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Question Number 166738 by ajfour last updated on 26/Feb/22

Commented by ajfour last updated on 26/Feb/22

$$Find{r}_{max}intermsofb.$$

Answered by mr W last updated on 27/Feb/22

$$P(a\cos \theta ,b\sin \theta )$$$$\frac{\sqrt{{(b\sin \theta -r)}^2-r^2}}{r}=\frac{b\sin \theta }{a\cos \theta }$$$$let\frac{b}{a}=\mu ⩽1,\frac{b}{r}=\xi$$$$\sqrt{{(\xi \sin \theta -1)}^2-1}=\frac{\mu \sin \theta }{\cos \theta }$$$$\xi =\frac{1}{\sin \theta }(1+\sqrt{\frac{{\mu }^2}{\frac{1}{{\sin }^2\theta }-1}+1})$$$$lets=\frac{1}{\sin \theta }$$$$\xi =s(1+\sqrt{\frac{{\mu }^2}{s^2-1}+1})$$$$\frac{d\xi }{ds}=1+\sqrt{\frac{{\mu }^2}{s^2-1}+1}-\frac{{\mu }^2{s}^2}{{(s^2-1)}^2\sqrt{\frac{{\mu }^2}{s^2-1}+1}}=0$$$${(s^2-1)}^2+\sqrt{\frac{{\mu }^2}{s^2-1}+1}={\mu }^2$$$$\sqrt{\frac{{\mu }^2}{s^2-1}+1}={\left(\frac{\mu }{s^2-1}\right)}^2-1$$$${\left(\frac{\mu }{s^2-1}\right)}^3-2\left(\frac{\mu }{s^2-1}\right)-\mu =0$$$$\mathrm{\Delta }=\frac{{\mu }^2}{4}-\frac{8}{27}⩽\frac{1}{4}-\frac{8}{27}=-\frac{5}{108}<0$$$$\frac{\mu }{s^2-1}=\frac{2\sqrt{6}}{3}\sin (\frac{\pi }{3}+\frac{1}{3}{\sin }^{-1}\frac{3\sqrt{6}\mu }{8})(validfor\mu ⩽\sqrt{\frac{32}{27}}=\frac{4\sqrt{6}}{9})$$$$if\mu >\frac{4\sqrt{6}}{9},$$$$\frac{\mu }{s^2-1}=\sqrt[3]{\sqrt{\frac{{\mu }^2}{4}-\frac{8}{27}}+\frac{\mu }{2}}-\sqrt[3]{\sqrt{\frac{{\mu }^2}{4}-\frac{8}{27}}-\frac{\mu }{2}}$$$$s=\frac{1}{\sin \theta }=\sqrt{1+\frac{\sqrt{6}\mu }{4\sin (\frac{\pi }{3}+\frac{1}{3}{\sin }^{-1}\frac{3\sqrt{6}\mu }{8})}}for\mu ⩽\frac{4\sqrt{6}}{9}$$$$s=\frac{1}{\sin \theta }=\sqrt{1+\frac{\mu }{\sqrt[3]{\sqrt{\frac{{\mu }^2}{4}-\frac{8}{27}}+\frac{\mu }{2}}-\sqrt[3]{\sqrt{\frac{{\mu }^2}{4}-\frac{8}{27}}-\frac{\mu }{2}}}}for\mu <\frac{4\sqrt{6}}{9}$$$${\xi }_{min}=\frac{b}{r_{max}}=s(1+\sqrt{\frac{{\mu }^2}{s^2-1}+1})$$$$$$$$examples:$$$$\mu =\frac{b}{a}=0.5\Rightarrow \frac{r_{max}}{b}\approx 0.3728$$$$\mu =\frac{b}{a}=2\Rightarrow \frac{r_{max}}{b}\approx 0.1987$$

Commented by mr W last updated on 27/Feb/22

Commented by mr W last updated on 27/Feb/22

Commented by mr W last updated on 27/Feb/22

Commented by ajfour last updated on 27/Feb/22

$$GreateffortnsolutionSir.$$$$Thanksalot.$$

Answered by ajfour last updated on 26/Feb/22

$$P(\cos \theta ,b\sin \theta )$$$$r+\frac{r}{\cos \varphi }=b\sin \theta$$$$\tan \varphi =b\tan \theta$$$$\Rightarrow r=\frac{b\sin \theta }{1+\sqrt{1+b^2\tan {}^2\theta }}$$$$\frac{b}{r}=\frac{1}{\sin \theta }+\sqrt{\frac{1}{\sin {}^2\theta }+\frac{b^2}{\cos {}^2\theta }}$$$$say\sin \theta =s$$$$\frac{b}{r}=\frac{1}{s}+\sqrt{\frac{1}{s^2}+\frac{b^2}{1-s^2}}$$$$\frac{d\left(b/r\right)}{ds}=-\frac{1}{s^2}+\frac{-\frac{1}{s^3}+\frac{b^{2}s}{{(1-s^2)}^2}}{\sqrt{\frac{1}{s^2}+\frac{1}{1-s^2}}}=0$$$$\Rightarrow \frac{1}{s^4}(\frac{1}{s^2}+\frac{1}{1-s^2})$$$$=\frac{1}{s^6}+\frac{b^4{s}^2}{{(1-s^2)}^4}-\frac{2b^2}{s^2{(1-s^2)}^2}$$$$\Rightarrow {(1-s^2)}^3+2b^2{s}^2{(1-s^2)}^2=b^4{s}^6$$$$\Rightarrow {(\frac{1}{s^2}-1)}^3+2b^2{(\frac{1}{s^2}-1)}^2-b^4=0$$$$say\frac{1}{\frac{1}{s^2}-1}=t\Rightarrow$$$$t^3-\frac{2t}{b^2}-\frac{1}{b^4}=0$$$$t={\{\frac{1}{2b^4}+\frac{1}{2b^4}\sqrt{1-\frac{32b^2}{27}}\}}^{1/3}$$$$+{\{\frac{1}{2b^4}-\frac{1}{2b^4}\sqrt{1-\frac{32b^2}{27}}\}}^{1/3}$$$$D>0if{b}^2<\frac{27}{32}$$$$$$$$$$

Answered by MJS_new last updated on 27/Feb/22

$$\mathrm{for}0⩽x⩽1\mathrm{and}\mathrm{the}\mathrm{triangle}\mathrm{has}\mathrm{the}$$$$\mathrm{sides}u,v,w$$$$u=v=\sqrt{b^2+(1-b^2)x^2}\mathrm{and}w=2x$$$$\Rightarrow \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{incircle}\mathrm{is}$$$$r=\frac{bx\sqrt{1-x^2}}{x+\sqrt{b^2+(1-b^2)x^2}}$$$$r^{\prime }=0\iff (1-b^2)x^4-2(1-b^2)x^2-b^2=x(x^2-2)\sqrt{b^2+(1-b^2)x^2}$$$$(\mathrm{obviously}\mathrm{this}\mathrm{is}\mathrm{always}\mathrm{true}\mathrm{for}x=1)$$$$\mathrm{squaring}\mathrm{both}\mathrm{sides}\mathrm{and}\mathrm{transforming}\mathrm{we}\mathrm{get}$$$$(x^2-1)(x^6-\frac{2-3b^2}{1-b^2}{x}^4-\frac{3b^2{x}^2}{1-b^2}+\frac{b^2}{1-b^2})=0$$$$\mathrm{and}\mathrm{we}\mathrm{can}\mathrm{always}\mathrm{find}\mathrm{an}\mathrm{exact}\mathrm{solution}$$$$\mathrm{for}\mathrm{any}\mathrm{given}b\mathrm{with}$$$$x=\sqrt{y}\wedge y=z+\frac{2-3b^2}{3(1-b^2)}$$$$\mathrm{and}$$$$z^3-\frac{4-3b^2}{3{(1-b^2)}^2}z-\frac{16-45b^2+27b^4}{27{(1-b^2)}^3}=0$$$$\Rightarrow$$$$0<b<1\vee 1<b<\frac{4\sqrt{6}}{9}$$$$z=-\frac{2\sqrt{4-3b^2}}{3(1-b^2)}\sin \left(\frac{1}{3}\arcsin \frac{16-45b^2+27b^4}{2{(4-3b^2)}^{3/2}}\right)$$$$b=1$$$$r^{\prime }\mathrm{reduces}\mathrm{to}{x}^3-2x+1=0\Rightarrow x=-\frac{1}{2}+\frac{\sqrt{5}}{2}$$$$\frac{4\sqrt{6}}{9}⩽b<\frac{2\sqrt{3}}{3}\vee b>\frac{2\sqrt{3}}{3}$$$$\mathrm{we}\mathrm{get}z\mathrm{with}{\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{Formula}$$$$b=\frac{2\sqrt{3}}{3}$$$$z^3+4=0\Rightarrow z=-2^{2/3}$$

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