Menu Close

Question-53071




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

Leave a Reply

Your email address will not be published. Required fields are marked *