Monday, October 1, 2012

Problem 809: Trisecting a Line Segment AB, Three Circles, Radius, Center

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 809.

Online Geometry Problem 809: Trisecting a Line Segment AB, Three Circles, Radius, Center.

5 comments:

  1. Let CD cut AB at I and BC cut DE at J
    Note that ABD, ABC and CEF are equilateral triangles
    We have DB//AC//CE => DBEC is a parallelogram
    And J is the midpoint of BC
    In triangle DBC, DJ and BI are medians => N is the centroid of DBC
    So BN=2.IN
    Due to symmetric we have AM=2.IN=MN=NB

    ReplyDelete
  2. Since FC//CB//AD, and FC = CB = AD,
    so FADC is a parallelogram,
    thus, FD and AC bisect each other.

    Join AD and DC. Consider ΔADC,
    DM bisects AC, and AB bisects DC.
    So M is the centroid of ΔADC.

    Let AB meets DC at H.
    Then AM:MH = 2:1.

    Similarly, HN:NB = 1:2.

    Since MH = HN, so AM:MN:NB = 1:1:1.
    As a result, AM = MN = NB.

    ReplyDelete
  3. Using vector to solve the problem.
    Let vector(AC) = u, vector(BC) = v.
    Then vector(DE) = 2u+v, vector(BA) = v-u.
    Now let
    vector(DN) = p*vector(DE) =p*(2u+v);
    vector(BN) = q*vector(BA) = q*(v-u).

    Since
    vector(DN) = vector(DB) + vector (BN)
    => p*(2u+v) = u + q*(v-u)
    => p = q = 1/3.

    Therefore BN = (1/3)BA
    Symmetrical on the other side, q.e.d.

    ReplyDelete
  4. Since Tr.s AMD and BMF are similar BM = 2AM. Similarly AN = 2BN and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Problem 809
    Is AB=AC=AD=BD=CE=CF=BC then <ABE=90.AD intersects EB at K then <DBK=30=<DKB.
    So DB=DK. But ΑD=DK and EB=BK so point N is centroid of triangle AEK.Then NB=AB/3.
    Similar AM=AB/3.Therefore AM=MN=NB.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete