More Symmetries

(This solution is just a generalization of the one given in the "More Tangents" problem.)

Consider the graphs of $y=x^{2n+1}$ and $y=\sqrt[2n+1]{x}$ . As these functions are inverses, they are reflections of each other across the line $y=x$ . However, they are both also odd functions. The combination of these two facts means that they are also reflections of each other across the line $y=-x$ , and therefore lines $AB$ and $A'B'$ are both perpendicular to $y=-x$ , i.e they both have slope 1. Likewise, lines $A''B''$ and $A'''B'''$ have slope -1.

Deriving $y=x^{2n+1}$ and setting this derivative equal to 1 will lead us to the coordinates of points $A$ and $B'$ .

$y' = (2n+1)x^{2n} = 1 \quad \Rightarrow x=\pm \dfrac{1}{\sqrt[2n]{2n+1}} \quad \Rightarrow y=\pm \dfrac{1}{(2n+1)\sqrt[2n]{2n+1}}$

Thus $A$ has coordinates $\left( \dfrac{-1}{\sqrt[2n]{2n+1}},\dfrac{-1}{(2n+1)\sqrt[2n]{2n+1}} \right)$ and $B'$ has coordinates $\left( \dfrac{1}{\sqrt[2n]{2n+1}},\dfrac{1}{(2n+1)\sqrt[2n]{2n+1}} \right)$ . $$ Using the same technique with the other three functions, we find the coordinates of all eight points of the octagon.

$A\left( \dfrac{-1}{\sqrt[2n]{2n+1}},\dfrac{-1}{(2n+1)\sqrt[2n]{2n+1}} \right); \quad A'\left( \dfrac{-1}{(2n+1)\sqrt[2n]{2n+1}},\dfrac{-1}{\sqrt[2n]{2n+1}} \right); \quad A''\left( \dfrac{1}{(2n+1)\sqrt[2n]{2n+1}},\dfrac{-1}{\sqrt[2n]{2n+1}} \right); \quad A'''\left( \dfrac{1}{\sqrt[2n]{2n+1}},\dfrac{-1}{(2n+1)\sqrt[2n]{2n+1}} \right)$

$B\left( \dfrac{1}{(2n+1)\sqrt[2n]{2n+1}},\dfrac{1}{\sqrt[2n]{2n+1}} \right); \quad B'\left( \dfrac{1}{\sqrt[2n]{2n+1}},\dfrac{1}{(2n+1)\sqrt[2n]{2n+1}} \right); \quad B''\left( \dfrac{-1}{\sqrt[2n]{2n+1}},\dfrac{1}{(2n+1)\sqrt[2n]{2n+1}} \right); \quad B'''\left( \dfrac{-1}{(2n+1)\sqrt[2n]{2n+1}},\dfrac{1}{\sqrt[2n]{2n+1}} \right)$ $$ The equation of line $AA'$ is clearly $x+y=\dfrac{-2(n+1)}{(2n+1)\sqrt[2n]{2n+1}}$ ; its $x$ -intercept is $\dfrac{-2(n+1)}{(2n+1)\sqrt[2n]{2n+1}}$ .

The equation of line $B''B'''$ is clearly $x-y=\dfrac{-2(n+1)}{(2n+1)\sqrt[2n]{2n+1}}$ ; its $x$ -intercept is also $\dfrac{-2(n+1)}{(2n+1)\sqrt[2n]{2n+1}}$ .

Thus lines $AA'$ and $B''B'''$ intersect on the negative half of the $x$ -axis, let's say at $\color{#20A900}P$ . $$ In the same way we can see that lines $AA'$ and $A''A'''$ intersect on the negative half of the $y$ -axis, say at $\color{#20A900}Q$ ; lines $A''A'''$ and $BB'$ intersect on the positive half of the $x$ -axis, let's say at $\color{#20A900}\color{#20A900}R$ ; and finally lines $BB'$ and $B''B'''$ intersect on the positive half of the $y$ -axis, let's say at $\color{#20A900}S$ .

So $PQRS$ must be a square. $PR$ is a diagonal of the square, and its length $d$ is clearly $\dfrac{4(n+1)}{(2n+1)\sqrt[2n]{2n+1}}$ . Then the area of $PQRS$ is $\dfrac{d^2}{2}=\dfrac{8(n+1)^2}{(2n+1)^2\sqrt[n]{2n+1}}$ . $$ To find the area of the octagon, we must remove the four corners of square $PQRS$ . Consider one of those four corners, the triangle $AB''P$ . It is a right isosceles triangle with hypotenuse of length $h=\dfrac{2}{(2n+1)\sqrt[2n]{2n+1}}$ . Then the area of triangle $AB''P$ is $\dfrac{h^2}{4}=\dfrac{1}{(2n+1)^2\sqrt[n]{2n+1}}$ . The other three corners are clearly the same size. $$ Thus we get

$A_n = \dfrac{8(n+1)^2}{(2n+1)^2\sqrt[n]{2n+1}} - 4 \left( \dfrac{1}{(2n+1)^2\sqrt[n]{2n+1}} \right) \\ = \dfrac{8(n+1)^2-4}{(2n+1)^2\sqrt[n]{2n+1}}$

Finally, and making use the fact that $\displaystyle\lim_{n\rightarrow \infty} \sqrt[n]{n} = 1$ , we have

$\displaystyle\lim_{n\rightarrow \infty} A_n = \displaystyle\lim_{n\rightarrow \infty} \dfrac{8(n+1)^2-4}{(2n+1)^2\sqrt[n]{2n+1}} = \dfrac{8}{4} = \boxed{2}$

Let $f(x) = x^{2n + 1}$ and $g(x) = x^{\frac{1}{2n + 1}} \implies \dfrac{d}{dx}(f(x))|_{x = a} = (2n + 1)a^{2n}$ and $\dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{(2n + 1)b^{\frac{2n}{2n + 1}}} \implies (2n + 1)a^{2n} = \dfrac{1}{(2n + 1)b^{\frac{2n}{2n + 1}}} \implies a = \pm\dfrac{1}{(2n + 1)^{\frac{1}{n}}b^{\frac{1}{2n + 1}}}$

Using point $A:(\dfrac{-1}{(2n + 1)^{\frac{1}{n}}b^{\frac{1}{2n + 1}}},\dfrac{-1}{(2n + 1)^{\frac{2n + 1}{n}}b})$ and $B: (b,b^{\frac{1}{2n + 1}}) \implies$
slope $m_{AB} = \dfrac{(2n + 1)^{\frac{2n + 1}{n}}b^{\frac{2(n + 1)}{2n + 1}} + 1}{(2n + 1)^2 b^{\frac{2n}{2n + 1}}((2n + 1)^{\frac{1}{n}}b^{\frac{2(n + 1)}{2n + 1}} + 1)}$ $= \dfrac{1}{(2n + 1)b^{\frac{2n}{2n + 1}}}$ $\implies b = \dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}}$ $\implies a = \dfrac{-1}{(2n + 1)^{\frac{1}{2n}}} \implies$

$A: ( \dfrac{-1}{(2n + 1)^{\frac{1}{2n}}}, \dfrac{-1}{(2n + 1)^{\frac{2n + 1}{2n}}})$ and $B: (\dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}}, \dfrac{1}{(2n + 1)^{\frac{1}{2n}}}) \implies m_{AB} = 1 \implies y = x + \dfrac{2n}{(2n + 1)^{\frac{2n + 1}{2n}}}$

Using the symmetry about the line $y = x \implies$

$A^{'}: ( \dfrac{-1}{(2n + 1)^{\frac{2n + 1}{2n}}}, \dfrac{-1}{(2n + 1)^{\frac{1}{2n}}})$ and $B^{'}: (\dfrac{1}{(2n + 1)^{\frac{1}{2n}}}, \dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}}) \implies m_{A^{'}B^{'}} = 1 \implies y = x - \dfrac{2n}{(2n + 1)^{\frac{2n + 1}{2n}}}$

Using the symmetry about the $y$ axis $\implies$

$A^{'} \rightarrow A^{''} = (\dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}}, \dfrac{-1}{(2n + 1)^{\frac{1}{2n}}})$ and $B^{'} \rightarrow B^{''} = (\dfrac{-1}{(2n + 1)^{\frac{1}{2n}}}, \dfrac{1}{(2n + 1)^{\frac{2n + 1}{2n}}})$

Using the symmetry about the line $y = -x \implies$

$A^{''} \rightarrow A^{'''} = (\dfrac{1}{(2n + 1)^{\frac{1}{2n}}}, \dfrac{-1}{(2n + 1)^{\frac{2n + 1}{2n}}})$ and $B^{''} \rightarrow B^{'''} = (\dfrac{-1}{(2n + 1)^{\frac{2n + 1)}{2n}}}, \dfrac{1}{(2n + 1)^{\frac{1}{2n}}})$ .

$AB^{''} = \dfrac{2}{(2n + 1)^{\frac{2n + 1}{2n}}} = A^{'}A^{''} = BB^{'''} = A^{'''}B{'}$

and,

$AA^{'} = \dfrac{2\sqrt{2}n}{(2n + 1)^{\frac{2n + 1}{2n}}} = BB^{'} = B^{''}B{'''} = A^{''}A{'''}$

For $\triangle{B^{'''}OB} \implies A_{\triangle{B^{'''}OB}} = \dfrac{1}{2}(BB^{'''})(OM^{*}) = \dfrac{1}{(2n + 1)^{\frac{n + 1}{n}}}$ , where $OM^{*} = \dfrac{1}{(2n + 1)^{\frac{1}{2n}}}$ .

For $\triangle{AOA^{'}}$ the midpoint $M$ of $AA^{'}$ is $M: (\dfrac{-(n + 1)}{(2n + 1)^{\frac{2n + 1}{2n}}}, \dfrac{-(n + 1)}{(2n + 1)^{\frac{2n + 1}{2n}}}) \implies OM = \dfrac{\sqrt{2}(n + 1)}{(2n + 1)^{\frac{2n + 1}{2n}}} \implies$

$A_\triangle{AOA^{'}} = \dfrac{1}{2}(AA')(OM) = \dfrac{2n(n + 1)}{(2n + 1)^{\frac{2n + 1}{n}}}$ .

$\implies A_{n} = 4(\triangle{AOA^{'}} + \triangle{B^{'''}OB}) = 4(\dfrac{2n^2 + 4n + 1}{(2n + 1)^2 (2n + 1)^{\frac{1}{n}}})$ .

For $\lim_{n \rightarrow \infty} \:\ A_{n}$ :

$y = \lim_{n \rightarrow \infty} (\dfrac{1}{2n + 1})^{\frac{1}{n}} \implies \ln(y) = \lim_{n \rightarrow \infty} \dfrac{1}{n}\ln(\dfrac{1}{2n + 1}) = \lim_{n \rightarrow \infty} \dfrac{-2}{(2n + 1)} = 0 \implies y = e^0 = 1$

$\implies \lim_{n \rightarrow \infty} A_{n} = \boxed{2}$ .

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Note: $\lim_{n \rightarrow \infty} A_{n} = A_{PQRS}$ .