Very "Strict" Triangle!

If the triangle has sides $a,b,c$ and semiperimeter $s = \tfrac12(a+ b+c)$ , we want $a+b-c,a+c-b,b+c-a \ge 5$ , and hence we want $2(s-a),2(s-b),2(s-c) \ge 5$ .If we write $\alpha \; = \; 2(s-a) \hspace{1cm} \beta \; = \; 2(s-b) \hspace{1cm} \gamma \;=\; 2(s-c)$ then we want $\alpha,\beta,\gamma$ to be integers that are all at least $5$ . We note that the area of the triangle is $\Delta \; = \; \sqrt{s(s-a)(s-b)(s-c)} \; = \; \tfrac14\sqrt{(\alpha+\beta+\gamma)\alpha\beta\gamma}$ while the perimeter of the triangle is $2s = \alpha + \beta + \gamma$ . Thus we want $\begin{aligned} \Delta & = \; 4s \\ \sqrt{(\alpha+\beta+\gamma)\alpha\beta\gamma} & = \; 8(\alpha+\beta+\gamma) \\ \alpha\beta\gamma & = \; 64(\alpha+\beta+\gamma) \end{aligned}$ We also require that $a = \tfrac12(\beta+\gamma)$ , $b = \tfrac12(\alpha+\gamma)$ , $c = \tfrac12(\alpha+\beta)$ all to be integers, and hence we require $\alpha,\beta,\gamma$ to all have the same parity. Since $\alpha\beta\gamma = 64(\alpha+\beta+\gamma)$ is even, we deduce that $\alpha,\beta,\gamma$ are all even, so we can write $\alpha = 2A$ , $\beta = 2B$ , $\gamma = 2C$ where $A,B,C$ are integers with $A,B,C \ge 3 \hspace{2cm} ABC \; = \; 16(A+B+C)$ Without loss of generality we can assume that $A \le B \le C$ . Then $ABC \le 48C$ , so that $AB \le 48$ . Thus we have $3 \le A \le 6$ and $A \le B \le \tfrac{48}{A}$ , so there are only a finitely many possible values for $A,B$ . Checking these, we obtain the following $8$ possible triples for $(A,B,C)$ : $\begin{array}{cccc} (3,6,72) & (3,7,32) & (3,8,22) & (3,12,12) \\ (4,5,36) & (4,6,20) & (4,8,12) & (6,7,8) \end{array}$ and hence the following $\boxed{8}$ triples for the triangle sides $(a,b,c)$ : $\begin{array}{cccc} (78,75,9) & (39,35,10) & (30,25,11) & (24,15,15) \\ (41,40,9) & (26,24,10) & (20,16,12) & (15,14,13) \end{array}$