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A triangle $A$$B$$C$ is integer sided and has inradius $1$.prove that it is right angled. Please help.give some hints please.

user435358
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3 Answers3

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Let the segments of tangents from the vertices to the incircle be $x,y,z$ (what is popularly known as the Ravi-substitution.

We have $(x+y), (y+z), (z+x)$ are all integers and so is $2(x+y+z)$ so $x+y+z = \frac{n}{2}, n \in \mathbb{N}$. Now its easy to see that $x,y,z$ are in the form $\frac{a}{2}, \frac{b}{2}, \frac{c}{2}$. Since $x+y, y+z, z+x \in \mathbb{N}, a,b,c$ have the same parity

We are given that $r=\frac{xyz}{x+y+z} =1 $ i.e. $abc = 4(a+b+c)$ from which its clear that $a,b,c$ are all even $\Rightarrow x,y,z \in \mathbb{N}$.

So we are looking for solutions in $\mathbb{N}$ for $xyz = x+y+z$.

WLOG $x \le y \le z$. Hence $xyz = x+y+z \le 3z \Rightarrow xy \le 3$

Now taking the cases $(x,y) = (1,1), (1,2), (1,3)$ we see that $x=1, y=2, z=3$ is the only solution.

This gives the sides as $3,4,5$ which is a Pythagorean Triplet and hence sides of a right angled triangle.

Hari Shankar
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My idea : $r = \Delta/s = 1 \implies \Delta = s \implies \Delta^2 = s^2 \implies (s-a)(s-b)(s-c) = s \implies (b+c-a)(c+a-b)(a+b-c) = 4(a+b+c) $.

If all $a,b,c$ are even or odd , then the parity doesn't satisfy. Thus that means $a+b+c$ is even. So $s$ is an integer. Thus by the above argument, we have $ s-a|s,s-b|s,s-c|s $. Now let $r_a,r_b,r_c$ be the exradii of the triangle $ABC$. We have our formula $r_a= \Delta/s-a = s/s-a $. Thus $r_a,r_b,r_c$ are integers. Now also we have the equation : $\frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c} = \frac{1}{r} \implies\frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c} = 1 $ and this has only one solution namely $ 2,3,6$ and hence you get the values of sides as $3,4,5$.

Chirantan Chowdhury
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Let $a$, $b$ and $c$ be sides-lengths of the triangle.

Thus, $$1=\frac{2S}{a+b+c}$$ or $$4(a+b+c)^2=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ or $$4(a+b+c)=(a+b-c)(a+c-b)(b+c-a).$$ Now, it's obvious that $a+b+c$ is an even number.

Thus, $a+b-c$, $a+c-b$ and $b+c-a$ are even positive numbers.

Let $a+b-c=2w$, $a+c-b=2v$ and $b+c-a=2u$.

Thus, we need to solve $$u+v+w=uvw$$ or $$\frac{1}{uv}+\frac{1}{uw}+\frac{1}{vw}=1.$$ Let $uv=z$, $uw=y$ and $vw=x$.

Hence, $x$, $y$ and $z$ are natural numbers such that $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.$$ This equation we can prove by the following way.

Let $x\leq y\leq z$.

Hence, $$1=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq\frac{3}{x},$$ which gives $x\leq3$.

$x=1$ is impossible and $x=2$ and $x=3$ give the following solutions:

$(2,3,6)$, $(2,4.4)$ and $(3,3,3)$.

Last two are impossible and $(x,y,z)=(2,3,6)$ gives $(u,v,w)=(3,2,1)$,

which gives $(a,b,c)=(3,4,5)$ and since $4^2+3^2=5^2,$ we are done!

Michael Rozenberg
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