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.