Question Number 52539 by ajfour last updated on 09/Jan/19
Answered by mr W last updated on 09/Jan/19
![p=((√([(b+c)^2 −a^2 ][a^2 −(b−c)^2 ]))/(2a)) r=((√([(b+c)^2 −a^2 ]bc))/(b+c)) m=((√(2(b^2 +c^2 )−a^2 ))/2) ⇒a^2 =2(b^2 +c^2 )−4m^2 ...(iii) ⇒(r^2 −bc)(b+c)^2 =2bc[2m^2 +2bc−(b+c)^2 ] ⇒8p^2 [(b+c)^2 −2m^2 −2bc)=[4m^2 +4bc−(b+c)^2 ][(b+c)^2 −4m^2 ] let X=b+c, Y=bc ⇒(r^2 −Y)X^2 =2Y(2m^2 +2Y−X^2 ) ...(i) ⇒8p^2 (X^2 −2Y−2m^2 )=(4m^2 +4Y−X^2 )(X^2 −4m^2 ) ...(ii) from (i): ⇒X^2 =((4Y(m^2 +Y))/(r^2 +Y)) ⇒p^2 [((2Y(m^2 +Y))/(r^2 +Y))−Y−m^2 ]=[m^2 +Y−((Y(m^2 +Y))/(r^2 +Y))][((Y(m^2 +Y))/(r^2 +Y))−m^2 ] ⇒Y=r^2 (√((m^2 −p^2 )/(r^2 −p^2 ))) ...(1) ⇒X=2(√((Y(m^2 +Y))/(r^2 +Y))) ...(2) with p=(√(15)), m=(√(31)), r=2(√6) ⇒Y=24(√((31−15)/(24−15)))=24×(4/3)=32(=bc) ⇒X=2(√((32(31+32))/(24+32)))=12(=b+c) ⇒b and c are roots of z^2 −12z+32=0 (z−8)(z−4)=0 ⇒b=8, c=4 (or b=4, c=8) from (iii): ⇒a=6](https://www.tinkutara.com/question/Q52568.png)
$${p}=\frac{\sqrt{\left[\left({b}+{c}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]\left[{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \right]}}{\mathrm{2}{a}} \\ $$$${r}=\frac{\sqrt{\left[\left({b}+{c}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]{bc}}}{{b}+{c}} \\ $$$${m}=\frac{\sqrt{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{4}{m}^{\mathrm{2}} \:\:\:…\left({iii}\right) \\ $$$$ \\ $$$$\Rightarrow\left({r}^{\mathrm{2}} −{bc}\right)\left({b}+{c}\right)^{\mathrm{2}} =\mathrm{2}{bc}\left[\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{bc}−\left({b}+{c}\right)^{\mathrm{2}} \right]\:\: \\ $$$$\Rightarrow\mathrm{8}{p}^{\mathrm{2}} \left[\left({b}+{c}\right)^{\mathrm{2}} −\mathrm{2}{m}^{\mathrm{2}} −\mathrm{2}{bc}\right)=\left[\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{bc}−\left({b}+{c}\right)^{\mathrm{2}} \right]\left[\left({b}+{c}\right)^{\mathrm{2}} −\mathrm{4}{m}^{\mathrm{2}} \right] \\ $$$$ \\ $$$${let}\:{X}={b}+{c},\:{Y}={bc} \\ $$$$\Rightarrow\left({r}^{\mathrm{2}} −{Y}\right){X}^{\mathrm{2}} =\mathrm{2}{Y}\left(\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{Y}−{X}^{\mathrm{2}} \right)\:\:\:…\left({i}\right) \\ $$$$\Rightarrow\mathrm{8}{p}^{\mathrm{2}} \left({X}^{\mathrm{2}} −\mathrm{2}{Y}−\mathrm{2}{m}^{\mathrm{2}} \right)=\left(\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{Y}−{X}^{\mathrm{2}} \right)\left({X}^{\mathrm{2}} −\mathrm{4}{m}^{\mathrm{2}} \right)\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$$\Rightarrow{X}^{\mathrm{2}} =\frac{\mathrm{4}{Y}\left({m}^{\mathrm{2}} +{Y}\right)}{{r}^{\mathrm{2}} +{Y}} \\ $$$$\Rightarrow{p}^{\mathrm{2}} \left[\frac{\mathrm{2}{Y}\left({m}^{\mathrm{2}} +{Y}\right)}{{r}^{\mathrm{2}} +{Y}}−{Y}−{m}^{\mathrm{2}} \right]=\left[{m}^{\mathrm{2}} +{Y}−\frac{{Y}\left({m}^{\mathrm{2}} +{Y}\right)}{{r}^{\mathrm{2}} +{Y}}\right]\left[\frac{{Y}\left({m}^{\mathrm{2}} +{Y}\right)}{{r}^{\mathrm{2}} +{Y}}−{m}^{\mathrm{2}} \right] \\ $$$$\Rightarrow{Y}={r}^{\mathrm{2}} \sqrt{\frac{{m}^{\mathrm{2}} −{p}^{\mathrm{2}} }{{r}^{\mathrm{2}} −{p}^{\mathrm{2}} }}\:\:\:…\left(\mathrm{1}\right) \\ $$$$\Rightarrow{X}=\mathrm{2}\sqrt{\frac{{Y}\left({m}^{\mathrm{2}} +{Y}\right)}{{r}^{\mathrm{2}} +{Y}}}\:\:\:…\left(\mathrm{2}\right) \\ $$$$ \\ $$$${with}\:{p}=\sqrt{\mathrm{15}},\:{m}=\sqrt{\mathrm{31}},\:{r}=\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\Rightarrow{Y}=\mathrm{24}\sqrt{\frac{\mathrm{31}−\mathrm{15}}{\mathrm{24}−\mathrm{15}}}=\mathrm{24}×\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{32}\left(={bc}\right) \\ $$$$\Rightarrow{X}=\mathrm{2}\sqrt{\frac{\mathrm{32}\left(\mathrm{31}+\mathrm{32}\right)}{\mathrm{24}+\mathrm{32}}}=\mathrm{12}\left(={b}+{c}\right) \\ $$$$\Rightarrow{b}\:{and}\:{c}\:{are}\:{roots}\:{of} \\ $$$${z}^{\mathrm{2}} −\mathrm{12}{z}+\mathrm{32}=\mathrm{0} \\ $$$$\left({z}−\mathrm{8}\right)\left({z}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{8},\:{c}=\mathrm{4}\:\left({or}\:{b}=\mathrm{4},\:{c}=\mathrm{8}\right) \\ $$$${from}\:\left({iii}\right): \\ $$$$\Rightarrow{a}=\mathrm{6} \\ $$
Commented by ajfour last updated on 10/Jan/19
$${correct}\:{answers}\:{sir}.\:{thank}\:{you}. \\ $$$$\left({i}\:{had}\:{created}\:{the}\:{question}\:{myself},\right. \\ $$$$\left.{solving}\:{is}\:{a}\:\:{tougher}\:{task}\:{i}\:{see}\:{now}\right). \\ $$
Commented by mr W last updated on 10/Jan/19
$${it}'{s}\:{a}\:{nice}\:{question}\:{sir}. \\ $$$${at}\:{the}\:{begining}\:{i}\:{thought}\:{the}\:{sides}\:{of} \\ $$$${the}\:{triangle}\:{could}\:{be}\:{not}\:{unique}.\:{and} \\ $$$${i}\:{thought}\:{it}'{s}\:{mayby}\:{impossible}\:{to}\:{determine} \\ $$$${the}\:{sides}\:{of}\:{the}\:{triangle}\:{with}\:{direct} \\ $$$${formula},\:{then}\:{i}\:{found}\:{this}\:{solution} \\ $$$${which}\:{can}\:{solve}\:{b}+{c}\:{and}\:{bc}\:{directly}. \\ $$$${with}\:{this}\:{method},\:{the}\:{sides}\:{of}\:{the} \\ $$$${triangle}\:{can}\:{all}\:{be}\:{calculated}\:{directly} \\ $$$${and}\:{exactly}. \\ $$
Commented by mr W last updated on 13/Jan/19
Answered by behi83417@gmail.com last updated on 09/Jan/19
![4m^2 =2(b^2 +c^2 )−a^2 =124(i) ⇒b^2 +c^2 =62+(a^2 /2) r=((2bc)/(b+c))cos(A/2)=((2bc)/(b+c))(√((p(p−a))/(bc))) (2/(b+c))(√(bcp(p−a)))=2(√6)⇒bcp(p−a)=6(b+c)^2 (ii) bc(((a+b+c)/2))(((b+c−a)/2))=6(b+c)^2 ⇒ bc[(b+c)^2 −a^2 ]=24(b+c)^2 bc[62+(a^2 /2)+2bc−a^2 ]=24(b+c)^2 =24(62+(a^2 /2)+2bc) ⇒bc(62+2bc−(a^2 /2))=24(62+(a^2 /2)+2bc) 62bc+2(bc)^2 −bc.(a^2 /2)=24(62+(a^2 /2)+2bc) ⇒(bc)^2 +(7−(a^2 /4)).bc−6a^2 −744=0 .....](https://www.tinkutara.com/question/Q52569.png)
$$\mathrm{4}{m}^{\mathrm{2}} =\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} =\mathrm{124}\left({i}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{62}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${r}=\frac{\mathrm{2}{bc}}{{b}+{c}}{cos}\frac{{A}}{\mathrm{2}}=\frac{\mathrm{2}{bc}}{{b}+{c}}\sqrt{\frac{{p}\left({p}−{a}\right)}{{bc}}} \\ $$$$\frac{\mathrm{2}}{{b}+{c}}\sqrt{{bcp}\left({p}−{a}\right)}=\mathrm{2}\sqrt{\mathrm{6}}\Rightarrow{bcp}\left({p}−{a}\right)=\mathrm{6}\left({b}+{c}\right)^{\mathrm{2}} \left({ii}\right) \\ $$$${bc}\left(\frac{{a}+{b}+{c}}{\mathrm{2}}\right)\left(\frac{{b}+{c}−{a}}{\mathrm{2}}\right)=\mathrm{6}\left({b}+{c}\right)^{\mathrm{2}} \Rightarrow \\ $$$${bc}\left[\left({b}+{c}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} \right]=\mathrm{24}\left({b}+{c}\right)^{\mathrm{2}} \\ $$$${bc}\left[\mathrm{62}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{bc}−{a}^{\mathrm{2}} \right]=\mathrm{24}\left({b}+{c}\right)^{\mathrm{2}} =\mathrm{24}\left(\mathrm{62}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{bc}\right) \\ $$$$\Rightarrow{bc}\left(\mathrm{62}+\mathrm{2}{bc}−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\right)=\mathrm{24}\left(\mathrm{62}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{bc}\right) \\ $$$$\mathrm{62}{bc}+\mathrm{2}\left({bc}\right)^{\mathrm{2}} −{bc}.\frac{{a}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{24}\left(\mathrm{62}+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}{bc}\right) \\ $$$$\Rightarrow\left({bc}\right)^{\mathrm{2}} +\left(\mathrm{7}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right).{bc}−\mathrm{6}{a}^{\mathrm{2}} −\mathrm{744}=\mathrm{0} \\ $$$$….. \\ $$