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Geometry word problem: the golden ratio

Triangle similarity, the Golden ratio, and art, all converge in this inspiring video! Created by Sal Khan.

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Video transcript

This right over here is a self-portrait that Rembrandt made in 1640, and what's really interesting about it is like other great artists like Leonardo da Vinci and Salvador Dali and many, many, many others, Rembrandt really cared about something called the golden ratio. And I've done whole videos about it. And it's this fascinating, fascinating, fascinating number that's usually denoted by the Greek letter phi. And if you were to expand it out, it's an irrational number, 1.61803, and it just goes on and on and on forever, but there's some very neat mathematical properties of phi, or the golden ratio. If you start with phi, and if you were to add to that, or actually let's start it this way. If you were to start with 1 and add to that 1 over phi. Let me write my phi a little bit better. You add to that 1 over phi, that gives you phi. So that's kind of a neat thing. Now, if you were to multiply both sides of this equation by phi, you get that, if you start with phi, and then if you add 1, you get phi squared. So it's a number you add 1, you get its square. These are all really, really neat things. It can even be written as a continued fraction. Phi could be rewritten as 1 plus 1 over 1 plus 1 over 1 plus 1 over, and we just go like that forever and ever and ever. That also gives you phi. So hopefully this gives you a little appreciation that this is a really cool number. And not only is it cool mathematically, but it shows up throughout nature, and it's something that artists have cared about because they believe that it helps define human beauty. And we see that Rembrandt really cared about it in this painting. And how can we tell? Well that's what we're going to analyze a little bit through this exercise in this video. We can construct a triangle. Obviously these triangles aren't part of his original painting. We superimposed these. But if you were to put a base of a triangle right where his arm is resting, and then if you were to have the two sides of the triangle outline his arms and shoulders and then meet at the tip right at the top of this arch, you would construct triangle ABD just like we have here. And then if you were to go to his eyes, and you could imagine human eyes are what we naturally look at, whether we're looking at a face or a painting of a face. If you look at his eyes, and if you were to draw a line there that's parallel, well, that really connects the eyes, and that's parallel to the BD right over here-- so let's call that segment PR right over there-- we'll see that this ratio, the ratio between this smaller triangle and this larger triangle, involves phi. So this is what we know, what we're being told about this painting, and this is quite fascinating. The ratio between the length of segment CD and BC is phi to 1. So you drop an altitude from this larger triangle, this ratio, the ratio of CD, the length of CD to BC, that's phi. So clearly, Rembrandt probably thought about this. Even more, we know that PR is parallel to BD. We've actually constructed it that way. So that is going to be parallel to that right over there. And so the next clue is what tells us that Rembrandt really thought about this. The ratio of AC to AQ. So AC is the altitude of the larger triangle. The ratio of that to AQ, which is the altitude of this top triangle, that ratio is phi plus 1 to 1, or you could even say that ratio is phi plus 1. So clearly, Rembrandt thought a lot about this. Now using all that information, let's just explore a little bit. Let's see if we can come up with an expression that is the ratio of the area of triangle ABD, so the area of the larger triangle, to the area of triangle APR. So that's this smaller triangle right up here. So we want to find the ratio of the area of the larger triangle to the area of the smaller triangle, and I want to see if we can do it in terms of phi, if we can come up with some expression here that only involves phi, or constant numbers, or manipulating phi in some way. So I encourage you to pause the video now and try to do that. So let's take it step by step. What is the area of a triangle? Well, the area of any triangle is 1/2 times base times height. So the area of triangle ABD we could write as 1/2 times our base. Our base is the length of segment BD. So 1/2 times BD. And what's our height? Well that's the length of segment AC. 1/2 times BD-- Maybe I'll do segment AC. Well, let me do it in the same color-- times the length of segment AC. Now what's the area? This is the area of triangle ABD. 1/2 base times height. Now what's the area of triangle APR? Well, it's going to be 1/2 times the length of our base, which is PR, segment PR, the length of that, times the height, which is, the height is segment AQ, so the length of segment AQ, we could just write it like that, times the length of segment AQ. So how can we simplify this a little bit? Well, we could divide the 1/2 by the 1/2. Those two cancel out. But what else do we know? Well, they gave us the ratio between AC and AQ. The ratio of AC to AQ right over here is phi plus 1 to 1. Or we could just say this is equal to phi. Or we could say this is just equal to phi plus 1. So let me rewrite this. Actually, let me write it this way. This is going to be equal to-- So we have the length of segment BD over the length of segment PR, and then this part right over here we can rewrite, this is equal to phi plus 1 over 1. So I'll just write it that way. So times phi plus 1 over 1. So what's the ratio of BD to PR? So the ratio of the base of the larger triangle to the base of the smaller triangle. So let's think about it a little bit. What might jump out at you is that the larger triangle and the smaller triangle, that they are similar triangles. They both obviously have angle A in common, and since PR is parallel to BD, we know that this angle corresponds to this angle. So these are going to be congruent angles. And we know that this angle corresponds to this angle right over here. So now we have three correspondingly angles are congruent. This is congruent to itself, which is in both triangles. This is congruent to this. This is congruent to that. You have three congruent angles, you're dealing with two similar triangles. And what's useful about similar triangles are the ratio between corresponding parts. Corresponding lengths of the corresponding parts of the similar triangles are going to be the same. And they gave us one of those ratios. They gave us the ratio of the altitude of the larger triangle to the altitude of the smaller triangle. AC to AQ is phi plus 1 to phi. But since this is true for one corresponding part of the similar triangles, this is true for any corresponding parts of the similar triangle, that the ratio is going to be phi plus 1 to 1. So the ratio of BD, the ratio of the base of the larger triangle to the base of the smaller one, that's also going to be phi plus 1 over 1. Let me just write it this way. This could also be rewritten as phi plus 1 over 1. So what does this simplify to? Well, we have phi plus 1 over 1 times phi plus 1 over 1. Well, we could just divide by 1. You're not changing the value. This is just going to be equal to, and we deserve a drum roll now. This is equal to phi plus 1 squared. So that was pretty neat. And I encourage you to even think about this because we already saw that phi plus 1 is equal to phi squared, and there's all sorts of weird, interesting ways you could continue to analyze this.