# The hardest problem on the hardest test

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.