1 00:00:03,200 --> 00:00:06,560 PROFESSOR: All right, so lecture 10 was about two main things, 2 00:00:06,560 --> 00:00:07,060 I guess. 3 00:00:07,060 --> 00:00:09,370 We had the conversion from folding states 4 00:00:09,370 --> 00:00:11,480 to folding motions, talked briefly about that. 5 00:00:11,480 --> 00:00:13,050 And then the bulk of the class was 6 00:00:13,050 --> 00:00:16,450 about Kempe and Kempe's universality theorem 7 00:00:16,450 --> 00:00:18,980 and the beginning of linkages. 8 00:00:18,980 --> 00:00:22,480 So let's start with an open problem about converting 9 00:00:22,480 --> 00:00:24,960 folded states to folding motions. 10 00:00:24,960 --> 00:00:26,390 This is a nice question. 11 00:00:26,390 --> 00:00:30,130 So suppose you have a sheet of paper like this one, 12 00:00:30,130 --> 00:00:33,970 but it has a hole in the middle. 13 00:00:33,970 --> 00:00:36,890 And then you construct some folded 14 00:00:36,890 --> 00:00:40,830 state of that piece of paper with a hole in it. 15 00:00:40,830 --> 00:00:42,720 And now you'd like to actually get there 16 00:00:42,720 --> 00:00:44,190 by continuous folding motion. 17 00:00:46,800 --> 00:00:48,880 So the question is why doesn't that work? 18 00:00:48,880 --> 00:00:52,100 What we know is that the same proof technique doesn't work. 19 00:00:52,100 --> 00:00:56,670 We don't necessarily know that it's impossible. 20 00:00:56,670 --> 00:00:59,820 That would be a nice problem to solve, actually. 21 00:00:59,820 --> 00:01:02,210 So what difference does the hole make? 22 00:01:02,210 --> 00:01:04,730 So this is the method we saw before. 23 00:01:04,730 --> 00:01:06,560 You imagine having some folded state, say, 24 00:01:06,560 --> 00:01:09,890 from a flat piece of paper to a crane. 25 00:01:09,890 --> 00:01:12,900 You roll up that piece of paper to a tiny triangle 26 00:01:12,900 --> 00:01:17,120 that maps to a nice almost flat portion of the crane. 27 00:01:17,120 --> 00:01:24,230 Then you basically play that motion backwards, 28 00:01:24,230 --> 00:01:28,760 but on the surface of the crane instead of on the flat sheet. 29 00:01:28,760 --> 00:01:31,840 And that always works for simple polygons, polygons 30 00:01:31,840 --> 00:01:34,280 without holes. 31 00:01:34,280 --> 00:01:37,630 If you have a hole in here, you could imagine just 32 00:01:37,630 --> 00:01:38,520 filling the hole. 33 00:01:38,520 --> 00:01:40,330 That's what the question suggested. 34 00:01:40,330 --> 00:01:42,440 Just fill the hole, do this thing, 35 00:01:42,440 --> 00:01:45,490 and then remember that the hole actually wasn't there. 36 00:01:45,490 --> 00:01:46,640 Erase it again. 37 00:01:46,640 --> 00:01:49,170 That should give you a motion. 38 00:01:49,170 --> 00:01:51,340 Erasing the hole is fine. 39 00:01:51,340 --> 00:01:54,180 The trouble is this part. 40 00:01:57,080 --> 00:02:01,760 So if I define a folded state of a piece of paper with a hole, 41 00:02:01,760 --> 00:02:03,690 it doesn't tell you where-- let's say 42 00:02:03,690 --> 00:02:06,590 there's a little hole here-- this mapping won't tell you 43 00:02:06,590 --> 00:02:09,919 where that hole goes in 3D. 44 00:02:09,919 --> 00:02:11,170 You have no idea. 45 00:02:11,170 --> 00:02:13,070 And in fact, it may be impossible to map 46 00:02:13,070 --> 00:02:16,230 the hole anywhere in this folded state that's valid. 47 00:02:16,230 --> 00:02:18,460 When you tear a piece of paper, new foldings 48 00:02:18,460 --> 00:02:22,820 become possible that were not otherwise possible. 49 00:02:22,820 --> 00:02:23,940 What's an example of that? 50 00:02:23,940 --> 00:02:28,150 When I do a big tear like this, now I 51 00:02:28,150 --> 00:02:31,070 can pull these points of paper apart. 52 00:02:31,070 --> 00:02:34,915 And it's impossible to fill this hole in in 3D. 53 00:02:34,915 --> 00:02:36,540 It's possible to fill the hole in here. 54 00:02:36,540 --> 00:02:38,280 It's just suturing. 55 00:02:38,280 --> 00:02:40,530 But when I separate things, that means 56 00:02:40,530 --> 00:02:43,430 the original sheet could not fold into this state. 57 00:02:43,430 --> 00:02:46,180 So you get new folded states when you have holes, 58 00:02:46,180 --> 00:02:49,560 that you cannot just patch the hole and hope to find a place 59 00:02:49,560 --> 00:02:52,250 that it folds over here. 60 00:02:52,250 --> 00:02:54,070 There are other issues. 61 00:02:54,070 --> 00:02:57,060 You could maybe patch it in with some stretchy material 62 00:02:57,060 --> 00:02:57,950 or something. 63 00:02:57,950 --> 00:03:01,030 I have one example in the notes where 64 00:03:01,030 --> 00:03:05,950 you-- suppose you have a tube of paper. 65 00:03:05,950 --> 00:03:09,560 So this is not a flat example, but it's an interesting example 66 00:03:09,560 --> 00:03:10,430 anyway. 67 00:03:10,430 --> 00:03:18,240 So let's say the outside of this tube is purple, 68 00:03:18,240 --> 00:03:21,210 and the inside is white. 69 00:03:21,210 --> 00:03:25,580 And one thing you can do with the tube is turn it inside out. 70 00:03:25,580 --> 00:03:32,290 So you can make the inside purple and the outside white. 71 00:03:35,580 --> 00:03:37,570 So this is possible with a tube of paper. 72 00:03:37,570 --> 00:03:40,450 If you imagine this as being a hole 73 00:03:40,450 --> 00:03:42,870 and the bottom side is also being a hole-- both of these 74 00:03:42,870 --> 00:03:46,410 are open right now-- then you could also 75 00:03:46,410 --> 00:03:50,500 imagine filling them in and getting a cube of paper. 76 00:03:50,500 --> 00:03:53,280 So in this case you'd get a cube that's entirely purple. 77 00:03:57,070 --> 00:04:00,070 This would be bad, because a cube cannot be turned inside 78 00:04:00,070 --> 00:04:01,346 out without self intersection. 79 00:04:01,346 --> 00:04:02,720 So this is an example where there 80 00:04:02,720 --> 00:04:05,020 is a folding motion without the holes filled in. 81 00:04:05,020 --> 00:04:07,950 There is not a folding motion when you fill the holes in. 82 00:04:07,950 --> 00:04:10,570 So I don't know what that says exactly about the problem, 83 00:04:10,570 --> 00:04:12,820 but it's some intuition why this is tricky business. 84 00:04:12,820 --> 00:04:15,850 That's for a polyhedron, of course. 85 00:04:15,850 --> 00:04:17,870 If you're just trying to take polygon with holes 86 00:04:17,870 --> 00:04:20,560 and fill it in somehow, I mean, maybe there's a way, 87 00:04:20,560 --> 00:04:23,402 but certainly the obvious way does not work. 88 00:04:23,402 --> 00:04:25,840 What else did have in my notes here? 89 00:04:25,840 --> 00:04:27,390 All right. 90 00:04:27,390 --> 00:04:30,660 Any questions about that open problem? 91 00:04:30,660 --> 00:04:34,620 Next question is about, it's a neat idea. 92 00:04:34,620 --> 00:04:38,250 We talked about linkages that have joints like this one where 93 00:04:38,250 --> 00:04:40,500 you must stay connected. 94 00:04:40,500 --> 00:04:47,040 And then we briefly also talked about a different kind of joint 95 00:04:47,040 --> 00:04:52,920 where this vertex was pinned right along another edge. 96 00:04:52,920 --> 00:04:54,730 And we showed you could simulate that 97 00:04:54,730 --> 00:04:58,170 by just making a zero area triangle here. 98 00:04:58,170 --> 00:05:00,750 So you can force these edges to come right 99 00:05:00,750 --> 00:05:04,120 at this point of that bar. 100 00:05:04,120 --> 00:05:06,340 Well, different idea is what if you 101 00:05:06,340 --> 00:05:09,660 allow this point to be able to slide along that bar? 102 00:05:09,660 --> 00:05:12,220 Is that some new kind of linkage that's more powerful 103 00:05:12,220 --> 00:05:13,100 or something? 104 00:05:13,100 --> 00:05:14,790 Turns out, no, it's not more powerful. 105 00:05:14,790 --> 00:05:16,625 You can simulate that too. 106 00:05:16,625 --> 00:05:17,750 So I thought I'd show that. 107 00:05:17,750 --> 00:05:18,780 I had to think about it. 108 00:05:18,780 --> 00:05:20,250 It's kind of fun. 109 00:05:20,250 --> 00:05:22,250 This would have made a good problem set problem, 110 00:05:22,250 --> 00:05:24,040 but I decided to cover it. 111 00:05:24,040 --> 00:05:30,300 So remember our good friend the Peaucellier linkage, 112 00:05:30,300 --> 00:05:40,010 which looks like this. 113 00:05:40,010 --> 00:05:43,800 So these guys are equidistant, and this is a rhombus. 114 00:05:43,800 --> 00:05:46,790 All the edge lengths are equal. 115 00:05:46,790 --> 00:05:49,870 Then this vertex lies along a straight line. 116 00:05:53,950 --> 00:05:55,550 And it has a limit. 117 00:05:55,550 --> 00:05:58,100 Let's say it can go this high and this low, 118 00:05:58,100 --> 00:05:59,840 something like that. 119 00:05:59,840 --> 00:06:02,030 We've seen that in animation before. 120 00:06:02,030 --> 00:06:10,270 So what I'm going to do is imagine this as being my bar, 121 00:06:10,270 --> 00:06:14,040 and this as my flexible point that can move along that bar. 122 00:06:14,040 --> 00:06:16,900 So here's the existing bar. 123 00:06:16,900 --> 00:06:20,070 And if I want to add a point that can' slide along the bar, 124 00:06:20,070 --> 00:06:23,800 I'm just going to attach this construction to this bar, 125 00:06:23,800 --> 00:06:25,970 and this is where things get a little bit messy. 126 00:06:30,720 --> 00:06:34,374 These are the guys that are normally rigidly on the ground, 127 00:06:34,374 --> 00:06:36,040 but instead of them being on the ground, 128 00:06:36,040 --> 00:06:38,560 I'm going to attach them here. 129 00:06:38,560 --> 00:06:42,490 So I'm going to attach this to that, this to that, this 130 00:06:42,490 --> 00:06:45,251 to that, this to that. 131 00:06:45,251 --> 00:06:46,750 And because there's two connections, 132 00:06:46,750 --> 00:06:48,583 this is at the intersection of two-- I mean, 133 00:06:48,583 --> 00:06:49,890 this is a rigid triangle. 134 00:06:49,890 --> 00:06:51,710 So these guys can't move anymore. 135 00:06:51,710 --> 00:06:53,720 But, well, they move relative to this edge. 136 00:06:53,720 --> 00:06:56,050 So however this edge moves in the plane 137 00:06:56,050 --> 00:07:00,190 or in space, whatever, I guess in the plane here, 138 00:07:00,190 --> 00:07:03,764 this guy will be forced to track along the bar. 139 00:07:03,764 --> 00:07:06,180 So if you don't worry about intersections, which we're not 140 00:07:06,180 --> 00:07:08,951 in this lecture, this construction 141 00:07:08,951 --> 00:07:10,450 you could attach to any bar and make 142 00:07:10,450 --> 00:07:12,150 a point that can slide along the bar. 143 00:07:12,150 --> 00:07:13,520 Kind of fun. 144 00:07:13,520 --> 00:07:15,700 So you see the power of Peaucellier linkages. 145 00:07:15,700 --> 00:07:19,740 You can do all sorts of fun constructions like this. 146 00:07:19,740 --> 00:07:22,111 You could make it just occupy a little portion 147 00:07:22,111 --> 00:07:23,360 of the bar, whatever you want. 148 00:07:23,360 --> 00:07:26,770 Just build the appropriate Peaucellier linkage. 149 00:07:26,770 --> 00:07:28,990 OK. 150 00:07:28,990 --> 00:07:30,470 So we're into linkages. 151 00:07:30,470 --> 00:07:35,570 Next we're into Kempe's universality theorem, or sort 152 00:07:35,570 --> 00:07:39,240 of Kempe's universality theorem that he almost proved. 153 00:07:39,240 --> 00:07:41,620 So one question just to review, there 154 00:07:41,620 --> 00:07:43,820 is this parallelogram and contra parallelogram. 155 00:07:43,820 --> 00:07:45,528 Pretty much all of his construction other 156 00:07:45,528 --> 00:07:50,020 than Peaucellier at the end are a mix of these two gadgets. 157 00:07:50,020 --> 00:07:52,840 And there's this issue that you could flip one into the other. 158 00:07:52,840 --> 00:07:55,390 Here we had, this was the translator gadget. 159 00:07:55,390 --> 00:07:59,430 If you have two parallelograms, you can collapse one of them, 160 00:07:59,430 --> 00:08:03,340 say, and then flip it out to be a contra parallelogram. 161 00:08:03,340 --> 00:08:05,390 If you don't do any bracing, this can happen. 162 00:08:05,390 --> 00:08:06,692 And this is bad. 163 00:08:06,692 --> 00:08:08,900 You can actually see a few ways in which this is bad. 164 00:08:08,900 --> 00:08:12,210 One is here, we have the parallelogram-- 165 00:08:12,210 --> 00:08:14,030 or, the point of this translator gadget 166 00:08:14,030 --> 00:08:18,360 was to preserve this angle alpha, the green angle here. 167 00:08:18,360 --> 00:08:21,310 It's supposed to be the same as the green angle here. 168 00:08:21,310 --> 00:08:23,490 But if you do this flip, it won't be anymore. 169 00:08:23,490 --> 00:08:26,250 It's the angle between this and horizontal. 170 00:08:26,250 --> 00:08:28,450 Right now this edge is almost horizontal. 171 00:08:28,450 --> 00:08:30,270 So the new angle's almost 0. 172 00:08:30,270 --> 00:08:32,260 Here it's not. 173 00:08:32,260 --> 00:08:34,289 Here's an example with contra parallelograms. 174 00:08:34,289 --> 00:08:37,049 This is our angle trisector. 175 00:08:37,049 --> 00:08:40,059 If you line up this big angle with something, 176 00:08:40,059 --> 00:08:43,159 you get the thirds of it, or vice versa. 177 00:08:43,159 --> 00:08:45,430 We actually wanted to use it to triple angles. 178 00:08:45,430 --> 00:08:47,100 This is Kempe's original drawing. 179 00:08:47,100 --> 00:08:50,360 And if, say, this outermost contra parallelogram 180 00:08:50,360 --> 00:08:54,280 flipped open and became a parallelogram like this blue, 181 00:08:54,280 --> 00:08:56,530 then you'd be in big trouble. 182 00:08:56,530 --> 00:09:00,000 These two angles no longer equal this third angle. 183 00:09:00,000 --> 00:09:02,630 So you'd no longer be tripling or trisecting. 184 00:09:02,630 --> 00:09:05,279 So that's why it's bad. 185 00:09:05,279 --> 00:09:07,070 Next question is, how did you fix it again? 186 00:09:10,290 --> 00:09:12,650 I mean, the parallelogram was easy. 187 00:09:12,650 --> 00:09:15,990 I don't think I need to review that using the construction we 188 00:09:15,990 --> 00:09:17,222 already talked about. 189 00:09:17,222 --> 00:09:19,680 Let me tell you a little bit about the contra parallelogram 190 00:09:19,680 --> 00:09:22,138 bracing, although it's very messy to prove that this works, 191 00:09:22,138 --> 00:09:24,360 so I don't want to spend too much time on it. 192 00:09:24,360 --> 00:09:27,750 The idea again was to take the midpoints of the four 193 00:09:27,750 --> 00:09:29,680 edges of the contra parallelogram. 194 00:09:29,680 --> 00:09:32,840 And first you prove those always remain colinear 195 00:09:32,840 --> 00:09:34,530 in the contra parallelogram state. 196 00:09:34,530 --> 00:09:36,750 They are not colinear in the parallelogram state, 197 00:09:36,750 --> 00:09:38,910 and that's kind of what's good about it. 198 00:09:38,910 --> 00:09:44,730 Then you find a magic point out here off the board called X, 199 00:09:44,730 --> 00:09:50,470 and X is going to be on this perpendicular bisector of PR. 200 00:09:50,470 --> 00:09:53,250 It's also the perpendicular bisector of SQ. 201 00:09:53,250 --> 00:09:57,500 Turns out this distance always equals this distance 202 00:09:57,500 --> 00:09:59,910 by the symmetry of contra parallelogram. 203 00:09:59,910 --> 00:10:01,490 That's actually really easy to see, 204 00:10:01,490 --> 00:10:04,810 because you have opposite edge links being equal. 205 00:10:04,810 --> 00:10:07,486 You get that symmetry. 206 00:10:07,486 --> 00:10:10,110 So it turns out that has to be a fairly specific point for this 207 00:10:10,110 --> 00:10:12,611 to work. 208 00:10:12,611 --> 00:10:13,110 All right. 209 00:10:13,110 --> 00:10:15,400 So what do we do next? 210 00:10:15,400 --> 00:10:18,010 And then we add these four bars. 211 00:10:18,010 --> 00:10:21,040 And the harder part of the claim is at that this thing still 212 00:10:21,040 --> 00:10:24,740 moves within an appropriately chosen X. 213 00:10:24,740 --> 00:10:27,990 I kind of don't want to get into that too much. 214 00:10:27,990 --> 00:10:32,980 The easier part to see is that you can no longer-- if X 215 00:10:32,980 --> 00:10:39,310 is sufficiently far down there, this is no longer possible. 216 00:10:39,310 --> 00:10:41,470 So let's prove that first. 217 00:10:41,470 --> 00:10:45,224 So let's see. 218 00:10:45,224 --> 00:10:46,640 If you look at these bars, the bar 219 00:10:46,640 --> 00:10:50,430 PX and RX have the same length. 220 00:10:50,430 --> 00:10:53,440 That means however you fold this thing, 221 00:10:53,440 --> 00:10:57,360 X must be on the perpendicular bisector of PR. 222 00:10:57,360 --> 00:10:58,990 Here, that's fine. 223 00:10:58,990 --> 00:11:04,300 Over here, the perpendicular bisector would be, I guess, 224 00:11:04,300 --> 00:11:07,610 some thing like this, I guess. 225 00:11:07,610 --> 00:11:08,110 All right? 226 00:11:08,110 --> 00:11:12,145 So the perpendicular bisector of PR some ray like that. 227 00:11:14,740 --> 00:11:15,600 OK. 228 00:11:15,600 --> 00:11:17,430 And simultaneously for the same reason, 229 00:11:17,430 --> 00:11:20,530 X must be on the perpendicular bisector of S 230 00:11:20,530 --> 00:11:24,950 and Q. S and Q are these opposite midpoints, 231 00:11:24,950 --> 00:11:27,850 and so you've got to be in some kind of perpendicular bisector 232 00:11:27,850 --> 00:11:28,420 here. 233 00:11:28,420 --> 00:11:30,919 If you have to be on both of those lines, that means in fact 234 00:11:30,919 --> 00:11:33,702 you must be at the center of this parallelogram. 235 00:11:33,702 --> 00:11:35,410 Or, I don't really need this as a center. 236 00:11:35,410 --> 00:11:37,760 It's some point inside the parallelogram. 237 00:11:37,760 --> 00:11:40,510 That's really bad for X. If these links are really long, 238 00:11:40,510 --> 00:11:45,790 say, longer than the perimeter of this linkage, then 239 00:11:45,790 --> 00:11:48,946 X has to be outside, because, yeah. 240 00:11:52,280 --> 00:11:55,005 If it has to be far from S, R, Q, and P, 241 00:11:55,005 --> 00:11:57,430 you can't be inside the polygon. 242 00:11:57,430 --> 00:11:57,930 OK? 243 00:11:57,930 --> 00:12:00,470 So provided these lengths are sufficiently long, say longer 244 00:12:00,470 --> 00:12:02,800 than the perimeter, there's no way X is inside, 245 00:12:02,800 --> 00:12:06,320 and yet in the parallelogram state it has to be inside. 246 00:12:06,320 --> 00:12:09,550 And so the parallelogram stays impossible. 247 00:12:09,550 --> 00:12:11,302 So that's the easy part of the proof. 248 00:12:11,302 --> 00:12:13,740 The tricky part is to get this thing 249 00:12:13,740 --> 00:12:19,290 to still fold when this is in the contra parallelogram state, 250 00:12:19,290 --> 00:12:22,740 that X is still OK here. 251 00:12:22,740 --> 00:12:26,256 And I'll just mention to convince you 252 00:12:26,256 --> 00:12:34,490 that it's tricky-- you set the length of the XS bar 253 00:12:34,490 --> 00:12:47,180 must be in square, this thing. 254 00:12:47,180 --> 00:12:51,170 XS bar squared must be the XP bar squared plus 1/4 AB squared 255 00:12:51,170 --> 00:12:52,501 minus AD squared. 256 00:12:52,501 --> 00:12:53,750 And I won't go into the proof. 257 00:12:53,750 --> 00:12:57,580 There's some details in the notes here. 258 00:12:57,580 --> 00:13:03,180 But what this says is that we're talking about XS versus XP. 259 00:13:03,180 --> 00:13:06,180 The other two are symmetric , so they have to be equal. 260 00:13:06,180 --> 00:13:08,440 So XS has to be a bit bigger than XP, 261 00:13:08,440 --> 00:13:11,430 and this says how much bigger. 262 00:13:11,430 --> 00:13:13,000 The formula says how much bigger. 263 00:13:13,000 --> 00:13:14,760 They can both still be very large, 264 00:13:14,760 --> 00:13:17,210 so we can still get the part that we need, 265 00:13:17,210 --> 00:13:19,060 but we need that they're actually 266 00:13:19,060 --> 00:13:22,390 related in this way for the whole thing to hold together. 267 00:13:22,390 --> 00:13:23,970 And I will leave it at that. 268 00:13:27,094 --> 00:13:28,510 Sorry, it's a little unsatisfying, 269 00:13:28,510 --> 00:13:30,556 but the details are just not that exciting. 270 00:13:30,556 --> 00:13:31,930 If you're interested in them, you 271 00:13:31,930 --> 00:13:35,180 can read Tim Abbot's master's thesis, 272 00:13:35,180 --> 00:13:38,490 which is on my web page. 273 00:13:38,490 --> 00:13:39,100 Cool. 274 00:13:39,100 --> 00:13:43,920 I wanted to briefly remind you about some of the project 275 00:13:43,920 --> 00:13:49,720 ideas for Kempe before we go to generalizations of Kempe. 276 00:13:49,720 --> 00:13:53,420 So one of them is to implement Kempe. 277 00:13:53,420 --> 00:13:56,860 It's never been implemented, as far as I know, in general form. 278 00:13:56,860 --> 00:13:59,060 And it would be interesting to actually see 279 00:13:59,060 --> 00:14:01,900 it happen in action, some version of it. 280 00:14:01,900 --> 00:14:04,310 There's a lot of different versions, 281 00:14:04,310 --> 00:14:08,580 but ideally with bracing, I guess. 282 00:14:08,580 --> 00:14:10,830 Another fun sort of more design project 283 00:14:10,830 --> 00:14:13,620 would be to design an alphabet and be 284 00:14:13,620 --> 00:14:16,140 able to make every letter of the alphabet with some linkage. 285 00:14:16,140 --> 00:14:17,780 That doesn't have to follow Kempe, 286 00:14:17,780 --> 00:14:20,220 but it would be in the spirit of signing your name. 287 00:14:20,220 --> 00:14:24,260 And I have here one example that's on the web. 288 00:14:24,260 --> 00:14:28,730 I'll show you the web page of making the letter C. Here's 289 00:14:28,730 --> 00:14:31,330 what it looks like in action. 290 00:14:31,330 --> 00:14:34,650 So it's just a four bar linkage, pretty simple. 291 00:14:34,650 --> 00:14:36,890 I mean, it's three bars plus this closing bar. 292 00:14:36,890 --> 00:14:38,848 And then you look at the midpoint of this edge, 293 00:14:38,848 --> 00:14:41,070 and it happens to trace out this kind of letter C. 294 00:14:41,070 --> 00:14:43,706 So if you had a pen there, that's what it would make. 295 00:14:43,706 --> 00:14:46,080 You could imagine just a whole bunch of these in sequence 296 00:14:46,080 --> 00:14:48,170 and be another kind of mathematical font 297 00:14:48,170 --> 00:14:50,970 which would be fun to have, so. 298 00:14:53,550 --> 00:14:56,025 A lot more, 25 open problems left to go. 299 00:14:58,640 --> 00:14:59,970 I think I could do a circle. 300 00:14:59,970 --> 00:15:02,030 I can do an O, so. 301 00:15:02,030 --> 00:15:06,250 24, and these [? fall ?] fast. 302 00:15:06,250 --> 00:15:07,940 Another direction would be to build 303 00:15:07,940 --> 00:15:10,500 some kind of sculpture inspired by Kempe. 304 00:15:10,500 --> 00:15:13,042 This is one by Arthur Ganson since called Faster! 305 00:15:13,042 --> 00:15:15,500 And if you've been to the MIT Museum, you may have seen it. 306 00:15:15,500 --> 00:15:18,110 Sometimes it's out, although I've never seen it running. 307 00:15:18,110 --> 00:15:20,651 But there's a video of it online if you want to check it out. 308 00:15:20,651 --> 00:15:21,670 So this is a device. 309 00:15:21,670 --> 00:15:23,350 It's a kind of push cart. 310 00:15:23,350 --> 00:15:25,350 It's a sculpture push cart. 311 00:15:25,350 --> 00:15:27,160 You have to run with it, and as you 312 00:15:27,160 --> 00:15:30,510 run, the wheels power these gears, 313 00:15:30,510 --> 00:15:33,000 and the pen there with the hand signs 314 00:15:33,000 --> 00:15:36,580 faster, as in you should push it faster. 315 00:15:36,580 --> 00:15:37,830 It's pretty crazy. 316 00:15:37,830 --> 00:15:40,460 Now, this could be done with Kempe, but in this case 317 00:15:40,460 --> 00:15:40,960 it's not. 318 00:15:40,960 --> 00:15:43,630 It's done with these weirdly shaped gears. 319 00:15:43,630 --> 00:15:47,630 And those weirdly shaped gears control the different axes 320 00:15:47,630 --> 00:15:48,960 of the pen. 321 00:15:48,960 --> 00:15:52,540 X, Y, Z, N, in and out. 322 00:15:52,540 --> 00:15:57,550 You can see it actually lifts up to do the exclamation point. 323 00:15:57,550 --> 00:16:00,930 So that's one sculpture inspired by Kempe. 324 00:16:00,930 --> 00:16:04,290 In general, there are lots of ideas for making sculptures out 325 00:16:04,290 --> 00:16:05,090 of linkages. 326 00:16:05,090 --> 00:16:07,650 Arthur Ganson is particularly cool in making 327 00:16:07,650 --> 00:16:11,250 linkage-like sculptures that move kinetically. 328 00:16:11,250 --> 00:16:13,651 If you haven't been to the MIT Museum to see his stuff, 329 00:16:13,651 --> 00:16:14,400 you really should. 330 00:16:14,400 --> 00:16:17,470 It's super cool. 331 00:16:17,470 --> 00:16:19,200 All right. 332 00:16:19,200 --> 00:16:22,635 So that's that. 333 00:16:25,370 --> 00:16:28,517 Next we go on to generalizations of Kempe. 334 00:16:28,517 --> 00:16:30,100 So there are few questions about this. 335 00:16:30,100 --> 00:16:33,560 One was higher dimensions, how exactly does that work? 336 00:16:33,560 --> 00:16:34,916 Particularly D equals 3. 337 00:16:34,916 --> 00:16:36,666 If you want to follow a surface, does that 338 00:16:36,666 --> 00:16:38,750 mean the linkage now has two degrees of freedom? 339 00:16:38,750 --> 00:16:40,791 The answer is yes. 340 00:16:40,791 --> 00:16:41,540 You get to choose. 341 00:16:41,540 --> 00:16:43,040 If you want to trace out a surface, 342 00:16:43,040 --> 00:16:44,700 you'll have to have two degrees of freedom. 343 00:16:44,700 --> 00:16:46,450 So it's not just turning a circular crank. 344 00:16:46,450 --> 00:16:49,270 It's more like a spherical crank, which is indeed 345 00:16:49,270 --> 00:16:53,750 what my hand can do relative to my elbow is move 346 00:16:53,750 --> 00:16:57,707 along a sphere, well, maybe a half sphere or something. 347 00:16:57,707 --> 00:16:59,040 So that's what's possible there. 348 00:16:59,040 --> 00:17:00,540 If you want to trace out a 3D curve, 349 00:17:00,540 --> 00:17:04,545 then you only have one degree of freedom, of course. 350 00:17:04,545 --> 00:17:08,359 I thought I'd briefly tell you a little bit about how 3D works, 351 00:17:08,359 --> 00:17:10,550 or show you one of the constructions, 352 00:17:10,550 --> 00:17:13,710 which is the Peaucellier linkage. 353 00:17:13,710 --> 00:17:24,880 So here is the 2D Peaucellier linkage. 354 00:17:28,990 --> 00:17:31,244 In 3D, this won't work. 355 00:17:31,244 --> 00:17:33,160 Well, it won't work in the sense that this guy 356 00:17:33,160 --> 00:17:36,010 is rather unconstrained. 357 00:17:36,010 --> 00:17:42,710 But if we add, let's see, it's a little hard to see. 358 00:17:42,710 --> 00:17:50,460 Imagine a plane here, a vertical plane through these points. 359 00:17:50,460 --> 00:17:52,430 So if this is in the board plan, I 360 00:17:52,430 --> 00:17:56,110 want to choose a point that's roughly here out of the plane. 361 00:17:56,110 --> 00:17:58,010 I'm going to draw that here. 362 00:17:58,010 --> 00:18:00,330 And then connect it up the same as before. 363 00:18:00,330 --> 00:18:01,500 So it's connected to here. 364 00:18:01,500 --> 00:18:02,940 It's connected to here. 365 00:18:02,940 --> 00:18:04,910 And it's connected to there. 366 00:18:04,910 --> 00:18:07,170 So it's just a third point just like these two. 367 00:18:07,170 --> 00:18:09,460 These two are symmetric, so this one's also symmetric. 368 00:18:09,460 --> 00:18:11,820 And all these lengths are equal if you put it 369 00:18:11,820 --> 00:18:14,280 at right C coordinate. 370 00:18:14,280 --> 00:18:16,810 Then this is like a 3D Peaucellier. 371 00:18:22,460 --> 00:18:24,410 I don't think Peaucellier invented it. 372 00:18:24,410 --> 00:18:27,200 Probably we did, but the result is 373 00:18:27,200 --> 00:18:31,455 that this point will lie on a plane out here. 374 00:18:31,455 --> 00:18:32,080 So that's cool. 375 00:18:32,080 --> 00:18:34,940 It's like the higher dimensional version of Peaucellier. 376 00:18:34,940 --> 00:18:38,420 Now if you actually want this point to move along a line, 377 00:18:38,420 --> 00:18:40,579 what would you do? 378 00:18:40,579 --> 00:18:41,870 AUDIENCE: Intersect two planes. 379 00:18:41,870 --> 00:18:43,880 PROFESSOR: Intersect two planes, exactly. 380 00:18:43,880 --> 00:18:46,400 Take two planes, intersect them. 381 00:18:46,400 --> 00:18:50,250 It is a line. 382 00:18:50,250 --> 00:18:52,524 There's the line of intersection. 383 00:18:52,524 --> 00:18:53,690 Generically, you get a line. 384 00:18:53,690 --> 00:18:55,960 Unless they're the same planes, you always get a line. 385 00:18:55,960 --> 00:18:57,626 So if you take two Peaucellier linkages, 386 00:18:57,626 --> 00:18:59,910 you overlap them at this one point. 387 00:18:59,910 --> 00:19:02,010 Then this point, on the one hand, 388 00:19:02,010 --> 00:19:03,862 these two points are pinned. 389 00:19:03,862 --> 00:19:05,070 We'll have to line one plane. 390 00:19:05,070 --> 00:19:06,710 On the other hand, I'll have to line another plane 391 00:19:06,710 --> 00:19:08,168 from the other Peaucellier linkage, 392 00:19:08,168 --> 00:19:10,960 so you can force it to stay on an actual line. 393 00:19:10,960 --> 00:19:12,740 And both of these gadgets are useful. 394 00:19:12,740 --> 00:19:14,680 Sometimes you want things to stay in planes. 395 00:19:14,680 --> 00:19:18,030 Sometimes you want things to stay on lines. 396 00:19:18,030 --> 00:19:19,990 And once you have this construction, in fact, 397 00:19:19,990 --> 00:19:22,160 you could build the old Kempe construction, 398 00:19:22,160 --> 00:19:25,120 and just put in a ton of 3D Peaucellier linkages 399 00:19:25,120 --> 00:19:27,100 to force everything to stay in the XY plane. 400 00:19:27,100 --> 00:19:29,600 And then anything you could do in two dimensions you can now 401 00:19:29,600 --> 00:19:31,190 do in three dimensions. 402 00:19:31,190 --> 00:19:32,920 So that's observation one. 403 00:19:32,920 --> 00:19:41,520 You can do Kempe in the Z equals 0 plane, the XY plane. 404 00:19:41,520 --> 00:19:43,770 And that's the basic idea for 3D. 405 00:19:43,770 --> 00:19:46,800 You do all the stuff we're used to doing in there, 406 00:19:46,800 --> 00:19:53,310 all the angle doubling, adding new addition, all these things, 407 00:19:53,310 --> 00:19:56,170 and then you just have to translate. 408 00:19:56,170 --> 00:19:57,650 You have some points in 3D. 409 00:19:57,650 --> 00:20:00,870 You need to measure their coordinates or the angles 410 00:20:00,870 --> 00:20:03,720 they form with other edges in 3D. 411 00:20:03,720 --> 00:20:06,630 You just map all those things into the XY plane, 412 00:20:06,630 --> 00:20:09,230 do your computation on the XY plane, and then map them back. 413 00:20:09,230 --> 00:20:13,486 I won't talk about that mapping, but it's not too hard. 414 00:20:13,486 --> 00:20:15,860 And once you can map anything you want into the XY plane, 415 00:20:15,860 --> 00:20:18,220 you could do your computation, map it back, 416 00:20:18,220 --> 00:20:20,600 and force your points in three dimensions 417 00:20:20,600 --> 00:20:23,260 to have whatever properties you need. 418 00:20:23,260 --> 00:20:26,550 So you can write down any polynomial now in X, Y, and Z 419 00:20:26,550 --> 00:20:31,950 and set that equal to 0 just like before. 420 00:20:31,950 --> 00:20:35,680 Do all the trig expansions you did before, 421 00:20:35,680 --> 00:20:40,280 and you can force a point to trace exactly that curve in 3D. 422 00:20:40,280 --> 00:20:43,330 So that's just a sketch of how 3D works. 423 00:20:43,330 --> 00:20:47,140 Skipping some details because there are messy. 424 00:20:47,140 --> 00:20:50,140 The idea is very simple. 425 00:20:50,140 --> 00:20:52,068 3D Peaucellier. 426 00:20:52,068 --> 00:20:53,490 All right. 427 00:20:53,490 --> 00:21:00,800 Next question is related to-- properties 428 00:21:00,800 --> 00:21:04,250 is mentioned in the lecture notes. 429 00:21:04,250 --> 00:21:06,370 So there's a couple versions of this question. 430 00:21:06,370 --> 00:21:09,190 They're asking the same things at different levels of detail. 431 00:21:09,190 --> 00:21:11,557 So what about curves not represented by a polynomial? 432 00:21:11,557 --> 00:21:13,390 I read this like, well, that's not possible. 433 00:21:13,390 --> 00:21:15,770 Everything you make out of linkages 434 00:21:15,770 --> 00:21:17,360 has to be represented by a polynomial. 435 00:21:17,360 --> 00:21:20,600 That's true, but what about piecewise polynomials? 436 00:21:20,600 --> 00:21:23,120 That is possible. 437 00:21:23,120 --> 00:21:25,810 So you can do piecewise cubic splines. 438 00:21:25,810 --> 00:21:29,137 If you've ever drawn a curve in a vector drawing program, 439 00:21:29,137 --> 00:21:29,970 you've used splines. 440 00:21:29,970 --> 00:21:31,460 So those are splines, and they're 441 00:21:31,460 --> 00:21:34,600 made up of little polynomial pieces, like maybe 442 00:21:34,600 --> 00:21:39,180 of a parabola here, and then you design it to transition say, 443 00:21:39,180 --> 00:21:43,770 C2 into another parabola, or then into some hyperbola, 444 00:21:43,770 --> 00:21:44,570 whatever. 445 00:21:44,570 --> 00:21:46,750 So these kinds of general curves, 446 00:21:46,750 --> 00:21:49,480 you can do great things with splines. 447 00:21:49,480 --> 00:21:53,680 Pretty much every curve you've seen on a computer is a spline. 448 00:21:53,680 --> 00:21:56,290 And this is much better than the Weierstrass approximation 449 00:21:56,290 --> 00:21:58,590 theorem which we talked about before, which 450 00:21:58,590 --> 00:22:03,130 says you design one polynomial that approximates 451 00:22:03,130 --> 00:22:05,390 an entire curve, like your signature. 452 00:22:05,390 --> 00:22:07,650 But when it does that, it'll be like this. 453 00:22:07,650 --> 00:22:12,780 It's a very ugly-- if you use various pieces of polynomials, 454 00:22:12,780 --> 00:22:15,310 say all cubic polynomials, you can 455 00:22:15,310 --> 00:22:17,360 get some really nice looking curves. 456 00:22:17,360 --> 00:22:21,220 So you can really reproduce your signature. 457 00:22:21,220 --> 00:22:24,280 And there is a theorem mentioned in the notes that 458 00:22:24,280 --> 00:22:34,290 says you can trace any semi algebraic set. 459 00:22:34,290 --> 00:22:36,030 So I want to define semi algebraic, 460 00:22:36,030 --> 00:22:39,535 because this is actually closely related to splines. 461 00:22:39,535 --> 00:22:40,660 It's a little more general. 462 00:22:46,270 --> 00:22:47,840 So what's a semi algebraic set? 463 00:22:50,860 --> 00:22:53,300 So here's an example of a semi algebraic set. 464 00:22:53,300 --> 00:23:00,680 You have some polynomial, let's say on XYZ, 465 00:23:00,680 --> 00:23:04,446 and you want to say this is greater than or equal to 0. 466 00:23:04,446 --> 00:23:05,570 So it's a little different. 467 00:23:05,570 --> 00:23:09,210 Before, we could set polynomials equal to 0. 468 00:23:09,210 --> 00:23:11,520 Now I can set them greater than or equal to 0. 469 00:23:11,520 --> 00:23:14,620 So that's the semi part of-- if we just have this, 470 00:23:14,620 --> 00:23:17,430 this is an algebraic set, essentially. 471 00:23:17,430 --> 00:23:21,850 Semi algebraic, you can have half spaces in some sense. 472 00:23:21,850 --> 00:23:28,530 And then you can also take unions and intersections 473 00:23:28,530 --> 00:23:30,490 to form an algebraic set. 474 00:23:30,490 --> 00:23:33,760 Also compliments, but it doesn't matter. 475 00:23:33,760 --> 00:23:35,330 So what does this mean? 476 00:23:35,330 --> 00:23:36,740 It means I can take all the stuff 477 00:23:36,740 --> 00:23:43,760 on one side of a polynomial, and then-- well, on one polynomial. 478 00:23:43,760 --> 00:23:45,510 Sorry, that's what I should do. 479 00:23:45,510 --> 00:23:48,520 So here's, let's say, my parabola. 480 00:23:48,520 --> 00:23:51,110 I can take this region, and then I 481 00:23:51,110 --> 00:23:54,225 could say, OK, well, let's take-- 482 00:23:54,225 --> 00:23:59,300 this gets messy to do-- I could take also 483 00:23:59,300 --> 00:24:02,190 all the stuff outside this polynomial. 484 00:24:02,190 --> 00:24:04,660 So that's some bigger region here. 485 00:24:04,660 --> 00:24:06,160 I could take the union of those. 486 00:24:06,160 --> 00:24:07,670 I could clip off parts. 487 00:24:07,670 --> 00:24:11,510 Basically I can construct a spline in particular, 488 00:24:11,510 --> 00:24:13,510 but in general I can do lots of different things 489 00:24:13,510 --> 00:24:19,270 by unions and intersections of these polynomial half spaces. 490 00:24:19,270 --> 00:24:21,400 So this lets you construct splines. 491 00:24:21,400 --> 00:24:23,320 It lets you piece together components, 492 00:24:23,320 --> 00:24:27,520 because, for example, I could take this curve. 493 00:24:27,520 --> 00:24:29,640 And then on one side intersect it 494 00:24:29,640 --> 00:24:32,440 with the other side Greater equal to 0 and lesser equal 495 00:24:32,440 --> 00:24:36,320 to zero, then I get exactly just this curve. 496 00:24:36,320 --> 00:24:41,690 And then I can, for example, cut to the left of this line. 497 00:24:41,690 --> 00:24:44,691 And then I'll have this curve, but only if it stops here. 498 00:24:44,691 --> 00:24:46,440 I could do the same thing with this piece, 499 00:24:46,440 --> 00:24:47,630 end up with this piece, and then take 500 00:24:47,630 --> 00:24:48,970 the union of those two pieces. 501 00:24:48,970 --> 00:24:50,710 So I can construct a spline. 502 00:24:50,710 --> 00:24:52,890 I can have regions, of course. 503 00:24:52,890 --> 00:24:54,890 Infinite area, finite area, whatever. 504 00:24:54,890 --> 00:24:57,000 So semi algebraic sets are very general. 505 00:24:57,000 --> 00:24:58,690 It's fairly easy to see that this 506 00:24:58,690 --> 00:25:03,310 is the most you could hope for, because in general, you 507 00:25:03,310 --> 00:25:06,320 look at a linkage, it's defined by polynomial equations. 508 00:25:06,320 --> 00:25:08,151 You say, well the square distance from P 509 00:25:08,151 --> 00:25:11,000 to Q equals L for various things. 510 00:25:11,000 --> 00:25:14,040 So you only have polynomial equations to work with. 511 00:25:14,040 --> 00:25:17,200 Because you have flexibility, you get inequations. 512 00:25:17,200 --> 00:25:20,700 Because we can do things like this. 513 00:25:20,700 --> 00:25:22,700 I mean, this distance is now less than 514 00:25:22,700 --> 00:25:26,360 or equal to the sum of these two lines. 515 00:25:26,360 --> 00:25:29,820 So semi algebraic sets are the best 516 00:25:29,820 --> 00:25:32,850 you could hope for, a bunch of polynomial inequalities. 517 00:25:32,850 --> 00:25:38,680 And in fact, every such semi algebraic set is possible. 518 00:25:38,680 --> 00:25:39,780 So how do you prove that? 519 00:25:39,780 --> 00:25:42,220 It's actually really easy. 520 00:25:42,220 --> 00:25:44,120 We've essentially already done P of XYZ 521 00:25:44,120 --> 00:25:47,630 is greater or equal to 0, because we saw in Kempe how 522 00:25:47,630 --> 00:25:49,640 to set something equal to 0. 523 00:25:49,640 --> 00:25:52,590 To do that, we used the Peaucellier linkage, 524 00:25:52,590 --> 00:25:55,160 which I will draw for the Nth time, 525 00:25:55,160 --> 00:25:56,760 because I need to modify it. 526 00:26:00,380 --> 00:26:02,640 So here's a Peaucellier linkage. 527 00:26:02,640 --> 00:26:06,810 It forces this point to lie on a straight line. 528 00:26:06,810 --> 00:26:11,850 If I add a joint here, and basically let this length 529 00:26:11,850 --> 00:26:16,150 get smaller if it wants to-- it can do things like this now-- 530 00:26:16,150 --> 00:26:19,450 I should draw it more like that. 531 00:26:19,450 --> 00:26:22,267 Maybe I should draw it scale, so that's maybe 532 00:26:22,267 --> 00:26:23,100 something like this. 533 00:26:23,100 --> 00:26:26,220 This point can now move anywhere on the segment. 534 00:26:26,220 --> 00:26:30,630 Then this guy will end up being able to make, well not exactly 535 00:26:30,630 --> 00:26:33,500 this half plane, but a region of it. 536 00:26:33,500 --> 00:26:36,080 A big enough region if you set it up right. 537 00:26:36,080 --> 00:26:38,000 And as you may recall, the x-coordinate 538 00:26:38,000 --> 00:26:41,210 here was the sum of all my trig terms. 539 00:26:41,210 --> 00:26:44,670 And I wanted that equal to 0 for this to happen, 540 00:26:44,670 --> 00:26:47,450 but if I wanted to make greater than or equal to 0, 541 00:26:47,450 --> 00:26:51,370 then I just needed to be to the right of that vertical line. 542 00:26:51,370 --> 00:26:54,140 And so that lets me take any polynomial 543 00:26:54,140 --> 00:26:55,670 and set it greater or equal to 0. 544 00:26:55,670 --> 00:27:01,010 So that's basically Kempe, except I used this modified 545 00:27:01,010 --> 00:27:05,870 Peaucellier that let's things go to the right a little bit. 546 00:27:05,870 --> 00:27:07,130 OK. 547 00:27:07,130 --> 00:27:08,362 Intersections are also easy. 548 00:27:08,362 --> 00:27:10,320 If I want to take the intersection of two sets, 549 00:27:10,320 --> 00:27:14,590 I just overlay those linkages and let 550 00:27:14,590 --> 00:27:18,330 them share the same point. 551 00:27:18,330 --> 00:27:21,390 I call this P that we're constraining. 552 00:27:21,390 --> 00:27:24,000 So that will apply multiple constraints to that same point, 553 00:27:24,000 --> 00:27:26,790 and so that is the intersection of those two semi algebraic 554 00:27:26,790 --> 00:27:28,470 sets. 555 00:27:28,470 --> 00:27:30,900 The one tricky part is unions, and this 556 00:27:30,900 --> 00:27:33,350 is what the question is asking about here. 557 00:27:33,350 --> 00:27:37,380 Intersection gadgets are clear, but what's the union gadget? 558 00:27:37,380 --> 00:27:42,160 Union gadget turns out to also be possible using Kempe. 559 00:27:42,160 --> 00:27:43,570 Kind of surprising. 560 00:27:43,570 --> 00:27:45,590 Let me show you how. 561 00:27:54,850 --> 00:28:01,700 So suppose you have linkage one constraining 562 00:28:01,700 --> 00:28:04,340 some point I'll call it P1 here. 563 00:28:04,340 --> 00:28:07,030 And you have another linkage two, 564 00:28:07,030 --> 00:28:10,210 constraining at some point P2. 565 00:28:10,210 --> 00:28:12,495 And what you'd like to do is build a new linkage that 566 00:28:12,495 --> 00:28:17,200 has some point P that either follows P1 or follows P2, 567 00:28:17,200 --> 00:28:19,630 so it takes the union of those two sets. 568 00:28:19,630 --> 00:28:23,140 So whatever L1 constructs for P1, whatever L2 constructs 569 00:28:23,140 --> 00:28:28,200 for P2, you want to be able to trace P1 or trace P2. 570 00:28:28,200 --> 00:28:33,690 And what we're going to do is build another box here, 571 00:28:33,690 --> 00:28:36,422 another linkage, which is going to be-- I 572 00:28:36,422 --> 00:28:38,680 won't write L. Got some more room here. 573 00:29:01,160 --> 00:29:04,142 It's going to be a Kempe construction 574 00:29:04,142 --> 00:29:04,975 for this polynomial. 575 00:29:08,640 --> 00:29:12,235 So this polynomial, it's a little different 576 00:29:12,235 --> 00:29:13,110 from what we've seen. 577 00:29:13,110 --> 00:29:15,380 And over here it's going to be point P. P here 578 00:29:15,380 --> 00:29:22,560 is X comma Y. P2 is X2Y2 and so on. 579 00:29:22,560 --> 00:29:24,370 If you want, as X1Y1. 580 00:29:24,370 --> 00:29:28,380 So this is an equation involving three points. 581 00:29:28,380 --> 00:29:31,530 In the past, we've only had polynomial equations 582 00:29:31,530 --> 00:29:33,020 involving one point. 583 00:29:33,020 --> 00:29:35,835 This is another generalization of Kempe which I may or may not 584 00:29:35,835 --> 00:29:38,970 have mentioned, but it's really easy. 585 00:29:38,970 --> 00:29:44,899 We had, what was it, a rhombus to represent a single point. 586 00:29:44,899 --> 00:29:47,440 You just have three of them, and now you've got three points. 587 00:29:47,440 --> 00:29:49,820 You could do the same trig identities, and so on, 588 00:29:49,820 --> 00:29:50,560 to expand out. 589 00:29:50,560 --> 00:29:53,070 You get, you might not call it a polynomial, 590 00:29:53,070 --> 00:29:55,120 you might call it a multinomial. 591 00:29:55,120 --> 00:29:57,800 Well, I guess it was already a multinomial in X and Y. 592 00:29:57,800 --> 00:30:02,430 Now it's a polynomial on XY, X1, Y1, X2. 593 00:30:02,430 --> 00:30:03,170 Y2. 594 00:30:03,170 --> 00:30:04,960 And that's exactly what we have here, 595 00:30:04,960 --> 00:30:09,260 various powers of those six variables. 596 00:30:09,260 --> 00:30:12,490 You expand them out just in the way you did before with trig, 597 00:30:12,490 --> 00:30:15,050 and you just need to be able to add up angles now, 598 00:30:15,050 --> 00:30:17,170 not just alpha and beta, but now there's 599 00:30:17,170 --> 00:30:21,180 six possible angles representing each of the x- y-coordinates. 600 00:30:21,180 --> 00:30:24,000 So I claim I can build this thing of Kempe. 601 00:30:28,971 --> 00:30:29,470 OK. 602 00:30:29,470 --> 00:30:31,070 And if I build this thing, basically I 603 00:30:31,070 --> 00:30:37,210 force X and Y to be either X1, X2 or Y1, Y2, 604 00:30:37,210 --> 00:30:41,390 and that lets you take the union of link [? which ?] L1 and L2. 605 00:30:41,390 --> 00:30:46,850 The new point P can either live at P1 or can live at P2. 606 00:30:46,850 --> 00:30:47,870 Questions about that? 607 00:30:50,856 --> 00:30:51,356 Yeah. 608 00:30:51,356 --> 00:30:53,064 AUDIENCE: Should some of those be pluses? 609 00:30:53,064 --> 00:30:55,340 PROFESSOR: Should some of those be pluses? 610 00:30:55,340 --> 00:30:56,834 AUDIENCE: [INAUDIBLE] looks like X 611 00:30:56,834 --> 00:30:59,654 equals X1 is enough to make that whole thing 0. 612 00:30:59,654 --> 00:31:00,320 PROFESSOR: Yeah. 613 00:31:00,320 --> 00:31:02,156 I was curious about this. 614 00:31:02,156 --> 00:31:04,208 Do you see a way to fix this? 615 00:31:04,208 --> 00:31:09,328 The worry here is X could be X1, and Y could be Y2 [INAUDIBLE]. 616 00:31:09,328 --> 00:31:11,272 So you have-- 617 00:31:11,272 --> 00:31:15,646 AUDIENCE: X minus X1 squared plus Y minus Y1 618 00:31:15,646 --> 00:31:18,562 squared all [INAUDIBLE] X minus X2 squared plus Y 619 00:31:18,562 --> 00:31:19,415 minus Y2 squared. 620 00:31:19,415 --> 00:31:20,080 PROFESSOR: Yes. 621 00:31:20,080 --> 00:31:24,560 Plus-- this is extremely ugly. 622 00:31:24,560 --> 00:31:27,210 OK, let me maybe rewrite it. 623 00:31:27,210 --> 00:31:29,736 Thanks. 624 00:31:29,736 --> 00:31:31,820 Yeah, as I was writing this I was like-- 625 00:31:31,820 --> 00:31:34,180 AUDIENCE: Squared in the wrong place. 626 00:31:34,180 --> 00:31:38,930 PROFESSOR: Squared plus Y minus Y1 squared. 627 00:31:38,930 --> 00:31:44,760 And then this thing times yeah, thanks. 628 00:31:44,760 --> 00:31:45,620 Same thing with 2's. 629 00:31:50,460 --> 00:31:51,450 Equals 0. 630 00:31:51,450 --> 00:31:53,640 So the product being equal to 0 means 631 00:31:53,640 --> 00:31:56,780 one of the two terms better equal 0. 632 00:31:56,780 --> 00:32:00,530 And in this case, if this equals 0, because of the squares 633 00:32:00,530 --> 00:32:02,260 it forces it to be non-negative, which 634 00:32:02,260 --> 00:32:04,602 means the only time the sum is equal to 0 635 00:32:04,602 --> 00:32:06,560 is when both of the terms are equal to 0, which 636 00:32:06,560 --> 00:32:09,510 means X equals X1 and Y equals Y1. 637 00:32:09,510 --> 00:32:14,120 So this plays the role of and here, 638 00:32:14,120 --> 00:32:16,760 and the product plays the role of or. 639 00:32:16,760 --> 00:32:22,180 So either XY equals X1Y1, or XY equals X2Y2. 640 00:32:22,180 --> 00:32:24,010 Thank you. 641 00:32:24,010 --> 00:32:26,050 Good fix. 642 00:32:26,050 --> 00:32:29,458 Just make a quick note of that. 643 00:32:29,458 --> 00:32:30,372 Do I have a pen? 644 00:32:30,372 --> 00:32:32,200 I won't. 645 00:32:32,200 --> 00:32:36,210 OK, so that is how you do intersection of two linkages, 646 00:32:36,210 --> 00:32:38,470 is just Kempe again. 647 00:32:38,470 --> 00:32:40,150 It's kind of cool. 648 00:32:40,150 --> 00:32:42,120 Any questions about that? 649 00:32:42,120 --> 00:32:43,890 Once you have unions, we've already 650 00:32:43,890 --> 00:32:45,520 done intersections in half spaces, 651 00:32:45,520 --> 00:32:47,549 then you can make any semi algebraic set. 652 00:32:47,549 --> 00:32:49,340 So what I think would be cool in particular 653 00:32:49,340 --> 00:32:52,890 is to implement Kempe force spline, say, quadratic splines. 654 00:32:52,890 --> 00:32:54,515 Because Kempe for quadratic polynomials 655 00:32:54,515 --> 00:32:56,950 is going to be pretty reasonable. 656 00:32:56,950 --> 00:33:01,060 You have to implement this union gadget to piece together 657 00:33:01,060 --> 00:33:02,720 the pieces, which is kind of messy, 658 00:33:02,720 --> 00:33:07,326 but in principle, you can piece together a bunch of polynomials 659 00:33:07,326 --> 00:33:08,700 and see what a spline looks like. 660 00:33:08,700 --> 00:33:10,090 You had a question. 661 00:33:10,090 --> 00:33:14,794 AUDIENCE: So if the curves you're taking in enough 662 00:33:14,794 --> 00:33:18,682 don't intersect to get from point to the other-- 663 00:33:18,682 --> 00:33:19,670 PROFESSOR: OK. 664 00:33:19,670 --> 00:33:20,210 Yeah. 665 00:33:20,210 --> 00:33:25,580 So you're asking about sort of continuous crank ability 666 00:33:25,580 --> 00:33:27,840 of Kempe constructions. 667 00:33:27,840 --> 00:33:31,090 And indeed, if you had two sets that were disjoint, 668 00:33:31,090 --> 00:33:34,980 there's no way to continuously go from one spot to the other. 669 00:33:34,980 --> 00:33:37,800 What this is saying is that the overall trace of this point-- 670 00:33:37,800 --> 00:33:40,480 if you look at all possible configurations, the linkage, 671 00:33:40,480 --> 00:33:42,970 and then just see where P goes, it 672 00:33:42,970 --> 00:33:46,810 will trace out both of those connected components. 673 00:33:46,810 --> 00:33:48,870 On the other hand, if these do overlap 674 00:33:48,870 --> 00:33:51,490 if there is a common point between these two linkages, 675 00:33:51,490 --> 00:33:53,800 then this will allow you to transition from one 676 00:33:53,800 --> 00:33:54,920 to the other. 677 00:33:54,920 --> 00:33:57,099 So right, if you're building a spline, 678 00:33:57,099 --> 00:33:58,890 presumably you know that they're connected, 679 00:33:58,890 --> 00:34:01,720 and you want to be careful of the way you union them together 680 00:34:01,720 --> 00:34:04,850 to make them possible by continuous motion. 681 00:34:04,850 --> 00:34:06,210 I think that's always possible. 682 00:34:09,159 --> 00:34:12,254 But you do have to be careful. 683 00:34:12,254 --> 00:34:12,920 Other questions? 684 00:34:16,455 --> 00:34:17,820 All right. 685 00:34:17,820 --> 00:34:20,480 I want to get the hypar gluing, but there's 686 00:34:20,480 --> 00:34:24,580 one more topic some people asked about, 687 00:34:24,580 --> 00:34:26,679 which are these origami axioms. 688 00:34:26,679 --> 00:34:30,900 I mentioned them very briefly at the end of lecture 10, 689 00:34:30,900 --> 00:34:32,599 and they sounded amazingly powerful. 690 00:34:32,599 --> 00:34:34,520 They let you solve all these things. 691 00:34:34,520 --> 00:34:36,770 The setting here is something called ruler and compass 692 00:34:36,770 --> 00:34:37,807 constructions. 693 00:34:37,807 --> 00:34:40,182 How many people here have heard of ruler or straight edge 694 00:34:40,182 --> 00:34:41,600 and compass? 695 00:34:41,600 --> 00:34:44,080 Almost everyone. 696 00:34:44,080 --> 00:34:46,420 A compasses is this gadget like this. 697 00:34:46,420 --> 00:34:47,949 You can draw circles with it. 698 00:34:47,949 --> 00:34:50,630 And there's a standard mathematical formulation 699 00:34:50,630 --> 00:34:53,214 of a straight edge and compass which is, if I have two points, 700 00:34:53,214 --> 00:34:54,880 I can draw a straight line through them. 701 00:34:54,880 --> 00:34:57,390 If I have two points, I can draw a circle through them 702 00:34:57,390 --> 00:34:58,290 like that. 703 00:34:58,290 --> 00:35:02,910 And if I have lines and circles, I can take their intersections. 704 00:35:02,910 --> 00:35:05,960 And if that's all you're allowed to do, 705 00:35:05,960 --> 00:35:09,610 then you can prove that if you look at the coordinates of all 706 00:35:09,610 --> 00:35:11,130 the points you make-- and let's say 707 00:35:11,130 --> 00:35:13,800 you start with one point which I'll call 0 comma 708 00:35:13,800 --> 00:35:16,970 0, another 0.1 comma 0. 709 00:35:16,970 --> 00:35:18,561 So I have the number 0 in one. 710 00:35:18,561 --> 00:35:21,060 Then the numbers you can make, the coordinates you can make, 711 00:35:21,060 --> 00:35:22,940 are everything you could make from 0 to 1 712 00:35:22,940 --> 00:35:27,284 by plus, minus, times, divide, and square root. 713 00:35:27,284 --> 00:35:28,700 You could do all those operations, 714 00:35:28,700 --> 00:35:30,296 and that's all you could do. 715 00:35:30,296 --> 00:35:31,670 And so what that means basically, 716 00:35:31,670 --> 00:35:35,630 is you could solve quadratic polynomials, but nothing more. 717 00:35:35,630 --> 00:35:39,720 And that's an old result from 1800s and implies things like 718 00:35:39,720 --> 00:35:42,042 you cannot trisect an angle. 719 00:35:42,042 --> 00:35:43,750 You can bisect an angle because that only 720 00:35:43,750 --> 00:35:45,520 involves quadratic stuff. 721 00:35:45,520 --> 00:35:48,050 You can't trisect an angle, for example, 60 degrees, 722 00:35:48,050 --> 00:35:51,160 because that involves solving subcubic, which is not 723 00:35:51,160 --> 00:35:53,870 possible by straight edge and compass. 724 00:35:53,870 --> 00:35:56,280 You cannot compute the cube root of 2. 725 00:35:56,280 --> 00:35:59,810 It's a cube doubling problem, so all these great things. 726 00:35:59,810 --> 00:36:04,950 Then came along Huzita in 1989, and at the same time, 727 00:36:04,950 --> 00:36:08,270 Jacques Justin in 1989, who you remember 728 00:36:08,270 --> 00:36:11,150 did Kawasaki's theorem and Maekawa's theorem. 729 00:36:11,150 --> 00:36:14,350 He also did this, all independently. 730 00:36:14,350 --> 00:36:17,760 So Huzita suggested these axioms for folding. 731 00:36:17,760 --> 00:36:19,930 If I have two points, I can fold a crease 732 00:36:19,930 --> 00:36:23,290 along those two points, that line. 733 00:36:23,290 --> 00:36:25,800 If I have two points, I can fold a point onto a point. 734 00:36:25,800 --> 00:36:27,650 That constructs a perpendicular bisector. 735 00:36:27,650 --> 00:36:30,330 If I have two lines, I can fold one line onto the other. 736 00:36:30,330 --> 00:36:33,620 That is the angular bisector. 737 00:36:33,620 --> 00:36:36,785 If I have a point and the line, I can fold the line onto itself 738 00:36:36,785 --> 00:36:40,920 , which forces the crease to be perpendicular and pass through 739 00:36:40,920 --> 00:36:41,700 this point. 740 00:36:41,700 --> 00:36:44,150 You see these in lots of origami diagrams. 741 00:36:44,150 --> 00:36:47,150 They let you find interesting lines. 742 00:36:47,150 --> 00:36:50,024 If I have two points on a line, I 743 00:36:50,024 --> 00:36:51,440 can fold this point onto this line 744 00:36:51,440 --> 00:36:53,220 while also folding through this point. 745 00:36:53,220 --> 00:36:56,056 That's a tangent of a parabola if you look at it correctly. 746 00:36:56,056 --> 00:36:57,680 And if I have two points and two lines, 747 00:36:57,680 --> 00:36:58,960 I can fold this point onto this line 748 00:36:58,960 --> 00:37:01,290 while simultaneously folding this point onto this line. 749 00:37:01,290 --> 00:37:03,206 There actually is four or eight different ways 750 00:37:03,206 --> 00:37:05,180 to do it in general, but you can find them 751 00:37:05,180 --> 00:37:07,180 all by just manipulating the paper. 752 00:37:07,180 --> 00:37:08,400 That's the claim. 753 00:37:08,400 --> 00:37:12,970 There's one other that Huzita missed, Justin saw, called, 754 00:37:12,970 --> 00:37:14,490 these days, Hatori's axiom, where 755 00:37:14,490 --> 00:37:18,610 you fold a point onto a line. 756 00:37:18,610 --> 00:37:22,450 And not shown here, I believe, is another point 757 00:37:22,450 --> 00:37:24,480 that you must fold through. 758 00:37:24,480 --> 00:37:27,010 No, that looks like this axiom. 759 00:37:27,010 --> 00:37:30,790 So I've forgotten what the differences is for Hatori. 760 00:37:30,790 --> 00:37:35,134 Oh, not drawn here is, there's an edge in the bottom. 761 00:37:35,134 --> 00:37:36,550 And you also want to fold the line 762 00:37:36,550 --> 00:37:37,924 onto itself, which means you have 763 00:37:37,924 --> 00:37:41,180 to be perpendicular to this black line down here. 764 00:37:43,900 --> 00:37:45,927 So that's sort of all you could imagine 765 00:37:45,927 --> 00:37:48,010 if you have these sort of points onto lines, lines 766 00:37:48,010 --> 00:37:50,210 onto points, lines onto lines. 767 00:37:50,210 --> 00:37:52,630 Axioms, if you enumerate them all, 768 00:37:52,630 --> 00:37:54,880 this is all of them for a single fold. 769 00:37:54,880 --> 00:37:56,650 And with these axioms you could prove 770 00:37:56,650 --> 00:37:58,500 you can solve any cubic polynomials. 771 00:37:58,500 --> 00:38:01,670 So basically you can also take cube roots. 772 00:38:01,670 --> 00:38:04,810 And so in particular, you can do things like trisect an angle. 773 00:38:04,810 --> 00:38:06,240 It's what's shown up here. 774 00:38:06,240 --> 00:38:07,940 Fairly small sequence. 775 00:38:07,940 --> 00:38:10,180 This was discovered in the 1970s. 776 00:38:10,180 --> 00:38:12,910 You can trisect any angle, divide it into thirds 777 00:38:12,910 --> 00:38:14,200 just by these kinds of folds. 778 00:38:14,200 --> 00:38:16,325 And the tricky operation here is folding two points 779 00:38:16,325 --> 00:38:18,180 onto two lines simultaneously. 780 00:38:18,180 --> 00:38:20,100 That's the third degree operation. 781 00:38:20,100 --> 00:38:23,560 Everything else you can do with ruler and compass. 782 00:38:23,560 --> 00:38:25,920 And you can also do things like double a cube. 783 00:38:25,920 --> 00:38:29,930 This is computing a cube root of 2 ratio. 784 00:38:29,930 --> 00:38:31,770 First you bisect you thing into thirds. 785 00:38:31,770 --> 00:38:34,960 This is easy to do, because we can divide by three. 786 00:38:34,960 --> 00:38:39,660 Then you fold these two points onto these two lines. 787 00:38:39,660 --> 00:38:43,260 And it turns out the ratio here between this y-coordinate 788 00:38:43,260 --> 00:38:45,290 and this y-coordinate is cube root of 3. 789 00:38:45,290 --> 00:38:47,560 A over B is cube root of 3. 790 00:38:47,560 --> 00:38:54,250 This is by Peter Messer in 1985. 791 00:38:54,250 --> 00:38:57,080 So that's kind of cool. 792 00:38:57,080 --> 00:38:58,920 But that's all you can do with single folds. 793 00:38:58,920 --> 00:39:02,100 The most you hope for is solving cubic equations. 794 00:39:02,100 --> 00:39:03,440 You can't quintisect an angle. 795 00:39:03,440 --> 00:39:05,690 You can't divide an angle in fifths. 796 00:39:05,690 --> 00:39:08,810 You can't compute the fifth root of 2, I assume. 797 00:39:08,810 --> 00:39:11,055 You can't do lots of things. 798 00:39:11,055 --> 00:39:13,680 But it's at least more powerful than straight edge and compass, 799 00:39:13,680 --> 00:39:14,770 which is cool. 800 00:39:14,770 --> 00:39:17,550 So then this was just for single fold operations. 801 00:39:17,550 --> 00:39:19,190 You could look at two fold operations, 802 00:39:19,190 --> 00:39:20,689 three fold operations where you make 803 00:39:20,689 --> 00:39:23,170 two folds and simultaneously align lots of things. 804 00:39:23,170 --> 00:39:26,980 With two folds you can-- oh, sorry, before I get that. 805 00:39:26,980 --> 00:39:29,810 There is a software called ReferenceFinder by Robert Lang, 806 00:39:29,810 --> 00:39:33,030 if you're curious about how to construct various things. 807 00:39:33,030 --> 00:39:36,070 So here is plugged in, I want to compute a third. 808 00:39:36,070 --> 00:39:37,920 And it just enumerates all possible things 809 00:39:37,920 --> 00:39:39,700 you could do with five or six folds. 810 00:39:39,700 --> 00:39:41,400 And if it finds an exact solution, 811 00:39:41,400 --> 00:39:42,540 it will put it at the top. 812 00:39:42,540 --> 00:39:44,590 It also finds approximate solutions 813 00:39:44,590 --> 00:39:46,140 which are practically useful. 814 00:39:46,140 --> 00:39:48,880 So this is a sequence of operations that from a square, 815 00:39:48,880 --> 00:39:55,460 you could find this point at 0 comma third, which is nice. 816 00:39:55,460 --> 00:39:58,010 When you have two folds, you can quintisect an angle. 817 00:39:58,010 --> 00:40:01,230 These are Robert Lang's diagrams for quintisecting a given 818 00:40:01,230 --> 00:40:02,770 angle. 819 00:40:02,770 --> 00:40:05,020 You can see at the very end here we have an angle 820 00:40:05,020 --> 00:40:08,560 and it's divided evenly into fifths. 821 00:40:08,560 --> 00:40:10,370 It's a bit complicated, and at some point 822 00:40:10,370 --> 00:40:12,750 it involves a two fold operation. 823 00:40:12,750 --> 00:40:13,584 It says here. 824 00:40:13,584 --> 00:40:14,750 Here's where it all happens. 825 00:40:14,750 --> 00:40:18,050 Your fold here, and simultaneously you fold here. 826 00:40:18,050 --> 00:40:22,150 And you have to align all these points and lines and things. 827 00:40:22,150 --> 00:40:23,180 And that's cool. 828 00:40:23,180 --> 00:40:27,190 I think with three folds, they can solve any quintic equation. 829 00:40:27,190 --> 00:40:29,970 That's Alperin and Lang. 830 00:40:29,970 --> 00:40:32,750 And then the culmination, which I mentioned briefly 831 00:40:32,750 --> 00:40:35,490 at the end of lecture 10, is if you allow end folds 832 00:40:35,490 --> 00:40:37,410 simultaneously, then you can solve 833 00:40:37,410 --> 00:40:39,490 a degree end polynomial [INAUDIBLE] 834 00:40:39,490 --> 00:40:43,520 order N. First you set up your piece of paper 835 00:40:43,520 --> 00:40:46,020 into all these independently manipulatable limbs. 836 00:40:46,020 --> 00:40:48,680 You mark off these coordinates, which 837 00:40:48,680 --> 00:40:51,125 are the lengths of the bars in your Kempe construction. 838 00:40:51,125 --> 00:40:52,500 And then you say, well you've got 839 00:40:52,500 --> 00:40:54,810 to fold so that this all these points align. 840 00:40:54,810 --> 00:40:56,830 That will construct a linkage state, 841 00:40:56,830 --> 00:40:59,120 and if you set Kempe up right, there 842 00:40:59,120 --> 00:41:01,544 will only be one state, which is the solution 843 00:41:01,544 --> 00:41:02,335 to your polynomial. 844 00:41:05,295 --> 00:41:06,420 So that's one way to do it. 845 00:41:06,420 --> 00:41:08,045 There are actually other ways to do it. 846 00:41:08,045 --> 00:41:09,890 Alperin and Lang have another solution. 847 00:41:09,890 --> 00:41:11,640 It's a little more simple, but this is fun 848 00:41:11,640 --> 00:41:13,630 because it uses Kempe. 849 00:41:13,630 --> 00:41:19,590 And that is a brief story of origami axioms. 850 00:41:19,590 --> 00:41:21,160 Any questions about that? 851 00:41:21,160 --> 00:41:23,080 There's a small chapter in the book. 852 00:41:23,080 --> 00:41:25,879 It's called Geometric Construction. 853 00:41:25,879 --> 00:41:28,170 I didn't prove anything here, because it's a little bit 854 00:41:28,170 --> 00:41:28,878 tedious to prove. 855 00:41:28,878 --> 00:41:31,740 You can solve any cubic, you can solve all these things. 856 00:41:31,740 --> 00:41:33,550 That's the most you could solve. 857 00:41:33,550 --> 00:41:36,110 But all these things have been completely characterized. 858 00:41:36,110 --> 00:41:38,600 I guess there isn't a complete characterization of, say, 859 00:41:38,600 --> 00:41:40,750 two fold axioms, exactly what you can make. 860 00:41:40,750 --> 00:41:42,480 That's still open. 861 00:41:42,480 --> 00:41:45,312 But the point is, as you add more folds, you get more power. 862 00:41:45,312 --> 00:41:46,770 Eventually you get all polynomials, 863 00:41:46,770 --> 00:41:48,090 which is the most you could hope for, 864 00:41:48,090 --> 00:41:49,715 for any kind of geometric construction. 865 00:41:53,880 --> 00:41:55,660 If there are no more questions, we 866 00:41:55,660 --> 00:41:58,180 resume our task of building something out 867 00:41:58,180 --> 00:42:00,940 of hyperbolic paraboloids. 868 00:42:00,940 --> 00:42:03,170 Remember this is the hat construction. 869 00:42:03,170 --> 00:42:06,800 If you have a square in your polyhedron, 870 00:42:06,800 --> 00:42:08,420 you take four hyperbolic paraboloids 871 00:42:08,420 --> 00:42:10,840 and join them together like this picture. 872 00:42:10,840 --> 00:42:13,137 And then this is going to represent one edge. 873 00:42:13,137 --> 00:42:14,720 Those two edges of the hypar are going 874 00:42:14,720 --> 00:42:17,540 to represent one edge of the square. 875 00:42:17,540 --> 00:42:19,750 And I was suggesting we make this shape, which 876 00:42:19,750 --> 00:42:21,020 we have enough hypars for. 877 00:42:21,020 --> 00:42:23,220 We just need to do some taping. 878 00:42:23,220 --> 00:42:25,120 This is the truncated tetrahedron. 879 00:42:25,120 --> 00:42:28,290 It's got four triangles, four hexagons. 880 00:42:28,290 --> 00:42:29,700 So we've already made-- last time 881 00:42:29,700 --> 00:42:31,025 we made the four triangles. 882 00:42:34,690 --> 00:42:37,990 These are the three hats, and they can be joined together 883 00:42:37,990 --> 00:42:40,700 to form a tetrahedron by themselves. 884 00:42:40,700 --> 00:42:44,747 We've got enough hypars to make the four hexagons, 885 00:42:44,747 --> 00:42:46,580 and then we just need to tape them together, 886 00:42:46,580 --> 00:42:49,010 and we will get the truncated tetrahedron. 887 00:42:49,010 --> 00:42:50,320 Who would like to help? 888 00:42:50,320 --> 00:42:51,110 Come on up. 889 00:42:55,164 --> 00:42:56,080 AUDIENCE: [INAUDIBLE]. 890 00:42:59,559 --> 00:43:00,553 AUDIENCE: [INAUDIBLE]. 891 00:43:00,553 --> 00:43:01,136 AUDIENCE: Yep. 892 00:43:04,750 --> 00:43:06,977 PROFESSOR: It will be something like this. 893 00:43:06,977 --> 00:43:07,518 AUDIENCE: OK. 894 00:43:10,150 --> 00:43:12,906 PROFESSOR: I think we need a whole other six, but. 895 00:43:12,906 --> 00:43:14,878 AUDIENCE: [INAUDIBLE] Is there another roll 896 00:43:14,878 --> 00:43:16,357 of tape hidden somewhere? 897 00:43:33,605 --> 00:43:34,105 [INAUDIBLE] 898 00:43:42,753 --> 00:43:44,794 STUDENT: And the other main thing is [INAUDIBLE]. 899 00:43:44,794 --> 00:43:46,877 STUDENT: [INAUDIBLE] these two on a triangle here. 900 00:43:52,460 --> 00:43:53,060 It's like 36. 901 00:43:55,760 --> 00:43:57,310 Sort of.