Y by centslordm, Mango247, Mango247
A square with side
is given on the plane. It is allowed to select any two of its vertices and move one of them to any distance in an arbitrary direction (within the plane), and move also the other vertices to the same distance in the opposite direction. After several such operations, we got a square congruent to the original one. Prove that the intersection area of the squares is greater than
.


This post has been edited 1 time. Last edited by parmenides51, Jun 10, 2021, 6:29 PM