The incenter and Euler line.

Question

It seems well known that the incenter of a triangle lies on the the Euler line if and only if the triangle is isosceles (or equilateral, but that is trivial). Searching the internet, I could not find a simple geometric proof of this fact. Can anyone provide such a proof? Also, when the incenter lies on the Euler line, does it do so in a set location? (For example, we know the centroid is a third of the way from the circumcenter to the orthocenter on the Euler line, does the incenter satisfy any nice ratios like that?)

Answer 1

One approach could be to use trilinear coordinates and show that the incentre at $1:1:1$ is usually not collinear with for example the circumcentre at $\cos A :\cos B :\cos C$ and the orthocentre at $\sec A :\sec B :\sec C$ by looking at the determinate

$$\begin{vmatrix}1&1&1\\ \cos A &\cos B &\cos C\\ \sec A &\sec B &\sec C\end{vmatrix}$$

which is non-zero unless at least two of $A$, $B$ and $C$ are equal.

Answer 2

I would like to present the result of W. Massaro in a slightly different manner.

It is known (see for example p. 82 of his document](http://www.journal-1.eu/2017/Grozdev-Okumura-Dekov-Euler-Line-pp.81-85.pdf)) that the Euler line of a triangle with sidelengths $a,b,c$ has barycentric equation:

$$(b^2-c^2)(b^2+c^2-a^2)x+(c^2-a^2)(c^2+a^2-b^2)y+(a^2-b^2)(a^2+b^2-c^2)z=0\tag{1}$$

As the barycentric coordinates of the incenter are $(x,y,z)=(a,b,c)$, plugging them into (1) results in an expression which, factorized (using a Computer Algebra System !), becomes:

$$(a-b)(b-c)(c-a)(a+b+c)^2$$

Nothing astonishing that we find back the expression in the numerator of the result of W. Massaro.