How do you prove Archimedes' Arbelos theorem?

If someone can manage to prove it, I'd appreciate it greatly.

Thanks,

Brad

I've been told you can prove this theorem by repeated application of Pythagoras' Theorem, however I haven't been told how, and haven't had the chance to sit down and have a go yet. You might like to give that a try...

I've done a proof using Pythag; so here goes. You may want to draw your own diagram, or print out the one above and draw on it.

Apart from the labels on the diagram, I will use S₁ and S₂ to refer to the semi-circles on AB and BC respectively, C₂ to refer to the circle labelled C₁ ', and R₁ and R₂ for the radii of C₁ and C₂ . Note that r and 1-r on the diagram refer to diameters rather than radii!

Ok. Now from the base semi-circle, draw a perpendicular line that goes through the centre of the circle C₁ .
From the centre of the base of S₁ , draw a second line that goes through the centre of C₁ . This line will have length r/2 + R₁ . The distance between the two lines at the base is r/2 - R₁ . Pythag then gives you a length for the perpendicular line from the base to the centre of C₁ as sqrt(2rR₁ )

Doing the same for C₂ with respect to S₂ allows the perpendicular line from the base to the centre of C₂ to be calculated as sqrt(2(1-r)R₂ ).

Now from the centre of the base line of the large semi-circle draw a line to the centre of C₁ . The length of this line will be 1/2 - R₁ . Furthermore, the distance from the perpendicular line whose length has been calculated above is r - 1/2 - R₁ . We now have three sides of a triangle. Using Pythagoras, we get
(1/2 - R₁ )² = (r-1/2 - R₁ )² + 2rR₁
So multiplying out, etc.,
r² - r + 2R₁ = 0

Now do likewise for C₂ : the length of the line from the centre of the large semi-circle will be 1/2 - R₂ , and the distance to the perpendicular through its centre will be R₂ + r - 1/2. Using Pythagoras again,
(1/2 - R₂ )² = (R₂ + r - 1/2)² + 2(1-r)R₂
so
r² - r + 2R₂ = 0

So R₁ = R₂ .

So the circles are equal.

Tom.

I think I have a new method of attack on the problem.

At today's Algebra and Geometry lecture (our fourth) the lecturer, Dr Peter Haynes, talked about Mobius transformations. These are transformations on complex numbers z and are always of the form:

z -> (az+b)/(cz+d)

This means that a complex number z is replaced with the complex number (az+b)/(cz+d).

Then he considered the special case a = c = 1, b = d = 0.

z -> 1/z

If you recall complex arithmetic, this means a number re^ix will be replaced by 1/r e^-ix .

In other words, relative to axes and an origin, the point will be reflected in the x - axis and its length from the origin change to its reciprocal.

So if we have a set of points, we can apply the transformation to the set of points to get a new set of points. These may have interesting properties.

For example consider a circle of radius 1. What happens when z -> 1/z is applied? Very simply we get an identical circle (can you prove this?)

What's more, if you have any line or circle and you apply the transformation, you always get a circle, UNLESS THE ORIGINAL LINE OR CIRCLE PASSES THROUGH THE ORIGIN! (In which case you get a line.) The algebra is a bit tedious here, so get back to me if you can't prove this.

No applications of this were given in the lecture, but perhaps we could use it to solve your little problem. Try applying this transformation to your diagram, taking the origin as the intersection point of the two smaller circles. Apply the transformation to all the lines and circles there, and tell me what you get. I just want to check I've got the result right before continuing. Then we are well on the way to proving the theorem.

Yours,

Michael

I think you can forget about the e^ix changing to e^-ix - all this does is turn the diagram upside down which doesn't make any difference. It is easiest to just do the following transformation:

For every point P replace it with a point Q such that OPQ are on the same line, and OP * OQ = 1.

Obviously circles + lines are still going to go to circles (unless they pass the origin). By the way, I forgot to say how to find which circles/lines they become - I guess the best way to do this is to consider individual points on the original circle/line, and see where they go to. Then form the resultant circle/line from these.

OK, I now have a complete solution using the special case of the Mobius transformation. I will have a go at writing it out shortly because it shows the problem in a different light. Of course Tom's method is probably the quickest possible, but using the special case of the Mobius transformation (a = c = 1, b = d = 0) may be able to solve other geometrical puzzles easily. Actually today's lecture is apparently the only one mentioning the Mobius transformation for the time being, but perhaps later they come back and do other geometrical stuff with it. Dan?

Thank you all for your help. I would be interested to hear your proof, Michael, however for the problem at hand, Tom's works quite nice. Your proof seems very interesting though, and would be nice to know for experience on other problems. I'm not sure that I can prove the bit about evry line forming a circle however, this seems somewhat counterintuitive.

Brad

Hi Brad - sorry for the delay. Eventually I'm going to need another diagram to explain the other method, but we can start off by showing what happens to circles and lines when we perform the transformation such that:

For each point P in the original figure we replace it with P' so that OP×OP' = 1 and OPP' are all on the same line. (Note that we specify that O never lies between P and P'.)

(This is the special case of the Mobius transformation.)

Let's start off with a line, and represent it by a complex number. Nice and simple:

z = 1 + ai

So this is a horizontal line (on the Argand diagram you move right to 1, then vertically up or down by an arbitrary amount a). Now if we calculate 1/z we get:

1/(1+ai) = (1 - ai)/(1+a² )

So the y-value is -a/(1+a² ) and the x value is 1/(1+a² )

y = -a/(1+a² )
x = 1/(1+a² )

so y² + x² = x

and:

y² + (x-1/2)² = 1/4

which is indeed a circle (whose circumference, incidentally, passes through the origin).

Actually 1/z isn't quite what we want because this is not only going to change the modulus to the reciprocal but it is also going to negate the argument of z. However this will only reflect it in x = 0, so it is still a circle (and still passes through the origin!).

So we've shown that if you apply the transformation to a horizontal line through (1,0), you get a circle. Now you'll be able to see that if it was a horizontal line passing through (a,0) for non-zero a, then you still get a circle (the whole graph will be scaled up by a factor of a, so the graph after the transform will simply be scaled down by a factor of a, therefore we still have a circle). Also by rotational symmetry, you can see that even if the line isn't horizontal it still becomes a circle.

There's only one case we haven't dealt with - if the line passes through the origin (note that the scaling argument now fails). So we have to deal with this separately. But I think you'll be able to see that if you have a line through the origin then it will map itself to an identical line under the transformation. Except technically there shouldn't be a point at the origin in the transformed diagram at all as this corresponds to a point at infinity in the original diagram, but we can ignore this for now.

To summarise what happens to lines under the transformation: If they go through the origin then they map onto another line (exactly the same one in fact). If they don't go through the origin then they become circles.

Then we have to see what happens to circles under the transformation. I'll leave that for the time being - any problems so far?

Yours,

Michael

I've been doing some of my own calculations, and I think I see why it must form a circle now. Before, I was thinking that the point would be on the opposite side of the origin. But, it of course is not,and when on the same side of the origin, it works out to be a circle. So, no major problems right now.

Thanks,

Brad