The sum of the three vectors determined by the medians of triangle directed from the vertices is

Options

A. 0
B. 1
C. `-1`
D. `1/3`

Answer

To find the sum of the three vectors determined by the medians of a triangle directed from the vertices, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Triangle and Medians**:
Let the vertices of the triangle be \( A \), \( B \), and \( C \). The midpoints of the sides \( BC \), \( CA \), and \( AB \) are denoted as \( D \), \( E \), and \( F \), respectively. The medians are the segments \( AD \), \( BE \), and \( CF \).

2. **Express the Medians as Vectors**:
- The median \( AD \) can be expressed as:
\[ \vec{AD} = \vec{AB} + \vec{BD} \]
Since \( D \) is the midpoint of \( BC \), we have:
\[ \vec{BD} = \frac{1}{2} \vec{BC} \]
Thus,
\[ \vec{AD} = \vec{AB} + \frac{1}{2} \vec{BC} \]

- For the median \( BE \):
\[ \vec{BE} = \vec{BC} + \vec{CE} \]
Here, \( E \) is the midpoint of \( CA \), so:
\[ \vec{CE} = \frac{1}{2} \vec{CA} \]
Therefore,
\[ \vec{BE} = \vec{BC} + \frac{1}{2} \vec{CA} \]

- For the median \( CF \):
\[ \vec{CF} = \vec{CA} + \vec{AF} \]
Since \( F \) is the midpoint of \( AB \):
\[ \vec{AF} = \frac{1}{2} \vec{AB} \]
Thus,
\[ \vec{CF} = \vec{CA} + \frac{1}{2} \vec{AB} \]

3. **Sum the Medians**:
Now, we need to sum the three medians:
\[ \vec{AD} + \vec{BE} + \vec{CF} \]

Substituting the expressions we derived:
\[ \vec{AD} + \vec{BE} + \vec{CF} = \left(\vec{AB} + \frac{1}{2} \vec{BC}\right) + \left(\vec{BC} + \frac{1}{2} \vec{CA}\right) + \left(\vec{CA} + \frac{1}{2} \vec{AB}\right) \]

Combining like terms:
\[ = \vec{AB} + \frac{1}{2} \vec{AB} + \vec{BC} + \frac{1}{2} \vec{BC} + \vec{CA} + \frac{1}{2} \vec{CA} \]
\[ = \left(1 + \frac{1}{2}\right) \vec{AB} + \left(1 + \frac{1}{2}\right) \vec{BC} + \left(1 + \frac{1}{2}\right) \vec{CA} \]
\[ = \frac{3}{2} \vec{AB} + \frac{3}{2} \vec{BC} + \frac{3}{2} \vec{CA} \]

4. **Factor Out Common Terms**:
\[ = \frac{3}{2} \left(\vec{AB} + \vec{BC} + \vec{CA}\right) \]

5. **Recognize the Vector Sum**:
The sum \( \vec{AB} + \vec{BC} + \vec{CA} \) represents a closed triangle path, which equals zero:
\[ \vec{AB} + \vec{BC} + \vec{CA} = \vec{0} \]

6. **Final Result**:
Therefore, the sum of the three medians is:
\[ \frac{3}{2} \cdot \vec{0} = \vec{0} \]

### Conclusion:
The sum of the three vectors determined by the medians of the triangle directed from the vertices is \( \vec{0} \).

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