9.4. Given a triangle with sides $a, b, c$. On its sides as diameters, semicircles are constructed outward, resulting in a figure $\Phi$ composed of the triangle and three semicircles. Find the diameter of $\Phi$ (the diameter of a set on a plane is the greatest distance between its points).
Answer: $\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$. Solution. Let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $B C, A C$, and $A B$ respectively. It is obvious that if $M N$ is the diameter of figure $\Phi$, then $M$ and $N$ are points on the boundary of $\Phi$ (otherwise, segment $M N$ could be extended to the boundary). There can be two cases: 1) $M$ and $N$ are points on one semicircle, say, constructed on $A B$, and 2) $M$ and $N$ belong to two different semicircles, say, constructed on $A B$ and $B C$. In the first case, it is clear that the length of segment $M N$ does not exceed the length of $A B$. In the second case, draw the line $C_{1} A_{1}$ and denote the points of intersection with the semicircles as $P$ and $Q$. Then $M N$ does not exceed the length of the broken line $M C_{1} A_{1} N$, and this length is equal to $P Q$. Thus, in the second case, $M N$ reaches

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the maximum value, equal to $P Q$, when $M$ coincides with $P$, and $N$ coincides with $Q$. The length of $P Q$ is equal to the semiperimeter of triangle $A B C$, since $P C_{1}=\frac{A B}{2}$, $A_{1} Q=\frac{B C}{2}$, and $A_{1} C_{1}=\frac{A C}{2}$ by the property of the midline of a triangle. Since the semiperimeter of the triangle is greater than any side (by the triangle inequality), we conclude that the second case actually applies, and there are exactly three positions of segment $M N$ where the length of $M N$ reaches the value of the semiperimeter of triangle $A B C$ (for each pair of semicircles).