Wednesday, June 20, 2012

Problem 769: Triangle, Circumcircle, Circumcenter, Orthocenter, Midpoint of arc, Parallel to chord, Congruence

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 769 details.

Online Geometry Problem 769: Triangle, Circumcircle, Circumcenter, Orthocenter, Midpoint of arc, Parallel to chord, Congruence.


  1. Partial Solution:

    Triangles AOM, COM are congruent,
    OE // AM, and OF // MC.
    So Triangles AEM,CFM have equal area.
    M lies on the bisector of angle ABC.
    So altitudes from M on bases
    AE and FC are of equal length.
    So AE = FC

  2. total solution was sent to the email
    Mixalis Tsourakakis from Greece

    1. Problem 769: Solution by Mixalis Tsourakakis from Greece.
      Thanks Mixalis.

  3. To Mixalis Tsourakakis

    I appreciate if you can translate your solution of problem 769 to English so that we can understand it

    Peter Tran

    1. Because i don't speak english very fluently i only send you the text of the solution that Antonio Gutierrez attached translated in english without any symbols. The text is in the same order. My solution is very detailed and this is why is so lengthy. I hope this was helpful.

      Page 1


      As known (proof is very simple) it applies that ΝΒΜ= Let’s assume that Α>C in that case ΝΒΜ= [If angle Α< angle C then ΝΒΜ= ]
      ΗΝC+HXC=180 thus ΑΗΝ=C
      ΑΗD=C+ΝΗD=C+ΝΒΜ (HD//BM)= C+ =
      S is the midpoint AC .MS mediator of ΑC (since ΑΜ=ΜC) and ΒCMS=Q .Then QA=QC and QAC=C.Also BAL=A-C and due to symmetry LCB=A.In addition MQ,AB,CL intersect at G.

      Page 2

      I define yellow triangles ΗDC ,OFC,OEG and blue triangles HAD,QOF,EOA. All yellow triangles are similar and all blue triangles are similar. Indeed:
      ΑΗD=AHN+NHD=C+NHD=C+NBM (HD//BM)= C+ =
      DHC=CHN-HDN=A- = (CHN=A because UHNA inscribable) and so HD is the bisector of ΑΗC .
      HAD=NBC(NABX inscribable)=(BN//QM)= OQF
      Since OF//MC, then
      Therefore OF is the bisector of triangle QOC. But QOF=OMC= (ABCM is inscribable and so ΑΜC=A+C). Thus also FOC=
      Since ΕΟ//ΑΜ,OF//MC then ΕΟF=AMC=A+C and since A+C+B=180 BEOF will be inscribable therefore ΟFC=BEO=EAM=A+ (CAM=ABM= )
      It applies …………………………. Therefore OE is the bisector of triangle GOA.

      Page 3

      However GOE= thus also ΕΟΑ=
      Also it is obvious that G= HCD (G+A= A+HCD=90 G= HCD)
      Finally ΑΕΟ= ΒFO=  (OF//MC) and ΗDC=BIC=A+
      triangle ΑΗD  triangle QOF since ΑΗD=QOF= and
      HAD= OQF
      Triangle QOF triangle EOA since EOA=  QOF= and AEO=QFO=
      Thus all blue triangles are similar
      Triangle HDC triangle OFC since FOC= FOC=CHD and ΟFC=ΗDC=A+
      Triangle HDC triangle GEO since G=UCA and DHC=GOE=
      Thus all yellow triangles are similar.

      Since ΗD is the bisector of triangle AHC it will be:
      Howerver triangle AHC triangle OQF and therefore
      triangle HDC triangle OFC and therefore
        
      Thus triangle FDC is isosceles, therefore DF=FC (1)
      triangle ΕΟΑ triangle HAD therefore
      triangle EOG triangle HDC thus


      Page 4

      Therefore triangle ΑΕD is isosceles which means that ΕΑ=ΕD (2)
      For triangle ΕΖΑ applies ΑΕΖ=Γ+ , ΕΑΖ=Α-Γ therefore ΕΖΑ= Γ+ and hence triangle ΑΕΖ is isosceles and therefore ΕΑ=ΖΑ. But obviously ΖΑ=FC (3)
      From (1),(2),(3) DF=FC =ΕΑ=ΕD

      Michael Tsourakakis from Greece