If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:8:23

CCSS.Math:

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 the golden ratio and I've done whole videos about it it's this fascinating fascinating fascinating number that's usually denoted by usually donated by the Greek letter Phi and if you were to expanded out it's an irrational number so you could six one eight zero three 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 if you were to add to that or actually let's start it this way if you were to start with one and add to that 1 over 1 over Phi let me write my file 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 the number you add 1 you get you get its square these are all really really neat things it can even be written as a continued fraction Phi could be re-written 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 word aren't part of his original painting we superimpose 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 kind of outline his arms and shoulders and then meet at the tip right at the top of this arch you would construct triangle a B D just like we have here and then if you were to go to his eyes and you could imagine human eyes or 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 it's parallel to this to the BD right over here so let's call that segment PR right over there we'll see that there's that this that this ratio the ratio between this smaller triangle and this larger triangle involves Phi so this is what we know that we're we're being told about this painting this is quite fascinating the ratio between seg the length of segment CD and BC is five to one so you drop an altitude from this larger triangle this ratio the ratio of CD length of CD to B see 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 this is also and though the next clue is what tells us that remember really thought about this the ratio of AC to aq so AC is the altitude of the larger triangle the ratio of that to a Q which is the altitude of this top triangle that ratio is five plus one to one or you could even say that ratio is Phi plus one so clearly Rembrandt thought a lot about this now using all of 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 ap are so that's this that's the smaller triangle right up here so we want to find the ratio of the area of the larger triangle to the area of area of the smaller triangle and I want to see if we can do it in terms of five if we can come up some with some expression here that only involves only involves spy or our 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 where the area of any triangle is 1/2 times base times height so the area of triangle abd we can 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 a/c 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 one half times our the length of our base which is PR segment PR the length of that times the height which is the height of 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 one half by the one half 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 the ratio of AC to aq right over here is 5 plus 1 2 Phi 2 1 or we could just say this is equal to 5 is it or we could say this is just equal to 5 plus one so let me rewrite this let me 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 5 plus 1 over 1 so I'll just write it that way so times 5 plus 1 over 1 so what's the ratio of BD to PR so bd2 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 in the smaller triangle that they are similar triangles they both obviously have angle a and 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 we know that this angle corresponds to this angle right over here so now we have three corresponding 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 they gave us the ratio of the altitude of the larger triangle to the altitude of the smaller triangle ac 2aq is five plus one to five but since this is true for one as 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 five plus one to one 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 five plus one over one so this is also this is also going to be Phi I could write this as let me just write it this way this this could also be rewritten as five plus one over one so what does this simplify to well we have five plus one over one times five plus one over one well we could just you know divide by one 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 one five plus one squared so that was pretty neat and I encourage you to even think about this because we already saw that five plus one is equal to Phi squared and there's all sorts of weird interesting ways that you can continue to analyze this