1. The problem is to find the centroid of a triangle given its vertices at points $A(1,2)$, $B(3,7)$, and $C(5,5)$. The centroid is the point where the three medians intersect and is located two-thirds of the distance from each vertex along the median. 2. The formula for the centroid $G$ of a triangle with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$ is: $$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$$ This formula comes from the fact that the centroid is the average of the coordinates of the vertices. 3. Calculate the $x$-coordinate of the centroid: $$x_G = \frac{1 + 3 + 5}{3} = \frac{9}{3} = 3$$ 4. Calculate the $y$-coordinate of the centroid: $$y_G = \frac{2 + 7 + 5}{3} = \frac{14}{3} \approx 4.67$$ 5. Therefore, the centroid $G$ of the triangle is at: $$G(3, \frac{14}{3})$$ 6. This point is two-thirds of the way along each median from the vertex to the midpoint of the opposite side, confirming the problem statement. Final answer: The centroid of the triangle is at $\boxed{\left(3, \frac{14}{3}\right)}$.