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estudier
 3 years ago
All Pythagorean triples where area = perimeter (numerically)?
estudier
 3 years ago
All Pythagorean triples where area = perimeter (numerically)?

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RadEn
 3 years ago
Best ResponseYou've already chosen the best response.15,12,13 using trial and error method ^^

CheetahBooks1
 3 years ago
Best ResponseYou've already chosen the best response.0what do you mean?

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1ok, general form tripel pythagorean number is : a=x b=(x^21)/2 c=(x^2+1)/2 with a<b<c (a,b,c are sides of a righttriangle) perimeter = area x + (x^21)/2 + (x^2+1)/2 = x(x^21)/2, simplify it we get : x^3  4x^2  5x =0 x(x^2  4x  5) = 0 x(x5)(x+1)=0 x=5 satifisfy it (so, a=5, b=12, c=13) with same way, for a=2x b=x^21 c=x^2+1 perimeter = area 2x+x^21+x^2+1 = 2x(x^21)/2, simplify it we get : x^3  2x^2  3x = 0 x(x^2  2x  3) =0 x(x3)(x+1)=0 x=3 satifisfy it (so, a=6, b=8, c=10) i have done for a=kx b=k(x^21) c=k(x^2+1) with k=3,4,5,6... there are x satisfy it but the result just reflication of on... and its rest not satisfy for x which make for a,b,c integer number

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1so, just 2 triple only...

CheetahBooks1
 3 years ago
Best ResponseYou've already chosen the best response.03,4,5 is a triple, too!

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1but, its area not equal its perimeter...

estudier
 3 years ago
Best ResponseYou've already chosen the best response.1you can have an easier algebra by eliminating z from x^2+y^2=z^2 and xy/2 = x + y +z leads to xy 4x 4y +8 = 0 > (x4)(y4) = 8 etc.

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1just combine of (x4)(y4) = 8, and we will get 2 triple only are : (x,y,z) = 5,12,13 and 6,8,10 but yours easier than me :)
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