11. Let $A B C D E F$ be a convex equilateral hexagon such that lines $B C, A D$, and $E F$ are parallel. Let $H$ be the orthocenter of triangle $A B D$. If the smallest interior angle of the hexagon is 4 degrees, determine the smallest angle of the triangle $H A D$ in degrees.
Answer: 3
Note that $A B C D$ and $D E F A$ are isosceles trapezoids, so $\angle B A D=\angle C D A$ and $\angle F A D=\angle E D A$. In order for the hexagon to be convex, the angles at $B, C, E$, and $F$ have to be obtuse, so $\angle A=\angle D=4^{\circ}$. Letting $s$ be a side length of the hexagon, $A D=A B \cos \angle B A D+B C+C D \cos \angle C D A=s(1+$ $2 \cos \angle B A D)$, so $\angle B A D$ is uniquely determined by $A D$. Since the same equation holds for trapezoid $D E F A$, it follows that $\angle B A D=\angle F A D=\angle C D A=\angle E D A=2^{\circ}$. Then $\angle B C D=180^{\circ}-2^{\circ}=178^{\circ}$. Since $\triangle B C D$ is isosceles, $\angle C D B=1^{\circ}$ and $\angle B D A=1^{\circ}$. (One may also note that $\angle B D A=1^{\circ}$ by observing that equal lengths $A B$ and $B C$ must intercept equal arcs on the circumcircle of isosceles trapezoid $A B C D)$.
Let $A^{\prime}, B^{\prime}$, and $D^{\prime}$ be the feet of the perpendiculars from $A, B$, and $D$ to $B D, D A$, and $A B$, respectively. Angle chasing yields
$$
\begin{aligned}
\angle A H D & =\angle A H B^{\prime}+\angle D H B^{\prime}=\left(90^{\circ}-\angle A^{\prime} A B^{\prime}\right)+\left(90^{\circ}-\angle D^{\prime} D B^{\prime}\right) \\
& =\angle B D A+\angle B A D=1^{\circ}+2^{\circ}=3^{\circ} \\
\angle H A D & =90^{\circ}-\angle A H B^{\prime}=89^{\circ} \\
\angle H D A & =90^{\circ}-\angle D H B^{\prime}=88^{\circ}
\end{aligned}
$$

Hence the smallest angle in $\triangle H A D$ is $3^{\circ}$.
It is faster, however, to draw the circumcircle of $D E F A$, and to note that since $H$ is the orthocenter of triangle $A B D, B$ is the orthocenter of triangle $H A D$. Then since $F$ is the reflection of $B$ across $A D$, quadrilateral $H A F D$ is cyclic, so $\angle A H D=\angle A D F+\angle D A F=1^{\circ}+2^{\circ}=3^{\circ}$, as desired.