Comparison of sum of the minimum distances from the vertices between two triangles

Question

1. Can I say that the sum of the minimum distances from the vertices in triangle ABC is less than the sum of the minimum distances from the vertices in triangle DEF, if given that the perimeter of triangle ABC is less than the perimeter of triangle DEF?

2. Can I say that the sum of the minimum distances from the vertices in triangle ABC is equal to the sum of the minimum distances from the vertices in triangle DEF, if given that the perimeter of triangle ABC is equal to the perimeter of triangle DEF?

Thanks ahead!

Answer 1

No and no.

In the depicted configuration, the triangles $F_1 F_2 A_1$ and $F_1 F_2 A_2$ have the same perimeter, but the lengths of their Steiner nets (i.e. the sum of distances from the vertices for the Fermat point of such triangles), given by $F_2 V_1$ and $F_1 V_2$, is not the same.$^{(*)}$

In general, if $ABC$ is a triangle with angles $\leq 120^\circ$ and $P$ is its Fermat point,

$$(PA+PB+PC)^2 = a^2+b^2-2ab\cos(C+60^\circ) = \color{red}{\frac{a^2+b^2+c^2}{2}+2\sqrt{3}\Delta}.$$

Given such identity, it is interesting to prove that in the vast majority of cases the length of the Steiner net is around the $55\%$ of the perimeter. However, we cannot state that if the perimeter of $T_1$ is greater than the perimeter of $T_2$, the same holds for the lengths of their Steiner nets.

$^{(*)}$ We are just stating that an arc of an ellipse is never an arc of circle, i.e. an ellipse is not a curve with constant curvature, kind of obvious.