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Two squares (Posted on 2006-01-05) Difficulty: 3 of 5
Construct a large square ABCD with AB at the top.

Next construct a smaller square (A'B'C'D') inside ABCD, with any orientation and centre and join the corresponding corners.

This divides the region between the squares into four.

We will name these divisions (and their areas):

(N)orth = ABB'A'
(E)ast = BCC'B'
(S)outh = CDD'C'
(W)est = DAA'D'.

Show that the areas of these regions satisfy the equality
N+S = E+W.

See The Solution Submitted by goFish    
Rating: 4.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 7

Let X and Y be complex numbers and O the origin. Define the
cross product
    X x Y = X"Y - XY" = 4i*Area(OXY)
where X" and Y" are the complex conjugates of X and Y, i is
the square root of minus 1, and triangle OXY is labeled
Let A = 0+0i, B = 1+0i, C = 1+1i, and D = 0+1i be the vertices
of the square ABCD.
Let A' = x+yi, B' = u+vi, 
    C' = B' + i*(B' - A'), and
    D' = A' + i*(B' - A')  be the vertices
of the square A'B'C'D'.
    4i*Area(ABB'A') = (B - A) x (B' - A) +
                      (B' - A) x (A' - A)
                    = (B' - A) x (A' - B)
Using MathCad we can show that
    4i*[Area(ABB'A')-Area(BCC'B')+Area(CDD'C')-Area(DAA'D')] =
    (B' - A) x (A' - B) -
    (C' - B) x (B' - C) +
    (D' - C) x (C' - D) -
    (A' - D) x (D' - A)     =    0
    Area(ABB'A') + AREA(CDD'C') = Area(BCC'B') + Area(DAA'D')
Note: For a simple graphical solution see

Edited on January 7, 2006, 2:40 pm
  Posted by Bractals on 2006-01-05 14:04:12

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