I've tried to use Heron's formula to approach the problem , but it doesn't make any sense .I also tried to guess the lengths and I got two triangles , one of them is (5,12,13) and the second is (6,8,10).

So,I hope you can help me to find out "Is there efficient way to solve this problem ?"

  • $\begingroup$ I deleted my post because I don't have a constructive answer (yet). There may be even more combinations that work! $\endgroup$ – imranfat Sep 22 '16 at 20:49
  • $\begingroup$ But something tells me that you should only look for right triangles. Reason is that your perimeter is always an integer, but your area will have a radical due to Heron's formula. But area of a right triangle is just half the product of the legs $\endgroup$ – imranfat Sep 22 '16 at 20:51
  • $\begingroup$ So , you mean that impossible to get other triangles satisfy the conditions , but actually , we have to prove that . $\endgroup$ – user371433 Sep 22 '16 at 20:57
  • $\begingroup$ If the triangle is not right, then if the sides are integers, the area will contain a square root. In case of a right triangle, if the legs are $a$ and $b$, we need to satisfy $ab=2a+2b+2\sqrt{a^2+b^2}$. Graphing in Desmos I can only come up with the solutions you provided, I have no formal proof $\endgroup$ – imranfat Sep 22 '16 at 21:02
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    $\begingroup$ There are non-right triangles with integer sides and integer area. Topic: Heronian triangles. $\endgroup$ – coffeemath Sep 22 '16 at 21:09

These are the Equable triangles. There are only five: $(5,12,13), (6,8,10), (6,25,29), (7,15,20), \text{ and }(9,10,17)$. The first two are right triangles, the others are not. I couldn't get the references in th Wikipedia article to work.

  • $\begingroup$ till now I didn't find a full solution for the problem.I hope you could help me if you get any extra information $\endgroup$ – user371433 Sep 22 '16 at 22:05
  • $\begingroup$ No, I found the Wikipedia link, which has two references. I tried to access them without success. The second might be available to those with University library privileges. $\endgroup$ – Ross Millikan Sep 22 '16 at 22:08

The radius of inner circle is 2. They are called Perfect triangle or Heronian triangle. There are many similar question on web.


Equal perimeter and area

Right triangle where the perimeter = area*k


Trying with Pythagorean triplets with sides $ (2mn, m^2-n^2,m^2+n^2),$

your condition leads to

$$ 2 m^2 + 2 mn = mn ( m^2-n^2) ; \quad n (m-n) = 2; $$


$$ n = (m + \sqrt{ m^2-8})/2 $$

which gives an infinite set including

$$ (m,n) = (3,2), \sqrt2 (2,1),.. $$

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    $\begingroup$ OP asked that the sides be integral, which requires that $m,n=3,2$ and gives the $5,12,13$ solution. The $6,8,10$ solution gets missed because it is not a primitive triangle. This proves there are no more primitive Pythagorean triangles. $\endgroup$ – Ross Millikan Sep 22 '16 at 21:16
  • $\begingroup$ The op never stated the triangle were right triangles. $\endgroup$ – fleablood Sep 22 '16 at 21:38
  • $\begingroup$ The Pythagorean triple solution does find the $\{6,8,10\}$ solution. We just have to allow the "twice-primitive" cases by letting both $m$ and $n$ be odd, then admit both possible signs for the square root. $\endgroup$ – Oscar Lanzi Sep 27 '16 at 9:56
  • $\begingroup$ Was just a particular case. $\endgroup$ – Narasimham Sep 27 '16 at 10:53

This a Pythagorean triple treatment of area to perimeter ratios where $\angle AB$ is right and $A^2+B^2=C^2$.

We can find Pythagorean triples, if they exist, for any ratio $R$ of area/perimeter by finding the $m,n$(s) that represent them using the following formula which includes a difined finite search for values of $m$. Whenever the $m$ $R$ combination yields a positive integer for $n$, we have the $m,n$ for a triple. We begin by solving the area/perimeter equation for $n$ where area=$D$ so as not to confuse it with $A,B,C$.

$$D=\frac{AB}{2}=\frac{(m^2-n^2)*2mn}{2}=mn(m^2-n^2)\quad P=(m^2-n^2)+2mn+(m^2+n^2)=2m^2+2mn$$

$$\frac{D}{P}=\frac{mn(m^2-n^2)}{2m^2+2mn}=\frac{mn(m-n)(m+n)}{2m(m+n)}=\frac{n(m-n)}{2}=R\qquad n^2-mn+2R=0$$


$$n=\frac{m\pm \sqrt{m^2-8R}}{2}\text{ where }\lceil\sqrt{8R}\space \rceil \le m \le 2R+1$$

Example: For $R=0.5\quad \sqrt{8*0.5}=2\le m \le 2*0.5+1=2$

$$n=\frac{2\pm \sqrt{2^2-8*0.5}}{2}=1\qquad f(2,1)=(3,4,5)$$ But you are asking about $R=1$ and there only two such Pythagorean triples to be found.

$$R=1\rightarrow 3\le m \le 3\quad f(3,2)=(5,12,13)\quad f(3,1)=(8,6,10)$$ For other ratios: $$R=1.5\rightarrow 4\le m \le 4\quad f(4,3)=(7,24,25)\quad f(4,1)=(15,8,17)$$

$$R=2\rightarrow 4\le m \le 5\quad f(4,2)=(12,16,20)\quad f(5,4)=(9,40,41)\quad f(5,1)=(24,10,26)$$

$$R=2.5\rightarrow 4\le m \le 6\quad f(6,5)=(11,60,61)\quad f(6,1)=(35,12,37)$$

$$R=3\rightarrow 4\le m \le 7\quad f(5,3)=(16,30,34)\quad f(5,2)=(21,20,29)\quad f(7,6)=(13,84,85)\quad f(7,1)=(48,14,50)$$

$$R=18\rightarrow 12\le m \le 37\quad f(12,6)=(108,144,180)\quad f(13,9)=(88,234,250)\quad f(13,4)=(153,104,185)\quad f(15,12)=(81,360,369)\quad f(15,3)=(216,90,234)\quad f(20,18)=(76,720,724)\quad f(20,2)=(396,80,404)\quad f(37,36)=(73,2664,2665)\quad f(37,1)=(1368,74,1370)$$

Aside: you can always find one primitive triple for a given $R$ if you let $(m,n)=(2R+1,2R)$.


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