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Question Number 71955 by Maclaurin Stickker last updated on 22/Oct/19

Commented by Maclaurin Stickker last updated on 22/Oct/19

$$Whatisthevalueof\frac{AB}{AC}?$$

Answered by mr W last updated on 22/Oct/19

$$AC=R$$$$BC=2r$$$$\frac{r}{R-r}=\sin 22.5°=\frac{(\sqrt{2}-1)\sqrt{2+\sqrt{2}}}{2}=\lambda$$$$r=\frac{\lambda }{1+\lambda }R$$$$AB=R-2r=\left(\frac{1-\lambda }{1+\lambda }\right)R$$$$\Rightarrow \frac{AB}{AC}=\frac{1-\lambda }{1+\lambda }=\frac{2-(\sqrt{2}-1)\sqrt{2+\sqrt{2}}}{2+(\sqrt{2}-1)\sqrt{2+\sqrt{2}}}\approx 0.446$$

Commented by Maclaurin Stickker last updated on 22/Oct/19

$$whatifitwas\frac{AB}{BC}?$$

Commented by mr W last updated on 22/Oct/19

$$BC=2r=\frac{2\lambda }{1+\lambda }$$$$\frac{AB}{BC}=\frac{1-\lambda }{2\lambda }=\frac{2-(\sqrt{2}-1)\sqrt{2+\sqrt{2}}}{2(\sqrt{2}-1)\sqrt{2+\sqrt{2}}}\approx 0.806$$

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