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

In summary, the given problem states that points P and Q divide a given interval AB in the ratio 1:λ, with point X lying on the circle with diameter PQ. The goal is to prove that the ratio of AX to XB is 1:λ. Through algebraic manipulation and simplification, it is determined that the only solution is λ=2. However, the question presumes that the ratio should hold for all values of λ, leading to the realization that there may be a mistake in the reasoning.
  • #1
The_Doctor
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0

Homework Statement


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

Homework Equations


None

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:
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  • #2
Don't worry I've realized my mistake.
 

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

1. How do you prove the ratio of AX:XB = 1:λ?

The ratio of AX:XB = 1:λ can be proved by using the theorem of intersecting chords. This theorem states that if two chords intersect within a circle, the product of the segments of one chord is equal to the product of the segments of the other chord. By applying this theorem and the given ratio, we can prove that AX:XB = 1:λ.

2. What is the significance of X being a point on a circle in this ratio?

The fact that X is a point on a circle is important because it allows us to use the properties of circles, such as the theorem of intersecting chords, to prove the given ratio. Without this information, we would not be able to accurately determine the ratio.

3. Can this ratio be proven for any point X on the circle?

Yes, this ratio can be proven for any point X on the circle. As long as X is a point on the circle, the theorem of intersecting chords can be applied to prove the ratio of AX:XB = 1:λ.

4. How does the value of λ affect the ratio of AX:XB?

The value of λ directly affects the ratio of AX:XB. As the value of λ changes, the ratio will also change accordingly. For example, if λ is equal to 2, the ratio of AX:XB will be 1:2. If λ is equal to 3, the ratio will be 1:3. This shows that the ratio is directly proportional to the value of λ.

5. Can this ratio be applied to other shapes besides circles?

No, this ratio is specific to points on a circle. It utilizes the properties of circles, such as intersecting chords, to prove the ratio. Other shapes may have different properties and theorems that would need to be applied to prove a similar ratio.

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