<s>[INST] 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 $24$
 [/INST] You know that $$r=\frac{Δ}{s}$$ Hence we get $$Δ=5$$
Also $$R=\frac{abc}{4Δ} $$ From this we get $$ abc=60$$
We also have a formula that 
$$\frac{4(s-a)(s-b)(s-c)}{abc}=\frac{r}{R}$$
Hence, on substituting values we get
$$(5-a)(5-b)(5-c)=5$$
Which reduces to 
$$125-50s+5\sum_{cyc} ab -abc = 5$$
Further on reducing we get 
$$\sum_{cyc} ab=38$$
Now 
$${(\frac{a+b+c}{2})}^2=s^2$$
$$\frac{a^2+b^2+c^2+2\sum ab}{4}=25$$
$$a^2+b^2+c^2=100-76=24$$
Hence $\sum a^2 =24$
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