a) For the given triangle $ABC$, all angles of which are less than $120^{\circ}$, find the point for which the sum of distances to the vertices is minimal.

b) Inside the triangle $ABC$, all angles of which are less than $120^{\circ}$, a point $O$ is taken from which its sides are seen at an angle of $120^{\circ}$. Prove that the sum of distances from point $O$ to the vertices is $\left(a^{2}+b^{2}+c^{2}\right) / 2+2 \sqrt{3} S$.
a) Let $O$ be an arbitrary point. When rotated by $60^{\circ}$ around point $A$, points $B$, $C$, and $O$ transition to some points $B^{\prime}$, $C^{\prime}$, and $O^{\prime}$ (see figure). Since $A O = O O^{\prime}$ and $O C = O^{\prime} C^{\prime}$, it follows that $B O + A O + C O = B O + O O^{\prime} + O^{\prime} C^{\prime}$. The length of the broken line $B O O^{\prime} C^{\prime}$ is minimal when this broken line is a segment, i.e., $\angle A O B = \angle A O^{\prime} C^{\prime} = \angle A O C = 120^{\circ}$. To construct the required point, one can use the result from problem $\underline{2.8}$. The required point inside triangle $A B C$ exists if and only if all its angles are less than $120^{\circ}$. If one of the angles is $120^{\circ}$, then the required point is the vertex of this angle.

Remark. For the case when the triangle has an angle greater than $120^{\circ}$, see problem $18.29 \mathrm{~B}$.

b) The sum of the distances from point $O$ to the vertices is equal to the length of the segment $B C^{\prime}$, obtained when solving part a). It is also clear that $\left(B C^{\prime}\right)^{2} = b^{2} + c^{2} - 2 b c \cos \left(\alpha + 60^{\circ}\right) = b^{2} + c^{2} - b c \cos \alpha + b c \sqrt{3} \sin \alpha = \left(a^{2} + b^{2} + c^{2}\right) / 2 + 2 \sqrt{3} S$.

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