Category: Geometry

Question Number 218606 by Spillover last updated on 12/Apr/25 Answered by A5T last updated on 13/Apr/25 Commented by A5T last updated on 13/Apr/25 $$\angle\mathrm{BAF}=\alpha\:\Rightarrow\:\angle\mathrm{ABF}=\alpha\:\mathrm{since}\:\mathrm{BF}=\mathrm{AF} \\…

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 $$\mathrm{CD}=\mathrm{BD}\:\wedge\:\angle\mathrm{CDB}=\mathrm{90}°\:\Rightarrow\:\mathrm{BC}=\mathrm{CD}\sqrt{\mathrm{2}}=\mathrm{2R} \\ $$$$\Rightarrow\mathrm{CD}=\mathrm{R}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Ptolemy}'\mathrm{s}\:\mathrm{theorem}:\:\mathrm{CD}×\mathrm{AB}+\mathrm{AC}×\mathrm{BD}=\mathrm{AD}×\mathrm{BC} \\ $$$$\Rightarrow\left(\mathrm{AB}+\mathrm{AC}\right)=\mathrm{AD}\sqrt{\mathrm{2}}…\left(\mathrm{i}\right) \\ $$$$\mathrm{AB}^{\mathrm{2}}…

Question Number 218494 by Spillover last updated on 10/Apr/25 Answered by vnm last updated on 10/Apr/25 $$\frac{{R}}{\:\sqrt{{R}^{\mathrm{2}} +\left(\mathrm{2}{R}\right)^{\mathrm{2}} }}=\frac{{r}}{\:\sqrt{{R}^{\mathrm{2}} +\left(\mathrm{2}{R}\right)^{\mathrm{2}} }−\left({R}+{r}\right)} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}=\frac{{r}}{\:{R}\sqrt{\mathrm{5}}−{R}−{r}}=\frac{{r}/{R}}{\:\sqrt{\mathrm{5}}−\mathrm{1}−{r}/{R}} \\ $$$$\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 $${R}=\sqrt{\frac{\mathrm{116}}{\pi}} \\ $$ Answered by Nicholas666 last updated on…