It All Came From Pencils 4

Suppose the interior angles of $p$ - pencilogon is $u$ , the interior angles of $q$ - pencilogon is $v$ ( $u$ , $v$ is a positive real number), it's not hard ot show that the interior angles of the pencilogon are $180^\circ-\frac{x}{2}$ . We have $180^\circ-\frac{\alpha}{2}=u$ $\alpha=2(180^\circ-u)$ $180^\circ-\frac{\beta}{2}=v$ $\beta=2(180^\circ-v)$ $\beta-\alpha=2 (u-v)=2k$ $\therefore u-v=k$ Hence, $\frac{\alpha}{\beta}=\frac{180^\circ-u}{180^\circ-v}=\frac{180°-62k}{180°-61k}$ $-(u+61k) 180°+61uk=-(v+62k) 180°+62vk$ $(u-v-k) 180°=(61u-62v)k$ $(61u-62v)k=0$ Since $k\neq0$ , $61u-62v=0$ and we conclude $\frac{u}{v}=\frac{62} {61}$ .

Since the ratio of the interior angles of $p$ - pencilogon to the interior angles of $q$ - pencilogon is $\frac{62}{61}$ , $\frac{\frac{(p-2)\times 180^\circ}{p}}{\frac{(q-2)\times 180^\circ}{q}}=\frac{62}{61}$ $\frac{q(p-2)}{p(q-2)}=\frac{62}{61}$ $61q(p-2)=62p(q-2)$ $61pq-122q=62pq-124p$ $122q=124p-pq=p(124-q)$

As $q>0$ , $122q>0$ , so $p(124-q)>0$ , $q<124$ , therefore, the maximum value of $q$ is $123$ .