Thursday, October 6, 2016

Geometry Problem 1273 Circle, Tangent, Secant, Midpoint, Isogonal Lines, Congruent Angles

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

Click the figure below to view more details of problem 1273.


Geometry Problem 1273 Circle, Tangent, Secant, Midpoint, Isogonal Lines, Congruent Angles.

10 comments:

  1. https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8

    Using pole and polar properties:
    Let DE and DM cut circle O at F and G,
    Let MA cut circle O at H
    We have OM. OA= OC^2=OD^2
    So A is the pole of BC w.r.t circle O
    And A and E are harmonic conjugate points of D and F or ( A,E, D,F)= -1
    Since ∠EMA= 90 so Apollonius circle diameter AE will pass through M
    So ME is a angle bisector of angle MDF.
    And ∠AMD=∠FMH=∠HMG => BC//GF
    So Arc BG=Arc CF and ∠BDM=∠EDC

    ReplyDelete
  2. https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8

    Using elementary geometry:
    Let DE and DM cut circle O at F and G,
    Let MA cut circle O at H
    We have OM. OA= OC^2=OD^2
    So OA/OF=OF/OM => triangle MOF similar to triangle FOA ( case SAS)
    And FA/FO=FM/MO….. (1)
    Similarly triangle MOD similar to triangle DOA ( case SAS)
    And DA/DO=MD/MO ….. (2)
    Divide (1) to (2) and note that OD=OF
    We have FA/DA=FM/MD => MA is a external angle bisector of angle DMF
    And ∠AMD=∠FMH=∠HMG => BC//GF
    So Arc BG= Arc CF and angle BDM= angle CDE

    ReplyDelete
  3. Problem 1273
    If AD and DM cut circle O at K and L.Then AB^2=AM.AO=AC^2=AD.AK. So MOKD is cyclic.
    OD^2=OB^2=OM.OA so triangle ADO is similar with triangle DMO.Therefore <OMD=<ODA
    or <AMD=<ODK=<OMK. So <KMC=<CMD =<KMD/2=<KOD/2=<KBD, then <CMD=<CBD+<BDL=<KMC=<CBD+<CBK. So <BDL=<KBC=<KDC.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  4. Reading thro the 2nd proof of Peter, give below a slightly modified approach

    Let AD meet the circle at P.
    Let < MAP = p, < PAC = q, < ACD = < DBC = < DPC = r and < ODP = < OPD = s

    Then OC^2 = OM. OA = OD^2 = OP^2
    So < MDO = < MPA = p hence OMDP is concyclic

    Therefore < AMD = s and so
    < BDM = 90-s-r .....(1)

    Now < DOC = 2r so < CDE = 90-r-s (since < ODP = s)

    Therefore < BDM = < CDE = 90-s-r

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. How about the converse "if < BDM = < CDE then M is the mid point"?

    ReplyDelete
  6. The symmedian passes through the intersection of the tangents to the circumcircle at the extremities of the opposite side, i.e. AD is symmedian of tr. DBC.

    ReplyDelete
  7. Extend AD to meet the circle O at X.Connect AMO and form the right triangle ABO
    In the triangle ABO
    AB^2 = AM.AO
    But AB^2 = AD.AX
    Hence D,M,O and X are concyclic points.

    Let Angle XDC = EDC = @ and ODX = $
    => Angle OXD = $ => OMD = 180-$ => AMD = $ => BMD = 90+$ -------------(1)
    Join OC and consider the isosceles triangle ODC
    We know Angle ODC = @+$ = OCD => Angle DOC = 180-2(@+$) => Angle DBC = DBM = DOC/2 = 90-(@+$)----------(2)

    Now consider the triangle BDM and from (1) and (2)
    Angle BDM = 180-(90+$)-(90-(@+$)) = @ = EDC

    ReplyDelete
  8. Let DM extended meet the circle at X and let ADE extended meet the circle at Y. Let AMO extended meet XY at Z.

    OM.OA = OB^2 = OD^2 = OY^2

    So < OAY = < MDO = < MYO hence OMDY is concyclic with OD = OY and so OM bisects < XMY.

    Easily MOZ is perpendicular to XY and so BC//XY

    Therefore BX = CY in isosceles trapezoid BCYX and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. It is also interesting to note that the angle DEO is bisected by BC

    ReplyDelete