Is there a geometrical proof for this locus (exterior of inner Steiner ellipse)?

Question

Let $ABC$ be a triangle.

Consider a random line $L$ in the plane, non parallel to any of the sides of the triangle, intersecting line $BC$ in $P$, line $AC$ in $Q$ and line $AB$ in $R$.

Question : show that the locus of points

$$G=\frac13(P+Q+R) \ \tag{1}$$

(center of weights of the 3 points) when line $L$ varies is the exterior of "Steiner inellipse" $S_I$.

($S_I$ is the ellipse internally tangent to the sides of $ABC$ in their midpoints).

$$\textit{Fig. 1 : Many different points $G$ : none is inside Steiner inellipse of ABC.}$$

I have an analytical proof (see below).

My question : is it possible to have a geometrical proof ?

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My answer :

Let the barycentric coordinates of $P,Q,R$ wrt triangle $ABC$ be :

$$\begin{cases}P&(&0,&p,&(1-p)&),\\Q&(&(1-q),&0,&q&)\\R&(&r,&(1-r),&0&)\end{cases}\tag{2}$$

(where $p,q,r$ can be positive or negative).

As $P,Q,R$ are aligned, we have ;

$$\det\pmatrix{0&(1-q)&r\\p&0&(1-r)\\(1-p)&q&0}=0\tag{3}$$

which is equivalent to :

$$(1-p)(1-q)(1-r)=-pqr\tag{3'}$$

or to :

$$r=\frac{(1-p)(1-q)}{1-p-q}\tag{3''}$$

Remark : $(3),(3'),(3'')$ are different ways to express Menelaus theorem under its signed version.

Point $G$ defined by (1) has the following barycentric coordinates :

$$G \begin{cases}x&=&(1-q+r)/3,\\ y&=&(1-r+p)/3,\\z&=&(1-p+q)/3\end{cases}\tag{4}$$

(sanity check : the sum of these bar. coord. is $1$).

Let us consider the following barycentric coordinates expression (i.e., with $x+y+z=1$) :

$$f(x,y,z)=x^2+y^2+z^2-2(xy+yz+zx)\tag{5}$$

We have (see here) :

$$f(x,y,z) \begin{cases}&=&0&S_I \ \text{(Steiner inellipse)}&(a) .\\&\le&0&\text{interior of} \ S_I&(b).\\&\ge&0&\text{exterior of} \ S_I&(c). \end{cases}\tag{6}$$

If we "plug" the coordinates of $G$ given by (4) into the equation $S_I$ given in (6a), taking into account the expression (3'') of $r$, we get :

$$f(x,y,z)=\underbrace{\frac{(p^2+q^2+pq-2p-2q+1)^2}{(1-p-q)^2}}_{ \ge 0}\tag{7}$$

As a consequence of (6c) and (7), the generic point $G$ is always external to $S_I$.

We can conclude that the locus of points $G$ is exactly the exterior of $S_I$ because this exterior is the union of ellipses $E_k$ with equations :

$$f(x,y,z)=k \tag{8}$$

where $k$ is any positive constant ; these ellipses are homothetic images of ellipse $S_I$, the center of homothety being the centroid, taking into account the fact that the RHS of (7) can take arbitrarily large values.

Please note that (8), in the particular case $k=1$, is the barycentric equation of the Steiner outellipse, circumscribed to triangle $ABC$.

Answer 1

I have just come up with a geometric proof, but it is not strictly synthetic as I use Thales' theorem for ratio chasing.

The idea

Here's how I thought about the problem.

At first, I picked a random straight line $\ell$ intersecting the three sidelines of triangle $ABC$ at $P, Q, R$. Then I moved $\ell$ without rotating it. I observed that the centroid $\dfrac{1}{3}(P + Q + R)$ varied on a tangent line of the Steiner inellipse as $\ell$ slided. Any point on any tangent line of an ellipse is not an interior point of the ellipse.

A complete solution of a locus problem consists of two parts: (1) Any point that satisfies a property $\mathscr{P}$ must belong to a set (locus) $\mathscr{S}$; (2) Any point of $\mathscr{S}$ satisfies the property $\mathscr{P}$.

In this problem, $\mathscr{S}$ consists of every point not in the interior of the Steiner inellipse excluding 12 points through which any tangent line to the Steiner inellipse is parallel/identical to a sideline of triangle $ABC$.

For the 1st part, we will show that $\dfrac{1}{3}(P + Q + R)$ lies on a tangent line of the Steiner inellipse.

For the 2nd part, we will construct a straight line $\ell$ through a point in $\mathscr{S}$ such that $\dfrac{1}{3}(P + Q + R)$ is identical to the given point.

The proof

The proof here consists of three steps.

Step 1. Let $\ell_{A}, \ell_{B}, \ell_{C}$ be the straight lines passing through $A, B, C$ and parallel to $\ell$. These straight lines meet $BC, CA, AB$ at $A_{0}, B_{0}, C_{0}$ and $X = \dfrac{2A + A_{0}}{3}$, $Y = \dfrac{2B + B_{0}}{3}$, $Z = \dfrac{2C + C_{0}}{3}$. Prove that $\dfrac{1}{3}(P + Q + R), X, Y, Z$ are collinear.

This is done by using the midpoints of $QR, BB_{0}, CC_{0}$ and applying ERIQ theorem several times.

ERIQ theorem: Given 4 points $A, B, C, D$.

$E \in AD, F \in BC$ such that $\dfrac{\overline{AE}}{\overline{AD}} = \dfrac{\overline{BF}}{\overline{BC}}$.

$K \in AB, M \in CD, L \in EF$ such that $\dfrac{\overline{AK}}{\overline{AB}} = \dfrac{\overline{DL}}{\overline{DC}} = \dfrac{\overline{EL}}{\overline{EF}}$.

Then $K, L, M$ are collinear and $\dfrac{\overline{AE}}{\overline{AD}} = \dfrac{\overline{BF}}{\overline{BC}} = \dfrac{\overline{KL}}{\overline{KM}}$.

Step 2. Let $A^{\prime}B^{\prime}C^{\prime}$ be the image of $ABC$ under the point reflection centered at the centroid of triangle $ABC$. A straight line $\Delta$ intersects $B^{\prime}C^{\prime}, C^{\prime}A^{\prime}, A^{\prime}B^{\prime}$ at $X, Y, Z$. Prove that $\Delta$ is tangent to the Steiner inellipse of triangle $ABC$ (and of triangle $A^{\prime}B^{\prime}C^{\prime}$ as they share the same Steiner inellipse) if and only if $BY \parallel CZ$.

$B^{\prime}C^{\prime}$ intersects $AC, AB$ at $A_{b}, A_{c}$; $C^{\prime}A^{\prime}$ intersects $BA, BC$ at $B_{c}, B_{a}$; $A^{\prime}B^{\prime}$ intersects $CB, CA$ at $C_{a}, C_{b}$.

$M$ is the reflection of $Z$ across $C_{a}$; $B_{a}M$ intersects $A_{c}C_{a}$ at $N$.

According to the Brianchon's theorem, $\Delta$ is tangent to the Steiner inellipse iff $A_{c}Z, A_{b}Y, B_{c}C_{b}$ are concurrent or parallel.

According to Thales' theorem, $A_{c}Z, A_{b}Y, B_{c}C_{b}$ are concurrent or parallel iff $\dfrac{\overline{A_{c}B_{c}}}{\overline{ZC_{b}}} = \dfrac{\overline{YB_{c}}}{\overline{A_{b}C_{b}}}$.

Apply Thales' theorem successively $$ \begin{align*} \dfrac{\overline{A_{c}B_{c}}}{\overline{ZC_{b}}} & = \dfrac{\overline{C_{b}C_{a}}}{\overline{ZC_{b}}} = \dfrac{\overline{C_{a}A^{\prime}}}{\overline{A^{\prime}M}} = \dfrac{\overline{NB_{a}}}{\overline{B_{a}M}} = \dfrac{\overline{NG}}{\overline{GC_{a}}}, \\ \dfrac{\overline{YB_{c}}}{\overline{A_{b}C_{b}}} & = \dfrac{\overline{YB_{c}}}{\overline{B_{c}B_{a}}}. \end{align*} $$

Hence $\dfrac{\overline{A_{c}B_{c}}}{\overline{ZC_{b}}} = \dfrac{\overline{YB_{c}}}{\overline{A_{b}C_{b}}}$ iff $\dfrac{\overline{YB_{c}}}{\overline{B_{c}B_{a}}} = \dfrac{\overline{NG}}{\overline{GC_{a}}}$.

$BY \parallel CZ$ iff $BY \parallel B_{a}M$. Moreover, $BY \parallel B_{a}M$ iff $\dfrac{\overline{YB_{c}}}{\overline{B_{c}B_{a}}} = \dfrac{\overline{NG}}{\overline{GC_{a}}}$.

Hence $\Delta$ is tangent to the Steiner inellipse iff $BY \parallel CZ$.

Similarly, one can show that $AX \parallel BY \parallel CZ$ iff $\Delta$ is tangent to the Steiner inellipse.

Step 3.

1st part. According to Step 1 and Step 2, $\dfrac{1}{3}(P + Q + R)$ lies on some tangent line $\tau$ of the Steiner inellipse such that $\tau$ intersects $BC, CA, AB$. Hence $\dfrac{1}{3}(P + Q + R) \in \mathscr{S}$.

2nd part. Pick an arbitrary point in $\mathscr{S}$. According to the definition of $\mathscr{S}$, there is a tangent line $\tau$ through the point that intersects $BC, CA, AB$.

Let $X, Y, Z$ be the intersections of $\tau$ and $B^{\prime}C^{\prime}, C^{\prime}A^{\prime}, A^{\prime}B^{\prime}$. $A_{0}, B_{0}, C_{0}$ are the intersections of $BC$ and $AX$, $CA$ and $BY$, $AB$ and $CZ$. Let $\ell$ be the straight line through the given point and parallel to $AX$. $\ell$ intersects $BC, CA, AB$ at $P, Q, R$.

The result in Step 2 implies $AX \parallel BY \parallel CZ$. The result in Step 1 implies the given point is identical to $\dfrac{1}{3}(P + Q + R)$.

Answer 2

Converting and expanding a comment, by request ...

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Applying an affine transformation, we can consider the ellipse the incircle of a circle (of radius $r$) of the equilateral triangle with vertices at $2r\,(\cos k\pi_3, \sin k\pi_3)$, for $k=0,1,2$ and with $\pi_3=\pi/3$.

Now, given $P = p\;(\cos\theta,\sin\theta)$, the perpendicular to $OP$ meets the triangle's sidelines at the points

$$\csc(\theta - 2k\pi_3) \left(\;-p\sin2k\pi_3 - r \sin\theta, p\cos2k\pi_3 + r \cos\theta\;\right)\qquad k=0,1,2 \tag{1}$$

Writing $Q$ for the centroid of these points, we find $$Q = \frac{1}{4\sin\theta \sin(\theta+\pi_3)\sin(\theta-\pi_3)}\left(\;r(\sin\theta,-\cos\theta)-p(\sin2\theta,\cos2\theta)\;\right) \tag{2}$$ (which in itself is rather interesting) and then $$|OQ|^2-r^2 = \left(\frac{p + r \cos3\theta}{4\sin\theta \sin(\theta+\pi_3) \sin(\theta-\pi_3)}\right)^2 \geq 0 \tag{3}$$ Thus, $Q$ never crosses to the incircle's interior. For $P$ itself strictly outside the incircle $(p>r)$, $Q$ is also strictly outside.