Is there any natural number triangle that inscribed circle's radius is $1$ except length $(a,b,c)=(3,4,5)$?

I found that there are no right triangle except $(3,4,5)$.

Thm. There are only one natural number right triangle that inscribed circle's radius is 1.

Proof)WLOG, $0<a<b<c$. Than $$a^2+b^2=c^2$$ $$a+b+c=ab$$. Because it is right triangle, $$a-1+b-1=c$$ .So, $$b=\frac{2(a-1)}{(a-2)}$$

  1. If $a$ is even, $b$ is not a natural number.Because if $a=2k$ ,$b=\frac{2k-1}{k-1}=2+\frac{1}{k-1}$. Only integer solution in $k=2$.

  2. If $a$ is odd, $a=2k+1$ ,$a-1$is even but $a-2$ is odd. So $b$ is not integer except $a-2=1$.

I want to know general case about this. Such triangle is exist or not exist. I think It doesn't exist.

I think about it and I got one. Check It for me wrong or not.

We could make $$(a,b,c) =(x+y,y+z,z+x)$$ and by Heron's formula radius $$1= \frac{\sqrt{xyz}}{\sqrt{x+y+z}}$$ so $$x+y+z=xyz$$ and we knows only solution of this equation is $(x,y,z)=(3,2,1)$.

  • $\begingroup$ @GReyes So there are no more triangle that inscribed circle is 1??? $\endgroup$
    – user366725
    Feb 17, 2019 at 9:39
  • $\begingroup$ Here is a list of formulas for the radius of the incircle. Most notably, the radius of the incircle is equal to twice the area of the triangle, divided by the perimeter of the triangle. So one possible place to start is to look for triangles with integer area. $\endgroup$
    – Arthur
    Feb 17, 2019 at 9:54

1 Answer 1


Say $a\leq b\leq c$. If $s=(a+b+c)/2$ then $$r={S\over s}$$ where $S$ is area. By Heron formula we have

$$S=s \implies \boxed{(s-a)(s-b)(s-c)=s}$$

then $$s-a\mid s\implies s-a\mid a \implies a = k(s-a)$$

so $$k(b+c)=a(2+k)$$ Since $b+c\geq 2a$ we get $$2ka\leq a(2+k)\implies k\leq 2$$

  • $k= 1$ we have $b+c=3a$ and $s=2a$, so $$(2a-b)(2a-c)=2\implies -2a^2+bc=2$$ since $c= 2a-b$ we get $$(2a-b)(b-a)=2$$ here we have 2 possibilities:

If $b-a = 1$ and $2a-b = 2$ we get $a=3$ and $b= 4$ so $c= 5$

If $b-a = 2$ and $2a-b = 1$ we get $a=3$ and $b= 5$ so $c= 4$

  • $k=2$ then $b+c=2a$ so $a=b=c$, so $(s-a)^3=s$ so $a^3/8 = 3a/2$ so $a^2 = 12$ which is impossibile.

So the only such triangle has sides $3,4,5$.

  • $\begingroup$ Rats! Would have been neat to see another. Oh well, +1. $\endgroup$ Feb 17, 2019 at 10:29

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