$[ABC]^2=\frac{a^2b^2c^2}{16R^2}=s(s-a)(s-b)(s-c)$ $16R^2s(s-a)(s-b)(s-c)=a^2b^2c^2$ $R|a^2b^2c^2$ WLOG let $R|a^2$ and so $R|a$ $2R=a/sinA$, $SinA=\frac{a}{2R}$ $a=R$ or $2R$ (if $a=3R$ or more, it is a contradiction as $sinA>1$) If $a=R$, $sina=1/2, cosA=\sqrt{3}/2$ by cos rule, $a^2=b^2+c^2-bc\sqrt{3}$ Contradiction as $a^2$ cant be an integer so a must be $2R$ in which case $sinA=1$ so $ABC$ is a right angle triangle