The Pentagon!

For more than 2500 years, the regular Pentagon has held a special place among the favorite shapes of mathematicians.  Pythagoras practically invented the idea of mathematician back around 500 BC (when he should have been spending his time trying to prove his famous theorem; actually there are almost-proofs, or demonstrations, that are attributed to him).  He and his friends organized the very earliest known secret society of math geeks, called the Pythagorean Society.  Back then people were very conservative, and would have killed you if you told them that numbers other than simple fractions existed, like, say Pi.  (There are stories that Pythagoreans were pushed off ferries once they were discovered.  At least today they would have merely prayed for your death.)  So what's the punchline?  The Pythagoreans used the regular Pentagon as their secret symbol.  To this day, many of the mathematics societies have a pentagon somewhere in their insignia, or at least a 5-pointed star.  (At left is the logo of the MAA.)

[Added later:  I just realized that the MAA logo has little or nothing to do with a pentagon!  It is a so-called Icosahedron, a polygon with 20 faces.  But, as it happens, if you join the centers of every neighboring pair of triangular faces together, you get a companion figure, called a dodecahedron, which has all pentagonal faces.  Even in the figure at right, you can discern a number of pentagonal caps; in fact, every point is the center of a pentagonal cap.  Anyway, I still feel silly about that mistake.]

Incidentally, Pythagoras’s Theorem does play a part in the celebration of the Pentagon below:  If you have a right triangle (one of its angles must be 90 degrees; in other words, it must look like a rectangle cut down its diagonal), if you build three squares on the triangle, one on each side, the area of the largest one will be equal to the areas of the two smaller squares added together.  Above right you see a well-known instance of this:  There is a right-triangle, and squares on each of its sides.  The two smaller squares have area 9 little squares and sixteen little squares, and the large square has area 25.  Note that 9+16=25.


Let's look at a regular Pentagon a little closely.  A regular polygon always has all its sides congruent to each other, and all its angles congruent to each other; if you were to rotate it carefully, we could never tell the difference.  A little arithmetic, and the fact that the angles of any triangle have to add up to 180 degrees, tell us how much each of the inside angles are.  Here's how:

1.  There are 5 angles at the center (one from each of the little triangles I made by joining the sides to the center).  Those five angles must all be equal to each other, and add up to 360 degrees.  So each one must be 360/5 = 72 degrees.

2.  Now, all the five little triangles are identical, and because each of them has two equal sides (the spokes from the center), the remaining two angles must be equal.   This is a basic fact about so-called isosceles triangles.  Consider one of the triangles; one angle is 72, all three must add up to 180, and the two unknown angles have to be equal.  That makes each of them 54 degrees.  So the inside angles --the wide, shallow angles of the pentagon itself, without the radial lines-- are 108 degrees each.


3.  Now, we look at diagonals.  The diagonals of a Pentagon are fascinating, for a reason that will emerge shortly (but will perhaps not impress everyone equally).  Let's name the points of the Pentagon A, B, C, D, E, starting at the top, and consider the diagonals AD and EC.  It seems reasonable that all the diagonals must be the same length.  Now look at triangle EDC.  We can do the same trick as before: all three angles must add up to 180; angle D is 108, we know, the other two must split the remaining 72 degrees between them equally, so they must each be 36.  Looking carefully at triangle CXD, we discover that angle XDC is 108 − 36 = 72.  The remaining angle must also be 72, so triangle CXD is isosceles!  The side CD is a, which is what I'm going to call the length of one side of the Pentagon.  So the segment CX must also be a.

The little triangle EXD is a scale model of EDC --we call them similar triangles.  Similar triangles have proportional sides; so CD/EX should equal CE/ED.

Now EX = EC - CX, which is the same as EC − a, since CX and CD are equal in length.  So our similar triangles equation can be re-written
a/(EC−a) = EC/a
which can be simplified by multiplying through by a(EC−a) to get a² = EC² − a·EC.

This is a fascinating equation, as promised.  Suppose we represent the unknown length of the diagonal EC by x; the equation becomes x² − ax − a² = 0. This is what we call a quadratic equation in x, since a is unknown but fixed.

Now, most of my readers are probably thinking: hey, we didn't escape from Algebra 2 just to have to come to your blog and solve quadratic equations for you.  So I will graciously solve the equation for you: the answer is .

What in the heck is that?  It turns out that this is the fabulous Golden Ratio.  It is the ratio of the longer dimension to the smaller dimension of a kind of rectangle which the renaissance Italians thought was perfect.  Why is it perfect?  Because the part of it which needs to be trimmed off to make it a square is itself a perfect rectangle.  The ratio between a diagonal of a Pentagon and its side is the same as the Golden Ratio.  (For those of you who are of an algebraic disposition: if you take the fraction above, duplicate it with a minus sign instead of the plus sign, and multiply it into the original Golden Ratio, you should get something interesting! Heh heh heh ...)

4.  Now that we know the length of the diagonal, we can actually build a Pentagon, using only ruler and compasses, if only, given a line segment of length a, we can build another line segment of length .

It is not too difficult, merely long and drawn out. To practice, we could start with trying to construct a segment whose length is the square-root of two times a.  We'll do that another time...

A.