11. The distances from some point $M$ to the vertices $A$ and $B$ of an equilateral triangle $ABC$ are $MA=2$ and $MB=3$. What can the distance $MC$ from point $M$ to the third vertex of the triangle be?
11. If $MA$, $MB$, and $MC$ are the distances from an arbitrary point $M$ on the plane to the vertices of an equilateral triangle $ABC$, then the segment $CM$ is not greater than the sum of the distances $AM$ and $BM$ and not less than their difference (see, for example, problem 134b in the book by D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom [13]); thus, in our case,

$$
MB - MA = 1 \leqslant MC \leqslant MB + MA = 5
$$

We leave it to the reader to verify independently that the distance $CM$ can take any value in the interval $1 \leqslant x \leqslant 5$, and in particular, can equal 5 and can equal 1.