Let $ABC$ be a triangle with all interior angles being less than $120$ degrees. Let $X$ be any point inside the triangle and let $XA=p$, $XC=q$, and $XB=r$.

Fermat challenged Torricelli to find the position of $X$ such that $p+q+r$ was minimised.

Torricelli was able to prove that if equilateral triangles $AOB$, $BNC$ and $AMC$ are constructed on each side of triangle $ABC$, the circumscribed circles of $AOB$, $BNC$, and $AMC$ will intersect at a single point, $T$, inside the triangle. Moreover he proved that $T$, called the Torricelli/Fermat point, minimises $p+q+r$. Even more remarkable, it can be shown that when the sum is minimised, $AN=BM=CO=p+q+r$ and that $AN$, $BM$ and $CO$ also intersect at $T$.

If the sum is minimised and $a,b,c,p,q$ and $r$ are all positive integers we shall call triangle $ABC$ a Torricelli triangle. For example, $a=399$, $b=455$, $c=511$ is an example of a Torricelli triangle, with $p+q+r=784$.

Find the sum of all distinct values of $p+q+r\le 120000$ for Torricelli triangles.