[alogo] 1. Fermat's point of the triangle

On the sides of an a triangle ABC with all angles less that 2π/3 construct equilateral triangles BCK, CAL, ABM.
a) Lines AK, BL, CM are all equal.
b) All these lines pass through a point F.
c) All the angles formed at F by these lines are 60 degrees.
d) The centers of the equilateral triangles form again an equilateral triangle UVW (see Napoleon.html ).
e) The point F minimizes the sum of distances f(P) = PA+PB+PC.

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For (a):  compare triangles AKC, LBC and show them equal. The second results from the first by turningit by an
angle of π/3 about C. This implies also that the angle of AK to BL is π/3. From the equality ofangles AFL and
ACL follows that AFCL is cyclic, this implying that KFC is also π/3, thus F is distinguished on BL by its
property to view BC under the angle BFC which is 2π/3. This implies easily all propertiesup to (d).
For (e): Take a point P, join with A, B and C. Turn the triangle APC about A by an angle of 60 degrees (figure
below). You want to minimize BP+AP+CP = BP+PQ+QL. But this is always shorter than the segment BL.

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[alogo] 2. The case A > 2π/3

The results are true with some minor changes also in the case one angle of  the triangle is > 120 degrees. Lines
AK, BL, CM are again equal but this time intersect at a point F on the extension of AK, for A>2π/3.
Also point F in this case is viewing BC under an angle of measure 2π/3, whereas the other sides are seen
under an angle of &p;i/3.

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[alogo] 3. The other Fermat point

The construction of the equilaterals can be carried out inwards, creating a second configuration analogous to
the more usually drawn. Again segments {AK, BL, CM} have equal lengths and pass through the same point,
which is viewing the sides under π/3 or 2π/3 angles. The point F' thus defined is the second Fermat
point of the triangle.

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[alogo] 4. Trilinear coordinates

The trilinear coordinates (x:y:z) of the Fermat's point F are easily calculated using the properties of the figure
proven above (see Trilinears.html ).

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Last cancelation of factors due to the equilateral ACL. Hence the trilinears of points F, and, working analogously
for F', are given by the ratios:

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Thus, it is seen that {F,F'} are isogonal conjugate (see Isogonal_Conjugation.html ) to the isodynamic points {J, J'} of
the triangle (see Isodynamic.html and for the calculation of trilinears of {J, J'} see ApolloniusCircles.html ).

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