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.

I got a headache and still watch the video.

Science student!!!

Its too easy🤪

What if the center is on the triangle/tetrahedron

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

Can you show us question 6?

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%

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.

It's kind of ironic that you can prove this in such a simple way with that animation (Ok, the animation takes a

lotor art, knowledge and work to come out that good, but I mean, that you can visually do it in a way thatanyonecan 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.……

Can I go home? Kkkk

I just fell in love with this channel. You make maths so beautiful

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?

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?

So does it scale up in powers of two as you add more dimensions?

I just want to study math without giving in a worry of exams…

Just me and math…

So, this type of probability is 1/2^n where n is the dimension of the problem

I have an issue with this problem. How do you define random?

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

A caption riddle at the end

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.

The answer generalizes to n dimensions as well?

The mitochondria is the powerhouse of the cell.

我都没看懂题….

Sir, I love you.

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

Viewing the circle as two coins is really genius ! I wonder…

The creator of the test, how smart is he ?

How will it work for a 4th dimension circle with 5 points ? 1/64 ?

in 2nd dimension probability is 1/4 so probability for every dimension is 1/(2 to the power n)

The difficulty sounds like a one night stand

MY favorite math video

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

Q: What is the probability of the tetrahedron containing the sphere?

Me: 1/8th

Q: Show your work!

Me:

Nervously reaches for my stress ballI love this channel and problem solving 🙂 I'm a programmer, so I find it very interesting to do more of this. Really love it 🙂

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?

Just a side note about these kinds of probability questions… it's important to know what particular

distributionin which things are chosen "at random." This kind of fallacy should be avoided: https://en.wikipedia.org/wiki/Bertrand_paradox_(probability)Could this be generalized to n dimensions? so that the probability is 1/2^n ?

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

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) .

You used fractions and probability to solve such a high level problem. Amazing.

Beautiful

I dont find this to be an acurate answer…

One dimension: P=1/2; two dimensions: P=1/4; three dimensions: P=1/8. Conjecture: four dimensions: P=1/16.

An absolute gem this is!

But 1st we have to establish that plotting software(ADVANCED) in our mind

NICE EXPLANATIONThe 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…

Sees question: WTF how

After video: 200 IQ is me.

Loved this. Great work!

1/2^3

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?

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

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)

THIS THREE ARE USED IN WORSHIP OF HOLINESS OF ANGELS BEFIRE THE THRONE ROOM

HOLY EMANUELA – South – Settled

HOLY EDONA – North – Settled

GRACE – 7 Churches of Asia

HOLY ARIEL – West – Settled

GRACE – more spiritual information required.

THIS FIVE ARE ALSO REPRESENT THE FIVE HOLES IN TWO HANDS TWO FEET AND HEART ❤️- FEET MOST LIKELY ARE NORTH AND WEST SINCE THEY WERE ONE ON TOP OF ANOTHER NORTH AND WEST SINCE ALSO EPHRAIM AND MANASSEH ARE ON THE BOTTOM OF ALL TRIBAL PROCESSION.

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.

Now solve it with Common Core Math 🙂 (You'll need to start by drawing 363 dots)

Following this rule, would it mean that if we go up one more dimension, the probability would be 1/16?

Im ganna to type something here in case that others think that I can't understand this

This is my absolute favorite video on the entire internet!

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

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.

This is really intresting. Yeah I’m a maths nerd

congrats to 2mil subs

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.

12.5% final answer

Allah akhbar

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

I love you. My brain is cumming

I didnt understand annnnyyyythhhinng!😭

Your methodology appeals to a wider audience.

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 😮

Easy when you have an unlimited supply of NZT-48. Now try with RedBull only….

Test:

What is cercunference of circle

Me: ?

Teacher: so you have chosen 0 grades

# gandalf meme

holy shit I guessed 1/8 at the start of the video, I'm so lucky lmfao

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

It's 1/8

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.

Very easy problem.

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%?

Amezing

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.

The fact that P3 ( in 2D ) and P4 (in Sphere) were chosen first and had no free will is depressing.

chance = 1/(2^d)

d = dimension

I simply love your videos.

7:00 True jaw dropping moment!

I don’t even know how I got to this point. but I’m still watching.

Математичка ты довольна?

That's a lot of work for flipping burgers.

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.

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

this,,,,,, this is so beautiful oh my god,,,,,,,

Kaun he ye log …kahe se athe he….?

But what is the probability after all? For me it is 255/256

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

I got anxiety from this

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

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.

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

Got the answer right but not sure how.

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 🙂

50/50 either it contains the center or it doesn't

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.

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.