Austria 2014

Since $ABC$ is acute-angled, we first note that $U$ must lie in the interior of $ABC$. Since $AU\perp M_B{M}_C$ and $M_{C}U\perp AB$, we have $\angle UAB=\angle M_B{M}_{C}U$, and since analogous results hold all around the perimeter of the figure, we can write $$\begin{array}{rl}\varphi & =\angle M_B{M}_{C}U=\angle UAB=\angle UBA=\angle M_A{M}_{C}U\\ \psi & =\angle M_C{M}_{A}U=\angle UBC=\angle UCB=\angle M_B{M}_{A}U\text{and}\\ \chi & =\angle M_A{M}_{B}U=\angle UCA=\angle UAC=\angle M_C{M}_{B}U,\end{array}$$ and the angles in $ABC$ can then be written as $$\angle CAB=\alpha =\varphi +\chi ,\angle ABC=\beta =\psi +\varphi \text{and}\angle BCA=\gamma =\chi +\psi$$ and the angles in $M_A{M}_B{M}_C$ as $$\angle M_A{M}_B{M}_C=2\psi ,\angle M_B{M}_C{M}_A=2\chi \text{and}\angle M_C{M}_A{M}_B=2\varphi$$ We now have three cases to consider.

Case 1: $\alpha =\varphi +\chi =2\varphi$. In this case, we have $\varphi =\chi$ and therefore $\beta =\varphi +\psi =\chi +\psi =\gamma$, and since the triangles are similar also $2\psi =2\chi (=2\varphi )$. This implies that the triangles are equilateral.

Case 2: $\alpha =\varphi +\chi =2\psi$. In this case, we have $90^{\circ }=\varphi +\psi +\chi =3\psi$ and therefore $\psi =30^{\circ }$. We then either have $\varphi +\psi =2\varphi$ and $\chi +\psi =2\chi$, which implies $\varphi =\psi =\chi$ or $\varphi +\psi =2\chi$ and $\chi +\psi =2\varphi$, which implies $\varphi +30^{\circ }=2\chi =4\varphi -60^{\circ }$ and therefore $\varphi =30^{\circ }=\chi =\psi$, and in either case the triangles are again equilateral.

Case 3: $\alpha =\varphi +\chi =2\chi$. In this final case, we again have $\varphi =\chi$, and as in case 1, an analogous argument again yields $2\psi =2\chi (=2\varphi )$, which again implies that the triangles are equilateral.

In all possible cases, we see that $ABC$ and $M_A{M}_B{M}_C$ can only be similar if $ABC$ is equilateral. $\square$