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September 18th, 2011, 11:02 PM  #1 
Newbie Joined: Sep 2011 Posts: 8 Thanks: 0  cos question hard
show that the ratio of a regular ngon inscribed in a circle to the area of a regular ngon circumscribing the same circle is cos(pi/n)^2 : 1.

September 18th, 2011, 11:58 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Re: cos question hard
Using the symmetry of the ngon, we may look at the ratio of the areas of right triangles who in pairs make up the isosceles triangles making up the ngon. 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 xaxis, 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 ngons are similar. We may use the square of the ratio of the apothems as the ratio of the areas of the ngons: 
September 19th, 2011, 05:46 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,788 Thanks: 1037 Math Focus: Elementary mathematics and beyond  Re: cos question hard
[attachment=0:30zhasz7]cos.jpg[/attachment:30zhasz7] 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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