Seventeenth ROMANIAN MASTER OF MATHEMATICS

To prove this, choose an arbitrary $i<n$, and let $A_i{A}_{j_1},A_i{A}_{j_2},\dots ,A_i{A}_{j_{10}}$ be the ten smallest numbers among the $A_i{A}_j$ with $j\ne i$, ordered so that $j_1<j_2<\cdots <j_{10}$. If $j_{10}<i$, then $i>10$, and the problem condition yields $$b_i=\underset{1\le k\le 10}{\max }{A}_i{A}_{j_k}=a_i\ge a_n\ge b_n,$$ which is even stronger than we need. Otherwise, set $j=j_{10}>i$ (in this case we also have $j_{10}>10$), and denote $m=b_i={\max }_{1\le k\le 10}{A}_i{A}_{j_k}$. By the problem condition, we have $a_j\ge a_n=b_n$. On the other hand, we have $$a_j\le \max \left(A_j{A}_i,\underset{1\le k\le 9}{\max }{A}_j{A}_{j_k}\right)\le \max \left(A_j{A}_i,\underset{1\le k\le 9}{\max }(A_j{A}_i+A_i{A}_{j_k})\right)\le 2m,$$ as $A_j{A}_i,A_i{A}_{j_k}\le m$. So $b_n\le a_j\le 2m=2b_i$, as desired.

Alternative solution. Let $d_j=f_{10}(j,n)$, $j=1,\dots ,n$. To prove that $2d_j\ge d_n$ for every $j=1,\dots ,n-1$, induct on $n$. Consider the base case, $n=11$. Note that each $d_j={\max }_{i\ne j}{A}_i{A}_j$, as $f_{10}$ is 10-variate. Let $d_{11}=A_{11}{A}_k$ for some index $k\le 10$. Clearly, $2d_k\ge 2A_k{A}_{11}=2d_{11}\ge d_{11}$ and if $j\ne k$ then $2d_j\ge A_j{A}_k+A_j{A}_{11}\ge A_k{A}_{11}=d_{11}$, by the triangle inequality. For the induction step, let $n>11$ and note that $$\underset{n\ge l\ge k}{\max }{f}_{10}(l,k-1)=\underset{n-1\ge l\ge k}{\max }{f}_{10}(l,k-1),$$ as both maxima are achieved at $l=k$, by the condition in the statement. Hence $A_1,A_2,\dots ,A_{n-1}$ also satisfy this condition. Let $d_j^{\prime }=f_{10}(j,n-1)$, $j=1,\dots ,\le n-1$. By the induction hypothesis, $2d_j^{\prime }\ge d_{n-1}^{\prime }$ for all $j\le n-2$. Note that $d_{n-1}^{\prime }=f_{10}(n-1,n-2)\ge f_{10}(n,n-2)\ge f_{10}(n,n-1)=d_n$; the first inequality holds by the condition in the statement for $k=n-1$ and the second because adding more variables to $f_{10}$ does not increase its value. Let now ${\Delta }_i$ be the closed disc (interior and boundary) of radius $d_i$, centred at $A_i$. By the definition of $d_i$, each ${\Delta }_i$ contains at least 11 points, of which at most 10 ($A_i$, inclusive) lie strictly inside. Finally, suppose, if possible, that $2d_j<d_n$ for some index $j<n$. If $A_j{A}_n$ is not among the first 10 distances from $A_j$ to the other points, then $d_j=d_j^{\prime }$ and this leads to a contradiction with the induction hypothesis. So $A_j{A}_n$ has to be among the first 10 distances from $A_j$ to the other points. This means that $d_j\ge A_j{A}_n$, so $d_n>2d_j\ge 2A_j{A}_n$. Hence ${\Delta }_j$ lies strictly inside ${\Delta }_n$. This is a contradiction, as ${\Delta }_j$ contains at least 11 points, whereas ${\Delta }_n$ contains at most 10 strictly inside. The conclusion follows.