- Trig challenge problem: area of a triangle
- Trig challenge problem: area of a hexagon
- Trig challenge problem: cosine of angle-sum
- Trig challenge problem: arithmetic progression
- Trig challenge problem: maximum value
- Trig challenge problem: multiple constraints
- Trig challenge problem: system of equations
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.
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- 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?(6 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.(15 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...(5 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)
- 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)
- You know your brain is wondering when you view the arrows as cars and realize that they have collision point D. :((1 vote)
- 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?(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)
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.