4. We will say that a point on the plane $(u, v)$ lies between the parabolas $y=f(x)$ and $y=g(x)$ if $f(u) \leqslant$ $v \leqslant g(u)$. Find the smallest real $p$, for which the following statement is true: any segment, the ends and the midpoint of which lie between the parabolas $y=x^{2}$ and $y=x^{2}+1$, lies entirely between the parabolas $y=x^{2}$ and $y=x^{2}+p$.
Answer: 9/8. Consider a segment $s$, the ends and midpoint of which lie between the original parabolas, and the segment itself intersects the parabola. Let it lie on the line $y=k x+\ell$. The two original parabolas cut three segments $A B, B C$ and $C D$ on it. From Vieta's theorem, it immediately follows that $A B=C D$. The ends of the segment $s$ lie on the segments $A B$ and $C D$, and the midpoint, without loss of generality, on the segment $C D$. It is easy to understand that the existence of such a segment is equivalent to the condition $B C \leqslant C D$. Calculating the abscissas $x_{B}=\left(k-\sqrt{k^{2}+4 \ell-4}\right) / 2, x_{C}=$ $\left(k+\sqrt{k^{2}+4 \ell-4}\right) / 2$ and $x_{D}=\left(k+\sqrt{k^{2}+4 \ell}\right) / 2$, and simplifying the inequality $x_{C}-x_{B} \leqslant x_{C}-x_{D}$, we obtain the condition $k^{2}+4 \ell \leqslant 9 / 2(*)$

We will prove that not only the segment $s$, but the entire line $y=k x+\ell$ lies below the parabola $y=x^{2}+9 / 8$. For this, it is sufficient to check that the discriminant of the equation $x^{2}-k x-\ell+9 / 8=0$ is less than or equal to 0, i.e., precisely $k^{2}+4 \ell-9 / 2 \leqslant 0$.

On the other hand, the line $y=9 / 8$ (i.e., $k=0, \ell=9 / 8$) satisfies the equality in condition (*). In this case, the required segment on this line has the form $B D$ with midpoint $C$. It definitely intersects any parabola $y=x^{2}+p$ for $p<9 / 8$.