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Question Number 76346 by Maclaurin Stickker last updated on 26/Dec/19

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$1.Findxasafunctionofbandc.$$$$2.Underwhatconditionsdoesthe$$$$problemadmitasolution?$$

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$Ifindx=\sqrt{\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}}$$$$I{didn}^{\prime }tunderstandthesecond$$$$question,couldanyoneexplain?$$

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$Theanswerofthebookis:$$$$\frac{2}{\sqrt{5}+1}⩽\frac{b}{c}⩽\frac{\sqrt{5}+1}{2}.HowcanIgetthis$$$$result?$$

Commented by MJS last updated on 26/Dec/19

$$3b^2{c}^2-b^4-c^4⩾0$$$$b^2+c^2\mp 2\sqrt{3b^2{c}^2-b^4-c^4}⩾0$$$$\mathrm{you}\mathrm{have}\mathrm{to}\mathrm{solve}\mathrm{these}$$

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$Thankyou,sir.$$

Answered by mr W last updated on 26/Dec/19

Commented by mr W last updated on 26/Dec/19

$$b\sin \theta =x$$$$b\cos \theta =\sqrt{c^2-x^2}+x$$$$\Rightarrow b^2=x^2+{(\sqrt{c^2-x^2}+x)}^2$$$$\Rightarrow b^2=x^2+c^2+2x\sqrt{c^2-x^2}$$$$\Rightarrow b^2-c^2-x^2=2x\sqrt{c^2-x^2}$$$$\Rightarrow {(b^2-c^2)}^2-2(b^2-c^2)x^2+x^4=4x^2(c^2-x^2)$$$$\Rightarrow 5x^4-2(b^2+c^2)x^2+{(b^2-c^2)}^2=0$$$$\Rightarrow x^2=\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}$$$$\Rightarrow x=\sqrt{\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}}$$$$$$$$b\cos \theta =\sqrt{c^2-b^2{\sin }^2\theta }+b\sin \theta$$$$b(\cos \theta -\sin \theta )=\sqrt{c^2-b^2{\sin }^2\theta }$$$$b^2(1-\sin 2\theta )=c^2-b^2{\sin }^2\theta$$$$\Rightarrow {\left(\frac{b}{c}\right)}^2=\frac{1}{1+{\sin }^2\theta -\sin 2\theta }$$$$\Rightarrow {\left(\frac{b}{c}\right)}^2=\frac{2}{3-(\cos 2\theta +2\sin 2\theta )}$$$$\Rightarrow {\left(\frac{b}{c}\right)}^2=\frac{2}{3-\sqrt{5}\cos (2\theta -\alpha )}$$$$with\alpha ={\tan }^{-1}2$$$$\Rightarrow {\left(\frac{b}{c}\right)}_{max}^2=\frac{2}{3-\sqrt{5}}={\left(\frac{\sqrt{5}+1}{2}\right)}^2$$$$\Rightarrow {\left(\frac{b}{c}\right)}_{max}=\frac{\sqrt{5}+1}{2}$$$$at2\theta -\alpha =0$$$$\Rightarrow \theta =\frac{\alpha }{2}=\frac{{\tan }^{-1}2}{2}=31.7°$$$$\Rightarrow {\left(\frac{b}{c}\right)}_{min}^2=\frac{2}{3+\sqrt{5}}={\left(\frac{2}{\sqrt{5}+1}\right)}^2$$$$\Rightarrow {\left(\frac{b}{c}\right)}_{min}=\frac{2}{\sqrt{5}+1}$$$$at2\theta -\alpha =180°$$$$\Rightarrow \theta =\frac{\alpha +180}{2}=\frac{{\tan }^{-1}2+180}{2}=121.7°$$$$\Rightarrow \frac{2}{\sqrt{5}+1}⩽\frac{b}{c}⩽\frac{\sqrt{5}+1}{2}$$

Commented by mr W last updated on 26/Dec/19

$$igotthisresulttoo,sir.$$

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$thanksforcorrecting$$

Commented by mr W last updated on 26/Dec/19

$$ipushedthe⋗buttontooearly\left(unintended\right)$$$$beforeicheckedmyworking,and$$$$youcommentedtoofast:)$$

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$Igot\mathit{x}byusingalgebra:$$$$h_x=\frac{2\sqrt{p(p-x)(p-b)(p-c)}}{x}$$$$p=\frac{b+c+x}{2}$$$$\frac{x^2}{2}=\sqrt{\left(\frac{b+c+x}{2}\right)\left(\frac{b+c-x}{2}\right)\left(\frac{-b+c+x}{2}\right)\left(\frac{b-c+x}{2}\right)}$$$$4x^4=(b^2+2bc+c^2-x^2)(x^2-c^2+2bc-b^2)$$$$4x^4=2b^2{c}^2-b^4-c^4-x^4+2b^2{x}^2+2c^2{x}^2$$$$5x^4=-{(b^2-c^2)}^2+x^2(2b^2+2c^2)$$$$5x^4-x^2(2b^2+2c^2)+{(b^2-c^2)}^2=0$$$$lety=x^2$$$$5y^2-y(2b^2+2c^2)+{(b^2-c^2)}^2=0$$$$y=\frac{2b^2+c^2\pm 4\sqrt{3b^2{c}^2-b^4-c^4}}{2.5}$$$$y=\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}$$$$x^2=\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}$$$$x=\sqrt{\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}}$$

Commented by Maclaurin Stickker last updated on 26/Dec/19

$$Theresultinthebookis$$$$x=\sqrt{\frac{b^2+c^2\pm 2\sqrt{3b^2{c}^2-b^4-c^4}}{5}}.$$$$$$

Commented by mr W last updated on 26/Dec/19

$$verynicesir!$$$$thispathistheonewhichialso$$$$thoughtofatfirst.butididdecide$$$$touseanotherpathbyusinga$$$$parameter(e.g.angle\theta )suchthat$$$$theparttwoofthequestionalsogets$$$$answered.$$

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