Monday, October 1, 2012

Problem 810: Trisecting a Line Segment AB, Two Circles, Radius, Center. Diameter, Chord

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

Click the figure below to see the complete problem 810.

Online Geometry Problem 810: Trisecting a Line Segment AB, Two Circles, Radius, Center. Diameter, Chord.


  1. Let CD cut AB at I
    Note that ABD, ABC and CBF are equilateral triangles
    And G is the midpoint of BC and AF
    In triangle DBC, DG and BI are medians => N is the centroid of DBC
    So BN=2.IN
    Due to symmetric we have AM=2.IN=MN=NB

  2. Since DE is diameter with center A,
    we have EA = AD.

    Since CADB is a rhombus, AC//DB,
    hence, EH = HB.

    Now in ΔEDB, A and H are mid-points,
    which means BA and DH are medians.

    Therefore, M is the centroid of ΔEAB.
    Thus, AM:MB = 1:2.

    Similarly, N is the centroid of ΔFAD.
    We also have AN:NB = 2:1.

    As a result, AM = MN = NB.

  3. Observe that triangle CAG, triangle AGB, triangle FGB are congruent and they are 30-60-90 triangles. (proof is skipped)

    Hence AF is the perpendicular bisector of BC.

    Now insert coordinate system with the midpoint of AB be O(0,0), and auxillarily let B(1,0).
    Then D = (0,-sqrt3) ; C = (0,sqrt3) ; G is the midpoint hence = (1/2, (sqrt3)/2)

    Line DG : y + sqrt3 = 3*sqrt3*x
    When DG cuts AB, y = 0, x = 1/3. => N(1/3,0)
    Hence ON is one-thrid of OB.

    Symmetrically on the other side,

  4. This is a proof using the result in problem 809.
    *(using the points in problem 809)
    Let the point where BC cuts DE named P.
    Then since BC = CE, and CE//BD,
    trianglePCE is congruent to trianglePBD.
    Hence PB=PC.
    So point P is acutally the point G in this problem.

  5. First ABC is an equilateral triangle and G is the midpoint of CB, H is the midpoint of CA.

    Let's call P the point where HG(extended) cuts BF.

    . HG=GP (because CBF is an equilateral triangle)

    . therefore using TALETE (HG || AB) follows that MN=NB and so MN=NB=AM (by simmetry)

  6. AB = 2AH = BD and since Tr.s AHM and BDM are similar BM = 2AM

    Similarly AN = 2BN and the result follows

    Sumith Peiris
    Sri Lanka