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Question Number 105474 by Don08q last updated on 29/Jul/20

Answered by ajfour last updated on 29/Jul/20

Commented by ajfour last updated on 29/Jul/20

$$BD=\sqrt{25+36-2\times 5\times 6\cos 30°}$$$$=\sqrt{61-30\sqrt{3}}$$$$BD=2R\sin 30°=R$$$$.......$$

Answered by 1549442205PVT last updated on 29/Jul/20

Commented by 1549442205PVT last updated on 01/Aug/20

$$\mathrm{Apply}\mathrm{the}\mathrm{cosine}\mathrm{theorem}\mathrm{for}\mathrm{\Delta }\mathrm{ABG}$$$$\mathrm{we}\mathrm{get}:\mathrm{BG}=\sqrt{6^2+5^2-2.6.5\cos 30°}=$$$$\sqrt{61-30\sqrt{3}}=\mathrm{R}(\mathrm{as}\mathrm{\Delta }\mathrm{OBG}\mathrm{is}\mathrm{isosceles}$$$$\mathrm{at}\mathrm{O}\mathrm{and}\widehat{\mathrm{BOG}}=2\widehat{\mathrm{BAG}}=60°).$$$$\mathrm{we}\mathrm{have}\mathrm{also}\mathrm{AC}=\frac{\mathrm{r}}{\sin 15°},\sin 15°=\frac{\sqrt{6}-\sqrt{2}}{4},\cos 15°=\frac{\sqrt{6}+\sqrt{2}}{4}$$$$\mathrm{Denote}\mathrm{by}\mathrm{I}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{side}\mathrm{AG}$$$$\mathrm{we}\mathrm{have}\mathrm{AI}=3\Rightarrow \mathrm{OI}=\sqrt{{\mathrm{R}}^2-9}.\mathrm{Set}$$$$\widehat{\mathrm{OAI}}=\phi \Rightarrow \cos \phi =\frac{3}{\mathrm{R}},\sin \phi =\frac{\sqrt{{\mathrm{R}}^2-9}}{\mathrm{R}}$$$$\mathrm{Apply}\mathrm{the}\mathrm{cosine}\mathrm{theorem}\mathrm{for}\mathrm{\Delta }\mathrm{AOC}:$$$$\mathrm{OC}=\sqrt{{\mathrm{R}}^2+{\mathrm{AC}}^2-2\mathrm{AC}.\mathrm{Rcos}(15°+\phi )}=$$$$\sqrt{{\mathrm{R}}^2+\frac{{\mathrm{r}}^2}{{\sin }^{2}15°}-\frac{2\mathrm{Rr}}{\sin 15°}\times (\cos 15°\cos \phi -\sin 15°\sin \phi )}$$$$\mathrm{From}\mathrm{the}\mathrm{condition}\mathrm{that}\mathrm{the}\mathrm{circle}\left(\mathrm{C}\right)\mathrm{at}\mathrm{D}$$$$\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{circle}\left(\mathrm{O}\right)\mathrm{we}\mathrm{have}$$$$\mathrm{OC}+\mathrm{CD}=\mathrm{R}\iff \mathrm{OC}=\mathrm{R}-\mathrm{r}.\mathrm{Hence},$$$$\sqrt{{\mathrm{R}}^2+\frac{{\mathrm{r}}^2}{{\sin }^{2}15°}-\frac{2\mathrm{Rr}}{\sin 15°}\times (\cos 15°\cos \phi -\sin 15°\sin \phi )}$$$$\mathrm{R}-\mathrm{r}.\mathrm{Squaring}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{above}\mathrm{equality}\mathrm{we}\mathrm{get}:$$$${\mathrm{R}}^2+\frac{{\mathrm{r}}^2}{{\sin }^{2}15°}-\frac{2\mathrm{Rr}}{\sin 15°}\times (\cos 15°\cos \phi -\sin 15°\sin \phi )={\mathrm{R}}^2-2\mathrm{Rr}+{\mathrm{r}}^2$$$$\iff \frac{{4\mathrm{r}}^2}{2-\sqrt{3}}-\frac{8\mathrm{Rr}}{\sqrt{6}-\sqrt{2}}\times [\frac{3(\sqrt{6}+\sqrt{2})}{4\mathrm{R}}-\frac{(\sqrt{6}-\sqrt{2})\sqrt{{\mathrm{R}}^2-9}}{4\mathrm{R}}]={\mathrm{r}}^2-2\mathrm{Rr}$$$$4(2+\sqrt{3}){\mathrm{r}}^2-\frac{\mathrm{r}}{2}(\sqrt{6}+\sqrt{2})[3(\sqrt{6}+\sqrt{2}-(\sqrt{6}-\sqrt{2})\sqrt{{\mathrm{R}}^2-9}]$$$$\iff (7+4\sqrt{3}){\mathrm{r}}^2-3(4+2\sqrt{3})\mathrm{r}+2\mathrm{r}\sqrt{{\mathrm{R}}^2-9}+2\mathrm{Rr}=0$$$$\iff (7+4\sqrt{3})\mathrm{r}+[2\mathrm{R}-3(4+2\sqrt{3})+2\sqrt{52-30\sqrt{3}})=0$$$$\iff \mathbf{r}=\frac{2\sqrt{61-30\sqrt{3}}-3(4+2\sqrt{3})+2\sqrt{52-30\sqrt{3}}}{-(7+4\sqrt{3})}\approx 1,148$$

Commented by Don08q last updated on 29/Jul/20

$$ThankyouSir$$

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