WEBVTT 00:00:03.560 --> 00:00:05.320 Do you guys know about the Putnam? 00:00:05.559 --> 00:00:08.160 It's a math competition for undergraduate students. 00:00:08.720 --> 00:00:11.203 It's a six-hour long test that just has 12 questions 00:00:11.202 --> 00:00:13.500 broken up into two different three-hour sessions. 00:00:14.060 --> 00:00:16.696 And each one of those questions is scored 1 to 10, 00:00:16.696 --> 00:00:18.920 so the highest possible score would be 120. 00:00:19.660 --> 00:00:24.060 And yet, despite the fact that the only students taking this thing each year are those 00:00:24.059 --> 00:00:28.359 who clearly are already pretty interested in math, the median score is around 1 or 2. 00:00:28.960 --> 00:00:30.740 So it's a hard test. 00:00:31.399 --> 00:00:33.964 And on each one of those sections of six questions, 00:00:33.965 --> 00:00:36.679 the problems tend to get harder as you go from 1 to 6, 00:00:36.679 --> 00:00:39.640 although of course difficulty is in the eye of the beholder. 00:00:40.060 --> 00:00:43.783 But the thing about those fives and sixes is that even though they're 00:00:43.783 --> 00:00:46.975 positioned as the hardest problems on a famously hard test, 00:00:46.975 --> 00:00:50.911 quite often these are the ones with the most elegant solutions available, 00:00:50.911 --> 00:00:55.380 some subtle shift in perspective that transforms it from very challenging to doable. 00:00:56.079 --> 00:00:58.439 Here I'm going to share with you one problem that came up 00:00:58.439 --> 00:01:00.759 as the sixth question on one of these tests a while back. 00:01:01.299 --> 00:01:04.560 And those of you who follow the channel know that rather than just jumping 00:01:04.560 --> 00:01:07.775 straight to the solution, which in this case would be surprisingly short, 00:01:07.775 --> 00:01:10.905 when possible I like to take the time to walk you through how you might 00:01:10.906 --> 00:01:14.080 have stumbled across the solution yourself, where the insight comes from. 00:01:14.500 --> 00:01:17.174 That is, make a video more about the problem-solving 00:01:17.174 --> 00:01:19.799 process than about the problem used to exemplify it. 00:01:20.439 --> 00:01:21.459 So anyway, here's the question. 00:01:21.760 --> 00:01:25.347 If you choose four random points on a sphere, and consider the 00:01:25.347 --> 00:01:28.025 tetrahedron with these points as its vertices, 00:01:28.025 --> 00:01:32.640 what is the probability that the center of the sphere is inside that tetrahedron? 00:01:33.560 --> 00:01:35.640 Go ahead, take a moment and kind of digest this question. 00:01:37.480 --> 00:01:42.118 You might start thinking about which of these tetrahedra contain the sphere's center, 00:01:42.117 --> 00:01:45.786 which ones don't, how you might systematically distinguish the two, 00:01:45.786 --> 00:01:48.159 and how do you approach a problem like this? 00:01:48.540 --> 00:01:49.500 Where do you even start? 00:01:51.019 --> 00:01:54.322 Well, it's usually a good idea to think about simpler cases, 00:01:54.322 --> 00:01:58.599 so let's knock things down to two dimensions, where you'll choose three random 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, 00:02:03.364 --> 00:02:03.960 P2, and P3. 00:02:04.459 --> 00:02:07.454 The question is, what's the probability that the triangle 00:02:07.454 --> 00:02:10.400 formed by these points contains the center of the circle? 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. 00:02:18.960 --> 00:02:22.021 So again, you ask, is there a way to simplify what's going on, 00:02:22.020 --> 00:02:25.179 get ourselves to some kind of foothold that we can build up from? 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. 00:02:31.419 --> 00:02:34.421 And when you do this, and play around with it in your mind, 00:02:34.421 --> 00:02:37.575 you might notice that there's a special region, a certain arc, 00:02:37.575 --> 00:02:41.479 where when P3 is in that arc, the triangle contains the center, otherwise not. 00:02:42.539 --> 00:02:46.337 Specifically, if you draw lines from P1 and P2 through the center, 00:02:46.337 --> 00:02:49.682 these lines divide up the circle into four different arcs, 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, 00:02:53.651 --> 00:02:55.239 the triangle has the center. 00:02:55.659 --> 00:02:58.180 If it's in any of the other arcs though, no luck. 00:03:01.039 --> 00:03:04.179 We're assuming here that all of the points of the circle are equally likely. 00:03:04.560 --> 00:03:07.699 So what is the probability that P3 lands in that arc? 00:03:08.780 --> 00:03:12.706 It's the length of that arc divided by the full circumference of the circle, 00:03:12.706 --> 00:03:15.359 the proportion of the circle that this arc makes up. 00:03:15.840 --> 00:03:16.860 So what is that proportion? 00:03:17.419 --> 00:03:19.819 Obviously that depends on where you put the first two points. 00:03:20.319 --> 00:03:22.674 I mean, if they're 90 degrees apart from each other, 00:03:22.674 --> 00:03:24.939 then the relevant arc is one quarter of the circle. 00:03:25.219 --> 00:03:29.076 But if those two points were farther apart, that proportion would be something closer 00:03:29.076 --> 00:03:32.979 to a half, and if they were really close together, that proportion gets closer to zero. 00:03:34.039 --> 00:03:35.400 So think about this for a moment. 00:03:35.900 --> 00:03:39.900 P1 and P2 are chosen randomly, with every point on the circle being equally likely. 00:03:40.699 --> 00:03:43.639 So what is the average size of this relevant arc? 00:03:46.180 --> 00:03:48.566 Maybe you imagine fixing P1 in place, and just 00:03:48.566 --> 00:03:50.800 considering all the places that P2 might be. 00:03:51.560 --> 00:03:54.562 All of the possible angles between these two lines, 00:03:54.562 --> 00:03:58.259 every angle from 0 degrees up to 180 degrees, is equally likely. 00:03:58.819 --> 00:04:02.840 So every proportion between 0 and 0.5 is equally likely, 00:04:02.841 --> 00:04:06.439 and that means that the average proportion is 0.25. 00:04:08.139 --> 00:04:12.014 So, if the average size of this arc is a quarter of the full circle, 00:04:12.014 --> 00:04:16.002 the average probability that the third point lands in it is a quarter, 00:04:16.002 --> 00:04:20.216 and that means that the overall probability that our triangle contains the 00:04:20.216 --> 00:04:21.338 center is a quarter. 00:04:26.519 --> 00:04:29.139 But can we extend this into the three-dimensional case? 00:04:29.800 --> 00:04:33.653 If you imagine three out of those four points just being fixed in place, 00:04:33.653 --> 00:04:37.032 which points of the sphere can the fourth one be on so that the 00:04:37.031 --> 00:04:40.199 tetrahedron that they form contain the center of the sphere? 00:04:41.699 --> 00:04:44.382 Just like before, let's go ahead and draw some lines from each 00:04:44.382 --> 00:04:46.980 of those fixed three points through the center of the sphere. 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. 00:04:53.300 --> 00:04:56.723 What these planes do, you might notice, is divide the sphere into 00:04:56.723 --> 00:05:00.459 eight different sections, each of which is a sort of spherical triangle. 00:05:01.240 --> 00:05:05.704 And our tetrahedron is only going to contain the center of the sphere if the 00:05:05.704 --> 00:05:10.459 fourth point is in the spherical triangle on the opposite side as the first three. 00:05:11.420 --> 00:05:14.846 Now, unlike the 2D case, it's pretty difficult to think about the 00:05:14.846 --> 00:05:18.480 average size of this section, as we let the initial three points vary. 00:05:21.220 --> 00:05:24.233 Those of you with some multivariable calculus under your belt might think, 00:05:24.233 --> 00:05:25.600 let's just try a surface integral. 00:05:26.079 --> 00:05:28.199 And by all means, pull out some paper and give it a try. 00:05:28.500 --> 00:05:29.620 But it's not easy. 00:05:30.060 --> 00:05:31.100 And of course it should be difficult. 00:05:31.300 --> 00:05:34.000 I mean, this is the sixth problem on a Putnam, what do you expect? 00:05:35.439 --> 00:05:38.519 And what do you even do with that? 00:05:39.060 --> 00:05:42.341 Well, one thing you can do is back up to the two-dimensional case and 00:05:42.341 --> 00:05:46.000 contemplate if there is a different way to think about the same answer we got. 00:05:46.819 --> 00:05:49.644 That answer, one-fourth, looks suspiciously clean, 00:05:49.644 --> 00:05:52.579 and raises the question of what that four represents. 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 00:05:57.377 --> 00:06:01.120 what's about to happen carries with it a broader lesson for mathematical problem solving. 00:06:01.800 --> 00:06:05.360 Think about those two lines we drew for p1 and p2 through the origin. 00:06:05.920 --> 00:06:07.900 They made the problem a lot easier to think about. 00:06:08.279 --> 00:06:11.600 And in general, whenever you've added something to the problem 00:06:11.600 --> 00:06:14.975 setup that makes it conceptually easier, see if you can reframe 00:06:14.975 --> 00:06:18.139 the entire question in terms of those things you just added. 00:06:18.819 --> 00:06:23.026 In this case, rather than thinking about choosing three points randomly, 00:06:23.026 --> 00:06:27.579 start by saying, choose two random lines that pass through the circle's center. 00:06:28.459 --> 00:06:31.661 For each line, there's two possible points it could correspond to, 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, 00:06:35.723 --> 00:06:38.639 and likewise for the other, which endpoint is going to be p2. 00:06:39.339 --> 00:06:43.343 Choosing a random line and flipping a coin like this is the same thing as choosing 00:06:43.343 --> 00:06:47.060 a random point on the circle, it just feels a little bit convoluted at first. 00:06:47.560 --> 00:06:50.064 But the reason for thinking about the random process this 00:06:50.064 --> 00:06:52.439 way is that things are actually about to become easier. 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, 00:06:58.286 --> 00:07:01.719 but imagine that it was chosen before you do the two coin flips. 00:07:02.560 --> 00:07:06.504 Because you see, once the two lines and the third point are set in stone, 00:07:06.504 --> 00:07:10.021 there's only four possibilities for where P1 and P2 might end up, 00:07:10.021 --> 00:07:13.060 based on those coin flips, each one being equally likely. 00:07:13.680 --> 00:07:18.581 But one and only one of those four outcomes leaves p1 and p2 on the opposite 00:07:18.581 --> 00:07:23.420 side of the circle as p3, with the triangle they form containing the center. 00:07:23.920 --> 00:07:27.617 So no matter where those two lines end up, and where that p3 ends up, 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 00:07:32.158 --> 00:07:32.740 the center. 00:07:35.300 --> 00:07:36.460 Now that's very subtle. 00:07:37.040 --> 00:07:40.761 Just by reframing how we think about the random process for choosing points, 00:07:40.761 --> 00:07:44.580 the answer 1 quarter popped out in a very different way from how it did before. 00:07:45.420 --> 00:07:50.560 And importantly, this style of argument generalizes seamlessly up into three dimensions. 00:07:51.639 --> 00:07:55.173 Again, instead of starting off by picking four random points, 00:07:55.173 --> 00:07:59.161 imagine choosing three random lines through the center of the sphere, 00:07:59.161 --> 00:08:01.099 and then some random point for p4. 00:08:03.019 --> 00:08:05.796 That first line passes through the sphere at two points, 00:08:05.797 --> 00:08:09.160 so flip a coin to decide which of those two points is going to be p1. 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. 00:08:15.139 --> 00:08:18.942 There's eight equally likely outcomes of those coin flips, 00:08:18.942 --> 00:08:22.485 but one and only one of them is going to place p1, p2, 00:08:22.485 --> 00:08:25.579 and p3 on the opposite side of the center as p4. 00:08:26.459 --> 00:08:29.402 So one and only one of these eight equally likely 00:08:29.403 --> 00:08:32.759 outcomes gives us a tetrahedron that contains the center. 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? 00:08:40.500 --> 00:08:43.504 This is a valid solution to the problem, but admittedly 00:08:43.504 --> 00:08:46.779 the way I've stated it so far rests on some visual intuition. 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 00:08:50.251 --> 00:08:52.984 rely on visual intuition, I've left a link in the description to one 00:08:52.985 --> 00:08:55.639 such write-up in the language of linear algebra, if you're curious. 00:08:56.299 --> 00:08:59.730 And this is pretty common in math, where having the key insight and 00:08:59.730 --> 00:09:03.617 understanding is one thing, but having the relevant background to articulate 00:09:03.618 --> 00:09:07.201 that understanding more formally is almost a separate muscle entirely, 00:09:07.201 --> 00:09:11.340 one that undergraduate math students kind of spend most of their time building up. 00:09:12.159 --> 00:09:14.860 But the main takeaway here is not the solution itself, 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 00:09:18.692 --> 00:09:20.019 were just left to solve it. 00:09:20.299 --> 00:09:22.519 Namely, just keep asking simpler versions of the 00:09:22.519 --> 00:09:24.740 question until you can get some kind of foothold. 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, 00:09:29.985 --> 00:09:33.539 see if you can reframe the whole question around that new construct. 00:09:35.600 --> 00:09:38.464 To close things off here, I've got another probability puzzle, 00:09:38.464 --> 00:09:40.920 one that comes from this video sponsor, brilliant.org. 00:09:41.460 --> 00:09:44.400 Suppose you have eight students sitting in a circle taking the Putnam. 00:09:44.860 --> 00:09:48.568 It's a hard test, so each student tries to cheat off of his neighbor, 00:09:48.568 --> 00:09:51.059 choosing randomly which neighbor to cheat from. 00:09:51.720 --> 00:09:56.080 Now circle all of the students that don't have somebody cheating off of their test. 00:09:56.639 --> 00:09:59.980 What is the expected number of such circled students? 00:10:00.980 --> 00:10:02.759 It's an interesting question, right? 00:10:03.480 --> 00:10:05.879 Brilliant.org is a site where you can practice your problem 00:10:05.879 --> 00:10:08.399 solving abilities with questions like this and many, many more. 00:10:08.799 --> 00:10:10.399 And that really is the best way to learn. 00:10:10.980 --> 00:10:14.269 You're going to find countless interesting questions curated in a pretty 00:10:14.269 --> 00:10:17.559 thoughtful way so that you really do come away better at problem solving. 00:10:18.000 --> 00:10:21.139 If you want more probability, they have a really good course on probability, 00:10:21.139 --> 00:10:23.625 but they've got all sorts of other math and science as well, 00:10:23.625 --> 00:10:26.480 so you're almost certainly going to find something that interests you. 00:10:27.019 --> 00:10:27.220 Me? 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, 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 00:10:36.360 --> 00:10:40.484 link can get 20% off their premium membership, which is the one I use, 00:10:40.484 --> 00:10:41.820 if you want to upgrade. 00:10:42.799 --> 00:10:45.435 Also if you're just itching to see a solution to this puzzle, 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 00:10:48.836 --> 00:10:52.196 other circumstances, I also left a link in the description that just jumps you 00:10:52.196 --> 00:10:53.259 straight to the solution.