posted by Anonymous .
ABCDEF is a regular hexagon. Square ABXY is constructed on the outside of the hexagon. Let M be the midpoint of DX. What is the measure (in degrees) of ∠CMD?
Geometrically, I have no idea. Algebraically, if we let
A = (0,0)
B = (1,0)
D = (1,√3)
X = (1,-1)
M = (1,(√3-1)/2)
C = (3/2,√3/2)
slope of MC is 1
slope of MD is √3/2
∠CMD = 60°-45° = 15°
Maybe by working with all those 60° and 30° angles in the figure this will become clear geometrically.
Well, I sort of get a different answer: ∠CMD = 45
Here's how I got it:
Let CD = x. It is trivial by law of cosine that: BD = x√3
Whence DM = (√3 + 1)/2 * x. Let us denote the midpoint of BD as K. Then it is easy to see that:
MK = DM - DK = x/2. Now it is trivial to end of to get:
<CMD = tan^(-1) 1 = 45.
Well, since the live period is long over I suppose we can discuss. In the future, please refrain from posting live brilliant questions :)