题目143：考查三角形中的托里切利点
Investigating the Torricelli point of a triangleLet 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.
题目：
令 ABC 为所有内角均小于 120 度的三角形，X 为三角形内部一点，令 XA=p，XB=q，XC=r。
费马曾挑战托里切利找到使得 p + q + r 最小的 X 点。
托里切利证明了：如果在三角形 ABC 的三条边上分别构造等边三角形 AOB，BNC，AMC，那么 AOB，BNC 和 AMC 的外接圆会相交于一点 T。此外他还证明了：T 点能够使得 p + q + r 最小，T 也被乘坐托里切利/费马点。更神奇的是，当这个和达到最小时， AN = BM = CO = p + q + r，并且 AN，BM 和 CO 也相交于 T。
如果和取到最小，而且 a， b， c， p， q 和 r 都是正整数，那么称三角形 ABC 为一个托里切利三角形。例如 a = 399, b = 455, c = 511 就是一个托里切利三角形，其中 p + q + r = 784。
求所有满足 p + q + r ≤ 120000 的不同的 p + q + r 之和。
托里切利三角形内费马点到任意2个顶点之间的夹角都是120°，换言之，b*b=p*p+q*q+p*q,a*a=q*q+r*r+q*r,c*c=p*p+r*r+p*r；检验所有的p q r看a,b,c是否为整数即可。
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