A formula for the Feuerbach point

Question

Did some calculations related to the Feuerbach point with a CAS and noticed the following fact:

Consider tangents to the unit circle in the complex plane, at points $z_k$, $k=1,2,3$, forming a triangle. The circle through the midpoints of this triangle ( the Euler circle) is tangent to the unit circle at the Feuerbach point $z$ given by the formula

$$z = \frac{z_1 z_2 + z_1 z_3 + z_2 z_3}{z_1 + z_2 + z_3}$$

Note also that if $z_k = \frac{1+ i t_k}{1-i t_k}$, then we have $z = \frac{1+ i t}{1-i t}$, where

$$t = \frac{t_1 + t_2 + t_3+ 3 t_1 t_2 t_3}{3 + t_1 t_2 + t_1 t_3 + t_2 t_3}$$

Since we have a fairly neat formula for the center of a circle passing through $3$ points, we can get from the above a formula for the Feuerbach point, given the contact points of the incircle, or the excircles.

This gotta be a known fact, but did not find it in other places, so any reference/source would be appreciated!

$\bf{Added:}$ This question has a formula similar to the second one.

Not an answer to your question but an interesting document here. — Jean Marie, Jan 12 '24 at 08:33
Probably a pure coincidence, I have found (almost) the same formula in this article for the slope $m_E$ of the Euler line... — Jean Marie, Jan 12 '24 at 09:16
It has taken me some time to realize that if $z_k=e^{i \alpha_k}=\frac{1+it}{1-it}$ then $t$ is nothing else than $\tan (\tfrac12 \alpha_k)$... — Jean Marie, Jan 12 '24 at 09:45
A connection : the theorem given here connecting Feuerbach point with the Euler line of a "twin triangle". — Jean Marie, Jan 12 '24 at 10:57
@JeanMarie: Thanks! I added another link in the question, check it out. I think you meant the link provided in the comments(?) — orangeskid, Jan 12 '24 at 14:57
FYI: Euclid @ groups.io is a geometry forum with many discussions about triangle centers. — Blue, Jan 12 '24 at 15:15
@Blue: Thank you! That looks like a pro forum indeed. I looked at some books that use complex numbers in geometry (Andrica & Andreescu) but the formulas are different. — orangeskid, Jan 12 '24 at 19:23

score 3 · Accepted Answer · answered Jan 17 '24 at 20:45

Found this neat question that helps solve the problem.

So let's start with $a$, $b$, $c$ complex numbers on the unit circle ($\bar a =1/a $, $\bar b =1/b $, $\bar c = 1/c$) The triangle formed by the tangents the the unit circle at these points has coordinates $\frac{2 b c}{b+c}$, $\frac{2 c a}{c+a}$, $\frac{2 a b}{a+b}$ and its midpoints are

$$ \frac{ c a}{c+a}+\frac{ a b}{a+b}\ (=\colon p)\\ \frac{ a b}{a+b}+\frac{ b c}{b+c}\ (=\colon p)\\ \frac{ b c}{b+c}+\frac{ c a}{c+a}\ (=\colon r)\\ $$

Now recall the equation of a circle through three points $p$, $q$, $r$:

$$\left| \begin{matrix} 1 & z& \bar z & |z|^2 \\ 1 & p& \bar p & |p|^2 \\ 1 & q& \bar q & |q|^2 \\ 1 & r& \bar r & |r|^2 \end{matrix} \right|=0$$

Such a circle has a unique point of intersection with the unit circle $\bar z = 1/z$ if and only if the equation

$$\left| \begin{matrix} 1 & z& \frac{1}{z} & 1 \\ 1 & p& \bar p & |p|^2 \\ 1 & q& \bar q & |q|^2 \\ 1 & r& \bar r & |r|^2 \end{matrix} \right|=0$$

Note that

$$ \bar p= \frac{1}{c+a}+ \frac{1}{a+b}\\ \bar q = \frac{1}{a+b}+\frac{1}{b+c}\\ \bar r = \frac{1}{b+c}+\frac{1}{c+a} $$

Now one calculates the second determinant and gets

$$\frac{(a-b)(b-c)(c-a)\cdot \left((a+b+c)z - (a b + b c + c a)\right)^2}{(a+b)^2(b+c)^2(c+a)^2 z}$$

Therefore, the unique solution of the equation is

$$z= \frac{a b + b c + c a}{a+b+c}$$

A formula for the Feuerbach point

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