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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 ?"

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  • $\begingroup$ 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
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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.

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

http://jwilson.coe.uga.edu/emt725/Perfect/sol.html

Equal perimeter and area

Right triangle where the perimeter = area*k

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

your condition leads to

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

or

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

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
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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 $A,B,C$.

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

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

$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{m\pm\sqrt{m^2-4*1*2R}}{2*1}$$

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

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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