## The Regular Pentagon

In Golden Rectangle & Golden Ratio page, I mentioned that:

The golden ratio shows up (as an exact fit) in mathematics in many unexpected places. The ratio shows up everywhere in the pentagram (five pointed star) and its circumscribed pentagon (shown on the left). a/b=(a+b)/a=(a+b+a)/(a+b)=phi (the golden ratio). Here we also see the two kinds (acute and obtuse) of golden triangles (I've painted two of them green). They are any of the isosceles triangles in this diagram. These are triangles which exhibit the golden ratio.

There are several ways to construct a pentagon (using compasses and straightedge) in a circle. They normally start with: (1) draw a diameter, (2) draw a diameter perpendicular to the first diameter, (3) bisect one of the radii (this step produces the Golden Ratio in several forms), (4) draw a line from the middle of this radius to an endpoint of the other diameter. After that step, there are several ways to continue. The way I do it is to (5) draw an arc (with its center at the midpoint of the radius) from this last endpoint of a diameter to the other diameter. This produces the length marked with two vertical marks in the diagram. If the radius is 1 then that length is 1/phi or 1-phi. Alternatively, if this length is 1, then the radius is phi. So that length is the side of a decagon. In the diagram, (6) I drew two sides of a decagon. (7) Connect two vertices of the decagon (skipping a vertex), and you have one side of a pentagon. (8) Draw the other four sides. I discovered this method on my own. But, I suspect that it is one of the standard methods.

Another method was discovered by Richmond in 1893. The first steps are as above. Then (5) bisect the angle that is at the bisection of the radius. (6) Where this angle bisector crosses the other diameter, draw a line parallel to the bisected radius. Now (7) draw one side of the pentagon, from where that parallel line crosses the circle to the endpoint of the diameter that is perpendicular to the bisected radius, as shown in the diagram. (8) Draw the other four sides.

Tie an overhand knot in a strip of paper, being careful to tighten it slowly, and flatten it as it gets tight. Many books claim that this produces a regular pentagon. See the diagram. It is fairly obvious that this is at least close to being a regular pentagon, as each side is slightly longer than the width of the paper strip. And the angles are roughly the same. But it is not obvious that it is perfect, rather than close.

By symmetry (the knot is the same from either side of the paper), or in other ways, we can show that angles E (of the pentagon) and B are congruent. Also angles C and D are congruent. It is not difficult to show that angle A is congruent to B (by the way angle A is folded, which makes the exterior angles congruent).

The diagram on the right shows the pentagon with the diagonals. Each of the diagonals is parallel to a side, because the strip of paper has parallel edges. We can show that all kinds of angles are congruent, by symmetry and by parallel lines. In particular, we can show that the blue triangle is isosceles. From there, we can show that the two red triangles are congruent. And from there, we can show that at each vertex of the pentagon, all three acute angles are congruent. From that we show that the five vertices of the pentagon are congruent. And we can easily show, by congruent triangles, that the five sides of the pentagon are congruent.

So the pentagon is indeed regular. I did not show each step of the proof. Some of the steps may take some thought.

Also see Home Plate, which is an irregular pentagon.

It is fairly easy to figure out the angles of a pentagon. Draw a central angle, as in the diagram at the left. This central angle is 1/5 of 360 degrees, or 72 degrees. The sum of the angles of a triangle is 180 degrees. And so the three angles of the isosceles triangle are 72 degrees, 54 degrees and 54 degrees. That makes each angle of the pentagon 108 degrees. The sum of the five angles is 108x5=540 degrees.

This method works for any regular polygon. But we can deduce a general formula for all polygons. At the right, we see a pentagon with three inscribed triangles. An n-gon (polygon with n sides) would contain n-2 triangles in a similar way. The sum of the angles of each of these triangles is 180 degrees. Each angle of the n-gon is the sum of angles of these triangles, and there are no angles of the triangles that are not part of angles of the n-gon. And so the sum of the angles of a regular n-gon is 180(n-2) degrees. This formula works for irregular polygons (convex or not); but dividing a nonconvex polygon (with some angles greater than 180 degrees) into triangles may not be quite as easy. If the n-gon is regular, then each angle is 180(n-2)/n degrees. You can check that these two formula work for triangles and squares and pentagons.

The diagonals split each angle of a regular pentagon into three identical angles. In this diagram, we see that quadrilateral AEDF is a rhombus (parallel sides and consecutive sides are congruent (equal)). And so, the diagonal AD bisects the angle EAF. And, by symmetry, all three angles are congruent (36 degrees).

Using the same diagram, I will now show that the golden ratio (See Golden Rectangle and Golden Ratio) shows up in a regular pentagon. Let's say that each side of our pentagon is 1. AF=1, because it is the side of our rhombus. Triangles ADF and BCF are similar. So, x/1=1/y. But y = x+1. So x = 1/(x+1). x(x+1) = 1. So x^2+x-1=0 [where x^2 is x squared]. Solving for x, using the Quadratic Formula, x = (-1 +/- sqr(5) )/2 [where sqr() is the positive square root function, and +/- means plus or minus]. One of these lengths is positive: x = (-1+sqr(5))/2. And y = (1+sqr(5))/2. And that is the golden ratio.

The great book Amazing Origami by Kunihiko Kasahara describes the following method as "the American method" for making a pentagon by paper folding. We start with a square piece of paper (which can be created from any odd shaped piece of paper, by folding). We fold it diagonally in half (the horizontal line in the diagram), then crease one side in half at point A. We then fold point B over to point A, forming line CD. The idea is that angle ACD (and its image BCD) is about 72 degrees. It is actually about 71.56 degrees, almost half a degree off. This is close enough for most purposes. The book goes on to fold and cut a pentagon with center at C. I won't do that here, as the pentagon is not exact.

The same book gives this method (on the left) as "the Japanese method" for folding a pentagon. We fold the square in half, as above. Then we crease the long diagonal of the square at C. We crease the upper right side twice, in half and then in fourth at A. We fold line BC over to point A, producing line CD (point B will coincide with point A). Then we bisect angle CDE, producing line CF. And angle DCF (and its image ECF) is supposed to be near 72 degrees. Actually it is 72.11 degrees, which is closer than the American method.

The same book produces a "regular pentagon template" by folding a regular hexagon, cutting along one radius, and overlapping one sector over its neighbor. This produces a short pentagonal pyramid. We can use a similar method to fold any regular polygon that we want, theoretically.

The book says, "Many people in the past have tried to fold a regular pentagon." Apparently we are not going to do it by making a pyramid, or by tying an overhand knot. I'm sure that people have actually succeeded, as it is fairly easy to fold a golden rectangle. And, it has actually been proved that anything that can be done with compasses and straightedge can be done by folding paper, and more. I'm trying to discover a method. If our paper were cut into a circle, we could use one of the construction methods shown above (which used compasses and straightedge).

OK, here is a method that seems to work, based upon our first construction (which produced a decagon). We crease our paper into four smaller squares, as shown. We crease the right center horizontal line in half, and fold the diagonal AB. We bisect angle BAC to find point D. CD is the length of our decagon in our first construction. We fold the point C onto point D producing the perpendicular bisector EF of CD. We fold point B onto line EF to find point E. EG is now one side of our pentagon. From there we can reproduce every side of the pentagon. All vertices will be within our original square.

In practice, this may not be any more accurate as our American and Japanese approximations, as we don't get very precise angles when we fold paper. But in theory, this method is exact.

That method suggests an improvement to our construction of a pentagon. See the diagram on the right. We start as before, constructing a golden ratio. Then we bisect the segment which I have marked with the little double vertical lines. Then we draw the perpendicular bisector from there, and we can draw one side of our pentagon (in red) without bothering with a decagon. Then we can draw the other four sides.

In Vietnam, I sometimes painted white stars on green vehicles, as the officers found out that I was an artist. Nobody could ever find a template. Not remembering how to construct a pentagon, I just experimented until I had a roughly regular pentagon, and then drew my star. Once I just painted the star freehand, starting slightly small, and adjusting it over and over until it looked regular. The person watching me do this was amazed that I could do that.