Can you solve it in less than a minute?

We start by placing an equilateral triangle at the origin so that $O$ is on the origin.

We know $\vec{OP} = \left( 2 - t \right) \vec{OA} + t\vec{OB}= 2\vec{OA} + t \left( \vec{OB} - \vec{OA} \right)$

From here we see that $\vec{OB} - \vec{OA}$ represents the vector that connects $B$ to $A$ and so the possibilities for $\vec{OP}$ are below.

In addition we know that the closest vector possible for $\vec{OP}$ is when a right angle is formed like so this also minmize between $\vec{OA}$ and $\vec{OP}$ .

And so if we indeed have a right angle and $\alpha = \frac{\pi}{3}$ then we know the distance between $\vec{OA}$ and $\vec{OP}$ is $\boxed{\frac{\sqrt{3} }{2}}$ .

Place $\triangle OAB$ so that $O$ is at the origin, $A$ is at $(1, 0)$ , and $B$ is at $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ . Let $P$ be at $(p_x, p_y)$ .

Then $\overrightarrow{OA} = (1, 0)$ , $\overrightarrow{OB} = (\frac{1}{2}, \frac{\sqrt{3}}{2})$ , $\overrightarrow{OP} = (p_x, p_y)$ , and $\overrightarrow{AP} = (p_x - 1, p_y)$ , and from the given equation $\overrightarrow{OP} = (2 - t) \overrightarrow{OA} + t \overrightarrow{OB}$ we can derive $p_x = 2 - \frac{1}{2}t$ and $p_y = \frac{\sqrt{3}}{2}t$ .

Therefore, $|\overrightarrow{AP}| = \sqrt{(p_x - 1)^2 + p_y^2} = \sqrt{(2 - \frac{1}{2}t - 1)^2 + (\frac{\sqrt{3}}{2}t)^2} = \sqrt{(t - \frac{1}{2})^2 + \frac{3}{4}}$ , which has a minimum value at $t = \frac{1}{2}$ of $\boxed{\frac{\sqrt{3}}{2}}$ .