Answer: three circles. Solution. Let $A C=4, A B=2, B C=3$ and let $C_{1}, A_{1}$ and $B_{1}$ be the midpoints of sides $A B, B C$ and $A C$ respectively. Note that angle $B$ is obtuse, since $A C^{2}>A B^{2}+B C^{2}$. Therefore, points $B$ and $B_{1}$ lie inside the circle of radius 1 with center at point $O$ - the midpoint of segment $C_{1} A_{1}$ (here we take into account that $\angle C_{1} B_{1} A_{1}=\angle B$ and that the length of the midline is half of $A C$). Thus, the parallelogram $B A_{1} B_{1} C_{1}$ is covered by a unit circle with center $O$. Two other unit circles with centers $M$ and $N$ at the midpoints of $A B_{1}$ and $B_{1} C$ will cover triangles $A C_{1} B_{1}$ and $B_{1} A_{1} C$ (here again we use the fact that angles $A C_{1} B_{1}$ and $B_{1} A_{1} C$ are equal to angle $B$, and thus are obtuse). We will now show that two unit circles are insufficient. Otherwise, their centers must coincide with points $M$ and $N$ (otherwise, segment $A C$ of length 4 could not be covered by two circles of diameter 2). But then point $B$ lies outside both of these circles, since otherwise

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the triangle inequality would be violated for triangle $A B M \quad(A B<A M+M B)$ or for triangle $N B C(B C<N B+N C)^{\circ}$.
9.4. What is the minimum number of unit-radius circles required to completely cover a triangle with sides $2 ; 3 ; 4$?