67th Czech and Slovak Mathematical Olympiad

Consider linear functions $f(x)=ax+b$, $g(x)=bx+a$. Since $a,b$ are distinct and positive, their graphs are two distinct lines with positive slope. As $f(1)=g(1)=a+b$, point $P=[1,a+b]$ is the intersection of these lines (Fig. 3).

Without loss of generality, assume $b>a$ (i.e. the line determined by $g$ is the “steeper” one). Then $f(x)>g(x)$ for $x<1$, whereas $f(x)<g(x)$ for $x>1$: indeed, $$f(x)-g(x)=(ax+b)-(bx+a)=(b-a)(1-x).$$

Fig. 3

Let $t=\lfloor a+b\rfloor$ and consider $x_1\le 1<x_2$ such that $g(x_1)=t$ and $g(x_2)=t+1$ (that is, $x_1=\frac{t-a}{b}$ and $x_2=\frac{t+1-a}{b}$). We claim that the interval $[x_1,x_2]$ has all the desired properties.

First, for any $x\in [x_1,x_2)$ we have $$t=g(x_1)\le \min \{f(x),g(x)\}\le \max \{f(x),g(x)\}<g(x_2)=t+1,$$ and thus $x$ is a solution to the equation.

Second, $$1=(t+1)-t=b{x}_2+a-(b{x}_1+a)=b(x_2-x_1),$$ and thus $x_2-x_1=1/b=1/\max \{a,b\}$ and the interval has the desired length.