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.

6 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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  4. Problem 599
    Form the equilateral triangle BHK (points M and C are inside the triangle BHK).Then <BKH=60=2.30=2.<BAH so K is circumcenter of triangle ABH.So AK=BK=BH=HK and
    MK=MH=MA then triangle BMK=triangle BMH.Therefore < MBH=<MBK=30.Then
    <BMA=30+15=45. Therefore <BMH=135.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  5. Problem 599 solution 2
    Form the equilateral triangle BHK (points Β is inside the triangle BHK).Then
    Triangle ABK=triangle ABH then <AKB=<AHB=15 and BK=BH,<BKH=<BHK=45
    so <KBH=90=<KMH.Then KBMH is cyclic.So <MBH=<MKH=30.Therefore
    <BMH=135.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  6. Drop a perpendicular BP to ACH. Obviously PM= a/2 where a = side of the
    regular hexagon and the square. But BP also = a/2 considering the 30-60-90
    triangle BCP.

    Hence BP = PM and so x = 135

    Sumith Peiris
    Moratuwa
    Sri Lanka

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