2022 China Team Selection Test for IMO

(1) We first prove the following.

Lemma: [Gauss-Lucas Theorem] All zeros of the polynomial $f^{\prime }(z)$ 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 $f^{\prime }(z)$ with multiplicities ${\alpha }_1-1,{\alpha }_2-1,\dots ,{\alpha }_n-1$, respectively. Say $Z$ is any other zero of $f^{\prime }(z)$, then $$\frac{f^{\prime }(Z)}{f(Z)}=\frac{{\alpha }_1}{Z-z_1}+\frac{{\alpha }_2}{Z-z_2}+\cdots +\frac{{\alpha }_n}{Z-z_n}=0.$$ That is, $$\frac{{\alpha }_1}{|Z-z_1{|}^2}(\bar{Z}-{\bar{z}}_1)+\frac{{\alpha }_2}{|Z-z_2{|}^2}(\bar{Z}-{\bar{z}}_2)+\cdots +\frac{{\alpha }_n}{|Z-z_n{|}^2}(\bar{Z}-{\bar{z}}_n)=0.$$ After taking complex conjugations, the equality becomes $$Z=\frac{\frac{{\alpha }_1}{|Z-z_1{|}^2}{z}_1+\frac{{\alpha }_2}{|Z-z_2{|}^2}{z}_2+\cdots +\frac{{\alpha }_n}{|Z-z_n{|}^2}{z}_n}{\frac{{\alpha }_1}{|Z-z_1{|}^2}+\frac{{\alpha }_2}{|Z-z_2{|}^2}+\cdots +\frac{{\alpha }_n}{|Z-z_n{|}^2}}.$$ 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 $$a_1{z}^{k_1}+a_2{z}^{k_2}+\cdots +a_n{z}^{k_n}=w(\ast )$$ 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 $$w{z}^{k_n}-a_1{z}^{k_n-k_1}-a_2{z}^{k_n-k_2}-\cdots -a_{n-1}{z}^{k_n-k_{n-1}}-a_n=0,(\ast \ast )$$ denoted by $M_0$, is greater than or equal to $1/(3mn)$.

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

By Vieta's theorem, we have $$M_0\ge M_1\ge \cdots \ge M_{n-s}\ge {\left[\frac{(k_n-k_s)\cdots (k_{s+1}-k_s)}{k_n{k}_{n-1}\cdots k_{s+1}}\cdot \left|\frac{a_s}{w}\right|\right]}^{1/k_s}.$$ 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 $$\left[\prod _{p=s+1}^n\left(1+\frac{1}{k_p-k_s}\right)\right]\cdot {\left(\frac{m}{|a_s|}\right)}^{1/k_s}\le 3mn.2$$ 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.