$\underline{\textbf{Claim:}}$ $AC=CM$ $\textit{Proof 1:}$ Let the midpoint of $CM$ be point $N$. Since $\angle OMA=90^{\circ}$ and $N\in (AMO)$, this means that $\angle ONA=90^{\circ}$. If $\overline{AN}\cap (ABC)=X$, since $ON\perp AN$, this implies that $N$ is the midpoint of $AX$. Now $NA=NX$ and $NC=NM$, so $ACXM$ is a parallelogram, so \[\angle CAM=\angle CAX+\angle XAM=\angle CBX+\angle CXA=\angle CBX+\angle CBA=\angle ABX=\angle AMN=\angle AMC\Longrightarrow CA=CM\]$\textit{Proof 2:}$ Let $S$ be the midpoint of $AC$. Now $AMONS$ is cyclic, but $SN$ is a midsegment in $\triangle AMC$, so $SN\parallel AM$ and $AMNS$ is cyclic, therefore $AMNS$ is an isosceles trapezoid. Now $AS=NM\Longrightarrow CA=CM$. Now, if $a=BC$, $b=AC=CM$, $c=AB$ we can use the median formula ($m_{c}=\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$) to get that: \[b^2=CM^2=\frac{1}{4}(2a^2+2b^2-c^2)\Longrightarrow c^2=2(a-b)(a+b)\]We want $a,b,c$ to be positive integers with minimal sum (the perimeter of $\triangle ABC$). If $\gcd(a,b)=d>1$, then $\left(\frac{a}{d},\frac{b}{d},\frac{c}{d}\right)$ are the sidelengths of a triangle which satisfies the conditions in the problem with smaller perimeter. Thus we can assume that $a,b,c$ are two by two coprime. Now we have two cases: $\textbf{Case 1:}$ $a-b=2s^2$, $a+b=4t^2$ for positive integers $s,t$. Then $a=s^2+2t^2$, $b=2t^2-s^2$, $c=4st$, so the perimeter is $4t^2+4st$. If $s=1,t=1$ we get $(a,b,c)=(3,1,4)$, impossible. If $s=1, t=2$, then $(a,b,c)=(9,7,8)$, so the least possible perimeter in this case is $24$. $\textbf{Case 2:}$ $a-b=4s^2$, $a+b=2t^2$ for positive integers $s,t$. Then $a=2s^2+t^2$, $b=t^2-2s^2$, $c=4st$, so the perimeter is $2t^2+4st$. If $s=1,t=2$ we get $(a,b,c)=(6,2,8)$, impossible. If $s=1, t=3$, then $(a,b,c)=(11,7,12)$, so the least possible perimeter in this case is $30$. In both cases, it should be briefly explained why those are the minimal perimeter cases though it's pretty obvious. Finally, the answer is $\boxed{24}$ achieved when $(a,b,c)=(9,7,8)$.

Although it may be intuitive to think that $\triangle ABC$ is acute, this is not necessarily the case. We can have $AC<BC$, $\angle C>90^{\circ}$ and $(AMO)$ bisecting $CM$. This, however, isn't a problem for the solution above.