Let $O(0,0), A(2,0)$ and $B\left(1, \frac{1}{\sqrt{3}}\right)$ be the vertices of a triangle. Let $R

Question

Let $O(0,0), A(2,0)$ and $B\left(1, \frac{1}{\sqrt{3}}ight)$ be the vertices of a triangle. Let $R$ be the region consisting of all those points $P$ inside $	riangle O A B$ which satisfy $d(P, O A) \leq \min \{d(P, O B)$, $d(P, A B)\}$, where $d$ denotes the distance from the point to the corresponding line. If the area of the region R is $m-\sqrt{n}$, find $m+n$

Professor PC Thomas Classes · Answer

The coordinates of the points are\ $O(0,0), A(2,0), B\left(1, \frac{1}{\sqrt{3}}ight)$\ The equations of the sides of $	riangle O A B$ are\ $O A: y=0 ; O B: x-\sqrt{3} y=0$\ $A B: x+\sqrt{3} y-2=0$\ Let the coordinates of the point $P$ be $(h, k)$.\ $	herefore \quad d(P, O A)=k$\ $d(P, O B)=\frac{h-k \sqrt{3}}{2} d(P, A B)=\frac{h+k \sqrt{3}-2}{2}$\ But from the given condition\ $d(P, O A) \leq \min \{d(P, O B), d(P, A B)\}$\ $\Rightarrow \quad k \leq \min \left\{\begin{array}{cc}\frac{h-k \sqrt{3}}{2}, & \frac{h+k \sqrt{3}-2}{2} \ =x & =y\end{array}ight\}$\ $	herefore \quad x-y=\frac{-2 k \sqrt{3}+2}{2}=1-k \sqrt{3}$ is positive\ Because $k$ being inside $	riangle O A B, k<\frac{1}{\sqrt{3}}$\ $\Rightarrow k \sqrt{3}<1$\ Since $x>y$\ $	herefore \quad k \leq \frac{h+k \sqrt{3}-2}{2} \Rightarrow h+(\sqrt{3}-2) k-2 \geq 0$\ Locus of $P(h, k)$ is $(2-\sqrt{3}) y-x \leq 0$\ Area of the region\ $=\frac{1}{2} 	imes 2 	imes(2-\sqrt{3})=(2-\sqrt{3})$ sq. units\