Project Euler Lab - Problem 143

#143 - Torricelli Triangles

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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.

This problem is taken from Project Euler, Problem 143.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=143. Published Friday, 2nd March 2007, 09:00 pm. Solved by 3,302 members at time of mirroring.

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