2025 International Mathematical Olympiad China National Team Selection Test

Let $L$ denote the maximum squared modulus of the polynomial $f(z)=a{z}^3+b{z}^2+cz+d$ on the unit circle. By choosing a unit complex number $s$ with $s^3=d/a$ and considering $f_1(z)=d^{-1}f(sz)$, we may assume without loss of generality that $a=d=1$. Thus we only need to consider polynomials of the form $f(z)=z^3+b{z}^2+cz+1$.

First consider the symmetric case $b=c=u+vi$. The polynomial factors as $f(z)=(1+z)(1+(b-1)z+z^2)$. For $z=x+yi$ on the unit circle ($x^2+y^2=1$), we have: $$\begin{array}{rl}|1+z{|}^2& =2+2x,\\ |1+(b-1)z+z^2{|}^2& =4x^2+(2x-1)(2u-2),\\ |f(z){|}^2& =8\left(x^3+u{x}^2+\frac{u-1}{2}x-\frac{u-1}{2}\right).\end{array}$$ Let $h(x)=x^3+u{x}^2+\frac{u-1}{2}x-\frac{u-1}{2}$. We seek to minimize ${\max }_{x\in [-1,1]}h(x)$. Heuristically, the minimum occurs when $h(x)$ has a double root at $x_0$ and a simple root at 1. Solving gives $u=2-\sqrt{5}$. Indeed, for this value: $$h(x)-(3-\sqrt{5})={\left(x+\frac{3-\sqrt{5}}{2}\right)}^2(x-1)\le 0,$$ $$\text{so }|f(z){|}^2\le 8(3-\sqrt{5})=(2\sqrt{5}-2{)}^2.$$ For the general case where $b=c=u+vi$, consider $$|f(z){|}^2=8h(x)=(2x+2)[(4x^2-4x+2)+2(2x-1)u]$$ evaluated at the "endpoint": $$|f(1){|}^2=8h(1)=8+8u,$$ and at the point $z=z_{\star }=\frac{\sqrt{5}-3}{2}+\frac{\sqrt{5}-1}{2}i\sqrt{5}$ (corresponding to a possible vertex of $h(x)$): $$|f(z_{\star }){|}^2=8h\left(\frac{\sqrt{5}-3}{2}\right)=(30\sqrt{5}-62)+(18-10\sqrt{5})u.$$ We have $$|f(z^{\ast }){|}^2+\frac{5\sqrt{5}-9}{4}\cdot |f(1){|}^2=40\sqrt{5}-80=\frac{5\sqrt{5}-5}{4}\cdot (2\sqrt{5}-2{)}^2.$$ ---

This shows that a certain weighted average of them equals $(2\sqrt{5}-2{)}^2$, therefore when $b=c$ we always have $$L\ge \max \{|f(1){|}^2,|f(z_{\star }){|}^2\}\ge (2\sqrt{5}-2{)}^2.$$ For general $b,c$, we still attempt to prove $L\ge (2\sqrt{5}-2{)}^2$. Let $\omega =\frac{-1+\sqrt{3}i}{2}$ be a primitive cube root of unity. Since $f(z)=z^3+b{z}^2+cz+1$, we have $f(\omega z)=z^3+{\omega }^{2}b{z}^2+\omega cz+1$ and $f({\omega }^{2}z)=z^3+\omega b{z}^2+{\omega }^{2}cz+1$, all corresponding to the same $L$ value. Moreover, $(b+c)+({\omega }^{2}b+\omega c)+(\omega b+{\omega }^{2}c)=0$, and at least one of these has real part $\le 0$. Without loss of generality, assume $\frac{b+c}{2}=u+vi$ with $u\le 0$. Since $|b|=|c|=1$, we have $\frac{c-b}{b+c}=wi$ being purely imaginary. We may assume $w\ge 0$ (otherwise consider the reciprocal polynomial $f_2(z)=z^3+c{z}^2+bz+1$ which has the same $L$ value). Thus $\frac{c-b}{2}=(u+vi)wi$, and $|\frac{b+c}{2}{|}^2+|\frac{c-b}{2}{|}^2=(u^2+v^2)(1+w^2)=1$. Now we can express: $$f(z)=z^3+1+\frac{b+c}{2}(z^2+z)+\frac{c-b}{2}(z-z^2)=z(z+1)\left[z+z^{-1}-1+u+vi+\frac{1-z}{1+z}\cdot wi\cdot (u+vi)\right].$$ Note that $\frac{1-z}{1+z}\cdot wi=t$ is real. Writing the unit complex number $z=x+yi$, we have $|z+1{|}^2=2x+2$, $z+z^{-1}-1=2x-1$, and $u^2+v^2=\frac{1}{1+w^2}$. Therefore: $$\begin{array}{rl}|f(z){|}^2& =|z+1{|}^2\cdot |z+z^{-1}-1+(u+vi)(1+t){|}^2\\ & =(2x+2)\left[(2x-1+u+ut{)}^2+(v+vt{)}^2\right]\\ & =(2x+2)\left[(2x-1{)}^2+(u^2+v^2)(1+2t+t^2)+2(2x-1)u+2(2x-1)ut\right]\\ & =(2x+2)\left[(4x^2-4x+2)+2(2x-1)u+\frac{2t+t^2-w^2}{1+w^2}+2(2x-1)ut\right].\end{array}$$ To prove $L\ge (2\sqrt{5}-2{)}^2$, we want to show: $$T=|f(z_{\star }){|}^2+\frac{5\sqrt{5}-9}{4}\cdot |f(1){|}^2-(40\sqrt{5}-80)\ge 0.$$ First, for $z=z_{\star }=\frac{\sqrt{5}-3}{2}+\frac{\sqrt{5}-1}{2}\sqrt{5}i$ (where $x=\frac{\sqrt{5}-3}{2}$ and $\frac{1-z}{1+z}=-\sqrt{5}i$), we have $t=\sqrt{5}w$. Comparing with the case $b=c$, consider: $$\begin{array}{rl}{T}_1& =|f(z_{\star }){|}^2-\left[(30\sqrt{5}-62)+(18-10\sqrt{5})u\right]\\ & =|f(z_{\star }){|}^2-(2x+2)\left[(4x^2-4x+2)+2(2x-1)u\right]\\ & =(2x+2)\left(\frac{2t+t^2-w^2}{1+w^2}+2(2x-1)ut\right)\\ & =(\sqrt{5}-1)\cdot \left(\frac{2\sqrt{5}w+(\sqrt{5}-1)w^2}{1+w^2}+2(\sqrt{5}-4)ut\right)\\ & \ge (\sqrt{5}-1)\cdot \frac{2\sqrt{5}w+(\sqrt{5}-1)w^2}{1+w^2}.\end{array}$$ ---

Second, since $f(1)=2+2(u+vi)$, we have $|f(1){|}^2=4+4u^2+4v^2+8u$, and: $$T_2=|f(1){|}^2-(8+8u)=4(u^2+v^2-1)=-\frac{4w^2}{1+w^2}.$$ Therefore: $$T=T_1+\frac{5\sqrt{5}-9}{4}{T}_2\ge (\sqrt{5}-1)\cdot \frac{2\sqrt[4]{5}w+(\sqrt{5}-1)w^2}{1+w^2}-(5\sqrt{5}-9)\cdot \frac{w^2}{1+w^2}.$$ We want $T\ge 0$, which holds if: $$2\sqrt[4]{5}(\sqrt{5}-1)w+((\sqrt{5}-1{)}^2-(5\sqrt{5}-9))w^2\ge 0,$$ i.e., when: $$0\le w\le \frac{2(\sqrt{5}-1)\sqrt[4]{5}}{7\sqrt{5}-15}=\frac{4+2\sqrt{5}}{\sqrt[4]{5}}\approx \mathrm{5.666.}$$ Through appropriate substitutions or transformations, we have ensured $u\le 0$ and $w\ge 0$. If $0\le w\le 2$, then $T\ge 0$, meaning a certain weighted average of $|f(z_{\star }){|}^2$ and $|f(1){|}^2$ is $\ge (2\sqrt{5}-2{)}^2$, so $L\ge (2\sqrt{5}-2{)}^2$. If $w\ge 2$, then $|f(1){|}^2=4+4u^2+4v^2+8u\le 4+4\cdot \frac{1}{1+w^2}\le 4.8$. Since: $$|f(1){|}^2+|f(\omega ){|}^2+|f({\omega }^2){|}^2=|2+b+c{|}^2+|2+\omega c+{\omega }^{2}b{|}^2+|2+{\omega }^{2}c+\omega b{|}^2=18,$$ we have: $$L\ge \frac{|f(\omega ){|}^2+|f({\omega }^2){|}^2}{2}\ge \frac{18-4.8}{2}=6.6>(2\sqrt{5}-2{)}^2.$$ Therefore, we always have $L\ge (2\sqrt{5}-2{)}^2$, and equality can be achieved. Thus, the minimal real number we seek is $M=2\sqrt{5}-2$. □