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 6×10 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 `