[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(180), B = dir(0), C = dir(135), I = incenter(A,B,C), M = (A+B)/2, P = extension(I,I+dir(0),C,C+dir(B--foot(B,C,M))), H = orthocenter(P,A,B), Q = extension(C,H,P,M), L = dir(90), T = intersectionpoint(I--I+dir(L--I)*100,unitcircle), J = extension(P,I,M,L), D = foot(I,A,B), D1 = 2I-D, K = foot(P,A,B), B1 = foot(A,B,H); draw(A--B--C--A^^unitcircle^^incircle(A,B,C), heavycyan); draw(circumcircle(M,C,T)^^C--P, purple); draw(C--H^^P--M, orange); draw(C--J--T, orange); draw(J--P, orange+dotted); draw(B1--P--H--B^^A--K, heavyred+dotted); dot("$A$", A, dir(225)); dot("$B$", B, dir(0)); dot("$C$", C, dir(90)); dot("$P$", P, dir(180)); dot("$I$", I, dir(0)); dot("$M$", M, dir(305)); dot("$H$", H, dir(240)); dot("$Q$", Q, dir(210)); dot("$T$", T, dir(225)); dot("$J$", J, dir(5)); dot("$D$", D, dir(285)); dot("$D'$", D1, dir(75)); dot("$K$", K, dir(190)); dot("$B_1$", B1, dir(255)); [/asy][/asy] Let $T$ be the $C$-mixtilinear intouch point, $D$ be the foot from $I$ to $\overline{AB}$, $D'$ be the reflection of $D$ over $I$. Let $J = \overline{AD'} \cap \overline{TD}$, and redefine $P$ to be the circumcenter of $(CIT)$. The following claim is true for any triangle, with $M$ replaced by the circumcenter. Claim: $I$ is the incenter of $\triangle CJT$; points $C$, $J$, $M$, $T$, $P$ are concyclic; and $P$, $J$ lie on the line through $I$ parallel to $\overline{AB}$. Proof. By well-known mixtilinear properties, the point $J$ lies on the line through $I$ parallel to $\overline{AB}$ and the perpendicular bisector of $\overline{AB}$. Then $\angle CJI = \angle TJI$ and $\angle TCI = \angle JCI$, so $I$ is the incenter of $\triangle CJT$. Because $\angle IJM = 90^\circ$ and $MC = MT$, $M$ is the midpoint of major arc $CJT$. By Fact 5, $P$ is the midpoint of minor arc $CT$. $\square$ It follows that $\overline{CP}$ is tangent to $(ABC)$ from $\measuredangle PCM = 90^\circ$, so the problem statement's characterization of $P$ is equivalent to ours. Claim: $Q = \overline{MP} \cap \overline{CT}$ lies on the incircle. Proof. By our previous claim, $Q$ is the midpoint of $\overline{CT}$. Then $$\angle IQC = \angle IMJ = \angle IDJ,$$meaning $Q$ is the reflection of $D$ over $\overline{TI}$, where the first equality is from this lemma. $\square$ Now, it suffices to show that $H$ lies on $\overline{CT}$, the radical axis of $(CJT)$ and $(ABC)$. Since $\overline{MP}$ and $\overline{AB}$ are diameters in the respective circles, we have $K = \overline{MA} \cap \overline{PH}$ lies on $(CJT)$ and $B_1 = \overline{PA} \cap \overline{BH}$ lies on $(ABC)$. Thus $HK \cdot HP = HB_1 \cdot HB$, and we are done. $\blacksquare$ Remark: For a related problem also involving the point $J$, see here (shameless advertisement oops).

$PB\cap (ABC)=X,PA\cap (ABC)=Y,PH\cap AB=Q,D=(I)\cap BC$. Let $Z$ be the foot of the altitude from $C$ to $PM$. We have $PQ\cdot PH=PC^2=PZ\cdot PM \implies \angle HZM=90 \implies H-Z-C \implies CH \cap PM=Z$. $\textbf{Claim:}$ $PCM \sim IDM$ $\textbf{Proof:}$ Wlog $a>b$. I will write briefly. $PC=abc/((a-b)(a+b+c)), DM=(a-b)/2$. $PC/CM=r/DM \leftrightarrow (a+b-c)/2=r=ab/(a+b+c)$ which is true. $\angle PCM=\angle IDM$. Now we have $\frac{ZI}{IM}=\frac{PZ}{PI}=\frac{PZ}{PC}=\frac{PC}{PM}=\frac{r}{IM} \implies r=ZI.\square$ *I hope someone has a nicer proof for similarity.

An easy problem! First, note that wrt the circle $ (ABC)$, line $CH$ is the polar of $P$.So the intersection of $CH$ and $PM$ is the inverse of $P$,there is only a little caculations left.

SerdarBozdag wrote: $PB\cap (ABC)=X,PA\cap (ABC)=Y,PH\cap AB=Q,D=(I)\cap BC$. Let $Z$ be the foot of the altitude from $C$ to $PM$. We have $PQ\cdot PH=PC^2=PZ\cdot PM \implies \angle HZM=90 \implies H-Z-C \implies CH \cap PM=Z$. $\textbf{Claim:}$ $PCM \sim IDM$ $\textbf{Proof:}$ Wlog $a>b$. I will write briefly. $PC=abc/((a-b)(a+b+c)), DM=(a-b)/2$. $PC/CM=r/DM \leftrightarrow (a+b-c)/2=r=ab/(a+b+c)$ which is true. $\angle PCM=\angle IDM$. Now we have $\frac{ZI}{IM}=\frac{PZ}{PI}=\frac{PZ}{PC}=\frac{PC}{PM}=\frac{r}{IM} \implies r=ZI.\square$ *I hope someone has a nicer proof for similarity. Hmm, may I ask why does the length of $PC$ look like this? Is this a trivial fact (and that i’m being stupid...) or some well-known lemma?

It is not trivial but I did not wanted to write. If $T=PC\cap AB, U=CI\cap AB$, then $TA/TB=b^2/a^2$. You can find $TA$ and $TB$ then you can find $TC$ by PoP. $CP/CT=CI/IU+1=CA/AD+1=CB/BD+1=(CA+CB)/AB+1$. With the last you can find $CT$.

@SerdarBozdag That doesn’t seem to make sense to me. Isn’t $\frac{CT}{CP}=\frac{IU}{CI}+1$? And why would $\frac{CA}{AD}=\frac{CB}{BD}$?

$\frac{CA}{AD}=\frac{CB}{BD}$ must be $\frac{CA}{AU}=\frac{CB}{BU}$. Then $\frac{CT}{CP}=\frac{IU}{CI}+1=\frac{UA}{AC}+1=\frac{UB}{BC}+1=\frac{AB}{AC+BC}+1$.

Mr.Trick wrote: An easy problem! First, note that wrt the circle $ (ABC)$, line $CH$ is the polar of $P$.So the intersection of $CH$ and $PM$ is the inverse of $P$,there is only a little caculations left. why could CH be the polar of P?

Let $CI$ meet circumcircle again at $L,O=LM\cap PI, X=CH\cap PM.$ Denote by $\gamma$ circle centered at $P$ and passing through $C,I,$ by $R,r$ we denote circumradius and inradius of $ABC$ respectively. We will prove the desired assertion in two parts. Part 1. Redefine $X$ as inverse of $M$ wrt $\gamma.$ Clearly inversion wrt $\gamma$ preserves $\odot (ABC),$ but $AH\cap PB,BH\cap AP\in \odot (ABC).$ Hence inversion maps $H$ onto $PH\cap AB\in \odot (PCM),$ what finishes this part. Part 2. Prove $|XI|=r.$ We can easily see that $|CI|=\sqrt{2}r=\sqrt{2}|OM|, |LI|=\sqrt{2}R=\sqrt{2}|ML|,$ hence ($CPMO$ is cyclic) $\angle OMI=90^\circ - \angle COI=\angle CPM,$ so in fact $CPM\sim OMI.$ But reflection of $I$ over $P$ lies on $\odot (COL),$ therefore $$|PI|\cdot |OI|=\frac{1}{2}|CI|\cdot |IL|=Rr\implies |XI|=\frac{|PI|\cdot |MI|}{|MP|}=|OM|=r \text{ } \blacksquare$$ Remark. Note that inversion wrt $\gamma$ also preserves nine-point circle, so $X$ is the Feuerbach point of $ABC.$

Why we can deduce <IQJ = <IMC from $Q$ is the midpoint of $\overline{CT}$

Mr.janes wrote: why could CH be the polar of P? It is known (just a direct consequence of Brokard's) that $H$ lies on the polar of $C$ wrt $\odot(M,MA) \equiv \odot(ABC)$. Now $C$ lies on polar of $P$ wrt $\odot(ABC)$ just because $PC$ is tangent to $\odot(ABC)$ at $C$.

Let the incircle touch $BC$ at $D$, $Q = AB \cap PH$, ray $AP$ meet $(ABC)$ again at $Y$, the projection of $C$ onto $PM$ be $X$, ray $CX$ meet $(ABC)$ again at $T$, ray $CI$ meet $(ABC)$ again at $M_c$, ray $M_cM$ meet $(ABC)$ again at $N$, $E = PI \cap M_cN$, and $F = AB \cap CE$. It's clear that $M_cN$ is the perpendicular bisector of $AB$, so $PI \perp M_cN$. Now, since $X$ is the midpoint of chord $CT$ and $M$ is the circumcenter of $ABC$, we have $$\angle PTM = \angle PCM = 90^{\circ} = \angle PQM = \angle PEM$$which implies $CPQTME$ is a cyclic hexagon with diameter $PM$. Because $$\angle BQP = 90^{\circ} = \angle BYA = \angle BYP$$we know that $BPQY$ is cyclic with diameter $BP$ and $B, H, Y$ are collinear. Hence, the Radical Axis Theorem on $(CPQTME),$ $(BPQY),$ $(ABC)$ yields $H \in CT$. As a result, it suffices to show $X = CH \cap PM$ lies on the incircle. Observe that $$\angle ICN = \angle M_cCN = 90^{\circ} = \angle IEN$$which means $CIEN$ is cyclic with diameter $IN$. Thus, $$\angle PIC = 180^{\circ} - \angle EIC = \angle ENC = \angle CNM_c = \angle PCM_c = \angle PCI$$yielding $PC = PI$. Now, since equal tangents gives $PT = PC = PI,$ the Incenter-Excenter Lemma implies $I$ is the incenter of $CET$. Because $EI$ is the internal bisector of $\angle CET$ and $EI \perp EM$, we know $EM$ is the external bisector of $\angle CET$. This yields $\angle DEM = \angle FEM$, so $MD = MF$ holds. Now, the Diameter of the Incircle Lemma implies that $F$ belongs to the $C$-excircle. Since the incenter condition gives $\angle TCI = \angle ICF,$ a well-known lemma yields $T$ as the $C$-Mixtilinear Incircle tangency point. Now, another well-known lemma implies $I, N, T$ are collinear. Because $DIEM$ is a rectangle, we have $$\angle DIM = \angle DEM = \angle FEM = \angle CEN = \angle CIN$$$$= 90^{\circ} - \angle CNI = 90^{\circ} - \angle CNT = 90^{\circ} - \angle PCT = \angle XPC$$so $DIM \sim XPC$. In addition, we know $$PI^2 = PC^2 = PM \cdot PX$$so $PXI \sim PIM$. Combining these two similarities yields $$\frac{IX}{IM} = \frac{PX}{PI} = \frac{PX}{PC} = \frac{ID}{IM}$$which gives $IX = ID$, as required. $\blacksquare$ Remarks: My GGB can be found here! By POP at $M$, we know $X$ is the $P$-Humpty point of $ABP$, so $HX \perp PM$. Combining this result with $CX \perp PM$ yields $\overline{CXH}$. So, there is actually no need for Radical Axes! Another Way To Prove $\overline{CXH}$: Let $K = AB \cap CT$. Inverting about $(ABC)$ send $BXKY$ to cyclic $BPQY,$ so the first quadrilateral is also cyclic. Now, Radical Axes on $(BXKY), (BPQY), (PQKX)$ suffices. Also, $\overline{TDE}$ follows from a basic mixtilinear lemma: $TC$ and $TD$ are isogonal in $\angle ATB$.

Solved with CyclicISLscelesTrapezoid and v4913.

we claim the desired concurrency point is the Feuerbach tangency point, which we call $F$. Let $G$ be inverse of $F$ wrt circumcircle $\Omega$. Call incircle $\omega$. Claim 1: $P=G$.

Claim 2: $\overline{CH}$ is the polar of $P$ wrt $\Omega$.

Therefore $\overline{CH},\overline{PM}$ meet at the Feuerbach point $F$ as desired.