Circular Reasoning
by nsato, Jun 26, 2010, 12:50 AM
"The opposite of a great truth is another great truth." --Neils Bohr
In today's post, we're going to be looking at circles. Lots and lots of circles, many of which are tangent to each other. And the classic tool for dealing with a lot of tangent circles is inversion.
In case you're not familiar with inversion, you can think of it as reflection through a circle. After choosing a fixed circle
to invert through, points go to points, and figures go to figures. In the left diagram below, we invert a red circle through circle to obtain a blue circle. In the right diagram below, we invert a red line through circle to obtain a blue circle.
[jump]More after the jump[/jump]
One remarkable property about inversion, as these diagrams suggest, is that the image of a circle or a line is also a circle or a line. In fact, you can think of a line as just a circle with an infinitely large radius. This statement may sound like a joke, but it's mathematically sound. (For example, if we make the radius of the circle we are inverting through infinitely large, then the operation of inversion through becomes reflection through a line. Hence, the description of inversion as reflection through a circle.)
The only other property about inversion that we need to know is that if two figures are tangent, then their images are also tangent. If you want to learn more about inversion, I have added some references at the end of this post which you can look up.
The classic problem that can be solved easily using inversion is Steiner's porism, which is described as follows. Suppose we have two circles, one inside the other. We draw another (red) circle that is tangent to both circles. We then construct a sequence of red circles, such that each red circle is tangent to both original circles, and the previous red circle in the sequence.
Suppose in the sequence of red circles, we find that some red circle is tangent to the first red circle that we drew, thus closing the sequence and forming a chain of tangent circles. Then in fact, we obtain a chain of tangent circles no matter where we draw the first red circle! This is the statement of Steiner's porism. The diagram below shows the original chain of red tangent circles, and another chain of purple tangent circles.
It turns out that for any pair of circles where one circle is inside the other, there exists an inversion that maps them to concentric circles. And since inversion preserves tangency, this inversion produces another chain of (blue) tangent circles.
In the case of concentric circles, it is obvious that if we obtain a chain of blue tangent circles for some initial blue circle, then we obtain a chain of blue tangent circles no matter where we draw the first blue circle. Hence, by inversion, the same result holds for the original diagram. In the right diagram above, the chain of blue circles can rotate, like ball bearings in a wheel. One can then imagine the chain of red circles, in the left diagram above, rotating as well.
Let's look at a simpler diagram involving tangent circles. We start with three mutually tangent circles. Then there are two circles that are tangent to all three circles. What can we say about these two circles?
For one, take the circumcircle of the three points of tangency among the three original circles. If we invert through this circumcircle, then the image of each original circle is the circle itself. Since inversion preserves tangency, the image of the red circle is the blue circle under this inversion.
What about the radii of these two circles? There is a formula that tells us exactly what the radii of these two circles are. According to Descartes' Circle Theorem, if four circles with radii
, 
, 
, and 
are mutually tangent, then
![\[\left( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} \right)^2 = 2 \left( \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r_4^2} \right).\]](//latex.artofproblemsolving.com/8/2/7/8271fccbb4a786be98738e2700f91c4c0b544f9c.png)
We can express this formula another way. A circle with radius
has curvature 
. Hence,
![\[(\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4)^2 = 2 (\kappa_1^2 + \kappa_2^2 + \kappa_3^2 + \kappa_4^2).\]](//latex.artofproblemsolving.com/b/6/5/b6522da4b6b177fd56c3a8df506ecb814e1097ae.png)
For example, in the diagram above, let the three radii of the three original circles be
, 
, and 
, so their curvatures are 
, 
, and 
. Plugging these values into Descartes' formula, we get
![\[(\kappa_4 + 18 + 27 + 23)^2 = 2 (\kappa_4^2 + 18^2 + 27^2 + 23^2),\]](//latex.artofproblemsolving.com/b/3/0/b3031eebae6fa68f3936648d52f2be2dcd5e44f2.png)
which simplifies as
![\[\kappa_4^2 - 136 \kappa_4 - 1460 = (\kappa_4 - 146)(\kappa_4 + 10) = 0.\]](//latex.artofproblemsolving.com/0/4/a/04a9ccfdb5afab0fb37bc6198c935ece7430cc45.png)
Hence,
or 
, which means the radii of the two circles are 
(which corresponds to the small red circle) and 
(which corresponds to the large blue circle). The radius being negative indicates that the large blue circle is internally tangent to the other three circles, whereas the small red circle is externally tangent to the other three circles. This is the convention that Descartes' formulas uses. We indicate these curvatures on the diagram.
We can continue playing the game of finding curvatures of tangent circles. For example, what is the curvature of the circle that is tangent to the circles with curvatures
, 27, and 23?
(Note that we can obtain this circle by taking the circumcircle of the three points of tangency among the circles with curvatures, 27, and 23, and inverting the circle with curvature 146 through this circumcircle.)
By Descartes' formula, the curvature of this circle satisfies the equation
![\[(\kappa - 10 + 27 + 23)^2 = 2 [\kappa^2 + (-10)^2 + 27^2 + 23^2],\]](//latex.artofproblemsolving.com/d/b/c/dbc88cf24095a9867f5f0269641beb9f6bb4c3e5.png)
which simplifies as
![\[\kappa^2 - 80 \kappa + 116 = 0.\]](//latex.artofproblemsolving.com/f/5/3/f5318b9c9017bc4de22ee4ca761c058f5d59a203.png)
We can factor this as
![\[(\kappa - 18)(\kappa - 62) = 0,\]](//latex.artofproblemsolving.com/1/8/e/18e0c13abe7dd8e4317a322e9bbc3637e2996d38.png)
so
or 
. The solution makes sense because we already have a circle with curvature 18 that is tangent to the three circles we are interested in, so the curvature of the circle we want is 62.
What is the curvature of the circle that is tangent to the circles, 18, and 23?
Using the same technique as above, we arrive at the quadratic equation
![\[\kappa^2 - 62 \kappa + 945 = 0.\]](//latex.artofproblemsolving.com/b/5/2/b52ad707be59f078e56d0ffc90ec3f4a6a91fde3.png)
We can factor this quadratic, but we already know one root, namely
, because the circle with curvature 27 is also tangent to the three circles we are interested in. By Vieta's Formulas, the sum of the roots is 62, so the other root is 
. This technique of using one root of a quadratic equation to find the other root of a quadratic equation is known as Vieta Jumping, or root flipping. Hence, the curvature of the circle we want is 35.
We can then fill in the circle indefinitely, through inversion, giving us an Apollonian gasket. By Vieta Jumping, we know that the curvature of every circle we produce is a positive integer.
Now, let's go back to Steiner's porism.
At the start, we remarked how a line can be seen as a circle with an infinitely large radius. So, we can send the radius of the outer circle to infinity so that it becomes a line. We then obtain a chain of circles, all of which are tangent to a circle and a line.
As before, the chain of circles can rotate around the circle, while keeping the circle fixed. Hence, for any such chain of circles, the distance from the center of the circle to the line is always the same.
Now let's go for broke - let's send the radius of the remaining circle to infinity as well! This gives us two lines, and a chain of congruent circles.
This diagram is not particularly interesting, so let's spice it up by filling in circles, like an Apollonian gasket.
If we look at the circles that are tangent to the bottom line, then we see something very interesting. As a reference, we consider the bottom line as a number line, and let the large circles be tangent to this line at the integers. We look at the portion of the line from 0 to 1.
We find that every circle that is tangent to the line is tangent at a rational number, and every rational number between 0 and 1 has a corresponding circle. Furthermore, the circle that is tangent at the rational number
(where 
and 
are relatively prime) has radius 
. Beautiful! (You can apply Descartes' formulas - the curvature of a line is simply 0.) Collectively, these circles are called Ford circles. We can ask many interesting questions regarding Ford circles. For example, given two rational numbers 
and 
between 0 and 1, when are their corresponding Ford circles tangent?
To close, we leave the reader with a problem: Each circle in a chain of
circles is tangent to a unit circle and a line. Let 
be the distance from the center of the circle to the line.
Show that
![\[d_n = \frac{1 + \sin^2 \frac{\pi}{n}}{1 - \sin^2 \frac{\pi}{n}}.\]](//latex.artofproblemsolving.com/e/8/1/e8123ec31f4eb67c1ba237b38fbc0c0946de0dcd.png)
References
Inversion
Inversion, MathWorld
Inversion in the Plane, Berkeley Math Circle
Inversion, IMO Compendium website
Geometry Revisited, Coxeter & Greitzer
Steiner Porism
Steiner Chain, MathWorld
Steiner's Porism, Cut The Knot
Descartes' Circle Theorem
Descartes' Theorem, Wikipedia
Beyond the Descartes Circle Theorem, Lagarias, Mallows, Wilks
Apollonian Gaskets
Apollonian Gasket, Wikipedia
Apollonian Circle Packings: Number Theory, Graham, Lagarias, Mallows, Wilks, Yan
Ford Circles
Ford Circle, Wikipedia
Ford Circle, MathWorld
USAMTS, Problem 5/3/19 and solution
In today's post, we're going to be looking at circles. Lots and lots of circles, many of which are tangent to each other. And the classic tool for dealing with a lot of tangent circles is inversion.
In case you're not familiar with inversion, you can think of it as reflection through a circle. After choosing a fixed circle
to invert through, points go to points, and figures go to figures. In the left diagram below, we invert a red circle through circle to obtain a blue circle. In the right diagram below, we invert a red line through circle to obtain a blue circle.[jump]More after the jump[/jump]
One remarkable property about inversion, as these diagrams suggest, is that the image of a circle or a line is also a circle or a line. In fact, you can think of a line as just a circle with an infinitely large radius. This statement may sound like a joke, but it's mathematically sound. (For example, if we make the radius of the circle
we are inverting through infinitely large, then the operation of inversion through becomes reflection through a line. Hence, the description of inversion as reflection through a circle.)The only other property about inversion that we need to know is that if two figures are tangent, then their images are also tangent. If you want to learn more about inversion, I have added some references at the end of this post which you can look up.
The classic problem that can be solved easily using inversion is Steiner's porism, which is described as follows. Suppose we have two circles, one inside the other. We draw another (red) circle that is tangent to both circles. We then construct a sequence of red circles, such that each red circle is tangent to both original circles, and the previous red circle in the sequence.
Suppose in the sequence of red circles, we find that some red circle is tangent to the first red circle that we drew, thus closing the sequence and forming a chain of tangent circles. Then in fact, we obtain a chain of tangent circles no matter where we draw the first red circle! This is the statement of Steiner's porism. The diagram below shows the original chain of red tangent circles, and another chain of purple tangent circles.
It turns out that for any pair of circles where one circle is inside the other, there exists an inversion that maps them to concentric circles. And since inversion preserves tangency, this inversion produces another chain of (blue) tangent circles.
In the case of concentric circles, it is obvious that if we obtain a chain of blue tangent circles for some initial blue circle, then we obtain a chain of blue tangent circles no matter where we draw the first blue circle. Hence, by inversion, the same result holds for the original diagram. In the right diagram above, the chain of blue circles can rotate, like ball bearings in a wheel. One can then imagine the chain of red circles, in the left diagram above, rotating as well.
Let's look at a simpler diagram involving tangent circles. We start with three mutually tangent circles. Then there are two circles that are tangent to all three circles. What can we say about these two circles?
For one, take the circumcircle of the three points of tangency among the three original circles. If we invert through this circumcircle, then the image of each original circle is the circle itself. Since inversion preserves tangency, the image of the red circle is the blue circle under this inversion.
What about the radii of these two circles? There is a formula that tells us exactly what the radii of these two circles are. According to Descartes' Circle Theorem, if four circles with radii
, 
, 
, and 
are mutually tangent, then![\[\left( \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} \right)^2 = 2 \left( \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r_4^2} \right).\]](http://latex.artofproblemsolving.com/8/2/7/8271fccbb4a786be98738e2700f91c4c0b544f9c.png)
We can express this formula another way. A circle with radius
has curvature 
. Hence,![\[(\kappa_1 + \kappa_2 + \kappa_3 + \kappa_4)^2 = 2 (\kappa_1^2 + \kappa_2^2 + \kappa_3^2 + \kappa_4^2).\]](http://latex.artofproblemsolving.com/b/6/5/b6522da4b6b177fd56c3a8df506ecb814e1097ae.png)
For example, in the diagram above, let the three radii of the three original circles be
, 
, and 
, so their curvatures are 
, 
, and 
. Plugging these values into Descartes' formula, we get![\[(\kappa_4 + 18 + 27 + 23)^2 = 2 (\kappa_4^2 + 18^2 + 27^2 + 23^2),\]](http://latex.artofproblemsolving.com/b/3/0/b3031eebae6fa68f3936648d52f2be2dcd5e44f2.png)
which simplifies as
![\[\kappa_4^2 - 136 \kappa_4 - 1460 = (\kappa_4 - 146)(\kappa_4 + 10) = 0.\]](http://latex.artofproblemsolving.com/0/4/a/04a9ccfdb5afab0fb37bc6198c935ece7430cc45.png)
Hence,
or 
, which means the radii of the two circles are 
(which corresponds to the small red circle) and 
(which corresponds to the large blue circle). The radius being negative indicates that the large blue circle is internally tangent to the other three circles, whereas the small red circle is externally tangent to the other three circles. This is the convention that Descartes' formulas uses. We indicate these curvatures on the diagram.We can continue playing the game of finding curvatures of tangent circles. For example, what is the curvature of the circle that is tangent to the circles with curvatures
, 27, and 23?(Note that we can obtain this circle by taking the circumcircle of the three points of tangency among the circles with curvatures
, 27, and 23, and inverting the circle with curvature 146 through this circumcircle.)By Descartes' formula, the curvature of this circle satisfies the equation
![\[(\kappa - 10 + 27 + 23)^2 = 2 [\kappa^2 + (-10)^2 + 27^2 + 23^2],\]](http://latex.artofproblemsolving.com/d/b/c/dbc88cf24095a9867f5f0269641beb9f6bb4c3e5.png)
which simplifies as
![\[\kappa^2 - 80 \kappa + 116 = 0.\]](http://latex.artofproblemsolving.com/f/5/3/f5318b9c9017bc4de22ee4ca761c058f5d59a203.png)
We can factor this as
![\[(\kappa - 18)(\kappa - 62) = 0,\]](http://latex.artofproblemsolving.com/1/8/e/18e0c13abe7dd8e4317a322e9bbc3637e2996d38.png)
so
or 
. The solution makes sense because we already have a circle with curvature 18 that is tangent to the three circles we are interested in, so the curvature of the circle we want is 62.What is the curvature of the circle that is tangent to the circles
, 18, and 23?Using the same technique as above, we arrive at the quadratic equation
![\[\kappa^2 - 62 \kappa + 945 = 0.\]](http://latex.artofproblemsolving.com/b/5/2/b52ad707be59f078e56d0ffc90ec3f4a6a91fde3.png)
We can factor this quadratic, but we already know one root, namely
, because the circle with curvature 27 is also tangent to the three circles we are interested in. By Vieta's Formulas, the sum of the roots is 62, so the other root is 
. This technique of using one root of a quadratic equation to find the other root of a quadratic equation is known as Vieta Jumping, or root flipping. Hence, the curvature of the circle we want is 35.We can then fill in the circle indefinitely, through inversion, giving us an Apollonian gasket. By Vieta Jumping, we know that the curvature of every circle we produce is a positive integer.
Now, let's go back to Steiner's porism.
At the start, we remarked how a line can be seen as a circle with an infinitely large radius. So, we can send the radius of the outer circle to infinity so that it becomes a line. We then obtain a chain of circles, all of which are tangent to a circle and a line.
As before, the chain of circles can rotate around the circle, while keeping the circle fixed. Hence, for any such chain of circles, the distance from the center of the circle to the line is always the same.
Now let's go for broke - let's send the radius of the remaining circle to infinity as well! This gives us two lines, and a chain of congruent circles.
This diagram is not particularly interesting, so let's spice it up by filling in circles, like an Apollonian gasket.
If we look at the circles that are tangent to the bottom line, then we see something very interesting. As a reference, we consider the bottom line as a number line, and let the large circles be tangent to this line at the integers. We look at the portion of the line from 0 to 1.
We find that every circle that is tangent to the line is tangent at a rational number, and every rational number between 0 and 1 has a corresponding circle. Furthermore, the circle that is tangent at the rational number
(where 
and 
are relatively prime) has radius 
. Beautiful! (You can apply Descartes' formulas - the curvature of a line is simply 0.) Collectively, these circles are called Ford circles. We can ask many interesting questions regarding Ford circles. For example, given two rational numbers 
and 
between 0 and 1, when are their corresponding Ford circles tangent?To close, we leave the reader with a problem: Each circle in a chain of
circles is tangent to a unit circle and a line. Let 
be the distance from the center of the circle to the line.Show that
![\[d_n = \frac{1 + \sin^2 \frac{\pi}{n}}{1 - \sin^2 \frac{\pi}{n}}.\]](http://latex.artofproblemsolving.com/e/8/1/e8123ec31f4eb67c1ba237b38fbc0c0946de0dcd.png)
References
Inversion
Inversion, MathWorld
Inversion in the Plane, Berkeley Math Circle
Inversion, IMO Compendium website
Geometry Revisited, Coxeter & Greitzer
Steiner Porism
Steiner Chain, MathWorld
Steiner's Porism, Cut The Knot
Descartes' Circle Theorem
Descartes' Theorem, Wikipedia
Beyond the Descartes Circle Theorem, Lagarias, Mallows, Wilks
Apollonian Gaskets
Apollonian Gasket, Wikipedia
Apollonian Circle Packings: Number Theory, Graham, Lagarias, Mallows, Wilks, Yan
Ford Circles
Ford Circle, Wikipedia
Ford Circle, MathWorld
USAMTS, Problem 5/3/19 and solution
This post has been edited 5 times. Last edited by levans, Sep 7, 2010, 5:17 PM
September 2012
May 2012