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Fermat−Torricelli theorem∗
pahio†
2013-03-22 4:04:51
Theorem (Fermat−Torricelli). Let all angles of a triangle ABC be
at most 120◦ . Then the inner point F of the triangle which makes the sum
AF +BF +CF as little as possible, is the point from which the angle of view of
every side is 120◦ .
Proof. Let’s perform the rotation of 60◦ about the point A. When P is the
image of the point C, the triangle ACP is equilateral and its angles are 60◦ .
Let F be any inner point of the triangle ABC and Q its image in the rotation.
We infer that if the sides of the triangle ABC are all seen from F in the angle
120◦ , then the points B, F , Q, P lie on the same line.
C
P
Q
F
B
A
Generally, the triangles AP Q and ACF are congruent, whence CF = QP .
From the equilateral triangles we obtain:
AF +BF +CF = F Q+BF +QP = BF QP
Here, the right hand side is minimal when the points B, F , Q, P are collinear,
in which case
∠CF A = ∠P QA = 180◦ −∠AQF = 120◦ ,
∠AF B = 180◦ −∠QF A = 120◦ ,
∠BF C = 360◦ −240◦ = 120◦ .
∗ hFermatTorricelliTheoremi
created: h2013-03-2i by: hpahioi version: h42604i Privacy
setting: h1i hTheoremi h51M04i h51F20i
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C
Q
F
P
B
A
Remark. The point F is called the Fermat point of the triangle ABC.
References
[1] Tero Harju: Geometria. Lyhyt kurssi. Matematiikan laitos. Turun
yliopisto, Turku (2007).
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