_4$ lies on $AB$, and then on $C_1$, $C_2$, $C_4$ gives the exsimilicenter of $C_1$ and $C_4$ is on $AB$, nad by the tangent is $C$. Then $CS_2$ is tangent to $C_4$ and we are done.