Constructions of equilateral triangles, squares, regular hexagons and regular octagons (8 sides) are quite simple. We cannot help but asking the question whether it is possible to use compass and straightedge to construct regular pentagons, regular heptagons (7 sides) and regular nonagons (9 sides) or not.
It turns out that there are many ways to construct regular pentagons, but it is not possible to construct regular heptagons nor regular nonagons by using just compass and straightedge. Today, we will look at regular pentagon construction. We leave the discussion of heptagon and nonagon construction for our later posts.
Let us now do a bit of analysis. In the following figure, we can see that if we can construct the point $H$, then from $H$ we can construct the points $N_3$ and $N_4$, and thus we can easily obtain the regular pentagon $N_1 N_2 N_3 N_4 N_5$.
So in order to find a way to construct the point $H$, firstly, we need to determine the value of $\cos{\frac{\pi}{5}}$.
Calculating $\cos{\frac{\pi}{5}}$
Observe that the angle $\frac{\pi}{5}$ has the following property $$2 \frac{\pi}{5} + 3 \frac{\pi}{5} = \pi$$ so if we let $x = \frac{\pi}{5}$ then $2 x + 3 x =\pi$, it means that $2x$ and $3x$ are two supplementary angles. Thus by basic trigonometry, we derive the following equation $$\cos{2x} = - \cos{3x}.$$
Double and triple angle formulas tell us that $$\cos{2 x} = 2 \cos^2{x} - 1,$$ $$\cos{3 x} = 4 \cos^3{x} - 3 \cos{x},$$
so our above equation becomes $$2 \cos^2{x} - 1= -(4 \cos^3{x} - 3 \cos{x})$$
$$4 \cos^3{x} + 2 \cos^2{x} - 3 \cos{x} - 1=0$$
Let $t = \cos{x}$, then we have the following cubic equation $$4 t^3 + 2 t^2 - 3 t - 1=0$$
It is not hard to see straight away that $t=-1$ is a root of the cubic equation, so we will factor this cubic as $$4 t^3 + 2 t^2 - 3 t - 1 = (t+1)(4t^2 - 2t-1)=0$$
Now solving the quadratic equation $$4t^2 - 2t-1=0$$ we obtain two roots of opposite signs $$t_{1,2} = \frac{1 \pm \sqrt{5}}{4}$$
Thus, our $\cos{\frac{\pi}{5}}$ is the positive root and we complete our calculation $$\cos{\frac{\pi}{5}} = \frac{1 + \sqrt{5}}{4}.$$
Using Pythagorean Theorem
In order to construct the segment $OH$ we need to construct a line segment of length $(1 + \sqrt{5}) r$ and then divide it into four equal parts.
In order to construct a segment of length $(1 + \sqrt{5}) r$, we need to construct a line segment of length $\sqrt{5} r$.
Now mentioning $\sqrt{5}$, we immediately recall the Pythagorean Theorem because $5 = 1^2 + 2^2$.
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| Pythagorean Theorem: $c^2 = a^2 + b^2$. |
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| $$(r)^2 + (2r)^2 = (\sqrt{5} r)^2$$ |
Construction of a regular pentagon
Based on the above analysis, we arrive with the following construction:
- Construct two perpendicular coordinate lines $x'Ox$, $y'Oy$;
- Take $O$ as the center, construct an arbitrary circle which intersects $Ox$, $Ox'$, $Oy$ at $X$, $N_1$, $Y$, respectively;
- Construct the point $A$ on $Oy$ so that $YA=YO$;
- Construct the point $B$ on $Ox$ so that $XB=XA$;
- Construct the midpoint $C$ of $OB$ and the midpoint $H$ of $OC$;
- Through $H$ construct a line perpendicular to $Ox$ which intersects the circle $(O)$ at $N_3$, $N_4$;
- Construct a circle centered at $N_1$ with radius equal to $N_3N_4$ which intersects the circle $(O)$ at $N_2$, $N_5$;
- Then $N_1N_2N_3N_4N_5$ is a regular pentagon.
In this construction, we can see that $$OA = 2r, ~~XA = \sqrt{5}r, ~~XB = \sqrt{5} r, ~~OB=(1+\sqrt{5})r, ~~OH=(1+\sqrt{5})r/4$$
Here we go, we have completed a compass-and-straightedge construction of a regular pentagon!
The immediate question we can ask is, for which values of $n$ we can construct a regular polygon with $n$ sides. Despite the simplicity of its statement, it is actually a very hard mathematics problem. For over more than a thousand years, this problem had remained unsolved. Gauss was the person who made the first breakthrough to solve the problem. We will continue this fascinating story in our future posts. Hope to see you then.