In: Math
Give a method for solving Fermat's Problem when a triangle has an agle greater than 120°.
(This solution is taken from wikipedia)
Without loss of generality suppose that the angle at A is ≥
120°. Construct the equilateral triangle AFB and for any point P in
Δ (except A itself) construct Q so that the triangle AQP is
equilateral and has the orientation shown. Then the triangle ABP is
a 60° rotation of the triangle AFQ about A so these two triangles
are congruent and it follows that d(P) = CP+PQ+QF which is simply
the length of the path CPQF. As P is constrained to lie within ABC,
by the dogleg rule the length of this path exceeds AC+AF = d(A).
Therefore, d(A) < d(P) for all P є Δ, P ≠ A. Now allow P to
range outside Δ. From above a point P' є Ω exists such that d(P')
< d(P) and as d(A) ≤ d (P') it follows that d(A) < d(P) for
all P outside Δ. Thus d(A) < d(P) for all P ≠ A which means that
A is the Fermat point of Δ. In other words, the Fermat
point lies at the obtuse angled vertex.