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: , , , , ,

8 comments:

  1. PravinMay 1, 2011 at 8:46 AM

    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°

    ReplyDelete
  2. EditorMay 2, 2011 at 5:23 PM

    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°.

    ReplyDelete
  3. JoaquinApril 5, 2012 at 12:10 PM

    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.

    ReplyDelete
  4. APOSTOLIS MANOLOUDISJune 25, 2016 at 1:41 PM

    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

    ReplyDelete
  5. APOSTOLIS MANOLOUDISJune 25, 2016 at 1:55 PM

    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

    ReplyDelete
  6. Sumith PeirisJune 26, 2016 at 10:59 PM

    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

    ReplyDelete
  7. AnonymousSeptember 24, 2018 at 6:11 PM

    extend AB,DC, intersect at K,
    so BCK is equilateral triangle;B is midpoint of AK;
    connect KH, so KCH is isosceles right triangle;
    M is midpoint of AH;
    so BM//KH;
    so angle AMB is 45;
    angle BMC is 135.

    ReplyDelete
  8. MarcoDecember 27, 2023 at 9:30 PM

    Let the length of the hexagon and square be a
    〖AC〗^2=a^2+a^2-2a^2 cos120
    〖AC〗^2=3a^2
    AC=a√3
    AH=AC+CH=a(√3+1)
    AM=AH/2=(a(√3+1))/2

    〖BM〗^2=〖BA〗^2+〖AM〗^2-2(BA)(AM)cos30
    〖BM〗^2=a^2+((a(√3+1))/2)^2-2(a)((a(√3+1))/2)√3/2
    BM^2=a^2+(a^2 (4+2√3))/4-(a^2 (3+√(3)))/2
    〖BM〗^2=a^2/2
    BM=a/√2

    MC=MH-CH=AH/2-CH=(a(√3-1))/2
    〖MC〗^2=(a^2 (4-2√(3)))/4=(a^2 (2-√(3)))/2

    cosx=(〖BM〗^2+〖MC〗^2-〖BC〗^2)/(2(BM)(MC))
    cosx=(a^2/2+(a^2 (2-√(3)))/2-a^2)/(2(a/√2)(a(√3-1)/2))
    cosx=((1-√3)/2)/((√3-1)/√2)
    cosx=-1/√2
    x=135

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.

Subscribe to: Post Comments (Atom)