More than 2000 years ago, Apollonius proposed this famous problem: given any three circles,
how can we construct another circle that is tangent to all three of them.
To Apollonius, there are only a couple of tools we can use, a straightedge and a compass.
A straightedge is like a ruler, but without any markings on it, so together with a pencil,
we can draw straight lines but not measure any distances. A compass, on the other hand,
allows you to draw a circle. For the purpose of this video, we are considering a non-collapsing
compass, which means you can lift it off the paper, drag it somewhere else, and
draw a circle with the same radius as before. So these are the only tools allowed when we proceed.
This problem of Apollonius looks easy, but you can try it and see that it isn’t
a trivial geometry problem. In fact, we will have two simplifications:
first, we want to only consider the case where the circles are in a so-called generic configuration,
which means there are no intersections between any two circles, and their interiors
don’t overlap. The other cases can be harder to tackle, and sometimes there isn’t even a solution.
The second simplification is that we only want to construct one circle tangent to
the three given ones, because in general, there would be more than 1 solutions to the problem,
but we just want one of them for this video. Apollonius himself actually solved this problem,
but his works were lost; but because this is such a famous problem, mathematicians developed
different solutions to the problem. This video will cover one of these solutions, but the more
important thing is the problem-solving technique demonstrated by the solution.
This solution starts off with a powerful tool in Euclidean geometry called circle inversion.
Suppose you have a circle, and you want to invert this blue point with respect to the circle,
then you will get another point. This point is not arbitrary, and it fulfils two properties:
firstly, the image A prime has to lie on the ray from the centre to A, secondly,
the product of the distances from the centre is the radius squared. If you invert a point
twice with respect to the same circle, you get the same point back. This is because
this property still holds, and so A is the inversion point of A prime.
One other thing to note is that inversion of a point can be done with straightedge and compasses.
If you are wondering, here is one way to do the construction. We first draw a perpendicular from A
to meet the circle at B, then draw a tangent to the circle at B. The reason it works is this
pair of similar triangles, from which you can deduce this relationship between the distances.
Something more interesting happens when you try to invert an entire circle with
respect to another circle. This means that we pick a point on the circle we want to invert,
then invert this particular point, and do this for every other point on the blue circle. You will
get another circle. Here is the reason why. First, draw a line connecting the centres of the circles.
You will then get two diametrically opposite points. Invert these two points separately,
then we will use the following fact from Euclidean geometry. Whenever you have a circle with two
diametrically opposite points, every other point will form a right angle with those two points.
The logic works the other way too. If a point forms a right angle with two other points,
then it must lie on the circle with those two points diametrically opposite to each other.
With that said, we can use the definition of what A’, B’, C’ are, which are the points A,
B, and C under inversion. This will allow you to deduce the two pairs of similar triangles,
and hence the right angle from C’ to the images of A and B,
which is enough to prove that the image is really a circle even after inversion.
So far, we have established that a circle completely inside the yellow one,
will be inverted to another circle completely outside, and vice versa (wait). If the circle
crosses the circle of inversion, its image would still be a circle, and they both
intersect the yellow circle at exactly the same two points. If, however, we are considering a
circle that passes through the centre of the yellow circle, then we will have a line instead.
And of course, the line will invert to the circle. Again, we can use similar triangles to prove that
the image is indeed a line, but more importantly, in all the cases that we have discussed, circles
or lines are inverted to some other circles or lines. Because 3 points uniquely define a circle,
2 points for a line, we only need to invert at most 3 points to get the image under inversion.
Since inversion of a point can be done with straightedge and compass, the inversion of
three points certainly can be done; and so inversion of a line or a circle can also be done.
Sidetrack a little bit: circles or lines mapping to circles or lines is just one interesting
property of circle inversion, another important property is that angles are preserved, but
orientations are reversed. What this means is that if I want to invert this angle with respect to the
circle, then we will get another angle. If we are measuring an angle between two curves, we take the
angle between the tangents at the intersection. It turns out that angles don’t change during
inversion. But if we also assign a sense of orientation by saying that we measure angles
counterclockwise, then after inversion, this angle will become a clockwise one, so orientation
is reversed. This concept is called anticonformal mapping, an important concept in complex analysis.
Anyway, that’s too much of a tangent. This is just to show that circle inversion is really important.
Just one small point to note before we tackle the problem of Apollonius. If we
expand all three given circles by exactly the same amount, then the yellow solution circle
will be squished by the same amount. What I mean is that if we increase the radius of
the three given circles by say 1 cm, then the solution circle will decrease in radius by 1 cm
and the centres of all circles are unchanged. This fact can be seen rather easily by joining
the centres together, but this easy fact turns out to be very important.
With all that said, let’s get on to this problem of Apollonius. First off,
we inflate all the three circles by the same amount, until two of the circles touch at a point.
We then use this point as a centre to draw a circle. The radius can be arbitrary, because
we are treating this circle as the circle of inversion. Because these two circles pass through
the centre of the blue circle, when we invert everything with respect to the blue circle, the
two circles on the top will be transformed into two lines. And the remaining circle just becomes
another circle under inversion. Let’s clear things up for a second. What we need to do now
is to find a circle tangent to the two lines and the inverted circle. This is easy to do, because
by being tangent to these two parallel lines, we already know that the solution circle has
to lie on the middle line, AND its radius being half the distance between the two green lines.
Hence, we know the distance between the centres of the green circle and the yellow solution circle.
We just need to construct a circle with that distance as the radius to see where the yellow
centre can be. I could have chosen the other point here as the centre of the yellow circle,
and it would yield another solution, but I chose the bottom one for this video.
After we have obtained the yellow circle, it’s time to invert everything back with respect to
the blue circle. The green lines and circles will be inverted back to the inflated circles.
And the yellow circle will be inverted to yet another circle. But this inverted
yellow circle is not any other circle; it is tangent to all the three inflated circles.
Finally, we just deflate the three white circles, and the solution circle
will correspondingly inflate, and we have now obtained a solution to the original problem!
All these might be a lot to take in, so let’s take a deep breath and summarise what we are doing. The
initial steps are transformations to the given circles. These transformations include, first
of all, the expansion of the circles, then we do an inversion, so that they become two parallel
lines and another circle. Then we solve for the problem at hand, which is to construct a circle
tangent to the green objects. After solving, we do all the previous transformations in reverse,
so we first do the inversion. Then we revert the expansion and do the shrinking of the white
circles, so the yellow circle will correspondingly expand to become THE solution circle.
This demonstrates one very important problem-solving technique. If the problem
at hand is really difficult, try transforming it into a simple problem to deal with. Then instead
of solving for the hard problem directly, we solve for the simple problem first. Then we do the
inverse transformation to get the solution to the hard problem. In the case of Apollonius problem,
the transformation we did is a combination of expansion of the given circles and inversion.
But the problem of Apollonius is not the only thing that can be dealt with this way.
For example, an ordinary differential equation with a forcing term
is generally difficult to solve, but if we take the Fourier transform,
the problem becomes much easier. However, this is well outside the scope of this video.
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