Wednesday, November 14, 2012

Problem 821: Adams Circle Theorem, Incenter, Incircle, Gergonne Point, Contact triangle, Parallel, Six Concyclic points, Concentric Circles

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

Click the figure below to see the complete problem 821.

Online Geometry  Problem 821: Adams Circle Theorem, Incenter, Incircle, Gergonne Point, Contact triangle, Parallel, Six Concyclic points, Concentric Circles.

3 comments:

  1. http://img651.imageshack.us/img651/1259/problem821.png

    Draw lines per attached sketch
    We have ∆CA’B’ and CA”’B” are isosceles => A’A””=B’B”
    From B draw line C1A1//AC
    This line cut B’A’, B’C’,GC”’ and GA”’ at A1, C1, C2 and A2 .
    Note that ∆CA’B’, ∆ BA’A1, ∆AB’C’ and ∆BC’C1 are isosceles
    So BC1=BC’=BA’=BA1 => BA1=BC1
    Since GA2//B’A1 and GC2//B’C1 we have BA2/BA1=BG/BB’=BC2/BC1
    B is the midpoint of A2C2 => B’ is the midpoint of B”B”’
    Triangle IB”B”’ is isosceles
    Similarly IC”C”’ and IA”A”’ are isosceles => IB”=IB”’=IC”=IC”’=IA”=IA”’
    So I is the center of circle A”A”’B”B”’C”C

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  2. Denote points P1,P2,P3,...P6 as the points where the parallel lines cut the triangles (starting from the right hand side of A)
    By the properties of incircle, AC'=AB' and since P2P5 parrllel to B'C', AP2 = AP5 hence B'P5 = C'P2.

    By symmetry, B'P5 = C'P2 = A'P3 = B'P6 = C'P1 = A'P4
    Therefore P1P2P5P6 forms an isosceles trapezium and hence concylic. Again by symmetry, the 6 points concyclic.

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  3. Hi Antonio, here is my proof using geogebra as dynamic geometry visualization.

    http://www.youtube.com/watch?v=EaH7FwwGNwI

    ReplyDelete