 I'm delighted to bring up a dear friend of both mine and the museum's, Dr. Joel Spencer who will introduce tonight's talk. Thank you Cindy. It's always really a pleasure to come back to MoMath and it's a particular pleasure to be at the Math Encounters. So mathematicians love games, go, chess, bridge, but all kinds of games. My personal favorites are sorry, always loved sorry, and my current favorite is Ticket to Ride, which is a wonderful game. And mathematicians love to analyze games, like Tic Tac Toe and End Dimensions, whatever. And some highly creative mathematicians like to create games. And Jim is a highly creative, highly inventive mathematician, and his world always has new interesting ideas, always with an intriguing geometry involved with it. His approaches to mathematics are always fascinating. Jim's talents were recognized early. He was Magna Cum Laude in Math as an undergraduate at Harvard, and after some seasoning at that other Cambridge across the pond, he received his PhD in Ergodic Theory from Berkeley. He's visited Harvard, Olin, Brandeis, Maryland, consulted at Microsoft, AT&T, has been on the faculty of University of Washington, and is currently professor at University of Massachusetts at Lowell. And my own personal connection with Jim is with his chip firing game, where there are piles of chips on, let's say, the lattice points in the plane, but it could be much more general than that. And then Jim has a way of moving them around in some strictly deterministic fashion that he's came up with. My contribution is, I call it the prop machine, to name it for him, and I think it's caught on. But it appears as though all the chips are moving randomly. And Jim's ideas have started an entire area of what's called quasi-random walks, walks that are totally deterministic, and yet they behave as though they're random walks. So how do mathematicians prove theorems? Well, Bertrand Russell described the finished product in his usual beautiful prose. Mathematics rightly viewed possesses not only truth but supreme beauty, a beauty cold in austere like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure and capable of a stern perfection such as only the greatest art can show. Well, yes, I agree, but doing mathematics is completely different. And I want to give a quote by somebody who was an inspiration for both myself and for Jim and for countless others. His name is Jean-Carlo Rota. He was a professor of mathematics at MIT. Also for a couple of years right here at the Rockefeller Institute. And he was a giant of 20th century combinatorics, particularly algebraic combinatorics, which in one way he really developed the field in a tremendous way. But at the same time mathematics is an intensely personal subject. And he was an inspirational figure. A person, if you listen to him talk, you wanted to work on his problems. And he was incredibly encouraging to young talent as Jim and I were a couple of years ago. So here is what he says about doing mathematics. A mathematician's work is mostly a tangle of guesswork, analogy, wistful thinking, and frustration, and proof, far from being the core of discovery, is more often than not a way of making sure that our minds are not playing tricks. Jim is an inspirational teacher, an inspirational expositor, and today he'll give us some insight into just how mathematicians work on mathematical problems. So please welcome Jim Prott. Thank you for those words, Joel, and thank you, Cindy and Tim, for inviting me. I've been looking forward to giving this talk for a very long time. So I'm guessing that nearly all of you walked through the entranceway to get here. How many of you noticed that you were walking through a red hexagon? Well, when you go back out, there's a cube right at the entranceway. It's got six glass sides, which we mathematicians prefer to call faces. And each of those faces has a diagonal stripe on it, which together form a regular hexagon tilted at a jaunty angle. This is one of the less obvious cross sections that you can get when slicing a cube in half. You can play with cross sections of a cube using that wall of fire over there, which is actually closed now because the museum is closed, but there's a portable version here which is on, which is called the Ring of Fire. So let me switch the display to that and let me take this cube and you can see a square cross section. A little lower? Okay. And if you want to see that hexagonal cross section, I mentioned here it is. So this girl in the picture has created a four-sided cross section of a cube in the Ring of Fire. So I'm going to talk about a theorem that's not an especially important theorem and it wasn't especially hard to prove, but I wanted to tell you about it because it was conceived right here a couple of years ago. Here's the kind of question that leads to the theorem. When you take a triangular pyramid, which mathematicians prefer to call a tetrahedron, and intersect it with a plane, what can you get? So one thing you can get is a triangle. Lower? Okay. Get a triangle. A fun challenge is to try to find a square cross section of this tetrahedron and I'll come back to that later. But it turns out you can get a triangle, you can get a square. Can you get a pentagon? How many of you think you can get a pentagonal cross section? It turns out that you can't because the tetrahedron has just four faces. One on the bottom and then these three tilted ones up here. Four faces total. And when you intersect a plane with this shape, that plane can cross each face at most once. And that intersection is a side of the cross section polygon. So since the most number of sides, oops there, I just answered the question. There you can see the square cross section. Yeah, let's switch that. So there we see a configuration in which the wall of fire is crossing each of the four sides of the tetrahedron. And so we get four sides of the intersection polygon. But you can't get more because it would have to cross five faces of this tetrahedron and it only has four faces. So now let's go up to the cube. When you intersect a cube with a plane, what can you get? Well, it's not hard to get three sides or four sides. Can you get five sides? Yeah, you can get five sides. You can even get six sides. Can the polygon of intersection have more than six sides? I see you shaking your heads, no. And it's because the cube has only six sides or as I prefer to say faces. And the plane of fire can only cross as many faces as there are and here there are only six. So if you cross the plane with all six of the faces at once, you can get a six-sided intersection but you can't get more than six. So now we know what a cross section can look like. Now we can ask another question. What is the average number of sides? Okay? But what does average mean when you're averaging over an infinite set, the set of all possible cross sections? So it turns out that the answer to such questions can depend subtly on what you mean by average. And this phenomenon shows up very dramatically in a famous probability paradox called Bertrand's Paradox. So here's a circle and an equilateral triangle inscribed in the circle. Let's call it side-length S. So some chords of the circle, like the red one, shown at the bottom, are shorter than S and others, like the green one that's almost a diameter in the middle of the picture, are longer than S. Joseph Bertrand asked, what's the probability that a random chord of the circle is longer than S? So how many of you think the answer is a half? Raise your hand. The answer is a third. Raise your hand. How many of you think the answer is a quarter? Raise your hand. Okay? So it turns out Bertrand proved that all of you who raised your hands are right. Okay? He proved that the answer is a half and he proved that the answer is a third and he proved that the answer is a quarter using different notions of the phrase random chord. So the big lesson of that era in the theory of geometric probability is that you have to be very precise about what you mean by random. So those of you who didn't raise your hand, you're right too, because the question doesn't have a single right answer. So I won't go into Bertrand's proofs, but to see why the word random is subtle, imagine picking a random point on Earth's surface by choosing a random latitude and a random longitude. If you do this on a transparent globe and you shift the poles a little bit, this is what you get. We do it a thousand times and plot the points on a hollow or glass sphere. You'd see a lot of clustering of the sample points around the poles and the sample points would be a lot more sparse near the equator. So here you see those clusters at those two points which aren't at the top and bottom because I wanted the clusters to be clearer, so move them off-center and you can see clusters at both poles because the sphere is transparent. So that doesn't look random. It doesn't look like what we think random should mean. The latitude band between 0 and 10 degrees north comprises much more of Earth's surface area-wise than the latitude band between 50 and 60 degrees north which in turn is much bigger than the latitude band between 80 and 90 degrees north and yet our system of choosing a random latitude is going to give equal numbers of points in all three of those latitudinal bands even though they aren't all occupying the same area. That's what's going on with that clustering. Those bands have the same width but they have very different lengths and so they have very different areas. So what's wrong with this scheme for picking a random point by choosing a random latitude between negative 90 and positive 90 and choosing a random longitude between 0 and 360 is that the spacing of the points isn't statistically uniform and we think random should mean uniform. More specifically, the probability that a randomly chosen point on the sphere lies in a particular region let's call that region R should be proportional to the area of the region R. It shouldn't depend on where R is whether it's near the equator or near one of the poles. So now we know what's wrong with this scheme. It's not so easy to come up with a good scheme. How to accomplish a way of choosing points should be statistically uniform but let's do it in a different setting where it's easy to do, namely a line segment. We want to choose a point on a line segment so that each piece of the line segment will have the same statistical properties as any other and we can do this with the probabilist's favorite tool, a fair coin. So let's consider a line segment of length 1. We want to choose a random point on it. I'm going to toss a coin a hundred times if I wanted to get more accuracy maybe I'd do it a thousand times but a hundred would be good enough for our purposes. Toss a coin a hundred times to generate a hundred bits where we use zero for heads and one for tails. Then I can use those zeros and ones to write down a real number between zero and one in binary just writing the digits, those bits, in order. So for example, if I toss heads followed by tails I'm going to write down point zero one and then maybe more bits after it according to the coin tosses that I get. Now if you've pre-selected some interval the probability that my number that I choose with these coin tosses will land in your interval is going to be proportioned to the length of the interval you've chosen. In fact, it will be equal to the length of the interval. Let me not prove that for all intervals. Let me just do it for one interval that illustrates the idea. I'll use the interval from a quarter to a half. That interval consists of the numbers which if you write them in binary begin point zero one dot dot dot. So my random number will land in your pre-selected interval if the first bit of my number is a zero and the second bit of my number is a one. But that corresponds to my first tossing heads on my first coin toss and then tossing tails on the second. Since this is a fair coin the probability of tossing heads is a half. The probability of tails is a half. And since a fair coin has no memory those events are statistically independent of each other so the probability of tossing heads first and then tails is going to be a half times a half which is a quarter. And a quarter is exactly the length of the interval that you pre-selected from a quarter to one half. That's one example of how this way of choosing a point at random has the desired properties of representing each sub-interval proportionately according to its length. So we can use coin tosses to find a random point on the number line between zero and one or more generally a random point in any finite length interval. And computers do this sort of thing all the time except instead of tossing coins they use pseudo-random number generators. So now let's go back to that question what is the average number of sides of a random slice? If we want to be respectable probabilists we'd say if you choose a slice at random what is the expected value of the number of sides? I won't give a technical definition of expected value which is also called mathematical expectation but it more or less corresponds to the intuitive idea of average value. I also want to clarify a few implicit assumptions. First of all, I want the wall of fire to be infinite, not finite like this disk over here. I'm going to call it the plane of fire just to highlight that. So I'm not interested in cases where no part of the cube is in the plane of fire so that the cross-section is the empty set. So I'm not interested in the cube being all the way out here or in some other part of the museum or out on the street. I only care about looking at the cross-section when it actually intersects the plane of fire. If just one point of the cube lies in the plane of fire that's fine. But if no points of the cube are in the plane of fire we rule that out. We're not interested in those positions. It's all unclear how to randomize the orientation of the cube, how to wobble it around all the different possible axes. So I'm not going to want to get into that. So instead I'm going to hold the orientation of this cube fixed while altering its position in space. So I'm allowed to translate it but not to rotate it. For instance, let's look at all the cross-sections formed when the body diagonal of the cube that joins one corner of the cube to the opposite corner is perpendicular to the wall of fire. So let's switch back. So I'm going to hold it like this and the cross-section is a point and it's a, thank you, it's a point, it's a triangle, it's a hexagon, it's a triangle, point, and now it's gone. So some of the intersections were triangles and some were hexagons. Are these two possibilities equally likely? Or do we see a different mixture of sometimes triangle, sometimes hexagon? So in the spirit of geometric probability we replace this by a geometric question. As we slide the cube through the wall at constant speed, what fraction of the time is the cross-section a triangle and what fraction of the time is the cross-section a hexagon? It's going to be helpful to look at things from the cube's point of view. So to imagine that the cube is still and the plane is what's doing the moving but moving at constant speed. I'm going to put the cube in standard position so that its corners, its eight corners are triples of coordinates each of which is a zero or one. So the point zero zero zero is down there at the lower left. The point one one one is at the upper right and all the other points have coordinates x equals zero or one, y equals zero or one and z equals zero or one. And the coming plane that I'm going to use is going to be perpendicular to the line segment joining zero comma zero comma zero to one comma one comma one. And those are planes of the form x plus y plus z equals some constant and that constant is going to change. We're going to imagine that constant changing at constant speed. So we're going to start with the plane x plus y plus z equals zero which intersects the cube at just one point the corner point zero zero zero. Now I'm going to move that cutting plane a fixed distance so now I'm looking at x plus y plus z equals zero point two five. So now the cross section is going to be triangular. Move the plane again by an equal amount so we get x plus y plus z equals zero point five zero. It's a larger triangle is the cross section. A larger triangle is the cross section. When we get to x plus y plus z equals one it's a big triangle. But now something different happens. If the plane continues to move the three sides of the triangle broaden out to become sides of their own while the sides of the triangle we had before start to get shorter. So we get a hexagon with three short sides and three long sides. If we now progress to x plus y plus z equals one point five zero the lengthening short sides become the same length as the shortening long sides of the hexagon. So we get a regular hexagon the same regular hexagon you walked through when you came in the entrance. Now we have x plus y plus z equals one point seven five the shorter sides that used to be the long sides are getting even shorter and vice versa for the other sides until finally we get to x plus y plus z equals two we're back at a triangle but the plane's journey isn't over yet we keep going to x plus y plus z equals two point two five we get a triangle smaller triangle by the way all these cross sections are equally spaced from each other so as you increase the right hand side of the equation x plus y plus z equals something as you change the right hand side at constant rate the plane is moving at a constant speed through space and finally we get to x plus y plus z equals three that's going to intersect the cube just at the point one comma one comma one so for the first third of the plane's journey we see a triangle for the next third we see a hexagon and for the last third we see a triangle so two thirds of the time the cross section has three sides and one third of the time the cross section has six sides so the average number of sides is two thirds times three plus one third times six which is four or here's another way to compute it which might be helpful to you during the exercise I'm going to give you during the intermission you say it's three four while that's the number of sides then the number of sides is six for an equal amount of time and then it's three again for an equal amount of time so the average over time is three plus six plus three divided by the number of bands into which we've divided our time interval, namely three so we get three plus six plus three over three which is four if you didn't know the answer was four you might still be able to guess it by doing computer experiments so what do I mean by that? you'd have the computer look at the intersection between the unit cube and the plane x plus y plus z equals t is chosen randomly between zero and three so I told you before how to sample from a segment or from an interval in the real line so you can use that method to choose a random number between zero and three choose the cutting plane x plus y plus z equals that number t have the computer compute the intersection as a polygon, figure out how many sides the polygon has and then keep track of what number you found as you do this experiment over and over and over again the law of large numbers says that if the computer does this enough times and outputs the average number of sides in the polygons it's generated so far the average will converge to the true average, four so if you do it a million times and you see 4.007 or something you'll think, hmm, that's suspiciously close to four maybe the true answer is four so now I'm going to replace the cube by a parallel of pipette which includes the cube as a special case so let me remind you what a parallel of pipette is it's like a cube except instead of having three perpendicular axes which are the same length in the case of a cube they could be in different directions and they can be of different lengths so if you imagine three different directions you can go you can take any two of them and form a parallelogram you take two copies of that parallelogram those will be two of the faces of the parallel of pipette and if you form all the different combinations of choosing two of those three different directions you're going to get not two but three pairs of opposite faces that are parallelograms so here's a parallelogram over here and there's a matching parallel parallelogram over here and those will be the six faces of your parallel of pipette so now I'd like you to imagine passing a parallel of pipette through the plane of fire and looking at the cross sections since the parallel of pipette has only six faces the same argument we saw before tells you that the cross section can't have more than six sides so you're always going to get a cross section with three, four, five or six sides and the question I'll ask is how many sides does it have on average now you're about to get a chance to explore this in a hands-on way but I can't give each of you your own parallel of pipette and your own private wall of fire so we're going to bring the problem down a dimension with parallel of pipettes that can actually be drawn not just approximately but exactly on a piece of paper so I'm going to replace the parallel of pipette by a stick figure on a piece of paper consisting of 12 line segments I'll call it a squashed parallel of pipette so this may look three-dimensional but it's not meant to it's just a bunch of line segments in a plane so this is a two-dimensional representation of a three-dimensional object where I've drawn some of the lines as dashed to imply that they're behind the other lines this is not a three-dimensional object in any sense it's just a bunch of line segments and instead of having a plane of fire we're going to have a line of fire here's a line of fire it's just a line sitting inside that plane so here's a line of fire and it intersects the squashed parallel of pipette in five points okay instead of imagining that the line of fire stays put and the squashed parallel of pipette moves through it let's imagine that the squashed parallel of pipette stays put while the line of fire moves past it so here's what that looks like okay so we're going to assume that the speed is uniform and we'll look at the number of points in the intersection between the squashed parallel of pipette and the line of fire this is going to be a function of time from the moment of first contact to the moment of last contact and we want to compute its average value but since the line of fire is moving at constant speed the average as a function of time can also be written as an average as an average over position which is what you're going to be doing on your sheets so you're going to be in teams of two and each team will have a clipboard and four markers so half of you have clipboards with graph paper and half of you have four markers and you're going to do some mathematical experiments for each experiment use the colored markers to design a squashed parallel of pipette consisting of four parallel and equal red segments four parallel and equal blue segments and four parallel and equal green segments sort of like this but different so edges that are the same color should be parallel and equal but edges that aren't the same color should not be parallel they can be equal in length if you want but they should not be parallel and we're going to want all the points to lie on vertical grid lines just to make it possible for you to do certain kinds of calculations but other than that, you can be creative those points, by the way, do not have to lie on the horizontal grid lines in your graph paper just on the vertical grid lines here's what you're going to do after you've drawn your squashed parallel of pipette for any vertical band lying between two consecutive vertical grid lines count how many different line segments from the squashed parallel of pipette cross that band since that tells you the number of intersections between the squashed parallel of pipette and the line of fire when the line of fire lies in that band let me show you what that means say the line of fire lies in the leftmost band between that leftmost grid line and the grid line to its right notice how that line intersects the squashed parallel of pipette in three points that's going to be true for all the positions of that line of fire within that band so with your black marker you're going to write down the number three let's move on to the next band if the line of fire lies anywhere in that band line of fire is always vertical parallel to those grid lines the line of fire will cross three of those segments that are the edges of the squashed parallel of pipette one red, one blue, and one green since it's going to cross three of them I've written the number three down below and likewise here if the line of fire falls in that band it's going to cross four edges of the squashed parallel of pipette so I write down the number four and for the next band five and for the next band six and then it goes five, four, three, three again so remember we only want to look at positions where the line of fire crosses the squashed parallel of pipette in at least one point I don't want to look at positions of the cube or out of the museum likewise I want the line of fire to actually cross the squashed parallel of pipette so all the numbers you write down should be positive if you write down a zero that means you're looking at a band in which the line of fire doesn't cross the squashed parallel of pipette at all and we've agreed not to count these positions now once you've computed a black number for each of these bands I want you to compute the average of the black numbers to find the average number of intersections between the stationary squashed parallel of pipette and the moving line of fire why do I want you to do that well there are nine equal width bands and the line of fire is going to spend one ninth of the time in each of those bands so if in one ninth of the time the line of fire lies in this band and it meets the squashed parallel of pipette in three points and one ninth of the time it intersects it in three points because it's in the second band it's four points, one ninth of the time it's five points, et cetera for each of the bands then to find the average number of points of intersection we should average those numbers so we'd average the numbers three, three, four, five, six, five, four, three and three and that will give you the average number of intersections between a line of fire moving at uniform speed and the squashed parallel of pipette that you've drawn now your picture could look different from mine for instance, here's a different one and you'll notice that I haven't written a zero underneath the left band or underneath the right band because remember we're not interested in positions of the line of fire that don't cross the squashed parallel of pipette but it does cross the other bands and so I've written three, three, six, six, three, three in those other bands so if people clear on what I'm asking you to do any questions? okay I'm going to give you about ten minutes to work on this in pairs feel free to ask me questions or to tell me what you found as I wander around the room so here's a suggestion for how you might draw one of these things which is choose a point draw a red edge draw a blue edge and draw a green edge and then fill in so for instance if you want to know how long this should be it has to be the same length as this if you draw a red edge from here it has to be forming a parallelogram with this thing over here so that's kind of one order in which you can draw it so that it will all be consistent yeah try drawing it with all the corners points or actually at grid lines, intersections so that you'd have like a point over here instead of over here and maybe over here instead of over here because then you can say oh that line is always going to be one over and three up and then just make sure you follow that consistently all the time that should help okay so I'm sorry to cut some of you off we're still working on this exercise so many of you notice that there could be some exceptional moments corresponding to some exceptional positions of the line of fire when the number of points of intersections suddenly dips down like here the line of fire crosses the squash parallel of pipette in six points but if we move the line of fire just far enough to the right so it's exactly half way from the left to the right there are only four points of intersection because the green is crossing the blue up above I'm sorry the red is crossing the green up above and the red is also crossing the green down below but there are only finally many of those exceptional moments and there are infinitely many moments when the cross sections aren't exceptional so we can ignore the exceptional moments in geometric probability terms a single point has length zero and any finite collection of points has total length zero so such a collection can be ignored now this gloss is over a fascinating topic namely non-empty sets of measure zero and what it means for an event to be possible but still have probability zero but that would take us too far out of our way also some of you notice that if one of the edges of the squash parallel of pipette is parallel to the line of fire move that line that you see in this picture a little bit to the right at the instant that the line of fire coincides with the vertical edge of the squash parallel of pipette the number of points of intersection is infinite because all the points on that edge lie in the intersection this is more problematic but again a rigorous discussion using geometric probability theory would give us a rationale for ignoring those exceptional moments because there are only finally many of them but the really important thing that happened during the intermission is that even though all of you had different black numbers when you took the average of your black numbers you got the same answer namely four and some of you got this answer more than once from multiple experiments so you found strong experimental evidence for conjecture if you pass a squash parallel of pipette through a line at constant speed or vice versa then the average number of points of intersection from the moment of first contact to the moment of last contact is four so let's prove it so here's the picture I showed you before it's going to be completely typical but we want to focus on a specific example so I'll use this one notice that what I've done below is I've taken the twelve colored edges of the squash parallel of pipette and sort of ignored the fact that they're diagonal and just laid them out horizontally so for example over here I have a red edge that passes through two bands I've replaced that by a red segment of length one and another red segment of length one passing through those two bands so I'm choosing to measure distance so that each band has width one and likewise this blue edge which passes through three bands is represented by these three segments and this red edge that passes through two bands is represented by these two over here so now let's think about what the numbers we've written down here represent this three represents the number of intervals of width one that appear below it this six represents the number of intervals of width one that appear below that and so on for the other numbers so when we're adding up the numbers three three four five six five four three three what we're doing is we're counting the number of intervals that appear in the bottom half of this picture and the way we're going to count them is by changing the way we look at them and the way I want you to look at them is the way I've sort of drawn them as being separated from each other so we have two red intervals of length two we have sorry four red intervals of length two four blue intervals of length three and four green intervals of length four and we want to add the lengths of those intervals so what we want to compute is four times the length of each red interval plus four times the length of each blue interval plus four times the length of each green interval but if you look at the top of the squash parallel to piped you'll see a red segment followed by a blue segment followed by a green segment that stretches with no overlaps from the left endpoint of the squash parallel to piped to the right endpoint correspondingly we can find a red segment of length two followed by a blue segment of length three followed by a green segment of length four which collectively span the whole picture from left to right with no gaps or overlaps that means that the sum of the red length and the blue length has got to be the number of bands because each band is crossed by exactly one of these three segments this red segment this blue segment and this green segment so the sum of the lengths of a red segment and a blue segment and a green segment must be the number of bands but we have not just one but four red segments and likewise four blue segments and four green segments so all the segments which I'll remind you is going to be equal to the sum of these numbers we're going to get four times the number of bands now remember we're not just interested in adding these numbers we're interested in averaging them so if the sum of those numbers as we've just shown is four times the number of bands and we divide that by the number of bands that leaves us with four so we've just shown that for this specific example the average of those black numbers must be four not by calculating it with arithmetic but by thinking about it conceptually and this conceptual approach is going to apply to any squashed parallel of the pipette we draw for the same reason there's always a way to move from the left most point to the right most point following a single red edge and a single green edge and a single blue edge not necessarily in that order so that means the sum of red plus blue plus green must equal the width of the polygon as we did before so that's the proof for two dimensions the same idea applied in three dimensions proves the wall of fire theorem which was conjectured and proved in October of 2017 it says if you pass a parallel of pipette through a plane at constant speed then the average number of sides in the polygon of intersection from the moment of first contact to the moment of last contact the difference is that when we move from two dimensions like what you were doing on your papers to three dimensions like I could be doing in the ring of fire instead of counting intersections between a squashed parallel of pipette and a line of fire living in the plane we're counting intersections between a true parallel of pipette and a plane of fire living in a three dimensional space but the method of proof is the same so now that I've described the result and shown you a proof of it in the history of this discovery including the story of the three proofs that I know so here's the chronology going back to 2017 October 7th was the kickoff of the global math project and while I was here I was playing with a wall of fire and began using about the average number of sides and a day later I posted the question on the math fun email forum sidebar on that math fun is an ancient mathematical email and I mean more than 20 years old for serious amateurs and playful professionals here are some recent topics that have been threads on this email forum that last one GOL means game of life that's Conway's game of life so if these topics appeal to you then please get in touch with me and I'll put you in touch with the moderator of the math fun forum anyway getting back to our story there was a flurry of email on October 8th as Alan Wexler, Dan Asimov, Victor Miller and others and I tried to figure out how to clarify the question and we hit on a scheme for varying the orientation and position of the cube that's a fairly natural one so then a week passed and then Keith Lynch who had implemented our definition of random cutting plane on a computer and done some computer experiments reported the results of an experiment showing that the average appeared to be about 4.0004 and he asked could the average be exactly 4 that same day Warren Smith said the answer was exactly 4 and gave a proof the first proof of the theorem three days later George Hart who along with Glenn Whitney was one of the founders of the museum and one of the creators of the wall of fire gave a different proof in trying to understand Hart's proof I came up with a similar proof of my own and that is the proof that I showed you I've tried to present it in as accessible a form as possible so that's not quite the language that I used in stating the proof but that's the essential idea George Hart's proof the second proof that was found was a lot like the proof I've shown you except instead of asking how many corners the cross section has on average he asked how many sides it has on average now of course that's going to give the same answer because the cross section is a polygon and a polygon always has as many corners as it has sides but if you think about where the corners and sides come from they come from different places a corner of the cross section comes from where the cutting plane intersects an edge of the cube whereas a side of the cross section comes from a place where the cutting plane cuts a face of the cube now of course you're going to get the same answer since each cross section has as many sides as corners but the method of calculation will be different so that was George Hart's proof Warren Smith's proof is completely different he invites us to picture not just a single cube passing through a plane but a whole infinite 3D grid of cubes passing through a plane so think about not just a single cube but a cube sitting next to it and a cube sitting above it and a cube sitting in front of it a regular latticework of cubes filling all of three dimensional space sliding that through the plane of fire that's hard to picture but let me give you an example with an animation let's look at the case where the body diagonals of all those cubes joining diametrically opposite points on a single cube are perpendicular to the plane of fire in that case the intersection of the whole grid of cubes with the plane of fire gives a picture that looks like what's called a trihexagonal tiling or a tagome lattice so here's a video made by the educational animator Lucas Vieira and I'll show you what you get as the plane of fire moves through that it's only a finite collection of cubes 4x4x4 but it should give you the idea of what happens for an infinite grid of cubes so we have two colored the cubes so you can see the pattern more clearly so the cross sections are also going to be two colored light blue and dark blue so you always see triangles and hexagons well that's not quite true let's just catch it just right well it didn't quite right see if I can catch it now still didn't quite catch it there are exceptional moments when the hexagons disappear and you just see triangles but most of the time you see triangles, hexagons, and squares and you have one hexagon for every two triangles so the average number of sides is the average of 3, 3, and 6 which is 4 the whole fact that makes Smith's proof work is that when you have infinitely many straight lines dividing a plane into bounded polygonal pieces the average number of pieces sorry the average number of sides of the pieces is always 4 as long as there are no triple intersection points what we were looking at before the intermission when we were moving just one cube through the plane of fire is essentially the same thing as following one of the regions in this movie and taking the time average of the number of sides it has from birth to death so a piece is born it lives, it shrinks, it's dead and you take the time average of the number of sides that corresponds to a single cube passing into and then out of the wall what Smith is doing is replacing the time average by a space average where we freeze time but view the current region is just part of a big tiling of the plane showing that the time average and space average are equal takes work I hope you can see how it might be true instead of going into the nice cities I'll show you a picture illustrating a related result pointed out to me by Olivier Bernhardt if you have a bunch of lines in the plane no 3 of which meet at a point if you take any quadrilateral formed by 4 of those lines and look at the pieces that all the other lines chop that in up into then the average number of sides of the pieces is 4 so here's 8 lines in the plane I've chosen 4 of them to form a square the other 4 lines chop that square up into 5 pieces but if you look at the pieces and take the average number of sides you get the average of 3,3,3,3 and 8 which is again 4 now so far we've talked about cubes and parallelepipeds parallelepipeds what about other shapes I don't know the average number of sides of a cross section of a tetrahedron but I do know that unlike the situation with a cube the orientation doesn't matter you get the answer 4 regardless if you hold the tetrahedron in a fixed orientation the average depends on the orientation that you picked that's because if you hold the tetrahedron corner facing you and pass it through the wall of fire it has 3 sides from beginning to end whereas if you hold the tetrahedron with its edge starting going through then it's going to be a rectangle with 4 sides all the way through so the average can be 3 4 for others and a kind of intermediate values for other orientations so if there is a nice answer for the average number of sides of the cross section the proof is not going to be as nice because it genuinely depends on averaging over different orientations what about other things like the regular dodecahedron this is going to be like the tetrahedron the number of sides that an average cross section has if you average over just the cross sections for planes in a particular orientation is going to depend on what orientation you pick so it's not going to be a simple story on the other hand if you choose a less well known picture the rhombic dodecahedron things work out very nicely if you pass a rhombic dodecahedron through the wall of fire the average number of sides of the cross section is 6 regardless of the orientation rhombic dodecahedron like parallel pipeds are examples of polyhedra called zonahedra this is an image created by George Hart showing some other examples of zonahedra it's a different way you could try to follow up on the wall of fire theorem go back to the cube and get more refined information about the probability that a random intersection has 3, 4, 5 or 6 sides this is an unsolved problem empirically Keith Lynch got the numbers shown below or rather over there but what are the exact values I don't know we can also ask other kinds of questions about a random cross section of a cube for instance what is the expected area if you fix the orientation of the cube then the average area of the cross sections equals the volume of the cube divided by the width of the cube in its direction of motion where the width is defined as the distance between two parallel tangent planes so in the case of a cube if we look at these two parallel planes tangent to the cube at two opposite faces well then the cross sectional area is always going to be the area of one of the faces and that area is indeed equal to the volume of the cube divided by the width in that direction because volume divided by side length equals area of the face but this assertion is also true if we held the cube in this direction and the two tangent planes were like this average area for cross sections in a particular orientation of the cutting plane will always equal the volume divided by the distance between the two tangent planes that's called the width of the cube in that direction but what if you don't fix the orientation of the cube what if you average over all orientations then it turns out that the average area of the cross sections equals the volume of the cube divided by something called the mean width of the cube which is the average of its width in all the different possible directions it turns out that the mean width of a cube of side length s is three halves times s so that implies that the average cross sectional area of a cube if you allow cutting planes in all orientations is going to be two thirds times s squared so that's a nice answer I'd like to know what the expected perimeter of a random cross section is but I don't know what it is so if you figure that out please let me know now about a year ago I was telling a friend a non-mathematician about the wall of fire problem and the question of what is the expected number of sides of a random cross section of a cube and he stopped me and very politely he said I don't mean to be rude but can you tell me why the question is interesting and his request stumped me in the moment and it was only later that I realized that in and of itself the question isn't interesting because we mathematicians ask questions all the time and often we don't know whether they're interesting because the criterion that a pure mathematician uses in judging whether a question is interesting is whether the answer is interesting for instance it's interesting and surprising that the average number of sides is such a nice number namely four but it's even more interesting that you can hold the orientation of the cube fixed and the answer is always that same number four and when a pure mathematician encounters something surprising like that that wasn't inherent in the question but seems to be part of the answer the mathematician scratches her or his head and says huh I wonder what's going on here so when something is true about cubes but false about tetrahedra and it's false for regular dodecahedra but true for rhombic dodecahedra we start to think wait a second there's something going on here and it's my job to figure out what it is so I would say that unlike applied mathematics where one of the drivers of what you do with your time as a mathematician is solving some real-world problem one of the main drivers of pure math is traveling in the direction of greatest surprise you've got three things that you know kind of routine but one of them isn't what you would have expected that's the thing you choose to delve more deeply into aside from the fact that the wall of fire theorem was conceived here I'm happy to be talking about it because it highlights a bunch of points about mathematics one of which is the collaborative nature of mathematics I wouldn't be here today telling you about these results if there hadn't been people on the math fund forum helping me clarify the question Keith Lynch during the computer experiment Warren Smith and George Hart figuring out proofs so it's a very collaborative enterprise this theorem also highlights the role of experimentation and in particular computer experimentation in guiding us to good conjectures which can intrigue us to the point of making us do the hard work of trying to figure out why they're true it also shows that mathematicians aren't satisfied with just a single proof of a result just as we learn more about a cube by examining it from different angles face on, edge on, corner on we learn more things about mathematical reality by looking for as many ways to prove things as we can lastly, the story of the theorem demonstrates that you can find new problems in math not just by reading the latest research articles in journals but also just by keeping your eyes open to mathematical patterns in the world around you maybe your next theorem is here in this museum waiting for you to find it thank you so if anybody has any questions you can raise your hand I'll bring the microphone to you is there any general relationship to the number of edges? between the number of edges and the expected number of sides in the cross section? yes there is a relationship but it involves other things as well so you have to know more information if that's the context in which you're asking this so there's other analogs of the Wall of Fire a theorem for other shapes and the number of edges definitely plays a role so is there anything you can say about a polyhedron and its dual with regard to these properties? that's an interesting question so a polyhedron in its dual well I can say that the dual of the cube is the octahedron and the octahedron is not a zonahedron so the method of proof that I've shown here is to apply but that doesn't mean that there isn't an interesting answer to the question of how many sides you have on average for an octahedron so I'd be very interested if anyone can explore that experimentally or with rigorous methods to see what the average number of sides of an octahedron is when you intersect it with a random plane that might be very interesting I don't know hi, I'm not a mathematician but can you tell me what that ring is, that Wall of Fire what is it? I believe it is a ring of lasers, I'm not sure how many but 12 lasers and they're set up in such a way that I can move my hand anywhere in there let's do it with a cylinder and you see a circle so somehow they've cleverly set this up so that the laser beams even though there's only 12 of them they create all the points inside this disk more or less evenly each of those lasers has a little lens that spreads the light out from being a single line into a little fan of light so adding them all up it creates a plane of light it's quite beautifully done look how close the cylinder is to the border you can certainly come up and have a close look but I'll do it so that it's on the screen of the side so a little higher I can move it around a lot and you don't see any change in that ring, it's just a solid ring which is the physical representation of the mathematical intersection of the surface of the cylinder with the plane that goes through this ring of fire and I can bring it pretty close to the boundary before that ring starts to break well, I don't know of a way to show that without going out of focus but anyway do you think the result is true if you have an irregular, whatever you call it cube or pipe anyway, it's kind of cuboid type thing but the edges are all different lengths because your picture would still be the same you'd still have these paths that all add up to the length from one point to the opposite diagonal point so maybe even if they're irregular it still has average fold that's a great question and I haven't thought that through but you can be sure I will later tonight kind of sounds like it this is weird but in the intermission I got a shape that all the all the numbers are 4 like literally 4, 4, 4, yeah yeah great interesting how that can happen yeah, it can happen and it's kind of weird why it's like that, like why is it with the red right on the line well, first I'll hold it up to see and then I'll try putting it in the document camera so you've chosen a parallel pipe head or squash parallel pipe head which has some vertical edges to it and so I think that's related to the fact that you got 4 going all the way through from left to right yeah there's something special about having vertical edges as opposed to horizontal edges because of the way we're measuring things from left to right in these pictures down here very good let's give another hand for our speaker