Saturday, September 17, 2011

#Geometry Problem 670: Triangle, 60 Degrees, Orthocenter, Incenter, Circumcenter, Congruence

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

Click the figure below to see the complete problem 670.

Online Geometry Problem 670: Triangle, 60 Degrees, Orthocenter, Incenter, Circumcenter, Congruence

3 comments:

  1. http://img835.imageshack.us/img835/1895/problem670.png
    Per the result of problem 662
    We have ( AHC)=(AIC)=(AOC)=120
    And 5 points A, H, I, O and C are co cyclic ( see picture)
    Since (B)=60 >> (BAH)=30
    Since (AOC)=120 >> (OAC)=30
    AI is and angle bisector of (BAC) >> AI is also an angle bisector of (HAO)
    Arc HI=arc IO and IH=IO
    Peter Tran

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  2. B = 60°, A + C = 120°.
    Let D be the midpoint of BC
    BH = 2 OD = 2R cos B = R = BO
    ∠HBA = 90° - A
    ∠OBC = ∠OCB = 90° - ∠BOC/2
    = 90° - A = ∠HBA
    Already BI bisects ∠ABC
    Follows BI bisects ∠HBO and
    ∆BIH ≡ ∆BIO
    ∴HI = IO

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  3. Let D be midpoint of BC and let CH extended meet AB at E.

    BE = a/2 = BD and < ABH = < OBD = 90-A
    Hence tr.s BHE and BDO are congruent ASA and so BH = BO.

    Now < HBI = < OBI = B/2 - (90-A)
    Hence tr.s HBI and OBI are congruent SAS

    Therefore HI = OI

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete