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Question Number 44950 by ajfour last updated on 06/Oct/18

Commented by ajfour last updated on 06/Oct/18

Commented by ajfour last updated on 06/Oct/18

$$GF=CF=x$$$$GC=BC\Rightarrow$$$$2x\cos \theta =x\sin 2\theta +1....\left(i\right)$$$$DH=1\Rightarrow DG\cos \theta =CP\cos \theta =1$$$$\Rightarrow x\sin 2\theta \cos \theta =1....\left(ii\right)$$$$from\left(i\right)\mathrm{\&}\left(ii\right)$$$$2x\cos \theta -x\sin 2\theta =x\sin 2\theta \cos \theta$$$$\Rightarrow 2\cos \theta =\sin 2\theta (1+\cos \theta )$$$$\Rightarrow \sin \theta (1+\cos \theta )=1$$$$(1-\cos {}^2\theta ){(1+\cos \theta )}^2=1$$$$let\cos \theta =t$$$$\Rightarrow (1-t^2){(1+t)}^2-1=0$$$$\Rightarrow -t^4-2t^3+2t=0$$$$\Rightarrow t(t^3+2t^2-2)=0$$$$\Rightarrow t_1=0,or{t}_2\approx 0.83929$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$fromeq.\left(ii\right)$$$$x=\frac{1}{\sin 2\theta \cos \theta }$$$$x_2=\frac{1}{2t_2^2\sqrt{1-t_2^2}}\approx 2.4013.$$

Commented by MrW3 last updated on 07/Oct/18

$$congratulationstoyourskillfor$$$$solvingcubicequationssir!$$

Commented by MJS last updated on 09/Oct/18

$$\mathrm{great}!$$$$\mathrm{just}\mathrm{a}\mathrm{calculation}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{last}\mathrm{line}.\mathrm{it}\mathrm{must}\mathrm{be}$$$$x_2\approx 1.30557$$

Answered by MJS last updated on 09/Oct/18

$$\mathrm{renaming}x\to o\mathrm{because}\mathrm{of}\mathrm{coordinate}\mathrm{method}$$$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)E=\left(\begin{array}{c}o\\ 0\end{array}\right)B=\left(\begin{array}{c}o+p\\ 0\end{array}\right)$$$$G=\left(\begin{array}{c}0\\ 1\end{array}\right)F=\left(\begin{array}{c}o\\ 1\end{array}\right)$$$$D=\left(\begin{array}{c}0\\ 1+q\end{array}\right)C=\left(\begin{array}{c}o+p\\ 1+q\end{array}\right)$$$$l_{CG}:y=\frac{q}{o+p}x+1$$$$l_{DH}:y=-\frac{o+p}{q}x+q+1$$$$H=l_{CG}\cap l_{DH}=\left(\begin{array}{c}\frac{(o+p)q^2}{{(o+p)}^2+q^2}\\ \frac{{(o+p)}^2+q^3+q^2}{{(o+p)}^2+q^2}\end{array}\right)$$$$\mid CG{\mid }^2-\mid BC{\mid }^2=0\Rightarrow q=\frac{{(o+p)}^2-1}{2}$$$$\mid DH\mid =1\Rightarrow q=\frac{o+p}{\sqrt{{(o+p)}^2-1}}$$$$\frac{{(o+p)}^2-1}{2}=\frac{o+p}{\sqrt{{(o+p)}^2-1}}$$$$o+p=t$$$$\frac{t^2-1}{2}=\frac{t}{\sqrt{t^2-1}}$$$$t^6-3t^4-t^2-1=0$$$$t=\sqrt{u}$$$$u^3-3u^2-u-1=0$$$$u=v+1$$$$v^3-4v-4=0$$$$v=\frac{1}{3}(\sqrt[3]{54+6\sqrt{33}}+\sqrt[3]{54-6\sqrt{33}})\approx 2.38298$$$$u\approx 3.38298$$$$t\approx 1.83929$$$$p\approx 1.83929-o$$$$q\approx 1.19149$$$$\mid CF\mid =o\Rightarrow p^2+q^2=o^2$$$${(t-o)}^2+q^2-o^2=0$$$$o=\frac{q^2+t^2}{2t}\approx 1.30557x=1.30557$$$$p=.533721$$

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