Question-53066

Facebook Tweet Pin

Question Number 53066 by ajfour last updated on 16/Jan/19

Commented by ajfour last updated on 17/Jan/19

$${Determine}\:{radius}\:{R}\:{of}\:{larger}\:{circle} \\ $$$${in}\:{terms}\:{of}\:{a},{b},{c}\:.\:\:\:\: \\ $$

Commented by ajfour last updated on 17/Jan/19

$${Sir}\:..\overset{{please}\:{try}\:{this}\:{even}!} {\equiv\gg} \\ $$

Commented by MJS last updated on 17/Jan/19

$$…\mathrm{I}'\mathrm{m}\:\mathrm{busy}\:\mathrm{trying}… \\ $$

Commented by ajfour last updated on 18/Jan/19

Commented by ajfour last updated on 18/Jan/19

R=(δ/(8(b+c−a)))+((a^2 (b+c−a))/(2δ)) δ=(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) please check it now Sir.

$${R}=\frac{\delta}{\mathrm{8}\left({b}+{c}−{a}\right)}+\frac{{a}^{\mathrm{2}} \left({b}+{c}−{a}\right)}{\mathrm{2}\delta} \\ $$$$\delta=\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$$${please}\:{check}\:{it}\:{now}\:{Sir}. \\ $$

Commented by ajfour last updated on 18/Jan/19

BK = R, BM=(a/2) ⇒ MK=(√(R^2 −(a^2 /4))) JN = r−(√(R^2 −(a^2 /4))) , KN = R−r Now r(cot θ+cot φ)=a r(cot θ+cot ψ)=c r(cot φ+cot ψ)=b ⇒ 2rcot θ = a+c−b rcot θ = ((c−b)/2)+(a/2) BP = rcot θ JK = MP = BP−(a/2) JK = rcot θ−(a/2) = ((c−b)/2) In △JKN KN^( 2) =JK^( 2) +JN^( 2) ⇒ (R−r)^2 =(((c−b)/2))^2 +(r−(√(R^2 −(a^2 /4))))^2 ⇒ R^2 −2rR+r^2 =(((c−b)/2))^2 +r^2 +R^2 −(a^2 /4) −2r(√(R^2 −(a^2 /4))) ⇒ 2r(R−(√(R^2 −(a^2 /4))))=(((a+c−b)(a+b−c))/4) let area(△ABC)=△ we know r(a+b+c)= 2△ , and △= (√(s(s−a)(s−b)(s−c))) If δ=(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) ⇒ 4△= δ = 2r(a+b+c) ⇒ r = (𝛅/(2(a+b+c))) 2r(R−(√(R^2 −(a^2 /4))))=(((a+c−b)(a+b−c))/4) gives 2r(R−(√(R^2 −(a^2 /4))) ) = (δ^2 /(4(a+b+c)(b+c−a))) ⇒ R−(√(R^2 −(a^2 /4))) = (δ/(4(b+c−a))) ⇒ R^2 −(a^2 /4)=[R−(δ/(4(b+c−a)))]^2 ⇒ ((Rδ)/(2(b+c−a)))=(a^2 /4)+(δ^2 /(16(b+c−a)^2 )) ⇒ R=((a^2 (b+c−a))/(2δ))+(δ/(8(b+c−a))) .

$${BK}\:=\:{R},\:{BM}=\frac{{a}}{\mathrm{2}} \\ $$$$\Rightarrow\:{MK}=\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\:\:{JN}\:=\:{r}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:\:,\:\:{KN}\:=\:{R}−{r} \\ $$$${Now}\:\:{r}\left(\mathrm{cot}\:\theta+\mathrm{cot}\:\phi\right)={a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{r}\left(\mathrm{cot}\:\theta+\mathrm{cot}\:\psi\right)={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{r}\left(\mathrm{cot}\:\phi+\mathrm{cot}\:\psi\right)={b} \\ $$$$\Rightarrow\:\:\mathrm{2}{r}\mathrm{cot}\:\theta\:=\:{a}+{c}−{b} \\ $$$$\:\:\:\:\:\:\:\:\:\:{r}\mathrm{cot}\:\theta\:=\:\frac{{c}−{b}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}} \\ $$$${BP}\:=\:{r}\mathrm{cot}\:\theta \\ $$$${JK}\:=\:{MP}\:=\:{BP}−\frac{{a}}{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:\:{JK}\:=\:{r}\mathrm{cot}\:\theta−\frac{{a}}{\mathrm{2}}\:=\:\frac{{c}−{b}}{\mathrm{2}} \\ $$$${In}\:\bigtriangleup{JKN}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{KN}^{\:\mathrm{2}} ={JK}^{\:\mathrm{2}} +{JN}^{\:\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({R}−{r}\right)^{\mathrm{2}} =\left(\frac{{c}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({r}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{R}^{\mathrm{2}} −\mathrm{2}{rR}+{r}^{\mathrm{2}} =\left(\frac{{c}−{b}}{\mathrm{2}}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} +{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{r}\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\Rightarrow\:\mathrm{2}{r}\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)=\frac{\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)}{\mathrm{4}} \\ $$$${let}\:{area}\left(\bigtriangleup{ABC}\right)=\bigtriangleup \\ $$$${we}\:{know}\:\:{r}\left({a}+{b}+{c}\right)=\:\mathrm{2}\bigtriangleup\:\:\:,\:{and} \\ $$$$\:\:\bigtriangleup=\:\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$${If}\:\:\delta=\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$$$\Rightarrow\:\mathrm{4}\bigtriangleup=\:\delta\:=\:\mathrm{2}{r}\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow\:\:\boldsymbol{{r}}\:=\:\frac{\boldsymbol{\delta}}{\mathrm{2}\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)} \\ $$$$\:\mathrm{2}{r}\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)=\frac{\left({a}+{c}−{b}\right)\left({a}+{b}−{c}\right)}{\mathrm{4}} \\ $$$${gives} \\ $$$$\:\mathrm{2}{r}\left({R}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:\right)\:=\:\frac{\delta^{\mathrm{2}} }{\mathrm{4}\left({a}+{b}+{c}\right)\left({b}+{c}−{a}\right)} \\ $$$$\Rightarrow\:{R}−\sqrt{{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\:=\:\frac{\delta}{\mathrm{4}\left({b}+{c}−{a}\right)} \\ $$$$\Rightarrow\:{R}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}=\left[{R}−\frac{\delta}{\mathrm{4}\left({b}+{c}−{a}\right)}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{{R}\delta}{\mathrm{2}\left({b}+{c}−{a}\right)}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{\delta^{\mathrm{2}} }{\mathrm{16}\left({b}+{c}−{a}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{R}=\frac{{a}^{\mathrm{2}} \left({b}+{c}−{a}\right)}{\mathrm{2}\delta}+\frac{\delta}{\mathrm{8}\left({b}+{c}−{a}\right)}\:. \\ $$

Commented by MJS last updated on 18/Jan/19

now it′s the same as mine. how did you get it?

$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{mine}.\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{it}? \\ $$

Commented by MJS last updated on 18/Jan/19

great! now try the same for the excircles

$$\mathrm{great}!\:\mathrm{now}\:\mathrm{try}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for}\:\mathrm{the}\:\mathrm{excircles} \\ $$

Commented by behi83417@gmail.com last updated on 19/Jan/19

R=((2a^2 (p−a))/(8S))+((4S)/(16(p−a)))=((a^2 +r_a ^2 )/(4r_a )).

$${R}=\frac{\mathrm{2}{a}^{\mathrm{2}} \left({p}−{a}\right)}{\mathrm{8}{S}}+\frac{\mathrm{4}{S}}{\mathrm{16}\left({p}−{a}\right)}=\frac{{a}^{\mathrm{2}} +{r}_{{a}} ^{\mathrm{2}} }{\mathrm{4}{r}_{{a}} }. \\ $$

Answered by MJS last updated on 18/Jan/19

coordinate method gives δ=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) A= (((−(δ/(2a)))),(((c^2 −b^2 )/(2a))) ) B= ((0),((−(a/2))) ) C= ((0),((a/2)) ) incircle center = (((−(δ/(2(a+b+c))))),(((c−b)/2)) ) radius = (δ/(2(a+b+c))) big circle center = ((((5a^3 −b^3 −c^3 −3a^2 (b+c)−b^2 (a−c)−c^2 (a−b)+2abc)/(8δ))),(0) ) radius = −((3a^3 +b^3 +c^3 −5a^2 (b+c)+b^2 (a−c)+c^2 (a−b)−2abc)/(8δ)) I couldn′t simplify... but I plotted several triangles with both circles and it seems to hold...

$$\mathrm{coordinate}\:\mathrm{method}\:\mathrm{gives} \\ $$$$\delta=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$${A}=\begin{pmatrix}{−\frac{\delta}{\mathrm{2}{a}}}\\{\frac{{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{a}}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{0}}\\{−\frac{{a}}{\mathrm{2}}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{\mathrm{0}}\\{\frac{{a}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{incircle} \\ $$$$\mathrm{center}\:=\begin{pmatrix}{−\frac{\delta}{\mathrm{2}\left({a}+{b}+{c}\right)}}\\{\frac{{c}−{b}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{radius}\:=\:\frac{\delta}{\mathrm{2}\left({a}+{b}+{c}\right)} \\ $$$$\mathrm{big}\:\mathrm{circle} \\ $$$$\mathrm{center}\:=\:\begin{pmatrix}{\frac{\mathrm{5}{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} −\mathrm{3}{a}^{\mathrm{2}} \left({b}+{c}\right)−{b}^{\mathrm{2}} \left({a}−{c}\right)−{c}^{\mathrm{2}} \left({a}−{b}\right)+\mathrm{2}{abc}}{\mathrm{8}\delta}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{radius}\:=\:−\frac{\mathrm{3}{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{5}{a}^{\mathrm{2}} \left({b}+{c}\right)+{b}^{\mathrm{2}} \left({a}−{c}\right)+{c}^{\mathrm{2}} \left({a}−{b}\right)−\mathrm{2}{abc}}{\mathrm{8}\delta} \\ $$$$\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{simplify}…\:\mathrm{but}\:\mathrm{I}\:\mathrm{plotted}\:\mathrm{several} \\ $$$$\mathrm{triangles}\:\mathrm{with}\:\mathrm{both}\:\mathrm{circles}\:\mathrm{and}\:\mathrm{it}\:\mathrm{seems}\:\mathrm{to} \\ $$$$\mathrm{hold}… \\ $$

Commented by MJS last updated on 18/Jan/19

you′re welcome. I like your way to solve it!

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}. \\ $$$$\mathrm{I}\:\mathrm{like}\:\mathrm{your}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}! \\ $$

Commented by ajfour last updated on 18/Jan/19