imc

The trick is to realize that the sum $AMC+AM+MC+CA$ is similar to the product $(A+1)(M+1)(C+1)$. If we multiply $(A+1)(M+1)(C+1)$, we get $$(A+1)(M+1)(C+1)=AMC+AM+AC+MC+A+M+C+1.$$ We know that $A+M+C=10$, therefore $(A+1)(M+1)(C+1)=(AMC+AM+MC+CA)+11$ and $$AMC+AM+MC+CA=(A+1)(M+1)(C+1)-11.$$ Now consider the maximal value of this expression. Suppose that some two of $A$, $M$, and $C$ differ by at least $2$. Then this triple $(A,M,C)$ is not optimal. (To see this, WLOG let $A\ge C+2.$ We can then increase the value of $(A+1)(M+1)(C+1)$ by changing $A\to A-1$ and $C\to C+1$.) Therefore the maximum is achieved when $(A,M,C)$ is a rotation of $(3,3,4)$. The value of $(A+1)(M+1)(C+1)$ in this case is $4\cdot 4\cdot 5=80,$ and thus the maximum of $AMC+AM+MC+CA$ is $80-11=\boxed{69}.$