A long time back, I wrote about some geometric options for a problem I had found on Greg Ross' Futility Closet. Shortly afterward I got a note from a blogger at "Five Triangles" who mentioned he had posted a very similar problem (above) about a year earlier.

What I especially enjoyed about his presentation of the problem was the obvious invitation to generalize the idea to regular polygons of more sides.

Each of the figures is simplified by the visual approach of rotating the inner polygon until it has its vertices at the midpoints of the sides of the larger. From there it is almost trivial that for the triangles, the smaller is 1/4 the larger (the logo of the five-triangles web site shows this clearly), and the smaller square is 1/2 the larger.(top image)

When you go to five sides the quick visual solutions disappear, but a generalization should offer itself to a clever trig student. If we assume the sides of the larger n-gon are each of unit length, then the area of the two polygons should be in the same ratio as the square of the side of the smaller polygon..... ( some were confused by this, the area of two similar polygons are in the ratio of the square of their corresponding side, but since we et the larger at one unit, its square is 1 square unit, and so the ratio of the two area is the square of the inner edge length over one.)....and a clever trig student looking at all those triangles (such as the blue FGB) formed between the two polygons should know a quick rule for finding the square of the side lengths of the inner polygon.... the beautiful extension of the Pythagorean theorem they know as the law of cosines.

Since each leg on the outside of the triangle is \(\frac{1}{2}\) unit, and the angle is \(\frac{\pi(n-2)}{n}\) it should be easy to determine that the square of the sides of the inner polygon is \(\frac {1}{2})^2 + (\frac {1}{2})^2 - 2 *(\frac {1}{2}*\frac {1}{2} * \cos(\frac{\pi(n-2)}{n})\) Or more simply, \(\frac {1}{2}(1-\cos(\frac{\pi(n-2)}{n}))\)

For values from n= 3 to 12 I came up with the following with the support of Wolframalpha:

Only the triangle, square, and hexagons produced rational roots, in convenient consecutive quarters for easy remembering. The decimal approximations clearly support the intuitive idea that the limit should approach one as n grows larger without bound. By the time you get to the hectogon, the ratio is ,9990. Fans of the "golden ratio will appreciate its appearance in the pentagon even if slightly camouflaged.

N Ratio

3 ..... 0.25

4 ..... 0.5

5 ..... 0.654508

6 ..... 0.75

7 ..... 0.811745

8 ..... 0.853553

9 ..... 0.883022

10 .... 0.904508

11 .... 0.920627

12 .... 0.933013

An interesting exploration, I think, for good trig students to explore. Enjoy

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