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This is question (b) from problem 11 in Chapter 8 and it has to do with approximating the area of the circle by inscribing appropriate polygons. If $P$ is a regular polygon inscribed inside a circle and $P'$ is the inscribed regular polygon with twice as many sides, then we are asked to prove that the difference between the area of the circle and the area of $P'$ is less than half the difference between the area of the circle and the area of $P$ (use Figure 7)

When I first encountered that problem I tried to go for an analytic solution. By limiting ourselves to the triangle formed by one side of $P$ and two radii as shown in Figure 7, we quickly arrive at the inequality $x-\sin{x}<\sin{x} - \sin{x}\cos{x}$. Solving the previous inequality is really hard and if we consider the fact that trig functions have not been introduced in the text yet, we may conclude that a geometric approach is required. Since I was unable to come up with one of my own, I decided to look at the solution.

Solution: Let $R_i$ be the area of region number $i$ in the attached figure. We must show that $R_2 < \frac{1}{2} (R_1 + R_2)$, or $R_2 < R_1$. This is clear, since $R_2 < R_2 + R_3 = R_1$.

My problem with this proof lies in how much it relies on the visual information of the diagram. It feels hand-wavy. I was wondering if another more convincing/rigorous geometric proof is possible.

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First note that if $(a,b)$ and $(c,d)$ are vectors in the plane and $\|(a,b)\| = \sqrt{a^2+b^2}$ denotes the length of the vector $(a,b)$, then $$ \|(a,b)+(c,d)\|\leq \|(a,b)\|+\|(c,d)\| $$ since $$ \begin{split} (a+c)^2+(b+d)^2 &= a^2+c^2 + 2(ac+bd) + b^2+d^2 \\ &\leq (a^2+c^2) +2\sqrt{a^2+b^2}\sqrt{c^2+d^2} + (b^2+d^2) \\ &= (\sqrt{a^2+b^2}+\sqrt{c^2+d^2})^2 \end{split} $$ where in the second line we used the Cauchy-Schwarz inequality, so equality holds if and only if the vectors are collinear.

The arc of the circle

Now suppose we choose coordinates so that the point Q at the bottom-right of the rectangle in the figure above (a pathetic attempt at re-drawing the picture in the solution referenced in the question) are $(1,0)$ and the coordinates of the point $P$ are $(a,b)$. Then arc of the circle between $P$ and $Q$ consists of points $(r,s)$ with $a<r<1$ and $0<s<b$, in other words the arc lies inside the rectangle divided into two triangles $R_1$ and $R_2$.

But now the points in the triangle $OPQ$ all have coordinates $s_1(a,b)+s_2(1,0)$ where $0\leq s_1+s_2\leq 1$, and hence -- apart from P and Q themselves -- are at distance $$ \|(s_1a+s_2,s_1b)\| < s_1\|(a,b)\|+s_2\|(1,0)\| = s_1+s_2 =1 $$ and hence lie inside the circle. But then the arc of the circle from $P$ to $Q$ must lie entirely in $R_2$, and so that portion of the area of the circle lying outside the triangle $R_1$ can be no larger than the area of $R_2$ which is equal to the area of $R_1$.

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