A problem by Victor Loh

Let the 3 vertices of the equilateral triangle be $A$ , $B$ and $C$ such that $\triangle ABC$ has sides $\overline{AB}$ , $\overline{BC}$ and $\overline{AC}$ . Let the center of the circle (and triangle) be $D$ . Connect $A$ , $B$ and $C$ to $D$ . Let $r$ denote the radius of the circle. It is easy to see that $\overline{AD}$ = $\overline{BD}$ = $\overline{CD}$ = $r$ . Also, $\angle BDC = \angle ADC = \angle ADB = 120 ^ \circ$ . Since $\triangle ABC$ has been divided into 3 equal triangles, each triangle = $\frac{1}{3}M$ .

We have

$\frac{1}{3}M = \frac{1}{2}r^{2}\sin 120 ^ \circ$

$M = \frac{3}{2}r^{2}\frac{\sqrt{3}}{2}$

$N = \pi r^{2}$

Then

$\frac{100M}{N}$ $=\frac{\frac{100 \times 3}{2}r^{2}\frac{\sqrt{3}}{2}}{\pi r^{2}}$

$=\frac{75\sqrt{3}}{3.14}$

$\frac{75\sqrt{3}}{3.14} = 41.3706...$

Hence, the closest integer is $\boxed{41}$ .