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.

Has anyone else noticed though that the teacher always does the first question which is so simple? Then the sec question is completely different? It’s like:

1) 2+2

2) Find the percent of A multiplied by the number of pears Sarah had in her pocket double the Pythagoras equilateral triangular theorem? Don’t forget about BODMAS btw -3-

Why do so many of these comments have nothing to do with the video.

YouTube I'm in pre calc I don't need to think about this yet

The YouTube algorithm has brought us all together again

Me out here tryin' to solve 2+2before he even started to explain I randomly said 12% which is extremely close to the actual answer 🙃

Me:*is bad at redstone*

Redstone engineer: it’s not that hard

Redstone:

Test:

Ryan solves the equation of 1+2 and gets 3 as his answer…Ryan’s answer is incorrect, explain howMe: niggaaa…..

I really love, that the video is black! It’s so good for my eyes!

be my math teacher

In class: 2×10=20

The test: I have 5 apples I ate 2

What is the mass of the total of all the stars in the solar system subtracted by the mass of the sun times by the circumference of the apple I ate?

I feel pathetic

Bro im 14 how tf am i supposed to know?

Me at 3AM, watching this for no reason.

My brain hurts.

I was watching some minecraft videos and slept with autoplay on why am I here

This is philosophy not math ima head out

…how did I get here from a Rainbow Six Siege compilation…?

Anyone else just start reading the comments after 2 minutes of the video?

I'm a freshman highschooler and I'm watching this. Please God help me pass this test

Beynim yandı

Lmao idk why I’m watching this if I knowww I don’t get it 😂

Before I get to the solution I’m gonna take a guess:

If it’s 1D the chance is 100%

2D = 50%

3D is 25% I think

So 4D must mean that it’s a 12.5 percent chance

This is easy I haven’t even watched the vid and I know the answer its 12.5% reply and I’ll explain

As soon as you put the problem in 2D I instantly got the answer for 2D. Then, after you explained the first way how to get to 2D, I’m like: what a minute, before we divide the circle in 4, now we divide the sphere in 8! Boom, easy! Amazing, I would never get the answer alone… Sometimes, we just need to think outside the box. Now, proving the answer, gosh no! Already had lots of intensive math in college, I’m good.

Very cool! 🙂

Today at the University we solved the exact same problem… and the variation where you pick halfpoints of tetrahedron's edges.

I solved it by myself before teacher showed the whole solution. I mean, we were thought to think this way on Probability and Statistics… very interesting 😀

Hardest problems:

6×9+6+9

9+10

Divide by 0

This shit is unnecessary

teacher: cmon guys i taught u this

So ummmmm… Whats the anwser?!

Class: periodic table

Homework: mole calculations

Test: create a device to warp reality and tear the fabric of space time all while dimension hopping. 2 marks

Hermione has entered the chatme: average at math with no future prospects in a math career

me after this video: Im So GoOd At MaTh GuYs

aku tuh ga ngerti kenapa aku disini.

I was watching Peppa Pig edits why is this is my recomend :,)

School : 2+2

Homework : 4+4

Exam : Jonny is four years old, the train is seven meters, calculate the mass of the sun.

Exam questions:

Michael has 4 apples. His sister recently gave birth. Calculate the mass of the sun, and divide by dolphin.

It's actually 2 pink 1 brown

;Putnam;

I hate math why am I watching this?

Wut

Walter O’Brien would of figured it out

The real question is:

When will we ever use this in real life?

Bruh

Thanks ADHD i got lost near the end

Long division back in grade 3 be like:

Long division back in grade 3 be like:

Imagine doing this without the the 3D model

The answer is 2

I think my maths teacher said she took something like this. She said to this day she remembers the first problem, “prove 1>0”

Anyone else wanna torture themselves by doing the test?

Im still struggling on y=mx+b 😂

That’s cool but I just got a 65 on my math midterm exam

Imma just sit down for a sec…

Anyone else watch this and not understand a word he says?

my dumb school gives us questions like these to solve in 20 min

Why am i here… Im 15 and i dont even know wht tetrahedron is

Huh-

Пфффф… Пусть теперь ЕГЭ попробует написать. (Я уже про профильную не говорю₽

The only putnam competition I know of is the 25th annual putnam county spelling bee 😬

Im in algebra wth is this bruH

Class: On most if not all graphs time will always be the independent variable located on the x axis

Test and project: Draw the schematics for building a time machine and then use all forms of engineering and applied physics to build a time machine. You have 40 minutes. Good luck

The test: Find X X= 4 points are randomly chosen on a sphere, what is the probability that the tetrahedron formed by connecting the points also contains the center of the sphere?

Me: Yes????

Class: 2+2= 4

Homework: 2+2+5=9

Test: Jimmy has 3 apples and has to arrive at the bus in 8 minutes, calculate the mass of the sun.

This is some good will hunting type shit

We just did a geometry test and this went to my recommendations

I feel so small brain

Why am I getting this in my recommendation Im not even in high school?

what

Classwork: When was George Washington born?

Homework: Where was the declension of independence signed?

Test: If Josh was born in 1889 and just bought an iPhone 11, who assassinated Kennedy?

What the fuck was this man talking about

When you’re Only In 7th grade and you think you can Solve all of these but ends up in a coma From thinking to much🙏😔Issa ok I can relate to this to you’re not Alone!!!!!!

I honestly am not fully processing anything he says.

I’m in the 7th grade, I’m not trying to get a PhD in math to understand what you are saying.

OK so I now have read about a hundred comments and they were all just memes complaining about the difficulty😅

Early guess 12.5%

Me: wtf how can i solve my math homework

Youtube recommendations: bish look who's struggling here lmao

My idealology was that technically speaking there is no limit to how precise you could get with those points, for example, .9999965468997. Literally just tapped random numbers, which makes that a random point, and considering that I could get as precise with those numbers as I want, their are an infinite number of solutions in which the center is within the figure, and an infinite number of solutions in which it doesn’t, which leads me to believe that the probability would be .50, or 50/50 chance. While this is most likely wrong, that’s just how I was viewing it, in case anyone was wondering😂

NGL I guessed 1/8 almost as soon as the problem was proposed

Bruh🤦🏻♂️ this is so easy to visualize but when am I ever gonna need this knowledge?

Everyone knows this aight the hardest problem… this is: Using the average size of the apple * 🍎 * determine the speed at which the universe spreads across space

Why am I watching this when I can barely do PEMDAS

What?Im in 10th grade…its 12:30 am and I have school tomorrow at 8:00 am. How did i get here?

Math: easy ln paper

Hard in test

IMPOSSIBLEon real lifeWhy am I here? I had trouble with 9th grade geometry

Pfft..Joe can solve this problem.

Mindblow!!

I searched up all the questions and tried to solve them

HELP ME"the test will be easy if you pay attention to the notes"

Hey! Are you a genius?

I solved it in less then 10 seconds, but I'm sure my logic might be wrong. I assume that the P1 can be casual, so 100%. Now I want the P2 in the same half of the sphere ( I wanted all the 4 points in the same "half"), so P2 50%. Now P3, again, in the same half or in the other one, again 50%. And last but not least, P4, again, in the same "half", 50%. In the end becomes, 1 x 1/2 x 1/2 x 1/2, so the probability is 1/8

Why was I sent HERE. I'M STUPID.

My english is to bad for that :O

Wow, YouTube think that I am a smart guy just because I watched alot of video clip scene about Big Bang Theory TV show on YouTube.

I think i lost brain cells rather than gaining them in this video.

IT'S BIG BRAIN TIMEI’m in seventh grade and I can sort of understand this…

why.I don't have thinkAah nice tnx for the answer