The hardest problem on the hardest test

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.

100 Replies to “The hardest problem on the hardest test”

  1. in the 2d case, i remember seeing multiple solutions to it based on the measure that you choose. great video anyways!

  2. An easy way to think about the circle:

    1st point doesnt matter where it is so just put one somewher

    2nd point is all that matters. If it is almost exactly across the 1. point the chance that the 3rd point is in the arc that matters is 49.9999..
    When the 2nd one is very close to the 1st one its close to zero but jot quite.

    Now i thought about the solution to the sum S=1-1+1-1+1-1… which comes out to be =0.5

    So you just use logic and conclude that the probability has to be in the middle. 25%

  3. Parece mentira que veo en los comentarios del video que gente he dedicado su tiempo a crear subtítulos para estos videos tan educativos en portugués, italiano, alemán, polaco, vietnamita, inglés y griego, pero nadie ha tenido el interés de hacerlos para español. Por eso es que nuestros estudiantes y países no progresan. Solo ven videos de reguetón.

  4. It's kind of ironic that you can prove this in such a simple way with that animation (Ok, the animation takes a lot or art, knowledge and work to come out that good, but I mean, that you can visually do it in a way that anyone can get it) yet a pure mathematician will not be happy with it and will not accept that as proof unless you put in in written with all that difficult nomenclature that few can understand.

  5. So, now I'm wondering if there is a way to generalize this to higher dimensions. That is, if you have an n-ball and choose n+1 points randomly on it's surface, what is the probability that the center of the n-ball is bounded within the n-tetrahedron created by those points?
    And would it generalize to being a probability of (1/2)^n?

  6. Im a little confused. In the first 2d problem, I can see how it is 1/4, seeing as there are two lines with four possible points for p1 and p2. In the 3d problem, wouldn’t it be 1/6 since there are 3 lines with 6 possible outcomes?

  7. If the probability was 1/4 in a 2D plane and a 3D plane had 8 quadrants. The probability was 1/8. Man , i think im falling in love with mathematics again

  8. I probably wouldn't have figured it out without this video, but I only needed up until the lines on the 2D part to figure it out. I don't know if I'm the only one who did this or if there's a lot of people.

  9. randomly chosen! riiight 😀 but nail P1 and P2 and …. how is P3 random if u want to complete the condition ?!?
    addendum – this is NOT a hard question , it BAD language describing THE problem

  10. I just hate the fact that these creaters put lot of hard work to make these scientific videos on youtube and people here comment about funny stuff. I didn't find any relative and serious comment. Is there no one here interested in math? Then why did you click this video

  11. Q: What is the probability of the tetrahedron containing the sphere?
    Me: 1/8th
    Q: Show your work!
    Me: Nervously reaches for my stress ball

  12. I love this channel and problem solving 🙂 I'm a programmer, so I find it very interesting to do more of this. Really love it 🙂

  13. I could approach to the step of 'fixing three points on a circle' and then I thought the answer would be the average width of all possible triangles on a circle. Is my thought wrong? and can I even possible to get the average width?

  14. Just a side note about these kinds of probability questions… it's important to know what particular distribution in which things are chosen "at random." This kind of fallacy should be avoided:

  15. Somehow I'm 13 and I managed to solve for the probability in the circle, and just did the same for the sphere… not right tho

  16. In case of circle, When two points (P1 and P2) will be on diametrically opposite side ain't the probability will be 1 as the center of the circle will always be on the line joining P1 and P2 and it doesn't depends where you put third point the center will always be there.
    (Just imagine two coincident lines) .

  17. The answer is much easier, isn't it? Once you get that in 2 dimensions the answer is 1/4 (I did it the first way which is easier…) which is 1/2 squared, then in three dimensions it must be 1/2 cubed = 1/8. I wish I could have figured this out when I took the Putnam but, of course, I didn't…

  18. Here's a harder one:
    If a tetrahedron inside a sphere contains the center of the sphere, what is the likelihood the divorce court judge will let me keep the kids?

  19. If you think about it, if this was a question in your high school calculus class. Imagine wasting your brain cells because you needed that class to graduate and work at amazon

  20. N=7

    MERCY – EAST in the court has aslo to do with Moses in Psalm 90 and Job 9:15 and much more like the head part God of Jakob – Twelve more in this parts.
    MERCY – Isaiah 23:12 more spiritual info required – Like Elija and more advanced foundation of Saphari like the feet footstool (enemies under)


    HOLY EMANUELA – South – Settled

    HOLY EDONA – North – Settled

    GRACE – 7 Churches of Asia

    HOLY ARIEL – West – Settled

    GRACE – more spiritual information required.


    John 1:16 GRACE UPON GRACE.

    So one grace is in the churches of Asia seven lamp stands also MERCY IS ON HIS THE LEFT. There is also another Grace that is in the heart and caused insult – Hebrews 10:29

    1 Corinthians 14:19 five wise virgins are in the church or in the court

    Yet more God of ABRAHAM

    God of ISAAC.

    10,000 words spiritual stars we not ready yet to start on those parts.

  21. just asking, but wouldnt there be a formula for this, 1/2^n where n is a positive integer for the number of lines for n>1

  22. At the first attempt, I would think about one of the edges of tetrahedron which should be equal to or contains radius of the sphere … starting off with lines passing through the centre of sphere is easier than referencing points at the first stage. It's my opinion.

  23. Another way to think about this problem that'll help reach the solution intuitively is to actually simplify it even further, from 2 dimension to 1. With a given line segment and any 2 random points located on it (except at the center), what is the probability the new line segment formed by connecting the two points intersects the center? 1/2, as the first point is always going to be on one side of the center or the other, and the second point will, half of the time, be on the opposite half of the line. Then if you can make the determination that the probability for the 2 dimensional case is 1/4, its easier to understand how extrapolating out to 3 dimensions, the probability reduces to 1/8.

  24. thought you mightve gone down to the 1D case and then started there, but I guess the 2D case was good enough to start from

  25. sometimes i hate math. Getting it explained everything feels so easy to repeat, but once you have to solve a complex question on your own your brain just goes out of the window 😮

  26. Today I learnt that my brain was specifically not made for Math. But the problem solving takeaways are still useful irl.

  27. The answer is 1/8. We can use the same analogy as the 2D case. Think of 3 lines connecting the center and the first three points of the tetrahedron. The separate the space into 8 sections (just similar as Cartisian coordinates), and the average intersection area of the sphere surface and one section is 1/8 of the total surface area of the sphere. Moreover, the probability of a spherical triangle to occupy each one of the eight sections is the same. Therefore the answer has to be 1/8. Note: I have verified this conclusion with Python codes.

  28. So, if we had the 1 dimensional version of a circle, it would be 1/2 or 50%, and a 4 dimensional version of a circle would be 1/16 or 6.25%?

  29. I see that you've done your best to make math interesting, but i'm sorry to say it- It's still not and probably never gonna be.

  30. The answer is simple.

    After bisecting the perpendicular plane of existence in which the center of the sphere is located, you can use the quadratic equation to find the angle coherent to the circumference of the line if that line was intersecting the triangulated plane of functionary solutions that resides on the diagonal domain of the range value between x and y, where x is the bisected circumference of the triangulated square within the sphere and where y is the radius of that circumfrentated square to the second power of the quadratic perpendicular line segment running along the base of the triangulated plane of functionary solutions.

  31. altough i dont see any imediate use for this i wonder if we can generalize this for n dimensions that is given an n dimensional sphere and n+1 random points on the surface the random polyhedron contains the center with probability 1/2^n

  32. well I figured out it was a quarter and I'm so fokin happy I can't describe it

  33. As with all advanced math I would appreciate this more if I could apply it to something and not over complicate it

  34. Interesting. The probability of 3 points being on one half of the sphere is also 1/8. I guess the assumption is that the tetrahedron can’t invert or that at least one point must go through the center axis.

  35. Missing the proof that the lines and choosing points process is always containing the whole sphere. that is crucial for the argument.

  36. you can think in a different way once you come up with an idea about lines. as the result In 1-dimension (on a line), you will have 50% chance (1/2^1), in 2-dimension (circle) you have 25% (1/2^2), in 3-dimension (sphere) 1/2^3=1/8. So the next level will be hypersphere with a 1/2^4=1/16 chance 🙂

  37. If you can tell me how many particles move when you take a step forwards at 2.3 MPH and you take that step for 0.732 seconds and the length of your legs are 1.1 meters then so be it.

  38. Why you are confusing public by very complicated calculations. The simple answer is Probability of triangle having centre of circle is area(triangle) ÷ area(circle) and similarly vol(tetrahedron) ÷ vol(sphere). I challenge proove me wrong.

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