Let $CBA =x^{o},AC=x ,BC=y$ Sine Relation $y*\tan(X) = x$ since $k =2 \pi$ $4xy=x^2+y^2$ $4*y^2*\tan(x) = y^2(\tan(x)^2+1)$ $4\tan(x) = \tan(x)^2+1$ $\tan(x) = 2-\sqrt{3},2+\sqrt{3}$ $x=15$ or ,$ x=75$ A,B,C = (15,75,90) , (75,15,90) now if $k=3$ $\frac{6*\tan(x)}{\pi} =\tan^2(x)+1$ let tan(x)=a $\frac{6*a}{\pi} = a^2+1$ Discriminant $0>(\frac{6}{\pi})^2-4 $ meaning tan(x) is unreal

I should have given in b) $3.14$ instead of $3$, it would have been more amusing. Synthetic: If $m$ is the $C$-median and $h$ is the $C$-altitude, then $S = mh$, $S_1 = \pi m^2$ and so $k = \pi \frac{m}{h}$. If $k=2\pi$, then $m = 2h$ and so the angles are $15^{\circ}, 75^{\circ}, 90^{\circ}$. In general $m\geq h$, so $k \geq \pi > 3.14 > 3$.