I think this problem from All Russian MO 2006.

Dear, have a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=406369 Sincerely Jean-Louis

How to prove that $I_B,I_C\in (MIN)$ I can prove only $I_BI_CMN$ concyclic

ThE-dArK-lOrD wrote: How to prove that $I_B,I_C\in (MIN)$ I can prove only $I_BI_CMN$ concyclic $\angle I_{B}AC=\frac{1}{2}(\beta+\gamma)=90-\frac{1}{2}\alpha$ and $\angle I_{B}IC=90-\frac{1}{2}\alpha$ thus $ICI_{B}A$ is cyclic thus $B_1I \cdot BI_{B}=AB_1 \cdot B_1C$ but $ANCM$ is cyclic thus $AB_1 \cdot B_1C=MB_1 \cdot B_1N$ so $MB_1 \cdot B_1N=B_1I \cdot B_1I_{B}$ thus $MINI_B$ is concylic, similarly $NIMI_C$ is cyclic thus $I_B,I_C\in (MIN)$ Rest: Denote by $K$ and $L$ intersections of $BI_B$ and $\odot (ABC)$, and $CI_C$ and $\odot (ABC)$ respectively. Then from well known fact we have $KA=KI$ but $\angle IAI_C=90^{o}$ so $K$ is midpoint of $II_A$. Similarly $L$ is midpoint of $II_B$. Thus $\Delta KIL \sim \Delta II_BI_C$ with coefficient $\frac{1}{2}$. Thus $R(\Delta II_BI_C) : R(\Delta KIL)=1:2$. But we have $\Delta KIL = \Delta KAL$ thus $R(\Delta KIL) = R(\Delta KAL)= R(\Delta ABC)$ from here we have $R(\Delta II_BI_C) : R(\Delta ABC)=1:2$ thus $R(\Delta MNI) : R(\Delta ABC)=1:2$

ThE-dArK-lOrD wrote: How to prove that $I_B,I_C\in (MIN)$ I can prove only $I_BI_CMN$ concyclic $B_1I\cdot B_1I_B=B_1A\cdot B_1C=B_1M\cdot B_1N $thus $MINI_B $ are concyclic idem for $I_C$ further $ I$ is the orthocenter of $I_AI_BI_C$ so the circumcircle of $ I_BI_CI $has the same radius as the circumcircle of $I_AI_BI_C$ but the latter has radius equal to $ 2R$ where $ R$ is the circumradius of $ABC$