5. [5] A piece of paper is folded in half. A second fold is made such that the angle marked below has measure $\phi\left(0^{\circ}<\phi<90^{\circ}\right)$, and a cut is made as shown below.

When the piece of paper is unfolded, the resulting hole is a polygon. Let $O$ be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at $O$, and also that $\angle X O Y=144^{\circ}$, where $X$ and $Y$ are the the vertices of the hole adjacent to $O$. Find the value(s) of $\phi$ (in degrees).
Answer: $81^{\circ}$ Try actually folding a piece of paper. We see that the cut out area is a kite, as shown below. The fold was made on $A C$, and then $B E$ and $D E$. Since $D C$ was folded onto $D A$, we have $\angle A D E=\angle C D E$.
Either $A$ or $C$ is the center of the circle. If it's $A$, then $\angle B A D=144^{\circ}$, so $\angle C A D=72^{\circ}$. Using $C A=D A$, we see that $\angle A C D=\angle A D C=54^{\circ}$. So $\angle E D A=27^{\circ}$, and thus $\phi=72^{\circ}+27^{\circ}=99^{\circ}$, which is inadmissible, as $\phi<90^{\circ}$.
So $C$ is the center of the circle. Then, $\angle C A D=\angle C D A=54^{\circ}, \angle A D E=27^{\circ}$, and $\phi=54^{\circ}+27^{\circ}=81^{\circ}$.