Exercise 15.2.10

Question

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The polar distances of the $5$-division points of the lemniscate satisfy the equation
$$0 = r_0(r_0^{24} + 50 r_0^{20} - 125 r_0^{16} + 300 r_0^{12} - 105 r_0^8 - 62 r_0^4 + 5).$$
This equation was first derived by Fagnano in 1718.
\be
\item[(a)] Show that the $r_0$ corresponding to the $10$-division points also satisfy this equation.
\item[(b)] Use Maple or Mathematica (or Sage!) to show that this equation factors as
$$0 = r_0(r_0^8 - 2 r_0^4 + 5)(r_0^{16}+52r_0^{12} - 26 r_0^8 - 12 r_0^4 + 1)$$
and that the only positive real solutions are
$$\sqrt[4]{-13 + 6 \sqrt{5} \pm 2 \sqrt{85-38 \sqrt{5}}}.$$
Explain (with a picture) how these solutions relate to the $5$- and $10$-division points.
\ee

Richard Ganaye · Accepted Answer

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\begin{proof} 
\item[(a)] Since $5$ is odd, the $5$-division points are roots of $uP_5(u^4)$ by Corollary 15.2.6.
We obtain $P_5$ with the Sage function given in Exercise 7:
$$P_5(u) = u^{6} + 50u^{5} - 125u^{4} + 300u^{3} - 105u^{2} - 62u + 5.$$
Therefore the polar distances $r_0$ of the $5$-divisions points of the lemniscate satisfy the equation
$$0 = r_0(r_0^{24} + 50 r_0^{20} - 125 r_0^{16} + 300 r_0^{12} - 105 r_0^8 - 62 r_0^4 + 5).$$

\bigskip

We have seen in Exercise 6 that the $10$-division points are the $5$-divisions points, together with the symmetric points about the $x$-axis, which have same polar distances. Therefore the polar distance of any $10$-division point is also a polar distance of a $5$-division point, thus verify the given equation (see figure in Exercise 6).

\item[(b)]  We saw in Exercise 7 that $P_5(u)$ factors as
$$P_5(u) = (u^{2} - 2u + 5) (u^{4} + 52u^{3} - 26u^{2} - 12u + 1).$$
Therefore the polar distances of the $5$-division points (and of the $10$-division points) satisfy
$$0 = r_0P_5(r_0^4) =r_0 (r_0^{8} - 2r_0^{4} + 5)  (r_0^{16} + 52r_0^{12} - 26r_0^{8} - 12r_0^4 + 1).$$
$r_0^{8} - 2r_0^{4} + 5 = (r_0^4-1)^2 + 4 >0$ thus $r_0^{8} - 2r_0^{4} + 5$ has no real root.

We obtain the positive roots of $u^{4} + 52u^{3} - 26u^{2} - 12u + 1$ with Sage:
\begin{verbatim}
u = var('u')
P = u^4+52*u^3-26*u^2-12*u+1;
S = P.solve(u)
S
\end{verbatim}
\begin{align*}\
[u = -6 \, \sqrt{5} - \frac{1}{2} \, \sqrt{608 \, \sqrt{5} + 1360} - 13, \
u = -6 \, \sqrt{5} + \frac{1}{2} \, \sqrt{608 \, \sqrt{5} + 1360}- 13, \
u = 6 \, \sqrt{5} - \frac{1}{2} \, \sqrt{-608 \, \sqrt{5} + 1360}- 13,\
 u = 6 \, \sqrt{5} + \frac{1}{2} \, \sqrt{-608 \, \sqrt{5} + 1360}- 13]
\end{align*}
\begin{verbatim}
[e.right().n() for e in S]
\end{verbatim}
$$\left[-52.4909612184115, -0.341854511585989, 0.0733810146911846,0.759434715306293ight]$$
\begin{verbatim}
S[2].right()^(1/4),S[3].right()^(1/4)
\end{verbatim}
$$\left({\left(6 \, \sqrt{5} - \frac{1}{2} \, \sqrt{-608 \, \sqrt{5} +
1360} - 13ight)}^{\frac{1}{4}}, {\left(6 \, \sqrt{5} + \frac{1}{2} \,
\sqrt{-608 \, \sqrt{5} + 1360} - 13ight)}^{\frac{1}{4}}ight)$$

Since $1360 = 1 6 	imes 85$, and $608 = 16 	imes 38$, we obtain the two positive solutions of the equation
$$\sqrt[4]{-13 + 6 \sqrt{5} \pm 2 \sqrt{85-38 \sqrt{5}}}.$$
Since there are only two $5$-division points $M_1,M_2$ in the right loop of the lemniscate, the 5 division points have polar distances (using $OM_1 > OM_2$)
\begin{align*}
&OM_0   = 0 \
&OM_1  = OM_4= \left( \sqrt[4]{-13 + 6 \sqrt{5} +2 \sqrt{85-38 \sqrt{5}}} ight),\
 &OM_2  = OM_3  = \left( \sqrt[4]{-13 + 6 \sqrt{5} - 2 \sqrt{85-38 \sqrt{5}}} ight).\
 \end{align*}
 (See the figure of Exercise 6). 
 
 By Proposition 15.1.1, all these points are constructible. The $10$-division points have same polar distances, and are also constructible.
\end{proof}

Exercise 15.2.10

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Exercise 15.2.10

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