ABC is a triangle,whose one vertex is moving along a circular path. Discuss the nature of the path, along which the centroid of the triangle is moving.


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Question Number 6640 by Rasheed Soomro last updated on 07/Jul/16

$ABC i s a t r i a n g l e, w h o s e o n e v e r t e x$ $i s m o v i n g a l o n g a c i r c u l a r p a t h .$ $D i s c u s s t h e n a t u r e o f t h e p a t h, a l o n g$ $w h i c h t h e c e n t r o i d o f t h e t r i a n g l e$ $i s m o v i n g .$
Commented by Rasheed Soomro last updated on 08/Jul/16

$I f A i n a △ ABC i s a m o v i n g p o i n t, t h e n$ $\overset{―}{BC} a n d o b j e c t s t i e d w i t h i t w i l l b e s t a t i c .$
Commented by Yozzii last updated on 07/Jul/16

$C e n t r o i d O m o v e s i n a c i r c l e e i t h e r$ $o u t s i d e, o n o r i n s i d e t h e c i r c l e a l o n g$ $w h i c h o n e o f A, B o r C m o v e s .$ $L e t A O b e t h e l i n e j o i n i n g t h e v e r t e x$ $A a n d c e n t r o i d O o f △ A B C . L e t N b e$ $t h e c e n t r e o f t h e c i r c l e o f r a d i u s r$ $a l o n g w h i c h A m o v e s . S u p p o s e r >∣ O A ∣ .$ $L e t θ = ∠ O A N w h e r e 0 ⩽ θ ⩽ π . L e t θ = π .$ $T h e n, O, A, N a r e c o l l i n e a r a n d t h e p o i n t O$ $m o v e s i n a c i r c l e o f r a d i u s r + ∣ A O ∣, c e n t r e N .$ $G e n e r a l l y, N, A a n d O f o r m a t r i a n g l e$ $w h e r e ∣ N O ∣ i s t h e r a d i u s o f t h e c i r c u l a r$ $p a t h o f O s i n c e t h e e n t i r e t r i a n g l e A B C$ $r o t a t e s a b o u t N; a l l p r o p e r t i e s o f △ A B C$ $a r e c o n s e r v e d u n d e r t h e m o t i o n o f A .$
Commented by Yozzii last updated on 07/Jul/16

$S o, ∣ N O ∣= \sqrt{r^{2} + ∣ A O ∣^{2} - 2 r ∣ A O ∣ c o s θ} .$ $I f O l i e s o n t h e c i r c l e j u s t a s A m o v e s$ $a l o n g i t, ∣ N O ∣= r \Rightarrow c o s θ = \frac{∣ A O ∣}{2 r} .$ $θ = {c o s}^{- 1} \frac{∣ A O ∣}{2 r}$ $S i n c e ∣ c o s θ ∣⩽ 1 \forall θ \in [0, π], \Rightarrow∣ A O ∣⩽ 2 r o r r ⩾ \frac{1}{2} ∣ A O ∣$ $S o, i f t h e t r i a n g l e i s t o b e p o s i t i o n e d$ $s o t h a t b o t h A a n d t h e c e n t r o i d O m o v e$ $o n t h e s a m e c i r c l e o f r a d i u s r i t i s n e c e s s a r y t h a t$ $r ⩾ 0.5 ∣ A O ∣ a n d θ = {c o s}^{- 1} \frac{∣ A O ∣}{2 r} .$ $I f θ = 0 \Rightarrow∣ N O ∣= r - ∣ O A ∣ s o O m o v e s$ $a l o n g a c i r c l e i n s i d e t h e c i r c l e f o r A .$
Commented by Rasheed Soomro last updated on 08/Jul/16

$N i c e!$ $A c t u a l l y, I t h i n k, e v e r y p o i n t t i e d w i t h t r i a n g l e,$ $(w h i c h i s n o t o n t h e o p p o s i t e s i d e o f m o v i n g v e r t e x)$ $i s m o v i n g i n c i r c u l a r p a t h (w h e t h e r t h e p o i n t s$ $a r e o n t h e t r i a n g l e o r i n s i d e t h e t r i a n g l e o r$ $o u t s i d e t h e t r i a n g l e .$
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