Let-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-such-that-R-is-the-centroid-of-some-triangle-

Question Number 6564 by Yozzii last updated on 03/Jul/16

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} \dots \dots \dots \dots \dots \dots \dots \dots . (i)

q : y = m_{q} x \Rightarrow - 2 - b_{2} = - m_{q} b_{1} \dots \dots \dots \dots \dots (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}}) \dots \dots \dots \dots \dots \dots . (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}}) \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots (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

T h a n k s! N i c e a p p r o a c h .

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