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

Commented by MJS last updated on 12/Jan/19

for b=a (circle) the equilateral triangle is the one with minimum perimeter. we can draw an equilateral triangle for a circle with r=a and then compress with the factor (a/b) [P= ((x),(y) ) → P′= ((x),(((b/a)y)) )] A′= (((acos α)),((bsin α)) ) B′= (((acos (α+120°))),((bsin (α+120°))) ) C′= (((acos (α−120°))),((bsin (α−120°))) ) now calculate the perimeter p(α) and find its min/max p′(α)=0 sorry I have no time at the moment

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

Commented by ajfour last updated on 12/Jan/19

The △ circumscribing the ellipse shall be of minimum perimeter in which of the two shown orientations?(if a>b) or can it be so in some other orientation(may be equilateral)? Find its sides in terms of a and b.

$${The}\:\bigtriangleup\:{circumscribing}\:{the}\:{ellipse} \\ $$$${shall}\:{be}\:{of}\:{minimum}\:{perimeter} \\ $$$${in}\:{which}\:{of}\:{the}\:{two}\:{shown} \\ $$$${orientations}?\left({if}\:{a}>{b}\right) \\ $$$${or}\:{can}\:{it}\:{be}\:{so}\:{in}\:{some}\:{other} \\ $$$${orientation}\left({may}\:{be}\:{equilateral}\right)? \\ $$$${Find}\:{its}\:{sides}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$

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

i think there is no other orientation than the two shown.

$${i}\:{think}\:{there}\:{is}\:{no}\:{other}\:{orientation} \\ $$$${than}\:{the}\:{two}\:{shown}. \\ $$

Commented by ajfour last updated on 12/Jan/19

please go ahead; sides of the △ Sir?

$${please}\:{go}\:{ahead};\:{sides}\:{of}\:{the}\:\bigtriangleup\:{Sir}? \\ $$

Commented by ajfour last updated on 12/Jan/19

wont the triangle sides become curvilinear by such compression, Sir ?

$${wont}\:{the}\:{triangle}\:{sides}\:{become} \\ $$$${curvilinear}\:{by}\:{such}\:{compression}, \\ $$$${Sir}\:? \\ $$

Commented by MJS last updated on 12/Jan/19

no. it′s linear y=kx+d ⇔ P= ((x),((kx+d)) ) → P′= ((x),(((b/a)(kx+d))) ) ⇔ y=((bk)/a)x+((bd)/a) straight lines stay straight lines

$$\mathrm{no}.\:\mathrm{it}'\mathrm{s}\:\mathrm{linear} \\ $$$${y}={kx}+{d}\:\Leftrightarrow\:{P}=\begin{pmatrix}{{x}}\\{{kx}+{d}}\end{pmatrix}\:\rightarrow\:{P}'=\begin{pmatrix}{{x}}\\{\frac{{b}}{{a}}\left({kx}+{d}\right)}\end{pmatrix}\:\Leftrightarrow\:{y}=\frac{{bk}}{{a}}{x}+\frac{{bd}}{{a}} \\ $$$$\mathrm{straight}\:\mathrm{lines}\:\mathrm{stay}\:\mathrm{straight}\:\mathrm{lines} \\ $$

Commented by ajfour last updated on 12/Jan/19

Thanks MjS Sir, i′ll have to practice this concept. Then i can appreciate your solution much more.

$${Thanks}\:{MjS}\:{Sir},\:{i}'{ll}\:{have}\:{to}\:{practice} \\ $$$${this}\:{concept}.\:\:{Then}\:{i}\:{can}\: \\ $$$${appreciate}\:{your}\:{solution}\:{much}\:{more}. \\ $$

Commented by MJS last updated on 12/Jan/19

anyway this doesn′t lead to the smallest possible perimeter... ellipses are strange figures

$$\mathrm{anyway}\:\mathrm{this}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{the}\:\mathrm{smallest} \\ $$$$\mathrm{possible}\:\mathrm{perimeter}...\:\mathrm{ellipses}\:\mathrm{are}\:\mathrm{strange} \\ $$$$\mathrm{figures} \\ $$

Commented by ajfour last updated on 12/Jan/19

do you mean some other orientation then shall give minimum perimeter, other than the two shown in question?

$${do}\:{you}\:{mean}\:{some}\:{other}\:{orientation} \\ $$$${then}\:{shall}\:{give}\:{minimum}\:{perimeter}, \\ $$$${other}\:{than}\:{the}\:{two}\:{shown}\:{in}\:{question}? \\ $$

Commented by MJS last updated on 12/Jan/19

I′m not sure. but MrW′s method leads to smaller perimeter than mine. so mine is not useful in this case

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

Commented by ajfour last updated on 26/Jan/19

this might be required again in this trapezium problem..thought of reviewing it!

$${this}\:{might}\:{be}\:{required}\:{again}\:{in}\:{this} \\ $$$${trapezium}\:{problem}..{thought}\:{of} \\ $$$${reviewing}\:{it}! \\ $$

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

let μ=(a/b) let ∠RPQ=θ, (1/t)=tan θ Q(0,h) eqn. of PQ: (x/t)+y=h due to tangency: (a^2 /t^2 )+b^2 =h^2 ⇒h=(b/t)(√(μ^2 +t^2 )) RP=2(b+h)(1/(tan θ))=2(b+h)t QP=(b+h)(1/(sin θ))=(b+h)(√(1+t^2 )) perimeter U=RP+2QP U=2(b+h)(t+(√(1+t^2 ))) U=2b[1+(√(1+((μ/t))^2 ))](t+(√(1+t^2 ))) (U/(2b))=f(t)=[1+(√(1+((μ/t))^2 ))](t+(√(1+t^2 ))) f′(t)=((2((μ/t))(−(μ/t^2 )))/(2(√(1+((μ/t))^2 ))))(t+(√(1+t^2 )))+[1+(√(1+((μ/t))^2 ))](1+((2t)/(2(√(1+t^2 )))))=0 (μ^2 /(t^3 (√(1+((μ/t))^2 ))))=[1+(√(1+((μ/t))^2 ))]((1/(√(1+t^2 )))) μ^2 (√(1+t^2 ))=t^3 [1+(μ^2 /t^2 )+(√(1+((μ/t))^2 ))] ⇒μ^2 ((√(1+t^2 ))−t)=t^2 (t+(√(μ^2 +t^2 ))) ⇒solve for t=t(μ)... example μ=(a/b)=2: ⇒t=0.8045⇒f(t)=7.6828⇒U=15.3656b with orientation 2: μ=(1/2) ⇒t=0.4022⇒f(t)=3.8414⇒U=7.6828a=15.3656b i.e. both orientations give the same miminum perimeter. RP=2b[1+(√(1+((μ/t))^2 ))]t=5.92b PQ=RQ=b[1+(√(1+((μ/t))^2 ))](√(1+t^2 ))=4.72b

$${let}\:\mu=\frac{{a}}{{b}} \\ $$$${let}\:\angle{RPQ}=\theta,\:\frac{\mathrm{1}}{{t}}=\mathrm{tan}\:\theta \\ $$$${Q}\left(\mathrm{0},{h}\right) \\ $$$${eqn}.\:{of}\:{PQ}: \\ $$$$\frac{{x}}{{t}}+{y}={h} \\ $$$${due}\:{to}\:{tangency}: \\ $$$$\frac{{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }+{b}^{\mathrm{2}} ={h}^{\mathrm{2}} \\ $$$$\Rightarrow{h}=\frac{{b}}{{t}}\sqrt{\mu^{\mathrm{2}} +{t}^{\mathrm{2}} } \\ $$$${RP}=\mathrm{2}\left({b}+{h}\right)\frac{\mathrm{1}}{\mathrm{tan}\:\theta}=\mathrm{2}\left({b}+{h}\right){t} \\ $$$${QP}=\left({b}+{h}\right)\frac{\mathrm{1}}{\mathrm{sin}\:\theta}=\left({b}+{h}\right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${perimeter}\:{U}={RP}+\mathrm{2}{QP} \\ $$$${U}=\mathrm{2}\left({b}+{h}\right)\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$${U}=\mathrm{2}{b}\left[\mathrm{1}+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right]\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$$\frac{{U}}{\mathrm{2}{b}}={f}\left({t}\right)=\left[\mathrm{1}+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right]\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$${f}'\left({t}\right)=\frac{\mathrm{2}\left(\frac{\mu}{{t}}\right)\left(−\frac{\mu}{{t}^{\mathrm{2}} }\right)}{\mathrm{2}\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)+\left[\mathrm{1}+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right]\left(\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{2}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\right)=\mathrm{0} \\ $$$$\frac{\mu^{\mathrm{2}} }{{t}^{\mathrm{3}} \sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }}=\left[\mathrm{1}+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right]\left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\right) \\ $$$$\mu^{\mathrm{2}} \sqrt{\mathrm{1}+{t}^{\mathrm{2}} }={t}^{\mathrm{3}} \left[\mathrm{1}+\frac{\mu^{\mathrm{2}} }{{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right] \\ $$$$\Rightarrow\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−{t}\right)={t}^{\mathrm{2}} \left({t}+\sqrt{\mu^{\mathrm{2}} +{t}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{solve}\:{for}\:{t}={t}\left(\mu\right)... \\ $$$$ \\ $$$${example}\:\mu=\frac{{a}}{{b}}=\mathrm{2}: \\ $$$$\Rightarrow{t}=\mathrm{0}.\mathrm{8045}\Rightarrow{f}\left({t}\right)=\mathrm{7}.\mathrm{6828}\Rightarrow{U}=\mathrm{15}.\mathrm{3656}{b} \\ $$$$ \\ $$$${with}\:{orientation}\:\mathrm{2}:\:\mu=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{0}.\mathrm{4022}\Rightarrow{f}\left({t}\right)=\mathrm{3}.\mathrm{8414}\Rightarrow{U}=\mathrm{7}.\mathrm{6828}{a}=\mathrm{15}.\mathrm{3656}{b} \\ $$$$ \\ $$$${i}.{e}.\:{both}\:{orientations}\:{give}\:{the}\:{same} \\ $$$${miminum}\:{perimeter}. \\ $$$${RP}=\mathrm{2}{b}\left[\mathrm{1}+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right]{t}=\mathrm{5}.\mathrm{92}{b} \\ $$$${PQ}={RQ}={b}\left[\mathrm{1}+\sqrt{\mathrm{1}+\left(\frac{\mu}{{t}}\right)^{\mathrm{2}} }\right]\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{4}.\mathrm{72}{b} \\ $$

Commented by ajfour last updated on 12/Jan/19

Very Nice! Understood Sir, i had to go along writing it.

$$\mathcal{V}{ery}\:\mathcal{N}{ice}!\:{Understood}\:{Sir},\:{i}\:{had}\:{to} \\ $$$${go}\:{along}\:{writing}\:{it}. \\ $$

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