36. [20] For a cubic polynomial $P(x)$ with complex roots $z_{1}, z_{2}, z_{3}$, let
$$
M(P)=\frac{\max \left(\left|z_{1}-z_{2}\right|,\left|z_{1}-z_{3}\right|,\left|z_{2}-z_{3}\right|\right)}{\min \left(\left|z_{1}-z_{2}\right|,\left|z_{1}-z_{3}\right|,\left|z_{2}-z_{3}\right|\right)} .
$$

Over all polynomials $P(x)=x^{3}+a x^{2}+b x+c$, where $a, b, c$ are nonnegative integers at most 100 and $P(x)$ has no repeated roots, the twentieth largest possible value of $M(P)$ is $m$. Estimate $A=\lfloor m\rfloor$. An estimate of $E$ earns $\max \left(0,\left\lfloor 20-20|3 \ln (A / E)|^{1 / 2}\right\rfloor\right)$ points.
Answer: 8097
Solution: Consider fixing $a$ and $b$. Then, we know that $P^{\prime}(x)=3 x^{2}+2 a x+b$, which has a root at approximately $r \approx-b / 2 a$, which is rather small compared to 100 . Then $P(r) \approx-b^{2} / 4 a$. Assuming that this is greater than about -100 , then the value of $c$ that produces the roots that are closest together is the closest integer to $-P(r)$ (the chance when this creates a double root is pretty rare. Let $-P(r)=c+s$, so that we can now assume that $s$ is uniformly distributed in $(-1 / 2,1 / 2)$. One can
show that the difference between these roots is about $2 \sqrt{|s| / a}$. Since these roots are rather small, by Vieta's formulas the other root is near $-a$, so $M(P)$ is about $\frac{1}{2} \sqrt{a^{3} /|s|}$.
It's clear from this discussion that $a$ needs to be reasonably large for $M(P)$ to be large. Thus the condition $P(r)>-100$ is satisfied close to all the time-we will henceforth ignore it.
Set some $L$ and let's consider the expected number of $P$ so that $M(P)>L$. Then, for a given $a, b$, we need $|s|<a^{3} /(2 L)^{2}$. Summing over all $a$ and $b$, we find the probability is $2 \cdot 100 \cdot 100^{4} / 4 \cdot 1 /(2 L)^{2}$. Setting this equal to 20 gives us
$$
L^{2}=\frac{10^{10}}{160} \Longrightarrow L \approx 7900 \text {. }
$$

This is good enough for 14 points.