Monday, November 11, 2013

Geometry Problem 934: Square, Circle, Center, Perpendicular, Metric Relations

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

Click on the figure below to enlarge it.

Online Geometry Problem 934: Square, Circle, Center, Perpendicular, Metric Relations

5 comments:

  1. Since ∡ CEA = ∡ CDA , A,B,C,D,E are concylic

    Let α be ∡CDE, then ∡CAE = α ; ∡ EAD = ∡ECD = 45 - α ;
    ∡FCD = ∡FCE - ∡ECD = 45 - (45 - α) = α =∡CDE
    So FC//DE
    Further ∡ACF = 45 - ∡FCD = 45 - α
    Let L be the length of the square,

    By sine law on △CDE,
    L/sin(135) = 1/sin(45 - α)

    By sine law on △ACF
    (sqrt2)L / sin(135) = x/sin(45 - α)

    Hence, x = sqrt2

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  2. Let AD=a, CE=r.

    Since ADEC is cyclic quadrilateral, by Ptolemy theorem,
    AC×DE + AD×CE = AE×CD
    √2 a + ar = a(x+r)
    x = √2

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  3. a geometric solution:
    http://bleaug.free.fr/gogeometry/934.png

    By construction, ABCDE are co-cyclic.
    Let G be co-cyclic such that AC crosses CG in F. By construction AFG=EFC=45°
    AD chord in ABCDEG circle implies ACD=AED=45°. Similarly, CD chord implies CAD=CGD=45°
    Hence ED // CG and GD // AE, therefore EDGF is a parallelogram.
    AF = GF.√2 = ED.√2 = √2

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  4. <ACF=<DCE=45-<FCD, and <CDE=<CAF due to cyclic quadrilateral. Therefore X/CF=1/CE, X=CF/CE=sqrt2

    ReplyDelete