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
$$\angle PCB = (180^\circ - 30^\circ) : 2 = 75^\circ \quad \text{and, finally,} \quad \angle DCP = 120^\circ - 75^\circ = 45^\circ.$$
So $\angle DCQ = \angle DCP$ which implies that $C$, $P$ and $Q$ lie on a line.

(Walther Janous) $\square$