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Show that the map $\Phi$ on triplets

$$(a,b,c) \overset{\Phi}{\mapsto} \left(\frac{ (b+c) a - 2 b c}{2 a -(b+c)}, \frac{(a+c)b-2 a c}{2 b-(a+c)}, \frac{(a+b)c- 2 a b}{2 c-(a+b)}\right)$$ is an involution, that is $\Phi \circ \Phi = \operatorname{Id}$.

Notes:

  1. The map $t \mapsto \frac{ (b+c) t - 2 b c}{2 t -(b+c)}$ is an involution with fixed points $b$, $c$.

  2. The map $\Phi$ is invariant under Mobius transformations of the components.

  3. If $a$, $b$, $c$ are complex numbers, we can get some geometric interpretations of the map $\Phi$.

  4. We need some obvious conditions on $a$, $b$, $c$ : distinct, and not forming an arithmetic progression.

  5. Should be known(?). Posted as reference.

Any feedback would be appreciated!

orangeskid
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1 Answers1

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Define the midpoint $m(u,v)$ and swivel $\phi_{u,v}(w)$ by

$$ m(u,v)=\frac{u+v}{2}, \quad \phi_{u,v}(w)=\frac{m(u,v)w-uv}{w-m(u,v)}. $$

It is easy to verify $\phi_{u,v}$ sends $u,v,m(u,v)$ to $u,v,\infty$ respectively. Mobius transformations act sharply $3$-transitively (the orientation-preserving ones, to be precise), so this uniquely characterizes $\phi_{u,v}$.

Indeed, from the classification of Mobius transformations, we can interpret $\phi_{u,v}$ as a $180^\circ$ "rotation" around poles $u$ and $v$ in bipolar coordinates. This doesn't require $u,v$ to be finite: in the case of $\phi_{\infty,v}(w)=2v-w$, "bipolar" coordinates are just polar coordinates, and this is a typical $180^\circ$ rotation of $w$ around $v$. Since the Apollonian circles of bipolar coordinates are determined by the two poles, and Mobius transformations stabilize the set of (generalized) circles, we can say that $g\circ\phi_{u,v}\circ g^{-1}=\phi_{gu,gv}$ for any other Mobius transformation $g$, by symmetry.

Then we can construct the "duality" map $\Phi(a,b,c)=\bigl(\phi_{b,c}(a),\phi_{c,a}(b),\phi_{a,b}(c)\bigr)$. It is defined for any $a,b,c$ distinct (note collinearity is allowed). If we extend the action of Mobius transformations to triples, we then have $g\circ\Phi\circ g^{-1}=\Phi$ for any transformation $g$. Since transformations act triply transitively, to verify $\Phi^2=I$ it suffices to show $\Phi^2$ fixes a particular triple, which we can choose to be the equilateral triangle $(a,b,c)=(1,\omega,\omega^2)$ (where $\omega$ is a primitive cube root of unity); it is easy to verify $\Phi(1,\omega,\omega^2)=(-1,-\omega,-\omega^2)$, and the result follows.

While I've noted a geometric interpretation of $\phi_{u,v}(w)$, I fail to see one for $\Phi(a,b,c)$ at the moment. It seems to be some kind of "dual" triangle,

coiso
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    Neat answer! So since $\phi_{u,v}$ is "invariantly" defined, so is $\Phi$. Now we check for the standard equilateral... So basically it does for $a$, $b$, $c$ in an invariant way what it does for the equilateral. Basically completes to a hexagram. Great! Hm..., I would characterize $\phi_{u,v}$ just the involution with fixed set point ${u,v}$, since we "don't know" about $\infty$. – orangeskid Apr 25 '23 at 16:59
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    So $(a, \phi_{b,c}(a), b,c)$ are harmonic conjugate. Getting clearer. That definitely happens for $(a,a',b,c)$ with $a'$ the opposite if $(a,b,c)$ equilateral. – orangeskid Apr 25 '23 at 17:14