Let φ={ l_1 , l_2 , l_3 } be a set of three, nonparallel lines that all meet at one point R in the plane. Investigate conditions on φ such that R is the centroid of some triangle.


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Question Number 6564 by Yozzii last updated on 03/Jul/16
$Let φ={ l_1 , l_2 , l_3 } be a set of three, nonparallel lines that all meet at one point R in the plane. Investigate conditions on φ such that R is the centroid of some triangle.$
$L e t ϕ = {l_{1}, l_{2}, l_{3}} b e a s e t o f t h r e e, n o n p a r a l l e l$ $l i n e s t h a t a l l m e e t a t o n e p o i n t R i n$ $t h e p l a n e . I n v e s t i g a t e c o n d i t i o n s o n ϕ$ $s u c h t h a t R i s t h e c e n t r o i d o f s o m e t r i a n g l e .$
Commented by Yozzii last updated on 03/Jul/16

$E q u i v a l e n t l y, p r o v e / d i s p r o v e t h a t$ $t h e r e e x i s t s a t r i a n g l e w h o s e c e n t r o i d$ $i s t h a t o f t h e p o i n t o f i n t e r s e c t i o n$ $o f a n y t h r e e c o n c u r r e n t n o n p a r a l l e l l i n e s$ $i n t h e p l a n e .$
Commented by Rasheed Soomro last updated on 03/Jul/16

$N o t r e l a t e d t o a b o v e$ $S h a r i n g : S e e intmath . com i f y o u {h a v e n}^{'} t$ $s e e n a l r e a d y . S u b s c r i b e f o r i n t m a t h n e w s l e t t e r .$ $P e r h a p s t h e r e i s s o m e t h i n g o f i n t e r s t f o r y o u .$
Commented by Yozzii last updated on 03/Jul/16

$I n t e r e s t i n g w e b s i t e . T h a n k s!$
Commented by Rasheed Soomro last updated on 05/Jul/16

Commented by Rasheed Soomro last updated on 05/Jul/16

$A m e d i a n i s d i v i d e d a t t h e c e n t r o i d i n$ $r a t i o 2 : 1 f r o m t h e v e r t e x o f t h e t r i a n g l e .$ $p, q a n d r a r e t h r e e g i v e n c o n c u r r e n t l i n e s$ $c u t t i n g a t t h e p o i n t O .$ $t a k e A a n d D o n a n y o f t h r e e l i n e s (s a y p h e r e)$ $i n s u c h a w a y t h a t$ $∣ AO ∣ : ∣ OD ∣ = 2 : 1$ $N o w i f A i s a v e r t e x o f t h e r e q u i r e d t r i a n g l e a n d D i s$ $m i d p o i n t o f i t s (A^{'} s) o p p o s i t e s i d e, t h e n AD i s$ $a m e d i a n o f t h e t r i a n g l e .$ $W e h a v e t o p r o v e n o w t h a t t h e r e e x i s t p o i n t s B a n d C$ $o n l i n e s r a n d q, s u c h t h a t D i s m i d p o i n t o f BC$ $i - e ∣ BD ∣=∣ DC ∣ a n d B, D, C a r e c o l l i n e a r .$ $c o n t i n u e$
	Answered by Rasheed Soomro last updated on 06/Jul/16

	Commented by Rasheed Soomro last updated on 07/Jul/16

	$A n a l y t i c a l A p p r o a c h$ $L e t o n e o f t h r e e g i v e n c o n c u r r e n t l i n e s (n a m e d p)$ $i s t a k e n a s y - a x i s a n d t h e p o i n t o f t h e i r c o n c u r r e n c y$ $i s t a k e n a s o r i g i n o f c o o r d i a t e s y s t e m .$ $S i n c e p i s a l o n g y - a x i s i t s e q u a t i o n i s$ $p : x = 0$ $A l s o q a n d r p a s s t h r o u g h O (0, 0) t h e i r e q u a t o n s$ $w i l l b e o f t b e f o r m y = m x$ $L e t q : y = m_{q} x a n d r : y = m_{r} x$ $[m_{q} a n d m_{r} a r e s l o p e s]$ $L e t a t r i a n g l e ABC i s s u c h t h a t$ $(i) A a n d D a r e o n l i n e p (y - a x i s)$ $(i i) ∣ AO ∣ : ∣ OD ∣ = 2 : 1$ $(i i i) D i s m i d p o i n t o f BC$ $∴ AD i s a m e d i a n o f t h e t r i a n g l e a n d O i s i t s$ $c e n t r o i d .$ $A s s u m e u n i t o f c o o r d i n a t e s y s t e m e q u a l t o ∣ OD ∣$ $t h e n i n t h e d i a g r a m a b o v e D = (0, - 1) a n d A = (0, 2)$ $L e t B = (b_{1}, b_{2}) a n d C = (c_{1}, c_{2})$ $∵ D (0, - 1) i s m i d p o i n t o f BC$ $∴ \frac{b_{1} + c_{1}}{2} = 0 \land \frac{b_{2} + c_{2}}{2} = - 1$ $O r c_{1} = - b_{1} \land c_{2} = - 2 - b_{2}$ $O r C = (c_{1}, c_{2}) = (- b_{1}, - 2 - b_{2})$ $I f B a n d C a r e o n g i v e n l i n e s r a n d q r e s p e c t i v e l y$ $T h e n c o o r d i n a t e s o f B a n d C w i l l s a t i s f y t h e e q u a t i o n s$ $o f r a n d q r e s p e c t i v e l y$ $r : y = m_{r} x \Rightarrow b_{2} = m_{r} b_{1} . . . . . . . . . . . . . . . . . . . . . . . . . (i)$ $q : y = m_{q} x \Rightarrow - 2 - b_{2} = - m_{q} b_{1} . . . . . . . . . . . . . . . (i i)$ $F r o m (i) a n d (i i)$ $- 2 = (m_{r} - m_{q}) b_{1} \Rightarrow b_{1} = \frac{- 2}{m_{r} - m_{q}}$ $b_{2} = m_{r} (\frac{- 2}{m_{r} - m_{q}}) = \frac{- 2 m_{r}}{m_{r} - m_{q}}$ $B = (b_{1}, b_{2}) = (\frac{- 2}{m_{r} - m_{q}}, \frac{- 2 m_{r}}{m_{r} - m_{q}}) . . . . . . . . . . . . . . . . . . . (i i i)$ $C = (- b_{1}, - 2 - b_{2}) = (\frac{2}{m_{r} - m_{q}}, \frac{2 m_{r}}{m_{r} - m_{q}} - 2)$ $C = (\frac{2}{m_{r} - m_{q}}, \frac{2 m_{q}}{m_{r} - m_{q}}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i v)$ $S o f r o m (i) a n d (i v)$ $i f m_{r} \neq m_{q}, B a n d C a r e o n r a n d q r e s p e c t i v e l y .$ $S o f i n a l l y I f a l l t h e t h r e e c o n c u r r e n t l i n e s h a v e$ $d i f f e r e n t s l o p e s t h e r e e x i s t a t r i a n g l e w h o s e$ $c e n t r o i d i s t h e p o i n t o f c o n c u r r e n c y o f g i v e n l i n e s$
	Commented by Yozzii last updated on 07/Jul/16

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