Friday, May 25, 2012

Geometry Problem 756: Four Circles Theorem

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

Using TracenPoche Interactive Dynamic Software. Step-by-Step construction, Manipulation, and animation.
Click the figure below to interact with the problem 756 details.

Online Geometry Problem 756: Four Circles Theorem Using TracenPoche Interactive Dynamic Software. Step-by-Step construction, Manipulation, and animation.

Posted by Antonio Gutierrez at 8:55 PM
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3 comments:

  1. W FungMay 25, 2012 at 11:51 PM

    Angle A'AD + Angle A'D'D = 180
    Angle A'AB+ Angle A'B'B= 180
    Angle DCC' + Angle DD'C' = 180
    Angle C'CB + Angle C'B'B = 180

    Summing up the four identities, the four angles = 540

    By considering A'D'C' and A'B'C', they add up would equal to 180, hence concylic.
    Q.E.D.

    ReplyDelete
  2. Peter TranMay 26, 2012 at 8:33 AM

    http://img703.imageshack.us/img703/7325/problem756.png

    Connect lines per attached sketch
    Let a1, a2, c1,c2, d’1, d’2, b’1, b’2 are values of angles showed on the sketch
    Since A’ADD’ is cyclic so d’1=180-a1
    Since D’DCC’ is cyclic so d’2=180-c1
    So m∠A’D’C’=d’1+d’2=360-a1-c2
    Similarly we also have m∠A’B’C’=b’1+b’2=360-a2-c2
    So m∠A’D’C’+m∠A’B’C’=(360-a1-a2)+ (360-c1-c2)=m∠BAD+m∠BCD=180
    So ∠A’D’C’ supplement to ∠A’B’C’ and A’, B’, C’ ,D’ are cocyclic

    ReplyDelete
  3. Peter TranMay 26, 2012 at 2:44 PM

    Different positions of A, B,C, D on circle O and different location of centers of circle O1, O2,O3 and O4 will generate different arrangement of quadrilateral A’B’C’D’ relatively to other quadrilaterals.
    Below is a typical case. This proving can be applied for general case
    Given: quadrilaterals ABCD, ABB’A’, AA’D’D, DD’C’C, BB’C’C are cyclic
    To prove : quadrilateral B’A’D’C’ is cyclic
    Sketch : http://img59.imageshack.us/img59/3852/problem756generalcase.png

    Connect lines per attached sketch
    Let d1 , d2 ,b1 ,b2 ,c’1 ,c’2, a’1 and a’2 are values of angles showed on the sketch
    Since qua. B’BCC’ is cyclic => b1+b2+c’1+c’2=180……. (1)
    Since qua. DD’C’C, BADC, and BB’A’A are cyclic so
    ..c’2=180-d1
    ..b2=180-d2
    ..b1=180-a’2
    Replace these values in (1) we get c’1=-360+d1+d2+a’2 …..(2)
    Since ADD’A’ is cyclic => m∠AA’D + m∠ADD’=180
    Or 360-a’1-a’2+360-d1-d2=180
    Or a’1= 540-d1-d2-a’2…….(3)
    Add (2) to (3) we get c’1+a’1= 180 => qua. B’A’D’C’ is cyclic

    In general for these quadrilaterals ABCD, ABB’A’, AA’D’D, DD’C’C, BB’C’C and B’A’D’C’,
    If 5 of these quadrilaterals are cyclic, then the remaining quadrilateral will be cyclic

    ReplyDelete

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