# The centroid of the Morley triangle of an isosceles triangle is collinear with the foci of the Steiner inellipse

I have a little question about geometry found by a friend:

Let $\triangle ABC$ be isosceles. Let $M$ be the (equilateral) Morley triangle determined by the trisectors of the angles of $\triangle ABC$, and let $E$ be the Steiner inellipse tangent to the edges of $\triangle ABC$ at their midpoints. Prove that centroid of $M$ is collinear with the foci of $E$.

I don't have any idea to prove this.

Thanks a lot.

For $\triangle ABC$ a "tall" isosceles, with vertex angle of measure less than $60^\circ$, the result is obvious by symmetry. (All points of interest lie on the appropriate altitude.) We can include the $60^\circ$ vertex angle ---that is, equilateral $\triangle ABC$--- in that case. (All points of interest coincide with the triangle's centroid.) For a "short" isosceles, the result simply isn't true, as shown.

To answer a question raised in a comment: In an isosceles triangle, the Morley center aligns with some axis of the Steiner inellipse. However, that isn't saying much, because everything's symmetric in an isosceles triangle.

In a scalene triangle, the Morley center need not align with either axis of the Steiner inellipse: • Thanks for your answer (+1) , I have a last and more general question : Is the centroid M collinear with the semi major or the semi minor axes ? Have a good day – user448747 Dec 26 '17 at 13:00
• @FatsWallers: I've edited my answer with a response to your broader question. – Blue Dec 26 '17 at 17:08

Hints for an algebraic approach.

First, Marden's theorem says that the foci of the Steiner inellipse are the roots of the derivative of the cubic polynomial, whose roots are the triangles vertices (Here, we are thinking of the coordinates of the triangles as complex numbers in the plane). So, if your triangle is isosceles, then you can assume your triangle has vertices $i, -a, a$ for some $a>0.$ Then, the foci points are easily found using the Marden's theorem.

Second, this Wikipedia entry says that the trilinear coordinates of Morley's triangle's centroid are: $$\cos\frac{A}{3}+2\cos\frac{B}{3}\cos\frac{C}{3}:\cos\frac{B}{3}+2\cos\frac{C}{3}\cos\frac{A}{3}:\cos\frac{C}{3}+2\cos\frac{B}{3}\cos\frac{A}{3}$$

This enables you to find the centroid's coordinates and then you can use your favorite coordinate/complex bashing method to establish the concurrence of the three points. Note that while the above looks daunting, you can easily calculate those cosines because your triangle now has sides: $$2a, \sqrt{a^2+1}, \sqrt{a^2+1} .$$

But from the looks of it, the calculations could become a little tedious.