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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 at that 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? =/ .
• 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.
• 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?
• 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.
• 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...
• 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.
• Could you also have solved this by complex bashing? I mean cis(theta) rotations of a complex number with respect to the origin.
• Nice video!!
Also how is 15x^5 multi plied by 6 = sin789
• You know your brain is wondering when you view the arrows as cars and realize that they have collision point D. :(
• 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?
• At , 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.