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September 18th, 2011, 11:02 PM   #1
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cos question hard

show that the ratio of a regular n-gon inscribed in a circle to the area of a regular n-gon circumscribing the same circle is cos(pi/n)^2 : 1.
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September 18th, 2011, 11:58 PM   #2
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Re: cos question hard

Using the symmetry of the n-gon, we may look at the ratio of the areas of right triangles who in pairs make up the isosceles triangles making up the n-gon.

First, without loss of generality, on the unit circle, draw an arc from ? = 0 to ?/n. Now draw a vertical line segment from the end of the arc to the x-axis, and from the point (1,0) draw a vertical line up to the point of intersection with the line y = tan(?/n)x, which is (1, tan(?/n)).

The smaller triangle has hypotenuse 1, opposite side sin(?/n), thus its adjacent side is cos(?/n). Thus its area is:

The larger triangle has adjacent side 1, and opposite side tan(?/n), thus its area is:

So the ratio of the smaller to the larger is:

Multiply through by

Another approach is the use the fact that the two n-gons are similar. We may use the square of the ratio of the apothems as the ratio of the areas of the n-gons:

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September 19th, 2011, 05:46 AM   #3
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Re: cos question hard


Let be the radius of the circle. Then the area of triangle ABC is

and the area of triangle BDE is

So the ratio of areas is

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