Question

Consider the triangle having vertices $O(0,0),A(4,0)$ and $B(2,2\sqrt{3})$. Let $P$ be an interior point inside $\Delta OAB.$ and $R$ be the region consisting of all those points $P$ which satisfy $OP\le \text{min}[BP,AP]$. Let the area of the region $R$ is $\alpha$. Then the value of $4\sqrt{3}\alpha$ is

Answer 1

Here, $C$ is the circumcenter of $\Delta OAB$.
For $OP\le OA,$ the point $P\text{must lie in/on}\Delta OBL$ (As the $BL$ is perpendicular bisector of OA)
Similarly, For $OP\le BP,$ the point $P\text{must lie in/on}\Delta OAN$

Required area $\alpha =\frac{1}{3}\text{Area of}\Delta OAB=\frac{4}{\sqrt{3}}\Rightarrow 4\sqrt{3}\alpha =16$