Tuesday, April 2, 2013

Problem 868: Cyclic Quadrilateral, Circle, Five Rectangles, Four Centers, Congruence

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

Click the figure below to see the complete problem 868.

Online Geometry Problem 868: Cyclic Quadrilateral, Circle, Five Rectangles, Four Centers, Congruence

Posted by Antonio Gutierrez at 8:46 PM
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7 comments:

  1. Jacob HA (EMK2000)April 2, 2013 at 10:54 PM

    From problem 866,

    ∠O1O2O3 = 90°

    Similarly, we have
    ∠O2O3O4 = 90°
    ∠O3O4O1 = 90°
    ∠O4O1O2 = 90°

    Hence, O1O2O3O4 is a rectangle.

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  2. AnonymousApril 6, 2013 at 4:52 AM

    ∠O1O2O3 = 90°
    Why???

    ReplyDelete
  3. UnknownSeptember 13, 2015 at 8:10 PM

    thank you
    for this problem, i passed my prelim. :-)

    ReplyDelete
  4. Ivan BazarovJuly 20, 2016 at 9:30 PM

    AO4=CO2, AO1=CO3, <GCK=360-90-90-(180-<BAD)=<BAD, <O1AB=<KCO3, <DAO4=<GCO2, значит треугольники O3CO2 и O1AO4 конгруэнтно друг друга, и O1O4=O2O3. Подобным образом O1O2=O3O4. Поскольку противоположные стороны четырехугольника O1O2O3O4 равны, он параллелограмм. <O1O4O3=<AO4D-<O1O4A-<O3O4D, <O1O2O3=(<BO2C=180-<AO4D)+(<O1O2B=<O3O4D)+(<O3O2C=<O1O4A), значит что <O1O4O3+<O1O2O3=180 и O1O2O3O4 вписанный четырехугольник. Если O1O2O3O4 одновременно параллелограмм и вписанный четырехугольник, то оно прямоугольник.

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  5. Ivan BazarovJuly 20, 2016 at 9:35 PM

    AO4 = CO2, AO1 = CO3, <GCK = 360-90-90- (180- <BAD) = <BAD, <O1AB = <KCO3, <DAO4 = <GCO2, then triangles O3CO2 O1AO4 and congruent to each other, and O1O4 = O2O3. Similarly O1O2 = O3O4. Since the opposite sides of the quadrilateral O1O2O3O4 are equal, it is parallelogram. <O1O4O3 = <AO4D- <O1O4A- <O3O4D, <O1O2O3 = (<BO2C = 180- <AO4D) + (<O1O2B = <O3O4D) + (<O3O2C = <O1O4A), means that <O1O4O3 + <O1O2O3 = 180 and O1O2O3O4 inscribed quadrilateral. If O1O2O3O4 simultaneously parallelogram and inscribed quadrilateral, it is a rectangle.

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  6. Peter TranJuly 21, 2016 at 6:27 PM

    Excellence solution !

    ReplyDelete
  7. Sumith PeirisJuly 22, 2016 at 12:19 PM

    Triangle AO4D ≡ BO2H hence < ADO4 = HBO2 =á say….(1)

    Triangle O1AB ≡ DO3J hence < O1BA = < O3DJ = < 90- O3DC = â say….(2)

    Hence < O1BO2 = 360 – B – (90-á) – â = 270 – B + á –â …(3)

    < O4DO3 = 180 – B + á + 90 - â = 270 - B + á –â….(4)

    From (3) and (4) < O1BO2 = < O1BO2 and so Triangles O4DO3 ≡ O1BO2 (SAS)

    Therefore O1O2 = O3O4.
    Similarly O2O3 can be shown to be = to O1O4 and so O1O2O3O4 is a //ogram.

    Let < BO2O1 = < DO4O3 = q and < CO2O3 = < AO4O1 = n

    So in this //ogram < O1O4O3 = 180 - 2á – n- q = < O1O2O3 = q+n+ 180 – 2
    (90-á) = 2á +n +q from which

    2(2á +n +q) = 180 and so 2á +n +q = 90 and therefore from Triangle AO4D, <
    O1O4O3 = 90

    So O1O2O3O4 is a //ogram having an angle = to 90 and is hence a rectangle.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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