Let be $A_0,A_1,A_2,A_3,A_4$ the vertices of the regular pentagon, in this order. For each $k\in\{0,1,2,3,4\}$, there are exactly $2$ isosceles triangles with the top vertex $A_k:\; A_kA_{k-1}A_{k+1}$ and $ A_kA_{k-2}A_{k+2}$ (we consider the indexes modulo $5$). Denote $a_k$ the number associated to the vertex $A_k$. WLOG, we can consider $\{a_0,a_1,a_2,a_3,a_4\}=\{1,2,3,4,5\}$. The isosceles triangles with the top vertex with the associated number $1$ are both successful ($1$ is the smallest number). The isosceles triangles with the top vertex with the associated number $5$ are both successful ($5$ is the greatest number). Exactly one of the isosceles triangles with the top vertex with the associated number $2$ is successful ($1$ is smaller than $2$, the other $3$ numbers are greater). Exactly one of the isosceles triangles with the top vertex with the associated number $4$ is successful ($5$ is greater than $4$, the other $3$ numbers are smaller). The isosceles triangles with the top vertex with the associated number $3$ can be both successful (the triangles with the numbers $312$, respectively $345$). Hence, the maximum possible number of successful triples is $2+2+1+1+2=8$. A configuration with this property is $(a_0,a_1,a_2,a_3,a_4)=(3,2,5,4,1)$.