P and the Triangle

The First Fermat point $P$ is such that $h_{ABC} = XC$ , where $ABX$ is an equilateral triangle. Since the area of $ABC$ is $1$ , given the length $AB$ , the third vertex $C$ could lie on a line parallel to $AB$ a distance of $\tfrac{2}{AB}$ away from $AB$ . It is clear that $XC$ will be least if $XC$ is perpendicular to $AB$ , in which case $AC = BC$ and the triangle $ABC$ is isosceles.

The same argument could be applied, calculating $h_{ABC}$ by constructing an equilateral triangle on $AC$ , and this will tell us that $h_{ABC}$ will be least when $ABC$ is equilateral. Then the least value of $h_{ABC}$ is twice the altitude of an equilateral triangle of area $1$ , which is $2\times3^{\frac14}$ . This makes the answer $2^4 \times 3 = \boxed{48}$ .

Let $A, B, C$ be areas of triangles formed by sides $(PA, PB)$ , $(PB, PC)$ , and $(PC, PA)$ respectively

Then, because $P$ is a Fermat point where $PA, PB, PC$ are all $120$ degrees apart, we have

$\dfrac{\sqrt{3}}{4}PA \cdot PB=A$
$\dfrac{\sqrt{3}}{4}PB \cdot PC=B$
$\dfrac{\sqrt{3}}{4}PC \cdot PA=C$

from which we can find the sum $PA+PB+PC$ , given that $A+B+C=1$

$PA+PB+PC=\dfrac{2}{{3}^{\frac{1}{4}}}(AB+BC+CA)$

Hence, to find the minimum, we can do partial differentials of

$AB+B(1-A-B)+(1-A-B)A=A+B-{A}^{2}-AB-{B}^{2}$

with respect to $A$ and $B$ , and setting both to $0$ to find

$1-2A-B=0$
$1-A-2B=0$

which means $A=B=C$ , and we have an equilateral triangle of area $1$ , and the minimum is

$PA+PB+PC={48}^{\frac{1}{4}}$