Cookies   I display ads to cover the expenses. See the privacy policy for more information. You can keep or reject the ads.

Video thumbnail
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. 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.