If we extend each edge past a corner at which it leaves the pentagon, the angle it makes with the next edge round is thus turn/2 −3.turn/10 = turn/5 again, as may independently be observed by noting that travellers, passing round a regular pentagonal path, turn through this angle five times before arriving back where they start, facing the same way, having turned round exactly once along the way. If we likewise extend the next-but-one edge round, past where it meets the intervening edge, our two extended edges shall meet, somewhat outside the pentagon. Together with the intervening edge, the two extensions form a triangle, with internal angles of turn/5 where each extension meets the other edge. This leaves turn/2 −2.turn/5 = turn/10 (a.k.a. 36°) in the outer corner, half the angle in the first two. If we extend all five edges at both ends, in this way, we form such a triangle outside each edge; the combination of these five triangles with their pentagon is known as a pentagram. (Some folk attach mystical significance to these; I am presently only concerned with their geometry.)

As there are an odd number of sides, there is a vertex opposite the middle of each edge, across the mid-point. We can construct a circle, centred on the mid-point, whose radius is the distance to a corner; such a circle passes through all the corners; each edge is a chord of the circle and, as previously noted, subtends turn/5 at the middle. A chord of a circle subtends, at the middle, twice the angle it subtends at a point on the circumference (on the same side of the arc as the middle); so the triangle formed by a corner and the opposite edge has an angle of turn/10 at the chosen corner and each angle at the opposite edge (dividing turn/2 −turn/10 = 2.turn/5 evenly between them) is once more turn/5. So this internal triangle, between and edge and the opposite vertex, has exactly the same geometry as the external triangle formed by extending the edge's two neighbours: these two triangles are each other's mirror images in the edge.

Each edge is thus bisected (perpendicularly) by the line from its opposite
vertex to the exterior point where its neighbours' extensions meet. This line
also bisects the angles at its two ends; each half-angle
is turn/20 (a.k.a. 18°), of which all the other
angles mentioned are multiples. So let q = turn/20, this half-angle: we then
see angles of

- q
- as the half-angle at each end of a line from the vertex opposite a given edge, via the mid-point of that edge, to the exterior vertex where the edge's neighbours' extensions meet

- 2.q
- as the angle between lines from a vertex to each end of the edge opposite it
- as the angle in which the extensions of an edge's neighbours meet
- as the angle that an edge subtends at any of the three vertices not on that edge

- 3.q
- as the angle between an edge and the ray from either of its ends to the mid-point of the pentagon

- 4.q
- as the exterior angle that one edge's extension, past one of its ends, makes with the edge it meets at that end
- as the angle, at the mid-point of the pentagon, between rays from it to each end of any given edge

- 6.q
- as the interior angle between two edges of the pentagon; the lines from the vertex in which they meet to the ends of the opposite edge cut this angle into three equal parts

- 8.q
- as the angle between an edge's extension, past one of its ends, and the line from that end to the vertex opposite the other edge that shares the given end.

The ratios of lengths of the various lines discussed are all determined by the Sin and Cos of these angles; all of which may be inferred from those of q. So let's try to work out s = Sin(q) and c = Cos(q). Applying the addition formulæ for Sin and Cos, and exploiting s.s +c.c = 1 from time to time, we obtain:

- Sin(2.q)
- = Sin(q+q) = Sin(q).Cos(q) +Cos(q).Sin(q)
- = 2.s.c

- Cos(2.q)
- = Cos(q+q) = Cos(q).Cos(q) −Sin(q).Sin(q)
- = c.c −s.s = 2.c.c −1 = 1 −2.s.s

- Sin(3.q)
- = Sin(2.q).Cos(q) +Cos(2.q).Sin(q)
- = 2.s.c.c +(2.c.c −1).s
- = s.(4.c.c −1)

- Cos(3.q)
- = Cos(2.q).Cos(q) −Sin(2.q).Sin(q)
- = (1 −2.s.s).c −2.s.c.s
- = c.(1 −4.s.s) = (4.c.c −3).c

- Sin(5.q)
- = Sin(2.q).Cos(3.q) +Cos(2.q).Sin(3.q)
- = 2.s.c.(4.c.c −3).c +(2.c.c −1).s.(4.c.c −1)
- = s.(8.c.c.c.c −6.c.c +8.c.c.c.c −2.c.c −4.c.c +1)
- = s.(16.c.c.c.c −12.c.c +1)

- Cos(5.q)
- = Cos(2.q).Cos(3.q) −Sin(2.q).Sin(3.q)
- = (2.c.c −1).(4.c.c −3).c −2.s.c.s.(4.c.c −1)
- = c.(8.c.c.c −10.c.c +3 +2.(c.c −1).(4.c.c −1))
- = c.(16.c.c.c.c −20.c.c +5)

and 5.q = turn/4 whose Cos and Sin are 0 and 1, respectively. Since q is strictly between zero and turn/4, c and s are both strictly between 0 and 1; in particular, c is non-zero and Cos(5.q) = 0 reduces to 0 = (4.c.c).(4.c.c) −5.(4.c.c) +5 = (4.c.c −5/2).(4.c.c −5/2) +5 −25/4, whence 4.c.c −5/2 = ±(√5)/2 so c = √((5 ±√5)/8). We can write 1/s = Sin(5.q)/s as (4.c.c).(4.c.c) −3.(4.c.c) +1 and 4.c.c = (5 ±√5)/2, so we get 4/s = (5 ±√5).(5 ±√5) −6.(5 ±√5) +4 = 25 ±10.√5 +5 −30 −6.(±√5) +4 = 4 ±4.√5 so s = 1/(1 ±√5) and we know s is (strictly) between 0 and 1, so positive, obliging our ± to be +, since 1 < √5. So c.c = (5 +√5)/8 and s = 1/(1 +√5). We can now duly obtain

- Sin(2.q)
- = 2.s.c = 2/(1 +√5).√((5 +√5)/8)
- = 1/√(2.(1 +1/√5))

- Cos(2.q)
- = 2.c.c −1 = (5 +√5)/4 −1
- = (1 +√5)/4

- Sin(3.q)
- = s.(4.c.c −1) = ((5 +√5)/2 −1)/(1 +√5)
- = (3 +√5)/2/(1 +√5)
- = (3 +√5).(−1 +√5)/2/(1 +√5)/(−1 +√5)
- = (5 −3 +2.√5)/8 = (1 +√5)/4
- = Cos(2.q)

- Cos(3.q)
- = (4.c.c −3).c = ((5 +√5)/2 −3).√((5 +√5)/8)
- = (√5 −1).(√5 +1)/√(2.(1 +1/√5))/4
- = 1/√(2.(1 +1/√5))
- = Sin(2.q)

Note that the identities Cos(2.q) = Sin(3.q) and Sin(2.q) = Cos(3.q) are natural consequences of 5.q = turn/4. (We could, indeed, have used these as our polynomial equations, in place of Sin(5.q) = 1 and Cos(5.q) = 0. However, each of these has a Sin, hence a factor of s, on one side but a Cos, hence no factor of s, on the other: so the analysis isn't materially easier. It is, none the less, good to verify that direct computation gets answers consistent with these equations.) By the same reasoning, we can obtain Cos(4.q) = s and Sin(4.q) = c. Note that (−1 +√5).(1 +√5) = 4, so we can re-write the manageable quantities (not involving square roots of expressions involving square roots) into the forms:

- Sin(q) = 1/(1 +√5) = (−1 +√5)/4 = Cos(4.q), and
- Cos(2.q) = (1 +√5)/4 = 1/(−1 +√5) = Sin(3.q)

In particular, this last is half the golden ratio

, g = (1
+√5)/2, for which g.g = g +1; the previous is half its inverse, s =
1/2/g. We have Sin(q).Cos(2.q) = 1/4 and Cos(2.q)/Sin(q) = g.g = g +1. Among
the uglier terms, g = 1/2/s = Cos(q)/Sin(2.q), which can also be written as
Sin(4.q)/Sin(2.q) or as Cos(q)/Cos(3.q). We can at least express the uglier
terms in slightly less ugly form using g:

- c = Cos(q) = Sin(4.q) = √((5 +√5)/2)/2 = √((√5).g)/2
- Sin(2.q) = Cos(3.q) = 1/√(2.(1 +1/√5)) = √((√5)/g)/2 = c/g

height

of the pentagon, from
a vertex to the centre of the opposite edge, is thus r.(Cos(2.q) +1) = r.(5
+√5)/4 = (√5).r.Cos(2.q). Its width

, the distance between
two non-adjacent vertices, is 2.r.Sin(4.q), which is g times the length of each
edge. The line L joining two such non-adjacent vertices has one of the
remaining vertices on one one side of it, with the mid-point and the other two
vertices on the other. The edge between these other two vertices is parallel to
L; the line from this edge's mid-point to the isolated opposite vertex is the
perpendicular bisector of L. L's mid-point is r.Cos(4.q) = r.Sin(q) from the
mid-point of the pentagon and thus r.(1 −Sin(q)) = r.(5 −√5)/4
= (√5).r.Sin(q) from the isolated vertex. The area of the pentagon is
then five times the area of the triangle formed between the centre and each
side:

- area = 5.r.r.√(g.√5)/4
- 5.√(g.√5)/4 ≈ 2.37764

Geometrically, there are two easy ways to construct √5. (There are many more ways, such as one involving a right triangle with hypotenuse √8 and one side √3.) If you have a square and extend one of its edges to double its length, the diagonal from its outer end to the furthest corner of the square has length √5 times the side of the square. If you have a circle and one of its diameters, use a third of this diameter's length as unit, draw a second circle centred on an end of the diameter, with two units as radius: either line from an intersection of the two circles to the far end of the diameter is √5 units long.

Once you've constructed a line of length √5 times some unit, you can extend it by that unit and use it as the diameter of a circle. (Equivalently, use it as the radius, extend again to make a diameter and switch to using twice the prior unit as unit hereafter.) Connecting any point on a circle (aside from the diameter's ends) to each end of a diameter gives you a right-angle triangle, with the diameter as hypotenuse. A circle of unit radius about an end of our 1 +√5 diameter meets our first circle in two points, one on each side of the diameter. Either resulting right-angle triangle's internal angle at the far end of the diameter has 1/(1 +√5) = s as its Sin, so this internal angle is q. Join the points where the two circles meet with a straight line to get one edge of a pentagon whose opposite vertex is the far end of the diameter.

Use the point where this new edge meets the diameter as the centre of a circle through the far end of the diameter and identify the point diametrically opposite, on this larger circle: this is the vertex of an exterior triangle of the pentagram associated with the pentagon. Connect this far point to each end of the first edge of the pentagon and extend until each line meets the circle whose diameter was 1 +√5. These are the next two edges round the pentagon; connect the far end of each to the far end of the diameter to complete the pentagon.

The two extended edges of the pentagon, that we constructed from the far vertex of a pentagram-triangle, cut the unit circle (that we used to construct the first edge of what I must now refer to as the big pentagon, that we've just completed) in two sides of a pentagon. Their ends further from the bigger pentagon are closer together: join them with a straight line. Join the other end of each to the point where the unit circle meets the 1 +√5 diameter. You now have a smaller pentagon inscribed in the unit circle, as well as the larger one inscribed in the (1 +√5)/2 circle.

Naturally, you can repeat the construction of ever smaller pentagons, each centred in a circle centred where the extended diameter cut the previous circle, through the ends of that circle's vertical edge. The length scale of each gets smaller by the factor 2/(√5 +1) = (√5 −1)/2.

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