The 35th Japanese Mathematical Olympiad

Let $a_1,a_2,\dots ,a_n$ be a beautiful sequence of length $n$. Take an integer $m$ such that $a_m=2025$. Then $m\ge 2$ since $a_1=0$. We first prove $n\le 13$. Let $t$ be an integer with $1\le t\le n-2$. By applying the third condition to $(i,j,k)=(t,t+1,n)$, we obtain $$\frac{a_t+a_n}{2}\le a_{t+1},$$ which implies $$a_n-a_t\ge 2(a_n-a_{t+1}).$$ If $m=n$, repeatedly applying this inequality gives $$2025=a_n-a_1\ge 2(a_n-a_2)\ge \cdots \ge 2^{n-2}(a_n-a_{n-1})\ge 2^{n-2},$$ and hence $n\le 12$. If $m<n$, we have $$a_n-a_1\ge 2(a_n-a_2)\ge \cdots \ge 2^{m-2}(a_n-a_{m-1})\ge 2^{m-1}(a_n-a_m),$$ and $$a_n-a_m\ge 2(a_n-a_{m+1})\ge 2^2(a_n-a_{m+2})\ge \cdots \ge 2^{n-m-1}(a_n-a_{n-1})\ge 2^{n-m-1}.$$ $$\begin{array}{rl}2025& =a_m-a_1=(a_n-a_1)-(a_n-a_m)\ge (2^{m-1}-1)(a_n-a_m)\\ & \ge (2^{m-1}-1)2^{n-m-1}\ge 2^{m-2}\cdot 2^{n-m-1}=2^{n-3}\end{array}$$ \quad (*) and so we obtain $n \le 13$. Let $n = 13$. Then the above argument shows $n > m$. From $(*)$, we have 2025 \ge (2^{m-1} - 1)2^{n-m-1} = 2^{n-2} - 2^{n-m-1} = 2^{11} - 2^{12-m}, which implies $2^{12-m} \ge 23$, and hence $2^{12-m} \ge 32$. Therefore, a_n = a_m + (a_n - a_m) \ge a_m + 2^{n-m-1} = 2025 + 2^{12-m} \ge 2057. Thus, we have $a_n \ge 2057$ when $n = 13$. Finally, we show that the sequence (a_1, a_2, \dots, a_{13}) = (0, 1033, 1545, 1801, 1929, 1993, 2025, 2041, 2049, 2053, 2055, 2056, 2057) $$isabeautifulsequenceoflength13.Thefirstandsecondconditionsareclearlysatisfied.Wenowverifythethirdcondition.Wehave$$ (a_{13} - a_1, a_{13} - a_2, \dots, a_{13} - a_{12}) = (2057, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1), so for all integers $t$ with $1 \le t \le 11$, we have $a_{13} - a_t \ge 2(a_{13} - a_{t+1})$. Therefore, for all integers $1 \le i < j < k \le 13$, we have $$\begin{array}{rl}{a}_j-\frac{a_i+a_k}{2}& \ge a_j-\frac{a_i+a_{13}}{2}=\frac{(a_{13}-a_i)-2(a_{13}-a_j)}{2}\\ & \ge \frac{(a_{13}-a_i)-(a_{13}-a_{j-1})}{2}=\frac{a_{j-1}-a_i}{2}\\ & \ge 0.\end{array}$$ $$Thisconfirmsthethirdcondition.Therefore,themaximumvalueofthelengthofabeautifulsequenceis13andforbeautifulsequences$a_1, a_2, \dots, a_{13}$oflength13,theminimumvalueof$a_{13}$ is 2057.