August 4, 2023
OC638 (from Crux Mathematicorum Vol. 49, No. 6)
Find all the real numbers \(x\) such that \[\frac{1}{[x]} + \frac{1}{[2x]} = \{x\} + \frac{1}{3}\]where \([x]\) denotes the integer part of \(x\) and \(\{x\} = x-[x]\). For example \([2.5] = 2\), \(\{2.5\} = 0.5\) and \([-1.7] = -2\), \(\{-1.7\} = 0.3\).
We notice that \(x\) must be positive since \(\{x\} + \frac{1}{3}\) is always positive. Also, we claim that \(x\) can’t be an integer. Otherwise, \([x] =x\) and \(\{x\}=0\). Then, we get:
\[\frac{1}{x}+\frac{1}{2x}=\frac{1}{3} \implies \frac{3}{2x}=\frac{1}{3} \implies x=4.5.\]
However, this gives a contradiction because 4.5 is not an integer.
Now, we know that \(0<\{x\}<{1}\) and can do the following casework.
Case 1: \([2x]=2[x]\) i.e. \(0<\{x\}<0.5\)
\[\frac{1}{[x]}+\frac{1}{2[x]}=\{x\}+\frac{1}{3} \implies\{x\}=\frac{3}{2[x]}-\frac{1}{3}=\frac{9-2[x]}{6[x]}\]
Now, we have the inequality
\[0<\frac{9-2[x]}{6[x]}<0.5 \implies 0<9-2[x]<3[x] \implies \frac{9}{5} <[x] <{1}{2}\]
Since \([x]\) is an integer, \([x]\) can only take \(2\), \(3\), or \(4\).
Now, we can plug these values of \([x]\) find their corresponding values of \(\{x\}\).
\[[x]=2\implies \{x\}=\frac{5}{12} \implies x=\frac{29}{12}\] \[[x]=3\implies \{x\}=\frac{1}{6} \implies x=\frac{19}{6}\] \[[x]=4\implies \{x\}=\frac{1}{24} \implies x=\frac{97}{24}\]
Case 2: \([2x]=2[x]+1\) i.e. \(0.5 \le \{x\} <1\)
\[\frac{1}{[x]}+\frac{1}{2[x]+1}=\{x\}+\frac{1}{3}\] \[\implies \frac{3[x]+1}{2[x]^2+[x]}=\{x\}+\frac{1}{3}\] \[\implies \{x\}=\frac{3[x]+1}{2[x]^2+[x]}-\frac{1}{3}\]
We get the inequality
\[\frac{1}{2} \le \frac{3[x]+1}{2[x]^2+[x]}-\frac{1}{3} <1\] \[\implies \frac{5}{6} \le \frac{3[x]+1}{2[x]^2+[x]} < \frac{4}{3}\] \[\implies \frac{5(2[x]^2+[x])}{6} \le 3[x]+1< 4(2[x]^2+24[x])\] \[\implies 30[x]^2+15[x] \le 54[x]+18<48[x]^2+24[x]\]
Now we solve the first part \(30[x]^2+15[x] \le 54[x]+18\). Using the quadratic formula, we get \[\frac{13- \sqrt{409}}{20} \le [x] \le \frac{13+\sqrt{409}}{20}.\] Since \([x]\) is an integer, \([x]\) can only be \(1\). However, before solving the other half of the inequality, let’s plug in our value of \([x]\) into the equation for \(\{x\}\), we get \[\{x\}=\frac{4}{3}-\frac{1}{3}=1.\]
This isn’t possible as \(\{x\}\) in this case is supposed to be in the range \((0.5, 1)\)! Therefore, there aren’t any solutions from this case.
In summary, the only solutions for \(x\) are \(\boxed{\frac{29}{12}}\), \(\boxed{\frac{19}{6}}\), and \(\boxed{\frac{97}{24}}\)