16. Let $A B C D$ be a quadrilateral inscribed in the unit circle such that $\angle B A D$ is 30 degrees. Let $m$ denote the minimum value of $C P+P Q+C Q$, where $P$ and $Q$ may be any points lying along rays $A B$ and $A D$, respectively. Determine the maximum value of $m$.
Answer: $\square$
For a fixed quadrilateral $A B C D$ as described, we first show that $m$, the minimum possible length of $C P+P Q+Q C$, equals the length of $A C$. Reflect $B, C$, and $P$ across line $A D$ to points $E, F$, and $R$, respectively, and then reflect $D$ and $F$ across $A E$ to points $G$ and $H$, respectively. These two reflections combine to give a $60^{\circ}$ rotation around $A$, so triangle $A C H$ is equilateral. It also follows that $R H$ is a $60^{\circ}$ rotation of $P C$ around $A$, so, in particular, these segments have the same length. Because $Q R=Q P$ by reflection,
$$
C P+P Q+Q C=C Q+Q R+R H .
$$

The latter is the length of a broken path $C Q R H$ from $C$ to $H$, and by the "shortest path is a straight line" principle, this total length is at least as long as $C H=C A$. (More directly, this follows from the triangle inequality: $(C Q+Q R)+R H \geq C R+R H \geq C H)$. Therefore, the lower bound $m \geq A C$ indeed holds. To see that this is actually an equality, note that choosing $Q$ as the intersection of segment $C H$ with ray $A D$, and choosing $P$ so that its reflection $R$ is the intersection of $C H$ with ray $A E$, aligns path $C Q R H$ with segment $C H$, thus obtaining the desired minimum $m=A C$.
We may conclude that the largest possible value of $m$ is the largest possible length of $A C$, namely 2 : the length of a diameter of the circle.