this beautiful solution is from my master Mr.Artin Spiridonoff: let $K$ be the reflection of $A$ to $BC$ and $Y$ the intersection point of $OB$ and $AD$. then we have: $M(O,B,Y,P)=B(K,D,M,X)=C(K,D,M,X)=A(O,B,Y,P)$. so $O$,$B$ and $P$ are collinear.

please explain what those this mean?? edit: i have understood it sorry, you have used a very nice lemma sounding like this if 2 fascicles with the same biraport have 3 comon points then they have 4 comon ponits

paul1703 wrote: please explain what those this mean?? edit: i have understood it sorry, you have used a very nice lemma sounding like this if 2 fascicles with the same biraport have 3 comon points then they have 4 comon ponits yes, exactly!

I solved this one using trig, the calculations are a bit long (but not so long though), so I'll just sketch my proof. Let $P_{1}$ be the point where $AN$ meets $OB$ and let $P_2$ be the point where the tangent of the circumcircle of $ABC$ at $M$ meets $OB$. We want to prove that $OP_1=OP_2$. Now, looking at triangles $OAP_1$ and $OMP_2$ we have \[tan \angle AP_1O= \frac{OA}{OP_1}\] \[sin\angle OP_2M= \frac{OM}{OP_2}\] We also have $OA=OM$, so if we prove that $tan \angle AP_1O=sin\angle OP_2M$ we are done. But it´s straightforward to check that $\angle OP_2M=\angle ABX$ and $\angle AP_1O=\angle ACX$. So if we prove $tan \angle ACX=sin\angle ABX$ we are done.... [Trig here] $\blacksquare$

goodar2006 wrote: $M(O,B,Y,P)=B(K,D,M,X)=C(K,D,M,X)=A(O,B,Y,P)$. so $O$,$B$ and $P$ are collinear. So how do you work with those fasicles?

SCP wrote: goodar2006 wrote: $M(O,B,Y,P)=B(K,D,M,X)=C(K,D,M,X)=A(O,B,Y,P)$. so $O$,$B$ and $P$ are collinear. So how do you work with those fasicles? I can't undrestand what you mean. since the harmonic ratios $M(O,B,Y,P)$ and $A(O,B,Y,P)$ are equal, we conclude $O$,$B$ and $P$ are collinear. whats the problem?

Solution : Lemma : Given points ABCD , AC intersect BD at P , AC is perpendicular to BD , PB = AP , CB || AD , then there exist point Q , such that Q is isogonal to P wrt ABCD Not hard to prove that projections of P on sides of ABCD are on circle with center at O , reflect P wrt O and get Q Use this Lemma to XBMC (ADC ~ BDM) , so there exist point D' , such that Angle MCD' = DCX , CXD' = DXB , BMD = D'MC = 45 Easy to see that ABDO is cyclic (ABDO) intersect NA at points Y , A after Reim's theorem we get that YD || NC Angle BYD = BAO = 45 = CXD' , DOB = DAB = D'CX DO intersect PM at point Z Angle OBZ = ZMO = 90 = XD'C Angle BOY = BAN = MCD' Easy to see that YZOB ~ MXCD' , so angle YZO = MXC = NXA = YDA Angle DZM = 90 - PMB = 90 - MCD = ANC = AYD After easy angle chasing we get that AYZM is cyclic P is radical center of (AYBDO) , (OBZM) , (AYZM) . done

goodar2006 wrote: this beautiful solution is from my master Mr.Artin Spiridonoff: let $K$ be the reflection of $A$ to $BC$ and $Y$ the intersection point of $OB$ and $AD$. then we have: $M(O,B,Y,P)=B(K,D,M,X)=C(K,D,M,X)=A(O,B,Y,P)$. so $O$,$B$ and $P$ are collinear. Any simple way to see that $M(O,B,Y,P)=B(K,D,M,X)$, or it's not trivial?

Here's a simple solution by harmonic quadrilateral Let MM'⊥BO, and M' is on⊙O,so we only need to prove that PO is perpenticular to MM',which means that AM'NMis a harmonic quadrilateral. So we only need to prove that C(A,N;M,M') is harmonic. Let H be the orthocenter of triangle ABC,as HD^2=MD^2=BD^2=DX·DA,so (A,D;H,M)is harmonic. So we only need to prove that C,H,M' is on a line. As ∠MCB=∠MAB=∠HCB,we are done.