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January 18th, 2012, 06:33 AM  #11 
Newbie Joined: Jan 2012 Posts: 9 Thanks: 0  Re: Help with some pretty hard geometry issues
Yes, I finally got it! I'll post what I did! So the first steps I followed were already mentioned by Denis. Thus we get our second equation (first being 72=a+b+c)  c=7(ba) Now I continued, to find an equation that would define b for me! a+b+7(ba)=72 simplified: 4b3a=36 b3a/4=9 b=9+3a/4 b=(36+3a)/4 From this point, I used some logic, and realised that a,b and c are all integers, thus we must find such an a that if divided by 4, would be an integer. Which would be of course any multiple of 4. Then I have made the following table (I subsituted the multiples of 4 into the following equations): The equation would make sense only if a+b>c, c>a, c>b, a<b, and of course a/4?N, I ticked the equations that met those conditions, and written in red the condition they didn't meet. 72= a b c 72= 4 12 56 [color=#BF0000] a+b<c[/color] 72= 8 15 49 [color=#BF0000] a+b<c[/color] 72= 12 18 42 [color=#BF0000] a+b<c [/color] 72= 16 21 35 [color=#00BF00] ?[/color] 72= 20 24 28 [color=#00BF00] ?[/color] 72= 24 27 21 [color=#BF0000] a>c b>c[/color] ..... all higher as would be a higher number than c, so no point to continue. Thus we get both solutions for this problem! Thank you all for helping me out with the first steps! 

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