$ I$ is incenter of the triangle $ \triangle MB_1C_1,$ isn't it ?

$ \mbox{If I is the incircle center of } \triangle{ABC}, \mbox{ then points: } A, B_1, C_1 \mbox{ and } I \mbox{ concyclic!}$ SOLUTION: $ \mbox{I is the circumcircle center of }\triangle{B_1MC_1}\Rightarrow$ ${ \Rightarrow I\in \mbox{midperpendicular line of the }[B_1C_1] \mbox{ and } (AI \mbox{ bisectrix of }\angle B_1AC_1} \mbox{ of } \triangle{AB_1C_1} \Rightarrow$ $ \Rightarrow I\in \mbox{ circumcircle of }\triangle{AB_1C_1}.$ REMARK: $ \mbox{The orthocenter H of }\triangle{ABC} \mbox{ not lie on the circumcirle of } \triangle{AB_1C_1}.$

Let $ \mathcal M_B \equiv (M, MB),$ $ \mathcal M_C \equiv (M, MC)$ be 2 concentric circles with common center $ M$ and radii $ MB, BC,$ cutting $ AB, AC$ again at $ B_1, C_1,$ respectively. Rays $ (MB_1, (MC_1$ cut $ \mathcal M_C, \mathcal M_B$ at $ C_2, B_2,$ respectively. $ B_1B_2C_1C_2$ is isosceles cyclic trapezoid with $ B_1B_2 \parallel C_1C_2,$ let $ (P)$ be its circumcircle, $ MP$ bisects the angle $ \angle B_1MC_1.$ Let $ K$ be the midpoint of $ B_1C_2,$ then $ MK = \frac{_1}{^2} (MB_1 + MC_2) = \frac{_1}{^2} (MB + MC) = \frac{_1}{^2} BC.$ Angles $ \angle BMC_1 = 2 \angle C$ and $ \angle CMB_1 = 2 \angle B$ are fixed, the right $ \triangle MKP$ just translates with the point $ M \in BC,$ $ P$ is on a line $ p$ parallel to $ BC.$ WLOG, assume $ MB \le MC.$ Since $ \angle B_1MC_1 = 2\angle B + 2\angle C - 180^\circ,$ then from the isosceles $ \triangle MC_1C_2,$ $ \angle MC_1C_2 = 90^\circ - \frac{_1}{^2} \angle B_1MC_1 = 180^\circ - (\angle B + \angle C) = \angle A$ $ \Longrightarrow$ $ A \in (P).$ For any $ M \in BC,$ the centers $ P \in p$ of the circles $ (P)$ are collinear and they all go through $ A,$ they all go through reflection of $ A$ in $ p \parallel BC,$ which lies on the A-altitude of the $ \triangle ABC.$ When $ M$ is the midpoint of $ BC,$ the circles $ \mathcal M_B, \mathcal M_C$ coincide and $ B_1, C_1$ become feet of the C-, B-altitudes of the $ \triangle ABC.$ The circumcircle $ (P)$ of the $ \triangle AB_1C_1$ then goes through the orthocenter $ H$ of the $ \triangle ABC$ on account of the right angles $ \angle AB_1H, \angle AC_1H,$ i.e., it cuts the A-altitude at $ H,$ so that the orthocenter $ H$ of the $ \triangle ABC$ is the other fixed point of the circles $ (P).$ As the incenter $ I$ of the $ \triangle ABC$ is different from $ A, H,$ exactly one circle from the pencil $ \left \{(P) | A, H \in (P)\right\}$ goes through $ I,$ for exactly one position of $ M \in BC.$ Therefore, $ I \not\in (P)$ in general. $ \angle B_1PC_1 = 2 \angle A$ and $ \angle B_1PC_1 + \angle B_1MC_1 = 2 \angle A + 2 \angle B + 2 \angle C - 180^\circ = 180^\circ$ $ \Longrightarrow$ $ MB_1PC_1$ is cyclic, $ P$ is on circumcircle of the $ \triangle MB_1C_1$ $ \Longrightarrow$ $ (P)$ goes through the incenter $ J$ of this triangle.

$ \mbox{Sorry! I have another problem solved... We design wrong to me: } \\ |B_1B| = |BM|(M\in[AB]); |C_1C| = |MC|(C_1\in[AC]).$

yetti wrote: $ I$ is incenter of the triangle $ \triangle MB_1C_1,$ isn't it ? Yes problem is now corrected by the moderators.

my solution is similar to yetti's Notice ∠B1MC1=180-2∠A C,H,C2 are collinear(∠HCC1=∠C2CC1=90-∠A) Then the problem is trivial

Attachments:

Omid Hatami wrote: Suppose that $ M$ is an arbitrary point on side $ BC$ of triangle $ ABC$. $ B_1,C_1$ are points on $ AB,AC$ such that $ MB = MB_1$ and $ MC = MC_1$. Suppose that $ H,I$ are orthocenter of triangle $ ABC$ and incenter of triangle $ MB_1C_1$. Prove that $ A,B_1,H,I,C_1$ lie on a circle. Let the altitude from $A$ to $BC$ intersect $BC$ at $Z$. Let the circumcircle of $\triangle AB_1C_1$ intersect $AZ$ at $A,H$. We will prove $H$ is the orthocentre of $\triangle ABC$. Let $AZ$ intersect the circumcircle of $\triangle ABC$ at $T$. Then it suffices to prove $HZ = ZT$. Let $X,Y$ be the midpoint of $BB_1, CC_1$ respectively, and let $P$ be the second intersection of the circumcircles of $\triangle ABC, \triangle AB_1C_1$. Then $P$ is the center of spiral similarity that takes $B \rightarrow C, B_1 \rightarrow C_1$. Hence it takes the midpoint of $BB_1$, that is, $X$, to the midpoint of $CC_1$, that is, $Y$. Now, the center of spiral similarity (which we already know is $P$) that takes $X \rightarrow Y$, $B \rightarrow C$ must lie on the circumcircles of $\triangle ABC, \triangle AXY$ (as $A = BX \cap CY$). Thus, $P,A,X,Y$ are concyclic. But $M,X,A,Y,Z$ all lie on the circle with diameter $AM$, hence $P,A,X,Z$ are concyclic. Now, $P$ is the center of spiral similarity that takes $B \rightarrow B_1$ and $H \rightarrow T$, hence it must take the midpoint of $BB_1$, that is $X$ to the midpoint of $HT$. But that would mean that the midpoint of $HT$ also lies on the circumcircle of $\triangle APX$, but the only intersections with the altitude of this circle are $A,Z$, hence $Z$ is the midpoint of $HT$.