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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 n-gon inscribed in a circle to the area of a regular n-gon 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 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: $\frac{1}{2}\cos$$\frac{\pi}{n}$$\sin$$\frac{\pi}{n }$$=\frac{\sin$$\frac{\pi}{n}$$\cos$$\frac{\pi}{n}$$}{2}$ The larger triangle has adjacent side 1, and opposite side tan(?/n), thus its area is: $\frac{1}{2}\cdot1\cdot\tan$$\frac{\pi}{n}$$=\frac{ \tan$$\frac{\pi}{n}$$}{2}$ So the ratio of the smaller to the larger is: $\frac{\sin$$\frac{\pi}{n}$$\cos$$\frac{\pi}{n}$$}{ 2}\,:\,\frac{\tan$$\frac{\pi}{n}$$}{2}$ Multiply through by $2\cot$$\frac{\pi}{n}$$$ $\cos^2$$\frac{\pi}{n}$$\,:\,1$ 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: $A_S\,:\,A_L$ $\frac{A_S}{A_L}\,:\,1$ $$$\cos\(\frac{\pi}{n}$$\)^2\,:\,1$ $\cos^2$$\frac{\pi}{n}$$\,:\,1$
September 19th, 2011, 05:46 AM   #3
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Re: cos question hard

[attachment=0:30zhasz7]cos.jpg[/attachment:30zhasz7]

$\theta\,=\,\frac{2\pi}{2n}\,=\,\frac{\pi}{n}$

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

$r\cos(\frac{\pi}{n})\,\cdot\,r\sin(\frac{\pi}{n})\ ,=\,r^2\cos(\frac{\pi}{n})\sin(\frac{\pi}{n})$

and the area of triangle BDE is

$r\,\cdot\,r\tan(\frac{\pi}{n})\,=\,r^2\tan(\frac{\ pi}{n})$

So the ratio of areas is

$r^2\cos(\frac{\pi}{n})\sin(\frac{\pi}{n})\,:\,r^2\ tan(\frac{\pi}{n})\,=\,\frac{r^2\cos(\frac{\pi}{n} )\sin(\frac{\pi}{n})}{r^2\tan(\frac{\pi}{n})}\,:\, 1\,=\,\cos^2(\frac{\pi}{n})\,:\,1,\,n\,\ge\,3,\,n\ ,\in\,\mathbb{N}$
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