I think I solved it, but I used some projective transformations. Here's a hint (and, at the same time, an extension of the text of the problem): Try to consider just two circles ($C_a$ and $C_b$, for example), and try to show that the center of the large circle, the intersection of $AA'$ and $BB'$ and the center of the considered square are collinear. This would solve the problem, and it would say something more about the intersection pt of $AA'$ and $BB'$ (and $CC'$ and $DD'$, of course). It really is a nice problem! It's just the kind of problem which makes me feel sorry for not being a contestant in the TST .

dear Grobber,could you post your solution (I 'm not good at projective transformation)?

Let's introduce some notations: Let $O_a$ and $O_b$ be the centers of $C_a$ and $C_b$ respectively. Assume these two circles touch $AB$ at $A_1$ and $B_1$. Let $O$ be the center of the large circle and let $P$ be the center of $ABCD$. Assume $OA'$ and $OB'$ cut $AB$ at $A_2$ and $B_2$ respectively. Let $X$ be $A'B' \cap AB$. It's easy to see that $X$ is the external homothety center of $C_a$ and $C_b$, so it also lies on $O_aO_b$. This means that the lines $A'B'$, $O_aO_b$, $A_2B_2$ and $OX$ are concurrent in $X$, so $(A',A_2;O_a,O)=(B',B_2;O_b,O) \ (*)$. Consider the homographic transformation from the pencil of lines through $A$ to the pencil of lines through $B$ which maps $AA'\rightarrow BB'$, $AO_a\rightarrow BO_b$ and $AO\rightarrow BO$. From $(*)$ we find that it also maps $AB=AA_2\rightarrow BB_2=AB$, so it has a fixed element, and it's well-known result that such transformations are perspectivities (at least that's what they're called in Romanian ), i.e. the locus of the intersection points of corresponding lines is a line, so $O=AO\cap BO$, $P=AO_a\cap BO_b$, $AA'\cap BB'$ are collinear, and that's exactly what we wanted. I know that there are some much simpler solutions (a contestant showed me something using Menelaus, for example), but I like this idea.

Well, I think my solution is more beautiful. You see M and N and P and Q are the points on the circle that tangents from them to circle are respectively parallel to AB,BC,CD,DA. The tangents from Q and M to intersect at A". The tangents from M and N to intersect at B". The tangents from N and P to intersect at C". The tangents from P and Q to intersect at D". Homothety from A' takes A to A".So AA' goes from A". So we must prove AA",BB",CC" and DD" are concurrent. But this is obvious because sides of ABCD and A"B"C"D" are parallel. And so theses lines are concurrent on a line connecting center of square and circle. (Center of A"B"C"D" is center of circle) ____________________________________________ No pain ,no gain.

I heard of a solution given by a contestant which used homothety and this might be it. It is indeed nicer than mine .

Hi guys In my solution I havn't used that $ABCD$ is a square ,this problem is true when $ABCD$ is just a circumscribed quadrilateral . So try this problem now,I thing now it is more interesting .

Tiks wrote: ...this problem is true when $ABCD$ is just a circumscribed quadrilateral... This is like telling the solution. Let $(I)$ be the incircle of the quadrilateral $ABCD.$ $A$ is the internal homothety center of the circles $\mathcal C_a, (I)$ and $A'$ is the external homothety center of the circles $\mathcal C_a, \Gamma.$ Therefore, the internal homothety center $H$ of the circles $\Gamma, (I)$ lies on $AA'.$ Similarly, $H$ lies on $BB', CC', DD',$ so that it is the concurrency point.

It suffices to prove that $AA', BB', CC'$ are concurrent. By Desargues' Theorem, we just need to show that $AB \cap A'B', BC \cap B'C', CA \cap C'A'$ are collinear. By Monge's Theorem, $AB \cap A'B'$ is the exsimilicenter of $\mathcal{C}_a, \mathcal{C}_b$ and $BC \cap B'C'$ is the exsimilicenter of $\mathcal{C}_b, \mathcal{C}_c.$ Let $\omega$ be the incircle of $ABCD.$ By Monge's Theorem on $\omega, \mathcal{C}_a, \mathcal{C}_c$, we know that $AC$ goes through the exsimilicenter of $\mathcal{C}_a, \mathcal{C}_c.$ By Monge's Theorem on $\Gamma, \mathcal{C}_a, \mathcal{C}_c$, we know that $A'C'$ also does. Hence, $AC \cap A'C'$ is the exsimilicenter of $\mathcal{C}_a, \mathcal{C}_c$, and Monge's Theorem on $\mathcal{C}_a, \mathcal{C}_b, \mathcal{C}_c$ finishes. $\square$