Let $O(0,0), A(3,0), B(3,2)$ and $C(0,2)$ be four vertices of a rectangle. Let $$d(P,OA)≤\min {\Bigl(d(P,AB),d(P,BC),d(P,OC)\Bigl)}$$ where $d$ denotes the distance of the point $P$ with the line segments given. Let $S$ denote the region consisting of all those points $P$ inside the rectangle $OABC$ which satisfies the above inequality. Find the area of region $S$.
The last time I asked this question(which was almost a week ago), this problem was put on hold and then closed. I am preparing for an entrance exam and I came across this problem in my book called Co-ordinate Geometry for Jee Mains and Advanced by Dr. S.K Goyal. I had put my best effort to solve this problem but I get a different answer $\frac{9}{5+\sqrt{13}} $. The answer given in my textbook is $4$.Any help for this problem is appreciated. Thanks in advance.
An answer of $4$ seems wrong. The entire rectangle has area $6$ and the set of points closer to the bottom edge than the top edge is half the rectangle. These are contradictory statements.
A quick calculation in my head gives an area of $2$ (a trapezoid of bases $1$ and $3$ and of height $1$).