Question Number 201146 by ajfour last updated on 30/Nov/23
Commented by ajfour last updated on 30/Nov/23
$${How}\:{far}\:{is}\:{J}\:{from}\:{center}\:{of}\:{circle}? \\ $$
Commented by mr W last updated on 01/Dec/23
Commented by ajfour last updated on 02/Dec/23
$${Thanks}\:{for}\:{the}\:{coining},\:{wont}\:{you} \\ $$$${attempt}\:{the}\:{question}\:{sir}? \\ $$
Commented by mr W last updated on 02/Dec/23
$${to}\:{get}\:{a}\:{general}\:{formula}\:{is}\:{very} \\ $$$${hard},\:\:{i}\:{think}. \\ $$
Answered by mr W last updated on 04/Dec/23
Commented by mr W last updated on 04/Dec/23
$${r}={radius}\:{of}\:{incircle} \\ $$$${r}=\frac{\mathrm{2}\Delta}{{a}+{b}+{c}} \\ $$$${v}+{w}={a} \\ $$$${w}+{u}={b} \\ $$$${u}+{v}={c} \\ $$$$\Rightarrow{u}+{v}+{w}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$$$\Rightarrow{u}=\frac{−{a}+{b}+{c}}{\mathrm{2}},\:{v}=\frac{{a}−{b}+{c}}{\mathrm{2}},\:{w}=\frac{{a}+{b}−{c}}{\mathrm{2}} \\ $$$${v}\:\mathrm{sin}\:{B}=\left({c}−{v}\:\mathrm{cos}\:{B}\right)\:\mathrm{tan}\:\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{{v}\:\mathrm{sin}\:{B}}{{c}−{v}\:\mathrm{cos}\:{B}} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{{w}\:\mathrm{sin}\:{C}}{{a}−{w}\:\mathrm{cos}\:{C}} \\ $$$$\phi=\pi−{B}−\alpha \\ $$$${DJ}\:\mathrm{sin}\:\phi=\left({v}−{DJ}\:\mathrm{cos}\:\phi\right)\:\mathrm{tan}\:\beta \\ $$$$\Rightarrow{DJ}=\frac{{v}\:\mathrm{tan}\:\beta}{\mathrm{sin}\:\phi+\mathrm{cos}\:\phi\:\mathrm{tan}\:\beta} \\ $$$$\Rightarrow{DJ}=\frac{{v}\:\mathrm{tan}\:\beta}{\mathrm{sin}\:\left(\alpha+{B}\right)−\mathrm{cos}\:\left(\alpha+{B}\right)\:\mathrm{tan}\:\beta} \\ $$$${JI}^{\mathrm{2}} ={DJ}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}\:{DJ}\:\mathrm{sin}\:\phi \\ $$$${JI}=\sqrt{{DJ}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}\:{DJ}\:\mathrm{sin}\:\left(\alpha+{B}\right)} \\ $$
Commented by ajfour last updated on 04/Dec/23
$${Thanks}\:{sir},\:{better},\:{but}\:{can}\:{be} \\ $$$${improved}\:{still},\:{i}\:{shall}\:{try}. \\ $$