# Problem 143: Investigating the Torricelli point of a triangle
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 ≤ 120000 for Torricelli
triangles.