HMMT-Feb 2015 geo, Problem 5

Let $I$ be the set of points $(x, y)$ in the Cartesian plane such that \[ x>\left(\frac{y^{4}}{9}+2015\right)^{1 / 4} \] Let $f(r)$ denote the area of the intersection of $I$ and the disk $x^{2}+y^{2} \leq r^{2}$ of radius $r>0$ centered at the origin $(0,0)$. Determine the minimum possible real number $L$ such that $f(r)<L r^{2}$ for all $r>0$.

Solution: Let $B(P, r)$ be the (closed) disc centered at $P$ with radius $r$. Note that for all $(x, y) \in I, x>0$, and $x>\left(\frac{y^{4}}{9}+2015\right)^{1 / 4}>\frac{|y|}{\sqrt{3}}$. Let $I^{\prime}=\{(x, y): x \sqrt{3}>|y|\}$. Then $I \subseteq I^{\prime}$ and the intersection of $I^{\prime}$ with $B((0,0), r)$ is $\frac{\pi}{3} r^{2}$, so the $f(r)=$ the area of $I \cap B((0,0), r)$ is also less than $\frac{\pi}{3} r^{2}$. Thus $L=\frac{\pi}{3}$ works. On the other hand, if $x>\frac{|y|}{\sqrt{3}}+7$, then $x>\frac{|y|}{\sqrt{3}}+7>\left(\left(\frac{|y|}{9}\right)^{4}+7^{4}\right)^{1 / 4}>\left(\frac{y^{4}}{9}+2015\right)^{1 / 4}$, which means that if $I^{\prime \prime}=\{(x, y):(x-7) \sqrt{3}>|y|\}$, then $I^{\prime \prime} \subseteq I^{\prime}$. However, for $r>7$, the area of $I^{\prime \prime} \cap B((7,0), r-7)$ is $\frac{\pi}{3}(r-7)^{2}$, and $I^{\prime \prime} \subseteq I, B((7,0), r-7) \subseteq B((0,0), r)$, which means that $f(r)>\frac{\pi}{3}(r-7)^{2}$ for all $r>7$, from which it is not hard to see that $L=\frac{\pi}{3}$ is the minimum possible $L$. Remark: The lines $y= \pm \sqrt{3} x$ are actually asymptotes for the graph of $9 x^{4}-y^{4}=2015$. The bulk of the problem generalizes to the curve $|\sqrt{3} x|^{\alpha}-|y|^{\alpha}=C$ (for a positive real $\alpha>0$ and any real $C$ ); the case $\alpha=0$ is the most familiar case of a hyperbola. $\fbox{\frac{\pi}{3}}$.

Answer: \frac{\pi}{3}