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Question-201146




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

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