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In my math textbook there's a demonstration for the calculus of the circumference of a circle that involves regular polygons inscribed in the circle, but I don't get it. The book gives the following formula: $$l_{2n}=\sqrt{2R^2 - R\sqrt{4R^2-l_{n}^2}}$$ Where $l_{n}$ is the side of a regular n-sided polygon inscribed in a circle with R radius. Then, somehow using that formula, the books tells me that: $$l_{4}=R\sqrt2$$ But how do I reach that number? What was substituted for what?

And what's the use of that formula? Suppose, for example, that a square is inscribed in a circle. I know the radius and want to find the side of the square. How would I apply the formula? It doesn't exist a polygon with 2 sides!

The book then applies the same formula to a polygon with 8 sides, then one with 16, then 32, then 64 and shows the larger the number of sides of a polygon, the closer its perimeter is to the circumference. I can understand that, but I don't understand how it can find a number such as $R\sqrt2$ for the side of square without knowing the side of a hypotetical 2-sided polygon.

I hope I've made my question clear.

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    $\begingroup$ Probably the author does not intend to say that the expression for $l_4$ follows from the formula. It does, sort of, since $l_2=2R$. That's a "degenerate" polygon, liable to cause confusion. But we can compute $l_4$ directly. The diagonal of the square is $2R$, so if $s$ is a side, then by the Pythagorean Theorem $s^2+s^2=4R^2$, so $s=R\sqrt{2}$. $\endgroup$
    – André Nicolas
    Oct 27, 2012 at 22:12

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Probably the author does not intend to say that the expression for $l_4$ follows from the formula. Actually, it does, sort of. We have $l_2=2R$, and substitution in the formula gives $l_4=R\sqrt{2}$. We have used the formula starting from a "degenerate" polygon. It happens to work, but is liable, for good reason, to cause confusion.

However, $l_4$ can be computed directly, using basic geometry. The diagonal of the square inscribed in the circle is $2R$. So if $s$ is the length of a side of the square, then by the Pythagorean Theorem $$s^2+s^2=(2R)^2=4R^2,$$ so $s^2=2R^2$ and therefore $s=R\sqrt{2}$.

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  • $\begingroup$ Could you please talk more about "degenerate polygons"? Or point me to resources about that? Because, although I'll not quote it here, the book explicitly tells that $l_{4}=R\sqrt2$ follows from that formula. Thanks! $\endgroup$
    – BeetleTheNeato
    Oct 27, 2012 at 22:22
  • $\begingroup$ Sorry, have no reference. If you plug $l_2=2R$ into the formula, you will note that the $R\sqrt{4R^2-l_2^2}$ term dies, and you get $l_4=\sqrt{2R^2}$, the right thing. The fact that the formula happens to work even in this degenerate case is not very interesting. Actually, if you look at the proof of the formula, it makes sense even in this case. But it is best to start where you have solid geometric grounding, namely at $3$ (which is what Archimedes did) or $4$. $\endgroup$
    – André Nicolas
    Oct 27, 2012 at 22:30
  • $\begingroup$ Sadly, the book proves the formula using trigonometry. It does mention that the same result can be achieved using square triangles but then it moves on to this method of calculating pi. I'd really like a proof using geometry because I have trouble picturing the figures. I'll search it. Thanks again. $\endgroup$
    – BeetleTheNeato
    Oct 27, 2012 at 22:37

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