Area bounded by Point $P$ in xy plane, If $\max\left\{\bf{PA+PB\;,PB+PC}\right\}\leq 2,$

Question

A point $P$ moves in $xy$ plane such that $\max\left\{\bf{PA+PB\;,PB+PC}\right\}\leq 2,$ Then Area

of the Regine Bounded by Point $\bf{P}$ is, If Coordinate of $A(0,0)\;\;,B(1,0)$ and $C(2,0)$ Given

$\bf{My\; Try::}$ Let Coordinate of Point $P(x,y)\;,$ Then We get

$$PA= \sqrt{x^2+y^2}\;\;\;\; ,PB=\sqrt{(x-1)^2+y^2}\;\;\;, PC = \sqrt{(x-2)^2+y^2}$$

Now Using Cases::

$$\bullet\; \max\left\{\bf{PA+PB\;,PB+PC}\right\}\leq 2\Rightarrow PA+PB\leq 2\;,$$ If $PA+PB>PB+PC\Rightarrow PA>PC$

$$\bullet\; \max\left\{\bf{PA+PB\;,PB+PC}\right\}\leq 2\Rightarrow PB+PC\leq 2\;,$$ If $PA+PB\leq PB+PC\Rightarrow PA\leq PC$

Now How can I solve after that, Help me

Thanks

Answer 1

Have you sketched it to get any ideas?

Each of $\bf{PA+PB}$$\leq2$ and $\bf{PB+PC}$$\leq2$ describes an ellipse with foci of $\bf{A}$ and $\bf{B}$ for the first one and $\bf{B}$ and $\bf{C}$ for the second one. They both have semi-major radius of $a=1$ (as $2a=PF_1+PF_2$) and $b=\frac{\sqrt3}{2}$ (as $f=\sqrt{a^2-b^2}$ and $f=\frac12$).

From here you should be able to work out the area. Comment if you are still not sure where to go.