An integral triangle is defined as a triangle whose sides are measurable in whole numbers. Find all integral triangles whose perimeter equals their area.

At first, I thought this would be one of the easier contest math problems but I have only been able to work out a couple of things so far. The area of a triangle and the perimeter can be related by the same variables only using herons formula.

So assuming p to be the semiperimeter we get $\sqrt{p(p-a)(p-b)(p-c)}$ = $2p$.
Now Assuming $p-a$ = $x$, $p-b$ = $y$, $p-c$ = $z$ we get,
$\sqrt{(x+y+z)(xyz)}$= $2(x+y+z)$.
After squaring and simplifying we get, $xyz$ = $4(x+y+z)$.

After this, however, I'm not sure how to go about solving this equation.

Any help would be much appreciated.

  • $\begingroup$ Isn't it generally $\sqrt{p(p-a)(p-b)(p-c)}$? $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 at 18:34
  • $\begingroup$ Okay, you take the value of p=a+b+c/2 and then plug in the values of a and b and c from the equations x=p-a, y=p-b, z=p-c and you get 2p=3p-(x+y+z) and then you simplify. $\endgroup$ – Prakhar Nagpal Dec 30 '17 at 18:59
  • $\begingroup$ Yes I deleted my comment after I found it myself. $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 at 19:00
  • $\begingroup$ sorry for the late reply, i couldn't find the envelope i had figured it out on so I did it again xD $\endgroup$ – Prakhar Nagpal Dec 30 '17 at 19:01
  • $\begingroup$ desmos.com/calculator/1screrlayu $\endgroup$ – Mohammad Zuhair Khan Dec 30 '17 at 19:02

Let us start with your equation $xyz = 4(x+y+z)$. However note that you need to first establish $x, y, z $ are indeed integers, or alternately that the perimeter has to be even, before you get here! Not hard to do, so leaving that for you.

WLOG let $x \leqslant y \leqslant z$. This gives $xyz = 4(x+y+z) \leqslant 12z \implies xy \leqslant 12 \implies x \in \{1, 2, 3\}$.

Case $x = 1$: Now $yz = 4(y+z +1) \implies (y-4)(z-4)=20$. As $20$ factorises into $(1, 20), (2, 10)$ or $(4, 5)$, correspondingly we get $(y, z) \in \{(5, 24), (6, 14), (8,9) \}$.

Case $x = 2$: Now $2yz = 4(y+z + 1) \implies (y-2)(z-2)=8$, which factorises into $(1, 8)$ or $(2, 4)$ giving $(y, z) \in \{(3, 10), (4, 6)\}$.

Case $x=3$: Now $3yz = 4(y+z+1) \implies (3y-4)(3z-4)=52$, which however by the same process does not give any new integer solution.

Now it remains to translate the solutions got back to $(a, b, c) = (y+z, z+x, x+y)$ to get the five integer triangles you seek.


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