ISS?

The simplest and most beautiful triangle in the world.

Brady Who?

The Bermuda Triangle.

Well, is it of the pipe of triangles?

Times thinking.

I couldn't think of another famous trying.

You mean unequal.

Actual drunk, equilateral triangle.

That's what I'm thinking of.

Well, there we go.

Unequal lateral triangle which inscribed in a circle.

Let the steak any other point on this circle, for example, we take a point right here and we connect it with the Vergis is of this circle.

All right, so we're going to label these things B, C and D.

Ready?

What do you think could possibly be true about these red segments?

It could be true that b and C are the same as day at it together or yes, Amazingly, it is actually true.

It's true.

Yes, be Blast C is indeed equal to D.

But not the factorial, not deep factorial.

Definitely just on exclamation point.

Now their ways to prove this without parliament.

But since we have worked so hard to prove Ptolemy, why not use it?

Do we see 1/4 lateral?

Yeah, absolutely.

I need those green sides.

I don't know what they are, but looks like calling them all aces.

Good, Because this is an equilateral triangle.

So biped Allah Mystere Um what do we have?

The product of the opposite sides.

A B and a C.

If I am those products, I must get the product of the diagonals, The green?

They are the noise.

A diagonal is d.

So we're going to get a G again with your petition.

Let's kill it.

We divide by a and we get exactly what we want it.

The two shorter segments add up to the long.

I remember spending a lot of time on this problem in middle school.

I did not know about dollar Mystere.

Um and I did not know about inversion.

It is possible to do this problem with similar triangles, but they're not all drawn.

You have toe come up in a very imaginative way with those similar triangles.

Soap.

Ptolemy's theory really wings in terms of elegance and simplicity.

Okay, we have now a regular Pentagon are 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 quadrilateral, not the Pentagon.

So we're going to select randomly.

Four of the Vergis is, 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're all of equal length because they all participate in these triangles to off who sides are green.

These are from the pentacle.

It tell you, right?

We need to label things.

Let's say, a four, the sides of the Pentagon and we have two more, but we don't care about them.

We label be all of those diagonals.

There is really nothing else in the Pentagon there, only the sides and the day.

Agnes, what is their issue?

If I take the diagonal, which looks is the larger when divided by the side?

Which number do you think, Brady?

Are we going to get more?

It's a ratio could be the most famous ratio.

Okay, let's not get ahead of the game.

Let's see what happens.

Soaped Alamein tells us about this bottom quadrilateral that if I multiply the opposite sides, so a times B and the other opposite sides a times, eh?

I'm going to get the product of the two day agonal switches three times.

Be excellent.

Let's put everything on the wild side B squared minus a B minus.

A squared is equal to zero.

All right.

Any mathematician looking at this says this is a homogeneous equation.

I can do miracles with it.

In other words, which is divided by a squared and see what that would give us.

She'll be squared over a squared minus.

Now, one of those days will cancel be over a minus one cause a squared divided by a squared to swan.

And that zero well, here is the ratio that we were interested in be 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 one plus or minus one squared, plus four times one divided by two or one plus or minus square with the five divided bite, you wait.

Hold on, hold on.

We still have two choices.

Is it possible that for some Pentagon's you get one of those roots?

And for other Pentagon's the other route?

I don't think so.

One of those roots is utterly impossible.

The smaller route, which is one minus square root of five over two, is negative, but the larger route use positive, and it is the famous golden ratio.

So what we have shown using Thomas is that if you take any day agonal of a regular Pentagon and you divided but any one of its sides, you're going to get the golden ratio.

I think our journey was worth it.

This video was a continuation off their epic video about the proof of Ptolemy's theory.

Um, there are links on the screen and down in the description to that video.

On other stuff you might find interesting.

No matter where you are on this railroad, you build your station s.

People from both villages will walk exactly the same distance.