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Question Number 46864 by ajfour last updated on 01/Nov/18

Commented by ajfour last updated on 02/Nov/18

$$Ifasphereofradius\mathit{R}touches$$$$thewallsatA,B,C;find\mathit{a},\mathit{b},\mathit{c}$$$$intermsof\alpha ,\beta ,\gamma andR.$$

Commented by ajfour last updated on 02/Nov/18

Answered by MrW3 last updated on 02/Nov/18

Commented by MrW3 last updated on 04/Nov/18

$$M=centerofsphere$$$$MA=MB=MC=R$$$$letH=OM$$$$OA=\sqrt{{OM}^2-{MA}^2}=\sqrt{H^2-R^2}$$$$OB=\sqrt{{OM}^2-{MB}^2}=\sqrt{H^2-R^2}$$$$OC=\sqrt{{OM}^2-{MC}^2}=\sqrt{H^2-R^2}$$$$\Rightarrow OA=OB=OC=h\left(say\right)$$$$$$$$\mathrm{\Delta }OBQ\equiv \mathrm{\Delta }OAQ$$$$\Rightarrow \mathrm{\angle }OQB=\mathrm{\angle }OQA=\varphi \left(say\right)$$$$similarly$$$$\Rightarrow \mathrm{\angle }ORA=\mathrm{\angle }ORC=\psi \left(say\right)$$$$\Rightarrow \mathrm{\angle }OPC=\mathrm{\angle }OPB=\phi \left(say\right)$$$$$$$$\gamma +\phi +\varphi =\pi$$$$\alpha +\varphi +\psi =\pi$$$$\beta +\psi +\phi =\pi$$$$\Rightarrow \alpha +\beta +\gamma +2(\phi +\varphi +\psi )=3\pi$$$$\Rightarrow \phi +\varphi +\psi =\frac{3\pi }{2}-\frac{\alpha +\beta +\gamma }{2}$$$$\Rightarrow \phi +\pi -\alpha =\frac{3\pi }{2}-\frac{\alpha +\beta +\gamma }{2}$$$$\Rightarrow \phi =\frac{\pi }{2}-\frac{-\alpha +\beta +\gamma }{2}$$$$\Rightarrow \varphi +\pi -\beta =\frac{3\pi }{2}-\frac{\alpha +\beta +\gamma }{2}$$$$\Rightarrow \varphi =\frac{\pi }{2}-\frac{\alpha -\beta +\gamma }{2}$$$$\Rightarrow \psi +\pi -\gamma =\frac{3\pi }{2}-\frac{\alpha +\beta +\gamma }{2}$$$$\Rightarrow \psi =\frac{\pi }{2}-\frac{\alpha +\beta -\gamma }{2}$$$$$$$$letQR=u,PQ=v,RP=w$$$$\frac{u}{\sin \alpha }=\frac{OQ}{\sin \mathrm{\angle }ORQ}=\frac{OP}{\sin \psi }=\frac{h}{\sin \varphi \sin \psi }$$$$\Rightarrow u=\frac{\sin \alpha }{\sin \varphi \sin \psi }h=\frac{\sin \alpha }{\cos \frac{\alpha -\beta +\gamma }{2}\cos \frac{\alpha +\beta -\gamma }{2}}h=\mu h$$$$\frac{v}{\sin \gamma }=\frac{OP}{\sin \mathrm{\angle }OQP}=\frac{OP}{\sin \varphi }=\frac{h}{\sin \phi \sin \varphi }$$$$\Rightarrow v=\frac{\sin \gamma }{\sin \phi \sin \varphi }h=\frac{\sin \gamma }{\cos \frac{-\alpha +\beta +\gamma }{2}\cos \frac{\alpha -\beta +\gamma }{2}}h=\nu h$$$$\frac{w}{\sin \beta }=\frac{OR}{\sin \mathrm{\angle }OPR}=\frac{OR}{\sin \phi }=\frac{h}{\sin \psi \sin \phi }$$$$\Rightarrow w=\frac{\sin \beta }{\sin \psi \sin \phi }h=\frac{\sin \beta }{\cos \frac{\alpha +\beta -\gamma }{2}\cos \frac{-\alpha +\beta +\gamma }{2}}h=\omega h$$$$$$$$in\mathrm{\Delta }PQR:$$$$theincirclewithradiusrtouchesthe$$$$triangleatA,B,C.$$$$\frac{r}{R}=\frac{h}{\sqrt{h^2+R^2}}$$$$\Rightarrow r=\frac{R}{\sqrt{1+{\left(\frac{R}{h}\right)}^2}}$$$$lets=\frac{u+v+w}{2}=\frac{\mu +\nu +\omega }{2}h$$$$A_{\mathrm{\Delta }PQR}=sr=\sqrt{s(s-u)(s-v)(s-w)}$$$$\Rightarrow r=\sqrt{\frac{(s-u)(s-v)(s-w)}{s}}$$$$\Rightarrow r=\frac{1}{2}\sqrt{\frac{(\mu +\nu -\omega )(\mu -\nu +\omega )(-\mu +\nu +\omega )}{\mu +\nu +\omega }}h=\lambda h$$$$r=\frac{R}{\sqrt{1+{\left(\frac{R}{h}\right)}^2}}=\lambda h$$$$\Rightarrow {\lambda }^2[1+{\left(\frac{h}{R}\right)}^2]=1$$$$\Rightarrow h=R\sqrt{\frac{1}{{\lambda }^2}-1}$$$$\Rightarrow r=\lambda h=R\sqrt{1-{\lambda }^2}$$$$$$$$u^2=v^2+w^2-2vw\cos \mathrm{\angle }RPQ$$$$\Rightarrow \cos \mathrm{\angle }RPQ=\frac{v^2+w^2-u^2}{2vw}=\frac{{\nu }^2+{\omega }^2-{\mu }^2}{2\nu \omega }$$$$a^2=2r^2(1+\cos \mathrm{\angle }RPQ)=r^2(2+\frac{{\nu }^2+{\omega }^2-{\mu }^2}{\nu \omega })$$$$\Rightarrow a=r\sqrt{2+\frac{{\nu }^2+{\omega }^2-{\mu }^2}{\nu \omega }}$$$$\Rightarrow a=R\sqrt{(1-{\lambda }^2)(2+\frac{{\nu }^2+{\omega }^2-{\mu }^2}{\nu \omega })}$$$$similarly$$$$\Rightarrow b=R\sqrt{(1-{\lambda }^2)(2+\frac{{\omega }^2+{\mu }^2-{\nu }^2}{\omega \mu })}$$$$\Rightarrow c=R\sqrt{(1-{\lambda }^2)(2+\frac{{\mu }^2+{\nu }^2-{\omega }^2}{\mu \nu })}$$$$with$$$$\mu =\frac{\sin \alpha }{\cos \frac{\alpha -\beta +\gamma }{2}\cos \frac{\alpha +\beta -\gamma }{2}}$$$$\nu =\frac{\sin \gamma }{\cos \frac{-\alpha +\beta +\gamma }{2}\cos \frac{\alpha -\beta +\gamma }{2}}$$$$\omega =\frac{\sin \beta }{\cos \frac{\alpha +\beta -\gamma }{2}\cos \frac{-\alpha +\beta +\gamma }{2}}$$$$\lambda =\frac{1}{2}\sqrt{\frac{(\mu +\nu -\omega )(\mu -\nu +\omega )(-\mu +\nu +\omega )}{\mu +\nu +\omega }}$$$$$$$$\frac{H}{R}=\frac{h}{r}=\frac{1}{\lambda }$$$$\Rightarrow OM=H=\frac{R}{\lambda }$$$$====================$$$$checkwith\alpha =\beta =\gamma =\frac{\pi }{2}:$$$$\mu =\nu =\omega =\frac{\sin \frac{\pi }{2}}{\cos \frac{\pi }{4}\cos \frac{\pi }{4}}=2$$$$\lambda =\frac{1}{2}\sqrt{\frac{2\times 2\times 2}{2+2+2}}=\frac{1}{\sqrt{3}}$$$$H=\frac{R}{\lambda }=\sqrt{3}R\Rightarrow correct$$$$a=b=c=R\sqrt{(1-\frac{1}{3})(2+\frac{2^2}{2\times 2})}=\sqrt{2}R\Rightarrow correct$$

Commented by ajfour last updated on 03/Nov/18

$$Thankyoutoomuch,Sir.$$$$letmesometimetofollow..$$

Commented by MrW3 last updated on 04/Nov/18

Commented by ajfour last updated on 04/Nov/18

$$Thislooksaverygreatsolution,$$$$Sir;needtogathermyselfto$$$$comprehendit..$$

Commented by MrW3 last updated on 04/Nov/18

$$thankyouforreviewingsir!$$

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