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  • Do you guys know about the Putnam? It's a math competition for undergraduate students.

  • It's 6 hours long and consists of 12 questions, broken up into two different 3-hour sessions.

  • With each question being scored on a 1-10 scale, the highest possible score is 120.

  • And yet, despite the fact that the only students taking it each year are those who are clearly

  • already pretty into math, given that they opt into such a test, the median score tends

  • to be around 1 or 2. So... it's a hard test. And on each section of 6 questions, the problems

  • tend to get harder as you go from 1 to 6, although of course difficulty is in the eye

  • of the beholder.

  • But the thing about the 5's and 6's 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 challenging

  • to simple. Here I'll share with you one problem which

  • came up as the 6th question on one of these tests a while back.

  • And those of you who follow the channel know that rather than just jumping straight to

  • the solution, which in this case will be surprisingly short, when possible I prefer to take the

  • time to walk through how you might stumble upon the solution yourself.

  • That is, make the video more about the problem-solving process than the particular problem used to

  • exemplify it.

  • So here's the question: If you choose 4 random points on a sphere, and consider the

  • tetrahedron which has these points as its vertices, what's the probability that the

  • center of the sphere is inside the tetrahedron? Take a moment to kind of digest the question.

  • You might start thinking about which of these tetrahedra contain the sphere's center,

  • which ones don't, and how you might systematically distinguish the two.

  • And...how do approach a problem like this, where do you even start?

  • Well, it's often a good idea to think about simpler cases, so let's bring things down

  • into 2 dimensions.

  • Suppose you choose three random points on a circle. It's always helpful to name things,

  • so let's call these guys P1, P2, and P3. What's the probability that the triangle

  • formed by these points contains the center of the circle?

  • It's certainly easier to visualize now, but it's still a hard question.

  • So again, you ask yourself if there's a way to simplify what's going on. We still

  • need a foothold, something to build up from. Maybe you imagine fixing P1 and P2 in place,

  • only letting P3 vary. In doing this, you might notice that there's

  • special region, a certain arc, where when P3 is in that arc, the triangle contains the

  • circle's center. Specifically, if you draw a lines from P1

  • and P2 through the center, these lines divide the circle into 4 different arcs. If P3 happens

  • to be in the one opposite P1 and P2, the triangle will contain the center. Otherwise, you're

  • out of luck.

  • We're assuming all points of the circle are equally likely, so what's the probability

  • that P3 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? This depends on the first two points.

  • If they are 90 degrees apart from each other, for example, the relevant arc is ¼ of the

  • circle. But if those two points are farther apart, the proportion might be closer to ½.

  • If they are really close, that proportion might be closer to 0.

  • Alright, think about this for a moment. If P1 and P2 are chosen randomly, with every

  • point on the circle being equally likely, what's the average size of the relevant

  • arc? Maybe you imagine fixing P1 in place, and

  • considering all the places that P2 might be. All of the possible angles between these two

  • lines, every angle from 0 degrees up to 180 degrees is equally likely, so every proportion

  • between 0 and 0.5 is equally likely, making the average proportion 0.25.

  • Since the average size of this arc is ¼ this full circle, the average probability that

  • the third point lands in it is ¼, meaning the overall probability of our triangle containing

  • the center is ¼. Try to extend to 3D

  • Great! Can we extend this to the 3d case? If you imagine 3 of your 4 points fixed in

  • place, which points of the sphere can that 4th point be on so that our tetrahedron contains

  • the sphere's center? As before, let's draw some lines from each

  • of our first 3 points through the center of the sphere. And it's also helpful if we

  • draw the planes determined by any pair of these lines.

  • These planes divide the sphere into 8 different sections, each of which is a sort of spherical

  • triangle. Our tetrahedron will only contain the center of the sphere if the fourth point

  • is in the section on the opposite side of our three points.

  • Now, unlike the 2d case, it's rather difficult to think about the average size of this section

  • as we let our initial 3 points vary. Those of you with some multivariable calculus

  • under your belt might think to 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, this

  • is the 6th problem on a Putnam!

  • But let's back up to the 2d case, and contemplate if there's a different way of thinking about

  • it. This answer we got, ¼, is suspiciously clean and raises the question of what that

  • 4 represents. One of the main reasons I wanted to make a

  • video on this problem is that what's about to happen carries a broader lesson for mathematical

  • problem-solving. These lines that we drew from P1 and P2 through

  • the origin made the problem easier to think about.

  • In general, whenever you've added something to your problem setup which makes things conceptually

  • easier, see if you can reframe the entire question in terms of the thing you just added.

  • In this case, rather than thinking about choosing 3 points randomly, start by saying choose

  • two random lines that pass through the circle's center.

  • For each line, there are two possible points they could correspond to, so flip a coin for

  • each to choose which of those will be P1 and P2.

  • Choosing a random line then flipping a coin like this is the same as choosing a random

  • point on the circle, with all points being equally likely, and at first it might seem

  • needlessly convoluted. But by making those lines the starting point of our random process

  • things actually become easier. We'll still think about P3 as just being

  • a random point on the circle, but imagine that it was chosen before you do the two coin

  • flips. Because you see, once the two lines and a

  • random point have been chosen, there are four possibilities for where P1 and P2 end up,

  • based on the coin flips, each one of which is equally likely. But one and only one of

  • those outcomes leaves P1 and P2 on the opposite side of the circle as P3, with the triangle

  • they form containing the center. So no matter what those two lines and P3 turned

  • out to be, it's always a ¼ chance that the coin flips will leave us with a triangle

  • containing the center. That's very subtle. Just by reframing how

  • we think of the random process for choosing these points, the answer ¼ popped in a different

  • way from before.

  • And importantly, this style of argument generalizes seamlessly to 3 dimensions.

  • Again, instead of starting off by picking 4 random points, imagine choosing 3 random

  • lines through the center, and then a random point for P4.

  • That first line passes through the sphere at 2 points, so flip a coin to decide which

  • of those two points is P1. Likewise, for each of the other lines flip a coin to decide where

  • P2 and P3 end up. There are 8 equally likely outcomes of these

  • coin flips, but one and only one of these outcomes will place P1, P2, and P3 on the

  • opposite side of the center from P4. So only one of these 8 equally likely outcomes

  • gives a tetrahedron containing the center. Isn't that elegant?

  • This is a valid solution, but admittedly the way I've stated it so far rests on some

  • visual intuition. I've left a link in the description to a

  • slightly more formal write-up of this same solution in the language of linear algebra

  • if you're curious. This is common in math, where having the key

  • insight and understanding is one thing, but having the relevant background to articulate

  • this understanding more formally is almost a separate muscle entirely, one which undergraduate

  • math students spend much of their time building up.

  • Lesson Now the main takeaway here is not the solution

  • itself, but how you might find the key insight if you were left to solve it. Namely, keep

  • asking simpler versions of the question until you can get some foothold, and if some added

  • construct proves to be useful, see if you can reframe the whole question around that

  • new construct.

Do you guys know about the Putnam? It's a math competition for undergraduate students.

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B1 US p1 p2 p3 circle sphere center

The hardest problem on the hardest test

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    Loukei posted on 2021/04/29
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