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Prove the ratio of AX:XB = 1:λ if X is a point on a circle

  1. Jun 10, 2012 #1
    1. The problem statement, all variables and given/known data
    The points P and Q divide a given interval AB internally and externally respectively in the ratio 1:λ. The point X lies on the circle with diameter PQ. Prove that AX:XB=1:λ
    2. Relevant equations
    None
    3. The attempt at a solution
    Basically, if we define the centre of PQ and the circle as O, and if Y is on PQ such that XY is perpendicular to PQ, we see that AX^2=AY^2+XY^2 and BX^2=BY^2+XY^2.

    AY=OA-OY, OA+OY, OY-OA and BY = OB-OY, OB+OY
    AY^2=(OA±OY)^2 and BY^2 = (OB±OY)^2

    depending on where Y is. We can define all lengths in terms of λ and AB. I can't be bothered to go into the details, but basically OA=[itex]\frac{\lambda^2-2}{2(\lambda+1)}AB[/itex] and OB=[itex]\frac{\lambda(\lambda+2)}{2(\lambda+1)}AB[/itex] and the radius =[itex]\frac{\lambda^2}{2(\lambda+1)}AB[/itex]

    OY^2+XY^2=radius^2. If we sub AY^2 and BY^2 into AX^2=AY^2+XY^2 and BX^2=BY^2+XY^2, expand and simplify, we get

    [itex]BX^2=AB\left(\frac{\lambda^2(\lambda+2)^2+\lambda^4}{4(\lambda+1)^2}AB\pm\frac{\lambda(\lambda+2)}{ \lambda +1}OY\right)[/itex]
    and
    [itex]AX^2=AB\left(\frac{(\lambda^2-2)^2+\lambda^4}{4(\lambda+1)^2}AB\pm\frac{\lambda^2-2}{ \lambda +1}OY\right)[/itex]

    Since OY is variable, we cannot combine AB and OY. Let's just consider when AX^2 will be in ratio with BY^2. It will only happen when [itex]\frac{\lambda^2(\lambda+2)^2+\lambda^4}{4(\lambda+1)^2}AB:\frac{(\lambda^2-2)^2+\lambda^4}{4(\lambda+1)^2}AB=\frac{\lambda( \lambda +2)}{ \lambda +1}OY:\frac{\lambda^2-2}{ \lambda +1}OY[/itex], because we cannot combine AB and OY, the coefficients must be in the same ratio in order for when AX^2 is divided by BY^2 for the fraction to cancel out, leaving the ratio. Thus,
    [itex]\frac{\lambda^2(\lambda+2)^2+\lambda^4}{(\lambda^2-2)^2+\lambda^4}=\frac{\lambda( \lambda +2)}{\lambda^2-2}[/itex].

    When we simplify this, we get [itex]( \lambda +1)^2( \lambda -2)=0[/itex] leaving the solution λ=-1, 2. λ=-1 isn't a solution, because if we sub this into the equations above, we are dividing by 0, so the only solution is λ=2. However, in the question, there is the presumption that AX:XB should be in ratio for all λ, because we have to prove that. Am I wrong? I'm fairly sure the question isn't, so I must be wrong! The only thing I'm dubious about is that the fractions have to be in ratio or maybe I made a silly mistake within the masses of algebra.

    Note: This question came from The Year 12 3 Unit Cambridge Textbook for Australia, if you're wondering
     
    Last edited: Jun 10, 2012
  2. jcsd
  3. Jun 11, 2012 #2
    Don't worry I've realised my mistake.
     
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