We'll talk about one theorem that I learned when I was in high school and the seemingly
miraculous methods to solve it which is my
Favorite thing from the training for the International math Olympiad back in Bulgaria
The theorem is called Ptolemy's theorem and the method will wait a few minutes before we get there
Well, let's start with something that everyone knows. What is the theorem Brady that you think everyone knows?
in geometry
BRADY: Pythagoras ZVEZDA: Absolutely. Well, what do you know about it?
BRADY:Even I know this one a squared plus B squared
Is equal to C squared
ZVEZDA: All people know this theorem, but do they know how to prove it?
So Ptolomeny's theorem is somehow related to this one
So who do you think is stronger ptolemy or Pythagoras?
Stronger stronger, they're both Greeks
Well Pythagoras is more famous
That is true, but they have to compete with each other and we shall see I need another
brown sheet already
So we're going to draw a large circle we need four points on this circle
They form our so-called cyclic quadrilateral
It just means they lie on the circle and you can also draw the diagonals
What would Ptolemy tell us about these six segments? It's very beautiful
It says if you multiply the opposite sides and add up those products you will get the product of the two diagonals
so a
B
times C D plus a
Times B C must be equal to the product of the two diagonals
AC times BT and this is what?
ptolemies theorem tells us and it doesn't matter where those points on the circle as
Long as they in this order he's gonna work
So this is a property
about points on a circle
Now, let's see I said that
Pythagoras in Ptolemies are
related
Stronger
If I can prove
Pythagoras knowing Ptolemies theorem the thalamus is stronger
Okay
So it would be like a parent or yeah soap Ptolemy would be a parent of Pythagoras
It is possible certainly
So if I draw a right triangle and then take exactly the same triangle and flip it over
What kind of a shape do you think Brady? I'm going to get mmm a rectangle
it is precisely a rectangle so we have
ABCD for the vertices now a rectangle is a very symmetric figure it has all right angles, obviously
So if two diagonals also happen to be equal then moreover they are
Intersection. Oh, is it equal distances from each vertex?
Mm-hmm, so they must be on a circle those points exactly but there soon as we have four points
On the circle. Guess what happens?
Ptolemy you say yes potala me would kick him. Yeah
So now let us let us backtrack a little bit we started with the right triangle. Let us name it sides a
B and the diagonal will be C
Okay, but then this side should be a and this side should we be in the other diagonal should also be C
Mm-hmm. Let's hit this picture with Ptolemies
so what do we have the product of the opposite sides a
times a plus
the product of the other opposite sides
Me times B must be equal to the product of the two diagonal C times C. Hmm
Does this look like a theorem we've seen before it looks good. That is precisely the Pythagorean theorem
Which means that ptolemy is stronger
Pythagorean theorem is a special case of Ptolemies theorem, but Ptolemies theorem can do a lot
And so we create
a new method today
Actually, I've known about this method for a long time that will solve Thomas theorem
So actually ptolemies can be proven in many different ways. If you google you will see lots of different proofs
You will see a proof in plane geometry
Which is beautiful, but very tricky to come up with you can see technical proofs using
trigonometry who likes those or
using even complex numbers very sophisticated
But none of these approaches really shows the true elegance and simplicity of Thomas theorem
None of them shows you why it is really true the new method that we will talk about today
called inversion in the plane
Looks originally like a mysterious black box
because it will turn our circle and
Ptolemy situation into something that the third grader can tackle
So it will turn this whole picture into something very simple only three points that are lined up
a1 b1 and c1
Hmm, what could a third grader tell me about this picture? What is true Brady?
I in Bay and then they and say makes a whole are correct
If I add the two shorter ones, I must get the longer one look back at Ptolemies theorem
We are doing something similar. We're adding two things on getting a third one. What?
indeed
This be the same situation which in a different world in an inverted world and the answer is yes
So what is inversion inversion is obviously some transformation of the plane it's same point some points to some other points
So let's think about other
transformations in the plane that we all know
So what would be one that you Brady have heard about like reflections and things that is that we doing correct?
Yeah, you can reflect across a line. Yeah, that's not so perfectly good transformation. How about this one?
That's called
rotation about a point
definitely
How about this one that's called translation
And there is another one when you click on a map to expand it to zoom it out or zoom in
That's what's called a rescaling or dilation. So suppose I draw an elephant
What do you think will the elephant look after each of those transformations?
Would it look like a rabbit or
Would it look like an elephant? It will keep its elephant nests, correct?
It may kind of flip-flop but it will still look pretty much like an elephant. Well, our inversion is a completely different
anymore because we saw how it's
Transforming it's a circular
picture into a linear one
So it moves or something horrible two shapes if you take your nice elephant and hit it with that inversion
What will turn out in the end is something that might look like a lot in Shu from the movie completely
Ridiculous, so we need to have control over the situation
What is going on? So for an inversion you need a circle into the boss. You can draw the circle
Anywhere in the plane you like and as big as you like ant we mark its center
Oh, let's say we take a point outside of this circle X. Where will this point go?
It's like a small
Algorithm here. You start to the point X. We connect it with the center. We draw a tangent
Through X until it hits the circle at T
And then we draw a perpendicular to all X so our point X if it is outside of the circle
will travel
into the circle into X Y this
Actually can happen on any point in the plane
For instance if we take something very close here
And we stiffy we connected to all we draw a changin and we drop a perpendicular
That will be where y1 will go if I take any point outside the circle, where do you think it will go?
It looks like we'll go inside the circle correct that can happen to any point outside the circle now
We have more points to worry about what happens if you start with a point inside the circle
well, we want to be consistent so
The logical thing is to reverse this algorithm if I take a point Z
well
I still need the ray through Xion all because everything important looks like it's gonna lie there
but then I have to backtrack and erect a
perpendicular at Z to this ray hitting the circle and
Finally drawing my tangent to the circle until it intersects our original array pinpoint
Z 1 so if you start inside the circle you go outside of the circle. That's what inversion does to you
Ok
Mmm, I think we're still missing a few points
Which points did I not talk about?
The ones on the circle, correct? So suppose I take a point
About u so U is on the circle. I'm going to treat it as if it is a point outside of the circle
So we're going to apply our first algorithm. We connected to all next if you remember
I'm supposed to draw a tangent to the circle through that point. Ok. Well, the point is already on the circle
So it's just kind of kind of easy and the tangent will intersect the circle in our next point. Oh, but that's boring to you
already
Ok, and next I have to drop a perpendicular to my line until it intersects that ray
Ok, I'm gonna drop a perpendicular to this line
But I'm already there so I literally have not moved from you
So any point on the circle is going to go into itself and that kind of makes sense
You will take outside of the plane thus infinite plane and you will squash it inside
It will take the inside you will
parade outside
Sprayed it all over but the points on the borderline are gonna stay fixed and not gonna move anywhere
What happens to the center?
The center looks like it's an inside point so we should apply our second algorithm, but can it possibly work?
That Center is a very strange point
So the first thing we have to do by our second algorithm if you remember Point Z
We need to connect it to the center. Well, how do we connect it to the center? It's already one point
How do you draw that? Right? Yeah, I don't have a ray here
So we are already kind of stuck with our algorithm
But maybe you'll say it doesn't matter which ray they just draw a random one. Okay. There you go
no problem next I'm supposed to erect a
Perpendicular certainly. No problem. I'm supposed to draw a tangent to the circle
Until Oh
We are in trouble. It's parallel. They parallel
There is no way these two lines are going to intersect so you may say oh it goes somewhere
It's infinity. Okay in that direction, but what happens if I chose a different direction?
It's gonna go in that direction at infinity for our purposes of proving ptolemies theorem
We don't need to even bother with the center of inversion
We will not define
Inversion at the center when you're asking. Oh
That that doesn't seem right. It should be defined everywhere
beg your pardon if you for example
Know anything about functions one of the first function that would look like this is f of X
Equals to 1 over X Brady. Can I plug in any number here? I think we're in trouble
We can't we'll do there
So this function is not defined at zero, but it's a perfectly good function and you can see it all over the place
so why not leave inversion not defined at the center now when I
In middle school, they used to tease us with the following question. How can you catch all lions in Africa?
Alright we'll build the fins are really strong fans around the circle
Where would you like to stay Brady outside the circle inside the circle? I want to be free on the outside. Oh
That's where the Lions are. So you're gonna make sure that inside the circle. There are no lines. I thought I was alone
Oh, well, okay only I want to be away from the law. Okay, so you stay inside this lion?
free
Circle and then you hit the world with inversion
So all the Lions outside they'll come in, but you will be safely outside and observing them
Well, I don't know if that's a worthy goal
But you see the excitement of fifth graders how much they could foresee the power of inversion
What are the basic properties of?
Inversion, we already saw a few of them points outside the circle go in points inside go out
But do they do this randomly what happens if I apply inversion twice in a row?
So point X from the outside is going to go into point x1 on the inside
But if I do this, once again, I'll simply reverse my algorithm and go back to X
So what happens is that?
points in the plane are
Sort of married and they flip-flop from inside and outside. They have their chosen
Partner and all they can do is flip flop
The only points that the singleton are those on the circle they stay put and oh
We don't touch it we agreed
So applying inversion twice in a row will simply not do anything to the plane
It will keep the points wherever they are. This is super important because
if you take your
Elephant and invert it to Allah didn't show what would happen to not ensure if you hit it with inversion it will come back
In resurrect itself as the end of they need another sheet
You're gonna ruin these bows
We saw that Ptolemies theorem was originally situated on a circle, but miraculously that was turned into a lie
Now I want to know what happens to lines in the plane, but now some lines will be more special than others
some lines will go through the center of inversion and other lines will go not through the center of
Inversions in different things will happen to them in each case now
Let us look at the first case where they go through the center if I take a point outside on this line, it will go
Inside the same line - its twin
exactly, and then we'll flip flop the outside part of this line will go inside and the
Inside is going to go out and the points on the circle will just stay put of course
we said we're not gonna touch the center of inversion so we don't care about it, but by and large they go to themselves, so
They will still be lines
Exactly the same lines. You're not gonna see any difference before and after inversion
So if I took all the points on the line and inverted them, I would draw the same line exactly
Well, that's pretty nothing happened to that line. But if the line is outside the circle
Or inside but most importantly not going through the center of the circle
something
Wild is gonna happen. Let's experiment
So I'm gonna take a point X on this line
Connect it to the center draw a tangent drop a perpendicular. All right. Well so far no surprises
I knew that X was going to go inside the circle. Mm-hmm, but let's experiment more now
I'm going to take another point Y draw a change n't and drop a perpendicular
This is y1 so far. I don't see any surprise. Do you think this line is gonna go into a line here?
Could it possibly go into a line? I don't think so
Because there is no line inside the circle. It's gonna be just a line segment at best
But is it a lies? Let us experiment we z draw attention and draw a perpendicular
We already see that
The points inside the circle and not anymore
Lying on the line on what shape do you think very did they lie? I reckon I can guess now
yeah, I guess well, I
Reckon, we're gonna get so cool. Yes, but what kind of a circle circles are of two types?
They either pass through the center of inversion or they don't
what did you guess will our circle pass through or will it not I
think
What I know because you told me oh, it doesn't map to anything. So Oh actually matters a lot
We just don't touch it. It's like the king who does not participate in the battle, but only oversees what's happening
Okay. Now I
Can draw maybe one more point but let's try to draw it far away from oh
Maybe that will give us a clue. What happens to points that are far away from all where do they go?
and the inversion close to O or
Far away from all now. Remember those lines in Africa the further way
The lines are from our fans the closer to something
They will go I am getting close to Oh closer and closer and closer and you can go infinitely far away
And you're gonna get infinitely close
to our so actually if you
Continue doing this you will clearly see a shape of a circle
Now, of course, my picture is drawn by hand. So it looks more like an ellipse, but definitely
You can see that that's not a line
So it is a theorem that a line that does not go through the center
Will map to a circle that goes through the center. Okay, and that is fundamental now
It looks like we still need a lot of ground to cover
But do we now how many circles are there with respect to inversion to?
They either pass through the center of an version or they don't
But I already have one the fastest. It's right here
What do you think Brady happens if I hit that circle wooden version you get your lawn back exactly
absolutely, because
Doing inversion twice is gonna get us back where we started
Let us summarize so far lines through the center go to themselves lines
not through the center go to circles through the center and
Conversely circles through the center go to lines not through the center. Are we missing one type of figure here? I
Think we are
and this is
Circles that don't go through the center of inversion a circle not through the center of inversion
Goes to another circle not through the center of inversion
And in fact, that's a theorem and it can be proven using pairs of similar triangles
Just like any of the other theorems in this talk. We know properties of inversion now, we're back to Thomas theorem
Let's see if we can prove it at our first attempt. So what I have here is our
cyclic quadrilateral in other words four points on a circle and we are trying to prove something about the sides and the
Diagonals as Palomas theorem says we want to simplify this picture. How can we simplify?
Obviously lines are preferable to circles they are simpler to work with
can I turn
This circle into a line
Yes with inversion, but I need to choose my
center of inversion
very carefully
because
Different circles will go to different figures
If you remember if a circle doesn't pass through the center of inversion
It's just going to go to another circle that doesn't pass through the center of inversion and that's no use to us
however, if a circle passes through the center of inversion
schooner miraculously turned into a line
through the center of inversion
So where should the center of inversion be if I want to make a line out of this circle? What do you think Brady?
Well, you've got infinite options
Yes, but there is one place where it must be
It must be somewhere on the original circle for Tommy so that the original circle
Passes through the center of inversion. Yeah. All right. So let me just take something symmetric here looks Oh
Excellent. Now you are thinking I should draw the circle of inversion
I don't have to actually all I care is about the shapes of figures after
Inversion not exactly where they are located
So at any rate
This circle through the center of inversion
will turn into a line not through the center so I can just be bold and
Draw a line not through the center of inversion
And declare that my circle goes there. Why not now?
We want to quickly figure out where do the points a G C and B go?
You may be thinking well now I need the circle of inversion in order to do all of these tangents
And all of these perpendiculars now we can be efficient. We can apply a little bit of logic for
Instance take point B. I know that point B has to go somewhere on this line because it is on our blue circle
okay, so it has to be somewhere but also I know that it lies on array OB because that was our
first step of the algorithm
For inversion, so it has to lie both on this ray and to now align L
So I think we figured out its here and the same way we can efficiently
Find the images of the other points so our cyclic quadrilateral
now gets
Flattened on to a line and I do want to give a warning here
I'm not saying that the size of the quadrilateral
Become line segments now, they will act become arcs of circles. Most likely I
Am saying that the vertices are now lined up on the line great
Do you think that's really the best we can do it inversion?
because a while ago I
Promise that inversion will make this picture
look
something that the third grader
Can cope with three points on a line and then we are adding up the two shortest segments to get the bigger one
Well, I mean you've got four so you could you could do
That's not for a third grader
Now there are too many segments floating about here
We really want simplicity and elegance
Okay, so we have to go back again to our choice for the center of inversion
Even though we are on the right track to choose it on the original thalamus circle meaning
We're just a little bit too sloppy where exactly to place if I want to eliminate one of those points
Where should all be on one of the points exactly because we're not gonna even touch it
We're gonna leave it there and forget about it as our center of inversion
So since we promised we're going to end up with points a 1 B 1 C 1 which point shall we lose?
G that's right. That is exactly
Where we want to place our center of inversion?
Now we have to cut and I have to draw another picture because I cannot just to copy. It's too messy
Our first attempt failed, but we don't give up the blue circle is now going to go into a line L
but we have only three points to map points B C and a so we've got a 1
B 1
and C 1
Okay, great. So that was what inversion promised us
We can get it to this simplified situation. And we do now that if I add the length
Segments the two shorter ones
I'm going to get the longer one a 1 B 1 plus B. See one is a 1 C 1 mmm
So, what did we talk about what happens to distances on the inversion?
No, we didn't but I actually need exactly to know what happens to the old
distances
Here, I have 6 of them that participate in Thomas theorem in terms of the new distances have a scale
exactly, and we know that inversion is not like
Dilation or a scaling of the plane. He does actually quite horrible things to points
For example, if you have two points that are very close to the center of inversion
Each of them is going to be mapped very very far away and that will enlarge the segments drastically
So we need to derive a couple of formulas that will tell us exactly what happens to distances on the inversion
So let's get started. This is
the circle of inversion a point X
Outside the circle there is something that we have
overlooked here
There is a segment that is missing which one it is segment ot. That is
super important
First of all, it's hour long
That's the radius of inversion
But there is another I tango every time you draw a tangent. It is perpendicular to the radius
So now when you look at this picture, you see a bunch of similar triangles
There are three triangles in this picture too small and one large and they are all similar because they have the same angles
I'm interested only in two of them in this small red one and in the large one. So, let's see what they will give us
triangle Oh
T
X1 is similar to triangle o X
T and you can check that the right angle so matching so what do I need from this? I need a ratio of segments all
t2o
x1 is equal to o
X2 all t all right big deal, but you know what? Oh, geez Shaymin, it's the radius so I can simplify this
To multiplying the two distances from the center Oh X and Oh X
1/2 the radius squared
Which is a chi squared and this is our first
distance formula in other words all the distance from the center times
New distance from the center is always fixed
It is R squared which actually explains the phenomena white points close to the center gonna be mapped very very far away
Because those two distances all the new have to balance each other to multiply to R squared
So if one of them is small the other one has to be large and vice versa. Okay, that's formula 1 now
We need formula 2
Right here so what we have here
Is another inversion?
We really need to know what happens to any segment and the inversion not just two segments
That contain the center because we found formulas for those but not two random other segments
so what we have here is point a that goes to a 1 and B goes to B 1 if you want the weird the
Circle of inversion is it's somewhere like this in between them, but who cares about that?
We don't need it. Now again. We see two triangles here
Do you think they're similar the answer is yes, first of all, they share a common angle. That's good news
And now what do we know about?
These distances from the center
we just had a formula if I multiply away or a 1 that will be the same as
OB or B 1 but that turns into a ratio Oh a to OB
is equal to o be
1/2 Oh a 1
Obi-wan kenobe, right
what we have here is a ratio of
sides
Plus an angle makes two similar triangles and we can now read
Defects of this similarity we can well leave off
Defects can we say that report not to leap leap?
When you are you're gathering the harvest we can okay, I can treat the effects. Okay?
We can read the effects of the similarity. First of all, what are the two triangles? We are talking about we have
Triangle Oh a 1 B 1 is similar to triangle. Oh
But it's not gonna be it's gonna be B because the points flip-flop in this ratio, okay
so now what I'm interested in
Only is the new distance between the points in terms of the old one
So let's use some ratios a 1 B 1
Over or a 1 is equal to a B
over
OB, all right that looks like a mess but we're gonna resolve it we're interested in the new distance a 1 V 1
So we solve for it. We push away one to the other side. We get a B
Times Oh a 1 over OB now. This is a complete mishmash of new and old distances
I have new distance a1 b1 here. I have all distances. I don't like this one. This is new
I want my new distance to be completely expressed in something that happened before inversion
So I have control over can we replace all a 1 the answer is? Yes
Because we had our first formula that if I multiply the distances to the center of inversion
This is always R squared
So from here all a 1 will be some ratio of the radius and all a and we can simply plug it in
and get
This looks like a messy messy formula, but we really need to understand what it says
all right, so it says new distance is equal to
its
predecessor the old distance times always R squared over
all the distances to the center, so
Everything is expressed in terms of the world before inversion
We are ready to finally prove ptolemies theorem. Thank
Ok
But you have to be excited how all of this misses stuff actually rusev
I know you're excited, but you're not the one who is to edit the video afterwards
Okay final attempt, we are proving to Ptolemy theorem, yeah, but we will apply our distance formula
three times in the inverted world
Inverted world means really this red line here where everything mapped on the inversion
So we have a 1 B 1 plus B 1 C 1 is a 1 C 1 now
These are new distances so we can rewrite them in terms of our formula
So new distances were the old distance times the radius squared over the old
Distances to the center, but the center is not all real its D
So we will be talking about
Distances from a and B to the center, which is D in other words a G
Times B that alone is a1 b1 and we continue in a similar fashion
We have all distance, BC
Times radius squared and then distance is true the centre which is G and finally AC
times R squared over
distances to the
center
Hmm Brady, what would you like to do when things repeat get rid of them? That's right
So what do I get rid of them? You must be applicable if those are squares, correct?
We divide everything
Inside by R squared and we miraculously understand that the radius of inversion did not matter here
We just got rid on it
Ok
Next who likes denominators. Can we get rid of them true? The answer is yes
We just have to find the common denominator those three fractions now, they share a few things
They have a G twice and C D twice and B D twice. So we will multiply everything by
Ad times BD times CT. That's the common denominator now if I multiply this fraction by
This product we will in essence
Cancel the whole denominator
And the only thing that will be left multiplied is CD so that CD must be multiplied by a B
So we end up with a b times c g. Does this remind us of something?
Ad times CD we have just multiplied the opposite sides of our cyclic quadrilateral
This looks promising
Let's see if the rest will match
Bc from the numerator what's missing here? I think only a d is missing ad
BC times ad opposite sides and that is equal to AC
one of our diagonals times
What's left
BD the other diagonal
We are done
We have just shown Ptolemies theorem
Is that easy?
Well, it wasn't an easy proof in the sense of the ideas that were used because inversion is definitely not easy
And it is not a natural thing
That one would just try on the spur of the moment
but the actual computations and the proofs really boiled down to similar triangles and
operations with fractions as well as
Realizing where to take our center of inversion. That was a key point
It was worth away
Now you ask. Is there anything else?
Useful that Ptolemies theorem can prove the answer is yes
Not only the Pythagorean theorem follows from Ptolemies theory
But a couple of really cool facts, let's look at them
Well, we've come this far why not watch the additional footage showing how Ptolemies theorem reveals a link between Pentagon's
and the golden ratio
So one of those rules is utterly impossible
We've also got a little bit of extra footage about inversion
In fact, this kind of property is used very often in solving problems with inversion
Thank you for bringing it up
And if that's not enough you can go back and watch number files
Classic video about epic circles links are in all the usual places on screen down in the video description