If in ,$r=1$,$R=3$ and $s=5$ then find value of $a^2+b^2+c^2$ If in $\triangle ABC$,$r=1$,$R=3$ and $s=5$, then find value of $a^2+b^2+c^2$
(where $r$=inradius,$R$=circumradius and $s$=semi-perimeter)
The answer given is $24$

$a+b+c=10$.
Let $S$ be an area of the triangle.
Thus, $$\frac{2S}{a+b+c}=1,$$ which gives $$S=5$$ or
$$\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}=5$$ or
$$(a+b-c)(a+c-b)(b+c-a)=40.$$
Now, by AM-GM
$$40=(a+b-c)(a+c-b)(b+c-a)\leq\left(\frac{a+b-c+a+c-b+b+c-a}{3}\right)^3=\frac{1000}{27},$$
which is impossible.
Thus, this triangle does not exist. 
