Alike Triangles
Sujeet Kumar
We define “Alike Triangle” for any root @ of any equation as isosceles triangle with vertical angle
equal to f(@).
For example equation x^3 – 3*x + 1=0 has roots 2sin(pi/18), 2sin(5pi/18) and -2sin(7pi/18).
So for root -2sin(7pi/18) corresponding alike triangle will be isosceles triangle with vertical angle
equal to
f(-2sin(7pi/18))=2*sin^-1(mod[{-2sin(7pi/18)}/2])=2*sin^-1(mod[-sin(7pi/18)])=2*sin^-1(sin(7pi/18))
=2*(7pi/18)
In the same way, “Alike Triangles” for equation x^3 – 3*x + 1=0 will be isosceles triangle with vertical
angles 20 degree, 100 degree and 140 degrre.
We have seen these triangles have some alike properties.(See my post with title Puzzling Geometry).
Here I am not giving exact definition of alike properties.
Q.Find alike triangles and their alike properties corresponding to following 2 equations-
x^2 -x -1=0 and x^3 – x^2 -2*x +1=0.
Sujeet Kumar
Equation x^2 -x – 1=0 has its roots as -2sin18 and 2sin 54.
So , corresponding alike triangles will be isosceles triangle with vertical angles 36 and 108..
In triangle with vertical angles 36, angle bisector BD will be equal to base BC ie BD=BC.
Another triangle is isosceles triangle with vertical angle 108.
There is change in sign of root.So, we will see the property of external angle bisector BD.
Again , we have BD=BC.
Equation x^3 – x^2 -2x +1 =0 has roots 2sin(5pi/14), 2sin(pi/14) and -2sin(3pi/14).
Alike triangles for this equation will be isosceles triangles with vertical angles 5pi/7 , pi/7 and 6pi/7.
Try to find alike properties for these triangles.