All possible rectangles are shown here. To prove the other ones impossible let's have a look at the boundary of the s-rectangles and its edge squares. We call an edge square covered by a given pentabolo, if the pentabolo covers the boundary common to the rectangle and the square. The following table shows the maximum number of edge squares covered by each pentabolo in a rectangle with height 3 or more and height 2 respectively. Be sure that no area smaller then five triangles is separated.

group | pieces | maximum number of covered edge squares | |
---|---|---|---|

height = 3 | height = 2 | ||

A | 3 | 3 | |

B | 2 | 3 | |

C | 2 | 2 | |

D | 1 | 2 | |

E | 1 | <= 1 | |

F | 0 | 0 |

Given a fixed number of pentaboloes we add the maximum numbers of covered edge squares starting with pentaboloes from group A. If the sum is lower than the number of egde squares of the rectangle it is impossible to build the rectangle.

rectangle | pieces | edge squares | edge squares covered at most |
---|---|---|---|

2 x 15 | 12 | 30 | 3x3(A)+2x3(B)+7x2(C)=29 |

2 x 20 | 16 | 40 | 3x3(A)+2x3(B)+7x2(C)+4x2(D)=37 |

2 x 25 | 20 | 50 | 3x3(A)+2x3(B)+7x2(C)+5x2(D)+3x1(E)=42 |

2 x 30 | 24 | 60 | 3x3(A)+2x3(B)+7x2(C)+5x2(D)+7x1(E)=46 |

2 x 35 | 28 | 70 | 3x3(A)+2x3(B)+7x2(C)+5x2(D)+10x1(E)+1x0(F)=49 |

3 x 20 | 24 | 42 | 3x3(A)+2x2(B)+7x2(C)+5x1(D)+7x1(E)=39 |

3 x 25 | 30 | 52 | 3x3(A)+2x2(B)+7x2(C)+5x1(D)+10x1(E)+3x0(F)=42 |

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