Posted many times before. It's also P2 of Turkey National Olympiad 2002. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14655 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52563 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=30539 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=52596 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=182301 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=332588 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=395162

teps wrote: Given is a triangle $\triangle ABC$ and points $M$ and $K$ on lines $AB$ and $CB$ such that $AM=AC=CK$. Prove that the length of the radius of the circumcircle of triangle $\triangle BKM$ is equal to the lenght $OI$, where $O$ and $I$ are centers of the circumcircle and the incircle of $\triangle ABC$, respectively. Also prove that $OI\perp MK$. Let $M'$ be intersection of the perpendicular bisector of $AM$ with the line passes through $M$ and parallel to $BI$, $K'$ is defined similarly; $D,E$ be orthogonal projections of $I$ on $BA,BC$ respectively. Let $O^*$ be the circumcenter of $\triangle M'IK'$. ============================== Firstly, we will prove that $\triangle IM'D = \triangle IK'E$. Since $\triangle AMM' = \triangle KCK'$, $AD=KE, MD=CE$ and $ID \bot AB, IE \bot BC$, it's easy to verify that $DM'=EK', \widehat{IDM'} = \widehat{IEK'}$. Thus $\triangle IM'D = \triangle IK'E$. ============================== Secondly, we will prove that $O^* \equiv O$. Let $Z$ be the midpoint of $BI$. Since $O^*$ is the center of $(IM'K')$, we obtain that $\triangle O^*M'K' \sim \triangle ZDE$. Note that $AM' \parallel ZD$ and $CK' \parallel ZE$, we have \[ \begin{aligned} (M'O^*,M'A) &\equiv (M'O^*,K'O^*) + (K'O^*,K'C) + (K'C,M'A) \\ &\equiv (O^*M',O^*K') + (K'O^*,K'C) + (ZE,ZD) \\ &\equiv (K'O^*,K'C) \pmod{\pi}. \end{aligned} \] Hence $\triangle O^*M'A = \triangle O^*K'C$ (because $M'A=K'C$). Then $\triangle O^*AC \sim \triangle O^*M'K' \sim \triangle ZDE \sim \triangle OAC$. Thus $O^* \equiv O$. ============================== Now, let $B'$ be intersection of $(O^*)$ with $BI$ ($B'$ differs from $I$). Hence $B'M' \parallel BM, B'K' \parallel BK$. Moreover, $M'K' \parallel MK$ and $M'K' = MK$ (since $MKM'K'$ is a parallelogram). Therefore, $R_{(BMK)} = R_{(B'M'K')} = OI$. From part 1 and 2, we deduce that $OI$ is the perpendicular bisector of $M'K'$. Thus $OI \bot M'K'$, equivalently, $OI \bot MK$. (Q.E.D.)

Attachments:

For the first part we've to show $MK^2=\frac {(R-2r)b^2}{4R}$ It's easy because using cosine rule we'll get $MK^2$ easily. For second part, We've to show $OKTD$ cyclic. Let $\angle {IOD}=x$. So $\frac {b-c}{2OI}=Sin x$ Also we've $2OI=\frac {b-c}{\angle {BKM}}$.... from $BKM$ So we get $\angle {BKM}=x$ hence done.

We will assume that $AC$ is the shortest side of the triangle, while other cases can be done similarly. The first part is easy by length chasing, and my solution is similar to the above one(s). A short, synthetic solution for Part 2: From the power of points $M,K$ wrt circumcircle of triangle $ABC$, we see that $OM^2 - OK^2 =MA\cdot MB - KC \cdot KB = (-b(c-b)) - (-b(a-b)) = b(a-c)$. Also, it is easy to see that as $ACK,ACM$ are isosceles triangles, $IA=IK , IC=IM$. So, $IM^2 - IK^2 = IC^2 - IA^2 = (s-c)^2 - (s- a)^2 = b(a-c)$, where the last but one equality follows from projecting $I$ onto $AC$. So, $IM^2 - IK^2 = OM^2 - OK^2$. So, we conclude that $OI\perp KM$. $\blacksquare$. EDIT: After opening the links on Luis Gonzales' post (#2), I realized that this is essentially Virgil Nicula's Solution.