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Question-52190

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Question Number 52190 by ajfour last updated on 04/Jan/19
Commented by ajfour last updated on 04/Jan/19
Or else find just θ.
$${Or}\:{else}\:{find}\:{just}\:\theta. \\ $$
Commented by MJS last updated on 04/Jan/19
putting R=1 I get θ=(π/4)+arcsin ((√2)/4)≈65.7048° maximum perimeter = 1+(√(3+(√7)))+(1/2)(√(10+2(√7)))≈5.33130
$$\mathrm{putting}\:{R}=\mathrm{1}\:\mathrm{I}\:\mathrm{get}\:\theta=\frac{\pi}{\mathrm{4}}+\mathrm{arcsin}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\approx\mathrm{65}.\mathrm{7048}° \\ $$$$\mathrm{maximum}\:\mathrm{perimeter}\:=\:\mathrm{1}+\sqrt{\mathrm{3}+\sqrt{\mathrm{7}}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{7}}}\approx\mathrm{5}.\mathrm{33130} \\ $$
Commented by ajfour last updated on 04/Jan/19
whats the equation, sir ?
$${whats}\:{the}\:{equation},\:{sir}\:? \\ $$
Answered by mr W last updated on 04/Jan/19
as shown in Q52061, calculus method is not the easiest way to solve such questions. the best way is to follow the nature directly. the optimum in mathematics is usually that what the nature does. symmetry is e.g. nature. if points A and C both lie on the circle, we know without “calculation” where the point B lies such that the length of path A−B−C is maximum. we can say it′s the symmetry, but we also know it′s the way a light ray in the nature will follow. therefore, without calculus, the path A−B−C has its maximum length, when OB is the bisector of ∠B. let α=∠OBA=∠OBC ⇒∠B=2α, ∠C=(π/2)+α, ∠A=π−2α−((π/2)+α)=(π/2)−3α BC=2R cos α ((BC)/(sin A))=((CA)/(sin B)) ⇒((2R cos α)/(cos 3α))=(R/(sin 2α)) ⇒cos 3α=2 sin 2α cos α ⇒4 cos^2 α−3=2 sin 2α ⇒2 cos 2α−1=2 sin 2α ⇒cos 2α−sin 2α=(1/2) ⇒sin (π/4)cos 2α−cos (π/4) sin 2α=((√2)/4) ⇒sin ((π/4)−2α)=((√2)/4) ⇒(π/4)−2α=sin^(−1) ((√2)/4) θ=(π/2)−2α=(π/4)+((π/4)−2α) ⇒θ=(π/4)+sin^(−1) ((√2)/4)
$${as}\:{shown}\:{in}\:{Q}\mathrm{52061},\:{calculus}\:{method} \\ $$$${is}\:{not}\:{the}\:{easiest}\:{way}\:{to}\:{solve}\:{such} \\ $$$${questions}.\:{the}\:{best}\:{way}\:{is}\:{to}\:{follow} \\ $$$${the}\:{nature}\:{directly}.\:{the}\:{optimum}\:{in} \\ $$$${mathematics}\:{is}\:{usually}\:{that}\:{what}\:{the} \\ $$$${nature}\:{does}.\:{symmetry}\:{is}\:{e}.{g}.\:{nature}. \\ $$$$ \\ $$$${if}\:{points}\:{A}\:{and}\:{C}\:{both}\:{lie}\:{on}\:{the}\:{circle}, \\ $$$${we}\:{know}\:{without}\:“{calculation}''\:{where} \\ $$$${the}\:{point}\:{B}\:{lies}\:{such}\:{that}\:{the}\:{length} \\ $$$${of}\:{path}\:{A}−{B}−{C}\:{is}\:{maximum}. \\ $$$${we}\:{can}\:{say}\:{it}'{s}\:{the}\:{symmetry},\:{but}\:{we} \\ $$$${also}\:{know}\:{it}'{s}\:{the}\:{way}\:{a}\:{light}\:{ray}\:{in} \\ $$$${the}\:{nature}\:{will}\:{follow}. \\ $$$${therefore},\:{without}\:{calculus},\:{the}\:{path} \\ $$$${A}−{B}−{C}\:{has}\:{its}\:{maximum}\:{length}, \\ $$$${when}\:{OB}\:{is}\:{the}\:{bisector}\:{of}\:\angle{B}. \\ $$$$ \\ $$$${let}\:\alpha=\angle{OBA}=\angle{OBC} \\ $$$$\Rightarrow\angle{B}=\mathrm{2}\alpha,\:\angle{C}=\frac{\pi}{\mathrm{2}}+\alpha,\:\angle{A}=\pi−\mathrm{2}\alpha−\left(\frac{\pi}{\mathrm{2}}+\alpha\right)=\frac{\pi}{\mathrm{2}}−\mathrm{3}\alpha \\ $$$${BC}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha \\ $$$$\frac{{BC}}{\mathrm{sin}\:{A}}=\frac{{CA}}{\mathrm{sin}\:{B}} \\ $$$$\Rightarrow\frac{\mathrm{2}{R}\:\mathrm{cos}\:\alpha}{\mathrm{cos}\:\mathrm{3}\alpha}=\frac{{R}}{\mathrm{sin}\:\mathrm{2}\alpha} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{3}\alpha=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\alpha−\mathrm{3}=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{1}=\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\mathrm{sin}\:\mathrm{2}\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{2}\alpha\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{4}}−\mathrm{2}\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\theta=\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha=\frac{\pi}{\mathrm{4}}+\left(\frac{\pi}{\mathrm{4}}−\mathrm{2}\alpha\right) \\ $$$$\Rightarrow\theta=\frac{\pi}{\mathrm{4}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 04/Jan/19
this result coincides with the result by calculus method as MJS sir has shown.
$${this}\:{result}\:{coincides}\:{with}\:{the}\:{result} \\ $$$${by}\:{calculus}\:{method}\:{as}\:{MJS}\:{sir}\:{has} \\ $$$${shown}. \\ $$
Answered by MJS last updated on 04/Jan/19
R=1 A= ((1),(0) ) C= ((0),(0) ) B= (((−cos θ)),((1+sin θ)) ) b=∣AC∣=1 c=∣AB∣=(√(3+2cos θ +2sin θ)) a=∣BC∣=(√(2+2sin θ)) p(θ)=1+(√(2+2sin θ))+(√(3+2cos θ +2sin θ)) p′(θ)=0 ((cos θ (√(6+4cos θ +4sin θ))+2(cos θ −sin θ)(√(1+sin θ)))/(2(√(1+sin θ))(√(3+2cos θ +2sin θ))))=0 cos θ =c sin θ =s ... s^3 −(2c−1)s^2 −2cs−((c^2 (2c+1))/2)=0 2c(2s^2 +2s+c^2 )=2s^3 +2s^2 −c^2 c^2 =1−s^2 2c(s^2 +2s+1)=2s^3 +3s^2 −1 2c(s+1)^2 =(s+1)^2 (2s−1) 2c=2s−1 1+2cos θ −2sin θ =0 1+2(√2)cos ((π/4)+θ)=0 θ=−(π/4)+arccos (−((√2)/4)) =(π/4)+arcsin ((√2)/4)
$${R}=\mathrm{1} \\ $$$${A}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{−\mathrm{cos}\:\theta}\\{\mathrm{1}+\mathrm{sin}\:\theta}\end{pmatrix} \\ $$$${b}=\mid{AC}\mid=\mathrm{1} \\ $$$${c}=\mid{AB}\mid=\sqrt{\mathrm{3}+\mathrm{2cos}\:\theta\:+\mathrm{2sin}\:\theta} \\ $$$${a}=\mid{BC}\mid=\sqrt{\mathrm{2}+\mathrm{2sin}\:\theta} \\ $$$${p}\left(\theta\right)=\mathrm{1}+\sqrt{\mathrm{2}+\mathrm{2sin}\:\theta}+\sqrt{\mathrm{3}+\mathrm{2cos}\:\theta\:+\mathrm{2sin}\:\theta} \\ $$$${p}'\left(\theta\right)=\mathrm{0} \\ $$$$\frac{\mathrm{cos}\:\theta\:\sqrt{\mathrm{6}+\mathrm{4cos}\:\theta\:+\mathrm{4sin}\:\theta}+\mathrm{2}\left(\mathrm{cos}\:\theta\:−\mathrm{sin}\:\theta\right)\sqrt{\mathrm{1}+\mathrm{sin}\:\theta}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{sin}\:\theta}\sqrt{\mathrm{3}+\mathrm{2cos}\:\theta\:+\mathrm{2sin}\:\theta}}=\mathrm{0} \\ $$$$\mathrm{cos}\:\theta\:={c} \\ $$$$\mathrm{sin}\:\theta\:={s} \\ $$$$… \\ $$$${s}^{\mathrm{3}} −\left(\mathrm{2}{c}−\mathrm{1}\right){s}^{\mathrm{2}} −\mathrm{2}{cs}−\frac{{c}^{\mathrm{2}} \left(\mathrm{2}{c}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{2}{c}\left(\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}{s}+{c}^{\mathrm{2}} \right)=\mathrm{2}{s}^{\mathrm{3}} +\mathrm{2}{s}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\mathrm{1}−{s}^{\mathrm{2}} \\ $$$$\mathrm{2}{c}\left({s}^{\mathrm{2}} +\mathrm{2}{s}+\mathrm{1}\right)=\mathrm{2}{s}^{\mathrm{3}} +\mathrm{3}{s}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{2}{c}\left({s}+\mathrm{1}\right)^{\mathrm{2}} =\left({s}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{s}−\mathrm{1}\right) \\ $$$$\mathrm{2}{c}=\mathrm{2}{s}−\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{2cos}\:\theta\:−\mathrm{2sin}\:\theta\:=\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)=\mathrm{0} \\ $$$$\theta=−\frac{\pi}{\mathrm{4}}+\mathrm{arccos}\:\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\:=\frac{\pi}{\mathrm{4}}+\mathrm{arcsin}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$
Commented by ajfour last updated on 04/Jan/19
Great Beauty Sir!
$${Great}\:{Beauty}\:{Sir}! \\ $$

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