Thursday, May 6, 2010

Fox 283

This one is updated. We received an excellent solution with geometric construction, that will later be our first construction problem. See ya later... http://www.8foxes.com/

6 comments:

  1. i think 15 degrees is a solution. do you have the proof newzad?

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  2. divide 3θ=2θ+θ
    use congruent and isosceles triangles
    you will find 90-θ=60+θ
    θ=15

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  3. there is a solution for every angle x<40
    let be GEF the in-triangle
    GEF is isoscele with angles 60+2x,60-x,60-x
    in G we must have x+60+2x<180
    construction
    ABC the equilateral triangle
    choose x<40 ;point G' on[AB]
    draw angle yG'B=x;yG'F'=60+2x,F'on [AC)
    then isoscele triangle G'F'E'
    line AE' intersects [BC] at E
    GFE is homothetic to G'F'E'
    we have the angles x at G,2x at E,3x at F.

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  4. with 40 the vertex G=A
    it works if we take the angle 3x with the line (AC) but not with the side [AC]

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