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Question Number 28303 by ajfour last updated on 24/Jan/18

If one line of the equation : ax^3 +bx^2 y+cxy^2 +dy^3 =0 bisects the angle between the the other two then prove (3a+c)^2 (bc+2cd−3ad)= (b+3d)^2 (bc+2ab−3ad) .

$${If}\:{one}\:{line}\:{of}\:{the}\:{equation}\:: \\ $$$$\boldsymbol{{ax}}^{\mathrm{3}} +\boldsymbol{{bx}}^{\mathrm{2}} \boldsymbol{{y}}+\boldsymbol{{cxy}}^{\mathrm{2}} +\boldsymbol{{dy}}^{\mathrm{3}} =\mathrm{0} \\ $$$${bisects}\:{the}\:{angle}\:{between}\:{the} \\ $$$${the}\:{other}\:{two}\:{then}\:{prove} \\ $$$$\left(\mathrm{3}{a}+{c}\right)^{\mathrm{2}} \left({bc}+\mathrm{2}{cd}−\mathrm{3}{ad}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}+\mathrm{3}{d}\right)^{\mathrm{2}} \left({bc}+\mathrm{2}{ab}−\mathrm{3}{ad}\right)\:. \\ $$

Answered by ajfour last updated on 24/Jan/18

let the lines be y=m_1 x, y=m_0 x, and y=m_2 x θ_0 =(((θ_1 +θ_2 )/2)) ⇒ ((2m_0 )/(1−m_0 ^2 ))=((m_1 +m_2 )/(1−m_1 m_2 )) ...(i) m_1 m_2 m_0 =−a/d ....(ii) (m_1 +m_2 )m_0 +m_1 m_2 =b/d ...(iii) m_1 +m_2 +m_0 =−c/d ....(iv) using (ii) and (iv) in (i) and (iii) to obtain two equations in m_0 : ((2m_0 )/(1−m_0 ^2 ))=−(((m_0 +(c/d)))/((1+(a/(m_0 d))))) ⇒ (2/(1−m_0 ^2 ))=− ((c+m_0 d)/(a+m_0 d)) .....(I) m_0 =−((((b/d)+(a/(m_0 d))))/((m_0 +(c/d)))) ⇒ m_0 ^2 =− ((a+m_0 b)/(c+m_0 d)) .....(II) (I)×(II) gives ((2m_0 ^2 )/(1−m_0 ^2 ))=((a+m_0 b)/(a+m_0 d)) adding 2 on bothsides (2/(1−m_0 ^2 ))=((3a+m_0 (b+2d))/(a+m_0 d)) .....(A) equating (I) with (A): (2/(1−m_0 ^2 ))=((3a+m_0 (b+2d))/(a+m_0 d))=−((c+m_0 d)/(a+m_0 d)) hence 3a+m_0 (b+2d)=−(c+m_0 d) m_0 =− ((3a+c)/(b+3d)) substituting in (II): (((3a+c)/(b+3d)))^2 =− ((a−(((3a+c)/(b+3d)))b)/(c−(((3a+c)/(b+3d)))d)) or (((3a+c)/(b+3d)))^2 =− ((3ad−2ab−bc)/(bc+2cd−3ad)) ⇒ (((3a+c)/(b+3d)))^2 = ((bc+2ab−3ad)/(bc+2cd−3ad)) .

$${let}\:{the}\:{lines}\:{be} \\ $$$${y}={m}_{\mathrm{1}} {x},\:{y}={m}_{\mathrm{0}} {x},\:{and}\:{y}={m}_{\mathrm{2}} {x} \\ $$$$\theta_{\mathrm{0}} =\left(\frac{\theta_{\mathrm{1}} +\theta_{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}{m}_{\mathrm{0}} }{\mathrm{1}−{m}_{\mathrm{0}} ^{\mathrm{2}} }=\frac{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }{\mathrm{1}−{m}_{\mathrm{1}} {m}_{\mathrm{2}} }\:\:\:\:...\left({i}\right) \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} {m}_{\mathrm{0}} =−{a}/{d}\:\:\:\:\:....\left({ii}\right) \\ $$$$\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right){m}_{\mathrm{0}} +{m}_{\mathrm{1}} {m}_{\mathrm{2}} ={b}/{d}\:\:\:...\left({iii}\right) \\ $$$${m}_{\mathrm{1}} +{m}_{\mathrm{2}} +{m}_{\mathrm{0}} =−{c}/{d}\:\:\:\:\:....\left({iv}\right) \\ $$$${using}\:\left({ii}\right)\:{and}\:\left({iv}\right)\:{in}\:\left({i}\right)\:{and}\:\left({iii}\right) \\ $$$${to}\:{obtain}\:{two}\:{equations}\:{in}\:{m}_{\mathrm{0}} : \\ $$$$\frac{\mathrm{2}{m}_{\mathrm{0}} }{\mathrm{1}−{m}_{\mathrm{0}} ^{\mathrm{2}} }=−\frac{\left({m}_{\mathrm{0}} +\frac{{c}}{{d}}\right)}{\left(\mathrm{1}+\frac{{a}}{{m}_{\mathrm{0}} {d}}\right)}\:\:\: \\ $$$$\Rightarrow\:\frac{\mathrm{2}}{\mathrm{1}−{m}_{\mathrm{0}} ^{\mathrm{2}} }=−\:\frac{{c}+{m}_{\mathrm{0}} {d}}{{a}+{m}_{\mathrm{0}} {d}}\:\:\:\:.....\left({I}\right) \\ $$$${m}_{\mathrm{0}} =−\frac{\left(\frac{{b}}{{d}}+\frac{{a}}{{m}_{\mathrm{0}} {d}}\right)}{\left({m}_{\mathrm{0}} +\frac{{c}}{{d}}\right)}\:\:\:\:\: \\ $$$$\Rightarrow\:\:\:\:{m}_{\mathrm{0}} ^{\mathrm{2}} =−\:\frac{{a}+{m}_{\mathrm{0}} {b}}{{c}+{m}_{\mathrm{0}} {d}}\:\:\:\:\:.....\left({II}\right) \\ $$$$\left({I}\right)×\left({II}\right)\:\:{gives} \\ $$$$\frac{\mathrm{2}{m}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{1}−{m}_{\mathrm{0}} ^{\mathrm{2}} }=\frac{{a}+{m}_{\mathrm{0}} {b}}{{a}+{m}_{\mathrm{0}} {d}} \\ $$$${adding}\:\mathrm{2}\:{on}\:{bothsides} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}−{m}_{\mathrm{0}} ^{\mathrm{2}} }=\frac{\mathrm{3}{a}+{m}_{\mathrm{0}} \left({b}+\mathrm{2}{d}\right)}{{a}+{m}_{\mathrm{0}} {d}}\:\:\:\:.....\left({A}\right) \\ $$$${equating}\:\left({I}\right)\:{with}\:\left({A}\right): \\ $$$$\frac{\mathrm{2}}{\mathrm{1}−\boldsymbol{{m}}_{\mathrm{0}} ^{\mathrm{2}} }=\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{m}}_{\mathrm{0}} \left(\boldsymbol{{b}}+\mathrm{2}\boldsymbol{{d}}\right)}{\boldsymbol{{a}}+\boldsymbol{{m}}_{\mathrm{0}} \boldsymbol{{d}}}=−\frac{\boldsymbol{{c}}+\boldsymbol{{m}}_{\mathrm{0}} \boldsymbol{{d}}}{\boldsymbol{{a}}+\boldsymbol{{m}}_{\mathrm{0}} \boldsymbol{{d}}} \\ $$$$\boldsymbol{{hence}} \\ $$$$\mathrm{3}{a}+{m}_{\mathrm{0}} \left({b}+\mathrm{2}{d}\right)=−\left({c}+{m}_{\mathrm{0}} {d}\right) \\ $$$$\boldsymbol{{m}}_{\mathrm{0}} =−\:\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}}{\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{d}}} \\ $$$${substituting}\:{in}\:\left({II}\right): \\ $$$$\left(\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}}{\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{d}}}\right)^{\mathrm{2}} =−\:\frac{\boldsymbol{{a}}−\left(\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}}{\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{d}}}\right)\boldsymbol{{b}}}{\boldsymbol{{c}}−\left(\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}}{\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{d}}}\right)\boldsymbol{{d}}} \\ $$$$\boldsymbol{{or}}\: \\ $$$$\left(\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}}{\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{d}}}\right)^{\mathrm{2}} =−\:\frac{\mathrm{3}\boldsymbol{{ad}}−\mathrm{2}\boldsymbol{{ab}}−\boldsymbol{{bc}}}{\boldsymbol{{bc}}+\mathrm{2}\boldsymbol{{cd}}−\mathrm{3}\boldsymbol{{ad}}}\: \\ $$$$\Rightarrow\:\:\left(\frac{\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}}{\boldsymbol{{b}}+\mathrm{3}\boldsymbol{{d}}}\right)^{\mathrm{2}} =\:\frac{\boldsymbol{{bc}}+\mathrm{2}\boldsymbol{{ab}}−\mathrm{3}\boldsymbol{{ad}}}{\boldsymbol{{bc}}+\mathrm{2}\boldsymbol{{cd}}−\mathrm{3}\boldsymbol{{ad}}}\:\:. \\ $$$$ \\ $$

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