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Question Number 111284 by Aina Samuel Temidayo last updated on 03/Sep/20
Find the maximum area of a triangle  whose vertices lie on a regular  hexagon of unit area.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{whose}\:\mathrm{vertices}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{a}\:\mathrm{regular} \\ $$$$\mathrm{hexagon}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{area}. \\ $$
Commented by mr W last updated on 03/Sep/20
Commented by mr W last updated on 03/Sep/20
max. triangle =((hexagon)/2)=(1/2)
$${max}.\:{triangle}\:=\frac{{hexagon}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
Just explain please.
$$\mathrm{Just}\:\mathrm{explain}\:\mathrm{please}. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
What about a well explanatory  solution? Thanks.
$$\mathrm{What}\:\mathrm{about}\:\mathrm{a}\:\mathrm{well}\:\mathrm{explanatory} \\ $$$$\mathrm{solution}?\:\mathrm{Thanks}. \\ $$
Commented by Her_Majesty last updated on 03/Sep/20
what is it you do not understand???
$${what}\:{is}\:{it}\:{you}\:{do}\:{not}\:{understand}??? \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
Isn′t there a proof? How do we know  definitely that the area of that  triangle is half of that of the hexagon?
$$\mathrm{Isn}'\mathrm{t}\:\mathrm{there}\:\mathrm{a}\:\mathrm{proof}?\:\mathrm{How}\:\mathrm{do}\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{definitely}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{that} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}? \\ $$
Commented by mr W last updated on 03/Sep/20
i just don′t want to waste my time for  obvious things.
$${i}\:{just}\:{don}'{t}\:{want}\:{to}\:{waste}\:{my}\:{time}\:{for} \\ $$$${obvious}\:{things}. \\ $$
Commented by Her_Majesty last updated on 03/Sep/20
it′s plain to see that the red lines cut the  small triangles in two, isn′t it?
$${it}'{s}\:{plain}\:{to}\:{see}\:{that}\:{the}\:{red}\:{lines}\:{cut}\:{the} \\ $$$${small}\:{triangles}\:{in}\:{two},\:{isn}'{t}\:{it}? \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
Not really. We can′t just assume that.
$$\mathrm{Not}\:\mathrm{really}.\:\mathrm{We}\:\mathrm{can}'\mathrm{t}\:\mathrm{just}\:\mathrm{assume}\:\mathrm{that}. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
How can it be proven and solved  mathematically?
$$\mathrm{How}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{proven}\:\mathrm{and}\:\mathrm{solved} \\ $$$$\mathrm{mathematically}? \\ $$
Commented by Her_Majesty last updated on 03/Sep/20
two equilateral squares form a rhombus  the diagonals each bisect the rhombus  the red line is the longer diagonal  we don′t need more
$${two}\:{equilateral}\:{squares}\:{form}\:{a}\:{rhombus} \\ $$$${the}\:{diagonals}\:{each}\:{bisect}\:{the}\:{rhombus} \\ $$$${the}\:{red}\:{line}\:{is}\:{the}\:{longer}\:{diagonal} \\ $$$${we}\:{don}'{t}\:{need}\:{more} \\ $$

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