Question Number 222436 by ajfour last updated on 26/Jun/25
Commented by ajfour last updated on 26/Jun/25
https://youtu.be/xLaPcyoBv2w?si=4zzxA2kFBwZVUHdh
Answered by mr W last updated on 27/Jun/25
$$\Delta{OAP}\sim\Delta{OPB} \\ $$$$\Rightarrow\frac{{a}}{{k}}=\frac{{r}}{{R}}\:\Rightarrow{a}=\frac{{kr}}{{R}} \\ $$$${ak}={r}\left({a}+{k}+\sqrt{{a}^{\mathrm{2}} +{k}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{k}={R}+{r}+\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} }={R}+{r}+\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$$\mathrm{15}=\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$$\Rightarrow{Rr}=\frac{\mathrm{15}}{\mathrm{2}} \\ $$$${R},\:{r}\:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} −\mathrm{8}{z}+\frac{\mathrm{15}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{R},\:{r}=\mathrm{4}\pm\frac{\sqrt{\mathrm{34}}}{\mathrm{2}} \\ $$$${k}=\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} −\mathrm{2}{Rr}} \\ $$$$\:\:\geqslant\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\left(\frac{\mathrm{8}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{8}+\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${k}=\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} −\mathrm{2}{Rr}}<\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} }=\mathrm{16} \\ $$$$\Rightarrow{k}\in\left[\mathrm{8}+\mathrm{4}\sqrt{\mathrm{2}},\:\mathrm{16}\right) \\ $$
Commented by mr W last updated on 27/Jun/25
