Question:

For lower triangular unipotent: $(A^n)_{21}=n\sqrt{a}=72$ and $(A^n)_{32}=n\sqrt{b}=72$ and $(A^n)_{31}=\binom{n}{2}a+n a\sqrt{a}/\sqrt{b}\cdot...$; actually for this form $A^n_{31}=\binom{n}{2}\sqrt{a}\cdot\sqrt{b}+na\sqrt{a}$... Let $(A)_{21}=\sqrt{a}$, $(A)_{31}=a\sqrt{a}$, $(A)_{32}=\sqrt{b}$. Pattern: $(A^n)_{21}=n\sqrt{a}$; $(A^n)_{32}=n\sqrt{b}$; $(A^n)_{31}=n\cdot a\sqrt{a}+\binom{n}{2}\sqrt{a}\sqrt{b}$. From data: $n\sqrt{a}=72$, $\binom{n}{2}\sqrt{ab}+na\sqrt{a}=3600$. And $(A^n)_{32}=n\sqrt{b}=72$. So $\sqrt{a}=\sqrt{b}$, $a=b$. Then $3600=n\cdot a\sqrt{a}+\binom{n}{2}a=na^{3/2}+\binom{n}{2}a$. $(n\sqrt{a})^2=72^2=5184=n^2 a$. $na^{3/2}=(n\sqrt{a})\cdot a=72a$. $\binom{n}{2}a=\frac{n(n-1)}{2}a$. $72a+\frac{n(n-1)}{2}a=3600$. Also $n\sqrt{a}=72\Rightarrow\sqrt{a}=72/n\Rightarrow a=5184/n^2$. $72\cdot(5184/n^2)+\frac{n(n-1)}{2}\cdot(5184/n^2)=3600$. $\frac{5184}{n^2}[72+\frac{n(n-1)}{2}]=3600$. $\frac{5184}{n^2}\cdot\frac{144+n^2-n}{2}=3600$. $5184(n^2-n+144)=7200n^2$. $5184n^2-5184n+5184\cdot 144=7200n^2$. $2016n^2+5184n-746496=0$. $n^2+\frac{5184}{2016}n-\frac{746496}{2016}=0$. $n^2+\frac{18}{7}n-\frac{2592}{7}=0$. $7n^2+18n-2592=0$. $n=\frac{-18\pm\sqrt{324+4\cdot 7\cdot 2592}}{14}=\frac{-18\pm\sqrt{324+72576}}{14}=\frac{-18\pm\sqrt{72900}}{14}=\frac{-18\pm 270}{14}$. $n=252/14=18$. Number of triangles with no side on 18-gon: $\frac{n(n-4)(n-5)}{6}=\frac{18\cdot 14\cdot 13}{6}=\frac{3276}{6}=546$.