Question Number 15828 by Tinkutara last updated on 14/Jun/17
$$\mathrm{In}\:\mathrm{a}\:\Delta{ABC},\:\mathrm{let}\:{M}_{{a}} ,\:{M}_{{b}} \:\mathrm{and}\:{M}_{{c}} \:\mathrm{denote} \\ $$$$\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{medians}, \\ $$$${s}\:=\:\frac{{M}_{{a}} \:+\:{M}_{{b}} \:+\:{M}_{{c}} }{\mathrm{2}}\:\:\mathrm{and}\:\Delta\:=\:\mathrm{ar}\left(\Delta{ABC}\right). \\ $$$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\Delta\:=\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{{s}\left({s}\:−\:{M}_{{a}} \right)\left({s}\:−\:{M}_{{b}} \right)\left({s}\:−\:{M}_{{c}} \right)} \\ $$
Answered by ajfour last updated on 14/Jun/17
Commented by ajfour last updated on 14/Jun/17
$${M}_{{a}} ={AP}\:\:\:\:{M}_{{b}} ={BQ}\:\:\:{M}_{{c}} ={CR} \\ $$$${A}\:{triangle}\:{with}\:{sides}\:{equal}\:{to} \\ $$$$\:\:\frac{\mathrm{2}}{\mathrm{3}}{M}_{{a}} ,\:\frac{\mathrm{2}}{\mathrm{3}}{M}_{{b}} ,\:\frac{\mathrm{2}}{\mathrm{3}}{M}_{{c}} \:\:{is}\:{for} \\ $$$${example}\:\Delta{GBL}.\:\left({All}\:{six}\:{triangles}\right. \\ $$$$\Delta{AGN},\:\Delta{BGN},\:\Delta{BGL},\:\Delta{CGL}, \\ $$$$\Delta{CGM},\:{and}\:\Delta{AGM}\:\:{of}\: \\ $$$$\:{hexagon}\:{ANBLCM}\:{have}\:{these} \\ $$$${lengths}\:{as}\:{their}\:{sides}. \\ $$$$\:\:\:\:{s}=\frac{{M}_{{a}} +{M}_{{b}} +{M}_{{c}} }{\mathrm{2}} \\ $$$$\:\:{let}\:\:{l}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{\mathrm{3}}{M}_{{a}} +\frac{\mathrm{2}}{\mathrm{3}}{M}_{{b}} +\frac{\mathrm{2}}{\mathrm{3}}{M}_{{c}} \right) \\ $$$$\:{then}\:\:{l}=\:\frac{\mathrm{2}}{\mathrm{3}}{s}\:\:\:\:\:\:\:\:\:\: \\ $$$$\:{Area}\:{of}\:{any}\:{such}\:\Delta\:{of}\:{the}\:{hexagon} \\ $$$$\:=\sqrt{{l}\left({l}−\frac{\mathrm{2}}{\mathrm{3}}{M}_{{a}} \right)\left({l}−\frac{\mathrm{2}}{\mathrm{3}}{M}_{{b}} \right)\left({l}−\frac{\mathrm{2}}{\mathrm{3}}{M}_{{c}} \right)}\: \\ $$$$\:{as}\:\:{l}=\frac{\mathrm{2}}{\mathrm{3}}{s}\:\:{we}\:{have} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{6}}\left({hexagon}\:{area}\right)\: \\ $$$$\:=\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} \sqrt{{s}\left({s}−{M}_{{a}} \right)\left({s}−{M}_{{b}} \right)\left({s}−{M}_{{c}} \right)}\: \\ $$$${hexagon}\:{area}\:=\mathrm{2}\left({area}\:{of}\:\Delta{ABC}\right) \\ $$$$\:\:\:\:\:=\mathrm{2}\Delta\:\:\:,\:\:\:\:{therefore} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{2}\Delta\right)=\frac{\mathrm{4}}{\mathrm{9}}\sqrt{{s}\left({s}−{M}_{{a}} \right)\left({s}−{M}_{{b}} \right)\left({s}−{M}_{{c}} \right)}\: \\ $$$$\:\Delta=\frac{\mathrm{4}}{\mathrm{3}}\sqrt{{s}\left({s}−{M}_{{a}} \right)\left({s}−{M}_{{b}} \left({s}−{M}_{{c}} \right)\right.}\:. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
$$\mathrm{How}\:\mathrm{are}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\Delta{GBL}\:\mathrm{two}-\mathrm{third} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{medians}\:\mathrm{of}\:\Delta{ABC}? \\ $$
Commented by ajfour last updated on 14/Jun/17
$${lines}\:{are}\:{constructed}\:{parallel}\:{to} \\ $$$${the}\:{medians}\:.. \\ $$$$\:{GB}=\frac{\mathrm{2}}{\mathrm{3}}{BQ}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{M}_{{b}} \\ $$$$\:{BL}=\frac{\mathrm{2}}{\mathrm{3}}{CG}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{M}_{{c}} \\ $$$$\:{GL}={BN}={AG}=\frac{\mathrm{2}}{\mathrm{3}}{AP}\:=\:\frac{\mathrm{2}}{\mathrm{3}}{M}_{{a}} \:. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by RasheedSoomro last updated on 14/Jun/17
$$\mathbb{V}{NiC}\mathcal{E}\:\:\:\mathbb{S}\mathrm{ir}! \\ $$