Do you guys know about the Putnam?
It’s a math competition for undergraduate students.
It’s a 6-hour long test that just has twelve questions
broken up into two different 3-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 tends to be 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 things about those 5s and 6s is that
even though they are 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.
Those of you who follow the channel,
you 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 even approach a problem like this, right?
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,
所以用P_1 P_2 P_3表示这三个点
so let’s call these guys P1, P2, and P3.
The question is
what’s the probability that the triangle formed
by these points contains the center of the circle?
I think you’ll agree it’s way easier to visualize now,
but it’s still a hard question.
So again, you ask is there a
way to simplify what’s going on,
get ourselves to some kind of foothold that we can build up from. Well,
maybe you imagine fixing P1 and P2 in place,
and only letting that third point vary.
And when you do this,
and you play around with it in your mind,
you might notice that there’s a special region, a certain arc,
where when P3 is in that arc,
the triangle contains the center, otherwise not. Specifically,
if you draw lines
from P1 and P2 through the center,
these lines divide up the circle into four different arcs.
And if P3 happens to be in the one
on the opposite side as P1 and P2,
the triangle has the center.
If it’s in any of the other arcs though, no luck.
We are assuming here that all of the points
of the circle are equally likely.
So what is 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?
Obviously that depends on where you put the first two points.
I mean, if they’re 90° apart from each other,
then the relevant arc is 1/4 of the circle.
But if those two points were farther apart,
that proportion would be something closer to 1/2.
And if they were really close together, that proportion gets closer to 0.
So think about this for a moment.
P1 and P2 are chosen randomly with every point
on the circle being equally likely.
So what is the average size of this relevant arc?
Maybe you imagine fixing P1 in place,
and just considering all the places that P2 might be.
All of the possible angles between these two lines,
every angle from 0° up to 180° is equally likely.
So every proportion between 0 and 0.5 is equally likely.
And that means that the average proportion is 0.25. So,
if the average size of this arc is 1/4
of the full circle,
the average probability that the third point lands in it is 1/4.
And that means that the overall probability
that our triangle contains the center is 1/4.
But can we extend this into the three-dimensional case?
If you imagine three
out of those fourth points just being fixed in place,
which points of the sphere can the fourth one be on
so that the tetrahedron that they form contain the center of the sphere?
Just like before, let’s go ahead and
draw some lines from each of those fixed 3
points through the center of the sphere.
And here, it’s also helpful
if we draw some planes that are determined by any pair of these lines.
Now what these planes do, you might notice,
is divide the sphere into eight different sections,
each of which is a sort of spherical triangle.
And our tetrahedron is only going
to contain the center of the sphere
if the fourth point is in the spherical triangle
on the opposite side as the first three.
Now unlike the 2D case,
it’s pretty difficult to think about the average size
of the section as we let the initial three points vary.
Those of you with some multivariable calculus
under your belt might think:
let’s just 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.
I mean, this is the sixth problem on a Putnam,
what do you expect? And…
what do you even do with that? Well,
one thing you can
do is back up to the two-dimensional case,
and contemplate if there is a different way to think
about the same answer that we got.
That answer”1/4″ look suspiciously clean,
and it raises the question of what that”4″ represents.
One of the main reasons I wanted to make a video
about this particular problem is that
what’s about to happen carries with it a broader lesson
for mathematical problem solving.
Think about those two lines that we drew
for P1 and P2 through the origin.
They made the problem a lot easier to think about.
And in general,
whenever you’ve added something to the problem set up
that makes it conceptually easier,
see if you can reframed the entire question
in terms of those things that you just added.
In this case, rather than thinking about choosing three points randomly,
start by saying:
choose two random lines that pass through the circle center;
for each line,
there are two possible points that could correspond to,
so just flip a coin for each one to choose which
of the endpoints is going to be P1;
and likewise, for the other which endpoint is going to be P2.
Choosing a random line and flipping a coin like this
is the same thing as choosing a random point on the circle.
It just feels a little bit convoluted at first.
But the reason
for thinking about the random process this way is that
things are actually about to become easier.
We’ll still think about that third point 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 that third point are set in stone,
there are only four possibilities
for where P1 and P2 might end up based on those coin flips,
each one being equally likely.
But one and only one of those four outcomes
leaves P1 and P2 on the opposite side
of the circle as P3
with the triangle that they form containing the center.
So no matter where those two lines end
up and where that P3 ends up,
it’s always a 1/4 chance
that the coin flips leave us with a triangle containing the center.
Now that’s very subtle.
Just by reframing how we think
about the random process for choosing points,
the answer 1/4 popped out in a very different way
from how it did before.
And importantly, this style of argument generalizes seamlessly up into three dimensions. Again,
instead of starting off by picking four random points,
imagine choosing three random lines through the center of the sphere,
and then some random point for P4.
That first line passes through the sphere at two points,
so flip a coin to decide which
of those two points is going to be P1. Likewise,
for each of the other lines,
flip a coin to decide where P2 and P3 end up.
Now there’s eight equally likely outcomes of those coin flips,
but one and only one of them is going
to place P1, P2 and P3 on the
opposite side of the center as P4.
So one and only one
一样的结果中 有且仅有一个 能让我们得到包含球心的四面体
of these eight equally likely outcomes gives us a tetrahedron the contains the center. Again,
it’s kind of subtle how that pops up to us; but,
isn’t that elegant?
This is a valid solution to the problem,
but admittedly, the way
that I’ve stated it so far rests on some visual intuition.
If you’re curious about how you might write it up
in a way that doesn’t rely on visual intuition,
I’ve left a link in the description
to one such write up in the language
of linear algebra if you’re curious.
And this is pretty common in math,
where having the key insight and understanding is one thing,
but having the relevant background to articulate that understanding more formally
is almost a separate muscle entirely,
one that undergraduate math students kind of spend most of their time building up.
But the main takeaway here is not the solution itself,
but how you might find that key insight if
it was put in front of you and you
were just left to solve it. Namely,
just keep asking simpler versions of the question
until you can get some kind of foothold.
And then when you do,
if there’s any kind of added construct that proves to be useful,
see if you can reframe the whole question around that new construct.
To close things off here, I’ve got another probability puzzle,
one that comes from this video sponsor – Brilliant.org.
Suppose that you have eight students sitting in a circle taking the Putnam.
It’s a hard test,
so each student tries to cheat off of his neighbor
choosing randomly which neighbor to cheat from.
Now circle all of the students that do
n’t have somebody cheating off of their test.
What is the expected number of such circled students?
It’s an interesting question, right?
Brilliant.org is a site where you can practice your problem solving abilities
with questions like this and many many more,
and that really is the best way to learn.
You are going to find countless
interesting questions curated in a pretty thoughtful way
so that you really do come away better at problem solving.
If you want more probability, they have a really good course on probability.
But they’ve got all sorts
of other math and science as well,
so you’re almost certainly going to find something that interests you. Me?
I’ve been a fan for a while.
And if you go to brilliant.org/3b1b,
it lets them know that you came from here.
And the first 256 of you to visit
that link can get 20 % off of their Premium Membership,
which is the one I use if you want to upgrade. Also,
if you’re just
itching to see a solution to this puzzle, which,
by the way,
uses a certain tactic in probability that’s useful in a lot of other circumstances,
I also left a link
in the description that just jumps you straight to the solution.