# Musing on constructing regular pentagon.
It is well-know that the regular pentagon is constructible. Though I've never done it myself. So I'd like to see how I might do it in a naive way.
Since we essentially need to construct the length $\cos 72\degree$, let us look at how $\cos 5x$ can be expressed: $$
\begin{align}
\cos 5x &= \Re (\cos x + i \sin x)^{5} \\
&= \cos^{5}x -10 \cos^{3}x \sin^{2}x + 5 \cos x \sin^{4}x \\
&= u^{5} - 10u^{3}(1-u^{2})+5 u (1-u^{2})^{2} \\
&= 16u^{5} -20u^{3} + 5u
\end{align}
$$
(Classic Chebyshev polynomial).
Now, if $5x = 90 \degree$, then $x = 360\degree / 20$, the angle of a regular icosagon. And if we can construct the angle of a regular icosagon, then we certainly can construct the angle of a regular pentagon by duplicating twice.
So, when $5x = 90\degree$, $u = \cos (360\degree / 20) = \cos 18\degree$ satisfies $$
0 = 16u^{5} - 20u^{3} +5u
$$I make this choice because we are just solving a polynomial root. And since $u \neq 0$, we need to solve $$
0 = 16u^{4} - 20 u ^{2} + 5.
$$
Set $w = 4u^{2}$, then we solve $$
0 = w^{2} -5u +5
$$
This gives $$
w = \frac{5 \pm \sqrt{25-20}}{2}= \frac{5 \pm \sqrt{5}}{2}
$$
So, $$
u = \pm \sqrt{\frac{5\pm\sqrt{5}}{8}}.
$$
Now we know $u = \cos (360\degree / 20) > 0$, so we take the outer positive sign, $$
u = \sqrt{\frac{5\pm\sqrt{5}}{8}}
$$Now which of these two is $u$? If we take $-$ sign, note $$
\frac{5 - \sqrt{5}}{8} < \frac{5-2}{8} = \frac{3}{8} = \frac{24}{64} < \frac{25}{64}
$$
and so $$
\sqrt{\frac{5-\sqrt{5}}{8}}< \sqrt{\frac{25}{64}} = \frac{5}{8} < \frac{\sqrt{3}}{2} = \frac{4 \sqrt{3}}{8} = \cos 30 \degree
$$(The latter is because $25 < 48$) This means $$
\sqrt{\frac{5-\sqrt{5}}{8}} = \cos \theta\degree
$$where $\theta > 30$, which is not the angle $360\degree / 20 = 18 \degree$ that we seek.
Hence $$
\cos 18\degree = \sqrt{\frac{5 + \sqrt{5}}{8}}
$$
Great, we can construct $\sqrt{5}$, and certainly $\sqrt{\frac{5+\sqrt{5}}{8}}$. Then by duplicating this angle twice, we get $72\degree$.
Alternatively, in a regular pentagon, if one connects two non-adjacent vertices with the center, we get an isosceles triangle where one angle is $18\degree$. Which we can construct the rest of the regular pentagon this way.
Now, this isn't the shortest. So can we do better?
By double angle formula, $$
\cos 2\theta = 2\cos^{2}\theta -1,
$$so $$
\cos 36\degree = 2 \left( \frac{5+\sqrt{5}}{8} \right) - 1 = \frac{1 + \sqrt{5}}{4}
$$
Much better! And $36\degree$ occurs as the angle of the isosceles triangle formed by connecting three consecutive vertices of a regular pentagon.
Hmm, if we apply double angle formula again, we get $$
\cos 72\degree = \frac{-1 + \sqrt{5}}{4}
$$
Both of these angles $72\degree$ and $36\degree$ are easier to construct than $18\degree$ that involves an iterated square root.
# A simple construction.
Take a regular pentagon, and we note the measures in the following figure:
![[regular-pentagon.png]]
Consider a regular pentagon inscribed in a circle of radius 1 in the left figure, with center $O$. Then we note that angle $\angle B A C = 36\degree$. And since triangle $\triangle CBA$ is a right triangle with $CA=2$, we see that $AB = 2 \cos 36\degree = \frac{1+\sqrt{5}}{2}$ .
Now in the figure on the right, also a circle of radius 1, but now with a circle of radius $1 / 2$ with center $P$ that is at the mid point from the center $O$ to the circumference. The line segment $AP = \sqrt{1 + \frac{1}{4}}$, and the segment $AQ = \frac{1}{2} + AP = \frac{1+\sqrt{5}}{2}$, which a length we desired for $AB$!
Hence all we need is to construct the figure on the right, join from $A$ to $P$ to meet $Q$, and $AQ$ gives the length of $AB$ that we need to construct the rest of the regular pentagon!