**Math Journeys, Volume 2
A Minor Intrusion, Part 1**

**"Uncle Bob" Mead**

I hope readers enjoyed our long walk through reversing products in the first math journey. For a change of pace, I’ve chosen geometry to be the realm for our next trek. A simple problem from the Mathematics Teacher [NCTM, 2003] inspired this next investigation. I state a variation of it here.

**The Launch.** Exactly two sides of a regular
pentagon lie inside another regular pentagon. How do their areas compare?

I named this investigation “A Minor Intrusion” because the diagram reminded me of a domestic scene – a little polygon sitting in daddy pentagon’s lap while daddy read the ... Stars and Stripes. Before we proceed, check your understanding of all the facts and the question. How will we compare the areas? Since we are not given measurements, we won’t be finding the difference. Perhaps the ratio of the areas will remain a constant and independent of any specific measures.

How shall we start? Where do we take this problem? Should we get some background info? Should we try an easier problem first? Is there a relationship between the areas no matter the number of sides, i.e., can we find a general solution? Hey, this is a math journey! Let’s do it all.

Since pentagon properties are less familiar to most than say, triangle, square, and hexagon properties, perhaps we’ll look at those first. I show a diagram for the analogous intrusion for 6 sides below.

See if you can draw the analogues for 3- and 4-sided cases. Don’t spend too much time on triangles because to get two sides of an equilateral triangle inside another, the triangles must be identical. Well then, that case has an answer – the ratio of areas is one to one. A diagram of the 4-sided case will follow shortly.

Note that both sketches above show the intrusion or overlap to be a triangle with the remainder of the minor polygon on the exterior of the parent. There is clearly some math work to be done on the cases for 4, 5, and 6 sides. Let’s take the case for hexagons first, or more precisely, let’s get the background info needed and then I’ll solve it. You, gentle reader, will be left to analyze the case for 4 sides.

We need to review some of the (very) basic stuff about those triangles familiar to us. Notice that in these cases, one side of the parent is a diagonal of the minor and also the long side of the overlapping triangle.

**Hexagon Intrusion.** The figure below utilizes
several facts. Regular polygons have a center and we’ve extended radii
to the vertices of the overlapping area DEB. This triangle is isosceles because
of the regularity of the minor polygon, and I’m calling the short sides
like ED an arbitrary unit one (1). The central triangles (DAB, etc.) have
three equal angles of 60 degrees because there are 6 of them around the center
A. We see that the obtuse angle EDB in the overlap is two 60s and that makes
the more acute angles 30 each.

So our overlapping triangle is shown here split by radius AD into two 30-60-90 right triangles. Their hypotenuses have been assigned a measure of one unit. The short leg FD is half of that. The longer leg is one-half the square root of three. Your next task is to draw an equilateral triangle of side 1, split it down the middle and use the Pythagorean Theorem to verify these ‘relative’ measures.

The side of our parent hexagon BE amounts to two of those long
legs or a full root 3. So the side measures are in a ratio of root 3 to one.
To avoid any further geometric entanglements we call upon another principle
– the areas will be in a ratio equal to the *square* of the side
ratio. Squaring root 3 and one, we get our first solution. The hexagons are
in a ratio of 3 to 1. Can you embellish the diagram above to make this ratio
more obvious? Would the lonely point on the right help?

The info you’ll need to solve the squares situation, barring any insights and shortcuts that occur to you, is that a 45-45-90 right triangle has sides in a ratio of root 2 to one. Bisect a square using a diagonal and use Pythagoras again to verify it. Your 4-sided sketch should look like this:

We’ll tackle the pentagons in Part 2.

Uncle Bob's Puzzle Corner | Get UBPMonthly FREE | Email Uncle Bob

© All rights reserved