My solution: Let $ \triangle DEF $ be the tangential triangle of $ \triangle ABC $ . From Cevian nest theorem we get $ DM, EN, FP $ are concurrent (i.e. $ \triangle DEF $ and $ \triangle MNP $ are perspective) , so from Desargue theorem we get $ X \equiv EF \cap NP, Y \equiv FD \cap PM, Z \equiv DE \cap MN $ are collinear . Q.E.D

By Menelaus' theorem we have to prove that $\frac{MZ.NX.PY}{NZ.PX.MY}=1$. But obviously $\frac{MZ}{NZ}=\frac{MC}{CN}.\frac{\sin Z\hat{C}M}{\sin Z\hat{C}N}=\frac{MC}{CN} .\frac{\sin\hat{A}}{\sin\hat{B}}$ so we just have to prove that $\frac{MC.NA.PB}{NC.PA.MB}=1$ which is by Cevian theorem equivalent to $AM$, $BN$ and $CP$ being concurrent, which is true. Is it correct? I didn't use the concurrency point being $O$ and it was easier than a TST.

It's very easy (for Romania) by Meneleus theorem, because we can prove easily $ \frac{XP}{XN}=\frac{AP}{AN} \cdot \frac{AB}{AC} $.

Here is another solution: Let the tangents to $\odot (ABC)$ at $B$ and $C$ meet at $Q.$ Define points $R$ and $S$ similarly. Now, we work in the frame of $\triangle QRS$! Notice that since $X \equiv NP \cap RS, \; Y \equiv PM \cap SQ, \; Z \equiv MN \cap QR$, the desired collinearity is equivalent to $\triangle QRS$ and $\triangle MNP$ being in perspective. By Desargues' Theorem, it suffices to show that $QM, \; RN, \; SP$ intersect at one point. We will in fact show that this concurrence point is the centroid of $\triangle QRS.$ To do so, we must draw upon a lemma: Lemma. Let $I$ be the incenter and $D, E, F$ be the points of tangency of the incircle with sides $\overline{BC}, \overline{CA}, \overline{AB}$ in $\triangle ABC.$ Then $ID, EF$, and the $A$-median meet at one point. Proof of Lemma. Let $M$ be the midpoint of side $\overline{BC}$ at suppose that $AM$ and $EF$ intersect at $X.$ We will prove that $X, I, D$ are collinear. Let $l$ be the line parallel to $BC$ passing through $A$ and let $Y \equiv EF \cap l.$ Finally, let $P_{\infty}$ be a point at infinity on line $BC.$ Then note that \[-1 = (B, C; M, P_{\infty}) = A(B, C; M, P_{\infty}) = (E, F; X, Y).\] Therefore, the division $(E, F; X, Y)$ is harmonic, so $Y$ lies on the polar of $X$ with respect to the incircle. In addition, since $AE$ and $AF$ are tangent to the incircle, $X$ lies on the polar of $A.$ Therefore, $A$ lies on the polar of $X.$ It follows that $AY$ is the polar of $X$ with respect to the incircle. Hence, $XI \perp l \implies XI \perp BC$, whence we deduce that $X, I, D$ are collinear, as desired. $\blacksquare$ Now, let's apply our lemma to the problem at hand. By the lemma, $QM, \; RN, \; SP$ are medians in $\triangle QRS.$ Therefore, they concur at one point, that is the centroid of $\triangle QRS.$ $\square$

Let $B^{'}$ be the antipode of $B$ wrt the $\odot ABC$. $(B,B^{'};A,C)=B(B,B^{'};A,C)=(Y,F;P,M)$ So $\frac{YP}{YM}\cdot \frac{FM}{FP}=\frac{c\cos\alpha}{a\cos\gamma}$ $\frac{FM}{FP}=\frac{\cos\alpha PB}{\cos\gamma MB}$ and so $\frac{YP}{YM}=\frac{MB\cdot c}{PB\cdot a}$, and from here we are done from Menelaus in $\triangle NPM$

Nice result, although too easy for Romanian TST . I'll post two solutions which I found (and also are the obvious ones) ComplexPhi wrote: Let $ABC$ a triangle and $O$ his circumcentre.The lines $OA$ and $BC$ intersect each other at $M$ ; the points $N$ and $P$ are defined in an analogous way.The tangent line in $A$ at the circumcircle of triangle $ABC$ intersect $NP$ in the point $X$ ; the points $Y$ and $Z$ are defined in an analogous way.Prove that the points $X$ , $Y$ and $Z$ are collinear. Solution 1: Note that by sine rule in $\triangle XPA$ and $\triangle XNA$, we obtain $\frac{XP}{XN} = \frac{AP}{AN} \cdot \frac{\sin{\gamma}}{\sin{\beta}}$. Thus, we have :$$ \prod_{\text{cyc}} \frac{XP}{XN} = \underbrace{\left(\prod_{\text{cyc}} \frac{AP}{AN}\right)}_{=1} \cdot \underbrace{\left(\prod_{\text{cyc}} \frac{\sin{\gamma}}{\sin{\beta}}\right)}_{=1}$$Hence, we are done by Menelaus' theorem on $\triangle MNP$. Solution 2: Let $\overline{BB} \cap\overline{CC} = J ; \overline{CC} \cap \overline{AA} = K ; \overline{AA} \cap \overline{BB} = L$.Now, in $\triangle JKL, \overline{JA} ,\overline{KB},\overline{CL}$ concur (at the symmedian point of $\triangle ABC$). Next, $\overline{AM} ; \overline{BN} ; \overline{CP}$ concur (at $O$). So, by the cevian-nest theorem we have that $\overline{JM} , \overline{KN} , \overline{LP}$ concur. This implies that $\triangle JKL$ and $\triangle MNP$ are perspective. So by Desargue's theorem $ \overline{JK} \cap \overline{MN} = Z ; \overline{KL} \cap \overline{NP} = X ; \overline{LJ} \cap \overline{PM} = Y$ are collinear.

Let $A',B',C'$ be the antipodes. Then from Pascal on $BAACC'B'$, we have that $X--B'--C'$,with this definition of $X$ ,one may finish via complex numbers or continue with Pascal....