[00:03] Do you guys know about the Putnam?
[00:05] It's a math competition for undergraduate students.
[00:08] It's a six-hour long test that just has 12 questions
[00:11] broken up into two different three-hour sessions.
[00:14] And each one of those questions is scored 1 to 10,
[00:16] so the highest possible score would be 120.
[00:19] And yet, despite the fact that the only students taking this thing each year are those
[00:24] who clearly are already pretty interested in math, the median score is around 1 or 2.
[00:28] So it's a hard test.
[00:31] And on each one of those sections of six questions,
[00:33] the problems tend to get harder as you go from 1 to 6,
[00:36] although of course difficulty is in the eye of the beholder.
[00:40] But the thing about those fives and sixes is that even though they're
[00:43] positioned as the hardest problems on a famously hard test,
[00:46] quite often these are the ones with the most elegant solutions available,
[00:50] some subtle shift in perspective that transforms it from very challenging to doable.
[00:56] Here I'm going to share with you one problem that came up
[00:58] as the sixth question on one of these tests a while back.
[01:01] And those of you who follow the channel know that rather than just jumping
[01:04] straight to the solution, which in this case would be surprisingly short,
[01:07] when possible I like to take the time to walk you through how you might
[01:10] have stumbled across the solution yourself, where the insight comes from.
[01:14] That is, make a video more about the problem-solving
[01:17] process than about the problem used to exemplify it.
[01:20] So anyway, here's the question.
[01:21] If you choose four random points on a sphere, and consider the
[01:25] tetrahedron with these points as its vertices,
[01:28] what is the probability that the center of the sphere is inside that tetrahedron?
[01:33] Go ahead, take a moment and kind of digest this question.
[01:37] You might start thinking about which of these tetrahedra contain the sphere's center,
[01:42] which ones don't, how you might systematically distinguish the two,
[01:45] and how do you approach a problem like this?
[01:48] Where do you even start?
[01:51] Well, it's usually a good idea to think about simpler cases,
[01:54] so let's knock things down to two dimensions, where you'll choose three random
[01:58] points on a circle, and it's always helpful to name things so let's call these guys P1,
[02:03] P2, and P3.
[02:04] The question is, what's the probability that the triangle
[02:07] formed by these points contains the center of the circle?
[02:14] I think you'll agree it's way easier to visualize now, but it's still a hard question.
[02:18] So again, you ask, is there a way to simplify what's going on,
[02:22] get ourselves to some kind of foothold that we can build up from?
[02:25] Well, maybe you imagine fixing P1 and P2 in place, and only letting that third point vary.
[02:31] And when you do this, and play around with it in your mind,
[02:34] you might notice that there's a special region, a certain arc,
[02:37] where when P3 is in that arc, the triangle contains the center, otherwise not.
[02:42] Specifically, if you draw lines from P1 and P2 through the center,
[02:46] these lines divide up the circle into four different arcs,
[02:49] and if P3 happens to be in the one on the opposite side as P1 and P2,
[02:53] the triangle has the center.
[02:55] If it's in any of the other arcs though, no luck.
[03:01] We're assuming here that all of the points of the circle are equally likely.
[03:04] So what is the probability that P3 lands in that arc?
[03:08] It's the length of that arc divided by the full circumference of the circle,
[03:12] the proportion of the circle that this arc makes up.
[03:15] So what is that proportion?
[03:17] Obviously that depends on where you put the first two points.
[03:20] I mean, if they're 90 degrees apart from each other,
[03:22] then the relevant arc is one quarter of the circle.
[03:25] But if those two points were farther apart, that proportion would be something closer
[03:29] to a half, and if they were really close together, that proportion gets closer to zero.
[03:34] So think about this for a moment.
[03:35] P1 and P2 are chosen randomly, with every point on the circle being equally likely.
[03:40] So what is the average size of this relevant arc?
[03:46] Maybe you imagine fixing P1 in place, and just
[03:48] considering all the places that P2 might be.
[03:51] All of the possible angles between these two lines,
[03:54] every angle from 0 degrees up to 180 degrees, is equally likely.
[03:58] So every proportion between 0 and 0.5 is equally likely,
[04:02] and that means that the average proportion is 0.25.
[04:08] So, if the average size of this arc is a quarter of the full circle,
[04:12] the average probability that the third point lands in it is a quarter,
[04:16] and that means that the overall probability that our triangle contains the
[04:20] center is a quarter.
[04:26] But can we extend this into the three-dimensional case?
[04:29] If you imagine three out of those four points just being fixed in place,
[04:33] which points of the sphere can the fourth one be on so that the
[04:37] tetrahedron that they form contain the center of the sphere?
[04:41] Just like before, let's go ahead and draw some lines from each
[04:44] of those fixed three points through the center of the sphere.
[04:47] It's also helpful if we draw some planes that are determined by any pair of these lines.
[04:53] What these planes do, you might notice, is divide the sphere into
[04:56] eight different sections, each of which is a sort of spherical triangle.
[05:01] And our tetrahedron is only going to contain the center of the sphere if the
[05:05] fourth point is in the spherical triangle on the opposite side as the first three.
[05:11] Now, unlike the 2D case, it's pretty difficult to think about the
[05:14] average size of this section, as we let the initial three points vary.
[05:21] Those of you with some multivariable calculus under your belt might think,
[05:24] let's just try a surface integral.
[05:26] And by all means, pull out some paper and give it a try.
[05:28] But it's not easy.
[05:30] And of course it should be difficult.
[05:31] I mean, this is the sixth problem on a Putnam, what do you expect?
[05:35] And what do you even do with that?
[05:39] Well, one thing you can do is back up to the two-dimensional case and
[05:42] contemplate if there is a different way to think about the same answer we got.
[05:46] That answer, one-fourth, looks suspiciously clean,
[05:49] and raises the question of what that four represents.
[05:53] One of the main reasons I wanted to make a video about this particular problem is that
[05:57] what's about to happen carries with it a broader lesson for mathematical problem solving.
[06:01] Think about those two lines we drew for p1 and p2 through the origin.
[06:05] They made the problem a lot easier to think about.
[06:08] And in general, whenever you've added something to the problem
[06:11] setup that makes it conceptually easier, see if you can reframe
[06:14] the entire question in terms of those things you just added.
[06:18] In this case, rather than thinking about choosing three points randomly,
[06:23] start by saying, choose two random lines that pass through the circle's center.
[06:28] For each line, there's two possible points it could correspond to,
[06:31] so just flip a coin for each one to choose which of the endpoints is going to be p1,
[06:35] and likewise for the other, which endpoint is going to be p2.
[06:39] Choosing a random line and flipping a coin like this is the same thing as choosing
[06:43] a random point on the circle, it just feels a little bit convoluted at first.
[06:47] But the reason for thinking about the random process this
[06:50] way is that things are actually about to become easier.
[06:53] We'll still think about that third point, p3, as just being a random point on the circle,
[06:58] but imagine that it was chosen before you do the two coin flips.
[07:02] Because you see, once the two lines and the third point are set in stone,
[07:06] there's only four possibilities for where P1 and P2 might end up,
[07:10] based on those coin flips, each one being equally likely.
[07:13] But one and only one of those four outcomes leaves p1 and p2 on the opposite
[07:18] side of the circle as p3, with the triangle they form containing the center.
[07:23] So no matter where those two lines end up, and where that p3 ends up,
[07:27] it's always a 1 fourth chance that the coin flips leave us with a triangle containing
[07:32] the center.
[07:35] Now that's very subtle.
[07:37] Just by reframing how we think about the random process for choosing points,
[07:40] the answer 1 quarter popped out in a very different way from how it did before.
[07:45] And importantly, this style of argument generalizes seamlessly up into three dimensions.
[07:51] Again, instead of starting off by picking four random points,
[07:55] imagine choosing three random lines through the center of the sphere,
[07:59] and then some random point for p4.
[08:03] That first line passes through the sphere at two points,
[08:05] so flip a coin to decide which of those two points is going to be p1.
[08:09] Likewise, for each of the other lines, flip a coin to decide where p2 and p3 end up.
[08:15] There's eight equally likely outcomes of those coin flips,
[08:18] but one and only one of them is going to place p1, p2,
[08:22] and p3 on the opposite side of the center as p4.
[08:26] So one and only one of these eight equally likely
[08:29] outcomes gives us a tetrahedron that contains the center.
[08:35] Again, it's kind of subtle how that pops out to us, but isn't that elegant?
[08:40] This is a valid solution to the problem, but admittedly
[08:43] the way I've stated it so far rests on some visual intuition.
[08:47] If you're curious about how you might write it up in a way that doesn't
[08:50] rely on visual intuition, I've left a link in the description to one
[08:52] such write-up in the language of linear algebra, if you're curious.
[08:56] And this is pretty common in math, where having the key insight and
[08:59] understanding is one thing, but having the relevant background to articulate
[09:03] that understanding more formally is almost a separate muscle entirely,
[09:07] one that undergraduate math students kind of spend most of their time building up.
[09:12] But the main takeaway here is not the solution itself,
[09:14] but how you might find that key insight if it was put in front of you and you
[09:18] were just left to solve it.
[09:20] Namely, just keep asking simpler versions of the
[09:22] question until you can get some kind of foothold.
[09:25] And then when you do, if there's any kind of added construct that proves to be useful,
[09:29] see if you can reframe the whole question around that new construct.
[09:35] To close things off here, I've got another probability puzzle,
[09:38] one that comes from this video sponsor, brilliant.org.
[09:41] Suppose you have eight students sitting in a circle taking the Putnam.
[09:44] It's a hard test, so each student tries to cheat off of his neighbor,
[09:48] choosing randomly which neighbor to cheat from.
[09:51] Now circle all of the students that don't have somebody cheating off of their test.
[09:56] What is the expected number of such circled students?
[10:00] It's an interesting question, right?
[10:03] Brilliant.org is a site where you can practice your problem
[10:05] solving abilities with questions like this and many, many more.
[10:08] And that really is the best way to learn.
[10:10] You're going to find countless interesting questions curated in a pretty
[10:14] thoughtful way so that you really do come away better at problem solving.
[10:18] If you want more probability, they have a really good course on probability,
[10:21] but they've got all sorts of other math and science as well,
[10:23] so you're almost certainly going to find something that interests you.
[10:27] Me?
[10:27] I've been a fan for a while, and if you go to brilliant.org slash 3b1b,
[10:31] it lets them know that you came from here, and the first 256 of you to visit that
[10:36] link can get 20% off their premium membership, which is the one I use,
[10:40] if you want to upgrade.
[10:42] Also if you're just itching to see a solution to this puzzle,
[10:45] which by the way uses a certain tactic in probability that's useful in a lot of
[10:48] other circumstances, I also left a link in the description that just jumps you
[10:52] straight to the solution.