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Topic: Alike Triangles
Replies: 1   Last Post: Dec 31, 2010 11:27 AM

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 Sujeet Kumar Posts: 190 From: Patna Registered: 10/21/10
Alike Triangles
Posted: Nov 29, 2010 10:15 AM

Consider the fucntion f(x)=2* sin^-1(mod[x/2]).

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.

Date Subject Author
11/29/10 Sujeet Kumar
12/31/10 Sujeet Kumar