Tiling a Polyomino at Scale 2 with a Tetromino and a Pentomino

  • Introduction
  • Table
  • Solutions

    • Introduction

    A tetromino is a figure made of 4 squares joined edge to edge. There are 5 such figures, not distinguishing reflections and rotations.

    A pentomino is a figure made of 5 squares joined edge to edge. There are 12 such figures, not distinguishing reflections and rotations.

    Here I study the problem of arranging copies of a tetromino and a pentomino to form some polyomino that has been scaled up by a factor of 2.

    See also

  • Tiling a Polyomino at Scale 2 with Two Pentominoes
  • Tiling a Polyiamond at Scale 2 with Two Hexiamonds
  • Tiling a Polyabolo at Scale 2 with Two Tetraboloes

    • Table of Results

    This table shows the areas of the smallest polyominoes at scale 2 that can be tiled by a tetromino and a pentomino, using at least one copy of each. If you find a smaller solution or solve an unsolved pair, please write.

    Bryce Herdt solved the pair 4N+5N without holes. My solution with a hole is derived from his solution without one.

    Bryce Herdt showed that 4N+5X and 4Q+5X are impossible.

    The green indexes are links to tilings by the specified tetromino alone. The blue indexes are links to tilings by the specified pentomino alone.

     FILNPTUVWXYZ
    I362424402436722872?2428
    L282424282436362836362828
    N?443222424363668?×3248
    Q?2424962464?24?×2824
    T363224362436363636682836

    Solutions

    So far as I know, these solutions use as few tiles as possible. They are not necessarily uniquely minimal.

    24 Cells

    28 Cells

    32 Cells

    36 Cells

    40 Cells

    44 Cells

    48 Cells

    64 Cells

    68 Cells

    72 Cells

    96 Cells

    224 Cells

    Holeless Variant

    Last revised 2025-10-06.

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    Col. George Sicherman [ HOME | MAIL ]