Image not found I'm sorry for my not perfect picture,it is look like $ A,I_2,C$ are collinear ,but it isn't true.Thank you.

I think you know that there is exits a circle, which is tangent to $ AC$, $ BD$ and $ (O)$ at $ E$, $ F$ and $ P$, respectively and the line $ EF$ is passes through the incenters $ I_1$, $ I_2$ of triangles $ ABC$, $ ABD$. Denote by $ X$, $ Y$, $ Z$, $ T$ be the middle points of arcs $ AC$, $ AD$, $ DB$, $ CB$. Applying Pascal's theorem for the cyclic hexagon $ PYBDAZ$, we have $ N = PY\cap DA$, $ I_2 = YB\cap AZ$ and $ F = BD\cap ZP$ are collinear. Similarly, $ M$, $ I_1$ and $ E$ are collinear. Hence we conclude that six points $ M$, $ N$, $ E$, $ F$, $ I_1$ and $ I_2$ are collinear. P.S. About the result $ EF$ passes through $ I_1$, $ I_2$, I think it's well-known (maybe it's called as Thebault theorem)

Here is an image for your nice solution: Image not found P.S: Yes,I know this solution,which use your facts above. I'll give a proof for them later.

April wrote: I think you know that there is exits a circle, which is tangent to $ AC$, $ BD$ and $ (O)$ at $ E$, $ F$ and $ P$, respectively I would like to see a proof of this!

Oops! Thank you very much dear nsato! This is exactly what I am wavering. Please give me a little time to check and try to prove it. And for now, I'll give another solution for this problem, to atone for one's mistake. Denote by $ X$ the intersection of $ PN$ with the circle $ \omega$ and $ I_2$ the intersection of $ BX$ with $ MN$. We have $ X$ is the middle point of the arc $ AD$ of the circle $ \omega$. Consider the common tangent $ Pt$ at $ P$ of two circles $ \omega$ and $ \zeta$. We have $ \angle I_2MP=\angle NPt=\angle I_2BP$, therefore the quadrilateral $ BPI_2M$ is cyclic. It implies that $ \angle PI_2B=\angle BMP=\angle MNP$, which follows that $ \triangle XI_2P\sim\triangle XPI_2$. Hence we conclude that $ XI_2^2=XN\cdot XP$ $ (1)$ On the other hand, we can easily to see that $ \triangle XDN\sim\triangle XPD$, so $ XD^2=XN\cdot XP$ $ (2)$ Combining $ (1)$ and $ (2)$, we get $ XI_2=XD$. And notice that $ X$ is the middle point of arc $ AD$ of the circumcircle of triangle $ ABD$ and $ I_2$ is a point lies on $ BX$, so we conclude that $ I_2$ is the incenter of triangle $ ABD$. Similarly, $ MN$ also passes through the incenter $ I_1$ of riangle $ ABC$. And our proof is completed. $ \bullet$ Hope that I didn't get any mistake in this proof . $ \bullet$ I believe that if $ ABCD$ is a given cyclic quadrilateral, then there always exits two circles $ \omega$ and $ \Omega$, where $ \omega$ is tangent to $ (O)$ - the circumcircle of quadrilateral $ ABCD$, at $ P$ and also to $ AC$, $ BD$, whereas the circle $ \Omega$ is also tangent to $ (O)$ at $ P$ and to $ AD$, $ BC$.

April wrote: $ \bullet$ I believe that if $ ABCD$ is a given cyclic quadrilateral, then there always exits two circles $ \omega$ and $ \Omega$, where $ \omega$ is tangent to $ (O)$ - the circumcircle of quadrilateral $ ABCD$, at $ P$ and also to $ AC$, $ BD$, whereas the circle $ \Omega$ is also tangent to $ (O)$ at $ P$ and to $ AD$, $ BC$. Yes,of course,it is obviously that there exist such circle $ \omega$.But we must prove that $ \omega$ is tangent to diagonals at points $ E,F$. Here is another lemma to this problem: Let $ ABCD$ be a cyclic quadrilateral and $ I_1,I_2$ are incenters of $ ABC$ and $ ABD$ respectively.Denote $ O\in AC\cap BD$,$ E=\in AC\cap I_1I_2$,$ F\in BD\cap I_1I_2$.Prove that $ OE=OF$. This problem is from All-Russian olympiad,i'll post a solution to it later.

Here is an image to your nice proof: Image not found To April: I think that it must be $ \triangle XI_2P\sim \triangle XNI_2$ in your proof,instead of $ \triangle XI_2P\sim \triangle XPI_2$. Thank you.Great proof.

Let $ R$ be the point of intersection $ AC$ and $ MN$.Prove that $ PR$ is an angle bisector of $ \angle APC$. Image not found

Nobody solved my last problem?

$ \frac {AN}{AP} = \frac {CP}{CM} =$constant$ (\sqrt{\frac{R-r}{R}}$ where $ R,r$ is radius of $ w,\zeta$). So $ \frac {PC}{AP} = \frac {CM}{AN}$. Let $ BC$ and $ AD$ meet $ X$. Since $ XM$,$ XN$is tangent line of $ \zeta$,$ XN = XM$. Hence $ sin\angle ANR = sin\angle CMR$ and $ sin\angle ARN = sin\angle CRM$. So by law of sin in triangle $ ANR,CRM,\frac {CM}{AN} = \frac {RC}{AR}$. Henc $ \frac {PC}{AP} = \frac {RC}{AR}$ Therefore, $ RP$ bisect $ \angle APC$

Wow it is a nice solution! but i don't know why AN/AP=CP/CM could you solve this more detail?

Partial Solution: Let $ PM$ and $ PN$ intersect $ \omega$ again at $ M'$ and $ N'$. Let $ P'$ be the midpoint of small arc $ AB$. Lemma 1: $ PP'$, $ AB$, and $ MN$ are concurrent. Proof: I don't have this yet, but I believe that it should be easy with an inversion. It is well known that $ M'$ and $ N'$ are the midpoints of small arcs $ BC$ and $ DA$, respectively. Apply Pascal on $ AM'PP'CB$ to get that $ I_1=AM'\cap P'C$, $ M=M'P\cap CB$ and $ X=PP'\cap BA$ are collinear. By symmetry of argument, $ I_2$, $ N$, and $ X$ are collinear. By lemma 1, $ X$, $ M$, and $ N$ are collinear, so $ I_2$ and $ I_1$ lie on $ XMN$, so $ M,N,I_1,I_2$ are collinear. I hope to come back with a proof for lemma 1.

In my post above, lemma 1 is easy to prove. If we look in triangle $ PAB$, $ PP'$ is the external angle bisector of the angle at $ P$. So if $ PP'\cap AB = X_1$, then $ BX_1: X_1A = PB: PA$ is easy to calculate. If we connect $ AD$ and $ BC$ to meet at $ Z$, then in triangle $ XAB$, we get that $ MN$ intersects $ AB$ at $ X_2$, we get $ BX_2: X_2A = BM: AN$ by menelaus. It is a pretty easy lemma to show that $ PB: PA = BM: AN$ (it is the same fact that Leonhard Euler uses in his solution to the second problem in this topic). Comment: This is a very interesting and (in my opinion) difficult problem. The reason is this: If we consider the quadrilateral $ ABCD$ fixed, there are two circles that are tangent to $ \omega$, $ BC$, and $ AD$. Point $ P$ can either lie on small arc $ AB$ or $ CD$. If $ P$ is on small arc $ CD$, then the problem is true. Another way to state this is, if $ PBCDA$ is convex, the problem is true, if $ PDABC$ is convex, the problem is not true. So when you use the methods that you use, they must differentiate between these two cases. Obviously, Pascal's theorem is projective, so it does not differentiate between these cases. The reason that my solution here works is because it implicitly invokes a sort of betweenness argument. If you look at point $ N$, it could potentially have two positions based on how I used it: in particular, there are two tangents from point $ A$ to the smaller circle. When I used menelaus, it differentiated between these two cases (because $ X_2$ would be on segment $ AB$ otherwise). I have not read April's post, but I'm sure the solution can be interpreted as I did above.

Proof of April's lemma is also here http://forumgeom.fau.edu/FG2003volume3/FG200325.pdf .

Here is my solution First, we construct the common targent Tx of two circles Hence, we have BMI1T is a cyclic quadrilatetal Let M is intersection of CI1 with (w). Easy to proof that MI1=MA=MX, so I1 be the incenter of triangle ABC.

Here is my solution First, we construct the common targent $ Tx$ of two circles Hence, we have $ BMI_{1}T$ is a cyclic quadrilatetal Let $ M$ is intersection of $ CI_{1}$ with $ (w)$. Easy to proof that $ MI_1=MA=MX$, so $ I_1$ be the incenter of triangle $ ABC$.

This problem isn't Chinese TST 2007 6th quiz P2. maybe orl have some mistakes.

Is there somebody who knows where this question comes from???