# Compass-and-straightedge

Every point of a solenoid can be constructed using compass alone. The process is very simple and requires a generatrix circle with centre coordinates $(n,0)$ and radius $n$, where $n$ is a natural number.

Once that the generatrix circle is drawn, center the compass on the origin and draw another circle with radius $r$ where $0\le r\in \mathrm{\mathbb{R}}\le 2n$. This circle is called goniometric circle.

Let us focus on the first and second quadrant of the Cartesian plane, that is the two quadrands where $y\ge 0$. The goniometric circle intersects the x-axis in two points, ${C}_{3}$ in quadrant II and ${C}_{1}$ in quadrant I, whereas it intersect the generatrix circle in ${C}_{2}$ in quadrant I. Of course it intersect the generatrix circle also in quadrant IV, but since the siluroid is symmetric with respect to the x-axis, we need only to build it on the first two quadrants and mirror it in the third and fourth one.

Now, let us center the compass in ${C}_{2}$ and in ${C}_{1}$, always with radius $r$, and draw the corresponding circles. They intersect each other in two points: the origin and $P$.

$P$ is a point of the main lobe of the siluroid.

You can likely obtain the secondary lobes by centering the compass in ${C}_{2}$ and in ${C}_{3}$, always with radius $r$. The two points of intersection are now $Q$ and the origin. $Q$ is a point of the secondary lobe lying in quadrant II.

## Demonstration

To demonstrate that $P$ is a point of a siluroid, let us focus on the isosceles triangle $\hat{O{C}_{2}{C}_{n}}$. That triangle has two sides equal to $n$ and the base equal to $r$. It has also the angles opposite to the two sides of the same length equal to $2\theta $. Therefore

[6] $$\frac{r}{2}=\u200a\u200an\u200a\mathrm{cos}\left(2\theta \right)$$Let us focus now on the isosceles triangle $\hat{OP{C}_{1}}$. That triangle has two sides equal to $r$ and the base equal to $\rho $. It has also the angles opposite to the two sides of the same length equal to $\theta $. Therefore

[7] $$\rho \u200a\mathrm{cos}\left(2\theta \right)=r\u200a\u200a(1+\mathrm{cos}(2\theta \left)\right)$$Thence, by substituting the $r$ in [7] by using [6] we obtain

[8] $$\rho \u200a=2n\u200a\mathrm{cos}\left(2\theta \right)\u200a\frac{1+\mathrm{cos}\left(2\theta \right)}{\mathrm{cos}\left(\theta \right)}$$However, since

[9] $$1+\mathrm{cos}\left(2\theta \right)=2\u200a{\mathrm{cos}}^{2}\left(\theta \right)$$we get to

[10] $$\rho =4n\mathrm{cos}\left(\theta \right)\mathrm{cos}\left(2\theta \right)$$that is, the polar equation of the siluroid.

**Q.E.D.**

A similar demonstration can be used to proof that $Q$ belongs to the siluroid too.