In triangle $OAB$, let $M$ be the midpoint of side $OA$, and let $N$ be the point that divides side $OB$ in the ratio $2:1$. Let $P$ be the intersection of line segments $AN$ and $BM$. Express $\vec{OP}$ in terms of $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$.

1. Problem Description

In triangle $OAB$, let $M$ be the midpoint of side $OA$, and let $N$ be the point that divides side $OB$ in the ratio $2:1$. Let $P$ be the intersection of line segments $AN$ and $BM$. Express $\vec{OP}$ in terms of $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$.

2. Solution Steps

Since $P$ lies on $AN$, we can express $\vec{OP}$ as a linear combination of $\vec{OA}$ and $\vec{ON}$:

$$\vec{OP} = (1 - s)\vec{OA} + s\vec{ON}$$

where $s$ is a scalar.

Since $N$ divides $OB$ in the ratio $2:1$, we have

$$\vec{ON} = \frac{2}{3}\vec{OB} = \frac{2}{3}\vec{b}$$

Thus,

$$\vec{OP} = (1 - s)\vec{a} + s\left(\frac{2}{3}\vec{b}\right) = (1 - s)\vec{a} + \frac{2s}{3}\vec{b}$$

Since $P$ lies on $BM$, we can express $\vec{OP}$ as a linear combination of $\vec{OB}$ and $\vec{OM}$:

$$\vec{OP} = (1 - t)\vec{OB} + t\vec{OM}$$

where $t$ is a scalar.

Since $M$ is the midpoint of $OA$, we have

$$\vec{OM} = \frac{1}{2}\vec{OA} = \frac{1}{2}\vec{a}$$

Thus,

$$\vec{OP} = (1 - t)\vec{b} + t\left(\frac{1}{2}\vec{a}\right) = \frac{t}{2}\vec{a} + (1 - t)\vec{b}$$

Now we have two expressions for $\vec{OP}$:

$$\vec{OP} = (1 - s)\vec{a} + \frac{2s}{3}\vec{b} = \frac{t}{2}\vec{a} + (1 - t)\vec{b}$$

Since $\vec{a}$ and $\vec{b}$ are linearly independent, we can equate the coefficients:

$$1 - s = \frac{t}{2}$$

$$\frac{2s}{3} = 1 - t$$

From the first equation, $t = 2(1 - s)$. Substituting this into the second equation, we get

$$\frac{2s}{3} = 1 - 2(1 - s) = 1 - 2 + 2s = 2s - 1$$

$$2s = 6s - 3$$

$$4s = 3$$

$$s = \frac{3}{4}$$

Now we can find $t$:

$$t = 2\left(1 - \frac{3}{4}\right) = 2\left(\frac{1}{4}\right) = \frac{1}{2}$$

Substituting $s = \frac{3}{4}$ into $\vec{OP} = (1 - s)\vec{a} + \frac{2s}{3}\vec{b}$, we get

$$\vec{OP} = \left(1 - \frac{3}{4}\right)\vec{a} + \frac{2(\frac{3}{4})}{3}\vec{b} = \frac{1}{4}\vec{a} + \frac{1}{2}\vec{b}$$

Alternatively, substituting $t = \frac{1}{2}$ into $\vec{OP} = \frac{t}{2}\vec{a} + (1 - t)\vec{b}$, we get

$$\vec{OP} = \frac{\frac{1}{2}}{2}\vec{a} + \left(1 - \frac{1}{2}\right)\vec{b} = \frac{1}{4}\vec{a} + \frac{1}{2}\vec{b}$$

3. Final Answer

$\vec{OP} = \frac{1}{4}\vec{a} + \frac{1}{2}\vec{b}$