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Question Number 52731 by ajfour last updated on 12/Jan/19

Commented by MJS last updated on 12/Jan/19

$$\mathrm{for}b=a\left(\mathrm{circle}\right)\mathrm{the}\mathrm{equilateral}\mathrm{triangle}\mathrm{is}$$$$\mathrm{the}\mathrm{one}\mathrm{with}\mathrm{minimum}\mathrm{perimeter}.$$$$\mathrm{we}\mathrm{can}\mathrm{draw}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{for}\mathrm{a}$$$$\mathrm{circle}\mathrm{with}r=a\mathrm{and}\mathrm{then}\mathrm{compress}\mathrm{with}$$$$\mathrm{the}\mathrm{factor}\frac{a}{b}[P=\left(\begin{array}{c}x\\ y\end{array}\right)\to P^{\prime }=\left(\begin{array}{c}x\\ \frac{b}{a}y\end{array}\right)]$$$$$$$$A^{\prime }=\left(\begin{array}{c}a\cos \alpha \\ b\sin \alpha \end{array}\right)B^{\prime }=\left(\begin{array}{c}a\cos (\alpha +120°)\\ b\sin (\alpha +120°)\end{array}\right)C^{\prime }=\left(\begin{array}{c}a\cos (\alpha -120°)\\ b\sin (\alpha -120°)\end{array}\right)$$$$\mathrm{now}\mathrm{calculate}\mathrm{the}\mathrm{perimeter}p\left(\alpha \right)\mathrm{and}\mathrm{find}$$$$\mathrm{its}\min /\max p^{\prime }\left(\alpha \right)=0$$$$\mathrm{sorry}\mathrm{I}\mathrm{have}\mathrm{no}\mathrm{time}\mathrm{at}\mathrm{the}\mathrm{moment}$$

Commented by ajfour last updated on 12/Jan/19

$$The△circumscribingtheellipse$$$$shallbeofminimumperimeter$$$$inwhichofthetwoshown$$$$orientations?(ifa>b)$$$$orcanitbesoinsomeother$$$$orientation\left(maybeequilateral\right)?$$$$Finditssidesintermsofaandb.$$

Commented by mr W last updated on 12/Jan/19

$$ithinkthereisnootherorientation$$$$thanthetwoshown.$$

Commented by ajfour last updated on 12/Jan/19

$$pleasegoahead;sidesofthe△Sir?$$

Commented by ajfour last updated on 12/Jan/19

$$wontthetrianglesidesbecome$$$$curvilinearbysuchcompression,$$$$Sir?$$

Commented by MJS last updated on 12/Jan/19

$$\mathrm{no}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{linear}$$$$y=kx+d\iff P=\left(\begin{array}{c}x\\ kx+d\end{array}\right)\to P^{\prime }=\left(\begin{array}{c}x\\ \frac{b}{a}(kx+d)\end{array}\right)\iff y=\frac{bk}{a}x+\frac{bd}{a}$$$$\mathrm{straight}\mathrm{lines}\mathrm{stay}\mathrm{straight}\mathrm{lines}$$

Commented by ajfour last updated on 12/Jan/19

$$ThanksMjSSir,i^{\prime }llhavetopractice$$$$thisconcept.Thenican$$$$appreciateyoursolutionmuchmore.$$

Commented by MJS last updated on 12/Jan/19

$$\mathrm{anyway}\mathrm{this}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{lead}\mathrm{to}\mathrm{the}\mathrm{smallest}$$$$\mathrm{possible}\mathrm{perimeter}...\mathrm{ellipses}\mathrm{are}\mathrm{strange}$$$$\mathrm{figures}$$

Commented by ajfour last updated on 12/Jan/19

$$doyoumeansomeotherorientation$$$$thenshallgiveminimumperimeter,$$$$otherthanthetwoshowninquestion?$$

Commented by MJS last updated on 12/Jan/19

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{sure}.\mathrm{but}{\mathrm{MrW}}^{\prime }\mathrm{s}\mathrm{method}\mathrm{leads}\mathrm{to}$$$$\mathrm{smaller}\mathrm{perimeter}\mathrm{than}\mathrm{mine}.\mathrm{so}\mathrm{mine}\mathrm{is}\mathrm{not}$$$$\mathrm{useful}\mathrm{in}\mathrm{this}\mathrm{case}$$

Commented by ajfour last updated on 26/Jan/19

$$thismightberequiredagaininthis$$$$trapeziumproblem..thoughtof$$$$reviewingit!$$

Answered by mr W last updated on 12/Jan/19

$$let\mu =\frac{a}{b}$$$$let\mathrm{\angle }RPQ=\theta ,\frac{1}{t}=\tan \theta$$$$Q(0,h)$$$$eqn.ofPQ:$$$$\frac{x}{t}+y=h$$$$duetotangency:$$$$\frac{a^2}{t^2}+b^2=h^2$$$$\Rightarrow h=\frac{b}{t}\sqrt{{\mu }^2+t^2}$$$$RP=2(b+h)\frac{1}{\tan \theta }=2(b+h)t$$$$QP=(b+h)\frac{1}{\sin \theta }=(b+h)\sqrt{1+t^2}$$$$perimeterU=RP+2QP$$$$U=2(b+h)(t+\sqrt{1+t^2})$$$$U=2b[1+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}](t+\sqrt{1+t^2})$$$$\frac{U}{2b}=f\left(t\right)=[1+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}](t+\sqrt{1+t^2})$$$$f^{\prime }\left(t\right)=\frac{2\left(\frac{\mu }{t}\right)(-\frac{\mu }{t^2})}{2\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}}(t+\sqrt{1+t^2})+[1+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}](1+\frac{2t}{2\sqrt{1+t^2}})=0$$$$\frac{{\mu }^2}{t^3\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}}=[1+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}]\left(\frac{1}{\sqrt{1+t^2}}\right)$$$${\mu }^2\sqrt{1+t^2}=t^3[1+\frac{{\mu }^2}{t^2}+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}]$$$$\Rightarrow {\mu }^2(\sqrt{1+t^2}-t)=t^2(t+\sqrt{{\mu }^2+t^2})$$$$\Rightarrow solvefort=t\left(\mu \right)...$$$$$$$$example\mu =\frac{a}{b}=2:$$$$\Rightarrow t=0.8045\Rightarrow f\left(t\right)=7.6828\Rightarrow U=15.3656b$$$$$$$$withorientation2:\mu =\frac{1}{2}$$$$\Rightarrow t=0.4022\Rightarrow f\left(t\right)=3.8414\Rightarrow U=7.6828a=15.3656b$$$$$$$$i.e.bothorientationsgivethesame$$$$miminumperimeter.$$$$RP=2b[1+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}]t=5.92b$$$$PQ=RQ=b[1+\sqrt{1+{\left(\frac{\mu }{t}\right)}^2}]\sqrt{1+t^2}=4.72b$$

Commented by ajfour last updated on 12/Jan/19

$$\mathcal{V}ery\mathcal{N}ice!UnderstoodSir,ihadto$$$$goalongwritingit.$$

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