Math Journeys, Volume 2
A Minor Intrusion, Part 2
In Part 1 we presented a touching domestic scene. A daddy pentagon reading the newspaper to a little one, pentagon that is, sitting in his lap. Then we had to go and make a math problem out of it. Sheesh! Here's the picture ...
... and here is the problem. If exactly two sides of a regular pentagon are contained in the interior of another regular pentagon, how do their areas compare?
The Journey Thus Far. We looked at 3 simpler cases to get a sense of the problem. We reasoned that if two sides of a regular triangle were inside another, the two would have to be congruent. The areas would be in a one-to-one ratio. Next, we found that in the case for six sides, the hexagons would be in a 3:1 ratio. The solution depended on the geometry of the 30-60-90 triangle. See Part 1, link below.
You, dear reader, were to analyze the case for four sides. If exactly two sides of a square are inside another square, it looks like this ...
... and the answer is that the areas are in a 2:1 ratio. This can be reasoned from the symmetries of the picture or by knowing that the side ratio is
√2 : 1, and that the areas will be in a ratio equal to the square of that one.
The Case for Five Sides. Well, so far we get nice numbers. One, 2, and 3 to one. Trouble is – five sides is between 4 and 6 (squares and hexagons) and we're running out of nice numbers. Sometimes it pays to anticipate. We'll get our solution from the diagram below.
Some readers might be comfortable using trigonometry, and we'll take that up momentarily. I'll argue from the similarity of the triangles ABC, the area of overlap, and CAD. These two have the same shape, equal angles, and their sides are in proportion which we write as
AB : AC :: AC : CD
... and if we set AB = 1 and AC = c, then it's ...
1/c = c/(c + 1)
... and readers of math history and/or Uncle Bob, will recognize c as the golden ratio defined by the ancient Greek geometers.
The golden ratio is irrational and approximately 1.618, and that is the ratio of sides c and 1. If we square the ratio, we get about 2.618 (not a coincidence really, but it's another whole journey we'll eschew for now), and that is how the areas compare.
We summarize in a little table
A Look Ahead. Do we see a trend in the data, and what kind of a trend is it? InPpart 3 we will solve for polygons with more sides and see where the results take us.
You, dear reader, have been relatively idle, but you can pave the way for our third leg in this journey. The angle at B above labeled theta is the interior angle. Use one of the "two Ds" to find its measure. D1 is dust off an old geom text. D2 is derive it, and find the formula for this angle in the case of any number of sides n.
You could apply trig to get our solution above also by using one of two methods. One way is to split ABC into two right triangles. A slicker way is to use the Law of Cosines on ABC in situ.
So situ down and get busy. Ha!
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