Tuesday, December 16, 2008

Archimedes' Book of Lemmas, Proposition #12

Problem 652: Diameter, chords, perpendicular, tangents
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #12 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 652

Archimedes' Book of Lemmas #12.
Continue reading at:
gogeometry.com/ArchBooLem12.htm

8 comments:

  1. http://img193.imageshack.us/img193/4671/problem652.png

    Connect OC , DE
    AD cut BF at L ( see picture)
    Quadrilateral LDFE is cyclic with LF as diameter ( Angles D=E = 90)
    Note that angle(DAE)= angle (CDE)
    So angle (DLE)= angle (DCO) ( angles complement to congruence angles)
    OC is perpendicular bisector of DE . On OC only center of circle LDFE have property (DLE)= ½ (DCE)
    So C is the center of circle LDFE and diameter LF pass through center C
    In Triangle ABL , F is the orthocenter so line LCF will perpendicular to AB
    Peter Tran

    ReplyDelete
    Replies
    1. Hi Peter!
      You mean "AD cut BE at L ( see picture)" ?

      Delete
  2. Peter Tran's solution suggests the following (perhaps easier) problem:

    "Let D, E be any two points on the semicircle on AB as diameter. Let AE, BD intersect at F; AD, BE intersect at L. Prove LF is perpendicular to AB."

    Pravin

    ReplyDelete
  3. Peter Tran's solution gives me this idea:
    To construct the tangents we draw the circle with diameter CO
    This cercle intersects the given semi circle at D and E
    he passes trough O midpoint of AB and the altitude feet E,D, he is the nine point circle of triangle ABL
    let be M the second intersection with AB
    M is an altitude foot
    CM perpendicular to AB (CO diameter)
    hence CM is an altitude and passes trough orthocenter F and M=G
    FC perpendicular to AB

    ReplyDelete
  4. Let AD, BE meet at M. MF is obviously perpendicular to AB

    Let DC meet MF at S and EC meet MF at T

    < SDM = < SMD = 90 - A so DS = SM ....(1)
    Similarly MT = TE .....(2)
    DC = CE......(3)

    (1) +(2) -(3) yields DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide

    So C must lie on MF which completes our proof

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Hello Sumith.
      In your proof, the sentence "DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide" seems to infer that SC, CT and TS should all be positive or equal to 0.
      How can we sure about that ?
      Best regards.

      Delete