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Question Number 58512 by ajfour last updated on 24/Apr/19

Commented by ajfour last updated on 24/Apr/19

$Find \frac{Area (△ PQR)}{Area (△ ABC)} in terms of$ $a, b, c .$
	Answered by mr W last updated on 24/Apr/19
	$let α=AR=AQ β=BP=BR γ=CP=CQ we have β+γ=a γ+α=b α+β=c ⇒α+β+γ=((a+b+c)/2)=s ⇒α=s−a=((−a+b+c)/2) ⇒β=s−b=((a−b+c)/2) ⇒γ=s−c=((a+b−c)/2) Δ_(ARQ) =((α^2 sin A)/2)=(α^2 /(bc))×((bc sin A)/2)=(α^2 /(bc))Δ_(ABC) =a(−a+b+c)^2 ×(Δ_(ABC) /(4abc)) similarly: Δ_(BPR) =b(a−b+c)^2 ×(Δ_(ABC) /(4abc)) Δ_(CQP) =c(a+b−c)^2 ×(Δ_(ABC) /(4abc)) Δ_(PQR) =Δ_(ABC) −{a(−a+b+c)^2 +b(a−b+c)^2 +c(a+b−c)^2 }×(Δ_(ABC) /(4abc)) Δ_(PQR) =Δ_(ABC) −{a(a+b+c−2a)^2 +b(a+b+c−2b)^2 +c(a+b+c−2c)^2 }×(Δ_(ABC) /(4abc)) Δ_(PQR) =Δ_(ABC) −{(a+b+c)^3 −4(a^2 +b^2 +c^2 )(a+b+c)+4(a^3 +b^3 +c^3 )}×(Δ_(ABC) /(4abc)) ⇒(Δ_(PQR) /Δ_(ABC) )=1−(((a+b+c)^3 −4(a^2 +b^2 +c^2 )(a+b+c)+4(a^3 +b^3 +c^3 ))/(4abc)) if a=b=c: (Δ_(PQR) /Δ_(ABC) )=1−((3^3 −4×3×3+4×3)/4)=1−(3/4)=(1/4)$
	$l e t$ $α = A R = A Q$ $β = B P = B R$ $γ = C P = C Q$ $w e h a v e$ $β + γ = a$ $γ + α = b$ $α + β = c$ $\Rightarrow α + β + γ = \frac{a + b + c}{2} = s$ $\Rightarrow α = s - a = \frac{- a + b + c}{2}$ $\Rightarrow β = s - b = \frac{a - b + c}{2}$ $\Rightarrow γ = s - c = \frac{a + b - c}{2}$ $Δ_{A R Q} = \frac{α^{2} \sin A}{2} = \frac{α^{2}}{b c} \times \frac{b c \sin A}{2} = \frac{α^{2}}{b c} Δ_{A B C} = a {(- a + b + c)}^{2} \times \frac{Δ_{A B C}}{4 a b c}$ $s i m i l a r l y :$ $Δ_{B P R} = b {(a - b + c)}^{2} \times \frac{Δ_{A B C}}{4 a b c}$ $Δ_{C Q P} = c {(a + b - c)}^{2} \times \frac{Δ_{A B C}}{4 a b c}$ $Δ_{P Q R} = Δ_{A B C} - {a {(- a + b + c)}^{2} + b {(a - b + c)}^{2} + c {(a + b - c)}^{2}} \times \frac{Δ_{A B C}}{4 a b c}$ $Δ_{P Q R} = Δ_{A B C} - {a {(a + b + c - 2 a)}^{2} + b {(a + b + c - 2 b)}^{2} + c {(a + b + c - 2 c)}^{2}} \times \frac{Δ_{A B C}}{4 a b c}$ $Δ_{P Q R} = Δ_{A B C} - {{(a + b + c)}^{3} - 4 (a^{2} + b^{2} + c^{2}) (a + b + c) + 4 (a^{3} + b^{3} + c^{3})} \times \frac{Δ_{A B C}}{4 a b c}$ $\Rightarrow \frac{Δ_{P Q R}}{Δ_{A B C}} = 1 - \frac{{(a + b + c)}^{3} - 4 (a^{2} + b^{2} + c^{2}) (a + b + c) + 4 (a^{3} + b^{3} + c^{3})}{4 a b c}$ $i f a = b = c :$ $\frac{Δ_{P Q R}}{Δ_{A B C}} = 1 - \frac{3^{3} - 4 \times 3 \times 3 + 4 \times 3}{4} = 1 - \frac{3}{4} = \frac{1}{4}$
	Commented by ajfour last updated on 25/Apr/19

	$Thank you Sir, excellent! Sir .$ $I m busy teaching class 8 - 12 in a$ $school and mis my time with the$ $forum; those were the days my Sir,$ $i thought {they}^{'} d never end . .$
	Commented by mr W last updated on 26/Apr/19

	$d e a r s i r : i c a n u n d e r s t a n d y o u v e r y$ $w e l l . t h e r e a l l i f e i s o u t s i d e o f f t h e$ $f o r u m, {i t}^{'} s m o r e i m p o r t a n t! i w i s h$ $y o u a l l t h e b e s t i n i t . w h e n y o u f i n d$ $t i m e, j u s t c o m e b a c k a n d l e t u s a d m i r e$ $y o u r q u e s t i o n s a n d a n s w e r s .$
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