# Solving the equation $8\Delta = \left( {b + c} \right)\left( {bc + 1} \right)$

In $$\Delta ABC$$, $$8\Delta = \left( {b + c} \right)\left( {bc + 1} \right)$$ then circumradius of is $$\Delta ABC$$ is ( where $$\Delta$$ denotes area of triangle and b, c are length of sides AC and AB respectively)

(1) $$\sqrt \Delta$$

(2) $$\frac{1}{{\sqrt {2\Delta } }}$$

(3) $$\sqrt {2\Delta }$$

(4) $$\frac{1}{{\sqrt \Delta }}$$

My approach is as follow $$R = \frac{{abc}}{{4\Delta }}$$ where R is the circumradius of the $$\Delta ABC$$

$$\Delta = \frac{1}{2}bc\sin A$$,

$$8\Delta = 4bc\sin A = \left( {b + c} \right)\left( {bc + 1} \right) \Rightarrow 4bc\sin A - bc\left( {b + c} \right) = \left( {b + c} \right)$$

Nor able to proceed further.

• Try the obvious example $b=c=\sin A=1$. Commented Sep 6, 2023 at 17:28
• @user10354138 yes this is what is mentioned but how did we arrive to this conclusion that is important Commented Sep 7, 2023 at 4:09

Rewriting everything in terms of $$R$$ and sine of the angles, the condition $$8\Delta=(b+c)(bc+1)$$ gives $$4R^2\sin B\sin C-8R\frac{\sin A\sin B\sin C}{\sin B+\sin C}+1=0.$$ As this quadratic in $$R$$ must have a positive root, we get $$\left(\frac{2\sin A\sin B\sin C}{\sin B+\sin C}\right)^2-\sin B\sin C\geq 0.$$ That is, $$4\sin^2A\sin B\sin C\geq (\sin B+\sin C)^2$$ Can you finish it from here?
! Apply AM-GM to $$(\sin B+\sin C)^2$$ and we have $$4\sin^2A\sin B\sin C\geq 4\sin B\sin C\geq 4\sin^2A\sin B\sin C$$
We have from the equation $$4\sin A = \left(\dfrac{1}{b}+\dfrac{1}{c} \right)(bc+1)$$
$$\Rightarrow 4 \sin A \ge \dfrac{2}{\sqrt{bc}} \times 2 \sqrt{bc} = 4$$
Hence $$\sin A =1 =b =c$$