Question Number 53071 by behi83417@gmail.com last updated on 16/Jan/19
Commented by behi83417@gmail.com last updated on 16/Jan/19
$${as}\:{shown}\:{in}\:{fig}: \\ $$$${prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\rightarrow\:\frac{\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}\:\:{outer}\:\:\boldsymbol{\mathrm{hexagon}}}{\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}\:\:{inner}\:\boldsymbol{\mathrm{triangle}}}\:\:\:\geqslant\:\mathrm{13} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
$${area}\:{ABC}\:{triangle}={S} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}{acsinB}=\frac{\mathrm{1}}{\mathrm{2}}{absinC}=\frac{\mathrm{1}}{\mathrm{2}}{bcsinA} \\ $$$${others}\:{three}\:{triangle}\:…\boldsymbol{{butterfly}}\:\boldsymbol{{wings}}\:\boldsymbol{{like}}.. \\ $$$$\boldsymbol{{are}}\:\boldsymbol{{also}}\:\boldsymbol{{S}}. \\ $$$${area}\:{of}\:{hexagon}=\mathrm{4}{S}+{P}+{Q}+{R} \\ $$$${now}\:{S}+{Q}=\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{c}\right)\left({a}+{b}\right){sinB} \\ $$$${S}+{P}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right)\left({a}+{b}\right){sinA} \\ $$$${S}+{R}=\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{c}\right)\left({a}+{c}\right){sinC} \\ $$$${now}\:{we}\:{know} \\ $$$$\frac{{sinA}}{{a}}=\frac{{sinB}}{{b}}=\frac{{sinC}}{{c}}={k}\left({say}\right) \\ $$$${so}\:\bigtriangleup{ABC}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}{acsinB}=\frac{{abck}}{\mathrm{2}} \\ $$$${area}\:{hexagon}=\mathrm{4}{S}+{P}+{Q}+{R} \\ $$$$=\left({S}+{P}\right)+\left({S}+{Q}\right)+\left({S}+{R}\right)+{S} \\ $$$$=\frac{\left({a}+{c}\right)\left({a}+{b}\right){ka}+\left({b}+{c}\right)\left({a}+{b}\right){kb}+\left({b}+{c}\right)\left({a}+{c}\right){kc}+{abck}}{\mathrm{2}} \\ $$$$=\frac{{k}}{\mathrm{2}}\left[\left({a}^{\mathrm{2}} +{ab}+{ac}+{bc}\right){a}+\left({ab}+{b}^{\mathrm{2}} +{ac}+{bc}\right){b}+\left({ab}+{bc}+{ac}+{c}^{\mathrm{2}} \right){c}+{abc}\right] \\ $$$$=\frac{{k}}{\mathrm{2}}\left[{a}^{\mathrm{3}} +{a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{abc}+{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} +{abc}+{b}^{\mathrm{2}} {c}+{abc}+{bc}^{\mathrm{2}} +{ac}^{\mathrm{2}} +{c}^{\mathrm{3}} +{abc}\right] \\ $$$$=\frac{{k}}{\mathrm{2}}\left[\mathrm{4}{abc}+{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{a}^{\mathrm{2}} \left({b}+{c}\right)+{b}^{\mathrm{2}} \left({a}+{c}\right)+{c}^{\mathrm{2}} \left({a}+{b}\right)\right] \\ $$$$=\mathrm{2}{abck}+\frac{{k}}{\mathrm{2}}\left[{a}^{\mathrm{2}} \left({a}+{b}+{c}\right)+{b}^{\mathrm{2}} \left({a}+{b}+{c}\right)+{c}^{\mathrm{2}} \left({a}+{b}+{c}\right)\right] \\ $$$$\:\:=\mathrm{2}{abck}+\frac{{k}}{\mathrm{2}}\left[\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right] \\ $$$${now}\:\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left({abc}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\geqslant\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${so}\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\mathrm{9}{abc} \\ $$$${so}\:{area}\:{of}\:{hexagon} \\ $$$$=\mathrm{2}{abck}+\frac{{k}}{\mathrm{2}}\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\mathrm{2}{abck}+\frac{\mathrm{9}{abck}}{\mathrm{2}} \\ $$$${so}\:{area}\:{of}\:{hexagon}\geqslant\frac{\mathrm{13}{abck}}{\mathrm{2}} \\ $$$${area}\:{of}\:{hexagon}\geqslant\mathrm{13}×{area}\:{of}\:\bigtriangleup{ABC} \\ $$$$\frac{{area}\:{of}\:{hexagon}}{\bigtriangleup{ABC}}\geqslant\mathrm{13}\:{proved} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
$${sir}\:{pls}\:{check}\:{i}\:{have}\:{proved}… \\ $$
Commented by behi83417@gmail.com last updated on 17/Jan/19
$${thank}\:{you}\:{very}\:{much}\:{sir}\:{tanmay}. \\ $$$${it}\:{is}\:{a}\:{nice}\:{and}\:{smart}\:{work}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19
$${thank}\:{you}\:{sir}… \\ $$