So, what is the simplest and most beautiful triangle in the world, Brady?
(Brady: Ohh...)
(The Bermuda Triangle?)
Well...
[Brady laughs] - Is it of the, type of triangle that I'm thinking of?
(Brady: I couldn't think of another famous triangle.)
(You mean an equilateral triangle.)
Equilateral triangle. That's what I'm thinking of.
Well, there we go. An equilateral triangle which, inscribed in a circle.
Let us take any other point on this circle.
For example, we take a point right here, and we connect it with the vertices of this circle.
Alright, so we're going to label these things b, c and d.
Brady, what do you think could possibly be true about these red segments?
(Brady: It *could* be true, that b and c are the same as d? Added together, or...)
- Yes.
Amazingly, it is actually true.
(Brady: That's true?)
- Yes! b + c is indeed, equal to d.
But why? - (Brady: Not "d" factorial?)
- Not d factorial, definitely, just an exclamation point.
[Chuckles] (Brady: Okay)
Now, there are ways to prove this without Ptolemy,
but since we have worked so hard to prove Ptolemy, why not use it?
Do we see a quadrilateral?
(Brady: Yeah!)
- Absolutely!
I need those green sides, I don't know what they are, but looks like...
calling them all as is good 'cause this is an equilateral triangle.
So by Ptolemy's theorem, what do we have?
The product of the opposite sides, ab and a,
if I add those products, I must get the product of the diagonals.
The green diagonal is a. The red diagonal is d. So we're going to get ad.
Hmm!
Again, we see repetition, let's kill it!
We divide by a, and we get exactly what we wanted.
The two shorter segments add up to the long one.
I remember spending a lot of time on this problem in middle school.
I did not know about Ptolemy's theorem, and I did not know about inversion.
It is possible to do this problem with similar triangles, but they are not all drawn;
you have to come up, in a very imaginative way, with those similar triangles.
So Ptolemy's theorem really wins in terms of elegance and simplicity.
Okay. We have now a regular pentagon, or, at least, my version, the closest I could draw by hand here.
Also, it is inscribed in a circle.
So what can Ptolemy tell us about this situation?
First of all, we want a quadrilateral, not a pentagon. So we are going to select randomly four of the vertices.
Let's say, the bottom four.
We now need to draw the remaining segments between them.
Those segments happen to be diagonals of the pentagon.
They are all of equal length, because they all participate in these triangles, two of whose sides are green - these are from the pentagon.
At any rate, we need to label things.
Let's say a for the sides of the pentagon,
and we have two more, but we don't care about them.
We label b, all of those diagonals.
There is really nothing else in a pentagon. There are only the sides and the diagonals.
What is their ratio?
If I take the diagonal which looks is the larger of the two and divide it by the side,
which number, do you think, Brady, are we going to get?
(Brady: I dunno!)
It's a ratio!
Could be the most famous ratio.
Okay, let's not get ahead of the game. Let's see what happens.
(Brady: Alright...)
So Ptolemy tells us about this bottom quadrilateral,
that if I multiply the opposite sides, so a times b,
and the other opposite sides, a times a,
I'm going to get the product of the two diagonals, which is b times b.
Excellent.
Let's put everything on one side:
b squared minus ab minus a squared is equal to zero.
Alright. Any mathematician looking at this says,
"Mmm, this is a homogeneous equation. I can do miracles with it."
In other words, let's just divide by a squared,
and see what that would give us.
So b squared over a squared minus...
- now one of those as will cancel -
...b over a, - 1 'cause a squared divided by a squared is one, and that's zero.
Hmm. Well here is the ratio that we were interested in: b over a.
If I write this ratio as x, we end up with a quadratic equation,
and when we solve it using the quadratic formula,
we get 1, plus or minus, 1 squared plus 4 times 1, divided by 2
or 1 plus or minus square root of 5 divided by 2.
Mm, wait, hold on- hold on! We still have two choices?!
Is it possible that for some pentagons you get one of those roots and for other pentagons the other root?
I don't think so.
One of those roots is utterly impossible.
The smaller root, which is one minus square root of five over two, is negative.
But the larger root is positive,
and it is the famous Golden Ratio.
So what we have shown, using Ptolemy's,
is that if you take any diagonal of a regular pentagon
and you divide it by any one of its sides, you're going to get the golden ratio.
I think our journey was worth it.
(Brady: This video was a continuation of our epic video about the proof of Ptolemy's theorem.)
(There are links on the screen and down in the description to that video and other stuff you might find interesting.)
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