 Welcome back to our lecture series, Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Misdine. It's good to have everyone. We are in lecture 32 in our series entitled, Hyperbolic Area and Distance. In this lecture, we want to talk about distance and area in hyperbolic geometry. In the previous lecture, we have seen and learned to understand the axioms of area, and how these axioms of area, so remember area has to be positive, congruent triangles have equal area, and we have that area as an additive function. We've seen that if the area axioms are combined with the axioms of Euclidean geometry, this produces the usual area formulas that we've seen in other geometry classes, like the area of a rectangle is length times width, and another formula is similar to that. What I want to do right now is explore, what does area look like in a hyperbolic geometry? Well, the area formula for a rectangle doesn't really make any sense here, because after all, rectangles don't exist in hyperbolic geometry, so we don't get length times width. But what about the area formula for a triangle? Area equals one-half base times height. We're used to this formula, and triangles including right triangles do exist in hyperbolic geometry. Could we use this as our area function for triangles? Hyperbolic speaking, it turns out we can't. This right here is a Euclidean formula. That is, if this formula was our formula for... Oh, I lost a letter there, Euclidean. If we use this area function for our hyperbolic triangles, that actually would force us to get the Euclidean parallel postulate. Turns out this area form is equivalent to EPP. Seems like everything is, isn't it? And so I want to give you an argument of why we can't use the traditional area formula. And so take the following situation. We're going to take two hyperbolic lines, call one of them L right here, and we're going to take a line perpendicular to L, and we're going to call that M. So we get something like this. This is M, and they are perpendicular. Take the point of intersection between L and M, we're going to call that A, like so. And so go some distance along, I guess, let me also point out, we're going to take a point C to be some point that's on M that's not on L, call that C. All right, now choose a point that's on L that's not on M so that the distance between A and this new point B is equal to one. And when I say distance, I mean the measure of this line segment, let that equal one, all right? And so with this, we can construct a natural triangle, like so we get this triangle ABC. And it is a right triangle, you'll notice. If we were to accept that the area of a triangle, if the area is equal to one half base times height, then we would see that this area is just going to be, well, the side length A was one, the side length AB was just one, we chose C to be whatever we want, right? And so this is just going to be AC over two, the area of that. And if you want to, we could choose AC, we could chose C so that distance was two. And therefore this gives us a unit triangle if we prefer. So what we want to do next is we're going to choose some more points on this same side of the line M so that we can keep on replicating this distance. So let's pick, let's call this first point, this first point B, we're going to call it B1, all right? And we're going to choose another point along this line, we're going to call it B2, so that the distance between B1 and B2 is itself one. Therefore the distance between A and B2 is two. Let's take a point B3, it's going to be the segment B2, B3 will have a measure of one, therefore the segment AB3 will have measure three. And we keep on doing this over and over and over again and we can do this by induction. We keep on making, choosing another point so that the measure of the segment is one, one, one, one. This will give us that the length of the segment ABN will itself be N. And so let's think about the associated hyperbolic, the hyperbolic triangles that we're forming from this construction. So we get a picture, it looks something like this. And we can label these triangles, we have triangle one, triangle two, triangle three, triangle four. And so I want you to notice that if we take the area, we'll take the area of the triangle, ABNC, the way that we've constructed this, if we get one half base times height, well, the one half is a constant. The base in every single situation is going to be BN minus one times BN. And then carry this down to the bottom right here. The height of each of these triangles is going to be AC again, right? And so the way we've constructed this, you end up with one half times one and we decide AC was two. So you end up with one as well. So the areas of each and every one of these triangles is equal to one. The first one is one, the second one's one, the third one's one. All the areas of these triangles are equal to one. Great. But also look at the, did I label that correctly? I'm sorry. Let me back up a little bit and fix a part of what I wrote down. The triangle, this shouldn't be the triangle ABNC. This should be the triangle B and minus one, BNC. That was the triangle that we're talking about right here. So we have like triangle one, triangle two, triangle three, triangle four. Each of those triangles has an area of one. All right. Now consider the triangle. Consider the triangle A, BNC. Well, the way that this diagram suggests to us is that there's a natural dissection of this triangle into the triangles BN minus one, BN, and C. So we can dissect this triangle in this way. And so the additivity of area is gonna tell us that the area of this triangle, ABNC should then equal N. We get N units of area right there. But what happens if we look at the defect of this triangle, right? So when we talk about the area, we've constructed a triangle whose area is gonna be N and we can make this get bigger, bigger, bigger by lengthening the base there. We get it bigger, bigger, bigger. But if we look at the defect, the defect is also an additive function that this decomposition, this dissection, partition of triangles right here also applies to the defect function. And so if we look at the defect of the triangle A, B, and C, right, this will equal the sum of the defects of all these little triangles of the triangles of the defect of triangle I there, right? Now, each of these triangles that we've constructed, triangle one, triangle two, triangle three, triangle four, they're not congruent triangles. They're not, but they all have an area equal to one. And since all the triangles have equal area, that actually makes them equivalent triangles. That is, there's some way of dissecting those triangles into smaller triangles, and we could repeat them together to make, so there's this correspondence between a smaller triangles right there. And equivalent polygons will have equal areas. So each of these triangles have equal area, they have equal area, but they have equal defect. And so we're gonna end up with N times the defect of the very first triangle, triangle one right there. All right? And so the defect is constant for each and every one of these things. And so consider what happens though, when we consider this defect, all right? So there's a little bit of a footnote going on, what's going on here. So if we consider this defect, the defect of our triangle A, B, and C, it's equal to N times the defect of the first triangle, triangle one. But we also know that defects are strictly less than 180 degrees in hyperbolic geometry. But consider this second piece right here. This defect, the defect D of triangle A, B, C, it is bounded itself as a positive number, right? That's what I wanted to say. It's a positive number, positive real number. And we're gonna multiply it by N. The thing is eventually there exists some large number, some large number N, such that capital N times the defect of triangle one, this will be greater than or equal to 180 degrees. And this is gonna give us a contradiction to our hyperbolic triangles, that eventually this thing will get too, too big. And therefore we get a contradiction to these properties of defect. So the classic formula, one half base times height, even though it would make sense that this is a possibility of an area function for hyperbolic geometry, we cannot calculate the area of a hyperbolic triangle using one half base times height. It turns out there's gonna have to be another way. So we're gonna end this part of the lecture in a cliffhanger, bum, bum, bum. Stay tuned till next time and to see exactly how do we get past this problem.