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Question Number 25871 by ajfour last updated on 16/Dec/17

Commented by ajfour last updated on 16/Dec/17

△ABC is equilateral and G its centroid. Parabolas AGB, BGC, and CGA each have their vertices at G. Find ratio of area in brown to the area of the △ABC .

$$\bigtriangleup{ABC}\:{is}\:{equilateral}\:{and}\:{G}\:{its} \\ $$$${centroid}.\:{Parabolas}\:{AGB},\:{BGC}, \\ $$$${and}\:{CGA}\:{each}\:{have}\:{their}\:{vertices} \\ $$$${at}\:{G}.\:{Find}\:{ratio}\:{of}\:{area}\:{in}\:{brown} \\ $$$${to}\:{the}\:{area}\:{of}\:{the}\:\bigtriangleup{ABC}\:. \\ $$

Commented by moxhix last updated on 16/Dec/17

put l=AB=BC=CD distance of G and BC=(1/3)×((√3)/2)l=(l/(2(√3))) find BGC by integration y=−ax(x−l)=−a(x−(l/2))^2 +((al^2 )/4) (l/(2(√3)))=((al^2 )/4) ∴a=(2/(l(√3))) BGC=∫_0 ^( l) −ax(x−l)dx=((al^3 )/6)=(l^2 /(3(√3))) △ABC=(1/2)l((√3)/2)l=(((√3)l^2 )/4) ((brown)/(△ABC))=((3BGC−△ABC)/(△ABC))=(3(l^2 /(3(√3)))−(((√3)l^2 )/4))/((((√3)l^2 )/4)) =(((√3)/3)−((√3)/4))/(((√3)/4)) =(1/3)

$${put}\:{l}={AB}={BC}={CD} \\ $$$${distance}\:{of}\:{G}\:{and}\:{BC}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{l}=\frac{{l}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$${find}\:{BGC}\:{by}\:{integration} \\ $$$${y}=−{ax}\left({x}−{l}\right)=−{a}\left({x}−\frac{{l}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{{al}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\frac{{l}}{\mathrm{2}\sqrt{\mathrm{3}}}=\frac{{al}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\therefore{a}=\frac{\mathrm{2}}{{l}\sqrt{\mathrm{3}}} \\ $$$${BGC}=\int_{\mathrm{0}} ^{\:{l}} −{ax}\left({x}−{l}\right){dx}=\frac{{al}^{\mathrm{3}} }{\mathrm{6}}=\frac{{l}^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}{l}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{l}=\frac{\sqrt{\mathrm{3}}{l}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$$\frac{{brown}}{\bigtriangleup{ABC}}=\frac{\mathrm{3}{BGC}−\bigtriangleup{ABC}}{\bigtriangleup{ABC}}=\left(\mathrm{3}\frac{{l}^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}−\frac{\sqrt{\mathrm{3}}{l}^{\mathrm{2}} }{\mathrm{4}}\right)/\left(\frac{\sqrt{\mathrm{3}}{l}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\:\:\:\:=\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)/\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by ajfour last updated on 16/Dec/17

Great Sir! Thank you too much.

$${Great}\:{Sir}!\:{Thank}\:{you}\:{too}\:{much}. \\ $$

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