Question Number 52806 by mr W last updated on 13/Jan/19
Commented by mr W last updated on 13/Jan/19
$${I}\:{found}\:{this}\:{unsolved}\:{question}\:{in} \\ $$$${Q}\mathrm{35115}\:\left({from}\:{ajfour}\:{sir}\right): \\ $$$${Find}\:{r}\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$
Commented by behi83417@gmail.com last updated on 14/Jan/19
Answered by MJS last updated on 14/Jan/19
![A= ((0),(0) ) B= ((c),(0) ) C= (((bcos α)),((bsin α)) ) tan (α/2) =k circumcircle center= ((m),(n) ); radius=R=(√(m^2 +n^2 )) blue circle center= ((o),((ko)) ); radius=r=ko (1) (x−m)^2 +(y−n)^2 −(m^2 +n^2 )=0 x^2 +y^2 −2mx−2ny=0 (2) (x−o)^2 +(y−ko)^2 −(ok)^2 =0 x^2 +y^2 −2ox−2koy+o^2 =0 (1)−(2) ⇒ y=((2(m−o)x+o^2 )/(2(ko−n))) insert in (1) or (2) ⇒ x^2 +ux+v=0 we need exactly one solution ⇒ ⇒ D=u^2 −4v=0 o^4 −((2(2k^2 n+2km+n))/k)o^3 −((4k^4 m^2 −8k^3 mn−4k^2 (m^2 +2n^2 )−8kmn−n^2 )/k^2 )o^2 +((4(2k^3 m^2 −4k^2 mn−k(2m^2 +n^2 )−mn)n)/k^2 )o−((4(k^2 m−2kn−m)mn^2 )/k^2 )=0 trying the factors of the constant we get o_1 =o_2 =(n/k) [not usable] o_3 =2kn+2m−2k(√(m^2 +n^2 )) o_4 =2kn+2m+2k(√(m^2 +n^2 )) now insert k=(((a+b−c)(a−b+c))/δ) m=(c/2) n=(((a^2 +b^2 −c^2 )c)/(2δ)) δ=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) R=((abc)/δ) o_3 =((2bc)/(a+b+c)) o_4 =((2bc)/(−a+b+c)) r_3 =((2(a+b−c)(a−b+c)bc)/((a+b+c)δ)) [blue circle] r_4 =((2(a+b−c)(a−b+c)bc)/((−a+b+c)δ)) [circle touching circumcircle from outside]](https://www.tinkutara.com/question/Q52894.png)
$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{b}\mathrm{cos}\:\alpha}\\{{b}\mathrm{sin}\:\alpha}\end{pmatrix}\:\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:={k} \\ $$$$ \\ $$$$\mathrm{circumcircle} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{center}=\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix};\:\mathrm{radius}={R}=\sqrt{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{blue}\:\mathrm{circle} \\ $$$$\:\:\:\:\:\mathrm{center}=\begin{pmatrix}{{o}}\\{{ko}}\end{pmatrix};\:\mathrm{radius}={r}={ko} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\:\left({x}−{m}\right)^{\mathrm{2}} +\left({y}−{n}\right)^{\mathrm{2}} −\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{mx}−\mathrm{2}{ny}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\left({x}−{o}\right)^{\mathrm{2}} +\left({y}−{ko}\right)^{\mathrm{2}} −\left({ok}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ox}−\mathrm{2}{koy}+{o}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:{y}=\frac{\mathrm{2}\left({m}−{o}\right){x}+{o}^{\mathrm{2}} }{\mathrm{2}\left({ko}−{n}\right)} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{or}\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} +{ux}+{v}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution}\:\Rightarrow \\ $$$$\Rightarrow\:{D}={u}^{\mathrm{2}} −\mathrm{4}{v}=\mathrm{0} \\ $$$${o}^{\mathrm{4}} −\frac{\mathrm{2}\left(\mathrm{2}{k}^{\mathrm{2}} {n}+\mathrm{2}{km}+{n}\right)}{{k}}{o}^{\mathrm{3}} −\frac{\mathrm{4}{k}^{\mathrm{4}} {m}^{\mathrm{2}} −\mathrm{8}{k}^{\mathrm{3}} {mn}−\mathrm{4}{k}^{\mathrm{2}} \left({m}^{\mathrm{2}} +\mathrm{2}{n}^{\mathrm{2}} \right)−\mathrm{8}{kmn}−{n}^{\mathrm{2}} }{{k}^{\mathrm{2}} }{o}^{\mathrm{2}} +\frac{\mathrm{4}\left(\mathrm{2}{k}^{\mathrm{3}} {m}^{\mathrm{2}} −\mathrm{4}{k}^{\mathrm{2}} {mn}−{k}\left(\mathrm{2}{m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)−{mn}\right){n}}{{k}^{\mathrm{2}} }{o}−\frac{\mathrm{4}\left({k}^{\mathrm{2}} {m}−\mathrm{2}{kn}−{m}\right){mn}^{\mathrm{2}} }{{k}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{we}\:\mathrm{get} \\ $$$${o}_{\mathrm{1}} ={o}_{\mathrm{2}} =\frac{{n}}{{k}}\:\left[\mathrm{not}\:\mathrm{usable}\right] \\ $$$${o}_{\mathrm{3}} =\mathrm{2}{kn}+\mathrm{2}{m}−\mathrm{2}{k}\sqrt{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$${o}_{\mathrm{4}} =\mathrm{2}{kn}+\mathrm{2}{m}+\mathrm{2}{k}\sqrt{{m}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$$\mathrm{now}\:\mathrm{insert} \\ $$$${k}=\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\delta} \\ $$$${m}=\frac{{c}}{\mathrm{2}} \\ $$$${n}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){c}}{\mathrm{2}\delta} \\ $$$$\delta=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$${R}=\frac{{abc}}{\delta} \\ $$$${o}_{\mathrm{3}} =\frac{\mathrm{2}{bc}}{{a}+{b}+{c}}\:\:{o}_{\mathrm{4}} =\frac{\mathrm{2}{bc}}{−{a}+{b}+{c}} \\ $$$${r}_{\mathrm{3}} =\frac{\mathrm{2}\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right){bc}}{\left({a}+{b}+{c}\right)\delta}\:\left[\mathrm{blue}\:\mathrm{circle}\right] \\ $$$${r}_{\mathrm{4}} =\frac{\mathrm{2}\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right){bc}}{\left(−{a}+{b}+{c}\right)\delta}\:\left[\mathrm{circle}\:\mathrm{touching}\:\mathrm{circumcircle}\:\mathrm{from}\:\mathrm{outside}\right] \\ $$
Commented by behi83417@gmail.com last updated on 15/Jan/19
$${thank}\:{you}\:{sir}\:{MJS}.{this}\:{is}\:{perfect}. \\ $$$${checking}\::{r}_{\mathrm{3}} \:\Rightarrow{right}\:{answer}. \\ $$
Commented by MJS last updated on 15/Jan/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{checking}\:\mathrm{Sir}\:\mathrm{Behi} \\ $$
Commented by mr W last updated on 15/Jan/19
$${r}={r}_{\mathrm{3}} =\frac{\mathrm{2}\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right){bc}}{\left({a}+{b}+{c}\right)\delta} \\ $$$${r}=\frac{\mathrm{2}{bc}\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)}} \\ $$$${r}=\frac{\mathrm{2}{bc}\sqrt{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}}{\left({a}+{b}+{c}\right)\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}} \\ $$$${r}=\frac{\mathrm{2}{bc}}{\left({a}+{b}+{c}\right)}\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)}{\left({a}+{b}+{c}\right)\left({b}+{c}−{a}\right)}} \\ $$$${this}\:{is}\:{correct}\:{sir}. \\ $$