 My Math Forum Help with some pretty hard geometry issues
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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(b-a) Now I continued, to find an equation that would define b for me! a+b+7(b-a)=72 simplified: 4b-3a=36 b-3a/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, ac 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!  Tags geometry, hard, issues, pretty ### how to find the perimeter of a triangle that circumscribes a circle

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