Friday, May 18, 2012

Problem 754: Equilateral Triangle, Center, Angle, 60 Degrees, Perimeter

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 754 details.

Online Geometry Problem 751: Parallelogram, Congruence, Intersecting Lines, Angle Bisector.

6 comments:

  1. http://img100.imageshack.us/img100/2228/problem754.png

    Let I is the incenter of triangle DBE ( see sketch)
    ∠DIE= ∠B+∠D/2+∠E/2= 60+ (180-60)/2= 120
    Since ∠DIE supplement to ∠DOE , quadrilateral DIEO is cyclic
    O is the intersecting point of angle bisector BI and circumcircle of DIE => O must be excenter of triangle DBE
    Let excircle touch DE at T .
    We have EQ=ET and DP=DT
    And perimeter of DBE=BP+BQ= AC

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  2. Rotate triangle ODB 120(degree) clockwise about O,
    it becomes triangle OD'B', where D' lies on BC and B'=C.

    Now since OD=OD', and angle D'OE=120(degree)-angle DOE=60(degree),
    thus triangle DOE is congruent to triangle D'OE.
    Hence, DE=D'E.

    Therefore, perimeter = BE+ED+DB = BE+ED'+D'B' = BC.

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  3. Replies
    1. Thank you very much!=)

      Delete
  4. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27343&p=133800

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  5. Problem 754
    Let the point F on the side BC (F is between the points B, C), such that BD=FC.Then
    Triangle DBO=triangleFCO (OB=OC,DB=FC, <DBO=<FCO=30).So OD=OF and <DOB=<COF.
    But <COB=120=<DOF.Then <EOF=60=<EOD.So triangleDOE=triangleEOF and DE=EF.So
    AC=BC=BE+EF+FC=BE+ED+DB.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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