All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Rhombus Rationale (Posted on 2013-08-11) Difficulty: 3 of 5
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

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

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
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information