Is there a quantitive measurement of how equilateral a triangle? Such as the following:
Triangle A(2,3,4)
Triangle B(3,3,3)
Triangle C(10,10,1)
Is there a measurement such that 0 is a triangle that doesn't satisfy the triangle inequality theorem and 1 is a perfect equilateral triangle, assuming that a triangle's side length range is not infinity but rather constrained between (0,c].
Therefore I should get:
Triangle A = some value between 0 - 1
Triangle B = 1
Triangle C = value close to 0
If R and r represent the circumradius and the in radius respectively then you may
consider the quanrity 1/(2-2r/R) which is 1 iff R=2r that is iff the triangle is equilateral.
an other possibility would be 1/(1+((a-b)^2+(b-c)^2+(c-a)^2)/(a+b+c)^2).
Maybe there is a simplest answer. Find first the three distances d1=d(A,B), d2=d(A,C), d3=d(B,C). Then compute the mean (i.e. the average) of the three distances, and then the standard deviation (which measures "how far" from the mean your data are, and in particular how far your data are from each other). If the triangle is equilateral, the standard deviation should be zero. Even better, if you divide the standard deviation by the mean, and multiply by 100, you can measure the deviation in percentage, which is clearer (such quotient is called the "coefficient of variation", in Statistics). Again, equilateral means that the coefficient of variation is exactly zero. Algorithmically, you can fix a tolerance (say, 0.1%, that is up to you and depends on how accurate you want to/can be), and decide that any triangle with coefficient of variation below that tolerance is very likely to be equilateral.
Dear Juan Gerardo Alcazar I think you are right, but you need some measure which is invariant when you apply homothetic transformation on the triangle, otherwise similar triangles would yield different results. If you do this you will see that the proposed answers are really different.
If you want the test to be scale invariant you can do the same thing, but with the angles, as P. Breuer suggests.
Dear Ahmed Fakeih would you add an explanation into your question? As it is written, it produces legitimate misunderstanding.
A,B,C seem to be vertices of a triangle. But are not. Apparently, each of A,B,C is a triangle, and the corresponding numbers are the lengths of its sides.
The square of the area is p(p-a)(p-b)(p-c) where p=(a+b+c)2. For an equilateral triangle, a=b=c=2p/3 that is 8p4/27. The expression 27(p-a)(p-b)(p-c)/p3 is maximized for an equilateral triangle and zero. It is negative if the three lengths cannot for a triangle (assuming all lengths given are positive), but the question was asked about triangles only.
Some of the answers to this intuitive question are very interesting as far as I am concerned.
What about taking the sum of the 3 differences in length (absolute value) and dividing the sum by 3? This is 0 if and only if the triangle is equilateral. May be this is equivalent to one of the earlier answer(s) I didn't read!
A little addition: I would further propose to divide the previously obtained answer (in my method) by the sum of the lengths of the sides of the triangle. This would ensure that similar triangles give the same quantitative deviation from "being equilateral".
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