Harvard-MIT Mathematics Tournament

Solution:

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 $A'$, $B'$, and $D'$ be the feet of the perpendiculars from $A$, $B$, and $D$ to $B D$, $D A$, and $A B$, respectively. Angle chasing yields

$$\begin{aligned} \angle A H D & = \angle A H B' + \angle D H B' = \left(90^{\circ} - \angle A' A B'\right) + \left(90^{\circ} - \angle D' D B'\right) \\ & = \angle B D A + \angle B A D = 1^{\circ} + 2^{\circ} = 3^{\circ} \\ \angle H A D & = 90^{\circ} - \angle A H B' = 89^{\circ} \\ \angle H D A & = 90^{\circ} - \angle D H B' = 88^{\circ} \end{aligned}$$

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.