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- Thread starter Loren Booda
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arildno

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Now, clearly we can form a radius sequence for each n-gon, where the radius for each n-gon [itex]R_{n}[/itex] is given by the formula:

[tex]R_{n}=\sqrt{\frac{2A}{n\sin(\frac{2\pi}{n})}}[/tex]

This value is probably needed to solve your problem in some manner.

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HallsofIvy

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Assuming you mean a sequence of polygon inscribed in a circle of radius R, each n-gon can be interpreted as n isosceles triangle with congruent sides of length R and angle between them of [itex]2\pi/n[/itex] which can then be divided into two right angles with angle [itex]\pi/n[/itex]. The base of each such triangle is [itex]2R sin(\pi/n)[/itex] and the height is [itex]R cos(\pi/n)[/itex] so the area of each triangle is [itex]R^2 sin(\pi/n) cos(\pi/n)[/itex] and the area of the entire n-gon is [itex]nR^2 sin(\pi/n) cos(\pi/n)[/itex].

Since you are asking about the area inside the circle NOT in the polygon, that would be [itex]\pi R^2- nR^2 sin(\pi/n) cos(\pi/n)[/itex] and the fraction of the area of the circle not occupied by the n-gon would be

[tex]\frac{\pi- n sin(\pi/n) cos(\pi/n)}{\pi}= 1-\frac{n}{\pi} sin(\pi/n)cos(\pi/n)[tex].

It's easy to see that the last term of that goes to 1 in the limit and the "fraction of the area of the circlee not occupied by the n-gon", of course, goes to 0.

I'm not sure what you mean by "fraction of the area not occupied by**successive polygons**".

Since you are asking about the area inside the circle NOT in the polygon, that would be [itex]\pi R^2- nR^2 sin(\pi/n) cos(\pi/n)[/itex] and the fraction of the area of the circle not occupied by the n-gon would be

[tex]\frac{\pi- n sin(\pi/n) cos(\pi/n)}{\pi}= 1-\frac{n}{\pi} sin(\pi/n)cos(\pi/n)[tex].

It's easy to see that the last term of that goes to 1 in the limit and the "fraction of the area of the circlee not occupied by the n-gon", of course, goes to 0.

I'm not sure what you mean by "fraction of the area not occupied by

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arildno

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However, he explicitly states that we are talking of polygons having the SAME area, that is their vertices lie on different circles!

Note therefore that the sequence of radii is decreasing, if I'm not mistaken.

Thus, there will be area bits left that is not covered by subsequent polygons.

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Sorry, by "a circle" I meant that a sequence of regular polygons of equal area and n sides, as n approaches infinity, approaches a circle of equal area.

arildno's formula, in an infinite series, might be used to determine my sequence - the fractional areas of regular polygons that are not included within successively sided, concentric, and bilaterally symmetric regular polygons of equal area.

I believe his second post captures the gist of what I am proposing.

HallsofIvy, would you repost your last formula?

arildno's formula, in an infinite series, might be used to determine my sequence - the fractional areas of regular polygons that are not included within successively sided, concentric, and bilaterally symmetric regular polygons of equal area.

I believe his second post captures the gist of what I am proposing.

HallsofIvy, would you repost your last formula?

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If a coaxial N+1 sided polygon is rotated with respect to the previous N sided regular polygon of the same area, would the amount of the N sided polygon that is not covered be changed?

However, he explicitly states that we are talking of polygons having the SAME area, that is their vertices lie on different circles!

Note therefore that the sequence of radii is decreasing, if I'm not mistaken.

Thus, there will be area bits left that is not covered by subsequent polygons.

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arildno

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Adding some, subtracting some..seems to become zero change..:blush:

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I dont think it is simple, for instance what if they were both pentagons instead of polygons with a different number of sides?Adding some, subtracting some..seems to become zero change..:blush:

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HallsofIvy

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[tex]\frac{\pi- n sin(\pi/n) cos(\pi/n)}{\pi}= 1-\frac{n}{\pi} sin(\pi/n)cos(\pi/n)[/tex].

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You also said concentric polygons, so I take your posts to specify that each polygon has the same center axis and that a perpendicular bisector of the bottom side the n+1 polygon coincides with a perpendicular bisector of the bottom side of the previous n sided polygon, though the distance to the bottom side is shorter with each subsequent polygon.

Still I have difficulty comming up with a plot of the respective polygons in polar coordinates. Is it sufficient for your purpose to just add up the area of the N-sided polygons that lie outside the radius of a circle of the same area?

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Some derivation of HallsofIvy's formula would probably do the trick, though. Thanks all for your patience.

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