Can you prove the claim given below?
Inspired by Conway circle theorem I have formulated the following claim:
Let $a,b,c$ be the side lengths and $\alpha,\beta,\gamma$ inside angles of the triangle $\Delta ABC$ . Let $P$ be a point on the extension of the segment $BA$ beyond $A$ such that $AP=\frac{a}{2}$ , let $Q$ be a point on the extension of the segment $CA$ beyond $A$ such that $AQ=\frac{a}{2}$ , let $R$ be a point on the extension of the segment $CB$ beyond $B$ such that $BR=\frac{b}{2}$ , let $S$ be a point on the extension of the segment $AB$ beyond $B$ such that $BS=\frac{b}{2}$ , let $T$ be a point on the extension of the segment $AC$ beyond $C$ such that $CT=\frac{c}{2}$ and let $U$ be a point on the extension of the segment $BC$ beyond $C$ such that $CU=\frac{c}{2}$ . Construct an ellipse $e$ through points $P,Q,R,S,T$ . If $\alpha \le \beta \le \gamma$ or $\beta \le \gamma \le \alpha$ or $ \gamma \le \alpha \le \beta$ then $U \in e$ .
GeoGebra applet that demonstrates this claim can be found here.
Best Answer
Given non-degenerate $\triangle ABC$, define points $A^+$, $B^+$, $C^+$, $A^-$, $B^-$, $C^-$ via $$\begin{align} A^+ = A + \alpha^+(B-A)\frac{a}{c} &\qquad A^-=A+\alpha^-(C-A)\frac{a}{b} \\[4pt] B^+ = B + \beta^+(C-B)\frac{b}{a} &\qquad B^-=B+ \beta^-(A-B)\frac{b}{c} \\[4pt] C^+ = C + \gamma^+(A-C)\frac{c}{b} &\qquad C^-=C+\gamma^-(B-C)\frac{c}{a} \\[4pt] \end{align} \tag{1}$$ for arbitrary values $\alpha^{\pm}$, $\beta^{\pm}$, $\gamma^{\pm}$. (That is, $A^+$ and $A^-$ are the translates of $A$ in directions $\overrightarrow{AB}$ and $\overrightarrow{AC}$ by signed distances $a\alpha^+$ and $a\alpha^-$, respectively.) Conway considers the case $\alpha^{\pm}=\beta^{\pm}=\gamma^{\pm}=-1$; OP considers $\alpha^{\pm}=\beta^{\pm}=\gamma^{\pm}=-1/2$. (In the cases where $\alpha^{\pm}=0$, $\beta^{\pm}=0$, or $\gamma^{\pm}=0$, some pair of the six points coincide with the corresponding vertex of the triangle.)
Via coordinates, it's not difficult (using, say, this determinant) to show that the points $A^\pm$, $B^\pm$, $C^\pm$ lie on a common conic (which may-or-may-not be an ellipse) if and only if $$\begin{align} 0 &= (a - (b^+ + c^-)) \; (b - (c^+ + a^-))\;(c - (a^+ + b^-)) \\[4pt] &\phantom{=}\cdot\left( \alpha^+ \beta^+ \gamma^+ (a - c^-) (b - a^-) (c - b^-) - \alpha^- \beta^- \gamma^- (a - b^+) (b - c^+) (c - a^+) \right) \end{align} \tag{2}$$ where $a^\pm := a\alpha^\pm$, $b^\pm := b \beta^\pm$, $c^\pm:= c \gamma^\pm$.
Each of the first three factors of $(2)$ corresponds to a trivial case where two of our six points coincide (as the translated distances of two vertices add to the length of the side between them). The interesting condition, therefore, is
$$\alpha^+ \beta^+ \gamma^+ (a - c^-) (b - a^-) (c - b^-) \;=\; \alpha^- \beta^- \gamma^- (a - b^+) (b - c^+) (c - a^+) \tag{$\star$}$$
For $\alpha^\pm=\beta^\pm=\gamma^\pm=:\lambda \neq 0$ (the zero case is trivial), this reduces to $$(1+\lambda)(a - b) (a - c) (b - c) = 0 \tag{$\star\star$}$$ Thus, for $\lambda=-1$, the six points lie on the ellipse, regardless of the shape of the original triangle; this is part of Conway's Theorem. (Showing that the conic is actually a circle in this case take a little more work.) For non-zero $\lambda\neq -1$ (in particular for OP's $\lambda=-1/2$), the six points lie on a common conic if and only if $\triangle ABC$ is isosceles. $\square$