# How to compute the area of the midpoint draw.

Give two closed domain $A_1,A_2$ of $\mathbb{R}^2$, the area of them is $S_1,S_2$. Give two points $P_1,P_2$, here $P_1 \in A_1, P_2 \in A_2$, let point $P$ is the midpoint of segment $P_1P_2$. Now, when $P_1, P_2$ move in $A_1,A_2$, the point $P$ will draw a domain $A$, how to compute the area of $A$?

A simple case, $A_1 = \{(x,y): x^2+y^2 \le 1\}, A_2 = \{(x,y): 6 \le x \le 8, 6 \le y \le 8\}$, it is easy to do. But can anything be said about the general case?

• If $A_1=A_2$ then $A$ will have the same area. If $A_2$ is a point $(S_2=0)$ then $A$ will have a quarter of the area of $A_1$. If $A_2$ is a straight line segment $(S_2=0)$ then I would have thought you could make $A$ have any area relative to that of $A_1$ by altering the length of $A_2$. Considering similar cases, I doubt there is any general formula which just uses $S_1$ and $S_2$ without considering the specifics of $A_1$ and $A_2$ – Henry Jun 20 '17 at 8:25
• Thank you. I try $A_1=\{(x,y):-c \le x,y \le c, c=\frac{\sqrt{\pi}}{2}\}$, $A_2$ as above. The area of $A_1,A_2$ as the same as above, but area of $A$ is different. – xunitc Jun 20 '17 at 10:26