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This goes on in Chapter 8, on least upper bounds and related topics. I have proven $(a),(b),(c)$.

The sketch is.

$(a)$ If $\{a_n\}$ is a sequence of positive terms such that $$a_{n+1}\leq a_n/2$$ then for every $\epsilon>0$ there exists an $n$ with $a_n<\epsilon$.

$(b)$ Let $P$ be a regular polygon inscribed in a circle. Let $P'$ be the regular polygon with twice the sides of $P$. Let $A$ be the area of $P$, $A'$ be the area of $P'$ and $C$ that of the circle. Prove

$$(C-P')\leq \frac 1 2 (C-P)$$

$(c)$ Prove there is a regular incsribed polygon with area as close as that to the circle.

This follows directly from $(a)$ and $(b)$.

Finally, he says

$(d)$ By using the fact the areas of two regular polygons with the same number of sides are in the same relation as the squares of their sides, prove the areas of two circles are in the same relation as the squares of their radii. Deduce this by showing that assuming otherwise leads to a contradiction. This should be achieved by inscribing adequate polygons.

Now, I'm not much of a geometer. Is "the areas of two regular polygons with the same number of sides are in the same relation as the squares of their sides" hard to prove? And how does " in the same relation as the squares of their sides" connect to " in the same relation as the squares of their radii."? Finally, could you hint about what adequate polygons should be inscribed? Or is he just asking to get to $1/2^n$ and use $(c)$?

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Once I saw a bumper sticker: What would Spivak do? –  Bill Dubuque Sep 12 '12 at 3:11
    
The area of a regular $n$-gon is $n$ times the area of a certain isosceles triangle. The area of that triangle is proportional to the square of its base. That gives first result that you asked about. –  Brian M. Scott Sep 12 '12 at 3:11
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Suppose otherwise, i.e. that there are two circles $C_1$ and $C_2$ with radii $r_1$ and $r_2$ and areas $A_1$ and $A_2$ such that $A_1/A_2\neq r_1^2/r_2^2$. You have already shown that there is a sequence $P_1^{(n)}$ of polygons with $2^n$ sides inscribed in $C_1$ whose areas converge to $A_1$ by (c), and similarly a sequence $P_2^{(n)}$ for $C_2$. Thus for sufficiently large $n$ the ratio of their areas is not $r_1^2/r_2^2$. But their side lengths are proportional to their radius, and their area is proportional to the square of the side lengths. Hence the ratio of their areas is $r_1^2/r_2^2$, a contradiction.

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