# Construction of Regular Polygons

## Regular Triangle

For some reason a regular triangle is more often called an equilateral triangle, though it is also the same thing as an equiangular triangle. I give here several constructions given in Euclid's Elements which are based on the method for constructing a regular triangle. Euclid I.1. To construct a regular triangle on a given line segment AB:- With radius equal to d(A, B) draw circles centred at A and B. They intersect at two points, C and C', one on each side of the line, each of which with A and B complete a regular triangle.

Euclid I.10. To bisect a given line segment AB:- Construct as above and join CC'. This line bisects AB at M. If d(A,B) is 1 then by applying the theorem of Pythagoras to the triangle ACM we find that the altitude CM of the regular triangle with unit side is √3/2.

Euclid I.11. To draw a perpendicular to a given line at a given point M:- Draw a circle centre M, to cut the line at A and B. Then construct the bisector, as above.

Euclid I.12. To draw a perpendicular to a given line from a point P outside it:- Draw a circle, centre P, to cut the line at A, B. Then construct the bisector as above.

## Vesica Piscis

The shape formed by the two circular arcs CAC' and CBC' was termed in mediaeval art a vesica piscis (literally fish bladder). The distance d(C,C') is √3, and the vesica piscis has proportions 1:√3. It follows that the rectangle containing the vesica piscis is of diameter 2, so its corners are on the unit circle centred at M. The circumcircles of the two regular triangles form an inner vesica, and similarly an outer vesica can be drawn, three times the linear size of the inner.

## Regular Tetragon: Square

The problem of squaring the circle, i.e. of constructing a square exactly equal in area to a given circle (or a circle equal in area to a given square) is one of the classic problems of geometry. It cannot be done exactly by purely Euclidean compass and straight-edge methods. However, the problems of inscribing a square in a circle or a circle in a square, or circumscribing a square about a circle or a circle about a square are much easier.

Euclid IV.6 To draw squares in a given circle:- Draw perpendicular diameters and join their ends.

Euclid IV.7 To draw squares round a given circle:- Draw perpendicular diameters and perpendiculars at the ends of the diameters.

Euclid IV.8 To draw the circle in a given square:- Draw the diagonals, these meet at the centre; perpendiculars from the centre to the sides are radii of the required circle.

Euclid IV.9 To draw the circle round a given square:- Draw diagonals, these are diameters of the required circle and meet at the centre.

## Regular Pentagon The triangle formed by a side and two radii of a pentagon is 54°-54°-72° and if the radius is R then the side is 2(R sin 36°). From the 36°-36°-108° golden triangle, with unit sides and base Ö5/2 + 1/2, sin 36° equals the altitude h = Ö[1 – 5/4 + 1/4)²] = Ö[5/2 – Ö5/2]/2. So side of inscribed pentagon = R Ö(5/2 – Ö5/2).

A simple but accurate construction given by H.E.Dudeney (Amusements in Mathematics, 1917 p.38) to inscribe a pentagon in a unit circle with a vertex at a given point A is shown. OA is a radius (length 1). OQ is perpendicular to OA and of length 1/2. Thus QA is of length Ö5/2. QR is equal to QA. Thus OR = Ö5/2 – 1/2. From the triangle AOR it follows that AR is Ö(5/2 – Ö5/2) the required length for a side of the pentagon.

Having constructed a pentagon the same vertices can be joined to give a pentagram. The triangles AC'E and BC'D are obtuse golden triangles and ADC, AD'E are acute golden triangles, i.e. the ratio of their sides is Ö5/2 + 1/2 = 1.618.... By bisecting the arcs we can construct a decagon or decagram. To construct a 15-gon is a matter of constructing a pentagon and a regular triangle with a common vertex.

The construction given by Euclid for a pentagon in a circle (Euclid IV.11) is a complex one in several stages, each referring back to a previous proposition. He constructs a triangle ADC which is isosceles and has its base angles twice the apex angle according to Euclid IV.10 by dividing a line AD at B' so that AD.B'D = (AB')², i.e. AB' is the geometric mean of AD and B'D. Then drawing a circular arc, centre A, through D and making DC = AB'. The division of the line is done according to Euclid II.11 by drawing a square ADA''D'', then bisecting AD'' at M, and finding N on D''A produced so that NM = MD, then finding B' so that AB' = AN.

To construct a regular pentagon on a given side AB. Draw BK perpendicular to AB and half the length. Draw AKH making KH - KB (half the given length). Then BH is the required radius of the circumcircle.

## Regular Hexagon

Euclid IV.15. To inscribe a regular hexagon in a given circle:- Step off round the circle at intervals equal to the radius. ## Regular Heptagon

Robert Dixon, Mathographics, 1987, describes three Euclidean constructions that approximate to a heptagon. The angle subtended at the centre by the sides of a heptagon is 360°/7 = 51.428571...° which cannot be constructed exactly by Eulidean1 construction.

(1) His first approximation is to take half the side of an inscribed regular triangle as the side; the chord of 120° in a unit circle = Ö3 and this is the side of the triangle. The angle whose chord is Ö3/2 is 51.317813...°. The difference of these two values is 0.110758° so about 0.8° short in seven stape round the circle. The ratio of the two values is 0.998 so the discrepancy is about 2 parts per thousand.

(2) His second approximation is to construct a right-angled triangle whose perpendicular sides are 4 and 5 units, and to take the angle opposite the 5-side as the central angle of the heptagon. The angle whose tangent is 5/4 is 51.340192°. This also gives a ratio of 0.998. The difference is 0.088°, slightly more than 0.6° short in seven strides of the compass.

(3) His best approximation, illustrated here, is more complex. He first constructs a six-petalled flower as in the construction for a hexagon. Then, with centre on one of the hexagonal vertices, he draws a circular arc to touch the two neighbouring petals. The radius of this arc is Ö3 – 1, and the angle with this length chord is 42.941403°. Adding this to 60° and dividing by two (i.e. bisecting the angle between one end-point of the arc just constructed and a hexagonal vertex) gives 51.470701°. This gives a ratio of 1.0008 and a difference of 0.04213°, exceeding 360° in seven strides of the compass by 0.3°, less than a thousandth part of the circle. ## Regular Octagon

An octagon can be constructed in a circle by bisecting the angles between a pair of perpendicular diagonals. The eight moves of a knight or other skew-leaper form an octagon, but in no case is this octagon regular. Seven diagonal steps approximate to ten orthogonal steps.

## Regular Enneagon

The angle subtended at the centre by the sides of an enneagon is 360°/9 = 40° which cannot be constructed exactly by Eulidean construction. Robert Dixon, Mathographics, 1987, describes two Euclidean constructions that approximate to an enneagon. For best accuracy the circle should first be divided into three parts, which can be done exactly, and then the approximate methods are applied to trisect the three arcs. (1) His first approximation is to draw a right angled triangle with perpendicular sides of 5 and 6 units. Then the angle opposite the 5-side is 39.805571°. In nine strides this would fall short by 1.75° which should be clearly visible.

(2) His best approximation uses the same method as for his best heptagon construction, using the radius Ö3 – 1 constructed there as the base of a right angled triangle, height 2 units, which therefore has an acute angle whose tangent is (Ö3 – 1)/2 which is 20.103909°. This is then used to step off twice from each trisection point. The difference of twice this angle from 40° is 0.207819° and the ratio is 1.0052. Thus six of the angles are half a percent too large and three are one percent too small. This angle also enables one to construct an approximate regular 18-gon.

## Regular Seventeen-gon

We have shown how to construct regular figures by bisection of 2, 4, 8, 16, 32, 64, ... sides. We can also construct figures of 3 sides and by bisection 6, 12, 24, 48, 96, ... We have also shown how to construct figures of 5 sides and by bisection 10, 20, 40, 80, ... Combining 3 and 5 we can form 15 and by bisection 30, 60, ...

p>The cases missing from this list are 7, 9, 11, 13, 14, which require non-Euclidean methods. However the 17-sided case can be constructed by Euclidean methods.