Question Number 102769 by ajfour last updated on 11/Jul/20
Commented by bramlex last updated on 11/Jul/20
Commented by bramlex last updated on 11/Jul/20
$$\left(\mathrm{1}\right)\mathrm{BG}\:=\:\sqrt{\frac{\mathrm{s}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{R}^{\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right)\mathrm{GD}\:= \\ $$$$\sqrt{\left(\mathrm{r}+\mathrm{R}\right)^{\mathrm{2}} −\mathrm{R}^{\mathrm{2}} }=\:\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{2rR}} \\ $$$$\left(\mathrm{3}\right)\mathrm{BD}\:=\:\sqrt{\left(\mathrm{r}+\mathrm{R}\right)^{\mathrm{2}} +\frac{\mathrm{s}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\mathrm{BD}\:=\:\mathrm{BG}\:+\:\mathrm{GD} \\ $$$$\sqrt{\left(\mathrm{r}+\mathrm{R}\right)^{\mathrm{2}} +\frac{\mathrm{s}^{\mathrm{2}} }{\mathrm{4}}}\:= \\ $$$$\sqrt{\frac{\mathrm{s}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{R}^{\mathrm{2}} }+\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{2rR}}\: \\ $$
Commented by ajfour last updated on 11/Jul/20
$${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral}\:{witb} \\ $$$${side}\:\boldsymbol{{s}},\:{find}\:{radii}\:{of}\:{both}\:{circles} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{s}}. \\ $$
Commented by bramlex last updated on 11/Jul/20
Commented by bramlex last updated on 11/Jul/20
$$\mathrm{AD}\:=\:\mathrm{x}+\mathrm{2r}+\mathrm{R}=\frac{\mathrm{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\:…\left(\mathrm{1}\right) \\ $$$$\mathrm{sin}\:\mathrm{30}^{\mathrm{o}} \:=\:\frac{\mathrm{r}}{\mathrm{r}+\mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}=\mathrm{r}…\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\&\left(\mathrm{2}\right)\Rightarrow\mathrm{3r}+\mathrm{R}=\frac{\mathrm{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\:…\left(\mathrm{3}\right) \\ $$$$\mathrm{r}+\mathrm{R}=\mathrm{R}\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{r}\:=\:\mathrm{R}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)…\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{3}\right)\&\left(\mathrm{4}\right)\:\Rightarrow\mathrm{3R}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{R}=\frac{\mathrm{s}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{R}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)=\frac{\mathrm{s}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\mathrm{R}=\frac{\mathrm{s}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)} \\ $$$$\mathrm{then}\:\mathrm{r}\:=\:\frac{\mathrm{s}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{2}\right)}\:.\overset{\bullet} {\smile}\mid\overset{\bullet} {\smile}\: \\ $$
Commented by ajfour last updated on 11/Jul/20
$${how}\:{come}\:{R}+{r}={R}\sqrt{\mathrm{2}}\:\:? \\ $$
Commented by bramlex last updated on 11/Jul/20
Commented by ajfour last updated on 11/Jul/20
$${the}\:{quadrilateral}\:{isn}'{t}\:{a}\: \\ $$$${parallelogram},\:\:{wrong}.. \\ $$
Commented by bramlex last updated on 11/Jul/20
$$\mathrm{no}. \\ $$
Answered by mr W last updated on 11/Jul/20
Commented by mr W last updated on 11/Jul/20
$${CF}=\sqrt{\frac{{s}^{\mathrm{2}} }{\mathrm{4}}−{R}^{\mathrm{2}} } \\ $$$$\frac{{OD}}{{OC}}=\frac{{OF}}{{CF}} \\ $$$$\Rightarrow{OD}=\frac{{sR}}{\:\sqrt{{s}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} }} \\ $$$${r}={OD}−{R}=\left(\frac{{s}}{\:\sqrt{{s}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} }}−\mathrm{1}\right){R} \\ $$$${OA}={R}+\mathrm{3}{r}=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}} \\ $$$${R}+\mathrm{3}\left(\frac{{sR}}{\:\sqrt{{s}^{\mathrm{2}} −\mathrm{4}{R}^{\mathrm{2}} }}−{R}\right)=\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}} \\ $$$${with}\:\lambda=\frac{{R}}{{s}} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }}=\mathrm{2}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\lambda} \\ $$$$\Rightarrow\mathrm{64}\lambda^{\mathrm{4}} +\mathrm{32}\sqrt{\mathrm{3}}\lambda^{\mathrm{3}} +\mathrm{32}\lambda^{\mathrm{2}} −\mathrm{8}\sqrt{\mathrm{3}}\lambda−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{{R}}{{s}}\approx\mathrm{0}.\mathrm{3642} \\ $$$$\Rightarrow\frac{{r}}{{s}}=\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}\lambda^{\mathrm{2}} }}−\mathrm{1}\right)\lambda\approx\mathrm{0}.\mathrm{1674} \\ $$
Commented by mr W last updated on 11/Jul/20
Commented by ajfour last updated on 11/Jul/20
$${Very}\:{Nice}\:{solution}\:{Sir};\:{Thanks}! \\ $$
Answered by 1549442205 last updated on 11/Jul/20
Commented by 1549442205 last updated on 11/Jul/20