5. In triangle $\triangle A B C$, the shortest side has a length of $5 \mathrm{~cm}$. For the heights of this triangle, it holds that $v_{b}=v_{a}+v_{c}$. Determine all possible integer lengths of all sides of the triangle measured in centimeters.

NATIONAL MATHEMATICS COMPETITIONPrimošten, April 3-5, 2017.
5. The shortest side corresponds to the longest height, obviously $b=5 \mathrm{~cm}$. Since $v_{a}=\frac{2 P}{a}, v_{b}=\frac{2 P}{b}$ and $v_{c}=\frac{2 P}{c}$, from $v_{b}=v_{a}+v_{c}$ it follows: $\frac{2 P}{b}=\frac{2 P}{a}+\frac{2 P}{c}, \frac{1}{b}=\frac{1}{a}+\frac{1}{c}, \frac{1}{5}=\frac{a+c}{a c}, a c=5 a+5 c$, $a c-5 c=5 a, c(a-5)=5 a$, hence $c=\frac{5 a}{a-5}$.

Further, $c=\frac{5 a-25+25}{a-5}=\frac{5 a-25}{a-5}+\frac{25}{a-5}=\frac{5(a-5)}{a-5}+\frac{25}{a-5}=5+\frac{25}{a-5}$.

The number $c$ will be an integer if $a-5$ is a divisor of the number 25, i.e., if $a-5$ is equal to 1, 5, or 25, or -1, -5, or -25.

$a-5=1 \Rightarrow a=6 \mathrm{~cm} \Rightarrow c=30 \mathrm{~cm}$ which is impossible due to the triangle inequality

$a-5=5 \Rightarrow a=10 \mathrm{~cm} \Rightarrow c=10 \mathrm{~cm}$

$a-5=25 \Rightarrow a=30 \mathrm{~cm} \Rightarrow c=6 \mathrm{~cm}$ which is impossible due to the triangle inequality

$a-5=-1 \Rightarrow a=4 \mathrm{~cm}$ which is impossible since the shortest side is $5 \mathrm{~cm}$

(or: $a-5=-1 \Rightarrow a=4 \mathrm{~cm} \Rightarrow c=-20$ which is impossible)

$a-5=-5 \Rightarrow a=0$ which is impossible

$a-5=-25 \Rightarrow a=-20$ which is impossible

Therefore, there is only one such triangle with side lengths $a=10 \mathrm{~cm}, b=5 \mathrm{~cm}$ and $c=10 \mathrm{~cm}$.

## NATIONAL MATHEMATICS COMPETITION  Primosten, April 3-5, 2017.  7th grade - solutions

HERE ARE SOME WAYS TO SOLVE THE PROBLEMS. IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING, THE MEMBER OF THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.