Let $P_0 = (0,0)$, $P_1 = (0,4)$, $P_2 = (4,0)$, $P_3 = (-4,-4)$, $P_4 = (2,4)$, $P_5 = (4,6)$ (or similar points). Find the region of all points closer to $P_0$ than to any of $P_1, P_2, P_3, P_4, P_5$, and compute its perimeter.
The region is bounded by the perpendicular bisectors of segments $P_0P_i$, $i=1,2,3,4,5$: $P_0P_1$ is bisected by $y=2$ $P_0P_2$ is bisected by $x=2$ $P_0P_3$ is bisected by $x+y=-2$ $P_0P_4$ is bisected by $x+y=3$ $P_0P_5$ is bisected by $x+y=5$ Clearly we can ignore $x+y=5$ because the bound provided by $x+y=3$ is stronger. The region has vertices $(2,-4), (-4,2), (2,1), (1,2)$. The perimeter is $10 + 7\sqrt{2}$.
Let $P_0 = (0,0)$, $P_1 = (0,4)$, $P_2 = (4,0)$, $P_3 = (-4,-4)$, $P_4 = (2,4)$, $P_5 = (4,6)$ (or similar points). Find the region of all points closer to $P_0$ than to any of $P_1, P_2, P_3, P_4, P_5$, and compute its perimeter.