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/
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.
http://www.8foxes.com/
try tetha=15
ReplyDeletei think 15 degrees is a solution. do you have the proof newzad?
ReplyDeletedivide 3θ=2θ+θ
ReplyDeleteuse congruent and isosceles triangles
you will find 90-θ=60+θ
θ=15
there is a solution for every angle x<40
ReplyDeletelet 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.
and 40
ReplyDeletewith 40 the vertex G=A
ReplyDeleteit works if we take the angle 3x with the line (AC) but not with the side [AC]
(θ,2θ,3θ) triangles exist.
ReplyDeleteSee some examples.
http://obiwan3.greater.jp/math_battle/html/0187.html