More Tangents

Consider the graphs of $y=x^3$ and $y=x^{1/3}$ . 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 $y=x^3$ and setting this derivative equal to 1 will lead us to the coordinates of points $A$ and $B'$ .

$y' = 3x^2 = 1 \Rightarrow x=\pm \dfrac{1}{\sqrt{3}} \Rightarrow y=\pm \dfrac{1}{3\sqrt{3}}$

Thus $A$ has coordinates $\left( \dfrac{-1}{\sqrt{3}},\dfrac{-1}{3\sqrt{3}} \right)$ and $B'$ has coordinates $\left( \dfrac{1}{\sqrt{3}},\dfrac{1}{3\sqrt{3}} \right)$ .

Using the same technique with the other three functions, we find the coordinates of all eight points of the octagon.

$A\left( \dfrac{-1}{\sqrt{3}},\dfrac{-1}{3\sqrt{3}} \right); \quad A'\left( \dfrac{-1}{3\sqrt{3}},\dfrac{-1}{\sqrt{3}} \right); \quad A''\left( \dfrac{1}{3\sqrt{3}},\dfrac{-1}{\sqrt{3}} \right); \quad A'''\left( \dfrac{1}{\sqrt{3}},\dfrac{-1}{3\sqrt{3}} \right)$

$B\left( \dfrac{1}{3\sqrt{3}},\dfrac{1}{\sqrt{3}} \right); \quad B'\left( \dfrac{1}{\sqrt{3}},\dfrac{1}{3\sqrt{3}} \right); \quad B''\left( \dfrac{-1}{\sqrt{3}},\dfrac{1}{3\sqrt{3}} \right); \quad B'''\left( \dfrac{-1}{3\sqrt{3}},\dfrac{1}{\sqrt{3}} \right)$

The equation of line $AA'$ is clearly $x+y=\dfrac{-4}{3\sqrt{3}}$ ; its $x$ -intercept is $\dfrac{-4}{3\sqrt{3}}$ . The equation of line $B''B'''$ is clearly $x-y=\dfrac{-4}{3\sqrt{3}}$ ; its $x$ -intercept is also $\dfrac{-4}{3\sqrt{3}}$ . 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 $\color{#20A900}S$ .

Then $PQRS$ must be a square. $PR$ is a diagonal of the square, and its length $d$ is clearly $\dfrac{8}{3\sqrt{3}}$ . Then we can find the area of $PQRS$ to be $\dfrac{d^2}{2}=\dfrac{32}{27}$ .

To find the area of the octagon in question, 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 $h=\dfrac{2}{3\sqrt{3}}$ . Then the area of triangle $AB''P$ is $\dfrac{h^2}{4}=\dfrac{1}{27}$ . The other three corners are clearly the same size.

Finally, the area of the octagon in question is $\dfrac{32}{27} - 4 \left( \dfrac{1}{27} \right) = \dfrac{28}{27} = \dfrac{2^2\cdot7}{3^3}$ . Thus $a=2, b=7, c=3$ and $a+b+c=\boxed{12}$

Let $f(x) = x^3$ and $g(x) = x^{\frac{1}{3}} \implies \dfrac{d}{dx}(f(x))|_{x = a} = 3a^2$ and $\dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{3b^{\frac{2}{3}}} \implies 3a^2 = \dfrac{1}{3b^{\frac{2}{3}}} \implies a = \pm\dfrac{1}{3b^{\frac{1}{3}}}$

Using point $A:(\dfrac{-1}{3b^{\frac{1}{3}}},\dfrac{-1}{27b})$ and $B: (b,b^{\frac{1}{3}}) \implies$ slope $m_{AB} = \dfrac{27b^{\frac{4}{3}} + 1}{9b^{\frac{2}{3}}(3b^{\frac{4}{3}} + 1)} = \dfrac{1}{3b^{\frac{2}{3}}}$ $\implies b = \dfrac{1}{3^{\frac{3}{2}}}$ $\implies a = \dfrac{-1}{3^{\frac{1}{2}}} \implies$

$A: ( \dfrac{-1}{3^{\frac{1}{2}}}, \dfrac{-1}{3^{\frac{3}{2}}})$ and $B: (\dfrac{1}{3^{\frac{3}{2}}}, \dfrac{1}{3^{\frac{1}{2}}}) \implies m_{AB} = 1 \implies y = x + \dfrac{2}{3^{\frac{3}{2}}}$

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

$A^{'}: ( \dfrac{-1}{3^{\frac{3}{2}}}, \dfrac{-1}{3^{\frac{1}{2}}})$ and $B^{'}: (\dfrac{1}{3^{\frac{1}{2}}}, \dfrac{1}{3^{\frac{3}{2}}}) \implies m_{A^{'}B^{'}} = 1 \implies y = x - \dfrac{2}{3^{\frac{3}{2}}}$

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

$A^{'} \rightarrow A^{''} = (\dfrac{1}{3^{\frac{3}{2}}}, \dfrac{-1}{3^{\frac{1}{2}}})$ and $B^{'} \rightarrow B^{''} = (\dfrac{-1}{3^{\frac{1}{2}}}, \dfrac{1}{3^{\frac{3}{2}}})$

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

$A^{''} \rightarrow A^{'''} = (\dfrac{1}{3^{\frac{1}{2}}}, \dfrac{-1}{3^{\frac{3}{2}}})$ and $B^{''} \rightarrow B^{'''} = (\dfrac{-1}{3^{\frac{3}{2}}}, \dfrac{1}{3^{\frac{1}{2}}})$ .

$AB^{''} = \dfrac{2}{3\sqrt{3}} = A^{'}A^{''} = BB^{'''} = A^{'''}B{'}$

and,

$AA^{'} = \dfrac{2}{3}\sqrt{\dfrac{2}{3}} = BB^{'} = B^{''}B{'''} = A^{''}A{'''}$

For $\triangle{B^{'''}OB} \implies A_{\triangle{B^{'''}OB}} = \dfrac{1}{2}(BB^{'''})(OM^{*}) = \dfrac{1}{2}(\dfrac{2}{3\sqrt{3}})(\dfrac{1}{\sqrt{3}}) = \dfrac{1}{9}$

For $\triangle{AOA^{'}}$ the midpoint $M$ of $AA^{'}$ is $M: (\dfrac{-2}{3\sqrt{3}},\dfrac{-2}{3\sqrt{3}}) \implies OM = \dfrac{2}{3}\sqrt{\dfrac{2}{3}} \implies$

$A_\triangle{AOA^{'}} = \dfrac{1}{2}(AA')(OM) = \dfrac{1}{2}(\dfrac{2}{3}\sqrt{\dfrac{2}{3}})^2 = \dfrac{4}{27}$

$\therefore A_{octagon} = 4(\dfrac{1}{9} + \dfrac{4}{27}) = 4(\dfrac{7}{27}) = \dfrac{2^2 * 7}{3^3} = \dfrac{a^a * b}{c^c} \implies a + b + c = \boxed{12}$ .