TITLE: Five Porismatic Equations.
QUESTION [8 upvotes]: Here is a really tough problem.
If
$$\boldsymbol{a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=0}$$
$$\boldsymbol{a\cos\gamma\cos\delta+b\sin\gamma\sin\delta+c=0}$$
$$\boldsymbol{a\cos\beta\cos\gamma+b\sin\beta\sin\gamma+c=0}$$
$$\boldsymbol{a\cos\delta\cos\epsilon+b\sin\delta\sin\epsilon+c=0}$$
$$\boldsymbol{a\cos\epsilon \cos\alpha+b\sin\epsilon\sin\alpha+c=0}$$
prove that
$$\boldsymbol{\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=
\left(\frac{1}{b}+\frac{1}{c}\right)
\left(\frac{1}{c}+\frac{1}{a}\right)
\left(\frac{1}{a}+\frac{1}{b}\right)}$$
where all angles are unequal and between $0$ and $2\pi$.
I cannot work out the algebra on this problem.
This is a system of porsimatic equations. Meaning that it only has distinct solutions if some condition on the variables holds.
The method is the following, in the case of a chain of three equations,
$$a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=0$$
$$a\cos\beta\cos\gamma+b\sin\beta\sin\gamma+c=0$$
$$a\cos\gamma\cos\alpha+b\sin\gamma\sin\alpha+c=0$$
We can show
$$\tan\frac{1}{2}(\alpha+\beta)=\frac{b}{a}\tan \gamma$$
either by setting an equation in variable $t$ with solutions, $\tan\frac{1}{2}\alpha$ and $\tan\frac{1}{2}\beta$ and using Vietas formulas or more straighforwardly solving for $\sin\gamma$ and $\cos\gamma$ to get $\tan\gamma$.
Similarly
$$\tan\frac{1}{2}(\alpha+\gamma)=\frac{b}{a}\tan \beta$$
$$\tan\frac{1}{2}(\gamma+\beta)=\frac{b}{a}\tan \alpha$$
and using these we get
$$\tan\frac{1}{2}(\alpha-\beta)=\frac{ab\sin(\alpha-\beta)}{a^2\cos\alpha\cos\beta+b^2\sin\alpha\sin\beta}$$ then using the definition of $\tan=\frac{\sin}{\cos}$ and dividing by $\sin\frac{1}{2}(\alpha-\beta)$ we see that
$$a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=c-\frac{ab}{a+b}$$
In the case of a chain of five equations we get the fomulas,
$$\tan\frac{1}{2}(\alpha+\gamma)=\frac{b}{a}\tan\beta$$
$$\tan\frac{1}{2}(\beta+\delta)=\frac{b}{a}\tan\gamma$$
$$\tan\frac{1}{2}(\gamma+\epsilon)=\frac{b}{a}\tan\delta$$
$$\tan\frac{1}{2}(\alpha+\delta)=\frac{b}{a}\tan\epsilon$$
$$\tan\frac{1}{2}(\epsilon+\beta)=\frac{b}{a}\tan\alpha$$
But how to proceed from there ?
I guess you want  $\tan\frac{1}{2}(\alpha-\epsilon)
$ as a function of $\alpha$ and $\epsilon$.

REPLY [3 votes]: As Somos has thought, the problem is related to Poncelet's Porism.
It is equivalent that there's a pentagon inscribe in $\Gamma_1:x^2+y^2=1$ and circumscribe on $\Gamma_2: \frac{x^2}{c-b}+\frac{y^2}{c-a}+\frac{1}{a+b}=ux^2+vy^2+w=0$
Generally equation $ax_1x_2+by_1y_2+c=0$ means pointer $(x_1,y_1)$ is at polar of conic section $ax^2+by^2+c=0$ for point $(x_2,y_2)$. When point $(x_1,y_1)$ is moving in one conic section, the polar encloses another conic section (here the $ux^2+vy^2+w=0$.
For example, assume a line L is circumscribe on $\Gamma_2$ at $P_0(x_0,y_0)$ and intersection $\Gamma_1$ at $Q_1(x_1,y_1),Q_2(x_2,y_2)$, we have
$$\begin{cases}ux_0^2+vy_0^2+w=0\\ux_0x_1+vy_0y_1+w=0\\ux_0x_2+vy_0y_2+w=0\\x_1^2+y_1^2=1\\x_2^2+y_2^2=1\end{cases}$$
So we have $y_1=\frac{-w-ux_0x_1}{vy_0}$ and $x_1^2+\left(\frac{-w-ux_0x_1}{vy_0}\right)^2=1$, which means
$$(v^2y_0^2+u^2x_0^2)x_1^2+2wux_0x_1+w^2-v^2y_0^2=0$$
Similarly we have $(v^2y_0^2+u^2x_0^2)x_2^2+2wux_0x_2+w^2-v^2y_0^2=0$ so that $x_1,x_2$ are two roots of equation $(v^2y_0^2+u^2x_0^2)X^2+2wux_0X+w^2-v^2y_0^2=0$
By Vieta's formula we have $x_1x_2=\frac{w^2-v^2y_0^2}{v^2y_0^2+u^2x_0^2}$
With similar approach we have $y_1y_2=\frac{w^2-u^2x_0^2}{v^2y_0^2+u^2x_0^2}$.
So $ax_1x_2+by_1y_2+c=\frac{(a+b)w^2+(c-a)v^2y_0^2+(c-b)u^2x_0^2}{v^2y_0^2+u^2x_0^2}=\frac{w+vy_0^2+ux_0^2}{v^2y_0^2+u^2x_0^2}=0$.
According to Poncelet's Porism, we could start from any points in $\Gamma_1$ to form the pentagon ABCDE so that we could choose $\alpha=0$ which is equivalent that A(1,0). so $\cos(\beta)=\cos(\epsilon)=-\frac ca$ which shows that $B(-\frac ca, \frac{\sqrt{a^2-c^2}}a),E(-\frac ca,-\frac{\sqrt{a^2-c^2}}a)$.
By symmetric, we have CD circumscribes on $\Gamma_2$ at $(h,0)=(\pm\sqrt{-\frac{w}{u}},0)$ so $h^2=-\frac wu=\frac{b-c}{a+b}$ so $C(h,\sqrt{1-h^2}), D(h,-\sqrt{1-h^2})$ or $C(h,-\sqrt{1-h^2}), D(h,\sqrt{1-h^2})$ or $\cos(\gamma)=\cos(\delta)=h$
Since $a\cos(\beta)\cos(\gamma)+b\sin(\beta)\sin(\gamma)+c=0$
we have $b^2\sin^2(\beta)\sin^2(\gamma)=(-c-a\cos(\beta)\cos(\gamma))^2$ that's
$$b^2(1-\frac{c^2}{a^2})(1-h^2)=c^2(1-h)^2$$
or
$$\frac{1-h}{1+h}=\frac{b^2}{c^2}(1-\frac{c^2}{a^2})$$
or
$$\frac{b-c}{a+b}=h^2=\left(\frac{b^2a^2-b^2c^2-a^2c^2}{b^2a^2-b^2c^2+a^2c^2}\right)^2$$
Finally we could simplify the formula into
$$(a-b)^2(a+b)c^3+ab(a+b)^2c^2-a^2b^2(a+b)c-a^3b^3=0$$
or we could rewrite it as
$$ \frac1{a^3}+\frac1{b^3}-\frac1{c^3}=(\frac1a+\frac1b)(\frac1a-\frac1c)(\frac1b-\frac1c)$$
Now we could choose some data to verify the formula. For example a=2, c=1, b=1.0356563731256918583980883131742380541 satisfy formula above (but not for that of
dezdichado
)
and we could construct one of the root as below
$$\begin{cases}
\alpha=0&\cos(\alpha)=1&\sin(\alpha)=0\\
\beta=2.0943951023931954923084289221863352561&\cos(\beta)=-0.5&\sin(\beta)=-0.47942553860420300027328793521557138809\\
\gamma=-1.4622047455839744317084407839834855485&\cos(\gamma)=0.10837828597562565035637625724555457401&\sin(\gamma)=-0.99410972590000622348950262653193401668\\
\delta=1.4622047455839744317084407839834855480&\cos(\delta)=0.10837828597562565035637625724555457446&\sin(\delta)=0.99410972590000622348950262653193401663\\
\epsilon=-2.0943951023931954923084289221863352561&\cos(\epsilon)=-0.5&\sin(\epsilon)=-0.86602540378443864676372317075293618346
\end{cases}$$