Thread: Area of Square in Circle

1. Re: Area of Square in Circle

Originally Posted by harpazo
Improve our bounds???
Yes, at the moment we have:

$\displaystyle 2<\pi<4$

Surely, we can narrow the gap...

2. Re: Area of Square in Circle

Originally Posted by MarkFL
Yes, at the moment we have:

$\displaystyle 2<\pi<4$

Surely, we can narrow the gap...
Mark,

I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?

3. Re: Area of Square in Circle

Originally Posted by harpazo
Mark,

I would like to keep in touch with you via email. I can PM my personal email to you. I would like to keep all your replies to my questions on file as reference notes. Is this ok with you?
I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.

Can you think of a way to improve the bounds on pi we have so far?

4. Re: Area of Square in Circle

Originally Posted by MarkFL
I prefer using a forum for mathematical discourse, primarily because LaTeX is available. I can barely keep up with emails as it is.

Can you think of a way to improve the bounds on pi we have so far?
How can we improve the bounds on pi here?

5. Re: Area of Square in Circle

An easy way to determine the area is to see the square as a rhombus with the circle's diameter as the rhombus' diagonal. That way you can get its area without using those pesky square roots.

6. Re: Area of Square in Circle

Originally Posted by harpazo
How can we improve the bounds on pi here?
is there a simple formula for the area of a regular polygon?

7. Re: Area of Square in Circle

Originally Posted by harpazo
How can we improve the bounds on pi here?
Suppose we use hexagons rather than squares:

Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:

$\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$

And the area $\displaystyle A_L$ of the larger hexagon is:

$\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$

What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:

$\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$

Using a computer, we find:

 $\displaystyle n$ $\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122

It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.

8. Re: Area of Square in Circle

Originally Posted by MarkFL
Suppose we use hexagons rather than squares:

Let the radius of the circle be 1, so that its area is $\displaystyle \pi$ units squared. The area $\displaystyle A_S$ of the smaller hexagon is:

$\displaystyle A_S=6\left(\frac{1}{2} \sin\left(\frac{2\pi}{6}\right)\right)= \frac{3\sqrt{3}}{2}\approx2.6$

And the area $\displaystyle A_L$ of the larger hexagon is:

$\displaystyle A_L=6\left(\frac{1}{2} \left(\frac{2}{\sqrt{3}}\right)^2\sin\left(\frac{2 \pi}{6}\right)\right)= 2\sqrt{3}\approx3.5$

What do you think would happen if we used $\displaystyle n$-gons having more and more sides? Let's let $\displaystyle A_n$ be the area of an $\displaystyle n$-gon circumscribed by the circle...we have:

$\displaystyle A_n=\frac{n}{2}\sin\left(\frac{2\pi}{n}\right)$

Using a computer, we find:

 $\displaystyle n$ $\displaystyle A_n$ 10 2.938926261462366 100 3.1395259764656687 1000 3.1415719827794755 10000 3.141592446881286 1000000 3.141592653569122

It appears that as $\displaystyle n\to\infty$ we have $\displaystyle A_n$ approaching some fixed finite value.
What a great explanation! Nicely done!

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