Solution:
It is clear from the table that only five such triangles are possible and they can be written in order as below:
Let the sides of the triangle be a,b,c and s= semi perimeter= (a+b+c)/2. It means a+b+c =2s.
Given that, Area of triangle = perimeter. Applying Heroin's formula,
√{ s(s-a) (s-b) (s-c)}= 2s
Squaring both sides we get
s(s-a) (s-b) (s-c)= 4s²
(s-a) (s-b) (s-c) = 4s
Let s-a=x, s-b= y, and s-c=z;
xyz = 4s
Adding the above three equations, we get
3s -(a+b+c) = x+y+z
3s-2s = x+y+z
s = x+y+z
then xyz= 4s= 4(x+y+z)
(x+y+z) = xyz/4
Let x ≤y ≤z, then
x+y+z ≤ 3z
xyz/4≤ 3z
xy ≤ 12 ...(A)
xyz = 4(x+y+z)= 4x+4y +4z
xyz-4z = 4x+4y
z(xy-4)= 4x+4y
Z= (4x+4y)/xy-4...(B)
Finally we get two useful Diophantine equations:
xy ≤ 12 ...(A)
(4x+4y)/xy-4...(B)
Luckily the inequality (A) is very much workable. We know x,y,x,a,b,c are all positive integers. So we can try for the solution now , as there are very few cases for the inequality xy ≤ 12 ...(A)
We are left with only two inequalities which can be now solved easily.
The possible values of x, y and z,s,a,b,c are tabulated as below:
It is clear from the table that only five such triangles are possible and they can be written in order as below:
(1) 6,8,10
(2) 5,12,23
(3) 9,10,17
(4) 7,15,20
(5) 6,25,29
Finally, let us equate area of a triangle with its perimeter.
Area = perimeter
rs= a+b+c
rs= 2s
Hence r=2. For all such triangles!
Additionally, the above formula is applicable to regular polygons also. For regular polygons also:
rs= a+b+c
rs= 2s
Hence r=2.
So for all such regular polygons also the inradius will remain 2!
So inradius of all these five triangles will be equal to 2.
In fact in any triangle or any regular polygon which has its area equal to its perimeter; inradius will be only 2. However only five such integer triangles are there and they do have the same sized incircle!