Taiwan IMO Selection Camp

Define $\alpha = \pi - \angle FAB = \pi - \angle CDE$, $\beta = \pi - \angle ABC = \pi - \angle DEF$, $\gamma = \pi - \angle BCD = \pi - \angle EFA$. Clearly $\alpha + \beta + \gamma = \pi$, and at least two of the three are acute. Let $O'$ be the intersection point of $B_1E_1$ and $C_1F_1$, and $O''$ the intersection point of $C_1F_1$ and $A_1D_1$ (where $C_1, F_1$ are the midpoints of $CD$ and $FA$ respectively). Further let $\alpha' = \angle F_1O''A_1 = \angle C_1O''D_1$, $\beta' = \angle A_1OB_1 = \angle D_1OE_1$, $\gamma' = \angle B_1O'C_1 = \angle E_1O'F_1$. Since we may translate $C_1F_1$ parallel to itself until it passes through the point $O$, and such a parallel translation does not change the angles of intersection between lines, we know that $\alpha' + \beta' + \gamma' = \pi$. Moreover, by the cyclic symmetry of the problem, the statement of the problem is equivalent to $\beta' = \angle D_1OE_1 = \frac{1}{2}\angle DEF = \frac{\pi-\beta}{2}$, or to $\alpha' = \frac{\pi-\alpha}{2}$, or to $\gamma' = \frac{\pi-\gamma}{2}$. Hence, without loss of generality, we may relabel the vertices cyclically without changing the problem, so that we may assume that $\alpha$ and $\beta$ are acute. Write $m = \tan \alpha$, $M = \tan \beta$; both $m, M$ are positive.

Now we set up a coordinate system: without loss of generality let $A \equiv (-d, -h)$, $B \equiv (d, -h)$, $D \equiv (d' + \Delta, h)$, $E \equiv (-d' + \Delta, h)$, where $d, d', h$ are positive numbers, and $d' \ge d$. After a careful algebraic computation (finding the equations of the lines $BC$, $CD$, $EF$, $FA$, computing their intersection points, and using the fact that the slopes of lines $BC$, $CD$ are $m, M$ respectively), we obtain
$$C \equiv \left( \frac{Md' + M\Delta + md + 2h}{M+m}, \frac{Mmd' + Mm\Delta - Mmd - Mh + mh}{M+m} \right),$$
$$F \equiv \left( -\frac{md' - m\Delta + Md + 2h}{M+m}, \frac{Mmd' - Mm\Delta - Mmd + Mh - mh}{M+m} \right).$$
Note that the $y$-coordinates of $C, F$ must lie in the interval $(-h, h)$ in order for the hexagon to be convex, and hence we have
$$\frac{2h}{m} > d' - d + \Delta > -\frac{2h}{M}, \quad \frac{2h}{M} > d' - d - \Delta > -\frac{2h}{m}.$$
After some further algebraic computation, and using the above results, we obtain
$$BC = \frac{M(d' - d + \Delta) + 2h}{M + m}\sqrt{m^2 + 1},$$
$$EF = \frac{2h - M(d' - d - \Delta)}{M + m}\sqrt{m^2 + 1},$$
$$CD = \frac{2h - m(d' - d + \Delta)}{M + m}\sqrt{M^2 + 1},$$
$$FA = \frac{m(d' - d - \Delta) + 2h}{M + m}\sqrt{M^2 + 1}.$$
Hence we have
$$\begin{aligned} d+d' &= \frac{AB+DE}{2} = \frac{BC+EF}{2} = \frac{CD+FA}{2} \\ &= \frac{2h+M\Delta}{M+m}\sqrt{m^2+1} = \frac{2h-m\Delta}{M+m}\sqrt{M^2+1}. \end{aligned}$$
From the last two terms of the above equality, we find that
$$\begin{aligned} \left(\frac{\Delta}{2h}\right)^2 + 2 \frac{Mm+1}{M-m} \left(\frac{\Delta}{2h}\right) - 1 &= 0, \\ \frac{\Delta}{2h} &= \frac{-Mm-1 + \sqrt{(M^2+1)(m^2+1)}}{M-m} = \tan \frac{\alpha - \beta}{2}, \end{aligned}$$
where here we assume that $M \neq m$; if $M = m$, then $\Delta = 0$, and we have $d+d' = \frac{h\sqrt{m^2+1}}{m} = \frac{h}{\sin \beta} = \frac{h}{\sin \alpha}$. Note that we discard the negative root, not only for the sake of continuity of the solution as $M \to m$, but also because, after some computation, one can see that $2h = (d+d')(\sin \alpha + \sin \beta)$ always holds, since $2h = BC \sin \beta + CD \sin \alpha = EF \sin \beta + FA \sin \alpha$.

We can now easily derive that
$$A_1 \equiv (0, -h), \quad D_1 \equiv (\Delta, h),$$
$$B_1 \equiv \left( \frac{Md' + Md + M\Delta + 2md + 2h}{2(M+m)}, \frac{Mmd' + Mm\Delta - Mmd + 2Mh}{2(M+m)} \right),$$
$$E_1 \equiv \left( -\frac{2md' - 2m\Delta - M\Delta + Md + Md' + 2h}{2(M+m)}, \frac{Mmd' - Mm\Delta - Mmd + 2Mh}{2(M+m)} \right).$$
$\beta'$ is also the angle between the vectors $\overrightarrow{A_1D_1}$ and $\overrightarrow{B_1E_1}$, where
$$\overrightarrow{A_1D_1} \equiv (\Delta, 2h),$$
$$\overrightarrow{B_1E_1} \equiv \left( -\frac{(M+m)(d+d') + 2h - m\Delta}{M+m}, -M\frac{m\Delta - 2h}{M+m} \right).$$
Since $\frac{2h}{\Delta} = \cot \frac{\alpha-\beta}{2}$, the vector $\overrightarrow{A_1D_1}$ makes an angle of size $\frac{\pi-\alpha+\beta}{2}$ with the horizontal axis. Furthermore,
$$\begin{aligned} \frac{M(m\Delta - 2h)}{(M+m)(d+d') + 2h - m\Delta} &= \frac{\tan \alpha (\tan \beta \tan \frac{\alpha-\beta}{2} - 1)}{\frac{\tan \alpha + \tan \beta}{\sin \alpha + \sin \beta} + 1 - \tan \beta \tan \frac{\alpha-\beta}{2}} \\ &= \frac{-\sin \alpha \cos \frac{\alpha+\beta}{2}}{\cos \frac{\alpha+\beta}{2}(1 + \cos \alpha)} = -\tan \frac{\alpha}{2}, \end{aligned}$$
that is, the vector $\overrightarrow{B_1E_1}$ makes an angle of $\pi - \frac{\alpha}{2}$ with the horizontal axis. Therefore, the angle formed between $\overrightarrow{B_1E_1}$ and $\overrightarrow{A_1D_1}$ is $\beta' = \pi - \frac{\alpha}{2} - \frac{\pi-\alpha+\beta}{2} = \frac{\pi-\beta}{2}$. This completes the proof.

Note that when $M = m$, $\overrightarrow{A_1D_1} \equiv (0, 2h)$, which makes an angle of $\frac{\pi}{2}$ with the horizontal axis; and $\overrightarrow{B_1E_1} \equiv \left( -\frac{m(d+d')+h}{m}, h \right)$, and since
$$-\frac{mh}{m(d+d')+h} = -\frac{\sin \beta}{1+\cos \beta} = -\tan \frac{\beta}{2},$$
we get that $\overrightarrow{B_1E_1}$ makes an angle of $\pi - \frac{\beta}{2}$ with the horizontal axis. Once again we have $\beta' = \pi - \frac{\beta}{2} - \frac{\pi-\beta}{2} = \frac{\pi-\beta}{2}$. This completes the proof.