Monday, October 13, 2014

Geometry Problem 1050: Regular Hexagon, Center, Any Point, Inside, Outside, Distance, Congruence, Metric Relations

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 1050.

Online Math: Geometry Problem 1050: Regular Hexagon, Center, Any Point, Inside, Outside, Distance, Congruence, Metric Relations, Polya's Mind Map.

Posted by Antonio Gutierrez at 8:52 AM
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5 comments:

  1. PravinOctober 13, 2014 at 6:27 PM

    Let each side of the hexagon be a.
    Let M,N be the midpoints of AF,CD resp.
    Required sum
    = (2PM^2 + a^2 / 2) + (2PO^2 + 2OE^2) + (2PN^2 + a^2 / 2)
    = 2(PM^2 + PN^2) + a^2 + 2(PO^2 + OE^2)
    = 2( 2PO^2 + 1/2 MN^2) + a^2 + 2PO^2 + 2a^2
    = 6PO^2 + MN^2 + 3a^2
    = 6PO^2 + 3a^2 + 3a^2
    = 6PO^2 + 6a^2

    ReplyDelete
  2. Jacob HA (EMK2000)October 13, 2014 at 11:23 PM

    Let ∠POC=x, ∠POD=y. Let AB=OA=R, PO=d.

    Using cosine law, we have
    PA² = R² + d² − 2Rd cos(x+120°)
    PB² = R² + d² − 2Rd cos(x+60°)
    PC² = R² + d² − 2Rd cos(x)
    PD² = R² + d² − 2Rd cos(y)
    PE² = R² + d² − 2Rd cos(y+60°)
    PF² = R² + d² − 2Rd cos(y+120°)

    Note that
    cos(x)+cos(x+60°)+cos(x+120°)
    = 2 cos(x+60°) cos(60°) + cos(x+60°)
    = 2 cos(x+60°)

    cos(y)+cos(y+60°)+cos(y+120°)
    = 2 cos(y+60°)
    = 2 cos(180°−(x+60°))
    = −2 cos(x+60°)

    Thus,
    cos(x)+cos(x+60°)+cos(x+120°) + cos(y)+cos(y+60°)+cos(y+120°) = 0

    Hence,
    PA²+PB²+PC²+PD²+PE²+PF² = 6R²+6d² = 6 AB²+6 PO²

    ReplyDelete
  3. APOSTOLIS MANOLOUDISSeptember 7, 2016 at 6:57 AM

    Problem 1050
    Is AB=BC=CD=DE=EF=FA=AO=BO=CO=DO=EO=FO.But PA^2+PB^2+PC^2+PD^2+PE^2+PF^2=(PA^2+PD^2)+(PB^2+PE^2)+(PC^2+PF^2)=
    (2PO^2+2AO^2)+(2PO^2+2BO^2)+(2PO^2+2CO^2)=6PO^2+6AB^2.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  4. AnonymousSeptember 8, 2016 at 7:49 AM

    Considering the geometrical structure in a 3-D plane.
    Join OF,OE and consider the triangles OEF & OEP

    OEP is right angle triangle
    => PE^2 = PO^2 + OE^2 ---- (1)

    In the triangle OEF, OE = OF and Angle EFO = 60 => OEF is equilateral triangle
    Hence OE=OF=FE ------- (2)

    Substituting (2) in (1) => PE^2 = PO^2 + FE^2

    Considering symmetry, PE=PF=PA=PB=PC=PD & EF=FA=AB=BC=CD=DE
    The result thus follows

    ReplyDelete
  5. Sumith PeirisOctober 20, 2017 at 11:39 AM

    Note that AD, BE and CF are concurent at O

    Use Apollonius Theorem in each of the triangles PAD, PBE, PCF and add. The result follows easily

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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