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Question Number 170010 by mr W last updated on 13/May/22

Commented by mr W last updated on 13/May/22

Given a triangle ΔABC with sides  a,b,c.  Find the area of the inscribed triangle  in terms of a,b,c,θ.

GivenatriangleΔABCwithsidesa,b,c.Findtheareaoftheinscribedtriangleintermsofa,b,c,θ.

Answered by mr W last updated on 15/May/22

Commented by mr W last updated on 15/May/22

we can see that  ΔQRP∼ΔABC  ⇒(p/c)=(q/a)=(r/b)=k, say  ((BP)/(sin (B+θ)))=((RP)/(sin B))  ⇒BP=((ka sin (B+θ))/(sin B))  ((PC)/(sin θ))=((PQ)/(sin C))  ⇒PC=((kb sin θ)/(sin C))  a=((ka sin (B+θ))/(sin B))+((kb sin θ)/(sin C))  ⇒a×((sin B−k sin (B+θ))/(sin B))=((kb sin θ)/(sin C))   (i)  similarly  ⇒b×((sin C−k sin (C+θ))/(sin C))=((kc sin θ)/(sin A))   (ii)  ⇒c×((sin A−k sin (A+θ))/(sin A))=((ka sin θ)/(sin B))   (iii)  (i)×(ii)×(iii):  [sin A−k sin (A+θ)][sin B−k sin (B+θ)][sin C−k sin (C+θ)]=sin^3  θ k^3     [sin^3  θ+sin (A+θ) sin (B+θ) sin (C+θ)]k^3   −[sin A sin (B+θ) sin (C+θ)+sin B sin (C+θ) sin (A+θ)+sin C sin (A+θ) sin (B+θ)]k^2   +[sin A sin B sin (C+θ)+sin (A+θ) sin B sin C+sin A sin B sin (C+θ)]k  −sin A sin B sin C=0    sin^3  θ+sin (A+θ) sin (B+θ) sin (C+θ)  =sin^3  θ+(sin A cos θ+cos A sin θ)(sin B cos θ+cos B sin θ)(sin C cos θ+cos C sin θ)  =sin^3  θ+[sin A sin B−cos C sin^2  θ+sin C sin θ cos θ](sin C cos θ+cos C sin θ)  =sin^3  θ+sin A sin B sin C cos θ− sin^3   θ+(cos A cos B cos C+1)sin θ   =sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ     sin A sin (B+θ) sin (C+θ)  =sin A (sin B cos θ+cos B sin θ)(sin C cos θ+cos C sin θ)  =sin A (sin B sin C cos^2  θ+sin A sin θ cos θ+cos B cos C sin^2  θ)  =sin A sin B sin C cos^2  θ+sin^2  A sin θ cos θ+(1/4)(sin 2A+sin 2B− sin 2C) sin^2  θ  Σ=3 sin A sin B sin C cos^2  θ+(sin^2  A+sin^2  B+sin^2  C)sin θ cos θ     +(3/4)(sin 2A+sin 2B+ sin 2C) sin^2  θ−(1/2)(sin 2A+sin 2B+sin 2C) sin^2  θ  =sin A sin B sin C (cos 2θ+2)+(cos A cos B cos  C+1) sin 2θ    sin A sin B sin (C+θ)  sin A sin B (sin C cos θ+cos C sin θ)  sin A sin B sin C cos θ+(1/2)(sin^2  A+sin^2  B+sin^2  C)sin θ−sin^2  C sin θ  Σ=3 sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ    {sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ}k^3    −{sin A sin B sin C (cos 2θ+2)+(cos A cos B cos  C+1) sin 2θ}k^2   +{3 sin A sin B sin C cos θ+(cos A cos B cos C+1) sin θ}k  −sin A sin B sin C=0  or  let λ=((cos A cos B cos C+1)/(sin A sin B sin C))=((a^2 +b^2 +c^2 )/(4Δ_(ABC) ))  (cos 𝛉+𝛌 sin 𝛉)k^3 −(2+cos 2𝛉+𝛌 sin 2𝛉)k^2   +(3 cos 𝛉+𝛌 sin 𝛉)k−1=0  ⇒ determinant (((k=(1/(cos 𝛉+𝛌 sin 𝛉)))))  (wow! i didn′t expect that the root   of this complicated cubic equation is   so brief!)    Δ_(PQR) =k^2 Δ_(ABC)   𝚫_(PQR) =(𝚫_(ABC) /((cos 𝛉+𝛌 sin 𝛉)^2 ))    special case: θ=(π/2)  ⇒k=(1/λ)=((sin A sin B sin C)/(cos A cos B cos C+1))  this is the same as we get in Q169815.

wecanseethatΔQRPΔABCpc=qa=rb=k,sayBPsin(B+θ)=RPsinBBP=kasin(B+θ)sinBPCsinθ=PQsinCPC=kbsinθsinCa=kasin(B+θ)sinB+kbsinθsinCa×sinBksin(B+θ)sinB=kbsinθsinC(i)similarlyb×sinCksin(C+θ)sinC=kcsinθsinA(ii)c×sinAksin(A+θ)sinA=kasinθsinB(iii)(i)×(ii)×(iii):[sinAksin(A+θ)][sinBksin(B+θ)][sinCksin(C+θ)]=sin3θk3[sin3θ+sin(A+θ)sin(B+θ)sin(C+θ)]k3[sinAsin(B+θ)sin(C+θ)+sinBsin(C+θ)sin(A+θ)+sinCsin(A+θ)sin(B+θ)]k2+[sinAsinBsin(C+θ)+sin(A+θ)sinBsinC+sinAsinBsin(C+θ)]ksinAsinBsinC=0sin3θ+sin(A+θ)sin(B+θ)sin(C+θ)=sin3θ+(sinAcosθ+cosAsinθ)(sinBcosθ+cosBsinθ)(sinCcosθ+cosCsinθ)=sin3θ+[sinAsinBcosCsin2θ+sinCsinθcosθ](sinCcosθ+cosCsinθ)=sin3θ+sinAsinBsinCcosθsin3θ+(cosAcosBcosC+1)sinθ=sinAsinBsinCcosθ+(cosAcosBcosC+1)sinθsinAsin(B+θ)sin(C+θ)=sinA(sinBcosθ+cosBsinθ)(sinCcosθ+cosCsinθ)=sinA(sinBsinCcos2θ+sinAsinθcosθ+cosBcosCsin2θ)=sinAsinBsinCcos2θ+sin2Asinθcosθ+14(sin2A+sin2Bsin2C)sin2θΣ=3sinAsinBsinCcos2θ+(sin2A+sin2B+sin2C)sinθcosθ+34(sin2A+sin2B+sin2C)sin2θ12(sin2A+sin2B+sin2C)sin2θ=sinAsinBsinC(cos2θ+2)+(cosAcosBcosC+1)sin2θsinAsinBsin(C+θ)sinAsinB(sinCcosθ+cosCsinθ)sinAsinBsinCcosθ+12(sin2A+sin2B+sin2C)sinθsin2CsinθΣ=3sinAsinBsinCcosθ+(cosAcosBcosC+1)sinθ{sinAsinBsinCcosθ+(cosAcosBcosC+1)sinθ}k3{sinAsinBsinC(cos2θ+2)+(cosAcosBcosC+1)sin2θ}k2+{3sinAsinBsinCcosθ+(cosAcosBcosC+1)sinθ}ksinAsinBsinC=0orletλ=cosAcosBcosC+1sinAsinBsinC=a2+b2+c24ΔABC(cosθ+λsinθ)k3(2+cos2θ+λsin2θ)k2+(3cosθ+λsinθ)k1=0k=1cosθ+λsinθ(wow!ididntexpectthattherootofthiscomplicatedcubicequationissobrief!)ΔPQR=k2ΔABCΔPQR=ΔABC(cosθ+λsinθ)2specialcase:θ=π2k=1λ=sinAsinBsinCcosAcosBcosC+1thisisthesameaswegetinQ169815.

Commented by Tawa11 last updated on 08/Oct/22

Great sir

Greatsir

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