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Question Number 158422 by ajfour last updated on 03/Nov/21

Commented by Tawa11 last updated on 04/Nov/21

$$\mathrm{Great}\mathrm{sir}$$

Answered by ajfour last updated on 05/Nov/21

$$Letrightbottomangle$$$$=\frac{\pi }{4}+2\theta$$$$\frac{b}{c}=\frac{\frac{b}{\sin (\frac{\pi }{8}+\theta )}}{\frac{b}{\sin (\frac{\pi }{8}+\theta )}+b+c}$$$$\Rightarrow \frac{c}{b}=1+(1+\frac{c}{b})\sin \alpha$$$$\Rightarrow \frac{c}{b}=\frac{1+\sin \alpha }{1-\sin \alpha }..\left(i\right)$$$$$$$$similarly$$$$\frac{a}{c}=\frac{\frac{a}{\sin (\frac{\pi }{8}-\theta )}}{\frac{a}{\sin (\frac{\pi }{8}-\theta )+a+c}}$$$$\Rightarrow \frac{c}{a}=\frac{1+\sin \beta }{1-\sin \beta }....\left(ii\right)$$$$\alpha +\beta =\frac{\pi }{4}....\left(I\right)$$$$\tan (\frac{\pi }{4}+2\theta )=p...\left(iv\right)$$$$c=a+b...\left(v\right)$$$$unknowns:$$$$\theta ,p,c,a,b$$$$........................................$$$$work:$$$$from\frac{a}{c}+\frac{b}{c}=1$$$$\frac{1-\sin \alpha }{1+\sin \alpha }+\frac{1-\sin \beta }{1+\sin \beta }=1$$$$\Rightarrow 2-2\sin \alpha \sin \beta$$$$=1+\sin \alpha +\sin \beta +\sin \alpha \sin \beta$$$$\Rightarrow 3\sin \alpha \sin \beta =1-(\sin \alpha +\sin \beta )$$$$\Rightarrow \frac{3}{2}\{\cos (\alpha -\beta )-\frac{1}{\sqrt{2}}\}=1-2\sin \left(\frac{\pi }{8}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$$$$-------------$$$$let\alpha -\beta =2\delta$$$$\frac{3}{2}(2\cos {}^2\delta -1-\frac{1}{\sqrt{2}})$$$$=1-2\sin \frac{\pi }{8}\cos \delta$$$$=1-2\lambda \cos \delta$$$$6\cos {}^2\delta +4\lambda \cos \delta -(5+\frac{3}{\sqrt{2}})=0$$$$\cos \delta =-\frac{\lambda }{3}+\sqrt{\frac{{\lambda }^2}{9}+\frac{(10+3\sqrt{2})}{12}}$$$$\cos \delta =-\frac{\sqrt{2-\sqrt{2}}}{6}+\sqrt{\frac{2-\sqrt{2}+30+9\sqrt{2}}{36}}$$$$6\cos \delta =\sqrt{32+8\sqrt{2}}-\sqrt{2-\sqrt{2}}$$$$\Rightarrow \alpha -\beta =2\delta ={2\cos }^{-1}\frac{k}{6}$$$$andas\alpha +\beta =\frac{\pi }{4}$$$$p=\tan (\frac{\pi }{4}\pm 2\theta )=\tan (\frac{\pi }{4}\pm 2\delta )$$$$p=\tan (\frac{\pi }{4}\pm {2\cos }^{-1}\frac{k}{6})$$$$-------------$$$$\mathit{p}=\tan \{\frac{\pi }{4}\pm {2\cos }^{-1}\left(\frac{\sqrt{32+8\sqrt{2}}-\sqrt{2-\sqrt{2}}}{6}\right)\}$$$$-------------$$$$$$

Commented by mr W last updated on 05/Nov/21

$$perfectsolution!$$

Answered by mr W last updated on 04/Nov/21

$$c+\frac{c}{c-b}\sqrt{{(c+b)}^2-{(c-b)}^2}=1$$$$\Rightarrow c+\frac{2c\sqrt{bc}}{c-b}=1$$$$\Rightarrow c+\frac{2c\sqrt{bc}}{a}=1...\left(i\right)$$$$c+\frac{c}{c-a}\sqrt{{(c+a)}^2-{(c-a)}^2}=p$$$$\Rightarrow c+\frac{2c\sqrt{ac}}{c-a}=p$$$$\Rightarrow c+\frac{2c\sqrt{ac}}{b}=p...\left(ii\right)$$$$\sqrt{1^2+p^2}=\frac{2c\sqrt{bc}}{a}+\frac{2c\sqrt{ac}}{b}$$$$\sqrt{1^2+p^2}=1+p-2c...\left(iii\right)$$$$from\left(iii\right):$$$$c=\frac{1+p-\sqrt{1+p^2}}{2}$$$$from\left(i\right):$$$$\frac{\sqrt{b}}{a}=\frac{1-c}{2c\sqrt{c}}$$$$from\left(ii\right):$$$$\frac{\sqrt{a}}{b}=\frac{p-c}{2c\sqrt{c}}$$$$\frac{1}{\sqrt{ab}}=\frac{(1-c)(p-c)}{4c^3}$$$$ab=\frac{16c^6}{{(1-c)}^2{(p-c)}^2}$$$${\left(\frac{b}{a}\right)}^{\frac{3}{2}}=\frac{1-c}{p-c}$$$$\frac{b}{a}={\left(\frac{1-c}{p-c}\right)}^{\frac{2}{3}}$$$$b^2=\frac{16c^6}{{(1-c)}^2{(p-c)}^2}{\left(\frac{1-c}{p-c}\right)}^{\frac{2}{3}}$$$$b=\frac{4c^3}{(1-u)(p-u)}{\left(\frac{1-c}{p-c}\right)}^{\frac{1}{3}}$$$$a=\frac{4c^3}{(1-c)(p-c)}{\left(\frac{p-c}{1-c}\right)}^{\frac{1}{3}}$$$$a+b=\frac{4c^3}{(1-4)(p-c)}[{\left(\frac{1-c}{p-c}\right)}^{\frac{1}{3}}+{\left(\frac{p-c}{1-c}\right)}^{\frac{1}{3}}]$$$$c=\frac{4c^3}{(1-c)(p-c)}[{\left(\frac{1-c}{p-c}\right)}^{\frac{1}{3}}+{\left(\frac{p-c}{1-c}\right)}^{\frac{1}{3}}]$$$$\frac{4c^2}{(1-c)(p-c)}[{\left(\frac{1-c}{p-c}\right)}^{\frac{1}{3}}+{\left(\frac{p-c}{1-c}\right)}^{\frac{1}{3}}]=1$$$$or$$$${(1-\frac{2}{1+\frac{\sqrt{1+p^2}}{p-1}})}^{\frac{1}{3}}+{(1-\frac{2}{1-\frac{\sqrt{1+p^2}}{p-1}})}^{\frac{1}{3}}=(\frac{1}{1+p-\sqrt{1+p^2}}-\frac{1}{2})(\frac{p}{1+p-\sqrt{1+p^2}}-\frac{1}{2})$$$$\Rightarrow p\approx 0.2970or3.3665$$

Commented by mr W last updated on 04/Nov/21

$$anotherway:$$$$\alpha =\frac{a}{c},\beta =\frac{b}{c}$$$$\tan \theta =\frac{p}{1}=p=\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}$$$$\tan \frac{\theta }{2}=\frac{\sqrt{1+p^2}-1}{p}$$$$\sin \frac{\theta }{2}=\frac{c-b}{c+b}=\frac{1-\beta }{1+\beta }$$$$\beta =\frac{2}{1+\sin \frac{\theta }{2}}-1$$$$\beta =\frac{2}{1+\frac{\sqrt{1+p^2}-1}{\sqrt{2p^2+2-2\sqrt{1+p^2}}}}-1$$$$\alpha =\frac{2}{1+\sin (\frac{\pi }{4}-\frac{\theta }{2})}-1$$$$\alpha =\frac{2}{1+\frac{p+1-\sqrt{1+p^2}}{\sqrt{2}\sqrt{2p^2+2-2\sqrt{1+p^2}}}}-1$$$$\alpha +\beta =1$$$$\frac{1}{1+\frac{p+1-\sqrt{1+p^2}}{2\sqrt{p^2+1-\sqrt{1+p^2}}}}+\frac{1}{1+\frac{\sqrt{1+p^2}-1}{\sqrt{2}\sqrt{p^2+1-\sqrt{1+p^2}}}}=\frac{3}{2}$$

Commented by MJS_new last updated on 04/Nov/21

$$\mathrm{this}\mathrm{last}\mathrm{blue}\mathrm{equation}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{exactly}$$$$\mathrm{but}\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{not}‘‘{\mathrm{nice}}^{″}$$

Commented by mr W last updated on 04/Nov/21

$$wow!i{didn}^{\prime }texpectthatitcanbesolved$$$$exactly.$$

Commented by MJS_new last updated on 04/Nov/21

$$\mathrm{I}\mathrm{will}\mathrm{type}\mathrm{the}\mathrm{exact}\mathrm{solution}\mathrm{as}\mathrm{soon}\mathrm{as}\mathrm{I}$$$$\mathrm{have}\mathrm{the}\mathrm{time}...$$

Commented by MJS_new last updated on 05/Nov/21

$$\mathrm{let}t=p+\sqrt{p^2+1}\Rightarrow p=\frac{t^2-1}{2t};{\mathrm{we}}^{\prime }\mathrm{re}\mathrm{looking}$$$$\mathrm{for}p>0\Rightarrow t>1$$$$\mathrm{the}\mathrm{blue}\mathrm{equation}\mathrm{now}\mathrm{becomes}$$$$\frac{\sqrt{t^2+1}}{1+\sqrt{t^2+1}}+\frac{2\sqrt{t^2+1}}{\sqrt{2}(t-1)+2\sqrt{t^2+1}}=\frac{3}{2}$$$$\mathrm{this}\mathrm{transforms}\mathrm{to}$$$$(\sqrt{2}t+2+\sqrt{2})\sqrt{t^2+1}=2t^2-3\sqrt{2}t+2+3\sqrt{2}$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}$$$$t^4+2(1-4\sqrt{2})t^3+(9+8\sqrt{2})t^2-8(3+\sqrt{2})t+8(1+\sqrt{2})=0$$$$\mathrm{let}u=t+\frac{1}{2}-2\sqrt{2}$$$$u^4-(\frac{81}{2}-20\sqrt{2})u^2+8(17-15\sqrt{2})u-\frac{3231}{16}+127\sqrt{2}=0$$$$\mathrm{we}\mathrm{can}\mathrm{get}$$$$(u^2-\alpha u-\beta )(u^2+\alpha u-\gamma )=0$$$$\mathrm{with}$$$$\alpha =2\sqrt{6-2\sqrt{2}}$$$$\beta =\frac{33}{4}-6\sqrt{2}+(8-3\sqrt{2})\sqrt{3-\sqrt{2}}$$$$\gamma =\frac{33}{4}-6\sqrt{2}-(8-3\sqrt{2})\sqrt{3-\sqrt{2}}$$$$\mathrm{the}\mathrm{real}\mathrm{solutions}\mathrm{of}\mathrm{this}\mathrm{are}$$$$u=\frac{\alpha }{2}\pm \frac{\sqrt{{\alpha }^2+4\beta }}{2}$$$$t=u-\frac{1}{2}+2\sqrt{2}$$$$p=\frac{t^2-1}{2t}$$$$\mathrm{I}\mathrm{get}\mathrm{for}{p}_1>1[p_2=p_1^{-1}]$$$$p_1=-\frac{2-7\sqrt{2}}{8}-\frac{(2-9\sqrt{2})\sqrt{3-\sqrt{2}}}{16}+\frac{8-2\sqrt{2}-(2-\sqrt{2})\sqrt{3-\sqrt{2}}}{16}\sqrt{57-32\sqrt{2}+4(8-3\sqrt{2})\sqrt{3-\sqrt{2}}}$$$$p_1\approx 3.36651284$$$$p_2\approx .297043275$$

Commented by mr W last updated on 05/Nov/21

$$reallymagic!thanksalot!$$

Commented by ajfour last updated on 05/Nov/21

$$thankyoumrWsirnMjSsir!$$

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