The semiperimeter of a triangle is S and its centroid is G. What is the distance between G and centroid of the triangle formed by mid points of the sides of the given triangle ?

Options

A. `5/3`
B. `5/6`
C. `5/18`
D. 0

Answer

To solve the problem, we need to determine the distance between the centroid \( G \) of a triangle \( ABC \) and the centroid of the triangle formed by the midpoints of the sides of triangle \( ABC \).

### Step-by-Step Solution:

1. **Understanding the Centroid**:
The centroid \( G \) of a triangle is the point where all three medians intersect. It divides each median in a 2:1 ratio.

2. **Identifying the Midpoints**:
Let \( D \), \( E \), and \( F \) be the midpoints of sides \( BC \), \( CA \), and \( AB \) respectively. The triangle formed by these midpoints is triangle \( DEF \).

3. **Finding the Centroid of Triangle DEF**:
The centroid of triangle \( DEF \) can be found using the coordinates of points \( D \), \( E \), and \( F \). The centroid \( G' \) of triangle \( DEF \) is given by:
\[ G' = \left( \frac{x_D + x_E + x_F}{3}, \frac{y_D + y_E + y_F}{3} \right) \]
where \( (x_D, y_D) \), \( (x_E, y_E) \), and \( (x_F, y_F) \) are the coordinates of points \( D \), \( E \), and \( F \).

4. **Relationship Between the Two Centroids**:
It is a known property that the centroid \( G' \) of triangle \( DEF \) is also located at the same point as the centroid \( G \) of triangle \( ABC \). This is because the centroid of the triangle formed by the midpoints of the sides of any triangle is always the same point as the centroid of the original triangle.

5. **Calculating the Distance**:
Since both centroids \( G \) and \( G' \) are the same point, the distance between them is:
\[ \text{Distance} = |G - G'| = 0 \]

### Final Answer:
The distance between the centroid \( G \) of triangle \( ABC \) and the centroid of triangle \( DEF \) formed by the midpoints of the sides of triangle \( ABC \) is **0**.