The 37th Korean Mathematical Olympiad Final Round

If $x_1=x_2=\cdots =x_{2024}=y_1=y_2=\cdots =y_{2024}=p$, then $p^2\ge 1-p$, that is, $p\ge \frac{1+\sqrt{5}}{2}$. So, $p$ must be at least $\frac{1+\sqrt{5}}{2}$. We prove that the condition holds for $p=\frac{1+\sqrt{5}}{2}$, which implies that the answer is $\frac{1+\sqrt{5}}{2}$. Let $p=\frac{1+\sqrt{5}}{2}$, and let $$S=\{(x_1,\dots ,x_{2024})\mid 0\le x_1\le \cdots \le x_{2024}\le 1,x_1+x_2+\cdots +x_{2024}=2024p\}.$$ For $X=(x_1,x_2,\dots ,x_{2024})$, $Y=(y_1,y_2,\dots ,y_{2024})\in S$, we regard $x_0=y_0=0$ for convenience, and let $$\begin{gathered} f(X,Y)=\sum _{i=1}^{2024}{x}_i(y_{2025-i}-y_{2024-i}) \\ =x_1(y_{2024}-y_{2023})+x_2(y_{2023}-y_{2022})+\cdots +x_{2023}(y_2-y_1)+x_{2024}{y}_1. \end{gathered}$$ Since $f(X,Y)=y_1(x_{2024}-x_{2023})+y_2(x_{2023}-x_{2022})+\cdots +y_{2023}(x_2-x_1)+y_{2024}{x}_1$, $f(X,Y)$ is symmetric with regard to $X$ and $Y$. If $x_1=x_2=\cdots =x_{k_1}=a_1$, $x_{k_1+1}=\cdots =x_{k_1+k_2}=a_2$, \dots, $x_{k_1+k_2+\cdots +k_{m-1}+1}=\cdots =x_n=a_m$ for $0\le a_1<a_2<\cdots <a_m\le 1$, we denote $X$ by $(a_1^{k_1},a_2^{k_2},\dots ,a_m^{k_m})$ where $k_m=2024-k_1-k_2-\cdots -k_{m-1}$, and let $l(X)=m$. And in this case, we have $$\begin{gathered} f(X,Y)=a_1(y_{2024}-y_{2024-k_1})+a_2(y_{2024-k_1}-y_{2024-k_1-k_2})+\cdots +a_m{y}_{k_m} \\ =a_1{Y}_1+a_2{Y}_2+\cdots +a_m{Y}_m \end{gathered}$$ where $Y_t=y_{2024-k_1-k_2-\cdots -k_{t-1}}-y_{2024-k_1-k_2-\cdots -k_t}=y_{k_t+k_{t+1}+\cdots +k_m}-y_{k_t+1+\cdots +k_m}$ for $t=1,2,\dots ,m$. Now we consider $X,Y\in S$ such that $f(X,Y)$ is minimized, and subject to that $l(X)+l(Y)$ is minimum. Let $X=(a_1^{k_1},a_2^{k_2},\dots ,a_m^{k_m})$. Lemma. There is no $1\le i<m$ such that $0<a_i,a_{i+1}<1$.

Proof. Suppose $0<a_i<a_{i+1}<1$ for some $i$. Case 1. $\frac{Y_i}{k_i}\le \frac{Y_{i+1}}{k_{i+1}}$. Let $a=\frac{k_i{a}_i+k_{i+1}{a}_{i+1}}{k_i+k_{i+1}}$. Then, $a_i<a<a_{i+1}$, and for $$X^{\prime }=(a_1^{k_1},\dots ,a_{i-1}^{k_{i-1}},a_{i-1}^{k_{i-1}+k_{i+1}},a_{i+2}^{k_{i+2}},\dots ,a_m^{k_m})$$ we have $$f(X,Y)-f(X^{\prime },Y)\ge 0$$ and $l(X^{\prime })=m-1<l(X)$, which yields a contradiction to the fact that $l(X)+l(Y)$ is minimum. Case 2. $\frac{Y_i}{k_i}>\frac{Y_{i+1}}{k_{i+1}}$. Choose a positive real number $e$ such that $e<a_i-a_{i-1}$ (if $i=1$, let $a_0=0$) and $a_{i+1}+\frac{k_{i}e}{k_{i+1}}<a_{i+2}$ (if $i=m-1$, let $a_{m+1}=1$). Let $a=a_i-e$ and $b=a_{i+1}+\frac{k_{i}e}{k_{i+1}}$, and consider $X^{\prime }=(a_1^{k_1},a_2^{k_2},\dots ,a_{i-1}^{k_{i-1}},a^{k_i},b^{k_{i+1}},a_{i+2}^{k_{i+2}},\dots ,a_m^{k_m})$. Since $k_{i}a+k_{i+1}b=k_i{a}_i+k_{i+1}{a}_{i+1}$, $X^{\prime }\in S$. However, $f(X,Y)-f(X,Y^{\prime })>0$, yielding a contradiction. Therefore there is no $1\le i<m$ such that $0<a_i,a_{i+1}<1$. $\square$ The above lemma also holds for $Y$. Thus, there exist non-negative integers $k_1,k_2,k_3,s_1,s_2,s_3$ and real numbers $0<a,b<1$ such that $$\bullet k_1+k_2+k_3=s_1+s_2+s_3=2024,$$ $$\bullet x_1=x_2=\cdots =x_{k_1}=0,x_{k_1+1}=\cdots =x_{k_1+k_2}=a,x_{k_1+k_2+1}=\cdots =x_{2024}=1$$ $$\bullet y_1=y_2=\cdots =y_{s_1}=0,y_{s_1+1}=\cdots =y_{s_1+s_2}=a,y_{s_1+s_2+1}=\cdots =y_{2024}=1$$ $$\bullet k_{2}a+k_3=s_{2}b+s_3=2024p.$$ We note that $k_1,k_2,k_3,s_1,s_2,s_3$ could be zero, and we have $$f(X,Y)=a{y}_{k_2+k_3}+(1-a)y_{k_3}=b{x}_{s_2+s_3}+(1-b)x_{s_3}.$$ If $k_2=0$, then $x_1+x_2+\cdots +x_{2024}=k_3\ne 2024p$ (since $k_3$ is an integer), so $k_2>0$. Thus, $k_3<k_{2}a+k_3=2024p$ and $k_1=2024-(k_2+k_3)<2024-(k_{2}a+k_3)=2024(1-p)$. That is, $$k_1<2024(1-p),k_2>0,k_3<2024p,$$ and similarly, $$s_1<2024(1-p),s_2>0,s_3<2024p.$$ Since $y_1=y_2=\cdots =y_{s_1}=0$, $y_{s_1+1}=\cdots =y_{s_1+s_2}=a$, and $y_{s_1+s_2+1}=\cdots =y_{2024}=1$, in order for $f(X,Y)=a{y}_{k_2+k_3}+(1-a)y_{k_3}$ to be minimized, the following must hold. (i) $y_{k_2+k_3+1}=\cdots =y_{2024}=1$ (ii) $y_{k_3+1}=\cdots =y_{k_2+k_3}$ (iii) $y_1=\cdots =y_{k_3}$ Similarly, the following must hold. (i') $x_{s_2+s_3+1}=\cdots =x_{2024}=1$ (ii') $x_{s_3+1}=\cdots =x_{s_2+s_3}$ (iii') $x_1=\cdots =x_{s_3}$ By (i) and (i'), $k_1\le s_3$ and $s_1\le k_3$. If $k_1<s_3$, then $y_{k_3+1}=1$ by (i) and (ii), so $k_1+k_2\le s_3<2024p$. Then by (iii), $k_3=s_2$, $y_{k_3}=b$ and $s_1=0$, $k_1+k_2=s_3$. So, $$\begin{array}{rl}a& =\frac{2024p-k_3}{k_2}=\frac{2024p-k_3}{2024-k_1-k_3}\ge \frac{2024p-k_3}{2024-k_3}=1-\frac{2024(1-p)}{2024-k_3}\\ b& =\frac{2024p-s_3}{s_2}=\frac{2024p-k_1-k_2}{k_3}=\frac{k_3-2024(1-p)}{k_3}=1-\frac{2024(1-p)}{k_3}\end{array}$$ Since $2024p>k_3=2024-(k_1+k_2)=2024-s_3>2024(1-p)$, we have $$\begin{array}{rl}f(X,Y)& =a{y}_{k_2+k_3}+(1-a)y_{k_3}=a+(1-a)b=1-(1-a)(1-b)\\ & =1-\frac{2024(1-p)}{2024-k_3}\cdot \frac{2024(1-p)}{k_3}\ge 1-p.\end{array}$$ So, we may assume that $k_1=s_3$ and by the symmetry, we may further assume that $s_1=k_3$. Then, $k_2=s_2$, and (i) $y_{k_2+k_3+1}=\cdots =y_{2024}=1$ (ii) $y_{k_3+1}=\cdots =y_{k_2+k_3}=b$ (iii) $y_1=\cdots =y_{k_3}=0$.