Puzzle: [Dudeney 1931]
  Can 3 squares with side-length 4 cover a square of side length 5?








































Hint:

No covering is possible with alligned sides of the three squares.

Three squares with side-length 1 we can cover a square of side length
  z = sqrt((sqrt(5)+1)/2) = 1.272
This is best possible.
























Solution:
  Dudeney only gives the diagramm.
  The solution given here is from [Torbijn].

  Let the three squares be of size 1.
  Let z be the size of the square to cover.
  Place one square in the bottom left corner of the z-square.

  +--------------------------------A
  |                              / | \
  |                            /   |  
  |                          /     |    \
  +------------------------/+      |
  |                      /  |      |       \
  |                    /    |      |
  |                  /      |      |          \
  |                /        |      |
  |              /          |      |             \
  |            B            |      |               D
  |                         |      |             /
  |               \         |      |           /
  |                         |      |         /
  |                  \      |      |       /
  |                         |      |     /
  |                     \   |      |   /
  |                         |      | /
  +-------------------------F------E
                                 /
                               C

              1
  A-------------------------D
  |                         |
  |   \                     |
  |                         |
  |        \                |
  |                         | y
  |          z  \           |
  |                         |
  |                  \      |
  |                         |
  |                       \ |
  |                         E
  |                   z-1  /| 1-y
  |                      /  |
  B---------------------F---C

  AB = BC = CD = DA = 1
  AE = z
  EF = z-1                 The width of the boarder to be covered.
  angle(AEF) = 90 degree
  ED = y

  From similar triangles it follows
    AE : AD = EF : EC
  that is  z : 1 = (z-1) : (1-y) so y = 1/z.
  Now AE^2 = AD^2 + DE^2 that is z^2 = y^2 + 1. Eliminating y gives

    z^4 - z^2 - 1 = 0,

  from which z^2 = (sqrt(5) + 1)/2.
  Notice that (sqrt(5) + 1)/2 represents the proportion of the Golden Section.

References:
- Henry Ernest Dudeney;
  Puzzles and Curious Problems,
  Thomas Nelson and Sons, Ltd., London 1931,
  Problem 219: Three Tablecloths
  (only shows the configuration, no calculations.)

- Bundeswettbewerb Mathematik 1996, 1. Runde, Aufgabe 1
  Kann man ein Quadrat der Seitenlaenge 5cm vollstaendig mit drei
  Quadraten der Seitenlaenge 4cm ueberdecken?
  Loesungen: Vier Beweise werden angegeben.
  Es gibt verschiedene Anordnungen.

- Pieter Torbijn,
  Ladder Competition, CFF Contest 10, Problem 4,
  CFF 27 (Dec. 1991), p20

- Erich Friedman;
  http://www.stetson.edu/~efriedma/mathmagic/0900.html
  Trevor Green was interested in using n squares of equal size 
  to cover the largest square possible.
--
http://www.mathematik.uni-bielefeld.de/~sillke/
mailto:Torsten.Sillke@uni-bielefeld.de