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Thread: Coordinate geometry problem

  1. #1
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    Question 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...
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  2. #2
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    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$.
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  3. #3
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    Thumbs up Re: Coordinate geometry problem

    Thank you so much!
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