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Question Number 15908 by mrW1 last updated on 15/Jun/17

Commented by mrW1 last updated on 15/Jun/17

$On each side of a triangle an equilateral$ $triangle is constructed . The centroids$ $of these triangles form a new triangle .$ $Prove that the new triangle is equilateral .$ $What is the length of its sides if the$ $original triangle has sides a, b, c ?$ $What is the area of the new triangle ?$
Commented by ajfour last updated on 16/Jun/17

$s i d e o f t h e e q u i l a t e r a l t r i a n g l e$ $f o r m e d i s :$ $s = \frac{1}{\sqrt{3}} \sqrt{a^{2} + b^{2} + 2 a b \cos (C - \frac{2 π}{3})}$ $t h e r e f o r e i t s a r e a i s :$ $A = \frac{1}{4 \sqrt{3}} [a^{2} + b^{2} + 2 a b \cos (C - \frac{2 π}{3})]$ $w h e r e C = \cos^{- 1} (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) .$ $p l e a s e c o n f i r m, S i r .$ $a n s w e r s e e m s s o m e w h a t m e s s y . .$
Commented by ajfour last updated on 16/Jun/17

$i s i t n o t c o r r e c t, s i r (m r W 1) ?$
Commented by mrW1 last updated on 16/Jun/17

$I try to go on with your result :$ $\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$ $\sin C = \frac{2 \sqrt{p^{'} (p^{'} - a) (p^{'} - b) (p^{'} - c)}}{ab}$ $with p^{'} = \frac{a + b + c}{2}$ $\cos (C - \frac{2 π}{3}) = \cos C \cos \frac{2 π}{3} + \sin C \sin \frac{2 π}{3}$ $= \frac{1}{2} (- \cos C + \sqrt{3} \sin C)$ $= \frac{1}{2} [- \frac{a^{2} + b^{2} - c^{2}}{2 a b} + 2 \sqrt{3} \frac{\sqrt{p^{'} (p^{'} - a) (p^{'} - b) (p^{'} - c)}}{ab}]$ $= \frac{1}{2 ab} [- \frac{a^{2} + b^{2} - c^{2}}{2} + 2 \sqrt{3} \sqrt{p^{'} (p^{'} - a) (p^{'} - b) (p^{'} - c)}]$ $s = \frac{1}{\sqrt{3}} \sqrt{a^{2} + b^{2} + 2 a b \cos (C - \frac{2 π}{3})}$ $= \frac{1}{\sqrt{3}} \sqrt{a^{2} + b^{2} - \frac{a^{2} + b^{2} - c^{2}}{2} + 2 \sqrt{3} \sqrt{p^{'} (p^{'} - a) (p^{'} - b) (p^{'} - c)}}$ $= \frac{1}{\sqrt{3}} \sqrt{\frac{a^{2} + b^{2} + c^{2}}{2} + 2 \sqrt{3} \sqrt{p^{'} (p^{'} - a) (p^{'} - b) (p^{'} - c)}}$ $= \sqrt{\frac{a^{2} + b^{2} + c^{2}}{6} + \frac{2 \sqrt{3}}{3} \sqrt{\frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{2 \times 2 \times 2 \times 2}}}$ $= \sqrt{\frac{a^{2} + b^{2} + c^{2} + \sqrt{3 (a + b + c) (b + c - a) (c + a - b) (a + b - c)}}{6}}$
Commented by mrW1 last updated on 16/Jun/17

$that means your answer is correct .$
Commented by mrW1 last updated on 16/Jun/17

$please show how you proceeded to$ $s = \frac{1}{\sqrt{3}} \sqrt{a^{2} + b^{2} + 2 a b \cos (C - \frac{2 π}{3})}$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17

$l e t : E^{'}, P^{'}, D^{'}, b e t h e 3 r d v e r t e x o f t r i a n g l e s \overset{}{:}$ ${C B E}^{'}, {A C P}^{'}, {A B D}^{'} . b e c a u s e o f t h i s$ $t r i a n g l e s a r e e q u i l a t e r a l, w e c a n$ $w r i t e f o r e a c h o f t h e m :$ $(ω^{2} + ω + 1 = 0, z : i s a c o m p l e x n u m b e r)$ $(\frac{1}{ω} = - (1 + ω), \frac{1}{ω^{2}} = - 1 - \frac{1}{ω} = - 1 + 1 + ω = ω)$ $Δ {B C E}^{'} : z_{B} + ω z_{E^{'}} + ω^{2} z_{C} = 0$ $Δ {A C P}^{'} : z_{A} + ω z_{C} + ω^{2} z_{P^{'^{'}}} = 0$ $Δ D^{'} B A : z_{D^{'}} + ω z_{B} + ω^{2} z_{A} = 0$ $z_{D} = \frac{1}{3} (z_{A} + z_{B} + z_{D^{'}}), c o r d i n a t e o f c e n t r o i e d .$ $n o w f o r t r i a n g l e : D E P w e c a n w r i t e :$ $z_{D} + ω z_{E} + ω^{2} z_{P} = \frac{1}{3} (z_{D^{'}} + z_{B} + z_{A}) + \frac{ω}{3} (z_{B} + z_{E^{'}} + z_{C}) +$ $\frac{ω^{2}}{3} (z_{A} + z_{C} + z_{P^{'}}) = \frac{1}{3} (z_{B} + ω z_{E^{'}} + ω^{2} z_{C}) +$ $+ \frac{1}{3} (z_{D^{'}} + ω z_{B} + ω^{2} z_{A}) + \frac{1}{3} (z_{A} + ω z_{C} + ω^{2} z_{P^{^{'}}}) =$ $= \frac{1}{3} (0 + 0 + 0) = 0 .$ $i . e : D \overset{Δ}{E P} i s e q u i l a t e r a l .$ $2) c o r d i n a t e o f c e n t r o i d o f : A \overset{Δ}{B C}$ $= \frac{1}{3} (z_{A} + z_{B} + z_{C})$ $f o r t r i a n g l e : D \overset{Δ}{E P} = \frac{1}{3} (z_{D} + z_{E} + z_{P}) =$ $= \frac{1}{3} (\frac{1}{3} (z_{D^{'}} + z_{B} + z_{A}) + \frac{1}{3} (z_{B} + z_{E^{'}} + z_{C}) + \frac{1}{3} (z_{A} + z_{C} + z_{P^{'}})) =$ $= \frac{1}{9} (2 z_{A} + 2 z_{B} + 2 z_{C} + z_{E^{'}} + z_{P^{'}} + z_{D^{'}}) =$ $= \frac{1}{9} (2 z_{A} + 2 z_{B} + 2 z_{C} - \frac{z_{B} + ω^{2} z_{C}}{ω} - \frac{z_{A} + ω z_{C}}{ω^{2}} - ω z_{B} - ω^{2} z_{A}) =$ $= \frac{1}{9} (2 z_{A} + 2 z_{B} + 2 z_{C} + (1 + ω) (z_{B} + ω^{2} z_{C}) - ω (z_{A} + ω z_{C}) - ω z_{B} - ω^{2} z_{A}) =$ $= \frac{1}{9} (2 z_{A} + 2 z_{B} + 2 z_{C} + z_{B} + ω^{2} z_{C} + ω z_{B} + ω^{3} z_{C} - ω z_{A} - ω^{2} z_{C} - ω z_{B} - ω^{2} z_{A}) =$ $= \frac{1}{9} (3 z_{A} + 3 z_{B} + 3 z_{C}) = \frac{1}{3} (z_{A} + z_{B} + z_{C}) .$ $i t m e a n s t h a t : t r i a n g l e s : A \overset{Δ}{B C} a n d D \overset{Δ}{E P}$ $h a v e t h e s a m e c e n t r o i e d p o i n t .$
	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17

	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17

	$d r a w o u t c i r c l e s o f c o n s t r a c t e d t r i a n g l e s .$ $l e t : O b e t h e a n o t h e r i n t e r s e c t i o n o f 2 c i r c l e s$ $w i t h A B & A C c h o r d s, r e s p e c t p o i n t : A .$ $w e h a v e : ∠ A O B = ∠ A O C = 120^{∙} .$ $s o : ∠ B O C = 120^{∙} a l s o . i . e t h e c i r c l e w i t h$ $c h o r d : B C, p a s s e s f r o m p o i n t O .$ $b u t X Y i s p e r p e n d i c u l a r t o c o m m o n$ $c h o r d : O C, a n d a l s o : X Z i s p e r p e n d i c u l a r$ $t o : O B . w e p r o v e t h a t : ∠ B O C = 120^{∙}$ $a n d i t m e a n s t h a t : ∠ X = 60^{∙} .$ $s i m i l a r l y w e c a n p r o v e t h a t : ∠ Y = ∠ Z = 60$ $i . e : X \overset{Δ}{Y Z} i s e q u i l a t e r a l . ◼$
	Commented by mrW1 last updated on 19/Jun/17

	$Very nice sir!$
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