Question Number 18847 by Tinkutara last updated on 30/Jul/17
$$\mathrm{In}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{usual} \\ $$$$\mathrm{notations}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{27}{r}^{\mathrm{2}} {R}}{{r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} }\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Answered by behi.8.3.4.1.7@gmail.com last updated on 31/Jul/17
$${in}\:{equilateral}\:{triangle}: \\ $$$$\left.\mathrm{1}\right){R}=\mathrm{2}{r}=\frac{{a}}{\mathrm{2}{sinA}}=\frac{{a}}{\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}={a}.\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right){r}_{\mathrm{1}} ={r}_{\mathrm{2}} ={r}_{\mathrm{3}} ={p}.{tg}\frac{{A}}{\mathrm{2}}=\frac{\mathrm{3}{a}}{\mathrm{2}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}={a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{27}{r}^{\mathrm{2}} {R}}{\Pi{r}_{\mathrm{1}} }=\frac{\mathrm{27}×\frac{{a}^{\mathrm{2}} }{\mathrm{12}}×{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{{a}^{\mathrm{3}} ×\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}}=\mathrm{2}\:\:.\blacksquare \\ $$
Commented by Tinkutara last updated on 31/Jul/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{behi}\:\mathrm{Sir}! \\ $$