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(A) Let the vertices of the triangle be $A$,$B$,and $C$ with position vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ respectively.The medians are vectors d

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(A) Let the vertices of the triangle be $A$,$B$,and $C$ with position vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ respectively.The medians are vectors directed from the vertices to the midpoints of the opposite sides.Let the midpoints of sides $BC$,$CA$,and $AB$ be $D$,$E$,and $F$ respectively.The position vectors of the midpoints are $\vec{d} = \frac{\vec{b} + \vec{c}}{2}$,$\vec{e} = \frac{\vec{a} + \vec{c}}{2}$,and $\vec{f} = \frac{\vec{a} + \vec{b}}{2}$.The vectors representing the medians are $\vec{AD} = \vec{d} - \vec{a} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2}$.Similarly,$\vec{BE} = \vec{e} - \vec{b} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2}$ and $\vec{CF} = \vec{f} - \vec{c} = \frac{\vec{a} + \vec{b} - 2\vec{c}}{2}$.The sum of these vectors is $\vec{AD} + \vec{BE} + \vec{CF} = \frac{(\vec{b} + \vec{c} - 2\vec{a}) + (\vec{a} + \vec{c} - 2\vec{b}) + (\vec{a} + \vec{b} - 2\vec{c})}{2}$.Simplifying the numerator: $(\vec{b} + \vec{c} - 2\vec{a} + \vec{a} + \vec{c} - 2\vec{b} + \vec{a} + \vec{b} - 2\vec{c}) = (\vec{a} - \vec{a}) + (\vec{b} - \vec{b}) + (\vec{c} - \vec{c}) = 0$.Thus,the sum is $\frac{0}{2} = 0$.

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

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