Quick and cute summer Sangaku #9

• The smallest possible circle will be tangent to the intersecting point between the curves $\boxed{y=x^{3}}$ and $\boxed{y=-x^{3}}$ , this point is the origin $\boxed{O(0;0)}$ .
• $\boxed{P(0;y_P)}$ is the center of the circle, then its radius is $\boxed{y_P}$
• The equation of the circle is : $\boxed{x²+(y-y_P)²-y_P²=0}$
• Since there is common point between $\boxed{x²+(y-y_P)²-y_P²=0}$ and $\boxed{y=x^{3}}$ , we can substitute $\boxed{y}$ with $\boxed{x²}$ :
• $\boxed{x²+(x²-y_P)²-y_P²=0}$ $<=>$ $\boxed{x^2+x^6-2\cdot \:y_P\cdot \:x^3=0}$ $<=>$ $\boxed{x^4-2xy_P+1=0}$
• The discriminant of this quartic equation needs to be equal to zero since there is tangency between the circle and the curve.
• $\boxed{Δ=-432y_P^4+256=0}$ $<=>$ $\boxed{y_P=\frac{2}{3^{\frac{3}{4}}}}$
• Then the area of the circle is $\boxed{\frac{4\pi }{\sqrt{27}}}$
• $\boxed{a+b=4+27=31}$

Let's take a calculus approach to Valentin's solution. If $x^2 + (y-r)^2 = r^2$ be tangent to the symmetric curves $y = x^3$ and $y =-x^3$ at the points $(0,0); (x_{0}, x^{3}_{0}); (-x_{0}, x^{3}_{0})$ , then WLOG the right-hand branch satisfies:

$x^2 + (x^3 - r)^2 = r^2 \Rightarrow x^2 + x^6 - 2rx^3 + r^2 = r^2 \Rightarrow \boxed{r(x) = \frac{x^3}{2} + \frac{1}{2x}}$ .

The first derivative equal to zero yields:

$r'(x) = 0 \Rightarrow \frac{3x^2}{2} - \frac{1}{2x^2} = 0 \Rightarrow x = \frac{1}{3^{1/4}}$ ;

and the second derivative at this critical point yields:

$r''(x) = 3x + \frac{1}{x^3} \Rightarrow r( \frac{1}{3^{1/4}}) > 0$ (hence a global minimum).

NOTE: By $y-$ axis symmetry, $x = -\frac{1}{3^{1/4}}$ is the critical point for the left-hand branch. Thus, the area of this minimal circle computes to $\pi \cdot [r(\frac{1}{3^{1/4}})]^2 = \boxed{\frac{4\pi}{\sqrt{27}}}$