The hardest problem on the hardest test

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 1 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 is around 1 or 2.

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 1 to 6,

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 going to 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 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 approach a problem like this?

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 P1,

P2, and P3.