Video: The Hardest Problem on the Hardest Test

Grant Sanderson • 3Blue1Brown • Boclips

The Hardest Problem on the Hardest Test

09:36

Video Transcript

Do you guys know about the Putnam? It’s a math competition for undergraduate students. It’s a six-hour-long test that just has 12 questions broken up into two different three-hour sessions. And each one of those questions is scored one to 10. So the highest possible score would be 120.

And yet, despite the fact that the only students taking this thing each year are those who clearly are already pretty interested in math, the median score tends to be around one or two. So it’s a hard test. And on each one of those sections of six questions, the problems tend to get harder as you go from one to six. Although, of course, difficulty is in the eye of the beholder.

But the thing about those fives and sixes is that even though they’re positioned as the hardest problems on a famously hard test, quite often, these are the ones with the most elegant solutions available. Some subtle shift in perspective that transforms it from very challenging to doable.

Here, I’m gonna share with you one problem that came up as the sixth question on one of these tests a while back. And those of you who follow the channel, you know that, rather than just jumping straight to the solution. Which, in this case, would be surprisingly short. When possible, I like to take the time to walk you through how you might have stumbled across the solution yourself, where the insight comes from. That is, make a video more about the problem-solving process than about the problem used to exemplify it.

So anyway, here’s the question. If you choose four random points on a sphere and consider the tetrahedron with these points as its vertices, what is the probability that the center of the sphere is inside that tetrahedron? Go ahead. Take a moment and kind of digest this question. You might start thinking about which of these tetrahedra contain the sphere’s center, which ones don’t, how you might systematically distinguish the two. And how do you even approach a problem like this, right? Where do you even start?

Well, it’s usually a good idea to think about simpler cases. So let’s knock things down to two dimensions, where you’ll choose three random points on a circle. And it’s always helpful to name things. So let’s call these guys 𝑃 one, 𝑃 two, and 𝑃 three.

The question is, what’s the probability that the triangle formed by these points contains the center of the circle? I think you’ll agree it’s way easier to visualize now. But it’s still a hard question. So again, you ask, is there a way to simplify what’s going on, get ourselves to some kind of foothold that we can build up from?

Well, maybe you imagine fixing 𝑃 one and 𝑃 two in place and only letting that third point vary. And when you do this and you play around with it in your mind, you might notice that there’s a special region, a certain arc. Where when 𝑃 three is in that arc, the triangle contains the center, otherwise not. Specifically, if you draw lines from 𝑃 one and 𝑃 two through the center, these lines divide up the circle into four different arcs. And if 𝑃 three happens to be in the one on the opposite side as 𝑃 one and 𝑃 two, the triangle has the center. If it’s in any of the other arcs though, no luck.

We’re assuming here that all of the points of the circle are equally likely. So what is the probability that 𝑃 three lands in that arc? It’s the length of that arc divided by the full circumference of the circle, the proportion of the circle that this arc makes up. So what is that proportion?

Obviously, that depends on where you put the first two points. I mean, if they’re 90 degrees apart from each other, then the relevant arc is one-quarter of the circle. But if those two points were farther apart, that proportion would be something closer to a half. And if they were really close together, that proportion gets closer to zero.

So think about this for a moment. 𝑃 one and 𝑃 two are chosen randomly, with every point on the circle being equally likely. So what is the average size of this relevant arc? Maybe you imagine fixing 𝑃 one in place and just considering all the places that 𝑃 two might be. All of the possible angles between these two lines, every angle from zero degrees up to 180 degrees is equally likely. So every proportion between zero and 0.5 is equally likely. And that means that the average proportion is 0.25.

So, if the average size of this arc is a quarter of the full circle, the average probability that the third point lands in it is a quarter. And that means that the overall probability that our triangle contains the center is a quarter. But can we extend this into the three-dimensional case?

If you imagine three out of those four points just being fixed in place, which points of the sphere can the fourth one be on so that the tetrahedron that they form contain the center of the sphere? Just like before, let’s go ahead and draw some lines from each of those fixed three points through the center of the sphere. And here, it’s also helpful if we draw some planes that are determined by any pair of these lines.

Now what these planes do, you might notice, is divide the sphere into eight different sections, each of which is a sort of spherical triangle. And our tetrahedron is only gonna contain the center of the sphere if the fourth point is in the spherical triangle on the opposite side as the first three.

Now unlike the 2D case, it’s pretty difficult to think about the average size of this section as we let the initial three points vary. Those of you with some multivariable calculus under your belt might think, “Let’s just try a surface integral.” And by all means, pull out some paper and give it a try. But it’s not easy. And of course, it should be difficult. I mean, this is the sixth problem on a Putnam. What do you expect? And what do you even do with that?

Well, one thing you can do is back up to the two-dimensional case and contemplate if there is a different way to think about the same answer that we got. That answer, one-fourth, looks suspiciously clean. And it raises the question of what that four represents.

One of the main reasons I wanted to make a video about this particular problem is that what’s about to happen carries with it a broader lesson for mathematical problem-solving. Think about those two lines that we drew for 𝑃 one and 𝑃 two through the origin. They made the problem a lot easier to think about. And in general, whenever you’ve added something to the problem setup that makes it conceptually easier. See if you can reframe the entire question in terms of those things that you just added.

In this case, rather than thinking about choosing three points randomly, start by saying, “Choose two random lines that pass through the circle center.” For each line, there’s two possible points that it could correspond to. So just flip a coin for each one to choose which of the endpoints is gonna be 𝑃 one. And likewise, for the other, which endpoint is gonna be 𝑃 two.

Choosing a random line and flipping a coin like this is the same thing as choosing a random point on the circle. It just feels a little bit convoluted at first. But the reason for thinking about the random process this way is that things are actually about to become easier. We’ll still think about that third point, 𝑃 three, as just being a random point on the circle. But imagine that it was chosen before you do the two coin flips. Cause you see, once the two lines and that third point are set in stone, there’s only four possibilities for where 𝑃 one and 𝑃 two might end up based on those coin flips. Each one being equally likely. But one and only one of those four outcomes leaves 𝑃 one and 𝑃 two on the opposite side of the circle as 𝑃 three, with the triangle that they form containing the center. So no matter where those two lines end up and where that 𝑃 three ends up, it’s always a one-fourth chance that the coin flips leave us with the triangle containing the center.

Now that’s very subtle. Just by reframing how we think about the random process for choosing points, the answer one-quarter popped out in a very different way from how it did before. And importantly, this style of argument generalizes seamlessly up into three dimensions. Again, instead of starting off by picking four random points, imagine choosing three random lines through the center of the sphere and then some random point for 𝑃 four. That first line passes through the sphere at two points. So flip a coin to decide which of those two points is gonna be 𝑃 one.

Likewise, for each of the other lines, flip a coin to decide where 𝑃 two and 𝑃 three end up. Now, there’s eight equally likely outcomes of those coin flips. But one and only one of them is gonna place 𝑃 one, 𝑃 two, and 𝑃 three on the opposite side of the center as 𝑃 four. So one and only one of these eight equally likely outcomes gives us a tetrahedron that contains the center.

Again, it’s kind of subtle how that pops out to us. But isn’t that elegant? This is a valid solution to the problem. But admittedly, the way that I’ve stated it so far rests on some visual intuition. If you’re curious about how you might write it up in a way that doesn’t rely on visual intuition, I’ve left a link in the description to one such write-up in the language of linear algebra, if you’re curious. And this is pretty common in math, where having the key insight and understanding is one thing. But having the relevant background to articulate that understanding more formally is almost a separate muscle entirely. One that undergraduate math students kind of spend most of their time building up.

But the main takeaway here is not the solution itself, but how you might find that key insight if it was put in front of you and you were just left to solve it. Namely, just keep asking simpler versions of the question until you can get some kind of foothold. And then when you do, if there’s any kind of added construct that proves to be useful, see if you can reframe the whole question around that new construct.

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