9.3. Find all triples of distinct natural numbers, the least common multiple of which is equal to their sum. The least common multiple of several numbers is the smallest natural number that is divisible by each of these numbers.
Answer. All triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n$.

Solution. Let the required numbers be $a<b<c$. By the condition, their least common multiple, which is equal to $a+b+c$, is divisible by $c$, so $c$ divides the sum $a+b<2 c$. Therefore, $a+b=c$. Next, $a+b+c=2 a+2 b$ is divisible by $b$, so $b$ divides $2 a<2 b$. Therefore, $2 a=b$ and $c=a+b=3 a$, and the natural number $a$ can be any.

On the other hand, for any triple of natural numbers of the form $a=n, b=2 n, c=3 n$, their least common multiple is $6 n$. Indeed, it must divide $3 n$, but it is not equal to $3 n$, because $3 n$ does not divide $2 n$, so it is not less than $3 n \cdot 2=6 n$. It remains to note that $6 n$ divides each of the numbers $a=n, b=2 n, c=3 n$.

Evaluation criteria. Only the example of triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n: 2$ points. Example for specific $n: 1$ point. Proved that $a+b=c: 2$ points.

Proved that $2 a=b: 2$ points.

9.4.. In triangle $\mathrm{ABC}$, point $M$ is the midpoint of side $\mathrm{BC}$, and $\mathrm{H}$ is the foot of the altitude dropped from vertex $\mathrm{B}$. It is known that angle $\mathrm{MCA}$ is twice the angle $\mathrm{MAC}$, and the length of $\mathrm{BC}$ is 10 cm. Find the length of segment $\mathrm{AH}$.

Answer. $\mathrm{AH}=5$ cm.

Solution. In the right triangle $\mathrm{BHC}$, segment $\mathrm{HM}$ is the median to the hypotenuse, so the length of $\mathrm{HM}$ is half the length of $\mathrm{BC}$, which is 5 cm.

Triangle $\mathrm{CHM}$ is isosceles with $\mathrm{HM}=\mathrm{MC}$, so angle $\mathrm{MHC}$ equals angle $\mathrm{MCH}=\mathrm{MCA}$ and is twice the angle $\mathrm{MAC}$.

Angle $\mathrm{MHC}$ is the exterior angle for triangle $\mathrm{AMH}$, its measure is equal to the sum of angles $\mathrm{MAH}$ and $\mathrm{AMH}$ and it is twice $\mathrm{MAH}=\mathrm{MAC}$. Therefore, angles $\mathrm{MAH}$ and $\mathrm{AMH}$ are equal, triangle $\mathrm{AMH}$ is isosceles, so $\mathrm{AH}=\mathrm{HM}=5$ cm.