 Welcome back everyone to our lecture series Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'm your professor Dr. Andrew Misaline. It's good to have you here today. In today's lecture, we're going to talk about the defects of triangles, particularly the defect of hyperbolic triangles. The topics of this lecture are loosely based upon section 6.4 from the textbook Roads to Geometry by Wallace and West entitled Hyperbolic Results about Polygons. We've seen previously in neutral geometry that if one triangle has an angle sum equal to 180 degrees, then all of them do. This was the all or nothing theorem, and that the existence of a 180 triangle is equivalence to the Euclidean Parallel Apostolate. On the other hand, in hyperbolic geometry, all triangles have an angle sum that is strictly less than 180 degrees. All of them have this defect. How much do they defect from this 180 degree measure? So given a triangle ABC, we call the defect of a D of ABC. So this is just the angle sum of the triangle subtracted from 180 degrees. Now, it's important to mention that the defects of hyperbolic triangles are not all equal to each other in hyperbolic geometry. This is important to recognize because in Euclidean geometry, since everyone has 180 degrees angle sum, we sometimes think that, oh, hyperbolic geometry, everyone has 179 degrees, or in a different hyperbolic geometry, everyone has 137 degrees. It's not like that at all. There will exist triangles with an angle sum of 137. There will be a triangle whose angle sum is 179, but they're not all equal to each other because if the defect of a triangle was constant throughout the whole geometry, notice in Euclidean geometry, the defect is always zero degrees. But if the defect was constant, all that is all the triangles have the same defect, then one actually could argue that every defect has to be zero degrees. I don't really want to go through the argument right now, but we'll see in a little bit that if your defect is positive, you can actually get arbitrarily smaller defects. And therefore, if they're all constant, that would violate some triangles getting smaller defects. That's the basic argument. So if all defects are constant, that actually is equivalent to Euclidean parallel postulate. Therefore, in hyperbolic geometry, there exist triangles with different distinct defects. And it turns out in hyperbolic geometry, you can get a defect almost arbitrarily close to 180 degrees, and you can get a defect arbitrarily close to zero degrees. We'll talk about this a little bit more in the future here. This kind of summarizes what we've seen here on the screen right now. Because of the all or nothing theorem and the scary Legendre theorem, we do know that the defect of a triangle has to be positive, that all hyperbolic triangles have an angle sum strictly less than 180 degrees. And so when we subtract that from 180, we get a positive number. And so when it comes to this defect, I mean, it's important to realize that the defect is gonna be 180 degrees minus the measure of angle A minus the measure of angle B minus the measure of angle C. And so this here can range anywhere from up to most 180, but at smallest zero. And these bounds are never actually obtained by any triangles in hyperbolic geometry. Now, another thing important to realize when it comes to this defect, so defect is always a positive quantity. Defects, when a triangle is congruent, when two triangles are congruent, so ABC is congruent to A prime, B prime, C prime, then their defects will be equal to each other. And this follows very quickly from this formula right here for the defect that if two triangles are congruent, their angle measures will be congruent. And therefore this difference from 180 will be identical. So defects are equal for congruent triangles. A third property of the defect that I wanna mention right now, this one's a little bit harder to prove. It's not too hard, but it's not just an automatic obvious statement. Oh yeah, the sky's blue. Congruence is preserved by defects. Or I should say the congruence preserves defects, things like that. This one takes a little bit more argument, but there's not much to it. So if we have a triangle ABC, A, B, and C, I really didn't mean the tail for A to be so huge there, but we're sticking with it. That's just how it is. Anyways, if we have a triangle ABC and we pick a point D that lives between A and B. This naturally dissects the triangle into two pieces. There's the ADC triangle right here and the BDC triangle right there. All right, and so as these are two triangles in their own rights, the two smaller triangles have defects and so does the larger triangle. And when it comes to the defect, we have what's called additivity of defects. That is the defect of the large triangle and is the sum of the two smaller defects right there. And so let's talk about why this is. Well, first of all, the angle ADC right here is supplementary to angle BDC. So the sum of the measures of those two angles would be 180 degrees. We're gonna use that later. And so labeling things a little bit, if we look at the triangle ADC, it has three angles. There's angle CAD, there is angle ADC and then there is angle ACD right here. So the defect of the triangle ADC will be 180 degrees measured. So track the measure of angle one and the measure of angle two and the measure of angle three. All right, now let's look at the triangle DBC. Well, its defect can be computed by taking DBC, we'll call that angle four. DBC, we'll call that angle five and DCB angle six. So we get four, five and six. Great. Now let's look at the larger triangle ABC. So triangle ABC, its defect will be computed from its three angle measure. So it has angle one in it. It has angle five in it. And it has angle C, which angle C is gonna be a combination of angles three and six. So we get 180 degrees minus the measure of angle one, five, three and six. So that's almost all the angles except for two and four seem to be missing. So remember, these angles are supplementary. So what we're gonna do is we're gonna take 180 degrees minus 180 degrees, right? So this is angle two and angle four. Since there's supplements, their sum is 180 degrees and adding 180 and then subtracting it doesn't change the equality. And then moving things around, we get back one, two, three. I guess that's actually one, three, two. And then here we have five, six, four. Maybe I should have tried to get those in a better order but Ka'Chink, the sum of the two defects equals the defect of the larger triangle. And so the defects are additive. And so what, just kind of summarize what we've done here is that the defect function is a positive function. It preserves congruence and it's additive. I want you to remember those three things right there. Positive, congruence, preservation, additivity. Positive, congruence, preservation, additivity. Positive, congruence, addit, or congruence, preservation and additivity. So if I say this enough, you will have no choice but to think about these as you close your eyes and go to sleep tonight, you will remember them always. That way, when you watch a future lecture, maybe in the not too distant future, I don't know. It could be coming up anytime soon. A little Easter egg will pop out to you, I think. These words, positive, congruence, preservation and additivity will come back to your mind when you see them and be like, aha, there they are. I found it. So now that, note that using the previous theorem sort of like a base case, if a triangle is actually dissected into in non-overlapping triangles, so let's draw a picture here. If we have a triangle and we dissect it into smaller triangles, kind of like we did in the previous picture, so you can cut it down like this once and then you do it again like this and then you do it again maybe like this and you do it again like this and you can keep on going, right? Always taking medians of triangles along the way. We can dissect any triangle into smaller, smaller, smaller pieces. So it's like we have triangle one, triangle two, triangle three, four, five, six, seven. I got eight pieces right here. Well, by induction, if it's additive for this triangle and then it's additive for this triangle and then it's keep on going step by step by step by induction, we actually get it's additive for an arbitrary number of non-overlapping triangles. So this idea is what we call a partition, partition of triangles. So we took our triangle, we dissected it into non-overlapping triangles. Non-overlapping triangles here, of course, means that if we take any triangle, say, if we take any triangle, triangle one or triangle I and triangle J, if we take the interiors of the triangle, not the boundary but the interior and we intersect this, this will always equal the empty set. So there's no overlap between them whatsoever. And that's the only requirement we're gonna place on these triangles to get a partition. They should probably be connected but admittedly we might allow for a disconnected shape in which case this notion is perfectly fine in that situation. So additivity actually works for an arbitrary partition of triangle, not just into two triangles. The reason I mentioned this is that because of the previous theorem, we can show that triangles can have an arbitrarily small defect, always positive of course. And to see this, consider the following type picture. We have a hyperbolic triangle ABC, right? And so we'll call this triangle ABC and then dissect it along some median like so and then take, well, I mean, the way we selected this midpoint D right here, so we have ABC again. Depending on how we chose the point D, one of these is gonna have a smaller defect than the other. I mean worst case scenario, this line cuts the triangle into two equal pieces that is in terms of defect. And so one of these things will have at least one half of the original defect. And so let's say that the ADC triangle has less than half of the original defect. So we get something like this. For which case then we can relabel it to like, okay, the smaller triangle we get A1, we get B1 and we get C1. I think you have the first one as A0, B0, C0. And then we can repeat this process, not just do it once, but we can do it again. So we take this triangle, we cut it right here. And so one of these two triangles, this one or this one will have smaller defect. So we get this one's now an A2, this is a B2 and we get C2. We can do this again. Where again, one of these triangles will have at most one half of the defect of the original one. So we get C3, A3, and then B3. And so we can keep on going dissecting this triangle smaller and smaller and smaller, getting its smaller and smaller defect. And so it's by induction, we do get that there will exist a triangle, A and BNCN, who defect as less than or equal to one over two to the end, the defect of the original triangle. And so if we apply now the Archimedean principle, that is eventually given any epsilon greater than zero, eventually there'll exist a triangle whose defect is arbitrarily small, like so. And so we can get defects that are as close and close to zeros as we want, because epsilon can be chosen smaller and smaller and smaller and smaller. And in some respect, we can also make this thing get bigger and bigger and bigger by sort of a similar type of argument that if we take a triangle, we can join a new triangle who has a positive defects that makes it get bigger. And in some essence, bigger triangles will have bigger defects, smaller triangles will have smaller defects. But what in the world does small even mean? This is again, another sort of little Easter egg that I want you to look for in a future lecture. What does small mean in terms of triangles? What's a big triangle? What's a small triangle? In some respect, this notion of defect is actually measuring the size of hyperbolic triangles. Again, we'll talk about this a little bit more in some future secret lecture. So with regard to defect, there's two theorems I wanna prove today and not these theorems themselves. Well, this theorem you see in front of you is not actually a theorem about defect. It's just the defect of a hyperbolic triangle is useful in proving this theorem. So the first condition is actually a triangle congruence for hyperbolic triangles, known as the angle, angle, angle criterion. If two hyperbolic triangles are congruent with all of their corresponding angles, then the triangles themselves are congruent. And so this is kind of like the side, side, side condition, side, angle, side. Angle, angle, angle. Which this might come very bizarre to us at first because in Euclidean geometry, this is not the case. We can construct incongruent triangles which have the same three angles. In Euclidean geometry, this is what we call similar triangles which has to do with proportionality and things. I mean, like every other calculus problem it feels like we do a story problem with like related rates or work problems or optimization. It feels like similar triangles are used all over the place. But yet, I'm telling you right now in hyperbolic geometry, if two triangles are similar, they're actually congruent to each other. Similarity has no place in hyperbolic geometry. This maybe comes as no surprise by you at this moment, but similar triangles, that is the existence of similar triangles is equivalent to the hyperbolic parallel, or the Euclidean parallel, oops. The angle, angle, angle is actually a triangle condition in hyperbolic geometry. And so to prove this one, you can see the proof in front of you. If you're welcome to pause the video and read through it on your own. What I'm gonna actually do is switch over to some progressive slides from a former student, Haley John. This can be found on my webpage if you want to take a look at them. Turns out this lecture that comes from 6.4 in Rhodes to geometry actually has two slides that we're gonna use. Let's take a look at the first one, the angle, angle, angle condition. And so let's say that we have two triangles, ABC and A prime, B prime, C prime, so that the corresponding angles are congruent. A is congruent to A prime, B is congruent to B prime, C is congruent to C prime. And because their angles are congruent, their defects will be equal. That's an important part of this proof. That's basically what this proof is gonna be based upon. Now, if any of the sides were congruent, like if the corresponding sides AB was congruent to A prime, B prime, we could use an angle side angle argument to show that they are congruent as triangles and would be done. So let's assume for the sake of contradiction that there is no correspondence between sides of the triangles. So the corresponding sides, one's always longer than the other. Well, we have a triangle with three sides and there's no congruent sides whatsoever. So it's gonna be true that one of the triangles is gonna have two sides longer than the other. I mean, it could be all three, but if you consider all the possibilities, one triangle will necessarily have two sides longer than the other. And without the loss of generality, let's assume the ABC triangle has two sides longer than A prime, B prime, C prime, and that AB is longer than A prime, B prime, and AC is longer than A prime, C prime. Well, because we have those longer sides, I said all that stuff that just got highlighted, because the one has longer sides, there exists points B double prime that sits between A and B so that E, A, B double prime is congruent to A prime, B prime. That's because AB was longer than A prime, B prime. And there also exists a point C double prime that sits between A and C so that AC double prime is congruent to A prime, C prime. So we guarantee, so this new triangle that's been built has two sides congruent to the other one. And since angle A is congruent to angle A prime, we get that the triangle AB double prime, C double prime is congruent to A prime, B prime, C prime by angle, side angle, side, excuse me. So we get that those two triangles are gonna be congruent to each other. And so what we're gonna do is we're gonna dissect the larger triangle ABC using this congruent triangle that we found. So if we introduce the segment B double prime C, we can dissect the triangle ABC into three triangles. One of those illustrated in blue is congruent to the triangle ABC all primed there. And so now we're gonna use the additivity of the defect that we did earlier because the triangle ABC is then broken up into a partition of three triangles and we apply additivity in that situation. So you can see in the picture, there's the blue, green and red triangle. And so the sum of those three defects will equal the sum of the original one there. And so remember because the two triangles ABC and A prime, B prime, C prime had all three of the same angle, the angle measures were corresponding there, their defects are equal. So the defect of one is equal to the other. But for the larger, so to speak triangle ABC, by additivity, its defect will add up to be the defect of the blue triangle, A, B double prime, C double prime, the defect of the green triangle, C, C double prime, B double prime, and the defect of the red triangle, C, B, B double prime. But if you cancel out the two congruent triangles, because remember, so we use additivity of defect, now we're gonna use the congruence preservation because the two blue triangles have the same defects, we can cancel them out and we get that the sum of the green triangle and the red triangle adds up to be zero, which violates the positive property of defects. And so this gives us a contradiction. And so it's kind of cute in this proof, we used all three properties of the defect I told you about earlier, the positivity, congruence preservation and additivity. I was reciting it earlier, so you memorized it. This is not why I did it. I mean, it was useful in this proof, but you will see these again in the future. So that concludes the proof of angle, angle, angle. Let's go back to our lecture here. You can see the blackboard one more time. And so to finish up this lecture, I wanna talk about polygons for a moment. What is a polygon? If we were to define a polygon, a polygon is basically just a union of adjacent, non-overlapping triangle. So we have some partition of triangles, whoops, partition of triangles. They're non-overlapping like we talked about before. And of course, we kind of wanna ignore the lines that go in the middle here. So we have some union of non-overlapping triangles, and this is what we mean by a polygon. And so you see something like here. And so illustrate here, yellow is an example of a pentagon. There are five sides. Did I count that right? One, two, three, four, five, great. We have a pentagon. When we talked about quadrilaterals previously, a quadrilateral, a four-sided polygon, well, quadrilaterals can always be dissected into two triangles. And so that's the idea we have right now. So a polygon in general is a union of a partition of triangles. And so because of that, we can define the defect for any polygon, because we define the defect of a polygon just to be the sum of the defects of all of the triangles involved in the union that's above. And so positivity is automatic, right? Because each of the individual defects are positive, the sum will be positive as well. In terms of congruence, well congruence of a polygon we have to be a little bit more careful about, but kind of like we did with congruence of quadrilaterals, you use the dissection to get you the congruence that you're looking for. And additivity extends naturally to this setting right here, that the defect is gonna still be positive, additive, and congruence preserving for all polygons. But before we get too far ahead of ourselves, it sort of begs the question, is this defect function well-defined for polygons if we define them by these unions of triangles and such? The issue kind of comes down to the following. What if we chose a different way of partitioning? What if we choose a different way of partitioning the polygon? So you'll see here, I've now partitioned my pentagon into one, two, three, four, five different triangles. And these triangles are not congruent to each other whatsoever. What can be said about that? I mean, these different triangles would have different defects. Is the sum gonna be invariant to these different partitions of triangles? Well, the good news is, yes, absolutely. If a polygon is partitioned into triangles in any manner, the defect of the polygon is equal to the sum of the defects of the component triangles. That is to say, the defect function is well-defined for general polygons. It doesn't matter which partition of triangles you use. So again, you can see the proof in front of you. Feel free to pause it if you want. But like I said before, I'm gonna switch back over to the slides that Haley made for us. And so we see theorem 643 right here. Honestly, there's not really a lot to say with the proof in terms of the pictures, but there's one really cool part I want to use these slides for. The proof comes from a combinatorial argument about a common refinement. Because the polygons, you see two examples in front of you, the same polygon, clearly congruent. The proof is gonna rely on a common refinement. Where a refinement of a partition of triangles is a partition of the original polygon such that each triangle in the partition is a union of adjacent non-overlapping triangles in the refinement R. So you can see that I have a partition of the first polygon into triangles. And the second, the triangle which is the polygon which is congruent to the first one, I have a different partition into triangles, all right? So we have two different partitions. How are these compatible? Well, what we're gonna do is we wanna partition the partition on the left and we're gonna dissect triangles on the right so that we all have the same thing. And so basically what we're gonna do, here's the cool part there, that was awesome. We're gonna overlay the two partitions of triangles. And so now you're gonna see a partition, but the intersections of some of these triangles aren't necessarily triangles. Like if you look in the very middle, there's actually a quadrilateral that can happen. And so one has to start considering like what are all the possible intersections one gets if you intersect two triangles? That is if they're allowed to overlap. A lot of times this could be like a triangle itself, but sometimes you get things like quadrilaterals. You can get some other concave, polygons and such, but those ones themselves that the intersection of two triangles will always be a polygon. And therefore it has a partition you can use. And so again, there's a combinatorial argument. There's a lot of cases to consider and I'm not gonna go through all the details of these things right here, but the details rely on intersections and symmetric differences of triangles which themselves can always be inductively, not even, not inductively, but each of these intersections and symmetric differences can be partitioned into triangles because themselves are polygons. And so it turns out to be like a double induction argument. I don't necessarily care all the details here. It's left for you to do if you are demanding of that. But as you can see on the screen right now, those shapes which were not the intersections that weren't triangles, you can always dissect those into triangles. So we have a common refinement of a common refinement of the original partitions. And this common refinement because of the additivity of the defect doesn't change the defect of the original partition. And since any two partitions have a common refinement and the refinement will have the same defect as the original two partitions did, those two partitions give you the same defect. And therefore the defect of polygons is well-defined. All right, so coming back here, that actually concludes our lecture for today. If you liked what you saw, please like this video, subscribe, leave a positive comment. If you are a troll, unless you can sing really well or you have colorful hair, I don't want to hear from it. On the other hand, let's have positive comments about geometry and such. Anyways, I hope you do remember or the Easter egg I gave you for this lecture. It will come up very, very soon. For my students, there is a homework question you have to do where I'm gonna ask you to compute the defect of triangles. But in the book, it says area, scratch out the word area and calculate defect of some hyperbolic triangles. You'll notice that switching the word area to defect is not too much of a stretch on how one does these questions. Anyways, and it's great having you. I will see you next time. If you want a download of the script for this lecture, you can find it in the descriptions of this video. Also, if you want to find a link to Haley's progressive slides, you can find that link in the description as well. See you next time.