Convex hexagon ABCDEF is equiangular but has no two sides the same length. Its sides in some order are 1, 2, 3, 4, 5 and 6 units long. If AB=1 and CD>BC, what are the lengths of BC, CD, DE, EF and FA?

Another convex hexagon is also equiangular and has sided measuring 1, X, 3, 4, 5, and 6 units long in that order going clockwise. What is the measure of X?

Since it is equiangular, that means the angle between two adjacent sides will always be 120 degrees.

Ok, it kind of makes sense to me that the 5 and 6 unit long sides should be opposite each other (so parallel). Otherwise two vertices of those two sides will be so far away from each other, that the other 4 smaller sides just couldn’t end up closing the hexagon.

Following this same thought process, I was pretty sure that the 1 and 2 unit long sides should be opposite each other. If they were too close, then we’d be making one area of the hexagon too tight, and the longer sides wouldn’t be able to close off.

And then it follows that the 3 and 4 unit long sides will be opposite as well.

I actually read the problem wrong for a second, and thought it said that BC>CD. So I thought "ok, well let’s make BC 6 then so that CD will have to be less than BC." Well, if I make this assumption, then it follows that DE will have to be 2, and EF will have to be 5. So I laid out the 1 and 6 units sides at a 120 degree angle to each other. Then I laid out the 2 and 5 unit sides at a 120 degree angle to each other, such that the 1 and 2 unit sides were parallel, and the 6 and 5 unit sides were parallel. I was then able to easily use the 3 and 4 unit sides to close the hexagon. So that clockwise I had 1, 6, 3, 2, 5, 4.

Then I realized I read the problem wrong, and CD<BC in my solution. Well, I flipped my hexagon over, and saw that clockwise I now had 1, 4, 5, 2, 3, 6. Well, now CD>BC, so this solution works. So:

BC = 4
CD = 5
DE = 2
EF = 3
FA = 6

That was a little lucky though. If I hadn’t found the solution right away, I think I would have done this:

Let’s just start with the idea that the 5 and 6 unit lengths should be opposite each other. So they are parallel. This means that some pairing of the 1, 2, 3 and 4 length sides must both have the same "height" (measured perpendicular to the 5 and 6 lengths). Well, if the 5 and 6 lengths are oriented horizontally (0 degrees to the x-axis), the other 4 sides must be at +60 or –60 degrees to the x-axis. Their heights, then, are L*sin(60) = L*sqrt(3)/2. So we have:

1: 1sqrt(3)/2
2: 2sqrt(3)/2
3: 3sqrt(3)/2
4: 4sqrt(3)/2

Well, the only pairing that works out is when you have 1 and 4 adjacent, and then 2 and 3 adjacent.

Ok, now what is horizontal distance between the two ends of each pair? Well, for the 4-1 pair, that distance is 4cos(60)-1cos(60) = 4/2-1/2 = 1.5. For the 2-3 pair, that distance is 3cos(60)-2cos(60) = 3/2-2/2 = 0.5. Remember that the difference between the 5 and 6 lengths is 1.

Somehow these 4 chunks must fit together. Somehow we must get the addition or subtraction of the 1-4 and 2-3 pairs horizontal distances to equal the difference between the 5 and 6 lengths. Why? Well, picture everything put together happily. Slide one of the pairs along the 5 length side until it touches the other pair. Now look at the other ends of the two pairs. Well, they used to be 6 units apart, but we moved them 5 units closer, so now the two ends are 1 unit apart.

So this means that +/- 1.5 +/- 0.5 = 1. Well, that’s easy to figure out. 1.5 – 0.5 = 1, or 1.5 – 0.5 – 1 = 0

What do the +/- parts of the equation mean? Well, if I define the horizontal distance of a pair as (horizontal distance of the top piece) – (horizontal distance of the bottom piece), this means that a + number occurs when the longer piece is on top, and a – number occurs when the shorter piece is on top.

How do we use this information to correctly assemble the chunks? Well, +1.5 for the 1-4 pair means that the 4 unit length is above the 1 unit length. The –0.5 of the 2-3 pair means that the 2 unit length is above the 3 unit length. And the –1 of the 5&6 unit lengths means that the 5 unit length is above the 6 unit length. It doesn’t matter which side (left or right) I put the two pairs because I will end up with a close hexagon either way, just a mirrored version of each other.

Ok, so this means to me that 5, 2, 3, 6, 1, 4 is a legal hexagon. In order to match the configuration given, we either have 1, 4, 5, 2, 3, 6 or 1, 6, 3, 2, 5, 4 from AB to FA. The only one where CD>BC is the 1, 4, 5, 2, 3, 6 configuration. So this matches the solution above:

BC = 4
CD = 5
DE = 2
EF = 3
FA = 6

Posted by nikki on 2004-10-15 15:39:38