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 Two squares (Posted on 2006-01-05)
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)

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 Solution | Comment 1 of 7
`Let X and Y be complex numbers and O the origin. Define thecross product`
`    X x Y = X"Y - XY" = 4i*Area(OXY)`
`where X" and Y" are the complex conjugates of X and Y, i isthe square root of minus 1, and triangle OXY is labeledcounterclockwise.`
`Let A = 0+0i, B = 1+0i, C = 1+1i, and D = 0+1i be the verticesof 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`
`Therefore,`
`    Area(ABB'A') + AREA(CDD'C') = Area(BCC'B') + Area(DAA'D')`
`Note: For a simple graphical solution see`
`      www.geocities.com/bractals/2sq.jpg`
` `

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

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