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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

$G i v e n a t r i a n g l e Δ A B C w i t h s i d e s$ $a, b, c .$ $F i n d t h e a r e a o f t h e i n s c r i b e d t r i a n g l e$ $i n t e r m s o f a, b, c, θ .$
	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.$
	$w e c a n s e e t h a t$ $Δ Q R P \sim Δ A B C$ $\Rightarrow \frac{p}{c} = \frac{q}{a} = \frac{r}{b} = k, s a y$ $\frac{B P}{\sin (B + θ)} = \frac{R P}{\sin B}$ $\Rightarrow B P = \frac{k a \sin (B + θ)}{\sin B}$ $\frac{P C}{\sin θ} = \frac{P Q}{\sin C}$ $\Rightarrow P C = \frac{k b \sin θ}{\sin C}$ $a = \frac{k a \sin (B + θ)}{\sin B} + \frac{k b \sin θ}{\sin C}$ $\Rightarrow a \times \frac{\sin B - k \sin (B + θ)}{\sin B} = \frac{k b \sin θ}{\sin C} (i)$ $s i m i l a r l y$ $\Rightarrow b \times \frac{\sin C - k \sin (C + θ)}{\sin C} = \frac{k c \sin θ}{\sin A} (i i)$ $\Rightarrow c \times \frac{\sin A - k \sin (A + θ)}{\sin A} = \frac{k a \sin θ}{\sin B} (i i i)$ $(i) \times (i i) \times (i i i) :$ $[\sin A - k \sin (A + θ)] [\sin B - k \sin (B + θ)] [\sin C - k \sin (C + θ)] = \sin^{3} θ k^{3}$ $[\sin^{3} θ + \sin (A + θ) \sin (B + θ) \sin (C + θ)] k^{3}$ $- [\sin A \sin (B + θ) \sin (C + θ) + \sin B \sin (C + θ) \sin (A + θ) + \sin C \sin (A + θ) \sin (B + θ)] k^{2}$ $+ [\sin A \sin B \sin (C + θ) + \sin (A + θ) \sin B \sin C + \sin A \sin B \sin (C + θ)] k$ $- \sin A \sin B \sin C = 0$ $\sin^{3} θ + \sin (A + θ) \sin (B + θ) \sin (C + θ)$ $= \sin^{3} θ + (\sin A \cos θ + \cos A \sin θ) (\sin B \cos θ + \cos B \sin θ) (\sin C \cos θ + \cos C \sin θ)$ $= \sin^{3} θ + [\sin A \sin B - \cos C \sin^{2} θ + \sin C \sin θ \cos θ] (\sin C \cos θ + \cos C \sin θ)$ $= \sin^{3} θ + \sin A \sin B \sin C \cos θ - \sin^{3} θ + (\cos A \cos B \cos C + 1) \sin θ$ $= \sin A \sin B \sin C \cos θ + (\cos A \cos B \cos C + 1) \sin θ$ $\sin A \sin (B + θ) \sin (C + θ)$ $= \sin A (\sin B \cos θ + \cos B \sin θ) (\sin C \cos θ + \cos C \sin θ)$ $= \sin A (\sin B \sin C \cos^{2} θ + \sin A \sin θ \cos θ + \cos B \cos C \sin^{2} θ)$ $= \sin A \sin B \sin C \cos^{2} θ + \sin^{2} A \sin θ \cos θ + \frac{1}{4} (\sin 2 A + \sin 2 B - \sin 2 C) \sin^{2} θ$ $Σ = 3 \sin A \sin B \sin C \cos^{2} θ + (\sin^{2} A + \sin^{2} B + \sin^{2} C) \sin θ \cos θ$ $+ \frac{3}{4} (\sin 2 A + \sin 2 B + \sin 2 C) \sin^{2} θ - \frac{1}{2} (\sin 2 A + \sin 2 B + \sin 2 C) \sin^{2} θ$ $= \sin A \sin B \sin C (\cos 2 θ + 2) + (\cos A \cos B \cos C + 1) \sin 2 θ$ $\sin A \sin B \sin (C + θ)$ $\sin A \sin B (\sin C \cos θ + \cos C \sin θ)$ $\sin A \sin B \sin C \cos θ + \frac{1}{2} (\sin^{2} A + \sin^{2} B + \sin^{2} C) \sin θ - \sin^{2} C \sin θ$ $Σ = 3 \sin A \sin B \sin C \cos θ + (\cos A \cos B \cos C + 1) \sin θ$ ${\sin A \sin B \sin C \cos θ + (\cos A \cos B \cos C + 1) \sin θ} k^{3}$ $- {\sin A \sin B \sin C (\cos 2 θ + 2) + (\cos A \cos B \cos C + 1) \sin 2 θ} k^{2}$ $+ {3 \sin A \sin B \sin C \cos θ + (\cos A \cos B \cos C + 1) \sin θ} k$ $- \sin A \sin B \sin C = 0$ $o r$ $l e t λ = \frac{\cos A \cos B \cos C + 1}{\sin A \sin B \sin C} = \frac{a^{2} + b^{2} + c^{2}}{4 Δ_{A B C}}$ $(\cos θ + λ \sin θ) k^{3} - (2 + \cos 2 θ + λ \sin 2 θ) k^{2}$ $+ (3 \cos θ + λ \sin θ) k - 1 = 0$ $\Rightarrow \begin{matrix} k = \frac{1}{\cos θ + λ \sin θ} \end{matrix}$ $(w o w! i {d i d n}^{'} t e x p e c t t h a t t h e r o o t$ $o f t h i s c o m p l i c a t e d c u b i c e q u a t i o n i s$ $s o b r i e f!)$ $Δ_{P Q R} = k^{2} Δ_{A B C}$ $Δ_{P Q R} = \frac{Δ_{A B C}}{{(\cos θ + λ \sin θ)}^{2}}$ $s p e c i a l c a s e : θ = \frac{π}{2}$ $\Rightarrow k = \frac{1}{λ} = \frac{\sin A \sin B \sin C}{\cos A \cos B \cos C + 1}$ $t h i s i s t h e s a m e a s w e g e t i n Q 169815 .$
	Commented by Tawa11 last updated on 08/Oct/22

	$Great sir$
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