Note that $PM$ bisects $\angle BPC$ by the Incenter-Excenter Lemma. So, $$\frac{PB}{BF} = \frac{PB}{BD} = \frac{PC}{CD} = \frac{PC}{CE},$$and also, $\angle PBD = \angle PBA = \angle PCA = \angle PCE$ so that $\Delta PBF$ = $\Delta PCE$ or $\angle PFB = \angle PEC$. Taking supplementary angles implies $\angle PFA = \angle PEA$ or $P$ lies on $(AFIE)$. As $AI$ is its diameter, we have $PA \perp PI$. (Note: $\Delta DEF$ is the contact triangle of $\Delta ABC$.) Challenges: 1. Prove that $P$,$I$, and the $A$-antipode are collinear. (OR) Show that $PI$ meets $AO$ on $(ABC)$. 2. Define $X = PD \cap BE$. Show that $B$,$X$,$I$,$C$, and $I_{A}$ are concyclic.

This is really really well-known but it's interesting to see it in 2010. Assume $AB<AC$. Obviously we have $BM^2=BI^2=MD\cdot MP$. So by tangent-secant angles: $\angle DIM = \angle IPM$ and $\angle MBD=\angle MPB$ so we have $\angle BPI=\angle BPM +\angle MPI =180^{\circ}-\angle ACB-\angle IPA$. We can find the angles $BPM$ and $MPI$ so we're done.

Dear Mathlinkers, https://artofproblemsolving.com/community/c4t48f4h1684227_one_property_of_the_incircle Sincerely Jean-Louis

As $P$ is the Sharky-Devil Point of $ABC$, we are done. $\square$ [asy][asy] defaultpen(fontsize(10)); size(220); pair A=dir(70); pair B=dir(210); pair C=dir(-30); pair I=incenter(A,B,C); pair M=dir(-90); pair D= foot(I,B,C); pair K=0.5*A+0.5*I; pair P=intersectionpoints(CP(K,A), unitcircle)[0]; dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$I$", I, dir(I)); dot("$M$", M, dir(M)); dot("$P$", P, dir(P)); draw(A--B--C--cycle); draw(A--I--P--cycle); draw(incircle(A,B,C), pink); draw(circumcircle(A,B,C), yellow); draw(CP(K,A), blue+dashed); draw(M--P, red); [/asy][/asy]