$$A_1, A_2, A_3, \ldots, A_{20}$$ are the vertices of a regular polygon of 20 sides. $$A_1A_2P_2P_1$$ is a square on $$A_1A_2$$ interior of the polygon. $$A_3Q_3Q_4Q_5A_4$$ is the regular pentagon drawn on side $$A_3A_4$$ interior of the polygon. Prove that $$Q_3$$ lies on the line joining $$A_3P_2$$.

(The printed paper had $$A_1A_2A_3P_1$$ for the square; the intended square is $$A_1A_2P_2P_1$$.) Each interior angle of the regular 20-gon is $$\frac{(20 - 2) \times 180^\circ}{20} = 162^\circ$$ and each interior angle of a regular pentagon is $$108^\circ$$. Since $$\angle A_1A_2P_2 = 90^\circ$$, we get $$\angle A_3A_2P_2 = 162^\circ - 90^\circ = 72^\circ$$, and as $$A_2P_2 = A_1A_2 = A_2A_3$$, triangle $$A_2A_3P_2$$ is isosceles with $$\angle A_2A_3P_2 = \frac{180^\circ - 72^\circ}{2} = 54^\circ$$. In the pentagon $$\angle Q_3A_3A_4 = 108^\circ$$, so $$\angle A_2A_3Q_3 = 162^\circ - 108^\circ = 54^\circ$$. Thus rays $$A_3P_2$$ and $$A_3Q_3$$ make the same angle with $$A_3A_2$$ on the same (interior) side, so they coincide and $$Q_3$$ lies on the line $$A_3P_2$$.