Posted by **Anonymous** on Monday, August 19, 2013 at 11:56am.

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?

- geometry -
**Steve**, Monday, August 19, 2013 at 1:59pm
Geometrically, I have no idea. Algebraically, if we let

A = (0,0)

B = (1,0)

then

D = (1,√3)

X = (1,-1)

so

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.

- geometry -
**exactly**, Wednesday, October 2, 2013 at 4:48am
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 :)

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