1 00:00:03,560 --> 00:00:05,320 Do you guys know about the Putnam? 2 00:00:05,559 --> 00:00:08,160 It's a math competition for undergraduate students. 3 00:00:08,720 --> 00:00:11,203 It's a six-hour long test that just has 12 questions 4 00:00:11,202 --> 00:00:13,500 broken up into two different three-hour sessions. 5 00:00:14,060 --> 00:00:16,696 And each one of those questions is scored 1 to 10, 6 00:00:16,696 --> 00:00:18,920 so the highest possible score would be 120. 7 00:00:19,660 --> 00:00:24,060 And yet, despite the fact that the only students taking this thing each year are those 8 00:00:24,059 --> 00:00:28,359 who clearly are already pretty interested in math, the median score is around 1 or 2. 9 00:00:28,960 --> 00:00:30,740 So it's a hard test. 10 00:00:31,399 --> 00:00:33,964 And on each one of those sections of six questions, 11 00:00:33,965 --> 00:00:36,679 the problems tend to get harder as you go from 1 to 6, 12 00:00:36,679 --> 00:00:39,640 although of course difficulty is in the eye of the beholder. 13 00:00:40,060 --> 00:00:43,783 But the thing about those fives and sixes is that even though they're 14 00:00:43,783 --> 00:00:46,975 positioned as the hardest problems on a famously hard test, 15 00:00:46,975 --> 00:00:50,911 quite often these are the ones with the most elegant solutions available, 16 00:00:50,911 --> 00:00:55,380 some subtle shift in perspective that transforms it from very challenging to doable. 17 00:00:56,079 --> 00:00:58,439 Here I'm going to share with you one problem that came up 18 00:00:58,439 --> 00:01:00,759 as the sixth question on one of these tests a while back. 19 00:01:01,299 --> 00:01:04,560 And those of you who follow the channel know that rather than just jumping 20 00:01:04,560 --> 00:01:07,775 straight to the solution, which in this case would be surprisingly short, 21 00:01:07,775 --> 00:01:10,905 when possible I like to take the time to walk you through how you might 22 00:01:10,906 --> 00:01:14,080 have stumbled across the solution yourself, where the insight comes from. 23 00:01:14,500 --> 00:01:17,174 That is, make a video more about the problem-solving 24 00:01:17,174 --> 00:01:19,799 process than about the problem used to exemplify it. 25 00:01:20,439 --> 00:01:21,459 So anyway, here's the question. 26 00:01:21,760 --> 00:01:25,347 If you choose four random points on a sphere, and consider the 27 00:01:25,347 --> 00:01:28,025 tetrahedron with these points as its vertices, 28 00:01:28,025 --> 00:01:32,640 what is the probability that the center of the sphere is inside that tetrahedron? 29 00:01:33,560 --> 00:01:35,640 Go ahead, take a moment and kind of digest this question. 30 00:01:37,480 --> 00:01:42,118 You might start thinking about which of these tetrahedra contain the sphere's center, 31 00:01:42,117 --> 00:01:45,786 which ones don't, how you might systematically distinguish the two, 32 00:01:45,786 --> 00:01:48,159 and how do you approach a problem like this? 33 00:01:48,540 --> 00:01:49,500 Where do you even start? 34 00:01:51,019 --> 00:01:54,322 Well, it's usually a good idea to think about simpler cases, 35 00:01:54,322 --> 00:01:58,599 so let's knock things down to two dimensions, where you'll choose three random 36 00:01:58,599 --> 00:02:03,364 points on a circle, and it's always helpful to name things so let's call these guys P1, 37 00:02:03,364 --> 00:02:03,960 P2, and P3. 38 00:02:04,459 --> 00:02:07,454 The question is, what's the probability that the triangle 39 00:02:07,454 --> 00:02:10,400 formed by these points contains the center of the circle? 40 00:02:14,460 --> 00:02:18,680 I think you'll agree it's way easier to visualize now, but it's still a hard question. 41 00:02:18,960 --> 00:02:22,021 So again, you ask, is there a way to simplify what's going on, 42 00:02:22,020 --> 00:02:25,179 get ourselves to some kind of foothold that we can build up from? 43 00:02:25,979 --> 00:02:30,799 Well, maybe you imagine fixing P1 and P2 in place, and only letting that third point vary. 44 00:02:31,419 --> 00:02:34,421 And when you do this, and play around with it in your mind, 45 00:02:34,421 --> 00:02:37,575 you might notice that there's a special region, a certain arc, 46 00:02:37,575 --> 00:02:41,479 where when P3 is in that arc, the triangle contains the center, otherwise not. 47 00:02:42,539 --> 00:02:46,337 Specifically, if you draw lines from P1 and P2 through the center, 48 00:02:46,337 --> 00:02:49,682 these lines divide up the circle into four different arcs, 49 00:02:49,682 --> 00:02:53,651 and if P3 happens to be in the one on the opposite side as P1 and P2, 50 00:02:53,651 --> 00:02:55,239 the triangle has the center. 51 00:02:55,659 --> 00:02:58,180 If it's in any of the other arcs though, no luck. 52 00:03:01,039 --> 00:03:04,179 We're assuming here that all of the points of the circle are equally likely. 53 00:03:04,560 --> 00:03:07,699 So what is the probability that P3 lands in that arc? 54 00:03:08,780 --> 00:03:12,706 It's the length of that arc divided by the full circumference of the circle, 55 00:03:12,706 --> 00:03:15,359 the proportion of the circle that this arc makes up. 56 00:03:15,840 --> 00:03:16,860 So what is that proportion? 57 00:03:17,419 --> 00:03:19,819 Obviously that depends on where you put the first two points. 58 00:03:20,319 --> 00:03:22,674 I mean, if they're 90 degrees apart from each other, 59 00:03:22,674 --> 00:03:24,939 then the relevant arc is one quarter of the circle. 60 00:03:25,219 --> 00:03:29,076 But if those two points were farther apart, that proportion would be something closer 61 00:03:29,076 --> 00:03:32,979 to a half, and if they were really close together, that proportion gets closer to zero. 62 00:03:34,039 --> 00:03:35,400 So think about this for a moment. 63 00:03:35,900 --> 00:03:39,900 P1 and P2 are chosen randomly, with every point on the circle being equally likely. 64 00:03:40,699 --> 00:03:43,639 So what is the average size of this relevant arc? 65 00:03:46,180 --> 00:03:48,566 Maybe you imagine fixing P1 in place, and just 66 00:03:48,566 --> 00:03:50,800 considering all the places that P2 might be. 67 00:03:51,560 --> 00:03:54,562 All of the possible angles between these two lines, 68 00:03:54,562 --> 00:03:58,259 every angle from 0 degrees up to 180 degrees, is equally likely. 69 00:03:58,819 --> 00:04:02,840 So every proportion between 0 and 0.5 is equally likely, 70 00:04:02,841 --> 00:04:06,439 and that means that the average proportion is 0.25. 71 00:04:08,139 --> 00:04:12,014 So, if the average size of this arc is a quarter of the full circle, 72 00:04:12,014 --> 00:04:16,002 the average probability that the third point lands in it is a quarter, 73 00:04:16,002 --> 00:04:20,216 and that means that the overall probability that our triangle contains the 74 00:04:20,216 --> 00:04:21,338 center is a quarter. 75 00:04:26,519 --> 00:04:29,139 But can we extend this into the three-dimensional case? 76 00:04:29,800 --> 00:04:33,653 If you imagine three out of those four points just being fixed in place, 77 00:04:33,653 --> 00:04:37,032 which points of the sphere can the fourth one be on so that the 78 00:04:37,031 --> 00:04:40,199 tetrahedron that they form contain the center of the sphere? 79 00:04:41,699 --> 00:04:44,382 Just like before, let's go ahead and draw some lines from each 80 00:04:44,382 --> 00:04:46,980 of those fixed three points through the center of the sphere. 81 00:04:47,500 --> 00:04:52,180 It's also helpful if we draw some planes that are determined by any pair of these lines. 82 00:04:53,300 --> 00:04:56,723 What these planes do, you might notice, is divide the sphere into 83 00:04:56,723 --> 00:05:00,459 eight different sections, each of which is a sort of spherical triangle. 84 00:05:01,240 --> 00:05:05,704 And our tetrahedron is only going to contain the center of the sphere if the 85 00:05:05,704 --> 00:05:10,459 fourth point is in the spherical triangle on the opposite side as the first three. 86 00:05:11,420 --> 00:05:14,846 Now, unlike the 2D case, it's pretty difficult to think about the 87 00:05:14,846 --> 00:05:18,480 average size of this section, as we let the initial three points vary. 88 00:05:21,220 --> 00:05:24,233 Those of you with some multivariable calculus under your belt might think, 89 00:05:24,233 --> 00:05:25,600 let's just try a surface integral. 90 00:05:26,079 --> 00:05:28,199 And by all means, pull out some paper and give it a try. 91 00:05:28,500 --> 00:05:29,620 But it's not easy. 92 00:05:30,060 --> 00:05:31,100 And of course it should be difficult. 93 00:05:31,300 --> 00:05:34,000 I mean, this is the sixth problem on a Putnam, what do you expect? 94 00:05:35,439 --> 00:05:38,519 And what do you even do with that? 95 00:05:39,060 --> 00:05:42,341 Well, one thing you can do is back up to the two-dimensional case and 96 00:05:42,341 --> 00:05:46,000 contemplate if there is a different way to think about the same answer we got. 97 00:05:46,819 --> 00:05:49,644 That answer, one-fourth, looks suspiciously clean, 98 00:05:49,644 --> 00:05:52,579 and raises the question of what that four represents. 99 00:05:53,720 --> 00:05:57,377 One of the main reasons I wanted to make a video about this particular problem is that 100 00:05:57,377 --> 00:06:01,120 what's about to happen carries with it a broader lesson for mathematical problem solving. 101 00:06:01,800 --> 00:06:05,360 Think about those two lines we drew for p1 and p2 through the origin. 102 00:06:05,920 --> 00:06:07,900 They made the problem a lot easier to think about. 103 00:06:08,279 --> 00:06:11,600 And in general, whenever you've added something to the problem 104 00:06:11,600 --> 00:06:14,975 setup that makes it conceptually easier, see if you can reframe 105 00:06:14,975 --> 00:06:18,139 the entire question in terms of those things you just added. 106 00:06:18,819 --> 00:06:23,026 In this case, rather than thinking about choosing three points randomly, 107 00:06:23,026 --> 00:06:27,579 start by saying, choose two random lines that pass through the circle's center. 108 00:06:28,459 --> 00:06:31,661 For each line, there's two possible points it could correspond to, 109 00:06:31,661 --> 00:06:35,723 so just flip a coin for each one to choose which of the endpoints is going to be p1, 110 00:06:35,723 --> 00:06:38,639 and likewise for the other, which endpoint is going to be p2. 111 00:06:39,339 --> 00:06:43,343 Choosing a random line and flipping a coin like this is the same thing as choosing 112 00:06:43,343 --> 00:06:47,060 a random point on the circle, it just feels a little bit convoluted at first. 113 00:06:47,560 --> 00:06:50,064 But the reason for thinking about the random process this 114 00:06:50,064 --> 00:06:52,439 way is that things are actually about to become easier. 115 00:06:53,459 --> 00:06:58,286 We'll still think about that third point, p3, as just being a random point on the circle, 116 00:06:58,286 --> 00:07:01,719 but imagine that it was chosen before you do the two coin flips. 117 00:07:02,560 --> 00:07:06,504 Because you see, once the two lines and the third point are set in stone, 118 00:07:06,504 --> 00:07:10,021 there's only four possibilities for where P1 and P2 might end up, 119 00:07:10,021 --> 00:07:13,060 based on those coin flips, each one being equally likely. 120 00:07:13,680 --> 00:07:18,581 But one and only one of those four outcomes leaves p1 and p2 on the opposite 121 00:07:18,581 --> 00:07:23,420 side of the circle as p3, with the triangle they form containing the center. 122 00:07:23,920 --> 00:07:27,617 So no matter where those two lines end up, and where that p3 ends up, 123 00:07:27,617 --> 00:07:32,158 it's always a 1 fourth chance that the coin flips leave us with a triangle containing 124 00:07:32,158 --> 00:07:32,740 the center. 125 00:07:35,300 --> 00:07:36,460 Now that's very subtle. 126 00:07:37,040 --> 00:07:40,761 Just by reframing how we think about the random process for choosing points, 127 00:07:40,761 --> 00:07:44,580 the answer 1 quarter popped out in a very different way from how it did before. 128 00:07:45,420 --> 00:07:50,560 And importantly, this style of argument generalizes seamlessly up into three dimensions. 129 00:07:51,639 --> 00:07:55,173 Again, instead of starting off by picking four random points, 130 00:07:55,173 --> 00:07:59,161 imagine choosing three random lines through the center of the sphere, 131 00:07:59,161 --> 00:08:01,099 and then some random point for p4. 132 00:08:03,019 --> 00:08:05,796 That first line passes through the sphere at two points, 133 00:08:05,797 --> 00:08:09,160 so flip a coin to decide which of those two points is going to be p1. 134 00:08:09,660 --> 00:08:14,660 Likewise, for each of the other lines, flip a coin to decide where p2 and p3 end up. 135 00:08:15,139 --> 00:08:18,942 There's eight equally likely outcomes of those coin flips, 136 00:08:18,942 --> 00:08:22,485 but one and only one of them is going to place p1, p2, 137 00:08:22,485 --> 00:08:25,579 and p3 on the opposite side of the center as p4. 138 00:08:26,459 --> 00:08:29,402 So one and only one of these eight equally likely 139 00:08:29,403 --> 00:08:32,759 outcomes gives us a tetrahedron that contains the center. 140 00:08:35,139 --> 00:08:38,340 Again, it's kind of subtle how that pops out to us, but isn't that elegant? 141 00:08:40,500 --> 00:08:43,504 This is a valid solution to the problem, but admittedly 142 00:08:43,504 --> 00:08:46,779 the way I've stated it so far rests on some visual intuition. 143 00:08:47,399 --> 00:08:50,251 If you're curious about how you might write it up in a way that doesn't 144 00:08:50,251 --> 00:08:52,984 rely on visual intuition, I've left a link in the description to one 145 00:08:52,985 --> 00:08:55,639 such write-up in the language of linear algebra, if you're curious. 146 00:08:56,299 --> 00:08:59,730 And this is pretty common in math, where having the key insight and 147 00:08:59,730 --> 00:09:03,617 understanding is one thing, but having the relevant background to articulate 148 00:09:03,618 --> 00:09:07,201 that understanding more formally is almost a separate muscle entirely, 149 00:09:07,201 --> 00:09:11,340 one that undergraduate math students kind of spend most of their time building up. 150 00:09:12,159 --> 00:09:14,860 But the main takeaway here is not the solution itself, 151 00:09:14,860 --> 00:09:18,692 but how you might find that key insight if it was put in front of you and you 152 00:09:18,692 --> 00:09:20,019 were just left to solve it. 153 00:09:20,299 --> 00:09:22,519 Namely, just keep asking simpler versions of the 154 00:09:22,519 --> 00:09:24,740 question until you can get some kind of foothold. 155 00:09:25,440 --> 00:09:29,986 And then when you do, if there's any kind of added construct that proves to be useful, 156 00:09:29,985 --> 00:09:33,539 see if you can reframe the whole question around that new construct. 157 00:09:35,600 --> 00:09:38,464 To close things off here, I've got another probability puzzle, 158 00:09:38,464 --> 00:09:40,920 one that comes from this video sponsor, brilliant.org. 159 00:09:41,460 --> 00:09:44,400 Suppose you have eight students sitting in a circle taking the Putnam. 160 00:09:44,860 --> 00:09:48,568 It's a hard test, so each student tries to cheat off of his neighbor, 161 00:09:48,568 --> 00:09:51,059 choosing randomly which neighbor to cheat from. 162 00:09:51,720 --> 00:09:56,080 Now circle all of the students that don't have somebody cheating off of their test. 163 00:09:56,639 --> 00:09:59,980 What is the expected number of such circled students? 164 00:10:00,980 --> 00:10:02,759 It's an interesting question, right? 165 00:10:03,480 --> 00:10:05,879 Brilliant.org is a site where you can practice your problem 166 00:10:05,879 --> 00:10:08,399 solving abilities with questions like this and many, many more. 167 00:10:08,799 --> 00:10:10,399 And that really is the best way to learn. 168 00:10:10,980 --> 00:10:14,269 You're going to find countless interesting questions curated in a pretty 169 00:10:14,269 --> 00:10:17,559 thoughtful way so that you really do come away better at problem solving. 170 00:10:18,000 --> 00:10:21,139 If you want more probability, they have a really good course on probability, 171 00:10:21,139 --> 00:10:23,625 but they've got all sorts of other math and science as well, 172 00:10:23,625 --> 00:10:26,480 so you're almost certainly going to find something that interests you. 173 00:10:27,019 --> 00:10:27,220 Me? 174 00:10:27,419 --> 00:10:31,599 I've been a fan for a while, and if you go to brilliant.org slash 3b1b, 175 00:10:31,600 --> 00:10:36,360 it lets them know that you came from here, and the first 256 of you to visit that 176 00:10:36,360 --> 00:10:40,484 link can get 20% off their premium membership, which is the one I use, 177 00:10:40,484 --> 00:10:41,820 if you want to upgrade. 178 00:10:42,799 --> 00:10:45,435 Also if you're just itching to see a solution to this puzzle, 179 00:10:45,436 --> 00:10:48,836 which by the way uses a certain tactic in probability that's useful in a lot of 180 00:10:48,836 --> 00:10:52,196 other circumstances, I also left a link in the description that just jumps you 181 00:10:52,196 --> 00:10:53,259 straight to the solution.