The game of Pentomino is a tiling puzzle game played on a grid. A Pentomino piece is a two dimensional shape of five non-diagonally connected tiles. There are exactly 12 unique pieces (ignoring rotation/reflection). The goal of the game is to fill the board with the available pieces without overlapping.
This question concerns point-symmetric boards (not necessarily rectangular) of 10 spaces, i.e. with room for exactly two pieces.
Filling such a board is, of course, trivial when using the same piece twice.
Using the Y-piece twice on a point-symmetric hourglass board: #### O# OOOO
However, it seems impossible to do when using two different pieces. Is this true? How could this be proved?
This is not the case for larger boards or larger pieces (more than 5 tiles). Of course, the standard 6x10 board using 12 Pentomino pieces is also a counter example when using more than two pieces. Small counter examples:
Three 5-tile Pentomino pieces (V, P and L pieces) on a 3x5 grid: vvvpp vLppp vLLLL The 5-tile pieces P, T and F on an oval-shaped point-symmetric grid. ppTF pppTFFF TTTF Two 8-tile pieces on a 4x4 grid: #### #OOO #OOO ##OO