## Polygons

In working on a proposal for a book about three-dimensional polyhedra last week, I needed to write a brief section on polygons.  I found that there were so many different types of polygons with such interesting properties, I thought it worthwhile to spend a day talking about them.  If you’ve never thought a lot about polygons before, you might be surprised how much there is to say about them….

So start by imagining a polygon — make it a pentagon, to be specific.  Try to imagine as many different types of pentagons as you can.

How many did you come up with?  I stopped counting when I reached 20….

Likely one of the first to come to mind was the regular pentagon — five equal sides at angles of 108° from each other.  Question:  did you only think of the vertices and edges, or did you include the interior as well?

Why consider this question?  An important geometrical concept is that of convexity.  A convex polygon property has the property that a line segment joining any two points in the polygon lies entirely within the polygon.

The two polygons on the left are convex, while the two on the right are not.  But note that for this definition to make any sense at all, a polygon must include all of its interior points.

Convex polygons have various properties.  For example, if you take the vertices of a convex polygon and imagine stretching a rubber band beyond the vertices and letting it snap back, the rubber band will describe the edges of the polygon.  See this Wikipedia article on convex polygons for more properties of convex polygons.

Did the edges of any of your pentagons cross each other, like the one on the left below?

In this picture, we indicate vertices with dots to illustrate that this is in fact a pentagon.  The points where the edges cross are not considered vertices of the polygon.  The polygon on the right is actually a nonconvex decagon, even though it bears a resemblance to the pentagon on the left.

But not so fast!  If you ask Mathematica to draw the polygon on the left with the five vertices in the order they are traversed when drawing the edges, here is what you get:

So what’s going on here?  Why is the pentagon empty in the middle?  When I gave the same instructions using Tikz in LaTeX (which is how I created the light blue pentagram shown above), the middle pentagon was filled in.

Some computer graphics programs use the even-odd rule when drawing self-intersecting polygons.  This may be thought of in a few ways.  First, if you imagine drawing a segment from a point in the interior pentagon to a point outside, you have to cross two edges of the pentagon, as shown above.  If you draw a segment from a point in one of the light red regions to a point outside, you only need to cross one edge.  Points which require crossing an even number of edges are not considered as being interior to the polygon.

Said another way, if you imagine drawing the pentagram, you will notice that you are actually going around the interior pentagon twice.  Any region traversed twice (or an even number of times) is not considered interior to the polygon.

Why would you want to color a polygon in this way?  There are mathematical reasons, but if you watch this video by Vi Hart all the way through, you’ll see some compelling visual evidence why you might want to do this.

We call polygons whose edges intersect each other self-intersecting or crossed polygons.  And as you’ve seen, including the interiors can be done in one of two different ways.

But wait!  What about this polygon?  Can you really have a polygon where a vertex actually lies on one of the edges?

Again, it all depends on the context.  I think you’re beginning to see that the question “What is a pentagon?” is actually a subtle question.  There are many features a pentagon might have which you likely would not have encountered in a typical high school geometry course, but which still merit some thought.

Up to now, we’ve just considered a polygon as a two-dimensional geometrical object.  What changes when you jump up to three dimensions?

Again, it all depends on your definition.  You might insist that a polygon must lie in a plane, but….

It is possible to specify a polygon by a list of points in three dimensions — just connect the points one by one, and you’ve got a polygon!  Of course with this definition, many things are possible — maybe you can repeat points, and maybe the points do not all lie in the same plane.

An interesting example of such a polygon is shown below, outlined in black.

It is called a Petrie polygon after the mathematician who first described it.  In this case, it is a hexagon — think of holding a cube by two opposite corners, and form a hexagon by the six edges which your fingers are not touching.

There is a Petrie polygon for every Platonic solid, and may be defined as follows:  it is a closed path of connected edges such that no three consecutive edges belong to the same face.  If you look at the figure above, you’ll find this is an alternative way to define a Petrie hexagon on a cube.

And if that isn’t enough, it is possible to define a polygon with an infinite number of sides!  Just imagine the following jagged segment continuing infinitely in both directions.

This is called an apeirogon, and may be used to study the tiling of the plane by squares, four meeting at each vertex of the tiling.

And we haven’t even begun to look at polygons in other geometries — spherical geometry, projective geometry, inversive geometry….

Suffice it to say that the world of polygons is much more than just doodling a few triangles, squares or pentagons.  It is always amazes me how such a simple idea — polygon — can be the source of such seemingly endless investigation!  And serve as another illustration of the seemingly infinite diversity within the universe of Geometry….

## Still more on: What is…Inversive Geometry?

Now for the final post on inversive geometry!  I’ve been generating some fascinating images, and I’d like to share a bit about how I make them.

In order to create such images in Mathematica, you need to go beyond the geometrical definition of inversion and use coordinate geometry.  Let’s take a moment to see how to do this.

Recall that P′, the inverse point of P, is that point on a ray drawn from the origin through P such that

$[OP]\cdot[OP']=1,$

where $[AB]$ denotes the distance from A to B.  Feel free to reread the previous two posts on inversive geometry for a refresher (here are links to the first post and the second post).

Now suppose that the point P has Cartesian coordinates $(x,y).$  Points on a ray drawn from the origin through P will then have coordinates $(kx, ky),$ where $k>0.$  Thus, we just need to find the right $k$ so that the point $P'=(kx,ky)$ satisfies the definition of an inverse point.

This is just a matter of a little algebra; the result is

$P'=\left(\dfrac{x}{x^2+y^2},\dfrac{y}{x^2+y^2}\right).$

What this means is that if you have an equation of a curve in terms of $x$ and $y,$ if you substitute $x/(x^2+y^2)$ everywhere you see $x,$ and substitute $y/(x^2+y^2)$ everywhere you see $y,$ you’ll get an equation for the inverse curve.

Let’s illustrate with a simple example — in general, the computer will be doing all the work, so we won’t need to actually do the algebra in practice.  We’ll look at the line $x=1.$  From our previous work, we know that the inverse curve must be a circle going through the origin.

Making the substitution just discussed, we get the equation

$\dfrac x{x^2+y^2}=1,$

which may be written (after completing the square) in the form

$\left(x-\dfrac12\right)^2+y^2=\dfrac14.$

It is not hard to see that this is in fact a circle which passes through the point $(0,0).$

Now we need to add one more step.  In the definition of an inverse point, we had the point $O$ being the origin with coordinates $(0,0).$  What if $O$ were some other point, say with coordinates $(a,b)$?

Let’s proceed incrementally.  Beginning with a point $(x,y),$ translate to the point $(x-a,y-b)$ so that the point $(a,b)$ now acts like the origin.  Now use the previous formula to invert:

$\left(\dfrac{x-a}{(x-a)^2+(y-b)^2},\dfrac{y-b}{(x-a)^2+(y-b)^2}\right).$

Finally, translate back:

$\left(a+\dfrac{x-a}{(x-a)^2+(y-b)^2},b+\dfrac{y-b}{(x-a)^2+(y-b)^2}\right).$

This is now the inverse of the point $(x,y)$ about the point $(a,b).$

So what you see in the above image is several copies of the parabola $y=x^2$ inverted about a series of equally spaced points along the line segment with endpoints $(1/2,-1/2)$ and $(3/2,1/2).$  This might seem a little arbitrary, but it takes quite a bit of experimentation to find a set of points to invert about in order to create an aesthetically pleasing image.

Of course there is another perspective on accomplishing the same task — just shift the parabolas first, invert about the origin, and then shift back.  This is geometrically equivalent (and the algebra is the same); it just depends on how you want to look at it.

Here is another image creating by inverting the parabola $y=x^2$ about points which lie on a circle.

And while we’re on the subject of inverting parabolas, let’s take a moment to discuss the cardioid example we looked at in our last conversation about inversion:

To prove that this construction of circles actually yields a cardioid, the trick is to take the inverse of a parabola about its focus.  If you do this, the tangent lines of the parabola will then invert to circles tangent to a cardioid.  I won’t go into all the details here, but I’ll outline how the proof goes using the following diagram.

Draw a line (shown in black) tangent to the blue parabola at its vertex; the inverse curves are shown in the same color, but dashed.  Note that the black circle must be tangent to the blue cardioid since the inverse black line is tangent to the inverse parabola.

The small red disk is the focus of the parabola.  Key to the proof is the property of the parabola that if you draw a line from the focus to a point on the black line and then bounce off at a right angle (the red lines), the resulting line is tangent to the parabola.  So the inverse of this line — the red dashed circle — must be tangent to the cardioid.

Since perpendicularity is preserved and the line from the focus inverts to itself (since we’re inverting about the focus), the red circle must be perpendicular to this line — meaning that the line from the focus in fact contains a diameter, and hence the center, of the red circle.  Then using properties of circles, you can show that all centers of circles formed in this way lie on a circle (shown dotted in purple) which is half the size of the black circle.  I’ll leave the details to you….

Finally, I’d like to show a few examples of using the other conic sections.  Here is an image with 80 inversions of an ellipse around centers which lie on a line segment.

And here is an example of 100 hyperbolas inverted around centers which lie on a line segment.  Since the tails of the branches of a hyperbola all go to infinity, they all meet at the same point when inverted.

So now you know how to work with geometrical inversion from an algebraic standpoint.  I hope seeing some of the fascinating images you can create will inspire you to try creating some yourself!

## Imagifractalous! 7: Fractal Binary Trees IV

Bridges 2017 is next week in Waterloo, Ontario, Canada!  We’re all looking forward to it…Nick and his parents are also going, and I’ll get to see a lot of new friends I met last year in Finland.

In preparation, I’ve been creating a portfolio of binary and ternary trees.  I’ve been coloring and highlighting the branches to make them easier to understand, so I thought I’d take the opportunity to share several of them.  Mathematics just as it’s happening….

Here is an example of what I’m talking about.

Recall that we’re generalizing the idea of “left” and “right” branches of a binary tree — rather that just rotating left or right (and possibly scaling), we’re allowing any affine transformation to correspond to “left” and “right.”  In order to avoid any confusion, let’s refer to the transformations as “0” and “1.”

In the tree above, the 0 branches are shown in black, and the 1 branches in red.  Notice that sometimes, the 0 branches seem to go to the right, other times, to the left.  This is a common occurrence when you allow the branching to be determined by more general transformations.

Here, you’ll notice that 0 and 11 take you to the same place — beginning with the trunk (the vertical black line at the right), if you follow one black branch, you’ll end up at the same blue node as you would if you followed two red branches.  In fact, no matter what blue node you start at, following one black branch will always take you to the same place as following two red branches.

Here is not the place to go into too many details, but let me give you the gist of what we’re studying.  If you denote the transformation which generates the black branches by $B_0$ and that which generate the red branches by $B_1,$ you will observe behavior like that in the above tree if

$B_0=B_1+B_1^2.$

Right now, most of the work I’m doing revolves around saying that I want to generate a tree with a certain property (like 0 and 11 going to the same node), and then solving the corresponding matrix equation to find interesting examples of trees with this property.  Usually the linear algebra involved is fairly complicated — feel free to comment if you’d like to know more.  As I said, this blog is not the place for all the details….

Below is another interesting example.

In this tree, 0 and 00 go to the same node.  What this means in terms of branching is that there can never be two consecutive black branches (again, 0 is black and 1 is red) — taking the additional 0 branch leaves you in exactly the same place.

Note that for this property to hold, there is no restriction on what the 1 transformation can be.  Above, the transformation $B_1$ is a scaled rotation by 60°.  This is what produces the spirals of red branches emanating from the nodes of the tree.

In this example, 0 and 01 go to the same node (0 is black, 1 is red).  Implications for the tree are that the red branches zigzag back and forth, getting shorter with each iteration.  Moreover, there aren’t any red branches drawn after a black branch (except the first time, off the end of the trunk of the tree).  This is because 01 — taking a red branch after a black branch — takes you to the same place as 0, which is taking a black branch.  In other words, the red branch doesn’t take you any further.

Here, 01 and 10 take you to the same node (again, 0 is black and 1 is red).  This is easy to see — there are many quadrilaterals in the tree formed by alternating red and black branches.  In addition, the iterations of this tree move progressively higher each time — so, for example, all the nodes at the top are at a depth of 4.

If you look at the number of new nodes at each level, you get a sequence which begins

2, 3, 6, 12, 24, 48, 96, 192, 384, 768, ….

After the second level, it seems that the number of new nodes doubles with each iteration.  I say “seems” since it is generally very difficult to prove that such a relationship continues up arbitrarily many levels.  Although it is easy to find the sequence of numbers of nodes for any tree using Mathematica, I have rigorous proofs of correctness for only three cases so far.

What makes the proofs so difficult?  The following tree is another example of a case where 01 and 10 go to the same node.

But the sequence of new nodes at each level is just 2, 3, 4, 5, 6, ….  (This is one case I’ve rigorously proved.)  The transformations $B_0$ and $B_1$ used to make this tree, in addition to forcing 01 and 10 to go to the same node, also have additional properties which force any two strings of the same length with the same number of 0’s (and therefore the same number of 1’s as well) to go to the same node.  So to prove the sequence of nodes in the previous case is 2, 3, 6, 12, 24, 48, …, you need to say which strings take you to the same nodes, and prove that this pattern continues no matter how far you go up — and that there are no “surprise” coincidences of other nodes as you proceed.

It is also possible to look at infinite strings going to the same node, as shown below.  The linear algebra gets quite a bit more involved in these cases.

Here, 0 and 1 do in fact correspond to going left and right, respectively.  The thicker black path corresponds to the string 011111…, that is, going left once, and then going to the right infinitely often, creating the spiral you see above.  The thicker green path corresponds to alternating 10101010… infinitely often, creating a zigzag path which approaches exactly the same point as the black spiral does.

Here is another example, where the spiral 011111… (in black) approaches the same point as the infinite zigzag 100100100100… (in purple).

These are some of my favorite examples.  But I should remark that once you know how to mathematically find transformations which produce trees with a desired property, it takes a lot of fiddling around to find aesthetically pleasing examples.

I hope these examples give you a better idea of the nature of our research.  I’ll update you on Bridges when I get back, and let you know about any interesting comments other participants make about our trees.  I leave on Wednesday, and will post pictures daily on my Twitter @cre8math if you’d like a daily dose of mathematical art!

## More on: What is…Projective Geometry?

This week, I thought I’d go a little deeper into the subject of projective geometry (feel free to reread last week’s post for a refresher).  Why? I think this is a good opportunity to discuss the idea of an algebraic model of a geometry.  Certainly using Cartesian coordinates in the Euclidean plane makes many geometrical problems easier to approach.

So what would a coordinate system in projective geometry look like?  The most commonly used system is homogeneous coordinates.  Let’s see how they work.

The difficulty with adding the line at infinity is that we need infinitely many points in addition to those already in the plane.  Perhaps you might imagine that adding an “infinite” coordinate might do the trick, but there is an alternative — and algebraically simpler — approach.

First, think of the Euclidean plane as being the plane z = 1 in three-dimensional Cartesian coordinates, so that every Euclidean point (xy) is given homogeneous coordinates (xy, 1).  But we don’t want to make the z-coordinate special — we want all the homogeneous coordinates to have similar interpretations.  We accomplish this by giving each point infinitely many names — specifically, for any k ≠ 0, the coordinates (kxkyk) describe the same point as (xy, 1).  Geometrically, you might think of this as saying that the line through  (xy, 1) (except the origin) is, in some sense, the point (xy, 1).

So if z ≠ 0, the point (xyz) in homogeneous coordinates is another name for the Euclidean point (x/zy/z).  But why would you want to do this?

The punch line is that we can now use z = 0 to indicate that a point is on the line at infinity!  We still keep the convention that if k ≠ 0, the homogeneous coordinates (kxky, 0) describe the same point as (xy, 0), as long as x and y are not both 0.

So our system of homogenous coordinates contains all points (xyz) such that not all three coordinates are 0.  Any point with z ≠ 0 corresponds to a Euclidean point, and any point with z = 0 corresponds to a point on the line at infinity.

Is it really worth all this trouble?  There are many interesting features of such a coordinate system — and I’d like to mention a few of them here, omitting the proofs.  There are many resources online that include all the details — one example is the classic Projective Geometry by Veblen and Young available free as an ebook.

Let’s start be looking at equations of lines in projective geometry.  In the Cartesian plane, we may represent a line in the form Ax + By + C = 0, where A, B, and C are not all zero.  In the projective plane, a line has the form form Ax + By + Cz = 0, where A, B, and C are not all zero.  Nice, isn’t it?  This form is also consistent with our system of having many names for a point:  if Ax + By + Cz = 0 and k ≠ 0, then also A(kx) + B(ky) + C(kz) = 0 as well. So no problems there.

But the really neat aspect of this representation is how you can use linear algebra in three dimensions to solve many problems in projective geometry.  This isn’t a post about linear algebra — so I’ll limit myself to just one example.  But in case you do know something about linear algebra, I think you should see how it can be used in projective geometry when homogeneous coordinate are used.

We’ll consider the problem of finding the intersection of two lines, say Ax + By + Cz = 0 and Dx + Ey + Fz = 0.  The form of these expressions should remind you of taking the dot product, so that we can rewrite these expressions as

$(A,B,C)\cdot(x,y,z)=(D,E,F)\cdot(x,y,z)=0.$

Interpreting these coordinates and coefficients as vectors in three-dimensional space, we observe that the common point (xyz) is simultaneously perpendicular to both (A, B, C) and (D, E, F), since both dot products are zero.  So (xyz) can be found using the cross-product

$(x,y,z)=(A,B,C)\times(D,E,F).$

Very nice!  Again, there are many such examples, but this is not the place to discuss them….

This algebraic model also suggests another geometric model for the projective plane besides adding a line at infinity to the Euclidean plane.  Begin with the surface of a sphere centered at the origin in three-dimensional Cartesian space, and observe that opposite points on the sphere have homogeneous coordinates that are different names for the same point in the projective plane.

So, in some sense, we have exactly twice as many points as we need — so we identify opposite points on this sphere, so that they are in fact the same point.  (You might recall a similar discussion about points in the post on spherical geometry.)  Thinking about it in another way, we might just consider the top hemisphere of the sphere with only half of the equator, including just one of the endpoints of the semicircle.

And while this model is fairly simple geometrically, it is important to point out that this does not mean that the projective plane lies inside three-dimensional space.  Once we have our hemisphere and semicircle, we have to think about it without any of the surrounding space.  This is not easy to do, but this type of thinking is necessary all the time when studying differential geometry, a topic for another time….

One last benefit to using homogenous coordinates:  it can easily be abstracted to any number of dimensions.  Do you want a projective space?  Just add a plane at infinity to three-dimensional Euclidean space!  Coordinates are easy — all points (x, y, z, w), with not all coordinates equal to 0.  When w ≠ 0, the point (x, y, z, w) corresponds to the Euclidean point (x/w, y/w, z/w), and when w = 0, the point is on the plane at infinity.

And clearly, there would be no need to stop at three dimensions — it is just as easy to create a projective space of four or more dimensions.

Finding a workable system of coordinates for a particular geometry is not always a simple matter — but finding a system that allows problems to be solved easily is often a key step to studying any type of geometry.  I hope this gives you some insight to yet another aspect of the diverse universe of so many different Geometries….

## What is…Projective Geometry?

The last type of geometry I discussed was inversive geometry, which is obtained by adding a point at infinity to the Euclidean plane.  Recall that as long as we had a consistent, useful model of this extended plane, it made perfect sense to define this new type of geometry.

Today, we’re going to add a line at infinity — creating what is called a projective geometry.  There are in fact many different types of projective geometries, but let’s just try to understand one at a time….

You might remember one important property of ω, the point at infinity in inversive geometry:  it was on every unbounded curve, and in particular, on every line. We need to be a bit more specific with our projective geometry, in the following sense.  Every line will intersect the line at infinity — denoted by λ — but not every line will intersect at the same point.

Consider all lines with some given slope m.  We then add one point to λ which lies on these lines, but no other lines.  In other words, each point on λ corresponds to an infinite family of parallel Euclidean lines — since now, with the addition of λ, there are no parallel lines in the projective plane.  The point at which two parallel (in the Euclidean sense) lines intersect is determined by their slope.

How does this differ from inversive geometry?  Well, in inversive geometry, if two lines had different slopes, they intersected in two points:  the usual finite point of intersection you learned about in algebra class, as well as ω.  But in projective geometry, two lines with different slopes intersect in only one point, since they intersect the line at infinity in two different points.

What this means is that any two distinct lines always intersect in exactly one point.  See the difference?  Parallel lines (in the Euclidean sense) intersect in a point on λ, and nonparallel (in the Euclidean sense) lines intersect in the same point they did in Euclidean geometry.

Now let’s look at the dual question (recall the discussion of duality, an important concept in spherical geometry):  what about a line between two points?  Two finite points generate a line, as usual.  If one point is finite and one lies on λ, the line generated is that line through the finite point with the slope corresponding to the point on λ.  And if both points are infinite, then the line through them is just λ.

Thus we have the following duality:  two distinct lines determine a unique point, and two distinct points determine a unique line.  Again, duality is an extremely powerful concept in geometry, so the fact that points and lines are dual concepts really is a legitimate justification for thinking about projective geometry.

Projective plane geometry is a broad subject — but some of my favorite objects to look at in projective geometry are the conic sections:  ellipses, parabolas, and hyperbolas.  This is because from the point of view of projective geometry, they are, in a sense which we’ll look at right now, all the same.

This sounds odd at first reading, but follow along.  The first question we need to consider is what points on λ lie on unbounded curves.  For lines, it’s easy — it’s just the point on λ corresponding to the slope of the line.

You can see from the image that as points on the parabola move further away from its vertex, the tangents have slopes of ever-increasing magnitude.  Determining the precise slopes is an easy exercise in calculus — but what is important is that the tangent lines approach, in slope, the axis of the parabola.  In this case the axis is vertical, but of course the parabola may be rotated.

What this means is that the parabola intersects the line at infinity in the point corresponding to the slope of its axis.  Again, the tangent lines on either side approach this slope, but from two directions — from above and from below.

Now hold this thought while we look at a hyperbola.

Let’s see how we travel along the hyperbola.  As we move to the upper right toward the open red circle, we are moving closer to the red dashed asymptote — meaning we approach the point on λ corresponding to the slope of the red asymptote.  But here is where it becomes interesting:  as we cross the line at infinity, we come back along the asymptote toward the filled red circle on the other branch of the hyperbola!

Then we continue moving along the left branch, getting continually closer to the blue asymptote, approaching that point on λ corresponding to the slope of the blue asymptote as we pass through the blue filled circle.  Then we cross the line at infinity again, and jump to the right branch of the hyperbola toward the open blue circle.

Can you see what this means?  In projective geometry, a hyperbola does not have two separate branches.  Adding the line at infinity allows us a means to jump between the two branches.

Let’s look at these scenarios from a slightly different perspective.  Suppose that somehow, we can actually draw the line at infinity.  It’s the blue dashed line in the figure below.  Easy.

Can you see where we’re going with this?  The left circle represents an ellipse, since it is a bounded curve and cannot intersect λ.  The middle circle represents a parabola — it’s a curve which intersects the line at infinity at exactly one point, and so is actually tangent to λ.  The right circle represents a hyperbola, since it intersects the line at infinity at two points.  It may look strange from this perspective, but this right circle really does represent the hyperbola illustrated above.

To summarize:  if a conic does not intersect the line at infinity, we call it a Euclidean ellipse, if it is tangent to λ, we call it a Euclidean parabola, and if it intersects λ in two points, we call it a Euclidean hyperbola.

Because you see, in projective geometry, there is no distinguished line.  Every line is just like any other.  So, in some sense, these is just one conic section in projective geometry!

What we have looked at today is what would be called an embedding of the Euclidean plane in the projective plane.  We saw how we could just add a line at infinity with certain properties, and our plane would behave “projectively.”  But from the perspective of projective geometry, you’ve just got conic sections and lines, with possibly 0, 1, or 2 intersection points as illustrated above.  A little mind-blowing, but absolutely true.

Of course we’ve only been able to scratch the surface of this really amazing geometrical world today.  Hopefully you will be inspired to learn a little bit more about it….

## More on: What is…Inversive Geometry?

The more I thought about it, the more I realized I just couldn’t stop at just one post on inversive geometry.  There is so much interesting geometry to discuss….

This image of a cardioid, made up of a collection of tangent circles, is one of my favorite geometrical constructions.  It’s created by beginning with a base circle (in black).  Choose any point on the circle (like the one in black), which we take to be the origin.

Now for any other point on the black circle (like the point in red), draw a circle with that point as center, and the radius as the distance from the origin, as in the figure.  If you do this several times, the circles you draw will envelope a cardioid.

Why do you get a cardioid?  It’s actually not too difficult to prove using inversive geometry.  We won’t go through all the details, but you’ll see enough to get the gist of the proof.

So let’s go!  There’s actually a lot going on once you’ve added the point at infinity, ω.  If you’ve forgotten about ω, you might want to look at Day091 to refresh your memory.  One property of ω which will be important to our discussion today is that ω lies on any curve which is unbounded.  And the simplest unbounded curve is just a line.

So imagine a line, and some point P on that line.  Now chose a direction long the line, and think about moving along the line in that direction.  What happens?

Well, eventually, you’ll start moving further and further away from the origin.  And as you keep moving further from the origin, you’re actually moving “closer” to the point at infinity.  Not closer in a numerical sense of the distance from ω, since you are always infinitely far away from ω.  But closer in the sense that the line “ends” at the point at infinity.  If ω lies on the line, you’ll need to able to get there — even it takes forever….

Yeah, a bit to wrap your head around.  But it gets even better!  Think back to where you started — the point P — and then move along the other direction along the line.  Keep going…as you approach the point at infinity from the other side of the line.

Do you see what’s happening?  Both ends of the line actually meet at ω!  Yes, the line is a closed curve in inversive geometry, like a circle is in Euclidean geometry.

We’ll come back to this point in a moment, but next we want to ask the following question:  what is the inverse curve of a line?  Recall that the inverse curve of a line is that curve obtained by taking all the individual points on a line, and then taking all of their inverse points.

Interestingly, the inverse curve turns out to be a circle — and not just any circle, but a circle which goes through the origin.  As the proof isn’t all that difficult, we’ll take a few moments to work through it.

So suppose the red circle centered at O has radius 1, and choose a point P.  Now draw the line through P such that the segment OP is perpendicular to the line, and consider another point Q on this line.  Thus, triangle OPQ has a right angle at P.

Since the point at infinity, ω, is on the line, its inverse point, ω′ = O, must lie on the inverse curve.  Thus, the inverse curve (shown in blue) must pass through the origin. Of course P and Q′, the inverse points of P and Q, must also lie on the inverse curve.

Then by the property of inverse points, $[OP]\cdot[OP']=[OQ]\cdot[OQ']=1,$ and so

$\dfrac{[OP]}{[OQ]}=\dfrac{[OQ']}{[OP']}.$

But because inverse points lie on the same ray through the origin, then $\angle POQ$ and $\angle Q'OP'$ must have the same measure, making triangles $\Delta POQ$ and $\Delta Q'OP'$ similar triangles.

What does this mean?  This implies that $\angle OQ'P'$ is also a right angle.  To summarize:  given any point Q on the line, its inverse point Q′  has the property that when joined to O and P′, a right triangle is formed with a right angle at Q′.

But because any triangle inscribed in a semicircle is a right triangle, this means that the set of all inverse points Q forms a circle with diameter OP′.  Neat!

Thinking back to ω being on the line, we see that as we begin at P and move either direction along the line toward the point at infinity, here’s what we’re doing on the inverse curve:  we’re starting at the point P′ and moving either direction around the circle toward the origin.  It really does make sound geometrical sense.

Will this work for any line?  If you think about it for a moment, you’ll find that it does…almost.  Our geometrical argument fails if the line actually goes through the origin.  But it shouldn’t be difficult to convince yourself that the inverse curve of a line going through the origin is actually the same line!  The two points where the line intersects the circle of inversion stay in the same place, but the other points switch in pairs.

Where are we going with all this?  It turns out that adding the point at infinity does, in some sense, make lines behave like circles — we can even imagine a line as a “circle with infinite radius.” And where is the center of this circle of infinite radius?  The point at infinity, of course….

As mentioned last week, the inverse curves of circles not passing through the origin are also circles not passing through the origin.  To summarize:

• Inverses of circles not passing through the origin are also circles not passing through the origin;
• Inverses of circles passing through the origin are lines not going through the origin (and vice versa);
• Inverses of lines through the origin are also lines through the origin, and each such line is its own inverse.

Now here’s the next step.  Given the close relationship between lines and circles in inversive geometry, we define a Circle in inversive geometry (capital “C”, like we did in spherical geometry) to be either a line or circle.  Then here’s the revised summary:

• Inverses of Circles are Circles.

Beautifully simple.  The ability to make simple statements like this with real geometrical meaning is one more reason why inversive geometry it so very interesting.

And with that, we’ll have to call it a week….  In the next (and final) post on inversive geometry, we’ll briefly look at the algebra of inverse curves, and finish the cardioid construction mentioned above.  Until then!

## What is…Inversive Geometry?

It’s been a while since we’ve looked at a new type of geometry, so today I’d like to introduce inversive geometry.  Let’s look at a few pictures first for some motivation.

We would like to consider the operation of inverting points, as illustrated above.  Suppose the red circle is a unit circle, the white dot is the origin, and consider the blue point P.  Now define P′ (the orange dot) to be the inverse point of P as follows:  draw a ray from the origin through P, and let P′ be that point on the ray such that

$[OP]\cdot[OP']=1,$

where $[AB]$ denotes the distance from A to B. That is, when you multiply the distances from the origin of a point and its inverse point, you get 1.

Now why would you want to do this?  It turns out this operation is actually very interesting, and has many unexpected consequences.  Consider the circles in the figure below.

Start with the blue circle.  Now for every point P on the blue circle, find its inverse point P′, and plot it in orange.  The result is actually another circle!  Yes, it works out exactly, although we won’t prove that here.  Note that since every point on the blue circle is outside the unit circle (that is, greater than a distance of 1 from the origin), every inverse point on the orange circle must be inside the unit circle (that is, less than a distance of 1 from the origin).

This is just one geometrical property of inversion, and there are many others.  I just want to suggest that the operation of inversion has many interesting properties.

If you pause a moment to think about this operation, you’ll notice there’s a little snag.  Not every point has an inverse point.  There’s just one point which is problematic here:  the origin.  According to the definition,

$[OO]\cdot[OO']=1.$

But the distance from the origin to itself is just 0, and it is not possible to multiply 0 by another number and get 1, since 0 times any real number is still 0.

Unless…could we somehow make the distance from O to O′ infinite?  It should be clear that O′ cannot be a point in the plane different from the origin, since any point in the plane has a finite distance to the origin (just use the Pythagorean theorem).

So how do we solve this problem?  We add another point to the usual Euclidean plane, called the point at infinity, which is usually denoted by the Greek letter ω.

You might be thinking, “Hey, wait a minute!  You can’t just add a point because you need one.  Where would you put it?  The plane already extends out to infinity as it is!”

In a sense, that’s correct.  But remember the idea of this thread — we’re exploring what geometry is.  When we looked at taxicab geometry, we just changed the distance, not the points.  And with spherical geometry, we looked at a very familiar geometrical object:  the surface of a sphere.  We can also change the points in our geometrical space.

How can we do this?  Though perhaps a simplification, we must do this consistently as long as it is interesting.

What does this mean?  In some sense, I can mathematically define lots of things.  Let’s suppose I want to define a new operation on numbers, like this:

$a\otimes b=a^2+37b.$

Wow, we’ve just created something likely no one has thought of before!  And yes, we probably have — and for good reason.  This operation is not very interesting at all — no nice properties like commutativity or associativity, no applications that I can think of (what are you going to do with the 37?).  But it is a perfectly legitimate arithmetical operation, defined for all real numbers a and b.

As I tried to suggest earlier, the operation of inversion is actually quite interesting.  So it would be very nice to have a definition which allowed every point to have an inverse point.

But if we want to add ω, we must do so consistently.  That is, the properties of ω cannot result in any contradictory statements or results.

It turns out that this is in fact possible — thought we do have to be careful.  Although we can’t go into all the details of adding the point at infinity to the plane, here are some important properties that ω has:

1. The distance from ω to any other point in the plane is infinite;
2. ω lies on any unbounded curve (like a line or a parabola, but not a circle, for example);
3. The inverse point of ω is the origin; that is, O′ = ω and ω′ = O.

So we can create an entirely consistent system of geometry which contains all the points in the Euclidean plane plus a point at infinity which is infinitely far away from every other point.  We usually call the Euclidean plane with ω added the extended plane.

With the new point, we can now say that every point in the extended plane has a unique inverse point.

This idea of adding a point at infinity is also important is other areas of mathematics.  In considering the complex plane — that is, the set of points of the form abi, where i is a solution of the equation $x^2+1=0$ — it is often useful to add the number ∞, so that division by 0 is now possible.  This results in the extended complex plane, and is very important in the study of complex analysis.

The point at infinity is also important in defining stereographic projection.  Here, a sphere is placed on the plane, so that the South pole is the origin of the plane.  The sphere is then projected onto the plane as follows:  for any point on the sphere, draw a ray from the North pole through that point, and see where it intersects the plane.

Where does the North pole get mapped to under this projection?  Take one guess:  the point at infinity!

So here is another new geometry!  As we get introduced to more and more various geometrical systems, I hope you will continue to deepen your intuition about the question, What is a geometry?