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Question-62970

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Question Number 62970 by ajfour last updated on 27/Jun/19
Commented by ajfour last updated on 28/Jun/19
Hope its correct and unique..
$${Hope}\:{its}\:\:{correct}\:{and}\:{unique}.. \\ $$
Answered by ajfour last updated on 27/Jun/19
Commented by ajfour last updated on 27/Jun/19
perimeter of quadrilateral s. s=asec θ+(b−a)sec φ+bsec ψ +atan θ+b^2 −a^2 +btan ψ ...(i) let area of quad ABCD be △. 2△=a^2 tan θ+b^2 tan ψ +(a+b)(b−a)tan φ ...(ii) tan φ=((b^2 −a^2 )/(b−a))=a+b ...(iii) ⇒ 2△=a^2 tan θ+b^2 tan ψ +(a+b)^2 (b−a) s=a(sec θ+tan θ)+b(sec ψ+tan ψ) +(b−a)(√(1+(a+b)^2 ))+b^2 −a^2 let sec ψ+tan ψ=t ⇒1+tan^2 ψ=t^2 −2ttan ψ+tan^2 ψ ⇒ tan ψ = ((t^2 −1)/(2t)) similarly if sec θ+tan θ =r ⇒ tan θ = ((r^2 −1)/(2r)) ; Now s= ar+bt+(b−a)(√(1+(a+b)^2 )) +b^2 −a^2 4△=((a^2 (r^2 −1))/r)+((b^2 (t^2 −1))/t) +2(a+b)^2 (b−a) let ar+bt = p & ar−bt = q ⇒ 2ar=p+q , 2bt=p−q hence 8△=a(p+q)+b(p−q) +((4a^3 )/(p+q))+((4b^3 )/(p−q))+4(a+b)^2 (b−a) while s=p+(b−a)(√(1+(a+b)^2 )) +b^2 −a^2 And for maximum △, (∂△/∂a)=(∂△/∂b)=(∂△/∂q) = 0 .....
$${perimeter}\:{of}\:{quadrilateral}\:\boldsymbol{{s}}. \\ $$$$\boldsymbol{{s}}={a}\mathrm{sec}\:\theta+\left({b}−{a}\right)\mathrm{sec}\:\phi+{b}\mathrm{sec}\:\psi \\ $$$$\:\:\:\:\:\:+{a}\mathrm{tan}\:\theta+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}\mathrm{tan}\:\psi\:\:\:…\left({i}\right) \\ $$$${let}\:{area}\:{of}\:{quad}\:{ABCD}\:{be}\:\bigtriangleup. \\ $$$$\mathrm{2}\bigtriangleup={a}^{\mathrm{2}} \mathrm{tan}\:\theta+{b}^{\mathrm{2}} \mathrm{tan}\:\psi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({a}+{b}\right)\left({b}−{a}\right)\mathrm{tan}\:\phi\:\:\:\:\:…\left({ii}\right) \\ $$$$\mathrm{tan}\:\phi=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}−{a}}={a}+{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\:\mathrm{2}\bigtriangleup={a}^{\mathrm{2}} \mathrm{tan}\:\theta+{b}^{\mathrm{2}} \mathrm{tan}\:\psi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({a}+{b}\right)^{\mathrm{2}} \left({b}−{a}\right) \\ $$$${s}={a}\left(\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\right)+{b}\left(\mathrm{sec}\:\psi+\mathrm{tan}\:\psi\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\left({b}−{a}\right)\sqrt{\mathrm{1}+\left({a}+{b}\right)^{\mathrm{2}} }+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${let}\:\:\mathrm{sec}\:\psi+\mathrm{tan}\:\psi={t} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \psi={t}^{\mathrm{2}} −\mathrm{2}{t}\mathrm{tan}\:\psi+\mathrm{tan}\:^{\mathrm{2}} \psi \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\psi\:=\:\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}} \\ $$$${similarly}\:{if}\:\: \\ $$$$\:\:\:\:\mathrm{sec}\:\theta+\mathrm{tan}\:\theta\:={r} \\ $$$$\Rightarrow\:\:\:\:\mathrm{tan}\:\theta\:=\:\frac{{r}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{r}}\:\:\:\:;\:{Now} \\ $$$$\:{s}=\:{ar}+{bt}+\left({b}−{a}\right)\sqrt{\mathrm{1}+\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\mathrm{4}\bigtriangleup=\frac{{a}^{\mathrm{2}} \left({r}^{\mathrm{2}} −\mathrm{1}\right)}{{r}}+\frac{{b}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)}{{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} \left({b}−{a}\right) \\ $$$${let}\:\:\:\:\:\:\:\:\:{ar}+{bt}\:=\:{p}\:\: \\ $$$$\&\:\:\:\:\:\:\:\:\:\:\:{ar}−{bt}\:=\:{q} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{ar}={p}+{q}\:\:,\:\:\mathrm{2}{bt}={p}−{q} \\ $$$${hence} \\ $$$$\:\:\:\:\mathrm{8}\bigtriangleup={a}\left({p}+{q}\right)+{b}\left({p}−{q}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{4}{a}^{\mathrm{3}} }{{p}+{q}}+\frac{\mathrm{4}{b}^{\mathrm{3}} }{{p}−{q}}+\mathrm{4}\left({a}+{b}\right)^{\mathrm{2}} \left({b}−{a}\right) \\ $$$${while}\:\:\:\boldsymbol{{s}}={p}+\left({b}−{a}\right)\sqrt{\mathrm{1}+\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \: \\ $$$${And}\:{for}\:{maximum}\:\bigtriangleup, \\ $$$$\:\:\:\frac{\partial\bigtriangleup}{\partial{a}}=\frac{\partial\bigtriangleup}{\partial{b}}=\frac{\partial\bigtriangleup}{\partial{q}}\:=\:\mathrm{0} \\ $$$$….. \\ $$
Commented by ajfour last updated on 28/Jun/19
Please help, couldn′t solve it; could the quadrilateral be cyclic for maximum area, under these conditions, could one of the vertices of quadrilateral be at origin? can one side of quadrilateral be zero, so that it is afterall a triangle then,..?
$${Please}\:{help},\:{couldn}'{t}\:{solve}\:{it}; \\ $$$${could}\:{the}\:{quadrilateral}\:{be} \\ $$$${cyclic}\:{for}\:{maximum}\:{area}, \\ $$$${under}\:{these}\:{conditions},\: \\ $$$${could}\:{one}\:{of}\:{the}\:{vertices}\:{of} \\ $$$${quadrilateral}\:{be}\:{at}\:{origin}? \\ $$$${can}\:{one}\:{side}\:{of}\:{quadrilateral} \\ $$$${be}\:{zero},\:{so}\:{that}\:{it}\:{is}\:{afterall}\:{a} \\ $$$${triangle}\:{then},..? \\ $$
Commented by mr W last updated on 28/Jun/19
i believe that the maximum quadrilateral exists, it mustn′t be cyclic . but the solution must be very hard. i′d suggest to try the case with a triangle instead of quadrilateral.
$${i}\:{believe}\:{that}\:{the}\:{maximum}\:{quadrilateral} \\ $$$${exists},\:{it}\:{mustn}'{t}\:{be}\:{cyclic}\:.\:{but}\:{the} \\ $$$${solution}\:{must}\:{be}\:{very}\:{hard}. \\ $$$${i}'{d}\:{suggest}\:{to}\:{try}\:{the}\:{case}\:{with}\:{a}\:{triangle} \\ $$$${instead}\:{of}\:{quadrilateral}. \\ $$

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