2. Equilateral Triangles

and so BA1CT is cyclic, i.e. T ∈ C(A1BC)
(b) We claim that C1, T, C are collinear points.
A T C = 120 ◦ , A T C1 = A BC1 = 60 ◦
giving AT C + AT C1 = 180 ◦ , i.e. C, T and C1
are collinear. Similarly A, T, A1 and B, T, B1 are
collinear.
(c) We claim that |CC1| = |T A| + |T B| + |T C|.
Since T ∈ C(AC1B) and AC1B is equilateral, then by
van Schooten’s theorem
|T C1| = |T A| + |T B|.
Thus |CC1| = |CT | + |T C1| = |T C| + |T A| + |T B|, as
required. Similarly for |AA1| and |BB1|.
(d) Now let M be any point in the plane of ABC. Then,
since ABC1 is equilateral:
|MC1| ≤ |MA| + |MB|
Thus |MA| + |MB| + |MC| ≥ |MC| + |MC1|
≥ |CC1|
= |T A| + |T B| + |T C|
So the point of a triangle which minimises the sum of the distances to the
three vertices is the Toricelli-Fermat point. One could ask the question of
weighted distances to the vertices and ask which point(s) minimise weighted
sums. This is the question we now investigate.
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