Nima Ahmadi Pour wrote: Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$. Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,BC$ at $M,N,L$. Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$. I think you meant that it touches $AB$ at $L$... Anyway, let $BI_a$ meet the circumcircle of $\triangle{ABC}$ at $K$. Then $K$ is the midpoint of the arc $\widehat{ABC}$ , thus $OK\perp{AC}$. Since $I_aN\perp{AC}$ and $OK=I_aN$ we conclude that $KI_a||ON$, but $LM\perp{KI_a}$ $\Longleftrightarrow$ $LM\perp{ON}$. Using similar arguments we obtain the conclusion.

Sorry, I just made it correct. Anyway, thanks for the notification.

I thought of $R=r_a \rightarrow R=p tg \frac{A}{2} \rightarrow |OM|^2-|ON|^2=|LM|^2-|LN|^2$ ...

An interesting remark. Denote the projection $D$ of the vertex $A$ on the opposite side $[BC]$. Then $\boxed {R=r_a\Longleftrightarrow D\in OI}\ .$

this problem is way too easy for IRAN tst. a solution very similar to sailor's but even simpler: M on BC, N on AC, L on AB. Suffices to prove OL perpendicular to MN (because then the similar argument that OM perpendicular to LN finishes off the problem neatly). MN is perpendicular to IC with I = excentre of A-excircle. So it suffices to prove a stronger result, i.e. ICOL is an iscoceles trapezium. OC = IL (since R=r_a), so it suffices to prove <OCI=<LIC. Simple angle chasing yields if <A = 2a,etc; then <OCI=<OCB+<ICB=90-2a+90-c=2b+c; <LIC=<LIB+<BIC = b + (b+c) = 2b+c, so we are done.

hey virgil nicula, do you have a good proof of the 'interesting remark' about D,I,O collinear when R= r_a? i cant do it without side-bashing or trig-bashing.

I have a solution: Let X be the excentre opposite A and D be the point (#A) where AX intersect (ABC). Because OD // MX ( perpendicular BC), OD=MX so ODXM is a parallelogram,AX perpendicular LN so OM perpen dicular LN. Let the internal angle bisector of BAC intersect (ABC) at E, EBX = 90, EA = EC. Let T the point of intersection of BX and(ABC) so TA = TC and TE passes O.Let B' the foot of the perpendicular on AC from T. TB' passes O Because OT // NX and OT= NX so TONX is a parallelogram so TX // ON hence ON perpendicular NM Hence O is the ortheocentre of MNL

Let $I_A$ be the $A$-excenter ; lines $I_AA,I_AB,I_AC$ meet $\odot(ABC)$ again at points $A',B',C'$, respectively. It isn't hard to note that: $I_A$ is the orthocenter of $\triangle A'B'C'$. corresponding sides of $\triangle A'B'C'$ and $\triangle MNL$ are parallel to each other and the two triangles have opposite orientations. Using the fact that $\triangle A'B'C'$ and $\triangle MNL$ have equal circumradius, we obtain there is a point $S$ such that reflection in $S$ (which we denote by $\mathcal R$) swaps $\{A',M\},\{B',N\},\{C',L\}$. Then as $O,I_A$ are corresponding cirucmenters of $\triangle A'B'C,\triangle MNL$, so $\mathcal R$ swaps $\{O,I_A\}$. Since $I_A$ was the orthocenter of $\triangle A'B'C'$, hence $O$ is the orthocenter of $\triangle MNL$. $\blacksquare$