A geometric property of the graph of $y = x^2$ Consider $n\geq 3$ and two points $A, B$ on the graph of $y = x^2$. Now choose points $P_1,...,P_{n-2}$ on this graph such that the area of the convex $n$-gon $AP_1...P_{n-2}B$ is maximum (they do exist). Let $S_n$ be the area of this $n$-gon and let $S$ be the area between the chord $AB$ and the graph. I want to show that $\frac {S_n}{S}$ is only a function of $n$ and is independent from (the position of) $A$ and $B$. 
 A: The area between $AB$ and the graph is just
$$\int_a^b a^2(1-\frac{x-a}{b-a})+b^2\frac{x-a}{b-a}-x^2=a^2x-\frac{a^2}{b-a}(\frac12 x^2-ax)+\frac{b^2}{b-a}(\frac 12x^2-ax)-\frac13x^3\bigg|_a^b$$
$$=a^2b-a^3+\frac12(b+a)(a-b)^2-\frac13 b^3+\frac13 a^3$$
$$=(a-b)(-a^2+\frac12(a^2-b^2)+\frac 13(a^2+ab+b^2))$$
$$=\frac 16(b-a)(a^2-2ab+b^2)=\frac16(b-a)^3$$
Now we note: maximizing the convex polygon is the same as minimizing the collection of smaller subarcs underneath them. If my points are $a_i$, with $a_0=a$ and $a_n=b$, we want to find the $a_i, 0<i<n$ that minimize
$$A=\frac 16\sum_{i=0}^{n-1}(a_{i+1}-a_i)^3$$
now we just use some simple calculus:
$$\frac{\partial A}{\partial a_i}=\frac 16\left(3(a_i-a_{i-1})^2-3(a_{i+1}-a_i)^2\right)$$
Solving $$\frac{\partial A}{\partial a_i}=0$$ gives
$$(a_i-a_{i-1})^2=(a_{i+1}-a_i)^2$$
Since we assume that $\forall i a_{i+1}>a_i$:
$$a_i-a_{i-1}=a_{i+1}-a_i$$
$$a_i=\frac12(a_{i+1}+a_{i-1})$$
In other words, the $x$-coordinate of each point is the average of its neighbors. This happens when the points are equally spaced on the $x$-axis. In this case, $a_i=a+(b-a)\frac in$, $a_{i+1}-a_i=\frac{b-a}n$, and the total area of these smaller arcs is
$$A=\frac 16\sum_{i=0}^{n-1}(\frac{b-a}n)^3=\frac 16\frac{(b-a)^3}{n^2}$$
The area of the polygon $S_n$ is then $S-A$:
$$\frac16(b-a)^3(1-\frac 1{n^2})$$
and we see that $\frac{S_n}S$ is independent of $a$ and $b$, and only dependent on $n$.
