All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Squeeze in II (Posted on 2024-04-01) Difficulty: 2 of 5
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.

  Submitted by Ady TZIDON    
No Rating
Solution: (Hide)
Answer: A= about 1.21
Solution:: If x and y are the sides of two squares with combined area 1, then x^2 + y^2 = 1. Suppose without loss of generality that x ≥ y. Then the shorter side of a rectangle containing both squares without overlap must be at least x, and the longer side must be at least x+y. Hence the desired value of A is the maximum of x(x + y). To find this maximum, we let x = cos θ, y = sin θ with θ ∈ [0, π/4].

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Puzzle ThoughtsK Sengupta2024-04-02 21:54:01
re(3): Second attemptJer2024-04-02 13:46:56
re(2): Second attemptBrian Smith2024-04-02 12:05:37
re: Second attemptJer2024-04-02 10:50:16
Second attemptLarry2024-04-01 11:41:01
re: solutionBrian Smith2024-04-01 11:31:08
SolutionsolutionLarry2024-04-01 11:03:11
Solutionproposed solutionCharlie2024-04-01 09:58:11
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information