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 Rhombus Rationale (Posted on 2013-08-11)
Each of the two diagonals and each of the four sides of a rhombus has integer lengths, satisfying the condition: x2 + y2 = 4z2, where x, y are the length of the diagonals and z being the length of each side.

Is the area of the rhombus always divisible by 24?
If so, prove it. Otherwise provide a counter example.

 No Solution Yet Submitted by K Sengupta No Rating

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Let x=2a and y=2b
Than a,b,z form a  Pythagorian triple .(PT).

A primitive PT has  inter alia two features:

• Exactly one of ab is divisible by 3
• Exactly one of ab is divisible by 4
So ab is 12*pq.- p,q integers.

The word exactly might be redundant in non primitive PT.

S=.5*XY= .5*4*ab=,5*4*12pq=24pq

Ergo :
the area of the "integer" rhombus
is always divisible by 24.

 Posted by Ady TZIDON on 2013-08-11 10:57:44

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