## Enumerating the Platonic Solids

The past few weeks, I outlined my approach to a series of lectures on polyhedra.  One of my constraints is that students will not have seen a lot of trigonometry yet, and will not have been exposed to three-dimensional Cartesian coordinates.  But there is Euler’s Formula!  I just finished a pair of lectures on the algebraic enumeration of the Platonic solids using Euler’s Formula, and I thought others might be interested as well.

As a reminder, Euler’s Formula states that if $V,$ $E,$ and $F$ are the number of vertices, edges, and faces, respectively, on a convex polyhedron, then

$V-E+F=2.$

How might we use this formula to enumerate the Platonic Solids?  We need to make sure we agree on what a Platonic Solid is:  a convex polyhedron with all the same regular polygon for faces, and with the same number meeting at each vertex.

To use this definition, we will define a few more variables:  let $p$ denote the number of sides on the regular polygons, and let $q$ denote the number of polygons meeting at each vertex of the Platonic solid.  (Those familiar with polyhedra will recognize these as the usual variables.)

The trick is to count the number of sides and vertices on all the polygons in two different ways.  For example, since there are $F$ polygons on the Platonic solid, each having $p$ sides, there are a total of $pF$ sides on all of the polygons.

But notice that when we build a cube from six squares, two sides of the squares meet at each edge of the cube.  This implies that $2E$ also counts all of the sides on the polygons.  Since we are counting the same thing in two different ways, we have

$pF=2E.$

We may similarly count all the vertices on the polygons as well.  Of course since a regular polygon with $p$ sides also has $p$ vertices, there are $pF$ vertices on all of the polygons.

But notice that when we put the squares together, three vertices from the squares meet at a vertex of the cube.  Thus, if there are $V$ vertices on a Platonic Solid, and if $q$ vertices of the polygons come together at each one, then it must be that $qV$ is the total number of vertices on all of the polygons.  Again, having counted the same thing in two different ways, we have

$pF=qV.$

Thus, so far we have

$V-E+F=2,\quad pF=2E,\quad pF=qV.$

Note that we have three equations in five variables here; in general, such a system has infinitely many solutions.  But we have additional constraints here — note that all variables are counting some feature of a Platonic Solid, and so all must be positive integers.

Also, since a regular polygon has at least three sides, we must have $p\ge3,$ and since at least three polygons must come together at the vertex of a convex polyhedron, we must also have $q\ge3.$

These additional constraints will guarantee a finite (as we know!) number of solutions.  So let’s go about solving this system.  The simplest approach is to solve the last two equations above for $E$ and $V$ and substitute into Euler’s Formula, yielding

$\dfrac{pF}q-\dfrac{pF}2+F=2.$

Now divide through by $F$ and observe that $F>0,$ so that

$\dfrac pq-\dfrac p2+1>0.$

Multiply through by $2q$ and rearrange terms, giving

$pq-2p-2q<0.$

How should we go about solving this inequality?  There’s a nice trick here:  add $4$ to both sides so that the left-hand side factors nicely:

$(p-2)(q-2)<4.$

Now we are almost done!  Since $p,q\ge3,$ then $p-2$ and $q-2$ must both be integers at least $1;$ but since their product must be less than $4,$ they can be at most $3.$

This directly implies that $p$ and $q$ must be $3, 4,$ or $5.$

This leaves only nine possibilities — but of course, not all options need be considered.  For example, if $p=q=5,$ then

$(p-2)(q-2)=9>4,$

and so does not represent a valid solution.  But when $p=3$ and $q=4,$ we have the octahedron, since $p=3$ means that the polygons on the Platonic Solid are equilateral triangles, and $q=4$ means that four triangles meet at each vertex.

So out of these nine possibilities to consider, there are just five options for $p$ and $q$ which satisfy the inequality $(p-2)(q-2)<4.$  And since each pair corresponds to a Platonic Solid, this implies that there are just five of them, as enumerated in the following table:

Actually, this implies that there are at most five Platonic Solids.  How do we know that twelve pentagons actually fit together exactly to form a regular dodecahedron?  A further argument is necessary here to be complete.  But for the purposes of my lectures, I just show images of these Platonic Solids, with the presumption that they do, in fact, exist.

Now keep in mind that in an earlier lecture, I enumerated the Platonic Solids using a geometrical approach; that is, by looking at those with triangular faces, square faces, etc.  I like the problem of enumerating the Platonic Solids since the geometric and algebraic methods are so different, and emphasize different aspects of the problem.  Further, both methods are fairly accessible to good algebra students.  The question of when to take an algebraic approach rather than a geometric approach to a geometry problem is frequently difficult for students to answer; hopefully, looking at this problem from both perspectives will give students more insight into this question.

## Teaching Three-Dimensional Geometry, III

This is the last of a three-part series on teaching three-dimensional geometry.  A few weeks ago, I had begun describing how I would go about putting together a series of about 20 online videos on 3D geometry, each lasting 5–7 minutes.  I just finished a discussion of buckyballs, and why regardless of the number of hexagonal faces on a buckyball, there are always exactly 12 pentagonal faces.

Euler’s Formula was key.  We’ll look at another application of Euler’s Formula, but before doing so, I’d like to point out that students at this level have not encountered Cartesian coordinates in three dimensions, and so I need to find things to talk about at an accessible level.

On to the truncation of polyhedra!  Again, we can apply Euler’s Formula, but it helps to think about the process systematically.  You can count the number of vertices, edges, and faces on a truncated cube, for example, one at a time — but little is gained from a brute force approach.  By thinking more geometrically, we would notice that each edge of the original cube contributes two vertices to the truncated cube, giving a total of 24 vertices.

We can continue on in this fashion, counting as efficiently as possible.  This sets the stage for a discussion of Archimedean solids in general.  A proof of the enumeration of the Archimedean solids is beyond the scope of a single lecture, but the important geometrical ideas can still be addressed.

This concludes the set of lectures on polyhedra in three dimensions.  Of course there is a lot more that can be said, but I need to make sure I get to some other topics.

Like spherical geometry, for instance, next on the slate.  There are two approaches one typically takes, depending how you define a point in spherical geometry.  There is a nice duality of theorems if you define a Point in this new geometry as a pair of antipodal points on a sphere, and a Line as a great circle on a sphere.  Thus two distinct Lines uniquely determine a Point, and two distinct Points uniquely determine a line.

This is a bit abstract for a first go at spherical geometry, so I plan to define a Point as just an ordinary point on a sphere, and a Line as a great circle.  Two points no longer uniquely determine a Line, since there are infinitely many Lines through two antipodal Points.

But still, there are lots of interesting things to discuss.  For example, there is no such thing as a pair parallel lines on a sphere:  two distinct Lines always intersect.

Triangles are also intriguing.  On the sphere, the sides are also angles, measured by the angle subtended at the center of the sphere.  So all together, there are six angular measures in any triangle.

Since students will not have had a lot of exposure to trigonometry at this point, I won’t discuss many of the neat spherical trigonometric formulas.  But still, there is the fact the angle sum of a spherical triangle is always greater than $180^\circ.$  And the fact that similarity and congruence on the sphere are the same concept, unlike in Euclidean geometry.  For example, if the angles in a Euclidean triangle are the same in pairs, the triangles are similar.  But on a sphere, if the angles of two spherical triangles measured the same in pairs, they would necessarily have to be congruent.

In other words, students are getting further exposure to non-Euclidean geometries.  (I did a lecture on inversive geometry in a previous section.)  One nice and accessible proof in spherical geometry is the proof that the area of a spherical triangle is proportional to its spherical excess — that is, how much the angle sum is greater than $180^\circ.$  So there will be something  I can talk about without needing to say the proof is too complicated to include….

The final topic I plan to address is higher-dimensional geometry.  The first natural go-to here is the hypercube.  Students are always intrigued by a fourth spatial dimension.  Ask a typical student who hasn’t been exposed to these ideas what the fourth dimension is, and the answer you invariably get is “time.”  So you have to do some work getting them to think outside of that box they’ve lived in for so long.

One thing I like about hypercubes is the different ways you can visualize them in two dimensions.

Viewed this way, you can see the black cube being moved along a direction perpendicular to itself to obtain the blue cube.  Of course the process is necessarily distorted since we’re looking at a static image.

This perspective highlights a pair of opposite cubes — the green one in the middle, and the outer shell — and the six cubes adjacent to both.

And this perspective is just aesthetically very pleasing, and also has the nice property that every one of the eight cubes looks exactly the same, except for a rotation.  Again, there won’t be any four-dimensional Cartesian coordinates, but still, there will be plenty to talk about.

I plan to wrap up the series with a discussion of volumes in higher dimensions.  As I mentioned last week, I’d like to discuss why you should avoid peeling a 100-dimensional potato….

Thinking by analogy, it is not difficult to motivate the fact that the volume of a sphere $n$ dimensions is of the form

$Kr^n.$

Now let’s look at peeling a potato in three dimensions, assuming it’s roughly spherical.  If you were a practiced potato peeler, maybe you could get away with the thickness of your potato peels being, say, just 1% of the radius of your potato.  This leaves the radius of your peeled potato as $0.99r,$ and calculating a simple ratio reveals that you’ve got $0.99^3\approx0.97$ of your potato left.

Extend this idea into higher dimensions.  If your potato-peeling expertise is as good in higher dimensions, you’ll have $0.99^n$ of your potato left, where $n$ is the number of dimensions of your potato.  Now $0.99^{100}\approx 0.366,$ so after you’ve peeled your potato, you’ve only got about one-third of it left!

What’s happening here is that as you go up in dimension, there is more volume near the surface of objects than there is near the center.  This is difficult to intuit from two and three dimensions, where it seems the opposite is the case.  Nonetheless, this discussion gives at least some intuition about volumes in higher dimensions.

And that’s it!  I’m looking forward to making these videos; I actually made my first set of slides today.  As usual, if I come across anything startling or unusual during the process, I’ll be sure to post about it!

## Teaching Three-Dimensional Geometry, II

A few weeks ago, I began a discussion of what I’d be presenting in a series of twenty (or so) 5—7 minute videos on three-dimensional geometry. I didn’t get very far then, so it’s time to continue….

So to recap a bit, I’ll begin with the usual cones/cylinders/spheres, looking at surface areas and contrasting flat surfaces with the surface of a sphere. Then on to a prelude to calculus by looking at the volume of a cone as a limiting case of a stack of circular disks.

Next, it’s on to polyhedra! A favorite topic of mine, certainly. Polyhedra are interesting, even from the very beginning, since there is still no accepted definition of what a polyhedron actually is. The exception is for convex polyhedra; a perfectly good definition of a convex polyhedron is the convex hull of a finite set of points not all lying in a single plane. Easy enough.

But once you move on to nonconvexity, uncertainties abound. For example, from a historical perspective, sometimes the object below was a polyhedron, and sometimes it wasn’t. Sounds odd, but whether or not you consider this object a polyhedron depends on how you look at the top “face,” which is a square with a smaller square removed from the center. Now is this “face” a polygon, or not? Many definitions of a polygon would exclude this geometrical object – which is problematic if you want to say that a polyhedron has polygons as faces.

So this brings us to a definition of a polygon, which is problematic in its own way – to see why, you can look at a previous post of mine on the definition of a polygon.  Now the point here is not to resolve the issue in an elementary lecture, but rather point out that mathematics is not “black-and-white,” as students tend to believe. Also, it provides a nice example of the importance of definitions in mathematics.

Now this would be discussed briefly in just one video. Next would be the (obligatory) Platonic solids – where else is there to begin? The simplest starting point is the geometric enumeration by looking at what types of polygons – and how many – can appear at any given vertex of a Platonic solid. This enumeration is straightforward enough.

Next, I plan on computing the volume of a regular tetrahedron using the usual $Bh/3$ formula. This is not really exciting in and of itself, but in the next lecture, I plan to find the volume of a regular tetrahedron by inscribing it in the usual way in the cube by joining alternate vertices.

Of course you get the same result. But for those of us who work a lot in three-dimensional space, we deeply understand the simple algebraic equation, $2 \times 4=8.$ What I’m referring to, specifically, is that the number of vertices on a three-dimensional simplex is half the number of vertices of a three-dimensional hypercube.

This simple fact is at the heart of any number of intriguing geometrical relationships between polyhedra in three dimensions. In particular, and quite importantly, the simplex and the cross-polytope together fill space. This relationship is at the heart of many architectural constructions in additional to generating other tilings of space with Archimedean solids. But most students have never seen this illustrated before, so I think it is important to include.

Then on to a geometry/algebra relationship: having enumerated the Platonic solids geometrically, how do we proceed to take an algebraic approach? A fairly direct way is to use Euler’s formula to find an algebraic enumeration.

No, I don’t intend to prove Euler’s formula; by far my favorite (and best!) is Legendre’s proof which involves projecting a polyhedron onto a sphere and looking at the areas of the spherical polygons created. This is a bit beyond the scope of this series of videos; there simply isn’t time for everything. But it is important to note the role that convexity plays here; yes, there are other formulas for polyhedra which are not essentially “spheres,” but this is not the place to discuss them.

Next, I want to talk about “buckyballs.” I still have somewhat of a pet peeve about the nomenclature – Buckminster Fuller did not invent the truncated icosahedron – and so the physicists who named this molecule were, in my opinion, polyhedrally rather naïve. But, sadly (as is the case so many times), they did not come to me first before making such a decision…

The polyhedrally interesting fact about buckyballs is this: if a polyhedron has just pentagonal and hexagonal faces, three meeting at every vertex, then there must be exactly twelve pentagons. Always.

Now I know that the polyhedrally savvy among you are well aware of this – but for those who aren’t, I’ll show you the beautiful and very short proof. Once you’ve seen the idea, I don’t think you’ll ever be able to forget it. It’s just remarkable – even with 123,456,789 hexagons, just 12 pentagons.

So let $P$ represent the number of pentagons on the buckyball, and $H$ represent the number of hexagons. Then the number of vertices $V$ is given by

$V=\dfrac{5P+6H}{3},$

since each pentagon contributes five vertices, each hexagon contributes six, and three vertices of the polygons meet at each vertex of the buckyball.

Moreover, the number of edges is given by

$E=\dfrac{5P+6H}{2},$

since the polygons on the buckyball meet edge-to-edge. Of course, $F=P+H,$ since the faces are just the pentagons and hexagons. Substitute these expressions into Euler’s formula

$V-E+F=2,$

and what happens? It turns out that $H$ cancels out, leaving $P=12!$

Amazes me every time. But what I like about this fact is that it is accessible just knowing Euler’s formula – no more advanced concepts are necessary.

And yes, there’s more! This is now Lecture #12 of my series, so I have a few more to describe to you. Until next time, when I caution you (rather strongly) against peeling a 100-dimensional potato….

## Teaching Three-Dimensional Geometry, I

I have recently had a rather unusual opportunity.  I’ve talked a bit over the last few months about my consulting work producing online videos for a flipped classroom; I’ve been working busily on the Geometry unit.

Now the last section of this unit is on three-dimensional geometry, and I’ve been given pretty free reign as to what to cover in this 20-lectures series of 5-7 minute videos.  And given my interest in polyhedra (which I could focus on exclusively with no shortage of things to discuss!), I felt I had a good start.

But the challenge was also to cover some traditional topics (cones, cylinders, spheres, etc.) — as well as more advanced topics — while not using mathematics beyond what I’ve used in the first several sections of the Geometry unit.

There is, of course, no “correct” answer to this problem.  But I thought I’d share how I’d approach this series of lectures, since geometry is such a passion of mine — and I know it is for many readers as well.  The process of reforming high school geometry courses is now well underway; I hope to contribute to this discussion with today’s post.

Where to start?  Cones and cylinders — a very traditional beginning.  But I thought I’d start with surface areas.  Now for cylinders, this is pretty straightforward.  It’s not much more difficult for cones, but the approach is less obvious than for cylinders.

Earlier in the unit, we derived the formula for the area of a sector of a circle, so finding the lateral surface area of a cone is a nice opportunity to revisit this topic.  And of course, finding the lateral surface area of a cylinder involves just finding the area of a rectangle.

Now what do both of these problems have in common?  Their solution implies that cones and cylinders are flat.  In other words, we reduce what is apparently a three-dimensional problem (the surface area of a three-dimensional object) to a two-dimensional problem.

This is in sharp contrast to finding the surface area of a sphere — you can’t flatten out a sphere.  In fact, the entire science of cartography has evolved specifically in response to this inability.

So this is a nice chance to introduce a little differential geometry!  And no, I don’t really intend to go into differential geometry in any detail — but why not take just a minute in a lecture involving spheres to comment on why the formulas for the surface areas of cones and cylinders are fairly easy to derive, and why — at this level — we’re just given the formula for the surface area of a sphere.

I try to mention such ideas as frequently as I can — pointing out contrasts and connections which go beyond the usual presentation.  Sure, it may be lost on many or most students, but it just may provide that small spark for another.

I think such comments also get at the idea that mathematics is not a series of problems with answers at the back of the book…on the face of it, there is no apparent reason for a student to think that finding the surface area of a cone would be simpler than finding the surface area of a sphere.  This discussion gets them thinking.

Next, I’m planning to discuss Archimedes’ inscription of a sphere in a cylinder (which involves the relative volumes).  This is a bit more straightforward, and it’s a nice way to bring in a little history.

I also plan to look at inscribing a sphere in a right circular cone whose slant height is the same as the diameter of the base, so that we can look at a two-dimensional cross-section to solve the problem.  In particular, this revisits the topic of incircles of triangles in a natural way — I find it more difficult to motivate why you’d want to find an incircle when looking at a strictly two-dimensional problem.

Now on to calculus!  Yes, calculus.  One great mystery for students is the presence of “1/3” in so many volume formulas.  There is always the glib response — the “3” is for “three” dimensions, like the “2” in “1/2 bh” is for “two” dimensions.

When deriving these formulas using integration, this is actually exactly a fairly solid explanation.  But for high school students who have yet to take calculus?

It is easy to approximate the volume of a right circular cone by stacking thin circular disks on top of each other.  If we let the disks get thinner and take more and more of them, we find the volume of the cone as limit of these approximations.  All you need is the sum

$\displaystyle\sum_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}6.$

I plan to prove that

$\displaystyle\sum_{k=1}^nk=\dfrac{n(n+1)}2,$

and then prove (or perhaps just suggest — I’m not sure yet) the formula for the sum of squares.

I think a fairly informal approach could be successful here.  But I do think such discussions are necessary — in calculus, I’ve routinely asked students why certain formulas they remember are true, and they struggle.  As a simple example, students can rarely tell me why the hypotenuse of a 30-60-90 triangle is twice as long as the shorter leg.

When teachers just give students formulas and ask them to plug numbers in to get answers to oversimplified word problems, of course there is a sense of mystery/confusion — where did these formulas come from?  I’m hoping that this discussion suggests that there is a lot more to mathematics than just a bunch of formulas to memorize.

As usual, I realize I have much more to say on this topic than I had originally supposed…I’ve only discussed up to the fifth lecture so far!  Since I have not had extensive experience teaching more traditional topics at the high school, it has been an interesting challenge to tackle the usual geometry topics in a way that grabs students’ attention.  It’s a challenge I enjoy, and of course I’ll have much more to say about it next week….

## Guest Blogger: Scott Kim, IV

Well, this is the last installment of Scott Kim’s blog post on transforming mathematics education!  These are all important issues, and when you think about them all at once, they seem insurmountable.  It takes each of us working one at a time in our local communities, as well as groups of us working together in broader communities, to effect a change.  What is crucial is that we not only discuss these issues, but we do something about them.  Those of us who participated in the discussion a month ago at the Bay Area Mathematical Artists Seminar are definitely interested in both discussing and doing.

Scott suggests we need to move past our differences and find constructive ways to act.  No, this isn’t easy.  But we need to do this to solve any problem, not just those surrounding mathematics education.  It’s time for some of us to start working on these issues, and many others of us to continue working.  We can’t just sit and watch, passively, any more.  It’s time to act.  What are you waiting for?

## Level 4. Resistance from SOCIETY (quarreling crew)

Sailing is a team sport. You can’t get where you want to go without a cooperative crew. Similarly, math education reform is a social issue. You can’t change how math is taught unless parents, teachers, administrators and policy makers are on board. Most adults cling to the way they were taught as if it were the only way to teach math, largely out of ignorance — they simply aren’t aware of other approaches.

Here are three ways society needs to change the way it thinks about math and math education in order for change to happen.

4a. Attitude. The United States has an attitude problem when it comes to math teachers. First, we underpay and under-respect teachers. And the situation is only getting worse as math graduates flock to lucrative high-tech jobs instead of the teaching profession. The book The Smartest Kids in the World and How They Got That Way describes how FInland turned their educational system around — they decided to pay teachers well, set high qualification standards, and give teachers considerable autonomy to teach however they think is best, with the remarkable result that student respect for teachers is extremely high.

Second, it is socially acceptable, even a badge of honor, to say that you were never good at math. You would never say the same thing about reading. Many people do not in fact read books, but no one would publicly brag that they were never good at reading. Our society supports the idea that parents should read to their kids at night, but perpetuates the idea that being no good at math is just fine.

Solution: respect teachers by paying them well, and value math literacy as much as we value reading literacy.

4b. Vision. The national conversation about math education in the United States is locked in a debate about whether we should teach the basics, or the concepts. As a result we see over the decades that the pendulum swings back and forth between No Child Left Behind and standardized testing on one extreme, and New Math and Common Core Math on the other extreme. As long as the pendulum keeps swinging, we will never settle on stable solution. The resolution, of course, is that we need both. In practice, schools that overemphasize rote math find that they must supplement with conceptual exercises, and schools that overemphasize conceptual understanding find that they must supplement with mechanical drill. We need both rote skills and conceptual understanding, just as kids learning to read need both the mechanical skills of grammar and vocabulary, and the conceptual skills of comprehension and argument construction.

Solution: We need a vision of math education that seamlessly integrates mechanical skills and conceptual understanding, in a way that works within the practical realities of teacher abilities and schoolday schedules. To form a vision, don’t just ask people what they want. A vision should go further than conventional wisdom. As Henry Ford is reported to have said (but probably didn’t), “If I had asked people what they wanted, they would have said faster horses.” Or as Steve Jobs did say, “It’s really hard to design products by focus groups. A lot of times, people don’t know what they want until you show it to them.”

4c. The will to act. As a child I grumbled about the educational system I found myself in. As a young adult I started attending math education conferences (regional meetings of the National Council of Teachers of Mathematics), and was astonished to find that all the thousands of teachers at the conference knew perfectly well what math education should look like — full of joyful constructive activities that challenged kids to play with ideas and think deeply. Yet they went back to their schools and largely continued business as usual. They knew what to do, but were unwilling or unable to act, except at a very small scale.

Solution: Yes, a journey of a thousand miles starts with a single step. And change is slow. But if we’re to get where we want to go, we need to think bigger. Assume that big long lasting change is possible, and in the long term, inevitable. As Margaret Mead said, “Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it’s the only thing that ever has.” I’m starting my small group. Others I know are starting theirs. What about you?

## Circle Geometry

Today, I thought I’d share a little more about things learned along the way with my curriculum consulting.  As I mentioned before, I’m creating a series of online lectures for the Geometry unit.  This past week, the section I was working on (and will still be working on into next week) is Circle Geometry.

As I also remarked earlier, I’m using the University of Chicago School Mathematics Project’s textbook on Geometry as a reference.  In this text are many theorems about the measure of the angle between two intersecting lines in terms of the measures of the intercepted arcs.

This image is certainly familiar.

The question I had to consider was how to organize all these results in a coherent 5–7-minute lecture.  It turns out that there was too much for just one lecture, so I did spread it out into two.  But I still needed a flow.

Although the results were not new to me, I had never taught this topic before.  My main experience teaching geometry at the high school level was designing and teaching a course on spherical trigonometry as it applies to studying polyhedra.  So this gave me an opportunity to stand back and just think about putting it all together.

I was happy with what I came up with — an approach which could be classified under “combinatorial geometry.”  I decided to pose the following question:

In other words, if you have two intersecting lines, and you draw a circle so that it intersects both lines, what configurations are possible?

Looked at in this way, there are just two considerations:  whether the intersection of the lines is outside, on, or inside the circle, and whether the lines are secant or tangent lines.

It’s not difficult to make the enumeration, so I’ll just give it briefly here.  There is only one configuration if the intersection of the lines lies inside the circle, since both lines must be secant lines.

When the intersection of the lines is on the circle, one of the lines may be tangent, although both cannot be since there is a unique tangent at any given point on a circle.  And when the intersection of the lines is outside the circle, zero, one, or two of the lines may be tangent to the circle.

This enumeration allows for a systematic approach.  If you’ve ever worked through find the angle measures, you know that starting with the arrangement in the upper right corner is the way to begin.  I won’t go through all the details, but I will just indicate that the following figure is all you need:

This simple case is analyzed by considering $\angle QOR$ as an angle exterior to $\Delta POQ.$  The analysis of all the other cases builds from this.

I decided to include a discussion not found in the UCSMP text — continuity.  Of course this is not a topic which can be rigorously discussed at say, the 10th-grade level.  But why not give students an intuition of the idea?

This series illustrates the case that the intersection of the lines is outside the circle, and one of the lines is tangent.  We look at this as the limiting case of a series of pairs of secant lines.

This argument depends upon the fact that the measurement of all arcs and angles varies continuously as $S$ moves around the circle.  While, as mentioned, this cannot be addressed rigorously, it is a very intuitive argument.  Moreover, there are many different software packages you could use to make an animation of this process, and display all the arc and angle measurements as $S$ moves around the circle.

There is no reason not to introduce this argument.  In my pair of lectures, I used more traditional geometrical arguments as well.  It doesn’t hurt students to be exposed to a wide range of proof ideas.

I summarize all of these results in the following graphic.

The measure of the angle indicated with the red dot is half the measure of the intercepted arc, or the sum/difference of the measures of the intercepts arcs, shown in red and blue.  An arc in blue indicates its measure is to be subtracted rather than added.  I was very happy with this graphic.  I think that if a student followed the lecture, they could state every result just by looking at it.

This also proved to be a great segue into looking at the power of a point.  I thought I’d begin with the figure in the upper left, proving the usual theorem using similar triangles.

And now for another continuity argument!

This is a nice way to see that the power of any point on the circle is 0.  It is also a nice contrast to the theorems about the angle between the intersecting lines:  when $PT$ and $RT$ eventually reach 0, you’re not able to conclude anything about a relationship between $QT$ and $ST.$

This means that there is no theorem relating lengths of segments for the two cases when the intersection of the lines lies on the circle.   I use the following graphic to indicate this, with two cases grayed out when the power of the point offers no conclusion.

Now all of these results are the usual ones found in high school geometry textbooks; nothing new here.  But for me, just having to step back and think about how to put them all together was a fun challenge.

Again, I am surprised at how much I’m learning even though I’m just putting together a few slides on elementary geometry.  The process of writing these lectures is an engaging one, and I hope the students who will eventually watch them will benefit from a perspective not found in more traditional textbooks.

## What Is…A Polygon?

A haven’t made a post in quite some time in my “What is a Geometry?” thread.  In working on my online lectures in the section on Polygons, I of course needed to define just what a polygon is.  This turned out to be a little more challenging than I had imagined.  I thought that the issues that arose would make this discussion an interesting continuation of the “What is a Geometry?” thread.

In general, I think that the Wikipedia does a good job with mathematics — but specifically, the definition of a polygon leaves quite a bit to be desired.  I’ll reproduce it here for you:

In elementary geometry, a polygon is a plane figure that is bounded by a finite chain of straight line segments closing in a loop to form a closed polygonal chain, or circuit.  These segments are called its edges or sides, and the points where two edges meet are the polygon’s vertices (singular: vertex) or corners.

OK, maybe not so bad of a start.  There are lots of examples given which fit this definition, but many which do not.  For example, this definition allows consecutive segments to lie on the same line, which is typically disallowed in most other definitions of polygons.

So maybe a clause may be added to the definition which does not allow this.  But then we encounter a polygon like this (I’m using screenshots from my lecture as illustrations):

My definition begins with a list of vertices — but the problem is still the same.  The vertex labeled “4” is on the edge joining the vertices labeled “1” and “2.”  Again, this is usually avoided.

And what about the following figure?

With the Wikipedia definition, a vertex can be an endpoint of more than two edges of a polygon.  Again, problematic.  There would be no way to distinguish this figure from a single polygon and two different triangles sharing a vertex.

Moreover, there is no condition saying that the straight line segments need to be distinct.  So the same segment might occur multiple times as an edge of a polygon.

None of these behaviors is illustrated anywhere on the Wikipedia page.  I’ve done some Wikipedia editing a while back, and would be interested in working on this page when I have more time to devote to such things.

So what is the fix?  I’m using the Geometry text of the University of Chicago School Mathematics Project as a reference, which is one of the most rigorous geometry texts around.  Here is their definition:

They remark that this is the definition used in 23 out of the 45 geometry text they surveyed.  And in fact, it is the rewrite of a definition in previous editions:  “A polygon is the union of segments in the same plane such that each segment intersects exactly two others, one at each of its endpoints.”  This definition was problematic, though, since by this definition, the following is actually a polygon!

Now this revised definition solves all of the problems above — but I couldn’t use it.  Why not?

One of the sections I’ll be writing lectures for is three-dimensional geometry — and (of course) I’ll be saying a lot about polyhedra in this section.  There are Platonic and Archimedean solids, as well as the Kepler-Poinsot polyhedra, like the small stellated dodecahedron shown below.

The faces of the small stellated dodecahedron are pentagrams, five meeting at each vertex.

But the UCSMP definition does not allow edges to cross.  Each edge meets exactly two others, at each of its endpoints.  So that means that an edge cannot cross another in its interior.

Now I just can’t talk about polyhedra without talking about nonconvex examples.  Sure, it is possible to talk about pentagrams with edges crossing as decagons without crossing edges.

But this would be the height of absurdity.  Besides the fact that none of the dozens of books and articles I’ve read on polyhedra in the past few decades ever do such a thing — and I’m sure none ever will.

So I had to go it alone.  I’ll share with you my definition — but I can’t say it’s the best.  The difficulty lies with being mathematically precise while still making the definition accessible to high school students.  Here it is:

A polygon is determined by a list of its vertices. Edges of the polygon connect adjacent vertices in the list, and there is also an edge connecting the last vertex in the list to the first one. All vertices in the list must be different. Finally, no three consecutive vertices of the polygon can lie on the same line, and no vertex can lie in the interior of another edge.

I don’t think this is too bad.  But there is still a subtle glitch, which I haven’t worked out yet, and which doesn’t necessarily need to be worked out at this level.  When I talk about triangles, for example, I allow cases where the sides have lengths 3, 4, and 7, for example.  But I qualify such a triangle by calling it a degenerate triangle.

Since a triangle is a polygon, a degenerate triangle should be a degenerate polygon, right?  The problem is that calling something a “degenerate polygon” gives the impression that it is actually some type of polygon.  But a degenerate triangle, by my definition, is not a polygon.  So when I use the term degenerate polygon, I’m not actually talking about a polygon….

So I’ll let you think this over.  I just wanted to share how surprised I was at how subtle the definition of something so “simple” could be.  An ordinary polygon.

If you find this sort of question intriguing, you might go online and research all the various definitions of polyhedron.  Convex polyhedra are easy to define (as are convex polygons), but when you get into the different types of behavior possible in the nonconvex cases, well, it becomes problematic.  In fact, no one, as far as I know, has ever come up with a satisfactory definition for “polyhedron.”  Might even do a blog post on that some day….