perplexus.info :: Geometry : Rhombus Rationale

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

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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 a, b is divisible by 3
Exactly one of a, b 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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