Exercise 1.12

Proof. Let $n$ be the number of sides of the regular polygon, $m$ the number of sides starting from a summit in the lattice, $\alpha$ the measure of the exterior angle, $\beta$ the measure of the interior angle (in radians) ( $\alpha +\beta =\pi$).

Then $\alpha =2\pi ∕n,\beta =\pi -2\pi ∕n$.

$\mathit{m\beta }=2\pi ,m(\pi -2\pi ∕n)=2\pi ,m(1-2∕n)=2$, so

As this equation is symmetric in $m,n$, we may suppose first $m\le n$.

In this case $1∕m\ge 1∕n$, so $2∕n\le 1∕2$ : $n\ge 4$.

If $n>6$, $1∕n<1∕6,1∕m=1∕2-1∕n>1∕2-1∕6=1∕3$, so $m<3,m\le 2$ : $m=1$ or $m=2$.

If $m=1$, $n<0$ : it is impossible. If $m=2$, $1∕n=0$ : also impossible. Therefore $n\le 6$ : $4\le n\le 5$. If $n=4,m=4$. if $n=5$, $n=10∕3$ : impossible. if $n=6$, $m=3$. Using the symmetry, the set of solutions of (1) is

corresponding with the usual lattices composed of equilateral triangles, squares or hexagons. □