In Delta ABC,8 Delta = (b + c)(bc + 1),then t | vedclass

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(A) Given $8 \Delta = (b + c)(bc + 1)$.Since $\Delta = \frac{1}{2} bc \sin A$,we have $8 \cdot \frac{1}{2} bc \sin A = (b + c)(bc + 1)$.$4 bc \sin A =

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$\sqrt{\Delta}$

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(A) Given $8 \Delta = (b + c)(bc + 1)$.Since $\Delta = \frac{1}{2} bc \sin A$,we have $8 \cdot \frac{1}{2} bc \sin A = (b + c)(bc + 1)$.$4 bc \sin A = (b + c)(bc + 1)$.Dividing by $bc$,we get $4 \sin A = (b + c) \left(1 + \frac{1}{bc}\right) = b + \frac{1}{b} + c + \frac{1}{c}$.By $AM$-$GM$ inequality,$b + \frac{1}{b} \ge 2$ and $c + \frac{1}{c} \ge 2$,so $b + \frac{1}{b} + c + \frac{1}{c} \ge 4$.Since $\sin A \le 1$,$4 \sin A \le 4$.Thus,equality holds only when $b = 1, c = 1$ and $\sin A = 1$,which means $A = 90^\circ$.In this case,the sides are $b = 1, c = 1$,and $a = \sqrt{b^2 + c^2} = \sqrt{2}$.The area $\Delta = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$.The circumradius $R = \frac{a}{2 \sin A} = \frac{\sqrt{2}}{2 \cdot 1} = \frac{1}{\sqrt{2}}$.Since $\Delta = \frac{1}{2}$,we have $\sqrt{\Delta} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,$R = \sqrt{\Delta}$.

In $\Delta ABC$,$8 \Delta = (b + c)(bc + 1)$,then the circumradius of $\Delta ABC$ is (where $\Delta$ denotes the area of the triangle and $b, c$ are the lengths of sides $AC$ and $AB$ respectively):

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