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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 -

    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.

  • geometry -

    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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