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.

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]