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Triangulation of Numbers (Posted on 2008-03-18) Difficulty: 3 of 5
       W------------------------X
       |                      * |
       |    A              *    |          
       |                O       |
       |             *  * *  B  |	 
       |          *     *  *    |
       |       *        *   *   |
       |    *           *    *  |
       | *      D       *  C  * |
       Z----------------Q-------Y
What is the minimum area of rectangle WXYZ if all lengths are whole numbers, as are the areas of the similar triangles, denoted by A, B, C & D?

See The Solution Submitted by brianjn    
Rating: 3.6667 (3 votes)

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Solution No Subject Comment 6 of 6 |
Let a = QY, b = OQ, c = OY, d = QZ, e = OZ, f = OX, g = XY, h = WX, i = XZ

Then a series of equations can be written using only the Pythagorean theorem and similar triangle theorem:
c = sqrt[a^2+b^2]
d = b^2/a
e = sqrt[d^2+b^2]
e = b*c/a
f = c*a/b
g = sqrt[c^2+f^2]
h = a+d
h = g*b/a
h = sqrt[c^2+e^2]
i = e + f
i = sqrt[g^2+h^2]

The system of equations is redundant and simplifies to:
c = sqrt[a^2+b^2]
d = b^2/a
e = b*c/a
f = a*c/b
g = c^2/b
h = c^2/a
i = c^3/(a*b)

For any integral Pythagorean triple (a,b,c), the parameterization gives a rational solution. To ensure an integral solution, each value needs to be multiplied by a*b. This results in:
c = sqrt[a^2+b^2]
QY = a^2*b
OQ = a*b^2
OY = a*b*c
QZ = b^3
OZ = b^2*c
OX = a^2*c
XY = a*c^2
WX = b*c^2
XZ = c^3

Using a=3, b=4, c=5 gives QY=36, OQ=48, OY=60, QZ=64, OZ=80, OX=45, XY=75, WX=100, XZ=125. The area of the rectangle is 7500, the area triangle A is 3750, the area of triangle B is 1350, the area of triangle C is 864, the area of triangle D is 1536.
  Posted by Brian Smith on 2008-03-20 00:56:39
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