Find the least number A such that for any two squares of combined area 1, a rectangle of area A exists such that the two squares can be packed in the rectangle (without interior
overlap).
You may assume that the sides of the squares are parallel to the sides of the
rectangle.
Case 1: If the 2 squares are equal, then each has side length √2/2 and area 1/2.
Side by side, the 2 fit into a rectangle √2/2 by 2*√2/2 with area 1.
Case 2: if one of the squares has side length 0, the other has side length 1, and they fit into a 1 by 1 square, and the area A is again 1.
In general:
One square has dimensions x by x.
The other has dimensions √(1-x^2) by √(1-x^2).
See Desmos graphic
https://www.desmos.com/calculator/2gfy5ohkqp
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Posted by Larry
on 2024-04-01 11:03:11 |
solution
