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- Thread starter Bushy
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- #1

- Feb 13, 2012

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... so that for n=2 is $\displaystyle A= \tan \frac{\pi}{2} = \infty$?... it seems a little improbable so that let's compute the base and the height of each triangle...Hi there, the problem says, an n-gon is circumscribed around a circle so the mid point of each side is tangent to the circle.

Prove the triangle consisting of one side of the n-gon and the sides from the end points to the middle of the circle has area

tan(pi/n)

Cheers!

$\displaystyle b = 2\ \sin \frac{\pi}{n}$

$\displaystyle h = \cos \frac{\pi}{n}$

... and the area is...

$\displaystyle A= \frac{b\ h}{2} = \sin \frac{\pi}{n}\ \cos \frac{\pi}{n}$

Kind regards

$\chi$ $\sigma$

P.S. I mistakenly undestood 'inscribed' instead of 'circumscribed' and that explains my answer... very sorry! ...

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- #3

- Jan 26, 2012

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Does the problem mention the radius of the circle? The area of the triangle you describe is going to have to depend on $r$. I could believe the result either if you were asked to show that the area is $r^{2} \tan \left( \tfrac{ \pi}{n} \right)$, or if the circle was assumed to have a radius of $1$.Hi there, the problem says, an n-gon is circumscribed around a circle so the mid point of each side is tangent to the circle.

Prove the triangle consisting of one side of the n-gon and the sides from the end points to the middle of the circle has area

tan(pi/n)

Cheers!

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- #5

- Jan 26, 2012

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So, imagine a sector of the circle, corresponding to the triangle whose area you want to find. What is its angle? Start labeling sides, and then form the area of the triangle. What does that give you? (I'd recommend drawing a picture!)

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The angle is 2*pi /n, I cannot find any side lengths.So, imagine a sector of the circle, corresponding to the triangle whose area you want to find. What is its angle? Start labeling sides, and then form the area of the triangle. What does that give you? (I'd recommend drawing a picture!)

If I halve the trianglefrom the centre of the circle the angle becomes pi/n and one of the side lengths becomes 1 from the radius. If we call the lengthof the tangent = AB then the other side length from the circle becomes

(1^2+(1/2 AB)^2)^(1/2) using the right angle.

Not sure if im heading off track here...

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- #7

- Jan 26, 2012

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You've definitely made some progress, but I don't think the Pythagorean Theorem is the best thing to do next. You don't, off-hand, know the length of the side from the point of tangency to the edge of the polygon side. So call it something. Do you now have a triangle you can work with?The angle is 2*pi /n, I cannot find any side lengths.

If I halve the trianglefrom the centre of the circle the angle becomes pi/n and one of the side lengths becomes 1 from the radius. If we call the lengthof the tangent = AB then the other side length from the circle becomes

(1^2+(1/2 AB)^2)^(1/2) using the right angle.

Not sure if im heading off track here...

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- #8

You've definitely made some progress, but I don't think the Pythagorean Theorem is the best thing to do next. You don't, off-hand, know the length of the side from the point of tangency to the edge of the polygon side. So call it something. Do you now have a triangle you can work with?

That length was not given, I have called it 1/2*AB in my attempt above.

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- #9

- Jan 26, 2012

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That seems an overly complicated name for it. I would just go with something like $y$. So you have a right triangle. The angle at the center of the polygon is $\pi/n$. The apothem is $1$. From that data, can you compute $y$? And then what is the area of the right triangle? And then what is the area of the whole triangle?That length was not given, I have called it 1/2*AB in my attempt above.