Bulgaria competition problems, 2022

Let $N$ and $P$ be the midpoints of $AC$ and $CM$, respectively. Then the pentagon $ANPOM$ is cyclic with $AM \parallel PN$, meaning that $\triangle CAM = 180^\circ - \angle ANP = 90^\circ - \angle PNO = 90^\circ - \angle PMO = \angle AMC$,

which is equivalent to $AC = CM$. Let $Q$ be the midpoint of $AM$ (clearly $CQ \perp AB$). Denote $AC = b$, $BC = a$, $AB = c$. We have $CM^2 - MQ^2 = CQ^2 = BC^2 - BQ^2$, i.e., $b^2 - \frac{c^2}{16} = a^2 - \frac{9c^2}{16}$, thus $2(a^2 - b^2) = c^2$. Obviously $c = 2k$ and $(a-b)(a+b) = 2k^2$, with the same parity of the factors in LHS. Therefore, $k$ should be even. If $k = 2$ then $a-b=2$, $a+b=4$, i.e., $a=3$, $b=1$, which together with $c=4$ fails the triangle inequality. When $k=4$, we have either $a-b=2$, $a+b=16$ (i.e., $a=9$, $b=7$, $c=8$, a valid configuration with a perimeter 24) or $a-b=4$, $a+b=8$ (i.e., $a=6$, $b=2$, again a non-valid configuration). For $k \ge 6$, we derive $c \ge 12$ and $a+b+c > 2c \ge 24$ from the triangle inequality. Thus, the answer is 24, and $AB = 8$, $AC = 7$, $BC = 9$.