# Question-77229

Question Number 77229 by Maclaurin Stickker last updated on 04/Jan/20
Commented by Maclaurin Stickker last updated on 04/Jan/20
$${ABC}\:{is}\:{a}\:{triangle}.\:{Radius}\:{circumferences} \\$$$${shown}\:{in}\:{the}\:{figure}\:{are}\:{drawn} \\$$$${internally}\:{to}\:{the}\:{triangle}\:{tanging} \\$$$${two}\:{of}\:{its}\:{sides}\:{and}\:{the}\:{incircle}. \\$$$${Determine}\:{the}\:{radius}\:{of}\:{the}\:{incircle}. \\$$$$\\$$
Answered by mr W last updated on 04/Jan/20
Commented by mr W last updated on 05/Jan/20
$${a}={y}+{z} \\$$$${b}={z}+{x} \\$$$${c}={x}+{y} \\$$$$\Rightarrow{x}+{y}+{z}=\frac{{a}+{b}+{c}}{\mathrm{2}}=\frac{{p}}{\mathrm{2}}={s} \\$$$$\Rightarrow{x}={s}−{a},\:{y}={s}−{b},\:{z}={s}−{c} \\$$$$\frac{{r}−{r}_{{b}} }{\:\sqrt{\left({r}+{r}_{{b}} \right)^{\mathrm{2}} −\left({r}−{r}_{{b}} \right)^{\mathrm{2}} }}=\frac{{r}}{{y}} \\$$$$\frac{{r}−{r}_{{b}} }{\mathrm{2}\sqrt{{rr}_{{b}} }}=\frac{{r}}{{y}} \\$$$$\Rightarrow{y}=\frac{\mathrm{2}{r}\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} } \\$$$$\Rightarrow{s}−{b}=\frac{\mathrm{2}{r}\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} } \\$$$${similarly} \\$$$$\Rightarrow{s}−{a}=\frac{\mathrm{2}{r}\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} } \\$$$$\Rightarrow{s}−{c}=\frac{\mathrm{2}{r}\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} } \\$$$$\Rightarrow\mathrm{3}{s}−\left({a}+{b}+{c}\right)=\mathrm{2}{r}\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right) \\$$$$\Rightarrow{s}=\mathrm{2}{r}\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right) \\$$$$\\$$$$\Delta=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\$$$$\Rightarrow\Delta=\sqrt{\mathrm{2}{r}\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)\left(\frac{\mathrm{2}{r}\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }\right)\left(\frac{\mathrm{2}{r}\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }\right)\left(\frac{\mathrm{2}{r}\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)} \\$$$$\Rightarrow\Delta=\mathrm{4}{r}^{\mathrm{2}} \sqrt{\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }\right)\left(\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }\right)\left(\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)} \\$$$$\Delta={sr}=\mathrm{2}{r}^{\mathrm{2}} \left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right) \\$$$$\Rightarrow\mathrm{2}{r}^{\mathrm{2}} \left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)=\mathrm{4}{r}^{\mathrm{2}} \sqrt{\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }\right)\left(\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }\right)\left(\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)} \\$$$$\Rightarrow\sqrt{\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }+\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }+\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }}=\mathrm{2}\sqrt{\left(\frac{\sqrt{{rr}_{{a}} }}{{r}−{r}_{{a}} }\right)\left(\frac{\sqrt{{rr}_{{b}} }}{{r}−{r}_{{b}} }\right)\left(\frac{\sqrt{{rr}_{{c}} }}{{r}−{r}_{{c}} }\right)} \\$$$$\Rightarrow\sqrt{\frac{\sqrt{{r}}}{{r}−\mathrm{1}}+\frac{\mathrm{2}\sqrt{{r}}}{{r}−\mathrm{4}}+\frac{\mathrm{3}\sqrt{{r}}}{{r}−\mathrm{9}}}=\mathrm{2}\sqrt{\left(\frac{\sqrt{{r}}}{{r}−\mathrm{1}}\right)\left(\frac{\mathrm{2}\sqrt{{r}}}{{r}−\mathrm{4}}\right)\left(\frac{\mathrm{3}\sqrt{{r}}}{{r}−\mathrm{9}}\right)} \\$$$${let}\:\lambda=\sqrt{{r}} \\$$$$\Rightarrow\sqrt{\frac{\mathrm{1}}{\lambda^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{2}}{\lambda^{\mathrm{2}} −\mathrm{4}}+\frac{\mathrm{3}}{\lambda^{\mathrm{2}} −\mathrm{9}}}=\mathrm{2}\lambda\sqrt{\frac{\mathrm{6}}{\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda^{\mathrm{2}} −\mathrm{4}\right)\left(\lambda^{\mathrm{2}} −\mathrm{9}\right)}} \\$$$$\Rightarrow\sqrt{\frac{\lambda^{\mathrm{4}} −\mathrm{13}\lambda^{\mathrm{2}} +\mathrm{36}}{\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda^{\mathrm{2}} −\mathrm{4}\right)\left(\lambda^{\mathrm{2}} −\mathrm{9}\right)}+\frac{\mathrm{2}\left(\lambda^{\mathrm{4}} −\mathrm{10}\lambda^{\mathrm{2}} +\mathrm{9}\right)}{\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda^{\mathrm{2}} −\mathrm{4}\right)\left(\lambda^{\mathrm{2}} −\mathrm{9}\right)}+\frac{\mathrm{3}\left(\lambda^{\mathrm{4}} −\mathrm{5}\lambda^{\mathrm{2}} +\mathrm{4}\right)}{\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda^{\mathrm{2}} −\mathrm{4}\right)\left(\lambda^{\mathrm{2}} −\mathrm{9}\right)}}=\mathrm{2}\lambda\sqrt{\frac{\mathrm{6}}{\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda^{\mathrm{2}} −\mathrm{4}\right)\left(\lambda^{\mathrm{2}} −\mathrm{9}\right)}} \\$$$$\sqrt{\mathrm{6}\lambda^{\mathrm{4}} −\mathrm{48}\lambda^{\mathrm{2}} +\mathrm{66}}=\mathrm{2}\lambda\sqrt{\mathrm{6}} \\$$$$\mathrm{6}\lambda^{\mathrm{4}} −\mathrm{48}\lambda^{\mathrm{2}} +\mathrm{66}=\mathrm{24}\lambda^{\mathrm{2}} \\$$$$\lambda^{\mathrm{4}} −\mathrm{12}\lambda^{\mathrm{2}} +\mathrm{11}=\mathrm{0} \\$$$$\left(\lambda^{\mathrm{2}} −\mathrm{1}\right)\left(\lambda^{\mathrm{2}} −\mathrm{11}\right)=\mathrm{0} \\$$$$\Rightarrow\lambda^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{r}=\mathrm{1}\:!\:{not}\:{good}! \\$$$$\Rightarrow\lambda^{\mathrm{2}} =\mathrm{11}\:\Rightarrow{r}=\mathrm{11} \\$$$${i}.{e}.\:{the}\:{radius}\:{of}\:{incircle}\:{is}\:\mathrm{11}. \\$$
Commented by Maclaurin Stickker last updated on 04/Jan/20
$${wow}!\:{perfect}\centerdot \\$$
Commented by mr W last updated on 04/Jan/20
$${it}\:{can}\:{be}\:{simplified}\:{to} \\$$$${r}=\sqrt{{r}_{{a}} {r}_{{b}} }+\sqrt{{r}_{{b}} {r}_{{c}} }+\sqrt{{r}_{{c}} {r}_{{a}} } \\$$$$=\sqrt{\mathrm{1}×\mathrm{4}}+\sqrt{\mathrm{4}×\mathrm{9}}+\sqrt{\mathrm{9}×\mathrm{1}} \\$$$$=\mathrm{2}+\mathrm{6}+\mathrm{3} \\$$$$=\mathrm{11} \\$$
Commented by otchereabdullai@gmail.com last updated on 04/Jan/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{fantastic}\:\mathrm{prof}\:\:\mathrm{w} \\$$
Commented by otchereabdullai@gmail.com last updated on 04/Jan/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{fantastic}\:\mathrm{prof}\:\:\mathrm{w} \\$$
Commented by otchereabdullai@gmail.com last updated on 04/Jan/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{fantastic}\:\mathrm{prof}\:\:\mathrm{w} \\$$
Commented by otchereabdullai@gmail.com last updated on 04/Jan/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{fantastic}\:\mathrm{prof}\:\:\mathrm{w} \\$$
Commented by jagoll last updated on 05/Jan/20
$$\mathrm{waw}\:\:\mathrm{amazing}\:\mathrm{sir}\: \\$$
Commented by peter frank last updated on 05/Jan/20
$${your}\:{genius} \\$$
Commented by mr W last updated on 06/Jan/20
$${see}\:{Q}\mathrm{77425}\:{for}\:{an}\:{other}\:{way}\:{to} \\$$$${prove}. \\$$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\$$