65th Czech and Slovak Mathematical Olympiad

Notice that no negative number $m$ satisfies the problem evidently (absolute value is non-negative number).

Now we interpret the problem geometrically. A graph of some function $y=x^2+ax+b$ lies in a horizontal strip between lines $y=+m$ and $y=-m$ and in the interval $(0,2)$. Our aim is to find the closest strip which contains the graph of such quadratic function in the interval $(0,2)$.



The function $$f(x)=(x-1{)}^2-\frac{1}{2}=x^2-2x+\frac{1}{2},$$ seems to be a good candidate for such the closest strip. Such function has $a=-2$, $b=\frac{1}{2}$ and it satisfies (to be shown below) inequalities $-\frac{1}{2}\le f(x)\le \frac{1}{2}$.

Really, this inequalities are equivalent to the inequalities $0\le (x-1{)}^2\le 1$, which are evidently fulfilled for $x\in (0,2)$. Quadratic function $f(x)=x^2-2x+\frac{1}{2}$ thus satisfies the conditions of the problem for $m=\frac{1}{2}$.

In the second part of the solution we will show that there is no quadratic function satisfying the problem for any $m<\frac{1}{2}$.

The crucial fact will be that at least one from differences $f(0)-f(1)$ and $f(2)-f(1)$ is greater or equal to $1$ for an arbitrary function $f(x)=x^2+ax+b$. This fact will imply that width of the closest strip will be greater or equal to $1$. This will exclude the values $m<\frac{1}{2}$. For $f(0)-f(1)\ge 1$ we obtain the desired estimate $2m\ge 1$ easily from the well-known triangle inequality $|a-b|\le |a|+|b|$: $$1\le |f(0)-f(1)|\le |f(0)|+|f(1)|\le 2m.$$ Similarly we estimate for $f(2)-f(1)\ge 1$.

Now it remains to verify at least one from inequalities $f(0)-f(1)\ge 1$ and $f(2)-f(1)\ge 1$ for arbitrary $f(x)=x^2+ax+b$. The values $$f(0)=b,f(1)=1+a+b,f(2)=4+2a+b,$$ yields $$f(0)-f(1)=-1-a\ge 1\iff a\le -2,$$ $$f(2)-f(1)=3+a\ge 1\iff a\ge -2.$$ So at least one from inequalities $f(0)-f(1)\ge 1$ and $f(2)-f(1)\ge 1$ is true (regardless of the choice $a,b$).

Conclusion. The desired minimal value of $m$ is $\frac{1}{2}$.