Every year, more than 100 countries send 6

of their brightest teenagers, or the occasional pre-pubescent prodigy, to represent them in

the International Math Olympiad, commonly known as the IMO. Considering that each country has its own

elaborate system of contests leading to their choice of 6 representatives, the IMO stands

as the culminating symbol for the surprisingly expansive and wonderful world which is contest

math. The contest is essentially a test, split over

two days, with three questions given over four and a half hours each day. The questions are all proofs, meaning you

don’t simply find some numerical answer, you have to discover and articulate a rigorous

line of reasoning to answer each difficult question, and they are scored on a scale from

0 to 7. Of interest to us today is the one from 2011,

with 563 total participants representing 101 countries. I know what your thinking, and the answer

is yes. Those do all happen to be prime numbers. But that’s not why this test was interesting,

out of these prime problem solvers, only one, Lisa Saurmann from Germany, got a perfect

score. The only thing standing between the next two

runners up and a perfect score that year was problem #2. This problem is beautiful, and despite evading

many of the world’s best mathematicians of their age, the solution is something which

anyone watching this video can understand. So let’s begin by reading it carefully. “Let S be a finite set of at least two points

in the plane” Mmkay, as you read a question it’s often

helpful to start drawing out an example for yourself. “Assume that no three points of S are collinear” In other words, you never have three points

that line up. So you can probably predict that the problem

involves drawing lines here in a way where three points on one line would mess things

up. “A windmill is a process that starts with

a line l going through a single point P in S. The line rotates clockwise about the pivot

P until the first time that the line meets some other point belonging to S”. Again, while reading it’s helpful to draw

out an example, so we’ve got this line pivoting around some point until it hits another. “This point, Q, takes over as the new pivot,

and the line now rotates clockwise about Q, until it next meets a point of S. This process

continues indefinitely”. Alright, that’s kind of fun. We keep rotating and changing the pivot. You can see why they call it a windmill process,

and you can also see why they specified that no three points lie on one line. You wouldn’t want to run into the ambiguity

where you don’t know what pivot to switch to. Okay, so what’s the question? “Show that we can choose a point P in S

and a line l going through P such that the resulting windmill uses each point of S as

a pivot infinitely many times.” Depending on your tolerance of puzzles for

puzzles’ sake, you might wonder why anyone would care about such a question. There’s a very good reason: I’d argue

that the act of solving this will make you better at math and other related fields, which

I’ll explain once you’ve seen the solution. But certainly on it surface this feels disconnected

from other parts of math. Other Olympiad problems involve some function

to analyze, or a numerical pattern to deduce, or perhaps a difficult counting setup or an

elaborate geometric construction, but problem 2 is an unusually pure puzzle. In some ways, that’s its charm: proving

that some initial condition will result in this windmill hitting all points infinitely

many times doesn’t test if you know a particular theorem, it tests if you can find a clever

perspective. But that blade cuts both ways; without resting

existing results from math, what could someone possibly study that would prepare them to

solve this? In fact, this brings us to the second unusual

thing about this problem: Based on the results, I’m guessing it turned out to be much harder

than the contest organizers expected. You see, typically, the three problems in

each day are supposed to get progressively harder. They’re all hard, of course, but problems

1 and 4 should be doable, problems 2 and 5 should be challenging, and problems 3 and

6…well they can be brutal. But take a look at how many of the 563 participants

that year got perfect scores on each problem: Only 22 got a perfect score for this question

number 2. By contrast, 170 got a perfect score on problem

5, which is supposed to be about the same difficulty, and more than twice as many, got

a perfect score for problem 3, which is supposed to be harder. You might notice that only six students got

full points for problem 6 that year, so by some measure that was the hardest problem

on the test. In fact, the way I introduced things earlier

was a little disingenuous, the full data would suggest problem 6 was the real clincher. But what’s strange is if you look at the

results of the only six students to solve problem 6, all of whom are clearly phenomenal,

world-class problem solvers, this windmill puzzle evaded five out of six of them. But again, this problem is hard not hard because

of the background knowledge it demands, it asks only for insight. So how do you approach something like this? The first step with any puzzle like this is

simply to play around to get a feel for it, and it’s always good to start simple and

slowly get more complicated from there. The simplest case would be two points, where

the line trades off between each point, so that works well enough. Adding a third point, it’s pretty clear

the line will just rotate around all of them. It might not be entirely clear how you’d

phrase that as a rigorous proof yet, but right now you’re just getting a feel for things. The fourth point is where things get interesting. In some places, your windmill will go around

the four points as it did with the triangle, but if we put it inside that triangle, then

it looks like our windmill never hits it. Looking back at the problem, it’s asking

you to show that for some starting position of the line, not any position, the process

will hit all the points infinitely many times. So for an example like this, if you start

with the line going through that troublesome middle point, what happens? And again, at this point you’re just playing

around, perhaps moving your pencil among dots you’ve drawn on your scratch paper, questions

of rigor and proof will come later. Here you’d see that your windmill does indeed

bounce off all the points as it goes through a cycle, and it ends up back where it started. The worry you might have is that in some large

set of points, where some are kind of inside the others, you might be able to start off

on the inside, but maybe the windmill process takes the line to the outside, where it will

be blocked off from those inner points. If you play around, and it might take some

time to draw out many examples and think this through, you might notice that when the line

starts off passing the middle of all your points, it tends to stay there; it never seems

to venture off to the outside. But can you guarantee that this always happens? Or rather, can you first make this idea of

starting off in the middle more rigorous, and from there, prove that all points will

be hit infinitely many times? As a general problem-solving tip, whenever

you have a vague idea that feels productive, find a way to be more exact about what you

mean, preferably putting numbers to it, then ask questions about those numbers. In our example, one way to formalize the idea

of a “middle” is to count how many points are on either side of the line. If you give the line some orientation, you

can reasonably talk about the left half, say coloring all points to the left of it blue,

and the right half, coloring all points to the right of it brown, and what it means for

the line to be in the middle is that there are as many blue points and brown points. For the moment, let’s say the total number

of points is odd, and the point the line passes through is colored white. So for example, if there were 11 points, then

you’d have five blue ones on the left of the line, and five brown ones on the right

of the line, and the one white one at the pivot. The case with an even number of points will

be similar, just slightly less symmetric. This gives you a new question to ask: What

happens to the number of blue points and brown points as the process plays out? In the example shown, you might notice that

it’s always 5 and 5, never changing. Playing with other examples, you’d find

the same is true. Take a moment to pause and see if you can

think about why that would be true. Well, the key is to think through what happens

as a line changes its pivot. Having given the line an orientation, we can

talk about which half of the line is “above” the pivot, and which is “below”. If the line hits a blue point to its left,

it must happen below the pivot, so as it changes pivots and continues rotating clockwise, the

old pivot, now above the new one, ends up to the left of the line, in the blue region. And entirely symmetrically, when it hits a

brown point, it happens above the pivot, meaning that old pivot ends up in the brown region. So no matter what, the number of points on

a given side of the line can’t change. When you lose a blue point, you gain a new

one; when you lose a brown point, you also gain a new one. Great, that’s insight #1. Now why would this imply that the line must

hit every point infinitely many times, no matter what weird set of points you dream

up? Well, think about letting this process go

until the line has turned 180 degrees. That means it’s parallel to its starting

position, and because it has to remain the case that half the points are on one side,

half are on the other, it must be passing through the same point it started on. Otherwise, if it ended up on some other point,

it would change the number on a given side. Additionally, since the line has rotated halfway

around, everything that was blue has become brown, and everything which was brown has

become blue, and the only way to change color is if you get hit by the line. So for our odd number case, that means that

after a half rotation, the line is back to where it started having hit all the other

points. So as time moves forward, it repeats that

exact same set of motions over and over, hitting all points infinitely many times. If there are an even number of points, we

need to alter the scheme slightly, but only slightly. Let’s say that in that case, the pivot counts

as a brown point. So we can still select an initial point so

that half the points are blue, all on the left, and half are brown, now either meaning

their either on the right or the pivot. The same argument applies that after a 180-degree

turn, all the points will have swapped color, although this time the line might be passing

through a different point after that first half turn, specifically one that used to be

blue. But after another 180 degrees the line must

be passing through the same point it started on. Why? Again, it must be parallel to its starting

position, and if it was passing through any other point, the number of points on a given

side would be different. So again, you have a cycle which hits all

points, and which ends in the same position where it started, so it must hit all points

infinitely many times. There are two important lessons to take away

from this puzzle, the first social, the second mathematical. Once you know this solution and turn it about

in your head a couple times, it’s very easy to fool yourself into thinking the problem

is easier than it is. Of course the number of points on a given

side stays constant, right? And of course when you start in the middle,

every point will switch sides after a half-turn, right? But the advantage of this problem coming from

the IMO is that we don’t have to rest on subjective statements, we have the data to

show that this is a genuinely hard problem, in that it evaded many students are demonstrably

able to solve hard problems. In math, it’s extremely hard to empathize

with what it feels like not to understand something. I was discussing this video with a former

coworker of mine from Khan Academy who worked a lot on their math exercises, and he pointed

out that across the wide variety of contributors he’d worked with, there was one constant. No one can tell how difficult their exercises

are. Knowing when math is hard is often way harder

than the math itself. This is important to keep in mind when teaching,

but it’s equally important to keep in mind when being taught. On are windmill puzzle, even if counting the

number of points on one side seems obvious in hindsight, you have to ask, given the vast

space of possible things to consider, why would anyone’s mind ever turn to this particular

idea? This brings us to the mathematical takeaway. What ultimately led to the solution was finding

something about this complex system which stays constant during its chaotic unfolding. This is a ubiquitous theme through math and

physics. We’re finding what’s called an “invariant”. Topologists do this when the count the holes

in a surface, physicists do this in defining the ideas of energy and momentum, or in special

relativity when they define proper time. As a student, it’s easy to take for granted

the definitions handed down to us, but the more puzzles you solve where the insight involves

an invariant, the more you appreciate that each of these definitions was once a clever

discovery. Terence Tao, one of the greatest modern mathematicians

and the world’s youngest IMO medalist, wrote that “Mathematical problems, or puzzles,

are important to real mathematics (like solving real-life problems), just as fables, stories,

and anecdotes are important to the young in understanding real life.” Sure, they’re contrived, but they carry

lessons relevant to useful problems you may actually need to solve one day. Maybe it seems silly to liken this windmill

puzzle to a fairytale; a mathematical Aesop summarizing that the moral of the story is

to seek quantities which stay constant. But some of you watching this will one day

face a problem where finding an invariant reveals a slick solution, and you might even

look like a genius for doing so. If a made-up windmill prepares you for a real

problem, who cares that it’s a fiction?

Well? Did you answer it or not? Or did i miss the answer? If there was one? What?

1:32 Two points IN the plane.

"The solution is something that anyone watching this video can understand."

You're giving me way too much credit, pal.

I've been told that i could do amazing things with my intellect and should apply for mensa and shit.

I dont understand any of this.

I knew i was retarded but not THAT retarded x(

Hello. I haven't gone past the problem explanation, and I came up with a solution(not a proof). We're talking about a finite set of points in space, so each point has a set of coordinates and none of them are off an infinite distance away. this means that the set has a perimeter, or that there is a set of points that make up a polygon with no outward-facing angles which contains the rest of the set of points. the remaining set of points has its own perimeter, and so on until we hit the center set. if we start out crossing the center set or on the center point, then we must have been able to take some path to get there center point. if the path rotating in one direction devolves into a pattern without hitting the starting point, then it cannot in the other direction and have ever hit the starting point. there is a finite number of combinations of points and parts of the space for the line to sweep, and windmilling will always hit the same point next if we are sweeping the same space from the same point in the same set of points, therefore there are not non-repeating paths and any path with a forward pattern must follow the same pattern reversed if the rotation is reversed. this also means that any combination of point and swept space cannot be shared by two paths around the same set of points. for each perimeter shape, there is some starting point and direction which allows us to sweep the entire interior of the perimeter. if we sweep the entire interior of a perimeter, then all of the space will have been swept, and all points will be hit in a repeating pattern. if we place a perimeter set inside of this perimeter set and start without the line crossing the inner perimeter, then the line will sweep the same area, except missing the space contained by the inner set. therefore if we start crossing the inner set the line cannot escape the inner set, and all of the space will be swept. if all of space is swept, then we can repeat this line of thinking to each further inner perimeter. a set with a center point instead of a center set can be swept by starting on the center point.

THE EARTH IS FLAT !! I DONT CARE WHAT YOU SAY !!!!!!!!!!!!!!