Do you have all the problems?If you have,could you post them in contest collections?Thanks

guangzhou-2015 wrote: Do you have all the problems?If you have,could you post them in contest collections?Thanks I am very lazy to post them). I can give problems in Russian. But however I will post soon. So you can solve this problem until I will post others)

Let $A'$ be the antipode of $ A$ ,$D$ be the point where the $A$-altitude hits the circumcircle we know that $A',M,P$ are collinear and $A'D \parallel BC$ .so $A'(B,C;M,D)=-1$ thus $-1=(BC,PH)\implies (CB,P'A)\stackrel{I_{A_1}}{=}(BC,PH)=-1$ where $P'$ is the inverse of P by the negative inversion of center $A_1$ but $(CB,AS)=-1$ then $P'=S$ which means that $P,A_1,S $ are collinear .in the other hand $-1=A_1(C,T ;S,A)=(MT,M',P_\infty )$ therefore $M'$ is the midpoint of $MT$ which is collinear with $A_1$ and $S$ . RH HAS

Any other synthetic solution?

PROF65 wrote: Let $A'$ be the antipode of $ A$ ,$D$ be the point where the $A$-altitude hits the circumcircle we know that $A',M,P$ are collinear and $A'D \parallel BC$ .so $A'(B,C;M,D)=-1$ thus $-1=(BC,PH)\implies (CB,P'A)=A_1(BC,PH)=-1$ but $(CB,AS)=-1$ then $P'=S$ which means that $P,A_1,S $ are collinear .in the other hand $-1=A_1(C,T ;S,A)=(MT,M',P_\infty )$ therefore $M'$ is the midpoint of $MT$ which is collinear with $A_1$ and $S$ . RH HAS $P,S,A_1$ are collinear because $APA_1M$ is cyclic

@fastlikearabbit no it s not the reason but as i ve mentioned we have two harmonic quadrilateral whith three identical points so the fourths are the same namely $P'=S$ but $P,A_1,P'$ are colllinear so $P,S,A_1$ are collinear.

Let $L=BC \cap B_1C_1$ and observe that $L$ lies on $AP$ by radical axis theorem for $\odot(ABC), \odot(AB_1C_1)$ and $\odot(BC)$. Let $H$ be the orthocenter of $ABC$ and $A_0$ be the reflection of $H$ in $BC$. Note that $ABCS$ is a harmonic quadrilateral and $$-1=(B, C; A_1, L)=A_0(B, C; A_1, L)=(B, C; A, A_0L \cap \odot(ABC)),$$so $A_0, L, S$ are collinear. It is known that an inversion at $\odot(M, MB=MC)$ sends $H$ to $P$ so $$\frac{BP}{PC}=\frac{BH}{HC}=\frac{BA_0}{CA_0},$$hence $PBCA_0$ is harmonic; so $T$ lies on $PA_0$. Since $HA_0 \parallel MT$ we get $P, A_1$ and the midpoint of $MT$ are collinear. Projecting from $S,$ $$-1=(B,C; A_1, L)=S(B,C;A_1, L)=(B,C; A_0,SA_1 \cap \odot(ABC)),$$so $P, A_1, S$ are collinear.

Let $A'$ be the reflection of $A$ over $BC$, and let $O'$ be the antipode of $A$ in $\odot(ABC)$. Reflecting over $BC$, we obtain that $\overline{MSA'}$ are collinear by observing that $S$ maps to the HM point with respect to $A$, and since $MT\parallel AA'$, we obtain $A_1, S$ and the midpoint of $\overline{MT}$ collinear. Since $\angle AA_1M=\angle AO'S=\tfrac{\pi}{2}$ and $\angle A_1AM=\angle SAO'$ by isogonal conjugacy, $A$ is the spiral center sending segment $\overline{A_1M}$ to segment $\overline{SO'}$ and so $A_1S$ passes through the second intersection of line $MO'$ with $\odot(ABC)$, which is easily seen to be $P$.

First, we show that $P,A_1,S$ are collinear. Let $U=AP\cap BC.$ By radical axis on $\odot (AC_1HB_1),\odot(B_1C_1A_1M)$ (9-point circle of $\triangle ABC$ and $\odot (ABC),$ $U$ lies on $B_1C_1.$ Now inversion about $U$ with radius $UB \cdot UC$ maps $A_1\to M,$ $P \to A.$ We claim that this inversion maps $S$ to $H_A=AA_1 \cap \odot (ABC).$ Indeed, $\sqrt{bc}$ inversion at $A$ maps $\overline{BA_1DC}$ to $\odot (H_ASCB),$ so radical axis on $\odot (H_ASCB), \odot (ABC), \odot (AC_1HB_1)$ gives the desired. Let $A'$ be the antipode of $A$ WRT $\odot (ABC).$ It is well known that $P,H,M,A'$ are collinear. Now $\angle UMH_A = \angle CMA' = \angle HMA_1 = \angle PAH = \angle UAH_A$ so $UAMH_A$ is cyclic, and inverting back yields $P, A_1, S$ collinear as desired. \ Now, we show that $A_1, S, X$ are collinear, where $X$ is the midpoint of $MT.$ Letting $D = AT\cap BC,$ it follows that $-1 = (T,D;S,A)$ since $BC$ is the polar of $T.$ Now $(T,D;S,A) \stackrel{A_1}{=}(T, M;A_1S \cap MT, AA_1 \cap MT).$ But $AA_1 \parallel BC,$ so $AA_1 \cap MT = P_{\infty}$ and we are done.