Austria competition problems, 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
$$\begin{aligned} \phi &= \angle M_B M_C U = \angle UAB = \angle UBA = \angle M_A M_C U \\ \psi &= \angle M_C M_A U = \angle UBC = \angle UCB = \angle M_B M_A U \quad \text{and} \\ \chi &= \angle M_A M_B U = \angle UCA = \angle UAC = \angle M_C M_B U, \end{aligned}$$
and the angles in $ABC$ can then be written as
$$\angle CAB = \alpha = \phi + \chi, \quad \angle ABC = \beta = \psi + \phi \quad \text{and} \quad \angle BCA = \gamma = \chi + \psi$$
and the angles in $M_A M_B M_C$ as
$$\angle M_A M_B M_C = 2\psi, \quad \angle M_B M_C M_A = 2\chi \quad \text{and} \quad \angle M_C M_A M_B = 2\phi$$
We now have three cases to consider.

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

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

Case 3: $\alpha = \phi + \chi = 2\chi$. In this final case, we again have $\phi = \chi$, and as in case 1, an analogous argument again yields $2\psi = 2\chi (= 2\phi)$, 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. $\square$