363. Given a triangle with sides $a, b$, and $c$. Determine the area of the largest regular triangle circumscribed around the given triangle, and the area of the smallest regular triangle inscribed in it.
363. There exist two families of equilateral triangles circumscribed around a given triangle (see problem II.305). Construct equilateral triangles $A B C_{1}, B C A_{1}, C A B_{1}$ on the sides of triangle $A B C$ outwardly and describe circles around them. The vertices of the triangles of the first family are located one each on these circles. Let $O_{1} O_{2} O_{3}$ be the centers of these circles ($\triangle O_{1} O_{2} O_{3}$ is equilateral, see problem II.304). The triangle with the largest area will have sides parallel to the sides of triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ (the secant passing through the intersection point of two circles has the greatest length when it is parallel to the line of centers, in which case its length is twice the distance between the centers). The area of the largest triangle will be $S_{0}=4 S_{O_{1} O_{2} O_{3}}=\frac{\sqrt{3}}{3}\left(\frac{a^{2}+b^{2}+c^{2}}{2}+2 S \sqrt{3}\right)$, where $S$ is the area of the given triangle (see the solution to problem II.305). For the triangles of the second family, the area of the largest will be smaller. Among the equilateral triangles inscribed in the given one, the one with the smallest area will have sides parallel to the sides of the largest circumscribed. This follows from the result of problem I.241. Its area is $S_{1}=S^{2} / S_{0}$. Thus, the area of the largest circumscribed equilateral triangle is $S_{0}=\frac{\sqrt{3}\left(a^{2}+b^{2}+c^{2}\right)}{6}+2 S$, and the smallest inscribed one is $-S_{1}=\frac{S^{2}}{S_{0}}$, where $S$ is the area of the given triangle.