delegat wrote: Let $ ABCD$ be a convex quadrilateral with $ BA$ different from $ BC$. Denote the incircles of triangles $ ABC$ and $ ADC$ by $ k_{1}$ and $ k_{2}$ respectively. Suppose that there exists a circle $ k$ tangent to ray $ BA$ beyond $ A$ and to the ray $ BC$ beyond $ C$, which is also tangent to the lines $ AD$ and $ CD$. Prove that the common external tangents to $ k_{1}$ and $ k_{2}$ intersects on $ k$. Author: Vladimir Shmarov, Russia I have not the best programm for images , but probably this will help:

here are some facts if i'm not mistaken... ABCD has an inscribed circle. The intersection point of the two common tangents lies on the diagonal BD.

So now we wait for the geometry experts of this forum to come and solve this problem. All I can say is that no one from Polish team has managed to solve this and we have some very strong geometers this year.

Albanian Eagle wrote: here are some facts if i'm not mistaken... ABCD has an inscribed circle. The intersection point of the two common tangents lies on the diagonal BD. I think you are wrong

shfdfzhjj wrote: Albanian Eagle wrote: here are some facts if i'm not mistaken... ABCD has an inscribed circle. The intersection point of the two common tangents lies on the diagonal BD. I think you are wrong To be more exact: you are wrong about that $ ABCD$ is circumscriptible (note that the bisectors of $ \angle{D}$ and $ \angle{B}$ don't even meet in the interior of $ ABCD$), but you are right about the fact that the two common internal tangents of $ k_{1}$ and $ k_{2}$ meet on $ BD$ (this follows from the 1-2 variant of the Monge-D'Alembert circle theorem applied for $ k_{1}$, $ k_{2}$ and $ k$; see the first part of my proof from http://www.mathlinks.ro/viewtopic.php?p=1186805#1186805. I'll be back with a proof soon.

why? i don't think so. the fact that ABCD has an inscribed circle follows from AB+CD=BC+AD. and then for the second fact apply monge's theorem to k_1, k_2 and the inscribed circle of ABCD. EDIT: oh ok oops! What I call the inscribed circle of ABCD is in fact circle k itself... (thanks pohoatza) EDIT2: where is Darij when you need him...

This problem is really nice, I like it. Here is my solution, it took me some time but I think it's correct. It's hard to draw a clear diagram, I hope this one's not too messy... (Unfortunately I am aware of the fact that it is. ) EDIT: I added a more detailed diagram for 'easy reading' (*cough*) Proof: Clearly, $ B$ and $ D$ are the external and internal centers of similitude of $ k;k_1$ and $ k;k_2$ respectively. The Monge-d'Alembert circle theorem now says that $ S$, the intersection point of $ AC$ and $ BD$, is the internal center of similitude of $ k_1$ and $ k_2$. Now, if we name $ D_1,D_2,B_1,B_2,Y,Z$ as in the figure, we have from Brianchon's theorem in a degenerate hexagon that $ B_1D_2,B_2D_1,YZ$ and $ BD$ are concurrent at a point, and we will call this point $ T$. Now we see that $ ZB,ZS,ZA,ZY$ is a harmonic bundle, so by looking at transversal $ BD$, we find that $ S$ and $ T$ divide $ B$ and $ D$ harmonically. If we call $ O$ the center of $ k$, we now have that $ OB,OS,OD,OT$ is a harmonic bundle. Now looking at the transversal which goes through the centers of $ k_1$ and $ k_2$, we find that $ U$, the intersection of $ OT$ and the line joining the centers of $ k_1$ and $ k_2$ is the external center of similitude we are looking for. We only have to prove that $ U$ lies on $ k$. Call $ U'$ the intersection of $ k$ and $ OT$, fartest away from $ AC$. Call $ B',D'$ the tangency points of $ k_1$ and $ k_2$ with $ AC$. Applying Monge-d'Alembert to $ k_1,k$ and the degenerate $ AC$ to obtain that $ B,B',U'$ are collinear. Similarly, we can prove that $ D,D',U'$ are collinear. From now on, we will also call $ E$ the intersection of $ AC$ and $ OT$. Using the degenerate form of Monge-d'Alembert on $ k_1,k,AC$, we obtain that $ B,B'',U$ are collinear, where $ B''$ is the antipode of $ B'$ in $ k_1$. Also, we can look at $ k_1,k_2,AC$, and obtain that $ U,D',B''$ are collinear. Combining these two results gives us that $ B,B'',D',U$ are collinear. Now, we know that $ U$ and $ S$ divide the centers of $ k_1$ and $ k_2$ harmonically. Projecting the four points on $ AC$ gives us that $ S$ and $ E$ divide $ B'$ and $ D'$ harmonically, and it follows that $ BB',BS,BD',BE$ is a harmonic bundle. Looking at the transversal $ OT$, we finally obtain that $ T$ and $ E$ divide $ U$ and $ U'$ harmonically, and because $ AC$ is the polar line of $ T$ wrt $ k$, we know that $ U$ lies on $ k$. Figures: Image not found And in detail: Image not found

JAN, is you IMO 2008 contestant? :

That's an impressive solution Jan. There is a very small typo: Jan wrote: Call $ B',C'$ the tangency points of $ k_1$ and $ k_2$ with $ AC$. It should be $ D'$ instead of $ C'$.

This problem is super difficult...is this problem G9 ???

I have also another proof. After reading Jan's proof, I feel we share quite similar points (and also you have two nice figures). So I will adopt your notation. Also, in principle, this proof is pretty closed to Jan's proof. So I will just sketch it briefly.

Let $ AB, CD$ meet at $ X$, $ AD, BC$ meet at $ Y$, let $ k$ meet $ AB, DC, AD, BC$ respecitvely at $ P,Q,R,S$. Using the tangent properties about $ k$ we obtain: $ BA + AD = BA + AR - DR = BP - DR = BS - DQ = BC + CQ - DQ = BC + CD$ Let $ k_1, k_2$ meet $ AC$ at $ J$ and $ L$ respectively. Using the incircles we have: $ AB + JC = BC + AJ$ and $ DA + LC = DC + LA$ Adding and using $ BA + AD = BC + CD$: $ JC + LC = AL + AJ$ Which quickly gives $ AL = JC$ Let the excircle of $ \triangle ABC$ on the side $ AC$ be $ k_3$, and the excircle of $ \triangle ADC$ on the side $ AC$ be $ k_4$. Hence $ k_3, k_4$ meets $ AC$ at $ L$ and $ J$ Now construct the tangent of $ k$ which is parallel to $ AC$ (and on the same side of $ k$ as $ AC$). Let that tangent meet $ k$ at $ Z$. Dilation about $ B$ takes $ k_3$ to $ k$ and $ L$ to $ Z$. Dilation (negative) about $ D$ takes $ k_4$ to $ k$ and $ J$ to $ Z$. Hence $ BL$ and $ DJ$ meet at $ Z$. Finally construct the two missing tangents to $ k_1$ and $ k_2$ which is parallel to $ AC$, let the points of tangency be $ M$ and $ N$ respectively. Similar dilation arguements shows that $ B,M,L,Z$ are collinear and $ D,N,J,Z$ are also collinear. Since $ JM$ and $ LN$ are parallel and are diameters of $ k_1$ and $ k_2$. $ JN$ and $ LM$ meet at the centre of dilation which takes $ k_1$ to $ k_2$, which we know is the point $ Z$. Hence $ Z$ is the intersection of the external common tangents, and from construction, $ Z$ lies on $ k$. $ Q.E.D$

Really amazing! Very ingenious and very elementary.

Some interesting results:$ PR,QS,BD,XY;AC,PS,QR,XY;PQ,RS,AC,BD$ are concurrent, respectively.

I have a feeling that this problem is very similar to IMO shortlist 2007 G8. They can both be done by the theorem on the three similitude centers.(Actually in the training camp in U.S we've done G8 in ISL 2007 so I think U.S can do well on this problem!)

I think that the configuration in the G8 is pretty different from the configuration of this problem. As Ivan's solution shows this problems relies on the properties which are analogical to the very simple (and known) properties of the circumscriptible quadrilateral. In the circumscriptible quadrilateral points of the tangency of incircles with the diagonal just coincide and here they are symmetric with respect to the midpoint, i.e. they are points of tangency of the suitable excircles. This seems to be the heart of the problem. I suspect that it should be possible to find a analogical result for the circumscriptible quadrilateral. The most similar problem I can think of is showing that if quadrilateral is circumscriptible then incenters of the triangles formed by every triple of vertices are concyclic. This is probably even harder but it deals with more familiar configuration when the circle is simply inscribed. By the way, Polish team also had known G8 before - some of the contestants even had solved it. So it seems it didn't help them much... Are there known any participants who solved this problem?

TomciO wrote: I think that the configuration in the G8 is pretty different from the configuration of this problem. As Ivan's solution shows this problems relies on the properties which are analogical to the very simple (and known) properties of the circumscriptible quadrilateral. In the circumscriptible quadrilateral points of the tangency of incircles with the diagonal just coincide and here they are symmetric with respect to the midpoint, i.e. they are points of tangency of the suitable excircles. This seems to be the heart of the problem. I suspect that it should be possible to find a analogical result for the circumscriptible quadrilateral. The most similar problem I can think of is showing that if quadrilateral is circumscriptible then incenters of the triangles formed by every triple of vertices are concyclic. This is probably even harder but it deals with more familiar configuration when the circle is simply inscribed. By the way, Polish team also had known G8 before - some of the contestants even had solved it. So it seems it didn't help them much... Are there known any participants who solved this problem? Maybe it's not that similar, but at least G8 helps me to solve this problem(at home). The key lemma is that the combination of dilation $ H_1$ which center at a point U and dilation $ H_2$ which center at a point V has center lies on UV.(This is also the key lemma in G8) Using notation in Ivan's solution. Observe dilation center at L send $ k_2$ to $ k_4$, and dilation center at B send $ k_4$ to $ k_1$. So by lemma the center of dilation that send $ k_1$ to $ k_2$ lies on BL. Similarly, J send $ k_1$ to $ k_3$, D send $ k_3$ to $ k_2$. So by lemma the center of dilation send $ k_1$ to $ k_2$ lies on JD. Now Define Z same as in Ivan's solution, we can then easily show that JD and BL both pass through Z. In fact, in ISL 2007 G8, the first step is to prove that we can put a incircle inside one of the quadrilaterals by segment chasing, then using dilations combinations and the lemma above to prove collinear.

The highest appreciation to Ivan!!! I like this solution I've also got a solution similar to Jan's solution(maybe more complex) with some algebra skills,which cost me over 3 hours. But after i accomplished,i didn't feel that proud because it could not show the spirit of this problem,and i once thought that it is merely a difficult problem but not a nice problem until i met Ivan's outstanding solution. My lack of pure geometric ideas and geometric inspiration is one of the reason why i could not get into our CHINESE IMO TEAM,even the training team. But i still believe that we are able to get the best result in this IMO.There's no dout that Ivan's solution is perfect,however,as i know,there are over 10 students who are strong enough to solve this problem in such a fantastic way.I'm waiting for the result. (Because it's the first time i come here,please don't laugh at my poor English )

JasonLiu wrote: however,as i know,there are over 10 students who are strong enough to solve this problem in such a fantastic way. could tell us how you knew that?

Orient $\overline{AC}$ to be horizontal. Call the north and south pole of a circle to be the point closest and furthest to $\overline{AC}$ respectively. Denote $N_{x}$ and $S_x$ to be the north and south poles for any circle $x$ respectively. Claim 1: $T_1A = T_2C$ where $T_1$ and $T_2$ are the tangents of $\omega_1$ and $\omega_2$ to $\overline{AC}$ respectively. Proof: We know that \[AT_1 = \frac{AB+AC-BC}{2} \text{ and } T_2C = \frac{DC+AC-AD}{2}\]So it suffices to show $AB+AD = BC+DC$, which is true by considering the tangents to $\omega$ and length chasing. $\square$ Therefore, $T_2$ is the excircle tangent point to $\overline{AC}$. $\square$ Claim 2: $B$, $N_{\omega_1}$, $N_{\omega_2}$, $N_\omega$ collinear. Proof: Since $T_2 = N_{\omega_2}$, this is true by two homotheties at $B$. One of them sends $\omega_1$ to the excircle and other sends $\omega_1$ to $\omega$. $\square$ Claim 3: $N_\omega, D, S_{\omega_2}, S_{\omega_1}$ collinear. Proof: Similar to the previous claim, a homothety at $D$ sending $\omega_2$ to the $D$-excircle of $\Delta ADC$ proves $D,S_{\omega_2},S_{\omega_1}$ collinear. To finish, take a negative homothety at $D$ sending $\omega_2$ to $\omega$. $\square$ These two claims imply there's a homothety at $N_w$ sending $\omega_2$ to $\omega_1$, as desired.

So much pain and suffering. Harmonic Solution: First we define points. Let $\triangle AB C$ and $\triangle ACD$ have incenters $I_1$ and $I_2$. Let $\overline{BA}$, $\overline{CA}$, $\overline{DA}$ and $\overline{DC}$ be tangent to $\omega$ at $E$, $F$, $G$, and $H$ respectively. Let $O$ be the center of $(EFGH)$, and define $Z = \overline{FH} \cap \overline{GE}$. Let $X$ and $Y$ be the insimilicenter and exsimilicenter of $\omega_1$ and $\omega_2$. Let $\overline{EF} \cap \overline{GH} = K$. Finally let $P$ be the intersection of the tangents from $E$ and $H$ to $\omega$. Similarly define $Q$ as the intersection of the tangents from $G$ and $F$. Also define $N$ as the foot from $O$ to $\overline{AC}$. Now note from La Hire's the polar of $K$ is $\overline{BD}$. Claim: $K \in \overline{AC}$. Proof. Complex bash with $(EFGH)$ as the unit circle. $\blacksquare$ Claim: $X \in \overline{BD}$ Proof. This follows from Monge d'Alembert on $\omega_1$, $\omega_2$ and $\omega$. $\blacksquare$ Claim: $X = \overline{EH} \cap \overline{FG}$. Proof. Complex bash with $(EFGH)$ as the unit circle. Compute $X' = \overline{EH} \cap \overline{FG}$ and then show that it is collinear with $B$ and $D$, and similarly with $A$ and $C$. $\blacksquare$ Then from La Hire's $P$ lies on the polar of $X$. Similarly $Q$ lies on the polar of $X$, hence the polar of $X$ is exactly $\overline{PQ}$. Claim: $K \in \overline{PQ}$ Proof. From La Hire's it suffices to show that $X$ lies on the polar of $K$. However this is exactly $\overline{BD}$ so we are done. $\blacksquare$ Claim: $\overline{OZ} \perp \overline{AC}$ Proof. Observe that this follows from Brokards on complete quadrilateral $EFGH$. We also find $Z \in \overline{BD}$ and $Z \in \overline{PQ}$ as a result of this Brokards application. $\blacksquare$ Now comes the tricky part, possibly incorrect. Claim: $\overline{OY} \perp \overline{AC}$ and hence $Y \in \overline{OZ}$. Proof. Note we have that $(I_1I_1,XY) = -1$. Then let $P_\infty$ denote the point at infinity along the line perpendicular to $\overline{AC}$. Then projecting through $P_\infty$ to $\overline{AC}$, $I_1$ and $I_2$ get sent to the incircle touch points of $\omega_1$ and $\omega_2$ say $S_1$ and $S_2$, $X$ stays fixed and $Y$ maps to the foot from $Y$ to $\overline{AC}$, say $N'$. It suffices to show that $N' \equiv N$, recalling that $N = \overline{OZ} \cap \overline{AC}$. Then it suffices to show that $- 1 = (S_1S_2, XN)$. Note $\frac{XS_1}{XS_2} = \frac{r_1}{r_2}$ where $r_1$ and $r_2$ denote the radii of $\omega_1$ and $\omega_2$ respectively. Then it suffices to show $\frac{NS_1}{NS_2} = \frac{r_1}{r_2}$. However as $S_1$ and $S_2$ are symmetric about the midpoint of $\overline{AC}$ it suffices to show that $\frac{NC}{NA} = \frac{r_1}{r_2}$. This is just a matter of computation (will edit in later). $\blacksquare$ Now assume for contradiction that $Y \not\in \omega$. Claim: $Y$ coincides with $Z$. Proof. Under inversion at $\omega$ we have $I_1 \mapsto I_1^*$, $I_2 \mapsto I_2^*$ etc. Then note $I_1^* = \overline{OI_1} \cap \overline{EF}$, $I_2^* = \overline{OI_2} \cap \overline{GH}$ and $X^* = \overline{OX} \cap \overline{PQ}$. Also we have $\angle KI_1^*O =90$. Similarly we also have $\angle KI_2^*O = 90$ and $\angle KX^*O = 90$. However then $O$, $I_1^*$, $I_2^*$, $X^*$, $K$ and $N$ are all concyclic. Then obviously $Y$ is forced to map to $N = \overline{OZ} \cap (OI_1^*I_2^*X^*KN)$ as $I_1I_2XY$ is a line, and $Y$ cannot map to $O$. $\blacksquare$ However then $Z$, $I_1$ and $I_2$ must always be collinear. Now looking at the polars of $I_1$ and $I_2$ as well as the pole of $I_1I_2$ we can derive a contradiction. I will also add a more normal finish tomorrow + a more elementary solution.

Let $\omega_1$ and $\omega_2$ be tangent to $\overline{AC}$ at $P_1$ and $P_2$ respectively. By "excircle Pitot", we have $AB+AD=CB+CD$, so $AP_1=\tfrac{1}{2}(BA+AC-BC)=\tfrac{1}{2}(DC+AC-DA)=CP_2$, hence $P_2$ is the $B$-extouch point of $\triangle BAC$ and $P_1$ is the $D$-extouch point of $\triangle DAC$. The intersection of the tangents is the exsimilicenter of $\omega_1$ and $\omega_2$, which may also be identified as the intersection of the line passing through the $P_1$-antipode and $P_2$ with the line passing through the $P_2$-antpide and $P_1$. It is well-known that the former line passes through $B$, while the latter passes through $D$. Let $\ell$ be the line tangent to $\omega$ that is parallel to $\overline{AC}$ and closer to $B$, and let $T$ be the tangency point between $\ell$ and $\omega$. A homothety at $B$ sending $\overline{AC}$ to $\ell$ sends $P_2$ to $T$, and a homothety at $D$ sending $\overline{AC}$ to $\ell$ sends $P_1$ to $T$, hence $T$ is the exsimilicenter, which lies on $\omega$. $\blacksquare$ Remark: I claim that this problem possibly has nothing to do with "art school" at all if you're aware of the $\overline{AC}$ tangency point fact for tangential quads (perhaps better known for when they have an incircle) beforehand, which is not that uncommon (shows up in at least two other problems on well-known contests)

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Let $\omega_1$ and $\omega_2$ be tangent to $\overline{AC}$ at $P$ and $Q$, with centers $I_1$ and $I_2$. Claim: $Q$ is the $B$-excircle tangency point in $\triangle AB C$. Proof. Note that from $AB + CD = AD + BC$ we find $AP + CQ = AQ + CP$ from common tangents, which implies $Q$ and $P$ are symmetric about the midpoint of $\overline{AC}$. However then our claim follows. $\blacksquare$ From similar reasoning $P$ is the $D$-excircle tangency point in $\triangle ACD$. Let the $B$-excircle in $\triangle A BC$ and $D$-excircle in $\triangle ACD$ be denoted by $\gamma_1$ and $\gamma_2$. Now orient such that $\overline{AC}$ is horizontal, with $B$ above $\overline{AC}$. Denote the north pole of $\omega_1$ and $\omega_2$ as the highest point on the respective circles. Similarly define the south pole. Claim: The north poles of $\omega$, $\omega_1$ and $\omega_2$ are collinear. Proof. From a homothety at $B$ mapping $\omega_1 \mapsto \gamma_2$ we find the north poles of $\omega_1$, $\gamma_2$ and $B$ are collinear. Similarly the homothety mapping $\omega_1 \mapsto \omega$ centered at $B$ gives that the north poles of $\omega_1$, $\omega$ and $B$ are collinear. Then our claim follows. $\blacksquare$ Similar reasoning allows us to conclude the south poles of $\omega_1$, $\omega_2$ and the north pole of $\omega$ are collinear. Then there exists a homothety at the north pole of $\omega$ mapping $\omega_2 \mapsto \omega_1$. Note that this center of homothety is in fact the intersection of the external tangents of $\omega_1$ and $\omega_2$, so we are done.

no way this is 40 mohs buh . If $f(P)$ denotes the length of the tangent from $P$ to $\omega$ then we have $AD-CD=(AD+f(D))-(CD+f(D))=f(A)-f(C)=-f(C)+f(A)=(f(B)-f(C))-(f(B)-f(A))=CB-AB.$ Now if we let $X,Y$ be the tangency points of $\omega_1,\omega_2$ on $AC$ respectively, we have $AY-CY=AD-CD=CB-AB=CX-AX,$ thus $AX=YC$ and $X,Y$ are the $D,B$ extouch points respectively. Now homotheties at $B$ take $\omega_1$ to the $B$ excircle to $\omega,$ so their topmost points with respect to $AC$ are collinear, that is, if $P$ is the topmost point on $\omega$ we have $PY$ passes through the topmost point $Q$ of $\omega_1.$ Similarly, by homothety at $D$ we see the bottommost points of $\omega_2$ and the $D$ excircle are collinear with $P,$ so $PX$ passes through the bottommost point $R$ of $\omega_2.$ Thus $P$ is the center of homothety taking $XQ$ to $RY,$ so it is the exsimilicenter of the circles with diameters $XQ,RY$ which are $\omega_1,\omega_2.$

Arrange the diagram so that $B$ is at the top. Then let $H_k$ be the point of tangency of the line parallel to $AC$ with $\omega_k$. Define $L_k$ similarly, with the parallel line closer to $D$. By Excircle Pitot, we have $AB + AD = BC + CD$ and it follows that $L_1$ and $H_2$ are reflections over the midpoint of $AC$ which implies that $H_2$ is the $B$-extouch point of $\triangle ABC$. Similarly, $L_1$ is the $D$-extouch point of $\triangle ACD$. Then by taking the homotheties $\omega_1 \to \omega$, and $\omega_1 \to$ the $B$-excircle wrt $\triangle ABC$, we find that $B - H_1 - H_2 - H$. Similarly, we can use homothety to find $L_1 - L_2 - D - H$. Now since $\mathcal{E}(\omega_2, \omega_1) = H$, we have the external tangents of $\omega_1$ and $\omega_2$ intersecting at $H$ which finishes. the diagram was annoying to draw but there was barely any art in this

Since when did external Pitot exist? Either way, nice problem. Let $T$ and $U$ be the tangency points of $\omega_1$ and $\omega_2$ with $AC,$ respectively. WLOG let $AC$ be horizontal with $B$ above $AC,$ and let $V$ be the bottom point of $\omega.$ Also set $T'$ to be the top point of $\omega_1.$ Finally let $X$ be the intersection of the external tangents of $\omega_1$ and $\omega_2$; we wish to show that $X$ lies on $\omega.$ By the external version of Pitot's theorem, the existence of $\omega$ tells us that $AB + AD = BC + CD,$ so $AB - BC = CD - AD.$ Thus $AT = \frac{AB + AC - BC}{2} = \frac{AC + CD - AD}{2} = CU.$ This implies that $B,T',U$ are collinear. We now consider the following homotheties. First, the homothety sending $\omega$ to $\omega_1$ sends $V$ to $T.$ Then, the homothety sending $\omega_1$ to $\omega_2$ sends $T$ to the bottom point $Y$ of $\omega_2.$ Finally, the homothety sending $\omega_2$ to $\omega$ must send $Y$ to the antipode $V'$ of $V$ with respect to $\omega$ because of composition of homotheties. Therefore, $V',D,Y$ are collinear, as are $B,T',U,V'$ since $B,T',V'$ are collinear, once again by homothety. On the other hand, since $AT = UC,$ we have that $T,Y,V'$ are collinear. Thus $T'U \cap TY = V'.$ However, $T'U \cap TY$ is just the exsimilicenter of $\omega_1$ and $\omega_2,$ which is just $X$! Therefore, $X = V',$ which lies on $\omega.$

Denote X, such that the tangent to $\omega$ trough X is parallel to AC. We want to show that PQ' and P'Q meet at X. If we do that we will be finished. By the homothety $\omega_1 \to \omega$ we have that B, Q', X are collinear. By the homothety $\omega_2 \to \omega$ we have that D, P', X are collinear $\Rightarrow$ we just reduced the problem to showing B, P, Q' and D, P', Q are collinear. B, P, Q' would be collinear if we showed P is the B-ex touch point in $\triangle ABC$, which is equivalent to showing AP = CQ, which is basically lemma 4.9. Now we do this similarly for the other collinearity of points D, P', Q. This means we proved PQ' and P'Q meet at X, which lies on $\omega$.

Define points and assume WLOG the configuration exists as shown below. A straightforward length chase forces $AB+AD = CB+CD$ (in reality a form of Pitot's), from which we can find $AE = CF$. This makes $BF$ the excentral-touchpoint cevian opposite $B$ in $\triangle ABC$. Notice that Monge's on $\omega_1$, $\omega_2$, and the $B$-excircle of $\triangle ABC$, say $\omega_B$ tells us the exsimilicenter of $\omega_1$ and $\omega_2$ lies on $BE'F$. Hence we claim the desired point is $BF \cap \omega$. To finish, we see the homothety at $B$ sending $\omega_B \mapsto \omega$ and $F \mapsto P$ implies the homothety at $D$ sending $\omega_2 \mapsto \omega$ also sends $F' \mapsto P$, so $P$, $D$, $F'$ collinear. However, we also have $D$, $F'$, $E$ collinear, giving the collinearity $P$, $F'$, $E$. Thus the homothety at $P$ maps corresponding diameters $EE'$, $FF'$ of $\omega_1$, $\omega_2$, so $P$ is the desired intersection of external tangents. $\blacksquare$ [asy][asy] size(350); defaultpen(linewidth(0.5)+fontsize(8)); pair A, B, C, D, E, F, E1, F1, P, I, I1, I2, IB, T1, T2, X1, X2, Y1, Y2; B = (0,0); I = (169,0); T1 = (144,60); T2 = (144,-60); X1 = 169+65*dir(160); X2 = 169+65*dir(220); A = extension(B, T1, X2, rotate(90,X2)*I); Y1 = extension(B, T1, X1, rotate(90,X1)*I); C = extension(B, T2, X1, rotate(90,X1)*I); Y2 = extension(B, T2, X2, rotate(90,X2)*I); D = extension(A, Y2, C, Y1); I1 = incenter(A,B,C); I2 = incenter(A,C,D); IB = extension(B,I1,C,rotate(90,C)*I1); E = foot(I1, A, C); F = foot(IB, A, C); E1 = 2*I1-E; F1 = 2*I2-F; P = intersectionpoint(B--1.5*F, circle(I,65)); draw(circle(I,65)^^incircle(A,B,C)^^incircle(A,C,D)); draw(circle(IB, length(IB-F)), red+linewidth(0.3)); draw(1.25*T1--B--1.25*T2^^Y2--A--C--Y1); draw(I--B--P--cycle^^E--E1^^F--F1, gray+linewidth(0.3)); dot("$A$", A, dir(120)); dot("$B$", B, dir(180)); dot("$C$", C, dir(240)); dot("$D$", D, dir(0)); dot("$E$", E, dir(180)); dot("$E'$", E1, dir(180)); dot("$F$", F, dir(240)); dot("$F'$", F1, dir(90)); dot("$P$", P, dir(330)); dot("$I$", I, dir(120)); dot("$I_1$", I1, dir(120)); dot("$I_2$", I2, dir(120)); dot("$I_B$", IB, dir(90)); [/asy][/asy]

Let the center of $k$ be $O$ and let $k$ touch $AB,CD,AD,BC$ at $W,X,Y,Z$ respectively. Now from Lenghts we notice that: \[ DA+AB=BW-AW+AY-DY=BZ-DX=BZ-CZ+CX-DX=BC+CD \]Now let $\omega_1, \omega_2$ touch $AC$ at $T_B,T_D$ respectively, from lenghts again note that: \[ CT_D=\frac{AC+CD-DA}{2}=\frac{AC+AB-BC}{2}=AT_B \]So if we let $\omega_B$ to be the $B$-excircle of $\triangle ABC$ and $\omega_D$ be the $D$-excircle of $\triangle ADC$ then we have that $\omega_B$ touches $AC$ at $T_D$ and $\omega_D$ touches $AC$ at $T_B$. Now let $k_1,k_2$ hit at $T$ then from Monge d'Alembert twice we get that $BT_D \cap DT_B=T$. Now consider homothety centered at $B$ from $\omega_B$ to $k$ which sends $AC \to \ell$, then we let $\ell$ touch $k$ at $T'$ and by the homothety we have $B,T_D,T'$ colinear, but note that homothety centered at $D$ sending $\omega_D \to k$ gives that $T_B,D,T'$ are colinear as well because in this homothety we also have $AC \to \ell$ because the condition is that $AC, \ell$ are on the same side of $O$ as a consequence of both excenters being on opposite sides w.r.t. $AC$ of their homothey centers and one of the homotheties being negative (so that $ABCD$ is convex, as we in fact have that $A,C$ lie in an ellipse with focus $B,D$ from the lenghts), so with all of this we can safely conclude that $T'=T$ which lies on $k$ thus we are done.

Very fun problem

Lemma: $ABCD$ is a cyclic quadrilateral with circumcircle $\omega$ and let $X$ be an arbitrary point on $\omega$. Let $K,L\in \omega$ such that $-1=(X,K;A,C)$ and $-1=(X,L;B,D)$. If $AD\cap BC=E$, then $DK,CL,XE$ are concurrent. Proof: Let's apply the projective transformation sending $AC\cap BD$ to the circumcenter of $\omega$. Then $ABCD$ is a rectangle. Work on the complex plane. Note that $a+c=0=b+d$. We have $k=\frac{a^2}{x},l=\frac{b^2}{x}$. Let $KD\cap CL=W$. \[w=\frac{-bk(l-a)+la(k-b)}{-bk+al}=\frac{x(a+b)+ab}{-x}\]\[(x+\frac{x(a+b)+ab}{x})(\frac{1}{a}+\frac{1}{b})=(\frac{1}{x}+\frac{\frac{1}{xa}+\frac{1}{xb}+\frac{1}{ab}}{\frac{1}{x}})(a+b)\]Thus, $XW,AD,BC$ are concurrent.$\square$ Lemma:$ABCD$ is a cyclic quadrilateral with circumcenter $O$ and $AD\cap BC=E,AB\cap CD=F,AC\cap BD=G$. Tangents to $(ABCD)$ at $A,C$ and $B,D$ and $A,B$ intersect at $P,Q,R$ respectively. If $I$ is the incenter of $\triangle PQR$, then prove that $OG$ and $EI$ intersect on $(ABCD)$. Proof: Let $OG$ meet $(ABCD)$ at $X,Y$ such that $Y,O,G,X$ lie on this order. Let $YF\cap (ABCD)=Z$. Note that if $M$ is the miquel point of quadrilateral $ABCD$, then $E,Y,X,F$ is an orthogonal system. Also we observe that $P,Q$ lie on $EF$ since they lie on the polar line of $G$. Define $J=(ZAP)\cap (ZBQ)$. We will show that $J$ is the incenter of $PQR$ and $E,J,X,Z$ are collinear which are sufficient. Let $XC\cap EF=H,XD\cap EF=I$. Claim of Lemma: $H,Q,B,Z$ and $I,P,A,Z$ are concyclic. Proof of Claim of Lemma: Since $YX\cap AC=G$ and $EF$ is the polar line of $G$, we see that $A,Y,H$ are collinear. Let $ZH$ meet $OG$ at $U$. Note that $H,F,A,Z$ are concyclic since under the inversion inf $Y$ with radius $\sqrt{YZ.YF}$, $EF$ and $(ABCD)$ swap. \[\measuredangle ZUO=\measuredangle HUM=90-\measuredangle FHZ=90-\measuredangle BAZ=\measuredangle ZBO\]Thus, $Z,B,O,U$ are concyclic. \[\measuredangle HZB=\measuredangle UZB=\measuredangle UOB=\measuredangle MOB=\measuredangle MQB=\measuredangle HQB\]Which gives that $H,B,Q,Z$ are concyclic. Similarily we observe that $I,A,P,Z$ are concyclic.$\square$ In order to prove the collinearity of $E,J,X$, we will show that $E$ lies on the radical axis of $(AJPZI)$ and $(BJQZH)$. We want to prove that there exists an involution $(E,EF_{\infty}),(I,P),(H,Q)$. Project this onto $\omega$ from $X$. If $PX\cap (ABCD)=K,QX\cap (ABCD)=L$, then we want to show that $DK,CL,EX$ are collinear which follows from the first Lemma.$\square$ Back to the problem, let $\omega$ be tangent to $BA,BC,DA,DC$ at $P,Q,S,R$ and $PS\cap QR=W$. Let $I_1,I_2,I$ be the circumcenters of $\omega_1,\omega_2,\omega$. By homothety, we see that $I$ lies on both $BI_1$ and $DI_2$. Let $AC\cap I_1I_2=T$. By Monge D'alambert, we conclude that the insimilicenter of $\omega,\omega_2$ and insimilicenter of $\omega_1,\omega_2$ and eximilicenter of $\omega,\omega_1$ are concurrent which is equavilent to the collinearity of $D,T,B$. By Pascal at $PPSRRQ$ we have $K,W,SR\cap PQ=U$ are collinear. Pascal at $SSPQQR$ gives $K,W,U$ are collinear thus, $K,L,W,U$ are collinear. By La Hire, polar line of $W$ with respect to $\omega$ must pass through $A,C$ hence it's $AC$. So the polar line of $T$ pass through $W$. Also $B,T,D$ are collinear so if we take their dulas with respect to $\omega$, those lines must be concurrent. Thus, polar of $T$ pass through $PQ\cap RS=U$ which implies $UW$ is the polar line of $T$. So $T,P,R$ and $T,S,Q$ are collinear. Pascal at $PPRSSQ$ yields $A,T,U$ are collinear and pascal at $RRSQQP$ gives $C,U,T$ are collinear. So $AC$ passes through $U$. Taking dual of concurrent lines $AC,RS,KL,PQ$ implies $B,T,D,W$ are collinear. Let $E$ be the eximilicenter of $\omega_1$ and $\omega_2$. \[(IE\cap BD,T;D,B)\overset{I}{=}(E,T;I_2,I_1)=-1=(W,T;D,B)\]Thus, $J,W,E$ are collinear. Now applying the second Lemma at quadrilateral $PQRS$ with circumcenter $I$ imply the result as desired.$\blacksquare$

We will start by defining a bajillion points: $N= AB \cap \omega$, $K = AB \cap DC$, $J= CD \cap \omega$, let $F$ be the center of homothety of $\omega_1$, and $\omega_2$, $M = \omega \cap AD$, $L = AD \cap BC$, $E = BC \cap \omega$, let $I_1$, $I_2$, and $I$ be the centers of $\omega_1$, $\omega_2$, $\omega_3$ respectively. $G$, and $H$ be the $AC$ intouch points of $\triangle ABC$, and $\triangle ADC$. We have $$EC+KJ+DJ = CJ + KJ + DM$$$$EC+KD = CK + DM$$$$EC+KD+MA = AN + CK+DM$$$$FB+EC+KD+AM = AN+ BE + CK + DM$$$$BN+EC+ KD + AM = AN+BE + CK + DM$$$$BN-AN- (BE-EC) = (CK-DK) - (AM-DM)$$$$AB-BC = DC-DA$$$$\tfrac{AB+AC-BC}{2} = \tfrac{AC+DC-AD}{2}$$$$AG = CH$$So $H$ is the $B$-extouch point of $\triangle ABC$. By homothety the tangent to $\omega$, and $BH \cap \omega$ (the intersection closer to $B$) is parallel to $AC$. By a well known incenter configuration the tangent to $\omega_1$ at $BH \cap \omega_1$ closer to $B$ is parallel to $AC$. Since the line through this point and $H$, or line $BH$. And the line through $GD$ which contains the point on $\omega_2$ such that the tangent to that point is parallel to $AC$ contains the point on $\omega$ that has tangent line parallel to $AC$ and is closer to $B$. However by homothety $GD \cap BH= F$ is the center of homothety between $\omega_1$, and $\omega_2$, and both of these lines contain the same point on $\omega$ and we are finished.