Second Round

An octagon can be subdivided into six triangles (see figure on the left). Together, the angles of those six triangles add up to the same number of degrees as the eight angles of the octagon. Since the angles of any triangle add up to $180$ degrees, this means that the eight angles of the octagon add up to $6\cdot 180^{\circ }=1080^{\circ }$. Hence, each of the angles of the regular octagon is $\frac{1}{8}\cdot 1080^{\circ }=135^{\circ }$.

We now consider the figure from the problem statement (see figure on the right). Line segment $BP$ bisects angle $ABC$, so $\angle ABP=\angle PBC=67{\frac{1}{2}}^{\circ }$. Since triangles $ABP$ and $BCP$ are isosceles (as $|AB|=|AP|$ and $|BC|=|CP|$), we also have $\angle APB=\angle BPC=67{\frac{1}{2}}^{\circ }$ and $\angle BAP=\angle BCP=180^{\circ }-135^{\circ }=45^{\circ }$.

In triangles $ABQ$ and $BCR$ all sides have the same length. These triangles are therefore equilateral and all angles are $60^{\circ }$. From this, we deduce that $\angle PAQ=\angle BAQ-\angle BAP=15^{\circ }$. In the same way, we find $\angle PCR=15^{\circ }$. Furthermore, triangles $PAQ$ and $PCR$ are isosceles (since $|AP|=|AQ|$ and $|CP|=|CR|$), so $\angle APQ=\angle BPC=67{\frac{1}{2}}^{\circ }$ and $\angle CPR=82{\frac{1}{2}}^{\circ }$.

By mirror symmetry, $PQ$ and $PR$ have the same length, so $PQR$ is an isosceles triangle with apex $P$. We have already determined all angles at $P$, except $\angle QPR$. We deduce that $$\begin{array}{rl}\angle QPR& =360^{\circ }-\angle APQ-\angle APB-\angle BPC-\angle CPR\\ & =360^{\circ }-2\cdot 67{\frac{1}{2}}^{\circ }-2\cdot 82{\frac{1}{2}}^{\circ }=60^{\circ }.\end{array}$$ From this and the fact that $PQR$ is isosceles, we directly conclude that $PQR$ is equilateral. $\square$