(1) Find the possible number of roots for the equation $|x + 1| + |x + 2| + |x + 3| = a$, where $x \in R$ and $a$ is parameter. (2) Let $\{ a_1, a_2, \ldots, a_n \}$ be an arithmetic progression, $n \in \mathbb{N}$, and satisfy the condition \[ \sum^{n}_{i=1}|a_i| = \sum^{n}_{i=1}|a_{i} + 1| = \sum^{n}_{i=1}|a_{i} - 2| = 507. \] Find the maximum value of $n$.
Problem
Source: CSEMO 2005-7
Tags: arithmetic sequence, algebra unsolved, algebra
21.07.2005 16:26
this question definitely had our team involved in a mathematical debate. nobody knew who was right. Alaban, our teammate told us there were no roots. Some answered 4, that's me and Jako... so need help here...
24.07.2005 12:46
For (1), draw the graph of $|x + 1| + |x + 2| + |x + 3| = y$, we find that there can be 0,1 or 2 roots. For (2), I think case division should work (eg. the difference $d = 0$, $1$ or more than $2$)
24.07.2005 14:14
For (2), $n=26$, right?
24.07.2005 16:20
enescu, you are right!
24.07.2005 20:45
Can someone post the solution to 2?
24.07.2005 21:08
First, observe that if $a_{1}<a_{2}<\ldots<a_{n}$ are real numbers and $f:\mathbb{R}\rightarrow\mathbb{R}$ is defined as \[ f\left( x\right) =\left| x-a_{1}\right| +\left| x-a_{2}\right| +\ldots+\left| x-a_{n}\right| , \] then $f$ is first decreasing, and then increasing. More precisely, if $n$ is odd, say $n=2k+1,$ then $f$ is strictly decreasing on the interval $(-\infty,a_{k}]$ and strictly increasing on $[a_{k},+\infty)$, while if $n$ is even, say $n=2k,$ then $f$ is strictly decreasing on the interval $(-\infty,a_{k}],$ constant on $[a_{k},a_{k+1}]$ and strictly increasing on $[a_{k+1},+\infty).$ Now, let WLOG $a_{1}\leq\ldots\leq a_{n}$ be our arithmetic sequence and $f$ defined as above. We see that $f(0)=f\left( -1\right) =f\left( 2\right)$, hence three values of $f$ are equal. This is possible only if $n$ is even (say $n=2k$) and $0,-1,2\in\lbrack a_{k},a_{k+1}].$ We deduce that the common difference $d$ is at least $3,$ and $a_{k}\leq-1,a_{k+1}\geq2.$ A short computation shows that in this case \[ 507=\sum\left| a_{i}\right| \geq3k^{2}, \] hence $k\leq13,$ that is, $n\leq26.$ Equality is obtained when $a_{1}=-37,a_{2}=-34,\ldots,a_{26}=38.$