AUT_ABooklet_2023

Figure 1: Problem 2

Solution:

Our strategy is to compute the angles $\angle DCQ$ and $\angle DCP$ to check that they are equal.

The interior angles of a regular hexagon equal $120^{\circ }$. The triangle $DEF$ is isosceles and therefore, we get $\angle DFE=\angle EDF=30^{\circ }$. This implies $\angle QDC=90^{\circ }$, and since the triangle $QDC$ is also isosceles, we also get $$\angle DCQ=45^{\circ }.$$ The triangle $CBP$ is isosceles and analogously to the above, we get $\angle CBP=\angle CBD=30^{\circ }$. Therefore, we obtain $$\angle PCB=(180^{\circ }-30^{\circ }):2=75^{\circ }\text{and, finally,}\angle DCP=120^{\circ }-75^{\circ }=45^{\circ }.$$ So $\angle DCQ=\angle DCP$ which implies that $C$, $P$ and $Q$ lie on a line.

(Walther Janous) $\square$