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## Question

A regular polygon has 1023 sides, each of length 1.

Is the area between the inscribed and circumscribed

circles of this polygon greater than 1?

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• 0

I get the radii of inscribed and circumscribed circles to be 162.814995 and 162.815763, respectively. Calculating the difference in areas, I get 1.5713, so yes, the area is greater than 1.

edit: more decimals

Edited by HoustonHokie
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• 0

...it appears that for any regular polygon the area between the circumscribed and inscribed circles will be equal to pi*s2/4, where s is the side of the polygon. So, for any regular polygon with s=1 the area in question is equal to pi/4 and is therefore less than 1.

Let r be the radius of the circumscribed circle and a be the apothem or the radius of the inscribed circle.

They are related to the side s and to each other by this formula (from Wikipedia http://en.wikipedia.org/wiki/Regular_polygon)

r = s/(2*sin(pi/n)) = a/cos(pi/n), where n is he number of sides.

The area of the circle is pi*r2, so the difference between the areas of 2 circles is

D = pi*r2 - pi*a2

Replacing a with r*cos(pi/n) we get

D = pi*r2 - pi*r2*cos2(pi/n) = pi*r2*(1-cos2(pi/n))

Knowing that 1 - cos2(x) = sin2(x) we can represent the above as

D = pi*r2*sin2(pi/n)

Now, lets put the value of r into the this equation and we get

D = pi*s2/4

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• 0 hmmm...

3.14159*(162.815763^2 -162.814995^2) = 3.14159*(26508.972681 -26508.722597) = 3.14159*(.250084) = .785661

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hmmm...

3.14159*(162.815763^2 -162.814995^2) = 3.14159*(26508.972681 -26508.722597) = 3.14159*(.250084) = .785661

oh, yeah. need to check my formulas - I combined area and perimeter into a formula that doesn't give any correct answer...

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• 0

Very interesting question. Forced me to do quite a little research (my geometry and trigonometry are pretty rusty).

The difference in area is less than 1.

the circumscribed circle's area is 273392545.50041819650470875183067 and the inscribed circle's area is 273392544.71502003310726044221501 leaving a difference of 0.78539816339744830961567197415016.

This is based on the radii of the circles being defined primary by the difference between the COSECANT of pi/1023 (circumscribed) versus the COTANGENT of pi/1023 (inscribed). Well, that and the length of the side times 0.5 times those things, which basically means 1/2 of what I said... Okay, I'm rambling now.

Edited by Smith
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• 0 I didn't use any trig to find my answer. I set up a right triangle with the radius of the inscribed circle the radius of the circumcribed circle and half the side length with the radius of the circumscribed circle as the hypotenuse. I meant to use trig to solve for the answer but a different method came to me. Let i be the radius of the inscribed circle and let c be the radius of the circumscribed circle. The difference in area is Pi*c^2-Pi*i^2. Now, by using the Pythagorean theorem, c^2=i^2+.5^2. By substitution Pi*c^2-Pi*i^2=Pi*(i^2+.5^2)-Pi*i^2. And Pi*(i^2+.5^2)-Pi*i^2=Pi*i^2+Pi*.5^2-Pi*i^2=Pi*.5^2. The difference in the areas of the two circles is Pi*.5^2 or approx .7854 which is not greater than 1.

How'd I do?

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• 0

philip1882, k-man, and rlgandy got it right. rlgandy had the simplest explanation.

I am a little confused (or maybe just hurt?) ... You say three answers were right, and they differ very slightly from one another, but not a fourth - which is similar to 22 significant digits to one of the three 'correct' answers? Why?

.785661 (philip1882)

approx .7854 (rglandy)

0.78539816339744830961566084581988 (k-man) (pi/4)

0.78539816339744830961567197415016 (smith) (???)

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• 0

I am a little confused (or maybe just hurt?) ... You say three answers were right, and they differ very slightly from one another, but not a fourth - which is similar to 22 significant digits to one of the three 'correct' answers? Why?

.785661 (philip1882)

approx .7854 (rglandy)

0.78539816339744830961566084581988 (k-man) (pi/4)

0.78539816339744830961567197415016 (smith) (???)

I'm so sorry, I must have answered too hastily. Yes, Smith surely got it right as well. Thanks for pointing it out to me.

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