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Transcript
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Do you guys know about the Putnam?
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It's a math competition for undergraduate students.
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It's a six-hour long test that just has 12 questions
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broken up into two different three-hour sessions.
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And each one of those questions is scored 1 to 10,
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so the highest possible score would be 120.
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And yet, despite the fact that the only students taking this thing each year are those
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who clearly are already pretty interested in math, the median score is around 1 or 2.
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0:28
So it's a hard test.
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And on each one of those sections of six questions,
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the problems tend to get harder as you go from 1 to 6,
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although of course difficulty is in the eye of the beholder.
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But the thing about those fives and sixes is that even though they're
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positioned as the hardest problems on a famously hard test,
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quite often these are the ones with the most elegant solutions available,
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some subtle shift in perspective that transforms it from very challenging to doable.
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Here I'm going to share with you one problem that came up
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as the sixth question on one of these tests a while back.
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1:01
And those of you who follow the channel know that rather than just jumping
1:04
straight to the solution, which in this case would be surprisingly short,
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when possible I like to take the time to walk you through how you might
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have stumbled across the solution yourself, where the insight comes from.
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That is, make a video more about the problem-solving
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process than about the problem used to exemplify it.
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So anyway, here's the question.
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If you choose four random points on a sphere, and consider the
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tetrahedron with these points as its vertices,
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what is the probability that the center of the sphere is inside that tetrahedron?
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Go ahead, take a moment and kind of digest this question.
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You might start thinking about which of these tetrahedra contain the sphere's center,
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which ones don't, how you might systematically distinguish the two,
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and how do you approach a problem like this?
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Where do you even start?
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Well, it's usually a good idea to think about simpler cases,
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so let's knock things down to two dimensions, where you'll choose three random
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points on a circle, and it's always helpful to name things so let's call these guys P1,
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P2, and P3.
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The question is, what's the probability that the triangle
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formed by these points contains the center of the circle?
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I think you'll agree it's way easier to visualize now, but it's still a hard question.
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So again, you ask, is there a way to simplify what's going on,
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get ourselves to some kind of foothold that we can build up from?
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Well, maybe you imagine fixing P1 and P2 in place, and only letting that third point vary.
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And when you do this, and play around with it in your mind,
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you might notice that there's a special region, a certain arc,
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where when P3 is in that arc, the triangle contains the center, otherwise not.
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Specifically, if you draw lines from P1 and P2 through the center,
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these lines divide up the circle into four different arcs,
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and if P3 happens to be in the one on the opposite side as P1 and P2,
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the triangle has the center.
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If it's in any of the other arcs though, no luck.
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We're assuming here that all of the points of the circle are equally likely.
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So what is the probability that P3 lands in that arc?
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It's the length of that arc divided by the full circumference of the circle,
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the proportion of the circle that this arc makes up.
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So what is that proportion?
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Obviously that depends on where you put the first two points.
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I mean, if they're 90 degrees apart from each other,
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then the relevant arc is one quarter of the circle.
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But if those two points were farther apart, that proportion would be something closer
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to a half, and if they were really close together, that proportion gets closer to zero.
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So think about this for a moment.
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P1 and P2 are chosen randomly, with every point on the circle being equally likely.
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So what is the average size of this relevant arc?
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Maybe you imagine fixing P1 in place, and just
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considering all the places that P2 might be.
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All of the possible angles between these two lines,
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every angle from 0 degrees up to 180 degrees, is equally likely.
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So every proportion between 0 and 0.5 is equally likely,
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and that means that the average proportion is 0.25.
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So, if the average size of this arc is a quarter of the full circle,
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the average probability that the third point lands in it is a quarter,
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and that means that the overall probability that our triangle contains the
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center is a quarter.
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But can we extend this into the three-dimensional case?
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If you imagine three out of those four points just being fixed in place,
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which points of the sphere can the fourth one be on so that the
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tetrahedron that they form contain the center of the sphere?
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Just like before, let's go ahead and draw some lines from each
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of those fixed three points through the center of the sphere.
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It's also helpful if we draw some planes that are determined by any pair of these lines.
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What these planes do, you might notice, is divide the sphere into
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eight different sections, each of which is a sort of spherical triangle.
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And our tetrahedron is only going to contain the center of the sphere if the
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fourth point is in the spherical triangle on the opposite side as the first three.
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Now, unlike the 2D case, it's pretty difficult to think about the
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average size of this section, as we let the initial three points vary.
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Those of you with some multivariable calculus under your belt might think,
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let's just try a surface integral.
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And by all means, pull out some paper and give it a try.
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But it's not easy.
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And of course it should be difficult.
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I mean, this is the sixth problem on a Putnam, what do you expect?
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And what do you even do with that?
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Well, one thing you can do is back up to the two-dimensional case and
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contemplate if there is a different way to think about the same answer we got.
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That answer, one-fourth, looks suspiciously clean,
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and raises the question of what that four represents.
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One of the main reasons I wanted to make a video about this particular problem is that
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what's about to happen carries with it a broader lesson for mathematical problem solving.
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Think about those two lines we drew for p1 and p2 through the origin.
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They made the problem a lot easier to think about.
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And in general, whenever you've added something to the problem
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setup that makes it conceptually easier, see if you can reframe
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the entire question in terms of those things you just added.
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In this case, rather than thinking about choosing three points randomly,
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start by saying, choose two random lines that pass through the circle's center.
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For each line, there's two possible points it could correspond to,
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so just flip a coin for each one to choose which of the endpoints is going to be p1,
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and likewise for the other, which endpoint is going to be p2.
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Choosing a random line and flipping a coin like this is the same thing as choosing
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a random point on the circle, it just feels a little bit convoluted at first.
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But the reason for thinking about the random process this
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way is that things are actually about to become easier.
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We'll still think about that third point, p3, as just being a random point on the circle,
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but imagine that it was chosen before you do the two coin flips.
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Because you see, once the two lines and the third point are set in stone,
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there's only four possibilities for where P1 and P2 might end up,
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based on those coin flips, each one being equally likely.
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But one and only one of those four outcomes leaves p1 and p2 on the opposite
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side of the circle as p3, with the triangle they form containing the center.
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So no matter where those two lines end up, and where that p3 ends up,
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it's always a 1 fourth chance that the coin flips leave us with a triangle containing
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the center.
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Now that's very subtle.
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Just by reframing how we think about the random process for choosing points,
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the answer 1 quarter popped out in a very different way from how it did before.
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And importantly, this style of argument generalizes seamlessly up into three dimensions.
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Again, instead of starting off by picking four random points,
7:55
imagine choosing three random lines through the center of the sphere,
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and then some random point for p4.
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That first line passes through the sphere at two points,
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so flip a coin to decide which of those two points is going to be p1.
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Likewise, for each of the other lines, flip a coin to decide where p2 and p3 end up.
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There's eight equally likely outcomes of those coin flips,
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but one and only one of them is going to place p1, p2,
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and p3 on the opposite side of the center as p4.
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So one and only one of these eight equally likely
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outcomes gives us a tetrahedron that contains the center.
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Again, it's kind of subtle how that pops out to us, but isn't that elegant?
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This is a valid solution to the problem, but admittedly
8:43
the way I've stated it so far rests on some visual intuition.
8:47
If you're curious about how you might write it up in a way that doesn't
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rely on visual intuition, I've left a link in the description to one
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such write-up in the language of linear algebra, if you're curious.
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And this is pretty common in math, where having the key insight and
8:59
understanding is one thing, but having the relevant background to articulate
9:03
that understanding more formally is almost a separate muscle entirely,
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one that undergraduate math students kind of spend most of their time building up.
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But the main takeaway here is not the solution itself,
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but how you might find that key insight if it was put in front of you and you
9:18
were just left to solve it.
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Namely, just keep asking simpler versions of the
9:22
question until you can get some kind of foothold.
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And then when you do, if there's any kind of added construct that proves to be useful,
9:29
see if you can reframe the whole question around that new construct.
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To close things off here, I've got another probability puzzle,
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one that comes from this video sponsor, brilliant.org.
9:41
Suppose you have eight students sitting in a circle taking the Putnam.
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It's a hard test, so each student tries to cheat off of his neighbor,
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choosing randomly which neighbor to cheat from.
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Now circle all of the students that don't have somebody cheating off of their test.
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What is the expected number of such circled students?
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It's an interesting question, right?
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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.
— end of transcript —
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