Patatiniuh

2022-07-05

For triangle ABC, find maximum value of $8+\frac{\mathrm{sin}2A+\mathrm{sin}2B+\mathrm{sin}2C}{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}$
Considering only the right hand part of the expression:
$\frac{\mathrm{sin}2A+\mathrm{sin}2B+\mathrm{sin}2C}{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}$
$\phantom{\rule{-10mm}{0ex}}=\frac{4\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}$

Kiana Cantu

Expert

Note that the area of the triangle $Area=\frac{abc}{4R}=\frac{1}{2}r\left(a+b+c\right)$. Then, per the sine rule
$\frac{4\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}=\frac{4\frac{abc}{8{R}^{3}}}{\frac{a+b+c}{2R}}=\frac{2r}{R}\le 1$
where, given the circumradius R, the inradius is the largest for equilateral triangle.

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