Saturday, February 21, 2015

Geometry Problem 1088: Triangles, Perpendicular, Concurrent Lines

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1088: Triangles, Perpendicular, Concurrent Lines.

2 comments:

  1. From A₁A₂BB₃ concyclic, we have ∠C₁A₁O₁=∠A₂BB₃.
    From A₁A₂CC₃ concyclic, we have ∠B₁A₁O₁=180°−∠A₂CC₃.
    From B₁B₂AA₃ concyclic, we have ∠C₁B₁O₁=∠B₂AA₃.
    From B₁B₂CC₃ concyclic, we have ∠A₁B₁O₁=∠B₂CC₃.
    From C₁C₂AA₃ concyclic, we have ∠B₁C₁O₁=∠C₂AA₃.
    From C₁C₂BB₃ concyclic, we have ∠A₁C₁O₁=180°−∠C₂BB₃.

    Since A₁A₂, B₁B₂, C₁C₂ concurrent at O₁,
    sin∠A₁B₁O₁ / sin∠C₁B₁O₁ × sin∠B₁C₁O₁ / sin∠A₁C₁O₁ × sin∠C₁A₁O₁ / sin∠B₁A₁O₁ = 1

    Therefore,
    sin∠B₂CC₃ / sin∠B₂AA₃ × sin∠C₂AA₃ / sin∠C₂BB₃ × sin∠A₂BB₃ / sin∠A₂CC₃ = 1

    Hence, AA₃, BB₃, CC₃ concurrent at O₂.

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  2. http://s11.postimg.org/jzprw7z7n/pro_1088.png
    Let a, a’, b, b’ and c, c’ are angles shown in the sketch
    Let (UV, YW) represent angle form by rays UV and YW
    We have (CB, Cz) = a …. ( A1, A2, C, C3 cocyclic)
    And ( CA,Cz) =b ….. ( C, C3, B2 and B1 cocyclic)
    Similarly (AC,Ax)=b’ and (Ax,AB)= c’
    And (BA,By)= c and ( By,BC)= a’

    Applying Ceva’s theorem in triangle A1B1C1 and rays A1A2, B1B2 and C1C2 we have .
    (sin(a)/sin(a’)) x( sin(c)/sin(c’)) x ( sin(b’)/sin(b))= 1
    In triangle ABC and rays Ax, By and Cz
    Verifying that (sin(CB, Cz)/sin(CA, Cz)) x( sin(AC, Ax)/sin(Ax, AB)) x ( sin(BA, By)/sin(By, BC))=
    (sin(a)/sin(a’)) x( sin(c)/sin(c’)) x ( sin(b’)/sin(b))= 1
    So Ax, By and Cz are concurred at O2 per converse of Ceva’s theorem

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