# Problem of the Month (June 2004)

This month we investigate the question "What is the smallest rectangle with aspect ratio r that contains n congruent rectangles with aspect ratio r?" In other words, for positive integers n, and r≥1, we define mn(r) to be the minimum value of m so that an m x mr rectangle contains n non-overlapping non-tilted 1 x r rectangles. Clearly if n is a perfect square, then mn(r) = √n with no wasted space.

What are the values of mn(r) for non-square values of n? A much harder question is to find Mn(r), the smallest value of m so that so that an m x mr rectangle contains n non-overlapping 1 x r rectangles that might be tilted. For example, when r > 1.9762, M2(r) < m2(r):

What are the values of Mn(r) for small values of n?

Jeremy Galvagni found mn(r) for n=3 and 5, and found most of n=6 and 7. Joseph DeVincentis found mn(r) for n=3 and 5.

Here are the piecewise definitions for mn(r) for small n:

 If r ≥ 1 r ≥ √2 r ≥ 2 n=2 then m2(r) = 2/r m2(r) = r m2(r) = 2

 If r ≥ 1 r ≥ 3/2 r ≥ √3 r ≥ 2 n=3 then m3(r) = 2 m3(r) = 3/r m3(r) = r m3(r) = 2

 If r ≥ 1 r ≥ (√13-1)/2 r ≥ (√5+1)/2 r ≥ 2 r ≥ √5 r ≥ 3 n=5 then m5(r) = 3/r m5(r) = r + 1 m5(r) = 2 + 1/r m5(r) = 5/r m5(r) = r m5(r) = 3

 If r ≥ 1 r ≥ √(3/2) r ≥ (√3+1)/2 r ≥ 3/2 r ≥ (√17-1)/2 r ≥ 2 r ≥ √6 r ≥ 3 n=6 then m6(r) = 3/r m6(r) = 2r m6(r) = 2 + 1/r m6(r) = 4/r m6(r) = r + 1 m6(r) = 6/r m6(r) = r m6(r) = 3

 If r ≥ 1 r ≥ 4/3 r ≥ √2 r ≥ 3/2 r ≥ √3 r ≥ 2 r ≥ 7/3 r ≥ √7 r ≥ 3 n=7 then m7(r) = 3 m7(r) = 4/r m7(r) = 2r m7(r) = 1 + 3/r m7(r) = r + 1 m7(r) = 3 m7(r) = 7/r m7(r) = r m7(r) = 3

 If r ≥ 1 r ≥ 4/3 r ≥ √2 r ≥ 3/2 r ≥ 8/3 r ≥ √8 r ≥ 3 n=8 then m8(r) = 3 m8(r) = 4/r m8(r) = 2r m8(r) = 3 m8(r) = 8/r m8(r) = r m8(r) = 3

 If r ≥ 1 r ≥ √5-1 r ≥ √2 r ≥ 3/2 r ≥ √(5/2) r ≥ (√17+3)/4 n=10 m10(r) = 4/r m10(r) = 2 + r m10(r) = 2 + 2/r m10(r) = 5/r m10(r) = 2r m10(r) = 3 + 1/r r ≥ 2 r ≥ (√13+1)/2 r ≥ 1+√2 r ≥ 3 r ≥ √10 r ≥ 4 then m10(r) = 2 + 3/r m10(r) = 1 + r m10(r) = 3 + 1/r m10(r) = 10/r m10(r) = r m10(r) = 4

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 6/7/04.