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Find all integer sided triangles whose area equals their perimeter?

Taking sides to be $e,f,g$ we have,

$$\sqrt{(e+f+g)(e+f-g)(e+g-f)(g+f-e)}=4(e+f+g)$$

how to proceed next ?

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Let $a,b,c$ be the lengths of the sides and $s=(a+b+c)/2$.

By Heron's formula, We have $$\sqrt{s(s-a)(s-b)(s-c)}=2s,$$ i.e. $$(-a+b+c)(a-b+c)(a+b-c)=16(a+b+c)\tag1$$ Here, let $x=-a+b+c,y=a-b+c,z=a+b-c$, then we have $$a=\frac{y+z}{2},\quad b=\frac{z+x}{2},\quad c=\frac{x+y}{2}\tag2$$ So, $(1)$ can be written as $$xyz=16(x+y+z)\tag3$$

Here, note that $x,y,z$ are positive integers and all even from $(2)(3)$. So, let $x=2X,y=2Y,z=2Z$, then $(3)$ can be written as $$XYZ=4(X+Y+Z).$$

We may suppose that $X\le Y\le Z$. So, $$XYZ\le 4\cdot 3Z=12Z\Rightarrow XY\le 12.$$ $$X^2\le XY\le 12\Rightarrow X^2\le 12\Rightarrow X\le 3.$$

  • For $X=1$, we have $(Y-4)(Z-4)=20\Rightarrow (Y,Z)=(5,24),(6,14),(8,9)$.

  • For $X=2$, we have $(Y-2)(Z-2)=8\Rightarrow (Y,Z)=(3,10),(4,6)$.

  • For $X=3$, we have $3YZ\le 4\cdot 3Z\Rightarrow Y\le 4$. If $Y=3$, then $9Z=4(6+Z)$. If $Y=4$, then $12Z=4(7+Z)$. Hence, there is no solution.

Thus, the answer is the followings :

$$(6,25,29),(7,15,20),(9,10,17),(5,12,13),(6,8,10).$$

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  • $\begingroup$ for X=1 why did you take (Y-4)(Z-4)=20 and X=2: (Y−2)(Z−2)=8 $\endgroup$ – shakunimama Aug 7 '15 at 10:50
  • $\begingroup$ @HeinzKloube: Because the form $(Y-4)(Z-4)=20$ is easy to solve : both $Y-4,Z-4$ are divisors of $20$ with $Y\le Z$. $\endgroup$ – mathlove Aug 7 '15 at 10:52
  • $\begingroup$ I am sorry! But I am still finding it difficult to understand. I mean, why 20 and how it is known that (Y-4),(Z-4) are divisors of 20 and same confusion I have for For X=2, we have (Y−2)(Z−2)=8⇒(Y,Z)=(3,10),(4,6). $\endgroup$ – shakunimama Aug 7 '15 at 11:00
  • $\begingroup$ @HeinzKloube: First of all, since $Y,Z$ are integers, both $Y-4$ and $Z-4$ are integers, right? From $(Y-4)(Z-4)=20$, we can have $Z-4=20/(Y-4)$. Since $Z-4$ is an integer, the right-hand-side also has to be an integer, which means that $Y-4$ has to be a divisor of $20$. Similarly, since we have $Y-4=20/(Z-4)$, we have that $Z-4$ has to be a divisor of $20$. Finally, from $1\le Y\le Z$, we have $(Y-4,Z-4)=(1,20),(2,10),(4,5)$, i.e. $(Y,Z)=(5,24),(6,14),(8,9)$. $\endgroup$ – mathlove Aug 7 '15 at 11:10
  • $\begingroup$ okay! So, its just an insitu math technique. I thought there is some sort of theorems.Just for knwoldge :: (Z-5)(Y-5)=20 can also be taken right ? $\endgroup$ – shakunimama Aug 7 '15 at 11:15

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