Angle chasing gives $\angle{PSM}=30º$ and $\angle{RSM}=75º$, so $\angle{PSR}=105º$. But then $PN=BN=NQ$, and $\angle{PNQ}=90º+60º=150º$, so $\angle{PQN}=15º$. But $NQ\perp{QR}$, so $\angle{PQR}=90º-15º=75º$. Thus, we have $\angle{PQR}+\angle{PSR}=75º+105º=180º$, so $PQRS$ is cyclic, as required.

posted for the image link

Attachments:

Points $M$ and $N$ are the midpoints of sides $AB$ and $BC$ of the square $ABCD$. According to the fgure, we have drawn a regular hexagon and a regular $12$-gon. The points $P, Q$ and $R$ are the centers of these three polygons. Prove that $PQRS$ is a cyclic quadrilateral. Proposed by Mahdi Etesamifard - Iran

The solution i found during the contest

Attachments:

ItsBesi wrote: I solved this problem for 15 minutes during the contest but writing the proof was a bit difficult for me in total it took me 40 minutes

Easy angle chasing gives us, $\angle PQR + \angle PSR=180^{\circ}$.

It's trivial if you deconstruct the do-decagon into equilateral triangles and squares. In fact $PQRS$ is an isosceles trapezoid (I think).

angle PSR = 105 and angle PQR = 75 which sum to 180 as desired

Note that $\angle SRQ = \angle SRB + \angle BRQ = 60^\circ + 15^\circ = 75^\circ,$ so it suffices to show that $\angle SPQ = 105^\circ.$ Note that $NP = NB = NQ,$ so $\angle PBQ = \frac{\angle BNQ}{2} = 30^\circ$, and also $\angle MPB = 45^\circ.$ Additionally $SM = MP$ and its easy to get $\angle SMP = 120^\circ$ so $\angle SPM = 30^\circ$ and so $\angle SPQ = 30 + 45 + 30 = 105,$ done