Problem 12.4. Find all triangles $A B C$ with integer sidelengths such that the side $A C$ is equal to the bisector of $\Varangle B A C$ and the perimeter of $\triangle A B C$ is equal to $10 p$, where $p$ is a prime number.

Oleg Mushkarov
12.4. Using the standard notation for the elements of $\triangle A B C$ we have $b^{2}=l_{a}^{2}=b c-\frac{a^{2} b c}{(b+c)^{2}}$. Hence

$$
a^{2} c=(c-b)(c+b)^{2}
$$

Let $\frac{a}{c}=\frac{m}{n},(m, n)=1$, and $\frac{b}{c}=\frac{r}{s},(r, s)=1$. Then (1) implies that

$$
\frac{m^{2}}{n^{2}}=\frac{(s-r)(s+r)^{2}}{s^{3}}
$$

Since both sides are irreducible fractions, we obtain $m^{2}=(s-r)(s+r)^{2}$ and $n^{2}=s^{3}$. The first equality shows that $s-r$ is a perfect square and the second one implies that $s$ is a perfect square. Set $s=t^{2}$ and $s-r=k^{2}$. Then $r=t^{2}-k^{2}, m=k\left(2 t^{2}-k^{2}\right)$ and $n=t^{3}$.

We now set $a=m x, c=n x, b=r y$ and $c=s y$. Then $n x=s y$, i.e. $y=t x$. Therefore $a=x k\left(2 t^{2}-k^{2}\right), b=x t\left(t^{2}-k^{2}\right)$ and $c=x t^{3}$, where $t>k$ and $(t, k)=1$. Moreover, one can easily check that these $a, b$ and $c$ satisfy the triangle inequality.

Now the condition $a+b+c=10 p$ becomes $x(k+t)\left(2 t^{2}-k^{2}\right)=10 p$. Note that $\left(k+t, 2 t^{2}-k^{2}\right)=1$. Then it is easy to see that we have only the following possibilities:

$$
\left|\begin{array}{l|l|l}
x=1 \\
k+t=5 \\
2 t^{2}-k^{2}=2 p
\end{array}, \quad\right| \begin{aligned}
& x=1 \\
& k+t=10 \\
& 2 t^{2}-k^{2}=p
\end{aligned}, \quad \begin{aligned}
& x=2 \\
& k+t=5 \\
& 2 t^{2}-k^{2}=p
\end{aligned}
$$

A direct verification shows that $(x, k, t)=(1,2,3),(x, k, t)=(1,3,7)$ and $(x, k, t)=(2,1,4)$. Therefore $(a, b, c)=(28,15,27),(a, b, c)=(267,280,343)$ and $(a, b, c)=(62,120,128)$.