A triangle OAB and a triangle OAC share one side OA. Further, they satisfy the following two conditions: (i) $\overrightarrow { \mathrm { OC } } = x \overrightarrow { \mathrm { OA } } + \frac { 1 } { 2 } \overrightarrow { \mathrm { OB } }$; (ii) the center of gravity G of the triangle OAC is on the segment AB. We are to find the value of $x$, and express $\overrightarrow { \mathrm { OG } }$ in terms of $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OB } }$. Let us denote the point of intersection of the segment OC and the segment AB by D. Then we have $$\overrightarrow { \mathrm { OD } } = \frac { x } { \mathbf { A } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { B } } { \mathbf { A } } \overrightarrow { \mathrm { OB } } .$$ Since D is on the segment AB, we have $x = \frac { \mathbf { D } } { \mathbf { E } }$. Hence we obtain $$\overrightarrow { \mathrm { OG } } = \frac { \mathbf { F } } { \mathbf { G } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { H } } { \mathbf { I } } \overrightarrow { \mathrm { OB } } .$$ In particular, when $\mathrm { OA } = 1 , \mathrm { OB } = 2$ and $\angle \mathrm { AOB } = 60 ^ { \circ }$, we have $\mathrm { OG } = \frac { \sqrt { \mathbf { J K } } } { \mathbf { L } }$.
A triangle OAB and a triangle OAC share one side OA. Further, they satisfy the following two conditions:
(i) $\overrightarrow { \mathrm { OC } } = x \overrightarrow { \mathrm { OA } } + \frac { 1 } { 2 } \overrightarrow { \mathrm { OB } }$;
(ii) the center of gravity G of the triangle OAC is on the segment AB.
We are to find the value of $x$, and express $\overrightarrow { \mathrm { OG } }$ in terms of $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OB } }$.
Let us denote the point of intersection of the segment OC and the segment AB by D. Then we have
$$\overrightarrow { \mathrm { OD } } = \frac { x } { \mathbf { A } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { B } } { \mathbf { A } } \overrightarrow { \mathrm { OB } } .$$
Since D is on the segment AB, we have $x = \frac { \mathbf { D } } { \mathbf { E } }$.
Hence we obtain
$$\overrightarrow { \mathrm { OG } } = \frac { \mathbf { F } } { \mathbf { G } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { H } } { \mathbf { I } } \overrightarrow { \mathrm { OB } } .$$
In particular, when $\mathrm { OA } = 1 , \mathrm { OB } = 2$ and $\angle \mathrm { AOB } = 60 ^ { \circ }$, we have $\mathrm { OG } = \frac { \sqrt { \mathbf { J K } } } { \mathbf { L } }$.