China Team Selection Test for IMO, 2022

(1) We first prove the following.

Lemma: [Gauss-Lucas Theorem] All zeros of the polynomial $f'(z)$ lie in the closure formed by zeros of $f(z)$.

Proof of Lemma: By the fundamental theorem of algebra, we know that $f(z)$ admits a decomposition:
$$f(z) = A(z - z_1)^{\alpha_1} (z - z_2)^{\alpha_2} \cdots (z - z_n)^{\alpha_n}.$$
Thus, $z_1, z_2, \dots, z_n$ are zeros of $f'(z)$ with multiplicities $\alpha_1 - 1, \alpha_2 - 1, \dots, \alpha_n - 1$, respectively. Say $Z$ is any other zero of $f'(z)$, then
$$\frac{f'(Z)}{f(Z)} = \frac{\alpha_1}{Z - z_1} + \frac{\alpha_2}{Z - z_2} + \dots + \frac{\alpha_n}{Z - z_n} = 0.$$
That is,
$$\frac{\alpha_1}{|Z - z_1|^2} (\bar{Z} - \bar{z}_1) + \frac{\alpha_2}{|Z - z_2|^2} (\bar{Z} - \bar{z}_2) + \dots + \frac{\alpha_n}{|Z - z_n|^2} (\bar{Z} - \bar{z}_n) = 0.$$
After taking complex conjugations, the equality becomes
$$Z = \frac{\frac{\alpha_1}{|Z-z_1|^2} z_1 + \frac{\alpha_2}{|Z-z_2|^2} z_2 + \dots + \frac{\alpha_n}{|Z-z_n|^2} z_n}{\frac{\alpha_1}{|Z-z_1|^2} + \frac{\alpha_2}{|Z-z_2|^2} + \dots + \frac{\alpha_n}{|Z-z_n|^2}}.$$
Hence $Z$ is a convex linear combination of $z_1, z_2, \dots, z_n$. This proves the lemma.

Back to the original problem. Gauss-Lucas Lemma implies that the convex hull of the zeros of the given equation contains the convex hull of the zeros of the following equation
$$20z^{19} - 63 = 0.$$
This is a regular 19-gon inscribed in a circle of radius $\left(\frac{63}{20}\right)^{1/19}$.

The inscribed circle of this regular 19-gon has a radius
$$\left(\frac{63}{20}\right)^{1/19} \cos \frac{\pi}{19} > \left(\frac{63}{20}\right)^{1/19} \left(1 - \frac{1}{2} \left(\frac{\pi}{19}\right)^2\right) > \left(\frac{63}{20}\right)^{1/19} \left(1 - \frac{1}{2} \left(\frac{1}{6}\right)^2\right) = \frac{71}{72} \left(\frac{63}{20}\right)^{1/19}.$$
On the other hand, since
$$\left(\frac{72}{71}\right)^{19} = \left(1 + \frac{1}{71}\right)^{19} < 1 + \frac{19}{71} + 18 \cdot C_{19}^2 \frac{1}{72^2} < 3 < \frac{63}{20},$$
the radius of the inscribed circle is larger than 1. This proves (1).

(2) We need to prove the following.

(A) If $a_1, a_2, \dots, a_n$ and $w$ satisfy the assumptions, then the minimal norm of the roots of
$$a_1 z^{k_1} + a_2 z^{k_2} + \dots + a_n z^{k_n} = w \quad (*)$$
is less than or equal to $3mn$.
This is equivalent to the following.

(B) If $a_1, a_2, \dots, a_n$ and $w$ satisfy the assumptions, then the maximal norm of the roots of
$$wz^{k_n} - a_1z^{k_n-k_1} - a_2z^{k_n-k_2} - \dots - a_{n-1}z^{k_n-k_{n-1}} - a_n = 0, \quad (**)$$
denoted by $M_0$, is greater than or equal to $1/(3mn)$.

By the lemma in (1), the maximal norm of roots of $(**)$ cannot be less than the maximal norm of roots of
$$w k_n z^{k_n-1} - a_1(k_n - k_1)z^{k_n-k_1-1} - \dots - a_{n-1}(k_n - k_{n-1})z^{k_n-k_{n-1}-1} = 0. \quad (1)$$
That is, the maximal norm, say $M_1$, of the roots of
$$w k_n z^{k_n-1} - a_1(k_n - k_1)z^{k_n-k_1-1} - a_2(k_{n-1} - k_2)z^{k_n-k_2-1} - \dots - a_{n-1}(k_n - k_{n-1}) = 0. \quad (1')$$
Beginning with $(1')$, we repeat the same process for $n-s$ times. It is known that the maximal norm of the roots decreases each time. Therefore,
$$M_0 \ge M_1 \ge \dots \ge M_{n-s},$$
where $M_{n-s}$ is the maximal norm of the roots of the equation
$$w k_n k_{n-1} \cdots k_{s+1} z^{k_s} - a_1(k_n - k_1)(k_{n-1} - k_1) \cdots (k_{s+1} - k_1) z^{k_s - k_1} \\ - \cdots - a_s(k_n - k_s) \cdots (k_{s+1} - k_s) = 0. \quad ((n-s)')$$

By Vieta's theorem, we have
$$M_0 \ge M_1 \ge \dots \ge M_{n-s} \ge \left[ \frac{(k_n - k_s) \cdots (k_{s+1} - k_s)}{k_n k_{n-1} \cdots k_{s+1}} \cdot \left| \frac{a_s}{w} \right| \right]^{1/k_s}.$$
Hence it suffices to show that there exists $s \in \{1, 2, \dots, n\}$ such that
$$\left[ \frac{k_n k_{n-1} \cdots k_{s+1}}{(k_n - k_s)(k_{n-1} - k_s) \cdots (k_{s+1} - k_s)} \right]^{1/k_s} \cdot \left[ \left| \frac{w}{a_s} \right| \right]^{1/k_s} \le 3mn.$$
Note that
$$\frac{k_p}{k_p - k_s} = 1 + \frac{k_s}{k_p - k_s} \le \left(1 + \frac{1}{k_p - k_s}\right)^{k_s}.$$
So it remains to show that there exists some $s$ such that
$$\left[ \prod_{p=s+1}^{n} \left( 1 + \frac{1}{k_p - k_s} \right) \right] \cdot \left( \frac{m}{|a_s|} \right)^{1/k_s} \le 3mn. \quad 2$$
We also make the following observations.

* Since all $k_j$'s are odd numbers, $k_p - k_s \ge 2(p-s)$. Therefore,
$$1 + \frac{1}{k_p - k_s} \le 1 + \frac{1}{2(p-s)} < \frac{p-s+1}{p-s};$$
* Since $\sum_{j=1}^n a_j = 1$, we must have $\sum_{j=1}^n |a_j| \ge 1$. Thus there must be $s$ satisfying $|a_s| \ge \frac{1}{2^s}$, or equivalently, $\frac{1}{|a_s|} \le 2^s$.

We separate the discussion into the following two cases.

(1) If there is $s$ (allowing $s=1$), such that $k_s \ge 3$ and $|a_s| \ge \frac{1}{2^s}$, then (note that $s \le 2s-1 \le k_s$)
$$\left[ \prod_{p=s+1}^{n} \left( 1 + \frac{1}{k_p - k_s} \right) \right] \cdot \left( \frac{m}{|a_s|} \right)^{1/k_s} < \left[ \prod_{p=s+1}^{n} \frac{p-s+1}{p-s} \right] \cdot m^{1/k_s} \cdot 2^{s/k_s} < n \cdot \sqrt[3]{m} \cdot 2 < 3mn.$$

(2) Otherwise, we must have $k_1 = 1$ and $|a_1| \ge \frac{1}{2}$. Hence
$$\left[ \prod_{p=2}^{n} \left( 1 + \frac{1}{k_p - k_1} \right) \right] \cdot \frac{m}{|a_1|} \le \left[ \prod_{p=2}^{n} \frac{2(p-1)+1}{2(p-1)} \right] \cdot m \cdot 2 < \sqrt{\prod_{k=1}^{2n-2} \frac{k+1}{k}} \cdot 63 \cdot 2 = \sqrt{2n-1} \cdot m \cdot 2 < 3mn.$$

This completes the proof.