Steiner's Inellipse in Barycentric Coordinates

Leo Giugiuc has communicated in a private mail a problem posed by Luis Gonzalez on AOPS.

Let $ABC$ be a triangle and $\mathscr{E}$ its Steiner inner ellipse. If $Q$ is any point on $\mathscr{E}$ and $x,$ $y,$ $z$ are the areas $[\Delta QBC],$ $[\Delta QCA],$ $[\Delta QAB],$ respectively, then

$x^2+y^2+z^2=2(xy+yz+zx).$

To remind, the Steiner inellipse is the only ellipse tangent to the sides of $\Delta ABC$ at the midpoints of its sides. The existence of Steiner's Inellipse has been established elsewhere. Note, that the parameters $x,y,z$ are popularly known as area or areal coordinates. These are in fact barycentric coordinates associated with $\Delta ABC.$ Thus, they may be considered as homogeneous coordinates, or normalized to have $x+y+z=1.$

There is an especially simple solution to the problem that exploits Joachimsthal's notations. Let $s$ be a homogeneous quadratic form

$s=Ax^2+By^2+Cz^2+2Dxy+2Eyz+2Fzx.$

A general equation of a conic in homogeneous coordinates is then $s=0.$ If $P=(x_1:y_1:z_1)$ is a point on a conic then the tangent to the conic at $P$ has the equation $s_1=0,$ where

$s_1=Ax_1x+By_1y+Cz_1z+D(x_1y+xy_1)+E(y_1z+yz_1)+F(z_1x+zx_1).$

Finally, the mid points of the sides of $\Delta ABC$ are, say, $A'=(0:1:1),$ $B'=(1:0:1),$ and $C'=(1:1:0).$

We are now in a position to establish the equation of Steiner's inellipse.

Proof

First of all we may set in turn $x_1=A',$ $x_1=B',$ $x_1=C'$ to obtain the following equations of the tangents to the ellipse through $A',$ $B',$ $C'$ at those points:

$By+Cz+Dx+Ez+Ey+Fx=0,\\ Ax+Cz+Dy+Ey+Fx+Fz=0,\\ Ax+By+Dy+Dx+Ez+Fz=0.$

The first of these equations is the equation of the side line $BC$ which is $x=0.$ Comparing the coefficients we get two equations: $B+E=0$ and $C+E=0.$ Thus, $B=C=-E.$ The second equation (which is $y=0)$ delivers $A+F=0$ and $C+F=0,$ i.e., $A=C=-F.$ Similarly, from the third equation we obtain $A=B=-D.$ Since $s$ is a homogeneous form we as well may take

$A=B=C=-D=-E=-F=1$

which gives the required equation.

Note that the fact that the ellipse passes through the midpoints $A',$ $B',$ $C'$ yields three equations:

$A+C+2F=0,\\ A+B+2D=0,\\ B+C+2E=0.$

It can be seen that the third tangent equation is redundant. It follows that an ellipse that passes through three of the midpoints and is tangent to two sides, it is also tangent to the third side.

Note: There is another ellipse associated with Steiner's name - Steiner's Circumellipse that is discussed on a separate page.


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