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### Course: Math for fun and glory > Unit 4

Lesson 1: 2003 AIME- 2003 AIME II problem 1
- 2003 AIME II problem 3
- Sum of factors of 27000
- Sum of factors 2
- 2003 AIME II problem 4 (part 1)
- 2003 AIME II problem 4 (part 2)
- 2003 AIME II problem 5
- 2003 AIME II problem 5 minor correction
- Area circumradius formula proof
- 2003 AIME II problem 6
- 2003 AIME II problem 7
- 2003 AIME II problem 8
- Sum of polynomial roots (proof)
- Sum of squares of polynomial roots
- 2003 AIME II problem 9
- 2003 AIME II problem 10
- Trig challenge problem: area of a triangle
- 2003 AIME II problem 12
- 2003 AIME II problem 13
- Trig challenge problem: area of a hexagon
- 2003 AIME II problem 15 (part 1)
- 2003 AIME II problem 15 (part 2)
- 2003 AIME II problem 15 (part 3)

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# Trig challenge problem: area of a hexagon

Sal solves a very complicated geometrical trig problem that appeared as problem 14 in the 2003 AIME II exam. Created by Sal Khan.

## Want to join the conversation?

- I don't understand how Sal can claim at2:45that at F y can't be 6. If B's b were greater, couldn't that allow y = 6 at F? I realise you'd have to get back to 4, but I can't see how that proves that B and F have to be at y = 2 and 4. In fact, I sketched a hexagon that seems to show otherwise. (
**Edit**: on second thought, it seems that making the required parallels means that going to y=6 from A makes it concave, but this isn't explained at all, and I can't prove it to myself.) Can someone explain that better? =/ .(20 votes)- If F was at y = 6 then segment AF (and all sides) would have to be at least 6 long. CD would be parallel to this, so it would have to go between y = 4 and y= 10, which are the only two points left that are at least 6 apart. DE is parallel to AB. But if D is at y = 10, this puts E at y =12. No good. D at y = 4 and C at y = 10 doesn't work either because then BC (and therefore all sides) would have to be at least 8 long. And it would also require DE to go between 4 and 6, which doesn't work because F would already be at 6.(8 votes)

- In the subtitle Sal says "Trigonometry and geometry to find the area of an equilateral (but not regular) hexagon". What is the difference between a regular and equilateral hexagon?(7 votes)
- A REGULAR HEXAGON has all its sides equal and its angles equal, while EQUILATERAL HEXAGON only has all its sides equal. There is also EQUIANGULAR HEXAGON which only has all its angles equal.(18 votes)

- I honestly don't get how he found out where some of the points of the hexagon are supposed to be (like how did he know that B was on the right side, that F was at four, and that E lay on the y axis and so on...(7 votes)
- B could be either to the left or right of the y axis.

You can see that E needs to be on the y axis because it's an equilateral hexagon. AF = EF, so the triangle AFE is going to be isosceles, thus the angles at the base are going to be equal. This means that the angle that AF forms with the y axis is equal to the angle FE forms with the y axis, and that must mean that E is on the y axis.

This is a wall of text, but Sal actually explains it (a bit faster) in the video.(6 votes)

- Could you also have solved this by complex bashing? I mean cis(theta) rotations of a complex number with respect to the origin.(3 votes)
- Nice video!!

Also how is 15x^5 multi plied by 6 = sin789(3 votes) - You know your brain is wondering when you view the arrows as cars and realize that they have collision point D. :((2 votes)
- It's really confusing.

AFE is Iscosceles triangle, and the angle F is 120

So the angle FAE and FEA both must be equal to 30

And angle FAB is 120

So angle EAB must be 90. And there's no place for Theta.

And from here I think 'Point E'must be in 2nd quadrant?(2 votes) - At4:18, how can you be certain that point E will take you back to the Y-Axis? Why CAN'T E have a positive x value (that is to the right of the Y-Axis)? Is there some sort of proof to that? Because I could imagine a Hexagon that fits all the criteria that have vertex E not on the Y-Axis. Please explain.(1 vote)
- We know that vertex E has to have a y-coordinate of 8. Sal already used up the y-coordinates 0, 2, 4, and 6, and if Sal wanted to get a y-coordinate of 10, he would need a side length of at least 6, but the side length is only 5.

So, that means that the change in y has to be 4. Since the total side length is 5, the change in x has to be 3, by the Pythagorean theorem. The change in x also has to be rightward, since if it were leftward A, F and E, would be collinear, and that can't happen.(3 votes)

- I think it's not very helpful to just throw point E at (0,10). Sal could've (or should've, as math teachers would say) proved that, since it's not a plain given fact(1 vote)
- I mean, if you really wanted to you can figure it out through point slope fform or whatnot. This whole video could've been simplified if Sal Khan knew the Shoelace Algorithm.(1 vote)

- I don't understand how Sal can claim at2:45that at F y can't be 6. If B's b were greater, couldn't that allow y = 6 at F? I realize you'd have to get back to 4, but I can't see how that proves that B and F have to be at y = 2 and 4. In fact, I sketched a hexagon that seems to show otherwise. (Edit: on second thought, it seems that making the required parallels means that going to y=6 from A makes it concave, but this isn't explained at all, and I can't prove it to myself.) Can someone explain that better? =/ . If they can, I would like to understand how you or another person got that answer. I would like to hear someones explanation on that. I did the exact same thing that Erick did approximately 4 years ago, and I understand his meaning, note the edit.(1 vote)

## Video transcript

Let A equal 0, 0 and
B equal lowercase b2 be points on the
coordinate plane. Let ABCDEF be a convex
equilateral hexagon. Convex means that
it's not concave. A concave hexagon
would look like this. That's 2 sides-- 3, 4, 5, 6. This would be a concave hexagon. So it's going to be popped out. And all the sides are
going to be equal. So it's an equilateral hexagon. They're not telling us that
it's a regular hexagon. So we don't know that
all of the angles are going to be the same. But all the sides
will be the same. Such that FAB is
equal to 120 degrees. Then they show us
a bunch of sides that are parallel to each other. And then, the y-coordinates
of its vertices are distinct elements of
the set 0, 2, 4, 6, 8, 10. The area of the hexagon could
be written in the form m square root of n, where m and n are
positive integers and n is not divisible by the
square of any prime. That's just a
fancy way of saying that we've simplified this
radical as much as possible. Find m plus n. So really the first
part-- let's just make sure we can
visualize this hexagon. So let me draw-- we know
1.1 vertex for short 0, 0. So let me draw my x-axis. That is my x-axis. Right over there. And then my y-axis. My y-axis would look like that. Y-axis. We know that the vertex
A sits at the point 0,0. That is the vertex A. Now, we know that
all of the vertices have y-coordinates that are
either 0, 2, 4, 6, 8, 10. And they are distinct
members of the set. Which means no 2 of the vertices
share the same y-coordinate. So they're not going to be
on the same horizontal line. So let me draw these horizontal
lines-- the x-axis is 0. Then you have y is equal to 2. Then you have 4, then you
have 6, and then you have 8, and then you have 10 up here. Now, B we already know. So first of all, we've
already used up to the 0 for A. A Is already
using up the 0. B uses of the 2-- they tell
us that the y-coordinate of B is 2. So we use that as well. Let me see if I can
draw B over here-- It sits on this
horizontal someplace. And the hexagon
has side length S, where we don't know
what that length is, but they're all the same. So let's just call
this S it's going to help me think about it. Now that I know that this
is equilateral equaliateral hexagon, all the sides are
going to be the same length. We're going to go out here
the coordinate B comma 2. We don't know what B is,
but that is our vertex B. Now, F is the other vertex
that is connected to A. F cannot sit on
this horizontal-- cannot sit on y is equal to 2. It can't sit on y is
equal to 6, because then this distance
would be super far. Clearly much further than
this distance over here. Or, actually, you
could have that. But then you would
you wouldn't be able to draw really
a convex hexagon. The next vertex is just going to
have to sit on this horizontal. So it's going to be S away. Maybe it will be
something like that. So let me draw it-- so
that is the next vertex. That is vertex F. Because
we're going A, B,C,D, E, F, and then back to A. Fair enough. Now what about vertex C? Well, vertex C can't
be on the 4 horizontal. So it's going to have to
be on the 6 horizontal. So vertex C is going to have
to be someplace like that. That's vertex C. And once again,
that length is S this length is S. Now what about vertex E? Can't be on the 6 horizontal,
already taken up by vertex C. So the 4 the 6 are taken up. So it has to be at
the 8 horizontal. And so this is length S. And we also know that we're
going back to the origin now. So this is the
vertex E right here. We know that we're going
back to the same x value, this is going to be
on the y-intercept. And the reason why
we know that is this is the length S and
in this is length S. And both of these
diagonals travel the same vertical distance. This base is 4. This base is 4. So you could kind of view
this as two right triangles. Both of them have base
4 and hypotenuse S and so they share this
side right over here. So this one goes out to the
left that distance and then this one's going to have
to come back that distance. Now, by the same
logic over here, this guy's going to
have to come back. So we can now use the
10 coordinate, the 10 y horizontal, or the
y-coordinate of 10. That's the only one we
haven't used yet for D. And since we came out when we
had a diagonal of length S, traveling 4 up, this
time, same logic-- we had a diagonal of length
S. It traveled up 4 over here and it moved out this distance. When we go back in the other
direction and travel up 4 you're going to go back
in the same direction. So this is going to be directly
on top of B. The coordinate for D is actually
going to be B comma 10. The y-coordinate here is 10. And there we have our hexagon. We're done drawing
our actual hexagon. And all this parallel line
information they told us-- AB is parallel to DE. And this is kind of obvious
here, BC is parallel to EF. And then they say CD
is parallel to FA. And the way that we
drew it, it looks pretty clear that is the case. Now, we need to
find the area, we need to find the
area of this hexagon. And it seems like a
good starting point would be to figure
out what S is. And to figure out
what S is, it's really going to be a
function of how much we've inclined this thing. So let's draw--
and you could see that since this isn't an
equally angular hexagon, that this is kind of skwed. We distorted it a little bit. But all the sides
are the same length. So let's just call this theta. Let's call that angle
right over there theta. And then they tell us that
angle FAB is 120 degrees. That is 120 degrees. So this angle over
here on the left, is going to be 180
minus 120 minus theta. So 180 minus 120 is 60. So this angle over
here 60 minus theta. Now the reason why I
did that because we have some information. We know that we
traveled up 4 over here. And we know that we
traveled up 2 over here. And maybe we can
use that information to solve for S. Because S
is the hypotenuse of both of these right triangles
that I just constructed. Let me draw them-- So this
right triangle right over here, I could draw it like this. So I have S. I have theta. And I have 2, that's this
right triangle right over here. This right triangle
looks like this. This angle is 60 minus theta. And this height over here is 4. So let's see what we can do
to solve for S. This triangle on the left, or it was
on the right over here, this triangle says
predict the sine of theta. The sine of theta is
equal to the opposite over the hypotenuse. It's equal to 2 over S. This triangle tells us that
the sine-- and remember this hypotenuse
over here is also S-- the sine of 60 minus
theta is equal to 4 over S. And if we want to set
these equal to each other, we can multiply this
guy by 2 on both sides. You could say 2 sine
of theta is equal to 4 over S. Sine of 60 minus 8
is also equal to 4 over S, so we could set them
equal to each other. So we have 2 sine of
theta is equal to sine of 60 minus theta. And then we could use some
of our trig identities, we know the sine of A
minus B is the same thing. The sine of A minus B, this is
equal to the sine of A times the cosine of B. Or I should
say theta in this case. So sine of 60 minus theta--
this is just a standard trig identity, it's called the
difference, sum and difference identity-- minus cosine of
60 times the sine of theta. And all this is equal
to 2 sine of theta. Well sine of 60 degrees-- This
is the square root of 3 over 2. Cosine of 60 degrees is 1/2. So we could add 1/2 sine
theta to both sides of this. And what we're going to get? So we're going to
add 1/2 sine theta. Then this guy's
going to go away. And then you add 1/2
sine theta to 2 sine of theta, which is
really 4/2 sine of theta. So that's going to be
5/2 the sine of theta. So I'm just adding 1/2
sine theta to this. So it's 5/2 sine of theta
is equal to square root of 3 over 2 cosine of theta. Right? I added 1/2 sine theta to both
sides of this to get this. I can multiply both sides
by 2 just to simplify it. So I get 5 sine of theta. 5 sine of theta is
equal to the square root of 3 cosine of theta. Now, I want to use the
identity sine squared of theta plus cosine squared
of theta is equal to 1. So let me just square
both sides of that. That'll also help us
with this radical. So we'll get 25 sine
squared of theta is equal to square
root of 3 squared is 3. Instead of writing
cosine squared of theta, let's just write-- that's 1
minus sine squared of theta. Right? Cosine squared theta is 1
minus sine squared of theta. I just squared both sides. Let me just write
what I just did-- I just squared to both sides. And so we get 25
sine squared theta is equal to 3 minus
3 sine squared theta. We can add 3 sine squared
theta to both sides-- We get 28 sine squared
theta is equal to 3. Or, that the sine squared
of theta-- homestretch-- is equal to 3 over 28. Or, we could even write
that sine of theta is equal to the square
root of 3 over 28. So it's equal to the
square root of 3 over 28. Now, we could simplify
that, 28 is 4 times 7, we could take it out. But that's good enough
for now, maybe we'll simplify it later if we have to. Sometimes these are
easier to deal with. So let's see over [INAUDIBLE]. So we have the sine of theta. And now we can relate that,
actually, to S over here. We know that, before I
messed with this thing, that the sine of theta is equal
to 2 over S. Or that S over 2 is equal to 1 over
sine of theta. Or that S is equal to 2
over the sine of theta. Well, we know what
sine of theta is, the square root of 3 over 28. So S is equal to 2
divided by sine of theta. That's like multiplying by
the inverse of sine of theta. So that's 2 times the
square root of 28 over 3. So we figured out our S. 2
times this thing over here. Now, given that we
know an S, let's see how we can
figure out the area. Well, what immediately
pops out is that we have this triangle
over here that has height, or I should say maybe its
base, if you view it sideways, its base is 8. And this distance
right over here, we should be able to figure out
using the Pythagorean theorem. Because we know that this
distance, right over here, is 4. We know that this distance,
the hypotenuse is S. So we could call this the
height of it, right over here. We could say that H squared
plus 4 squared plus 16 is equal to the hypotenuse
squared, is equal to S squared. S squared S is this
thing over here. So if we want to square S it
becomes 4 times 28 over 3. And we just subtract
16 from both sides. So H is equal to
4 times 28 over 3, minus, if I want
to write 16 over 3. Or if I want it, 16 to
something over three, that's going to be
minus 48 over 3. And let's see, I don't want to
have to multiply 4 times 28. So you could write
48 as 4 times 12. So this numerator is going to be
4 times 28 minus 12 over-- now, remember that's H squared. I should say-- H
squared is going to be 4 times 28
minus 12 over 3. Which is equal to
4 times 16 over 3. Which is equal to 64 over 3. That's H squared. So H is going to be the
square root of that, which is 8 over the
square root of 3. So H right over here is 8
over the square root of 3. So if I want to find the area
of this whole thing over here-- well first, let's find the
area of this small thing right over here. That's just going
to be H times 4. So it's going to be-- well,
I could do it either way. But let's just say this
is H times 4 times 1/2. So it's going to be 2 times--
so the area of this triangle-- let me do it in blue right
here-- this triangle's area is going to be H, which is 8 over
the square root of 3 times 4 times 1/2. So this guy right
over here is just going to be 2 times 8
over the square root of 3, or that's going to be 16
over the square root of 3. So this guy over here is 16
over the square root of 3. Now we have a bunch of
it, we have this guy, and now this guy's going to have
the exact same area, right over there. And then you have
this guy who's going to have the exact
same area again. Same exact logic, same exact
logic, same base, same height. They're actually congruent. So you have 4 of
these triangles. So you are going
to multiply by 4 If you want the
area of this area that I've are initiated in. So 4 times this. We're at 64 over the
square root of 3. Now the only area we
have left to figure out is the area of
this parallelogram. in the middle. Now, we know the base
of the parallelogram. The base of this
parallelogram is 8. We just have to
figure out its height. And once again, we can use
the Pythagorean theorem. So I'll call this, I
dunno, we already used H, well I'll use H again. But you just have to
remember that this is a different height over here. This base over here
is of length 2. I know it's hard to read now. So we can now write that H
squared plus 4 plus 2 squared is equal to S squared. Now we already figured out
what S squared was in the past. It's 4 times 28 over 3. Let's subtract 4
from both sides. You subtract 4 there. So minus 12 over 3. And now let's see, 12 is
the same thing as 4 times 3. So this is equal to
4 times 28 minus 3. So that's 4 times 25 over 3. Which is equal to 100 over 3. That's H squared. So this H is going to be equal
to the square root of this, which is 10 over the
square root of 3. So this distance
right over here is 10 over the square root of 3. So if we want the area
of this parallelogram, it's going to be
that height, times the base of the parallelogram. So the parallelogram is going
to be 8 times 10 square roots of 3, or 80 square roots of 3. 80-- oh no, let me be
very careful-- this was 10 over the square
root of 3 is this height. So the whole area
of this paralelagram is 8 times 10 over
the square root of 3. It's 80 over the
square root of 3. So our entire area now, if
we add everything together, we have 64 square roots of
3 for these 4 triangles, plus 80 over square root of 3. So let's add it together. So we have 80 over square roots
of 3 for the parallelogram, plus 64 over square roots
of 3 for the triangle parts. And this is equal to 144
over the square root of 3. We can rationalize
the denominator. So times the square root of
3, over the square root of 3. And in the denominator,
now, we're going to get a 3. 144 over 3 is going to be what? That's 48. Right? Three times 40 is
120, 3 times 8 is 24. So it's going to be
48 square roots of 3 for the area of
our entire hexagon. And so, we have in the form--
48 square roots of 3-- So if you want to find m plus
n is 48 plus 3, which is 51. That was a tiring problem. My brain started to
fry near the end of it. I had trouble keeping
track of things. Anyway, hopefully
you enjoyed that.