## Task B-1.4.

The length of $\overline{A B}$ is sequentially divided from vertex $A$ by points $T_{1}, T_{2}, T_{3}, \ldots, T_{2016}$ into 2017 equal parts. If the coordinates of points $T_{3}(5,-1)$ and $T_{4}(8,-3)$ are known, determine the coordinates of points $A$ and $B$. Can the difference in coordinates for any of the points $T_{1}, T_{2}, T_{3}, \ldots, T_{2016}$ be 2016? If such points exist, determine their coordinates.
## Solution.

Each of the points $T_{k}$ is the midpoint of the segment $\overline{T_{k-1} T_{k+1}}$. From $x_{T_{3}}=\frac{1}{2}\left(x_{T_{2}}+x_{T_{4}}\right)$ and $y_{T_{3}}=\frac{1}{2}\left(y_{T_{2}}+y_{T_{4}}\right)$, we get $T_{2}(2,1)$. Similarly, we get $A(-4,5)$. We can reach the same result by considering the distance between the points.

If all points are equally spaced, then the difference in the abscissas (or ordinates) of all pairs of consecutive points is always the same number. Thus, the abscissas of all points differ by 3 and are $-4,-1,2,5,8,11, \ldots$, and the ordinates differ by -2 and are $5,3,1,-1,-3,-5, \ldots$ Generally, the abscissa of point $T_{k}$ is of the form: $-4+3 k$, for $k \in\{0,1,2,3, \ldots, 2017\}$, and the ordinate of point $T_{k}$ is of the form: $5+k \cdot(-2)$, for $k \in\{0,1,2,3, \ldots, 2017\}$.

Using the given conclusion, we get the coordinates of point $B=T_{2017}: B(6047,-4029)$.

Consider the difference in the coordinates of some point $T_{k}$.

$$
\begin{gathered}
|(5-2 k)-(-4+3 k)|=2016 \\
|9-5 k|=2016 \\
9-5 k=2016 \quad \text { or } \quad 9-5 k=-2016
\end{gathered}
$$

The number $k$ is a natural number, so only the second equality is possible, i.e., $5 k=2025, k=405$.

The desired point is $T_{405}(1211,-805)$.