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This right over here[br]is a self-portrait

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that Rembrandt made[br]in 1640, and what's

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really interesting about it[br]is like other great artists

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like Leonardo da Vinci[br]and Salvador Dali

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and many, many, many[br]others, Rembrandt really

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cared about something[br]called the golden ratio.

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And I've done whole[br]videos about it.

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And it's this fascinating,[br]fascinating, fascinating number

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that's usually denoted[br]by the Greek letter phi.

0:00:27.195,0:00:28.570
And if you were[br]to expand it out,

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it's an irrational[br]number, 1.61803,

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and it just goes on[br]and on and on forever,

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but there's some very neat[br]mathematical properties of phi,

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or the golden ratio.

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If you start with phi, and[br]if you were to add to that,

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or actually let's[br]start it this way.

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If you were to start with 1[br]and add to that 1 over phi.

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Let me write my phi[br]a little bit better.

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You add to that 1 over[br]phi, that gives you phi.

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So that's kind of a neat thing.

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Now, if you were to multiply[br]both sides of this equation

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by phi, you get that,[br]if you start with phi,

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and then if you add 1,[br]you get phi squared.

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So it's a number you add[br]1, you get its square.

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These are all really,[br]really neat things.

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It can even be written[br]as a continued fraction.

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Phi could be rewritten as 1 plus[br]1 over 1 plus 1 over 1 plus 1

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over, and we just go like that[br]forever and ever and ever.

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That also gives you phi.

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So hopefully this gives[br]you a little appreciation

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that this is a[br]really cool number.

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And not only is it[br]cool mathematically,

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but it shows up[br]throughout nature,

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and it's something that artists[br]have cared about because they

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believe that it helps[br]define human beauty.

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And we see that Rembrandt[br]really cared about it

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in this painting.

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And how can we tell?

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Well that's what we're going[br]to analyze a little bit

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through this exercise[br]in this video.

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We can construct a triangle.

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Obviously these triangles aren't[br]part of his original painting.

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We superimposed these.

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But if you were to put a[br]base of a triangle right

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where his arm is[br]resting, and then

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if you were to have the[br]two sides of the triangle

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outline his arms and shoulders[br]and then meet at the tip right

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at the top of this arch, you[br]would construct triangle ABD

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just like we have here.

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And then if you were[br]to go to his eyes,

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and you could imagine[br]human eyes are

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what we naturally look at,[br]whether we're looking at a face

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or a painting of a face.

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If you look at his eyes, and if[br]you were to draw a line there

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that's parallel, well, that[br]really connects the eyes,

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and that's parallel to the[br]BD right over here-- so let's

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call that segment PR[br]right over there--

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we'll see that this ratio,[br]the ratio between this smaller

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triangle and this larger[br]triangle, involves phi.

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So this is what we[br]know, what we're

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being told about this painting,[br]and this is quite fascinating.

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The ratio between the length of[br]segment CD and BC is phi to 1.

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So you drop an altitude[br]from this larger triangle,

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this ratio, the ratio of CD, the[br]length of CD to BC, that's phi.

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So clearly, Rembrandt[br]probably thought about this.

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Even more, we know that[br]PR is parallel to BD.

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We've actually[br]constructed it that way.

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So that is going to be parallel[br]to that right over there.

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And so the next clue[br]is what tells us

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that Rembrandt really[br]thought about this.

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The ratio of AC to AQ.

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So AC is the altitude[br]of the larger triangle.

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The ratio of that[br]to AQ, which is

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the altitude of this top[br]triangle, that ratio is phi

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plus 1 to 1, or you could even[br]say that ratio is phi plus 1.

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So clearly, Rembrandt[br]thought a lot about this.

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Now using all that[br]information, let's

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just explore a little bit.

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Let's see if we can come[br]up with an expression that

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is the ratio of the[br]area of triangle ABD,

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so the area of the[br]larger triangle,

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to the area of triangle APR.

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So that's this smaller[br]triangle right up here.

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So we want to find the ratio of[br]the area of the larger triangle

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to the area of the[br]smaller triangle,

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and I want to see if we[br]can do it in terms of phi,

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if we can come up with some[br]expression here that only

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involves phi, or[br]constant numbers,

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or manipulating phi in some way.

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So I encourage you to pause the[br]video now and try to do that.

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So let's take it step by step.

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What is the area of a triangle?

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Well, the area of any triangle[br]is 1/2 times base times height.

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So the area of[br]triangle ABD we could

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write as 1/2 times our base.

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Our base is the[br]length of segment BD.

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So 1/2 times BD.

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And what's our height?

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Well that's the[br]length of segment AC.

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1/2 times BD-- Maybe[br]I'll do segment AC.

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Well, let me do it[br]in the same color--

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times the length of segment AC.

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Now what's the area?

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This is the area[br]of triangle ABD.

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1/2 base times height.

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Now what's the area[br]of triangle APR?

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Well, it's going to be 1/2 times[br]the length of our base, which

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is PR, segment PR,[br]the length of that,

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times the height, which is,[br]the height is segment AQ,

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so the length of segment[br]AQ, we could just

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write it like that, times[br]the length of segment AQ.

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So how can we simplify[br]this a little bit?

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Well, we could divide[br]the 1/2 by the 1/2.

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Those two cancel out.

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But what else do we know?

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Well, they gave us the[br]ratio between AC and AQ.

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The ratio of AC to AQ right[br]over here is phi plus 1 to 1.

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Or we could just say[br]this is equal to phi.

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Or we could say this is[br]just equal to phi plus 1.

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So let me rewrite this.

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Actually, let me[br]write it this way.

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This is going to[br]be equal to-- So we

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have the length of segment BD[br]over the length of segment PR,

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and then this part right[br]over here we can rewrite,

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this is equal to[br]phi plus 1 over 1.

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So I'll just write it that way.

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So times phi plus 1 over 1.

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So what's the ratio of BD to PR?

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So the ratio of the base of[br]the larger triangle to the base

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of the smaller triangle.

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So let's think about[br]it a little bit.

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What might jump out at you[br]is that the larger triangle

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and the smaller triangle, that[br]they are similar triangles.

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They both obviously[br]have angle A in common,

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and since PR is[br]parallel to BD, we

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know that this angle[br]corresponds to this angle.

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So these are going to[br]be congruent angles.

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And we know that this[br]angle corresponds

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to this angle right over here.

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So now we have three[br]correspondingly angles

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are congruent.

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This is congruent to itself,[br]which is in both triangles.

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This is congruent to this.

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This is congruent to that.

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You have three[br]congruent angles, you're

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dealing with two[br]similar triangles.

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And what's useful[br]about similar triangles

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are the ratio between[br]corresponding parts.

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Corresponding lengths of[br]the corresponding parts

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of the similar triangles[br]are going to be the same.

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And they gave us[br]one of those ratios.

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They gave us the ratio of the[br]altitude of the larger triangle

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to the altitude of[br]the smaller triangle.

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AC to AQ is phi plus 1 to phi.

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But since this is true[br]for one corresponding part

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of the similar[br]triangles, this is

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true for any corresponding[br]parts of the similar triangle,

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that the ratio is going[br]to be phi plus 1 to 1.

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So the ratio of BD, the ratio of[br]the base of the larger triangle

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to the base of the[br]smaller one, that's

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also going to be[br]phi plus 1 over 1.

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Let me just write it this way.

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This could also be rewritten[br]as phi plus 1 over 1.

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So what does this simplify to?

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Well, we have phi plus 1 over[br]1 times phi plus 1 over 1.

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Well, we could just divide by 1.

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You're not changing the value.

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This is just going[br]to be equal to,

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and we deserve a drum roll now.

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This is equal to[br]phi plus 1 squared.

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So that was pretty neat.

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And I encourage you to even[br]think about this because we

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already saw that phi plus[br]1 is equal to phi squared,

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and there's all sorts of[br]weird, interesting ways

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you could continue[br]to analyze this.