Construction of the regular pentagon

Four of the seven regular polygons with fewer than ten sides are easy to construct with compass and straightedge: the topic of Proposition I.1 of Euclid's Elements is the construction of the equilateral triangle, and bisecting an angle of an equilateral triangle inscribed in a circle gives the regular hexagon; similarly, the square is constructed by erecting the perpendicular bisector to a line (bisecting the 2-gon), and bisecting an angle of a square inscribed in a circle produces the regular octagon. The regular heptagon and regular nonagon are impossible to construct with compass and straightedge.

The difficulty of constructing the regular pentagon is somewhere between "easy" and "impossible": there is no obvious way to do it. The construction most commonly described online comes from Herbert Richmond's 1893 paper "A Construction for a Regular Polygon of Seventeen Sides". This construction has the advantage of being very short. Unfortunately, it also has the disadvantage of seeming unmotivated. Here is a construction based on Proposition IV.11 of Euclid's Elements that I hope you will find intuitive and motivated by the important properties of the regular pentagon.

A regular polygon can be thought of as being composed of several smaller polygons, not necessarily disjoint. A decomposition of a regular polygon into smaller parts may suggest a construction of that polygon: for example, a regular hexagon can be broken up into six equilateral triangles, so it can be made by constructing six equilateral triangles around a central point; the same hexagon can also be thought of as the convex hull of an equilateral triangle rotated by \(\pi/3\) and superimposed on itself, suggesting the construction by bisection I mentioned in the first paragraph.

What is a suggestive decomposition of the regular pentagon? Take a vertex of a regular pentagon and draw its two diagonals. These two diagonals and the side of the pentagon connecting the diagonals define a special isoceles triangle called the golden triangle. The convex hull of the superposition of five golden triangles each rotated by different increments of \(2\pi/5\) from \(2\pi/5\) to \(2\pi\) is a regular pentagram inscribed in a regular pentagon. Interestingly, the five segments surrounding the center of the pentagram make up a regular pentagon themselves (of course, a regular pentagon that has been shrunk and inverted through the center of the circle). This smaller regular pentagon is the intersection of the five golden triangles; the first regular pentagon itself must also be an intersection of five golden triangles from a larger regular pentagon, and so on. (We can imagine the first pentagon we constructed as having been produced by cutting the larger pentagon down to a pentagram and then folding in each of its points; we could then repeat the process to construct a smaller pentagon, and so on.) Regular pentagons are self-similar (similar to a proper locus of themselves).

The golden triangle receives its name from its most important property. Figure 1 shows two attached congruent regular pentagons. The length of a side is denoted \(s\); the length of a diagonal, \(d\). The two golden triangles in this figure are similar, so \(\frac{d}{s}=\frac{d+s}{d}\), or \((\frac{d}{s})^2=\frac{d}{s}+1\). \(\frac{d}{s}\), the ratio of the diagonal of a regular pentagon to its side, or the ratio of either long side of a golden triangle to its short side, is equal to the positive solution of this equation, φ, also known as the golden ratio.φ comes from the first letter of the name of the Greek sculptor Phidias, who supposedly used the golden ratio in his statues. φ has many interesting mathematical properties, some related to self-similarity.

A regular pentagon's diagonal and side, or a golden triangle's long side and short side, are in the golden ratio
Figure 1. A golden triangle's long side and short side are in the golden ratio

It is simple to construct φ, which is equal to \(\frac{1+\sqrt{5}}{2}\), or the sum of half a unit length and the hypotenuse of a right triangle with sides of half a unit length and a unit length. Confirm that AE/AB = φ in Figure 2.

Construction of phi
Figure 2. Construction of φ

Constructing the regular pentagon is now simple (Figure 3). First, make a golden triangle: construct the perpendicular bisector to AB through point D. Transfer length AE to an intersection F with the perpendicular bisector. AFB is a golden triangle. Next, construct the circumcenter of this triangle G and construct a circle centered on G with radius GF. Step around the circle with a compass fixed at length AB (effectively rotating the golden triangle!) and connect the dots to finish the regular pentagon.

Construction of the regular pentagon
Figure 3. Construction of the regular pentagon