Question Number 202925 by mr W last updated on 05/Jan/24
Commented by mr W last updated on 05/Jan/24
![[Q201146 reposted] find IJ=?](https://www.tinkutara.com/question/Q202926.png)
$$\left[{Q}\mathrm{201146}\:{reposted}\right] \\ $$$${find}\:{IJ}=? \\ $$
Commented by MathematicalUser2357 last updated on 09/Jan/24
$${q}\:{num}\:\mathrm{201146} \\ $$
Answered by mr W last updated on 06/Jan/24
![p=((a+b+c)/2) (semi−perimeter) r=radius of incircle R=radius of circumcircle S=area of ΔABC =pr=((abc)/(4R))=(√(p(p−a)(p−b)(p−c))) AE=AF=p−a BD=BF=p−b CD=CE=p−c AI^2 =(p−a)^2 +r^2 in ΔABC: aAD^2 =(p−b)b^2 +(p−c)c^2 −a(p−b)(p−c) AD^2 =(((p−b)b^2 +(p−c)c^2 )/a)−(p−b)(p−c) AD^2 =(((p−a)a^2 +(p−b)b^2 +(p−c)c^2 −abc)/a)+(p−a)^2 AD^2 =((p(a^2 +b^2 +c^2 )−(a^3 +b^3 +c^3 )−abc)/a)+(p−a)^2 AD^2 =((p(2p^2 −2r^2 −8Rr)−(2p^3 −6r^2 p−12Rrp)−abc)/a)+(p−a)^2 AD^2 =((4r^2 p+4Rrp−abc)/a)+(p−a)^2 AD^2 =((4pr^2 +4RS−4RS)/a)+(p−a)^2 AD^2 =((4pr^2 )/a)+(p−a)^2 ⇒AD=(√(((4pr^2 )/a)+(p−a)^2 )) ((AJ)/(JD))×((BD)/(BC))×((CE)/(EA))=1 ((AJ)/(JD))×((p−b)/a)×((p−c)/(p−a))=1 ⇒((AJ)/(JD))=((a(p−a))/((p−b)(p−c))) ⇒AJ=((a(p−a))/(a(p−a)+(p−b)(p−c)))×AD =((4a(p−a))/(2(ab+bc+ca)−(a^2 +b^2 +c^2 )))×AD =((4a(p−a))/(2(p^2 +r^2 +4Rr)−(2p^2 −2r^2 −8Rr)))×AD =((a(p−a))/(r^2 +4Rr))×AD ⇒JD=(((p−b)(p−c))/(a(p−a)+(p−b)(p−c)))×AD =(((p−b)(p−c))/(r^2 +4Rr))×AD in ΔIAD: AD×IJ^2 =AJ×DI^2 +JD×AI^2 −AD×AJ×JD IJ^2 =((AJ)/(AD))×DI^2 +((JD)/(AD))×AI^2 −AJ×JD IJ^2 =((a(p−a))/(r^2 +4Rr))×r^2 +(((p−b)(p−c))/(r^2 +4Rr))×[r^2 +(p−a)^2 ]−((a(p−a))/(r^2 +4Rr))×(((p−b)(p−c))/(r^2 +4Rr))×AD^2 IJ^2 =((a(p−a)+(p−b)(p−c))/(r^2 +4Rr))×r^2 +(((p−a)^2 (p−b)(p−c))/(r^2 +4Rr))−((a(p−a)(p−b)(p−c))/((r^2 +4Rr)^2 ))×AD^2 IJ^2 =((r^2 +4Rr)/(r^2 +4Rr))×r^2 +(((p−a)S^2 )/((r^2 +4Rr)p))−((aS^2 )/((r^2 +4Rr)^2 p))×AD^2 IJ^2 =r^2 +((pr^2 )/((r^2 +4Rr)^2 ))[(p−a)(r^2 +4Rr)−a×AD^2 ] IJ^2 =r^2 +(p/((r+4R)^2 ))[(p−a)(r^2 +4Rr)−4pr^2 −a(p−a)^2 ] IJ^2 =r^2 +(p/((r+4R)^2 ))[(p−a)(r^2 +4Rr−ap+a^2 )−4pr^2 ] IJ^2 =r^2 +(p/((r+4R)^2 ))[(p−a)(bc−(p−a)p)−4pr^2 ] IJ^2 =r^2 +(p^2 /(4(r+4R)^2 ))[2(ab+bc+ca)−(a^2 +b^2 +c^2 )−16Rr−16r^2 ] IJ^2 =r^2 +(p^2 /(4(r+4R)^2 ))[2p^2 +2r^2 +8Rr−2p^2 +2r^2 +8Rr−16Rr−16r^2 ] IJ^2 =r^2 −((3p^2 r^2 )/((r+4R)^2 )) ⇒IJ=r(√(1−((3p^2 )/((r+4R)^2 ))))](https://www.tinkutara.com/question/Q202927.png)
$${p}=\frac{{a}+{b}+{c}}{\mathrm{2}}\:\left({semi}−{perimeter}\right) \\ $$$${r}={radius}\:{of}\:{incircle} \\ $$$${R}={radius}\:{of}\:{circumcircle} \\ $$$${S}={area}\:{of}\:\Delta{ABC} \\ $$$$\:\:\:={pr}=\frac{{abc}}{\mathrm{4}{R}}=\sqrt{{p}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)} \\ $$$$ \\ $$$${AE}={AF}={p}−{a} \\ $$$${BD}={BF}={p}−{b} \\ $$$${CD}={CE}={p}−{c} \\ $$$$ \\ $$$${AI}^{\mathrm{2}} =\left({p}−{a}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$ \\ $$$${in}\:\Delta{ABC}: \\ $$$${aAD}^{\mathrm{2}} =\left({p}−{b}\right){b}^{\mathrm{2}} +\left({p}−{c}\right){c}^{\mathrm{2}} −{a}\left({p}−{b}\right)\left({p}−{c}\right) \\ $$$${AD}^{\mathrm{2}} =\frac{\left({p}−{b}\right){b}^{\mathrm{2}} +\left({p}−{c}\right){c}^{\mathrm{2}} }{{a}}−\left({p}−{b}\right)\left({p}−{c}\right) \\ $$$${AD}^{\mathrm{2}} =\frac{\left({p}−{a}\right){a}^{\mathrm{2}} +\left({p}−{b}\right){b}^{\mathrm{2}} +\left({p}−{c}\right){c}^{\mathrm{2}} −{abc}}{{a}}+\left({p}−{a}\right)^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} =\frac{{p}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)−{abc}}{{a}}+\left({p}−{a}\right)^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} =\frac{{p}\left(\mathrm{2}{p}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} −\mathrm{8}{Rr}\right)−\left(\mathrm{2}{p}^{\mathrm{3}} −\mathrm{6}{r}^{\mathrm{2}} {p}−\mathrm{12}{Rrp}\right)−{abc}}{{a}}+\left({p}−{a}\right)^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} =\frac{\mathrm{4}{r}^{\mathrm{2}} {p}+\mathrm{4}{Rrp}−{abc}}{{a}}+\left({p}−{a}\right)^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} =\frac{\mathrm{4}{pr}^{\mathrm{2}} +\mathrm{4}{RS}−\mathrm{4}{RS}}{{a}}+\left({p}−{a}\right)^{\mathrm{2}} \\ $$$${AD}^{\mathrm{2}} =\frac{\mathrm{4}{pr}^{\mathrm{2}} }{{a}}+\left({p}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{AD}=\sqrt{\frac{\mathrm{4}{pr}^{\mathrm{2}} }{{a}}+\left({p}−{a}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{{AJ}}{{JD}}×\frac{{BD}}{{BC}}×\frac{{CE}}{{EA}}=\mathrm{1} \\ $$$$\frac{{AJ}}{{JD}}×\frac{{p}−{b}}{{a}}×\frac{{p}−{c}}{{p}−{a}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{AJ}}{{JD}}=\frac{{a}\left({p}−{a}\right)}{\left({p}−{b}\right)\left({p}−{c}\right)} \\ $$$$\Rightarrow{AJ}=\frac{{a}\left({p}−{a}\right)}{{a}\left({p}−{a}\right)+\left({p}−{b}\right)\left({p}−{c}\right)}×{AD} \\ $$$$\:\:\:\:=\frac{\mathrm{4}{a}\left({p}−{a}\right)}{\mathrm{2}\left({ab}+{bc}+{ca}\right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}×{AD} \\ $$$$\:\:\:\:=\frac{\mathrm{4}{a}\left({p}−{a}\right)}{\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{4}{Rr}\right)−\left(\mathrm{2}{p}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} −\mathrm{8}{Rr}\right)}×{AD} \\ $$$$\:\:\:\:=\frac{{a}\left({p}−{a}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×{AD} \\ $$$$\Rightarrow{JD}=\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{a}\left({p}−{a}\right)+\left({p}−{b}\right)\left({p}−{c}\right)}×{AD} \\ $$$$\:\:\:\:=\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×{AD} \\ $$$$ \\ $$$${in}\:\Delta{IAD}: \\ $$$${AD}×{IJ}^{\mathrm{2}} ={AJ}×{DI}^{\mathrm{2}} +{JD}×{AI}^{\mathrm{2}} −{AD}×{AJ}×{JD} \\ $$$${IJ}^{\mathrm{2}} =\frac{{AJ}}{{AD}}×{DI}^{\mathrm{2}} +\frac{{JD}}{{AD}}×{AI}^{\mathrm{2}} −{AJ}×{JD} \\ $$$${IJ}^{\mathrm{2}} =\frac{{a}\left({p}−{a}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×{r}^{\mathrm{2}} +\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×\left[{r}^{\mathrm{2}} +\left({p}−{a}\right)^{\mathrm{2}} \right]−\frac{{a}\left({p}−{a}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×{AD}^{\mathrm{2}} \\ $$$${IJ}^{\mathrm{2}} =\frac{{a}\left({p}−{a}\right)+\left({p}−{b}\right)\left({p}−{c}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×{r}^{\mathrm{2}} +\frac{\left({p}−{a}\right)^{\mathrm{2}} \left({p}−{b}\right)\left({p}−{c}\right)}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}−\frac{{a}\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)}{\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}\right)^{\mathrm{2}} }×{AD}^{\mathrm{2}} \\ $$$${IJ}^{\mathrm{2}} =\frac{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}{{r}^{\mathrm{2}} +\mathrm{4}{Rr}}×{r}^{\mathrm{2}} +\frac{\left({p}−{a}\right){S}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}\right){p}}−\frac{{aS}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}\right)^{\mathrm{2}} {p}}×{AD}^{\mathrm{2}} \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{pr}^{\mathrm{2}} }{\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}\right)^{\mathrm{2}} }\left[\left({p}−{a}\right)\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}\right)−{a}×{AD}^{\mathrm{2}} \right] \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{p}}{\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} }\left[\left({p}−{a}\right)\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}\right)−\mathrm{4}{pr}^{\mathrm{2}} −{a}\left({p}−{a}\right)^{\mathrm{2}} \right] \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{p}}{\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} }\left[\left({p}−{a}\right)\left({r}^{\mathrm{2}} +\mathrm{4}{Rr}−{ap}+{a}^{\mathrm{2}} \right)−\mathrm{4}{pr}^{\mathrm{2}} \right] \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{p}}{\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} }\left[\left({p}−{a}\right)\left({bc}−\left({p}−{a}\right){p}\right)−\mathrm{4}{pr}^{\mathrm{2}} \right] \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} }{\mathrm{4}\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} }\left[\mathrm{2}\left({ab}+{bc}+{ca}\right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{16}{Rr}−\mathrm{16}{r}^{\mathrm{2}} \right] \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} }{\mathrm{4}\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} }\left[\mathrm{2}{p}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} +\mathrm{8}{Rr}−\mathrm{2}{p}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} +\mathrm{8}{Rr}−\mathrm{16}{Rr}−\mathrm{16}{r}^{\mathrm{2}} \right] \\ $$$${IJ}^{\mathrm{2}} ={r}^{\mathrm{2}} −\frac{\mathrm{3}{p}^{\mathrm{2}} {r}^{\mathrm{2}} }{\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{IJ}={r}\sqrt{\mathrm{1}−\frac{\mathrm{3}{p}^{\mathrm{2}} }{\left({r}+\mathrm{4}{R}\right)^{\mathrm{2}} }} \\ $$
Commented by mr W last updated on 06/Jan/24
$${yes},\:{you}\:{did}.\:{i}\:{didn}'{t}\:\:{expect}\:{that}\:{the} \\ $$$${formula}\:{can}\:{be}\:{so}\:{compact}. \\ $$
Commented by ajfour last updated on 06/Jan/24
$${I}\:{remember}\:{i}\:{had}\:{the}\:{hope}\:{n} \\ $$$${intuition}\:{that}\:{something}\:{compact} \\ $$$${n}\:{beautiful}\:{dhould}\:{result}.\:{Thanks}! \\ $$