Category: Geometry

Question Number 218525 by Spillover last updated on 11/Apr/25 Answered by mr W last updated on 11/Apr/25 Commented by mr W last updated on 11/Apr/25…

Question Number 218527 by Spillover last updated on 11/Apr/25 Answered by A5T last updated on 12/Apr/25 $$\mathrm{CD}=\mathrm{BD}\wedge \mathrm{\angle }\mathrm{CDB}=90°\Rightarrow \mathrm{BC}=\mathrm{CD}\sqrt{2}=2\mathrm{R}$$$$\Rightarrow \mathrm{CD}=\mathrm{R}\sqrt{2}$$$${\mathrm{Ptolemy}}^{\prime }\mathrm{s}\mathrm{theorem}:\mathrm{CD}\times \mathrm{AB}+\mathrm{AC}\times \mathrm{BD}=\mathrm{AD}\times \mathrm{BC}$$$$\Rightarrow (\mathrm{AB}+\mathrm{AC})=\mathrm{AD}\sqrt{2}\dots \left(\mathrm{i}\right)$$$$\mathrm{AB}^{\mathrm{2}}…

Question Number 218494 by Spillover last updated on 10/Apr/25 Answered by vnm last updated on 10/Apr/25 $$\frac{R}{\sqrt{R^2+{\left(2R\right)}^2}}=\frac{r}{\sqrt{R^2+{\left(2R\right)}^2}-(R+r)}$$$$\frac{1}{\sqrt{5}}=\frac{r}{R\sqrt{5}-R-r}=\frac{r/R}{\sqrt{5}-1-r/R}$$$$\sqrt{\mathrm{5}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{{r}/{R}}−\mathrm{1}…

Question Number 218491 by Spillover last updated on 10/Apr/25 Commented by Nicholas666 last updated on 10/Apr/25 $$R=\sqrt{\frac{116}{\pi }}$$ Answered by Nicholas666 last updated on…

Question Number 218456 by Spillover last updated on 10/Apr/25 Answered by vnm last updated on 10/Apr/25 $$\mathrm{d}\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{centers}$$$$\phi \mathrm{is}\mathcal{L}\cancel{\underbrace{}}$$ Commented by Spillover…