Define the **aspect ratio** of a rectangle to be the ratio of the smaller side to the bigger side. So the aspect ratio of a square is 1, and all other rectangles have aspect ratios smaller than 1. We want to tile a square with rectangles with areas 1, 2, 3, . . . n so that the smallest aspect ratio of the rectangles is as large as possible. Call this ratio **R(n)**.

The sequence R(n) starts 1, 1/3, 1/2, . . . . What are the values of R(n) for larger n? Can you show that R(n) converges to 1 as n → ∞?

The sequence S(n) starts 2, 4, 6, 8, 11, . . . . What are the values of S(n) for larger n? Does S(n) / n^{3/2} converge as n → ∞?

The sequence T(n) starts 1/2, (3-√5) / 2, . . . What are the values of T(n) for larger n? How fast does T(n)->0 as n → ∞?

**Lemma:** If R(n) ≥√(k/(k+1)), the rectangles 1,2, . . . , k do not share a complete edge with any other rectangle.

**Corollary 1:** If rectangle k shares a complete edge with another rectangle,
R(n) < √(k/(k+1)).

He uses these to get upper bounds on R(n) under the simplifying assumption that any time a rectangle is surrounded by 4 other rectangles, in such a way that none of them share an edge (he calls these windmills), these 5 rectangles form a rectangle. These upper bounds increase to 1 as n gets large.

Here are the best solutions found so far for R(n):

1 Trivial | 1/3 = .333+ Trivial | 1/2 = .5 Trivial | 8/15 = .533+ Joe DeVincentis | 64/135 = .474+ Erich Friedman |

64/105 = .609+ Erich Friedman | .614+ Maurizio Morandi | .620+ Maurizio Morandi | 15680/23409 = .669+ Erich Friedman | .671+ Maurizio Morandi |

17457/25600 = .681+ Maurizio Morandi | 147/208 = .706+ Erich Friedman | 2548/3721 = .684+ Erich Friedman |

**Conjecture:** Let S*(n) = √A(n). Then S(n) is equal to S*(n) when S*(n)^{2} - S*(n) ≥ A(n) (and when n=4), and S*(n)+1 otherwise.

If this is true, A(n) = 2/3 * n^{3} + O(n^{2}), S(n) = √(2/3) * n^{3/2} + O(n), so S(n) / n^{3/2} goes to √(2/3) as n gets large.

Here are the best known results for small n:

S(1)=2 Trivial | S(2)=4 Trivial | S(3)=6 Erich Friedman | S(4)=8 Erich Friedman |

S(5)=11 Erich Friedman | S(6)=14 Erich Friedman | S(7)=18? Erich Friedman | S(8)=21 Erich Friedman |

S(9)=25? Erich Friedman | S(10)=29? Erich Friedman | S(11)=33? Erich Friedman | S(12)=37 Erich Friedman |

S(13)=42? Erich Friedman | S(14)=46 Erich Friedman | S(15)=51 Erich Friedman |

T(1) = 1/2 = .5 Trivial | T(2) = (3-√5)/2 = .382+ Proved by Erich Friedman | T(3) = .308+ Proved by Sasha Ravsky | T(4) = 1/4 = .25 Proved by Sasha Ravsky |

.211+ ≤ T(5) ≤ 2/9 = .222+ Found by Erich Friedman | .179+ ≤ T(6) ≤ .196+ Found by Erich Friedman | .174+ ≤ T(7) ≤ .173+ Found by Erich Friedman | .159+ ≤ T(8) ≤ .158+ Found by Erich Friedman |

.146+ ≤ T(9) ≤ .145+ Found by Erich Friedman | .129+ ≤ T(10) ≤ .146+ Found by Sasha Ravsky | .105+ ≤ T(11) ≤ .146+ Found by Sasha Ravsky | .083+ ≤ T(12) ≤ .125+ Found by Sasha Ravsky |

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 7/9/07.