Saturday, April 30, 2011

Geometry Problem 599: Regular Hexagon, Square, Midpoint, Angle, Degrees

Geometry Problem
Click the figure below to see the complete problem 599.

Geometry Problem 599: Regular Hexagon, Square, Midpoint, Angle, Degrees.

Posted by Antonio Gutierrez at 7:33 AM
Labels: , , , , ,

3 comments:

  1. Drop BN perpendicular to AC
    w.l.o.g. let each side of the hexagon be 2
    Easy to see
    BN = 1,
    AN = √3,
    AC = 2√3
    AM = AH/2 = (AC + CH)/2 = (2√3 + 2)/2 = √3 + 1
    NM = AM – AN = 1
    ∆BNM is right angled isosceles,∠BMN = 45°
    x = 135°

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  2. it is a lot easier!!!! if you see <AHB=15° you can remember the special triangle 30-15-135.
    median traced to the largest side forma angle 45° and x=135°.

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  3. Let be O the center of square. Note that OM || AD and BC=OM. Thus triangles CMO and MCB are congruents. We know angle OCH = 45º, thus x = angle MCB=135º.

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