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Question-217358

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Question Number 217358 by mnjuly1970 last updated on 11/Mar/25
Answered by mr W last updated on 12/Mar/25
let Δ=area of triangle ΔABC we have Δ=((√(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )−(a^4 +b^4 +c^4 )))/4) Δ=((ca sin B)/2)=((ab sin C)/2)=((bc sin A)/2) ((xc sin ω)/2)+((ya sin ω)/2)+((zb sin ω)/2)=Δ ⇒(xc+ya+zb)sin ω=2Δ ...(i) y^2 =x^2 +c^2 −2xc cos ω z^2 =y^2 +a^2 −2ya cos ω x^2 =z^2 +b^2 −2zb cos ω ⇒(xc+ya+zb)cos ω=((a^2 +b^2 +c^2 )/2) ...(ii) (i)/(ii): ⇒tan ω=((4Δ)/(a^2 +b^2 +c^2 )) sin ω=((4Δ)/( (√((4Δ)^2 +(a^2 +b^2 +c^2 )^2 )))) =((4Δ)/( (√(2(a^2 b^2 +b^2 c^2 +c^2 a^2 )−(a^4 +b^4 +c^4 )+a^4 +b^4 +c^4 +2(a^2 b^2 +b^2 c^2 +c^2 a^2 ))))) =((2Δ)/( (√(a^2 b^2 +b^2 c^2 +c^2 a^2 )))) ⇒((sin ω)/(2Δ))=(1/( (√(a^2 b^2 +b^2 c^2 +c^2 a^2 )))) in ΔPAB we have (y/(sin ω))=(c/(sin (ω+B−ω)))=(c/(sin B))=((c^2 a)/(ca sin B))=((c^2 a)/(2Δ)) ⇒y=((c^2 a sin ω)/(2Δ))=((c^2 a)/( (√(a^2 b^2 +b^2 c^2 +c^2 a^2 )))) similarly z=((a^2 b sin ω)/(2Δ))=((a^2 b)/( (√(a^2 b^2 +b^2 c^2 +c^2 a^2 )))) x=((b^2 c sin ω)/(2Δ))=((b^2 c)/( (√(a^2 b^2 +b^2 c^2 +c^2 a^2 )))) ⇒x+y+z=((a^2 b+b^2 c+c^2 a)/( (√(a^2 b^2 +b^2 c^2 +c^2 a^2 )))) ✓
$${let}\:\Delta={area}\:{of}\:{triangle}\:\Delta{ABC} \\ $$$${we}\:{have} \\ $$$$\Delta=\frac{\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}}{\mathrm{4}} \\ $$$$\Delta=\frac{{ca}\:\mathrm{sin}\:{B}}{\mathrm{2}}=\frac{{ab}\:\mathrm{sin}\:{C}}{\mathrm{2}}=\frac{{bc}\:\mathrm{sin}\:{A}}{\mathrm{2}} \\ $$$$ \\ $$$$\frac{{xc}\:\mathrm{sin}\:\omega}{\mathrm{2}}+\frac{{ya}\:\mathrm{sin}\:\omega}{\mathrm{2}}+\frac{{zb}\:\mathrm{sin}\:\omega}{\mathrm{2}}=\Delta \\ $$$$\Rightarrow\left({xc}+{ya}+{zb}\right)\mathrm{sin}\:\omega=\mathrm{2}\Delta\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${y}^{\mathrm{2}} ={x}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{xc}\:\mathrm{cos}\:\omega \\ $$$${z}^{\mathrm{2}} ={y}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ya}\:\mathrm{cos}\:\omega \\ $$$${x}^{\mathrm{2}} ={z}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{zb}\:\mathrm{cos}\:\omega \\ $$$$\Rightarrow\left({xc}+{ya}+{zb}\right)\mathrm{cos}\:\omega=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\Rightarrow\mathrm{tan}\:\omega=\frac{\mathrm{4}\Delta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\mathrm{sin}\:\omega=\frac{\mathrm{4}\Delta}{\:\sqrt{\left(\mathrm{4}\Delta\right)^{\mathrm{2}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}\Delta}{\:\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)+{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{2}\Delta}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\omega}{\mathrm{2}\Delta}=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} }} \\ $$$$ \\ $$$${in}\:\Delta{PAB}\:{we}\:{have} \\ $$$$\frac{{y}}{\mathrm{sin}\:\omega}=\frac{{c}}{\mathrm{sin}\:\left(\omega+{B}−\omega\right)}=\frac{{c}}{\mathrm{sin}\:{B}}=\frac{{c}^{\mathrm{2}} {a}}{{ca}\:\mathrm{sin}\:{B}}=\frac{{c}^{\mathrm{2}} {a}}{\mathrm{2}\Delta} \\ $$$$\Rightarrow{y}=\frac{{c}^{\mathrm{2}} {a}\:\mathrm{sin}\:\omega}{\mathrm{2}\Delta}=\frac{{c}^{\mathrm{2}} {a}}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} }} \\ $$$${similarly} \\ $$$${z}=\frac{{a}^{\mathrm{2}} {b}\:\mathrm{sin}\:\omega}{\mathrm{2}\Delta}=\frac{{a}^{\mathrm{2}} {b}}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} }} \\ $$$${x}=\frac{{b}^{\mathrm{2}} {c}\:\mathrm{sin}\:\omega}{\mathrm{2}\Delta}=\frac{{b}^{\mathrm{2}} {c}}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} }} \\ $$$$\Rightarrow{x}+{y}+{z}=\frac{{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {c}+{c}^{\mathrm{2}} {a}}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} }}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 12/Mar/25
$$\:\: \\ $$

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