"Include the triangle with the least sum of squares of the sides in the circle."
Use Lagrangian $$L = ((x_{1}^1 - x_{1}^2)^2 + (x_{2}^1 - x_{2}^2)^2) +
((x_{1}^1 - x_{1}^3)^2 + (x_{2}^1 - x_{2}^3)^2) + ((x_{1}^2 - x_{1}^3)^2 + (x_{2}^2 - x_{2}^3)^2) + \sum\limits_{i=1}^n \lambda_{i}( (x_{1}^i + x_{1}^i)^2 )$$
and then standard scheme to solve ...
However, in a book I find next answer: $$X = R^2, |x^1 - x^2|^2 + |x^2 - x^3|^2 + |x^1 - x^3|^2 \rightarrow inf; |x^i|^2 = 1, x^i = (x_{1}^i, x_{2}^i), i = 1,2,3$$
- What is useful of this form?
I suggest something. Am I right?
Seems $|x^1 - x^2|^2 = (x^1 -x^2)(x^1 -x^2)^T$.
Derivative of Lagrangian by $x^1$ will be: $2x^1|x^1-x^2| + 2x^1|x^1-x^3| + 2\lambda_{1} x^1 = 0 $
And derivatives of Lagrangian by any $\lambda_{i}$: $|x^i|^2 - 1 = 0$
Therefore $|x^1|^2=|x^2|^2=|x^3|^2$. It means that triangle is equilateral...means angle between sides is 60 degree. Knowing that, can find points of triangle.
