Vector intersection

By Anonymous on Tuesday, February 27, 2001 - 09:31 pm :

A,B,&C are non colinear points in a plane.
C' is mid pt of AB
A' is mid pt of BC
B' is mid pt of CA
Show that the lines passing through A & A', B & B',
C & C' all meet at a point.

By Kerwin Hui (Kwkh2) on Tuesday, February 27, 2001 - 09:47 pm :

Let

O

be a fixed point, and let

a

b

c

denote the vectors

O A

O B

O C

respectively. Then,

vector $A A' = \frac{1}{2} (b + c) - a$

so equation of the line $A A'$ is

$r_{1} = a + λ_{1} (b + c - 2 a) = (1 - 2 λ_{1}) a + λ_{1} b + λ_{1} c$

Similar for $B B'$ , $C C'$ and check that $λ_{1} = λ_{2} = λ_{3} = 1 / 3$ gives the required point.

Alternatively we can get apurely geometric proof as follows: Join the lines $A B$ , $B C$ , $C A$ , $A A'$ , $B B'$ , $A' B'$ and let $G$ be the intersection of $A A'$ and $B B'$ . Then, by elementary geometry, we see that

$A G / G A' = BG / B G' = 2$ , hence result.

Kerwin