# Fermat's Point applies to isosceles triangles

Fermat's Point applies to equilateral triangles. Recently as I searched isosceles triangles on Wolfram Mathworld, I learnt that the same principle applies to similar isosceles triangles.

Besides the fact that the total distance from the three vertices of the triangle to Fermat's Point is the minimum possible, how is Fermat's Point different with the points formed by the isosceles triangles? Why is Fermat's point recorded in the Encyclopedia of Triangle Centers as X(13), while the other points formed by the isosceles triangles are omitted? In the construction OP mentions, the resulting point depends upon the exact shape of the isosceles triangles erected on the sides; that shape is determined by the vertex angle of the triangle in question. We can say, then that the resulting point is "parameterized" by that angle. The Fermat point is the point corresponding to the parameter $$\pi/3$$.

The Encyclopedia of Triangle Centers doesn't seem to have much capacity for parameterized centers. This is not unreasonable, as any parameterized family likely has uncountably-many members. However, the ETC does have a few entries related to the isosceles triangle construction.

Consider the following, which is straightforward (though a little tedious) to prove with coordinate methods ...

Given $$\triangle ABC$$, let $$D$$, $$E$$, $$F$$ be vertices of isosceles triangles erected upon respective sides $$\overline{BC}$$, $$\overline{CA}$$, $$\overline{AB}$$. Then $$\overleftrightarrow{AD}$$, $$\overleftrightarrow{BE}$$, $$\overleftrightarrow{CF}$$ concur at the point with trilinear coordinates $$\csc(A\pm\theta) : \csc(B\pm\theta):\csc(C\pm\theta) \tag{\star}$$ where "$$\pm$$" is "$$+$$" if the erected triangles are external to $$\triangle ABC$$, and "$$-$$" if they overlap.

As an immediate corollary:

If $$D$$, $$E$$, $$F$$ are centers of regular $$n$$-gons erected upon the sides of $$\triangle ABC$$ then $$\overleftrightarrow{AD}$$, $$\overleftrightarrow{BE}$$, $$\overleftrightarrow{CF}$$ concur at a point with trilinear coordinates as in $$(\star)$$, with $$\theta = 2\pi/n$$. (The "$$\pm$$" condition corresponds to whether the $$n$$-gons are external to $$\triangle ABC$$.)

The ETC identifies some centers corresponding to points constructed via the corollary.

$$\begin{array}{r|l|l|l} n & + & - & \text{ETC name} \\ \hline 6 & X(13) & & \text{1st Isogonic Center (sometimes Fermat Point)} \\ 6 & & X(14) & \text{2nd Isogonic Center} \\ 4 & X(485) & X(486) & \text{Outer/Inner Vecten Point}\\ 5 & X(5402) & X(3382) & \text{\csc(A\pm\pi/5) Point} \\ 8 & X(3373) & X(3388) & \text{\csc(A\pm\pi/8) Point} \\ 10 & X(3370) & X(3397) & \text{\sec(A\mp 2\pi/5) Point}\\ 12 & X(3366) & X(3392) & \text{\sec(A - 5\pi/12) Point; \csc(A-\pi/12) Point} \end{array}$$

As ETC notes, these points appear in Bernard Gilbert's "Table 38", which lists points with trilinear coordinates $$f(A+\theta):f(B+\theta):f(C+\theta)$$ for trig functions $$f$$.

All of these points lie on the Keipert Hyperbola. As noted in MathWorld's entry,

Kiepert (1869) showed that the lines connecting the vertices of the given triangle and the corresponding peaks of the isosceles triangles concur. The locus of this point as the base angle varies [...] is a rectangular hyperbola known as the Kiepert hyperbola.

So, the hyperbola is a geometric compendium of the uncountably-infinite family of such points.

Because you can vary the heights of the isosceles triangles and get different points, whereas triangle centres must depend solely on the base triangle itself. In the limit of isosceles triangle heights approaching zero, the centroid is obtained.