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?

First part: 1, 4, 5, 2, 3, 6

Break the 2-6 lengths into their components (sin and cos of 120, we'll mess with signs later)
2: 1, 2*(root(3)/2)
3: 1.5, 3*...
4: 2, 4*...
5: 2.5, 5*...
6: 3, 6*...

I set AB to be the top of the figure (horizontal)

The vertical parts (root(3)/2) of BCD and EFG must add up to be equal. Only options are 36:45, 26:35, and 25:34.

Now, the horizontal distances must add up 1+(BC-CD)+(AF-FE)=ED.

Three possibilities (clockwise): 1,6,2,4,3,5 : 1,4,5,2,3,6 : 1,5,3,4,2,6 : 1,6,3,2,5,4

The only one with CD>BC is 1,4,5,2,3,6


SECOND PART

We're given the left half (5,6) so the vertical part is 11*root(3)/2. That leaves 8 for X. Confirming the horizontal distances work as well:
1+(4-3/2)+(3-5/2) = 4 : confirmed

Rajal

Posted by Rajal on 2004-10-15 15:02:36