Ukrainian Mathematical Competitions

Let's consider the smallest tangential square, that is circumscribed around the equilateral triangle $\triangle ABC$ (Fig. 46). And the side of the square contains the side of the triangle $AC$. It's clear, that it will be a square $ADEC$ and the edge $B$ is located inside the square. Then let's make a small turn of the square around the edge $A$. Let it be the square $ATUV$. So, the edge $C$ will be inside the new square. It's clear from the Fig. 46, if to continue the line $AC$ till the intersection with $UV$ at some point $C'$, then at $\triangle AVC'$ $AC'$ is a hypotenuse, so, $AC' > AV = AC$. That's why we can make a small compression of the $ATUV$ (homothety with the center at the point $A$ and coefficient $k < 1$, but close enough to $1$), after which the area of the square will become smaller, but it will still cover the triangle $\triangle ABC$.

Fig. 46