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Question Number 35195 by ajfour last updated on 16/May/18

Commented by ajfour last updated on 16/May/18

$$Q.35178carriedforward..$$

Answered by ajfour last updated on 18/May/18

$$letstakethecommondiameter$$$$asouryaxis.LandTonxaxis.$$$$furtherletAP=l,\mathrm{\angle }ALT=\theta$$$$\alpha =\frac{\mathrm{\angle }A}{2}$$$$A(l\cos \theta ,r+l\sin \theta )isoncircum-$$$$circle,so$$$$l^2\cos {}^2\theta +{(r+l\sin \theta -R)}^2=R^2..\left(\mathit{i}\right)$$$$MR=OP\cos (\theta +\alpha -\frac{\pi }{2})$$$$\Rightarrow l\cos \alpha -\frac{b}{2}=(R-r)\sin (\theta +\alpha )$$$$....\left(\mathit{i}\mathit{i}\right)$$$$NS=OP\cos (\frac{\pi }{2}-(\theta -\alpha ))$$$$\Rightarrow l\cos \alpha -\frac{c}{2}=(R-r)\sin (\theta -\alpha )$$$$.....\left(\mathit{i}\mathit{i}\mathit{i}\right)$$$$\Rightarrow 4(R-r)\cos \theta \sin \alpha =c-b...\left(a\right)$$$$and$$$${(R-r)}^2(1+3\sin {}^2\theta )\cos {}^2\alpha -$$$$(R-r)(b+c)\sin \theta \cos \alpha +$$$${\left(\frac{b+c}{4}\right)}^2-R^2\cos {}^2\alpha =0....\left(b\right)$$$$whereR=\frac{a}{2\sin 2\alpha }=\frac{b}{2\sin 2\beta }=\frac{c}{2\sin 2\gamma }$$$$\alpha =\frac{\mathrm{\angle }A}{2},\beta =\frac{\mathrm{\angle }B}{2},\gamma =\frac{\mathrm{\angle }C}{2}.$$$$\sin \theta ,\cos \theta needbeeliminatedamongst$$$$theequations\left(a\right),and\left(b\right)!$$

Commented by MJS last updated on 21/May/18

$$...\mathrm{this}\mathrm{is}\mathrm{possible}\mathrm{I}\mathrm{think}...$$

Commented by ajfour last updated on 21/May/18

$$thanks.PleasesolveQ.35646$$

Commented by MJS last updated on 22/May/18

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{work},\mathrm{Sir}\mathrm{Aifour},{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{glad}$$$$\mathrm{to}\mathrm{add}\mathrm{my}\mathrm{part}:$$$$\left(a\right)$$$$\Rightarrow \cos \theta =\frac{c-b}{4(R-r)\sin \alpha };\sin \theta =\sqrt{1-{\cos }^2\theta }$$$$\left(b\right)$$$$\Rightarrow r^4+\mathcal{B}{r}^3+\mathcal{C}{r}^2+\mathcal{D}r+\mathcal{E}=0$$$$\mathrm{but}\mathrm{with}\mathrm{the}{1}^{\mathrm{st}}\mathrm{step}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{it}\mathrm{leads}\mathrm{to}$$$$r=s-\frac{\mathcal{B}}{4}$$$$s^4+\mathcal{P}{s}^2+\mathcal{Q}=0$$$$s_1=-\sqrt{-\frac{\mathcal{P}}{2}-\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}}$$$$s_2=-\sqrt{-\frac{\mathcal{P}}{2}+\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}}$$$$s_3=\sqrt{-\frac{\mathcal{P}}{2}-\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}}$$$$s_4=\sqrt{-\frac{\mathcal{P}}{2}+\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}}$$$$r_i=s_i-\frac{\mathcal{B}}{4}$$$$R=\frac{abc}{\delta };\sin \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{2}};\cos \frac{\alpha }{2}=\sqrt{\frac{1+\cos \alpha }{2}};$$$$\cos \alpha =\frac{-a^2+b^2+c^2}{2bc};\delta =\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$$$$\mathcal{B}=-\frac{4abc}{\delta }=-4R$$$$\mathcal{P}=\frac{bc(a^4-a^2(b^2+6bc+c^2)+{\delta }^2)}{4{\delta }^2}$$$$\mathcal{Q}=\frac{a^2{b}^3{c}^3{(b-c)}^2(a+b+c)(-a+b+c)}{4{\delta }^2}$$$$\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}=\frac{bc{(b+c)}^2}{8(a+b+c)(-a+b+c)}$$$$\sqrt{-\frac{\mathcal{P}}{2}-\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}}=\frac{(b-c)\sqrt{bc}}{2\sqrt{(a+b-c)(a-b+c)}}$$$$\sqrt{-\frac{\mathcal{P}}{2}+\frac{\sqrt{{\mathcal{P}}^2-4\mathcal{Q}}}{2}}=\frac{abc}{\delta }=R$$$$r_1=R-\frac{(b-c)\sqrt{bc}}{2\sqrt{(a+b-c)(a-b+c)}}$$$$r_2=R-R=0$$$$r_3=R+\frac{(b-c)\sqrt{bc}}{2\sqrt{(a+b-c)(a-b+c)}}$$$$r_4=R+R=2R$$$$$$$$r_1\mathrm{and}{r}_3\mathrm{are}\mathrm{the}\mathrm{same}\mathrm{if}\mathrm{we}\mathrm{interchange}$$$$b\mathrm{and}c.\mathrm{Could}\mathrm{the}\mathrm{lower}\mathrm{value}\mathrm{be}\mathrm{the}\mathrm{radius}$$$$\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{touching}\mathrm{the}\mathrm{circumcircle}\mathrm{inside}$$$$\mathrm{and}\mathrm{the}\mathrm{higher}\mathrm{value}\mathrm{be}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}$$$$\mathrm{circle}\mathrm{touching}\mathrm{it}\mathrm{from}\mathrm{outside}?$$$$r_2=0\mathrm{is}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{identical}$$$$\mathrm{with}\mathrm{the}\mathrm{point}A\mathrm{of}\mathrm{the}\mathrm{triangle}.$$$$\mathrm{But}\mathrm{what}\mathrm{about}{r}_4=2R?$$$$\mathrm{Please}\mathrm{try}\mathrm{to}\mathrm{construct}\mathrm{all}\mathrm{of}\mathrm{them}\mathrm{for}\mathrm{a}$$$$\mathrm{given}\mathrm{triangle},\mathrm{using}\mathrm{my}\mathrm{formulas}.{\mathrm{I}}^{\prime }\mathrm{m}$$$$\mathrm{afraid}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{the}\mathrm{time}\mathrm{to}\mathrm{do}\mathrm{it}\mathrm{myself},$$$$\mathrm{and}\mathrm{let}\mathrm{me}\mathrm{know}\mathrm{if}\mathrm{we}\mathrm{made}\mathrm{any}\mathrm{mistake}...$$$$$$$$...{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}\mathrm{for}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}...$$$$...{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}\mathrm{for}b=c...$$

Commented by ajfour last updated on 22/May/18

$$Thanksfortheeffort,letme$$$$comprehenditall,surelyits$$$$gonnatakeawhile!$$

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