The unit circle is divided into twelve equal parts, and the twelve dividing points are joined to the circle's centre, producing twelve rays. Starting from one of the dividing points a segment is drawn perpendicular to the next ray in the clockwise sense; from the foot of this perpendicular another perpendicular segment is drawn to the next ray, and so on to the center of the circle (note the diagram only shows part of the total curve). What is the limit of the sum of the lengths of these segments?

The classic solution approach is to see that each central angle is 30

^{o}and so the hypotenuse of each succeeding triangle is the cosine of 30^{o}times the previous. So the largest of the spiraling segments is opposite a 30^{o}with a hypotenuse of one, and must have a lenth of 1/2. Each succeeding length is the previous length times . These form a geometric series with the n^{th}segment being .
The sum of an infinite sequence with a common ratio less than one is given by and in this case that becomes .

A really pretty solution was given that can be illustrated with some simple rotations..

If we look at just the first two triangles to make it simple...

And then rotate the second (30 degrees in this cae) about the connection with the previous segment we can align them in a vertical column...

This can be continued with each following segment and aligning the segments to produce a triangle like the one below. [Students see this better, it seems, if you describe it as happening from the center out. Note that we get a right triangle with the right angle at top of the first segment created by drawing a one unit segment parallel side from the origin. The sum of the segments is now shown to be one leg of a triangle whose other leg is one. So the lengths must sum to the tan(75

^{o}). I call that pretty cool geometry.
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