Thread: Coordinate geometry problem

1. Coordinate geometry problem

This is a question from an old University of London UK A-Level question. I have solved most of the question but remain stuck on the last part (the angle/centroid) part (highlighted in red).

Here is the question:

The position vectors of the points A, B, C with respect to the origin O are a, b, c respectively. If OA is perpendicular to BC, and OB is perpendicular to CA, show that OC is perpendicular to AB, and that OA2 + BC2 = OB2 + CA2 = OC2 + AB2.

Show that the plane through BC perpendicular to OA meets the plane through AB perpendicular to OC in a line that lies in the plane through OB perpendicular to CA.

If this line passes through the centroid of the triangle AOC, show that the angle AOC is pi/3 radians.

Here is my solution to the first (blue) parts of the question. Help with the final part greatly appreciated!

OA is perpendicular to BC, so a.(c - b) = 0
OB is perpendicular to CA, so b.(a - c) = 0
From these equations, we have

a.c - a.b = 0
b.a - b.c = 0

Adding these two expressions gives

a.c - a.b + b.a - b.c = 0
a.c - b.c = 0
so c.(a - b) = 0

Hence, OC is perpendicular to AB.

OA2 + BC2 = a2 + (c - b)2 = a2 + b2 + c2 - 2b.c
OB2 + CA2 = b2 + (a - c)2 = a2 + b2 + c2 - 2a.c
OC2 + AB2 = c2 + (b - a)2 = a2 + b2 + c2 - 2b.a

But b.c = b.a, so OC2 + AB2 = OA2 + BC2
and a.c = a.b, so OB2 + CA2 = OC2 + AB2

Hence OA2 + BC2 = OB2 + CA2 = OC2 + AB2

The plane through BC perpendicular to OA has equation

r.a = b.a

and the plane through AB perpendicular to OC has equation

r.c = a.c

The line of intersection of these planes is perpendicular to both a and c, i.e. it is parallel to CA. We know that OB is perpendicular to CA, so the equation of the line of intersection is given by

r = b + t(c - a)

Then I grind to a halt...

2. Re: Coordinate geometry problem

Note that the above line passes through B (when t=0). If the line passes through the centroid (as well as B), since centroid of AOC lies on the same plane as O, A, and C (and B), centroid = B is held.
The center of AC is $(\boldsymbol{a}+\boldsymbol{c})/2$ and thus, the centroid is $\boldsymbol{b}=(\boldsymbol{a}+\boldsymbol{c})/3$ (centroid dissects median with 2:1 ratio).
Let $\angle OAC=\alpha$. Then $\cos \alpha = \frac{\boldsymbol{a}.\boldsymbol{c}}{ac}$.
Since $\boldsymbol{a}.\boldsymbol{c}=\boldsymbol{a} . \boldsymbol{b}$ (proved in the beginning),
$\cos \alpha = \frac{\boldsymbol{a}.\boldsymbol{c}}{ac} = \frac{\boldsymbol{a}.\boldsymbol{b}}{ac} = \frac{\boldsymbol{a}.(\boldsymbol{a}+\boldsymbol{c })}{3ac} = \frac{a}{3c} + \frac{\boldsymbol{a}.\boldsymbol{c}}{3ac} = \frac{a}{3c} + \frac{1}{3}\cos \alpha$ i.e. $\cos\alpha = \frac{a}{2c}$.
Similarly, using $\boldsymbol{a}.\boldsymbol{c}=\boldsymbol{b} . \boldsymbol{c}$, we can derive $\cos\alpha = \frac{c}{2a}$.
Multiply those two will ends up $\cos^2\alpha = \frac{1}{4}$, i.e. $\cos\alpha = 1/2$ since $\frac{a}{2c}>0$.

3. Re: Coordinate geometry problem

Thank you so much!