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: $a+b+c=10$.
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. 
A: 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 $\sum a^2 =24$
A: I shall attempt to do a step-by-step approach.  You can refer to this for one of the formulas.  Hence, the area of the triangle would be $1 * 5 = 5$.  Using another formula here, $$R = abc/4rs$$.  Hence, $$3 = \frac{abc}{20}$$, so $abc = 60$.  Since $s= 5$, perimeter $$a + b + c$$ = 10.
We have:
$$abc = 60.$$
And $$a + b + c = 10$$
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$$.
Consequently, using Heron's formula, we have $$5^2= 5(5-a)(5-b)(5-c)$$, so we get $$(5-a)(5-b)(5-c) = 5$$.  We have $$125 - 25a - 25b - 25c + 5ab + 5ac + 5bc -abc = 5$$.  Substituting 1 and 2, we have $$125 - 250 + 5(ab + ac + bc) - 60 = 5$$.  This means $$ab + ac + bc = 38$$.  Substitute this into $$(a + b + c)^2= a^2 + b^2 + c^2 + 2(ab + bc + ac)$$, we get $$100 = a^2 + b^2 + c^2 + 2*38$$.  Hence, $$a^2 + b^2 + c^2 = 100 - 76 = 24$$.
Note:
Any helpful user please help me format my code to look like simultaneous equations.  Thanks.
