Question Number 221102 by fantastic last updated on 24/May/25
Answered by mehdee7396 last updated on 25/May/25
$${AB}=\mathrm{2}\sqrt{{ar}}\:\:\&\:\:{BC}=\mathrm{2}\sqrt{{br}}\:\:\:\&\:\:\:{AC}=\mathrm{2}\sqrt{{ab}} \\ $$$$\Rightarrow\sqrt{{ab}}=\left(\sqrt{{a}}+\sqrt{{b}}\right)\sqrt{{r}\:} \\ $$$$\Rightarrow{r}=\frac{{ab}}{{a}+{b}+\mathrm{2}\sqrt{{ab}}}\:\:{or}\:\:\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{a}}}+\frac{\mathrm{1}}{\:\sqrt{{b}}}\: \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 24/May/25
Commented by fantastic last updated on 25/May/25
Answered by alvan545 last updated on 25/May/25
$$\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\sqrt{{a}}+\sqrt{{b}}}{\:\sqrt{{ab}}} \\ $$$$\sqrt{{r}}\:=\:\frac{\sqrt{{ab}}}{\:\sqrt{{a}}+\sqrt{{b}}} \\ $$$$\:\:{r}\:=\:\frac{{ab}}{{a}+{b}+\mathrm{2}\sqrt{{ab}}} \\ $$
Answered by mr W last updated on 25/May/25
$$\left(\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{r}^{\mathrm{2}} }−\mathrm{2}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\frac{\mathrm{1}}{{r}}+\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\mathrm{2}\sqrt{\frac{\mathrm{1}}{{ab}}}=\left(\frac{\mathrm{1}}{\:\sqrt{{a}}}+\frac{\mathrm{1}}{\:\sqrt{{b}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{a}}}+\frac{\mathrm{1}}{\:\sqrt{{b}}} \\ $$