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## Solving modeling problems with similar & congruent triangles

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

# Geometry word problem: the golden ratio

CCSS Math: HSG.SRT.B.5

## 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.