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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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