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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.

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

2 Answers 2

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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 $$ 4\sin^2A\sin B\sin C\geq 4\sin B\sin C\geq 4\sin^2A\sin B\sin C $$

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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 $\sin A =1 =b =c$

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