Let $I$ be the incenter of triangle $ABC$, $O_1$ and $O_2$ the incenters of triangles $ABD$ and $ACD$ respectively. Clearly, $B -O_1-I$ and $C-O_2-I$. By the angle bisector theorem applied in triangle $ABI$ with bisector $AO_1$ we get \[AI = \frac{AB \cdot IO_1}{BO_1}\] Similarly, in triangle $ACI$ we get \[AI = \frac{AC \cdot IO_2}{CO_2}\] Hence \[ \frac{AB \cdot IO_1}{BO_1} = \frac{AC \cdot IO_2}{CO_2} => \frac{BO_1 \cdot IO_2}{IO_1 \cdot CO_2} = \frac{AB}{AC}= \frac{BD}{DC}\] By ceva's theorem, the problem is equivalent to showing \[\frac{AM\cdot BD \cdot CN}{BM\cdot CD \cdot AN} = 1 \leftrightarrow \frac{BD}{DC} = \frac{BM\cdot AN}{AM\cdot CN} \] Note that \[\frac{AN}{AM} = \frac{\sin \angle AMN}{\sin \angle ANM}\] Also, notice that $\angle BO_1M = \angle IO_1O_2 $ and $\angle NO_2C = \angle IO_2O_1$ and therefore by sine law on triangles $BMO_1$ and $CNO_2$ we get \[BM =\frac{\sin \angle IO_1O_2 \cdot BO_1}{\sin \angle AMN}\] and \[CN =\frac{\sin \angle IO_2O_1 \cdot CO_2}{\sin \angle ANM} \] and therefore we obtain \[\frac{BM}{CN} = \frac{\sin \angle IO_1O_2 \cdot BO_1 \cdot \sin \angle ANM}{\sin \angle IO_2O_1 \cdot CO_2 \cdot \sin \angle AMN}= \frac{IO_2 \cdot BO_1 \cdot \sin \angle ANM}{IO_1 \cdot CO_2 \cdot \sin \angle AMN} \] Hence \[\frac{BM\cdot AN}{AM\cdot CN} = \frac{IO_2 \cdot BO_1}{IO_1 \cdot CO_2 } = \frac{BD}{DC} \] and we are done.

Let $I,U,V$ be the incenters of $\triangle ABC,\triangle ABD,\triangle ACD.$ Internal bisectors of $\angle DAB,$ $\angle DAC$ cut $BC$ at $P,Q$ and external bisector of $\angle BAC$ cuts $BC$ at $E.$ Let $EU$ cut $AB,AC,AQ$ at $M',N',V'.$ Since $AI,AE$ also bisect $\angle UAV',$ it follows that $A(U,V',I,E)=-1$ $\Longrightarrow$ $I(U,V',D,E)=-1.$ But, $I(B,C,D,E)=-1,$ thus $IV' \equiv IC'$ $\Longrightarrow$ $V \equiv V',$ $M \equiv M',$ and $N \equiv N'.$ Therefore, if $K \equiv BN \cap CM,$ we have $A(B,C,K,E)=-1$ $\Longrightarrow$ $K \in AD.$

We denote as $I,\ U,\ V,$ the incenters of the triangles $\vartriangle ABC,\ \vartriangle ABD,\ \vartriangle ACD$ respectively and let be the point $T\equiv AD\cap MN.$ Because of $DU$ bisects the angle $\angle ADB$ and $DU\perp DV,$ we conclude that the points $S\equiv BC\cap MN,\ U,\ T,\ V,$ are in Harmonic Division. So, the pencil $I.SUTV$ is also Harmonic and then, we have that the points $S,\ B,\ D,\ C,$ are in Harmonic Division. Hence, from the complete quadrilateral $AMKNBC,$ where $K\equiv BN\cap CM,$ we conclude that $K$ lies on $AD$ and the proof is completed. Kostas Vittas.

this problem was shortlisted for the Iberoamerican olympiad 2004, held in Spain... i guess the spanish guys submitted it again for this olympiad

where i can find Mediterran Mathematical Olympiad 2013?

Dear Mathlinkers, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=515962 Sincerely Jean-Louis