Saturday, May 19, 2012

Problem 755: Equilateral Triangle, Circumcircle, Chord, Arc, Midpoints, Perpendicular

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 755 details.

Online Geometry Problem 755: Equilateral Triangle, Circumcircle, Chord, Arc, Midpoints, Perpendicular.

4 comments:

  1. http://img407.imageshack.us/img407/8651/problem755.png

    Connect OE and FC
    A,O and E are collinear …..(F is midpoint of arc BC)
    OA=FC=Radius of circle
    OE ⊥EA …(E is midpoint of DA)
    ∠DCF=∠EAO… ( face same arc DBF)
    ∆FGC congruence to ∆ OEA (case ASA) => GC=EA
    ∠DAB =∠DCB …(face same arc DB)
    ∆BGC congruence to ∆ BEA… ( case ASA)=> BG=BE and ∠GBC=∠EBA
    So ∠EBG=∠ABC=60 => ∆EBG is equilateral

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  2. Let T be the rotation 60° clockwise about B.

    Then
    T(C) = A
    T(F) = O

    Since ∠ADC = ∠ABC = 60°,
    so T(line DC) = line DA

    Let T(G) = G', then G' lies on DA.

    Since FG⊥DC, we have OG'⊥DA,
    hence, G' = E.

    As a result, T(G) = E, T(BG) = BE.
    Hence, ∆BEG is equilateral.

    ReplyDelete
  3. Let O be the origin, z(P) be the complex number representing point P.
    Let z(C)=1, then z(B)=ω, z(A)=ω^2, z(F)=-ω^2.

    Let z(D)=ε, then z(E)=(ε+ω^2)/2.

    Since FG⊥DC, let [z(D)-z(G)]/[z(F)-z(G)]=ki.
    k = |z(D)-z(G)| / |z(F)-z(G)]| = √3
    because ∠DFG = 60°.

    Now [z(D)-z(G)]/[z(F)-z(G)] = i √3
    ε - z(G) = i √3 [-ω^2 - z(G)] = -i√3 ω^2 - i√3 z(G)
    [1 - i√3] z(G) = ε + i√3 ω^2
    -2ω z(G) = ε + i√3 ω^2
    z(G) = -1/2 ω^2 [ε + i√3 ω^2] = -1/2 ω^2 ε - i√3/2 ω

    Consider
    z(B) + ω z(E) + ω^2 z(G)
    = ω + ω(ε+ω^2)/2 + ω^2 [-1/2 ω^2 ε - i√3/2 ω]
    = ω + 1/2 ωε + 1/2 - 1/2 ωε - i√3/2
    = ω - ω
    = 0

    As a result, ∆BEG is equilateral.

    ReplyDelete
  4. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27405&p=134056

    ReplyDelete