Constructing a pentagon with ruler and compass

This page explains how to construct a regular pentagon with a ruler (an unmarked straight edge) and a compass.

We construct a pentagon by inscribing it in a circle of radius 1. Five similar triangles are formed by the vertices of the pentagon and the centre of the circle. The angles at the centre of the circle must be 360° divided by 5, 72°. So these triangles can be constructed from two right angled triangles each subtending an angle of 36°, and so the length of the outer side of the triangles is \(2\sin 36°\).

What is \(\sin36°\)?

Euler's equation \(e^{i\theta}=\cos\theta+i\sin\theta\) gives \(\sin\theta=\left(e^{i\theta}-e^{-i\theta}\right)/2i\). Using the binomial expansion \((a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\) \[ \sin (5\theta) = \sin^5\theta-10\cos^2\theta\sin^3\theta+5\cos^4\theta\sin\theta \] Thus, writing \(s\equiv\sin36°\), \[ \sin (180°) = s^5-10(1-s^2)s^3+5(1-s^2)^2s, \] and since \(\sin(180°) = 0\), \[ s^5-10(1-s^2)s^3+5(1-s^2)^2s=0. \] Since \(s\neq0\), divide a factor of \(s\) out and rearrange to give \(16s^4-20s^2+5=0\). This is a quadratic in \(s^2\), and its solution is \(s^2=\left(20\pm\sqrt{400-360}\right)/32=(5\pm\sqrt5)/8\). The positive solution here corresponds to 72°, which is the other solution to the quadratic, because \(\sin(5\times72°)=\sin360°=0\), so finally we have \[ \sin 36°=\sqrt{(5-\sqrt5)/8}. \] and the length of one edge of the pentagon, \(L\), is twice this, \(L=\sqrt{(5-\sqrt5)/2}\).

Constructing \(\sqrt5\)

We can construct square roots of numbers like 5 with a ruler and compass by constructing perpendicular lines, and then using Pythagoras's theorem for right angled triangles to get the square root by measuring the hypotenuse.

First, draw a diameter of the circle by drawing a line through the centre of the circle.

Then construct a line perpendicular (at right angles) to the line. Set the compass to a width between the radius and the diameter of the circle. In the following diagram the compass is set to 1.2 times the radius of the circle. Draw arcs at each of the intersections of the line with the circle, then join the intersections of these arcs to bisect the diameter of the circle with a perpendicular line through the centre. In the diagram, the red lines represent lines drawn by a compass, and the circles mark where the needle of the compass is placed.

To get to \(\sqrt5/2\), we need something with a length of 1/2 the radius of the circle so that the hypotenuse of the triangle will be \(\sqrt{1^2+(1/2)^2}=\sqrt5/2\), so now we bisect a radius of the circle using the compass at the centre and at the right hand intersection of the line and the circle.

The blue line is the hypotenuse of a right angled triangle with sides of length 1 and 1/2, and thus has length \(\sqrt5/2\).

Constructing \(L=\sqrt{(5-\sqrt5)/2}\)

Now we set the compass needle at the point where the blue line meets the horizontal diameter, and adjust its length so that the nib is at the intersection of the circle and the vertical line. The radius of the arc subtended is \(\sqrt{5}/2\), and we draw an arc downwards to the horizontal line. The point where it meets the line is at a distance \(\sqrt{5}/2-1/2\) from the centre of the circle.

Using Pythagoras's theorem with the vertical radius, the length of the green line is \(\sqrt{1^2+(\sqrt5/2-1/2)^2}=\sqrt{(5-\sqrt5)/2}=L\), the length we want. Put the compass's needle on the upper intersection of the circle and the vertical line, and the nib on the point of intersection of the previous arc with the horizontal line, and draw an arc from there. The points this arc meets the circle are the left and right upper vertices of the pentagon.

Then keeping the compass the same spacing and putting the needle on the vertices and drawing arcs we can construct the other two vertices of the pentagon.