HMMT February, 2020

Solution:

For each $i$, let $\tau_{i}$ be the transformation of the plane which is rotation by $\theta_{i}$ counterclockwise about $B_{i}$. Recall that a composition of rotations is a rotation or translation, and that the angles of rotation add. Consider the composition $\tau_{2020} \circ \tau_{2019} \circ \cdots \circ \tau_{1}$, with total rotation angle $1010 \pi$. This must be a translation because $1010 \pi=505(2 \pi)$. Also note that the composition sends $A_{1}$ to itself because $\tau_{i}\left(A_{i}\right)=A_{i+1}$. Therefore it is the identity. Now Jason can identify the map $\tau_{1}$ as $\tau_{2}^{-1} \circ \tau_{3}^{-1} \circ \cdots \circ \tau_{2020}^{-1}$, and $B_{1}$ is the unique fixed point of this map.

Solution 2:

Fix an arbitrary coordinate system. For $1 \leq k \leq 2020$, let $a_{k}, b_{k}$ be the complex numbers corresponding to $A_{k}, B_{k}$. The given condition translates to

$$e^{i \theta_{k}}\left(b_{k}-a_{k}\right)=\left(b_{k}-a_{k+1}\right)$$

$$\left(e^{i \theta_{k}}-1\right) b_{k}=e^{i \theta_{k}} a_{k}-a_{k+1}$$

$$\left(e^{-i\left(\theta_{k-1}+\cdots+\theta_{1}\right)}-e^{-i\left(\theta_{k}+\cdots+\theta_{1}\right)}\right) b_{k}=e^{-i\left(\theta_{k-1}+\cdots+\theta_{1}\right)} a_{k}-e^{-i\left(\theta_{k}+\cdots+\theta_{1}\right)} a_{k+1} .$$

Summing over all $k$, and using the fact that
$$e^{-i\left(\theta_{1}+\cdots+\theta_{2020}\right)}=1,$$
we see that the right hand side cancels to $0$, thus
$$\sum_{k=1}^{2020}\left(e^{-i\left(\theta_{k-1}+\cdots+\theta_{1}\right)}-e^{-i\left(\theta_{k}+\cdots+\theta_{1}\right)}\right) b_{k}=0 .$$

Jason knows $b_{2}, \ldots, b_{2020}$ and all the $\theta_{i}$, so the equation above is a linear equation in $b_{1}$. We finish by noting that the coefficient of $b_{1}$ is $1-e^{-i \theta_{1}}$ which is non-zero, as $\theta_{1} \in (0, \pi)$. Thus Jason can solve for $b_{1}$ uniquely.

Solution 3:

Let $A_{1} A_{2} \cdots A_{2020}$ and $\tilde{A}_{1} \tilde{A}_{2} \cdots \tilde{A}_{2020}$ be two 2020-gons that satisfy the conditions in the problem statement, and let $B_{k}, \tilde{B}_{k}$ be the points Alan would construct with respect to these two polygons. It suffices to show that if $B_{k}=\tilde{B}_{k}$ for $k=2,3, \ldots, 2020$, then $B_{1}=\tilde{B}_{1}$.
For $2 \leq k \leq 2020$, we note that

$$A_{k} B_{k}=A_{k+1} B_{k}, \quad \tilde{A}_{k} B_{k}=\tilde{A}_{k+1} B_{k}$$

Furthermore, we have the equality of directed angles $\angle A_{k} B_{k} A_{k+1}=\angle \tilde{A}_{k} B_{k} \tilde{A}_{k+1}=\theta_{k}$, therefore $\angle A_{k} B_{k} \tilde{A}_{k}=\angle A_{k+1} B_{k} \tilde{A}_{k+1}$. This implies the congruence $\triangle A_{k} B_{k} \tilde{A}_{k} \cong \triangle A_{k+1} B_{k} \tilde{A}_{k+1}$.
The congruence shows that $A_{k} \tilde{A}_{k}=A_{k+1} \tilde{A}_{k+1}$; furthermore, the angle from the directed segment $\overrightarrow{A_{k} \tilde{A}_{k}}$ to $\overrightarrow{A_{k+1} \tilde{A}_{k+1}}$ is $\theta_{k}$ counterclockwise. This holds for $k=2,3, \ldots, 2020$; we conclude that $A_{1} \tilde{A}_{1}=A_{2} \tilde{A}_{2}$, and the angle from the directed segments $\overrightarrow{A_{1}} \tilde{A}_{1}$ to $\overrightarrow{A_{2}} \tilde{A}_{2}$ is

$$-\sum_{k=2}^{2020} \theta_{k}=\theta_{1}-1010 \pi=\theta_{1}$$
counterclockwise.

Finally we observe that $A_{1} B_{1}=A_{2} B_{1}$, and the angle from the directed segment $\overrightarrow{A_{1} B_{1}}$ to $\overrightarrow{A_{2} B_{1}}$ is $\theta_{1}$ counterclockwise. This implies $\angle B_{1} A_{1} \tilde{A}_{1}=\angle B_{1} A_{2} \tilde{A}_{2}$, so $\triangle A_{1} B_{1} \tilde{A}_{1} \cong \triangle A_{2} B_{1} \tilde{A}_{2}$. Thus $\tilde{A}_{1} B_{1}=\tilde{A}_{2} B_{1}$, and the angle from $\overrightarrow{\tilde{A}_{1} B_{1}}$ to $\overrightarrow{\tilde{A}_{2} B_{1}}$ is $\theta_{1}$ counterclockwise. We conclude that $B_{1}=\tilde{B}_{1}$.