Task B-3.3. Matko and his friends, playing by the seashore, near a palm tree (about 20 meters from the sea), dug up a small box in which a carefully wrapped papyrus was found. They unrolled the papyrus, on which it was written: "The palm tree where you stand is the origin of a coordinate system whose x-axis is parallel to the seashore. Start from the palm tree, with the sea behind you, and move 5 units in the direction of the slope $\frac{4}{3}$. You will reach point $A$. From point $A$, move perpendicularly to the previous direction to reach point $B$, which lies on the line $y=-\frac{1}{3} x+\frac{10}{3}$. Point $C$ is on the line $x-2 y-10=0$ and the sum of its distances to points $A$ and $B$ is the smallest possible. The treasure is buried at the centroid of triangle $A B C$."

Find the coordinates of the point where the treasure is buried.
## Solution.

Point $A$ is obviously $(3,4)$ since $5^{2}=3^{2}+4^{2}$.

A line perpendicular to $y=\frac{4}{3} x$ has a slope of $-\frac{3}{4}$ and passes through point $A$, so its equation is $y=-\frac{3}{4} x+\frac{25}{4}$.

Point $B$ is the intersection of that line and the line $y=-\frac{1}{3} x+\frac{10}{3}$, i.e., $B(7,1)$.

If $|A C|+|B C|$ must be minimized, we will first determine the point $A^{\prime}$ which is symmetric to point $A$ with respect to the line $p \ldots x-2 y-10=0$.

![](https://cdn.mathpix.com/cropped/2024_05_30_c2de659189058fc448a1g-13.jpg?height=927&width=1203&top_left_y=201&top_left_x=431)

The equation of the line perpendicular to $p$ with a slope of -2, passing through point $A$, is $y=-2 x+10$. The intersection of the obtained line and line $p$ is point $P(6,-2)$, the midpoint of segment $A A^{\prime}$.

We have

$$
\begin{aligned}
6 & =\frac{3+x^{\prime}}{2}, & x^{\prime} & =9 \\
-2 & =\frac{4+y^{\prime}}{2}, & y^{\prime} & =-8
\end{aligned}
$$

Thus, $A^{\prime}(9,-8)$.

The line through points $B$ and $A^{\prime}$ is $y=-\frac{9}{2} x+\frac{65}{2}$ and intersects line $p$ at point $C\left(\frac{15}{2},-\frac{5}{4}\right)$, which satisfies the given condition, as points $B, C$, and $A^{\prime}$ lie on the same line and $|A C|=\left|A^{\prime} C\right|$.

(If the student finds the point symmetric to point $B$, then $P(8,-1), B^{\prime}(9,-3)$, and the line $\left.A B^{\prime} \ldots y=-\frac{7}{6} x+\frac{45}{6}\right)$

The coordinates of the centroid of the triangle are

$$
x=\frac{x_{A}+x_{B}+x_{C}}{3}=\frac{35}{6}, \quad y=\frac{y_{A}+y_{B}+y_{C}}{3}=\frac{5}{4}
$$

Thus, the treasure is at point $T\left(\frac{35}{6}, \frac{5}{4}\right)$.