Austria 2010

Note first that the function $f_n(x)$ satisfies $f_n(x)=f_n(n+1-x)$: $$f_n(n+1-x)=\sum _{k=1}^n|n+1-x-k|=\sum _{k=1}^n|x-(n+1-k)|=\sum _{k=1}^n|x-k|=f_n(x)$$ by reversing the order of summation. Let us first consider the case $x<1$: then, $x-k<0$ for every $k\ge 1$ and thus $$f_n(x)=\sum _{k=1}^n(k-x)=\frac{n(n+1)}{2}-nx>\frac{n(n+1)}{2}-n=\frac{n(n-1)}{2}\ge \frac{10\cdot 9}{2}=45>41,$$ so this case can be excluded. By symmetry (identity above), we can exclude $x>n$ as well. Thus we are left with $1\le x\le n$. Suppose that $x\in [\ell ,\ell +1]$ for some integer $\ell$ with $1\le \ell \le n-1$. Then $x-k\le 0$ for $k\ge \ell +1$ and $x-k\ge 0$ for $k\le \ell$, and we obtain $$\begin{array}{rl}{f}_n(x)& =\sum _{k=1}^{\ell }(x-k)+\sum _{k=\ell +1}^n(k-x)=\ell x-\frac{\ell (\ell +1)}{2}+\frac{(n-\ell )(n+\ell +1)}{2}-(n-\ell )x\\ & =\frac{n(n+1)}{2}-\ell (\ell +1)+(2\ell -n)x.\end{array}$$ This shows that $f_n(x)$ is strictly decreasing on $[\ell ,\ell +1]$ if $\ell <\frac{n}{2}$, constant on $[\frac{n}{2},\frac{n}{2}+1]$ (if $n$ is even) and strictly increasing on $[\ell ,\ell +1]$ if $\ell >\frac{n}{2}$. We conclude: If $n$ is even, then $f_n(x)$ is strictly decreasing on $[1,\frac{n}{2}]$, constant on $[\frac{n}{2},\frac{n}{2}+1]$ and strictly increasing on $[\frac{n}{2}+1,n]$. If $n$ is odd, then $f_n(x)$ is strictly decreasing on $[1,\frac{n+1}{2}]$ and strictly increasing on $[\frac{n+1}{2},n]$. We see that the minimum of $f_n(x)$ is always attained at $m=\lfloor \frac{n+1}{2}\rfloor$. Now we can complete squares in the above to obtain $$\begin{array}{rl}{f}_n(m)& =\frac{n(n+1)}{2}-m(m+1)+(2m-n)m=\frac{n(n+1)}{2}+m^2-(n+1)m\\ & =\frac{n(n+1)}{2}+{\left(m-\frac{n+1}{2}\right)}^2-{\left(\frac{n+1}{2}\right)}^2\ge \frac{n(n+1)}{2}-{\left(\frac{n+1}{2}\right)}^2=\frac{n^2-1}{4}.\end{array}$$ If $n\ge 13$, then this implies $f_n(x)\ge f_n(m)\ge \frac{13^2-1}{4}=42>41$ for all $x$, so that there is no solution. It remains to consider $n\in \{10,11,12\}$. For $n=10$, we obtain from above that $$f_{10}\left(\frac{3}{2}\right)=55-2-8\cdot \frac{3}{2}=41$$ and by the symmetry property $f_{10}\left(\frac{19}{2}\right)=41$. In view of our monotonicity considerations, $f_{10}(x)\ge 41$ for $x\le \frac{3}{2}$ and $x\ge \frac{19}{2}$ and $f_{10}(x)<41$ on the remaining interval. So we find that the solution set in this case is $\left(\frac{3}{2},\frac{19}{2}\right)$. For $n=11$, we have $$f_{11}\left(\frac{19}{7}\right)=66-6-7\cdot \frac{19}{7}=41$$ and $f_{11}\left(\frac{65}{7}\right)=41$ by symmetry. The same argument as before shows that the solution set is $\left(\frac{19}{7},\frac{65}{7}\right)$ in this case. For $n=12$, we have $$f_{12}\left(\frac{17}{4}\right)=78-20-4\cdot \frac{17}{4}=41$$ and $f_{12}\left(\frac{35}{4}\right)=41$ by symmetry. Hence we obtain the solution set $\left(\frac{17}{4},\frac{35}{4}\right)$ in this case. Let us summarize the solutions: $\frac{3}{2}<x<\frac{19}{2}$ for $n=10$, $\frac{19}{7}<x<\frac{65}{7}$ for $n=11$, $\frac{17}{4}<x<\frac{35}{4}$ for $n=12$, * no solutions if $n\ge 13$.