Problem - 3043
Let $P$ be a point inside a unit square $ABCD$. Find the minimal value of $AP+BP+CP$
P should be the Fermat point of $ABC$. Hence, by the Law of Sines, the desired result is
$$2\cdot\frac{\sin 45^\circ}{\sin 120^\circ}+\frac{\sin 15^\circ}{\sin 120^\circ}=\boxed{\frac{\sqrt{6}+\sqrt{2}}{2}}$$