# A number characterizing the deviation of a triangle from the regular triangle

Given a triangle $$\Delta$$ with sides of length $$a\le b\le c$$, consider the number $$q=\frac{a^4+b^4+c^4}{(a^2+b^2+c^2)^2}$$ and observe that $$\frac13\le q\le\frac12$$ and the extremal values of $$q$$ characterize some geometric properties of the triangle $$\Delta$$. Namely:

$$\bullet$$ $$q=\frac13$$ if and only if $$a=b=c$$ (which means that the triangle $$\Delta$$ is regular);

$$\bullet$$ $$q=\frac12$$ if and only if $$c=a+b$$ (which means that the triangle $$\Delta$$ is degenerated).

I am writing a paper (in applications of math to Electric Engineering) where the number $$q$$ is applied for evaluation of the deviation of a triangle (describing the quality of 3-phase electric energy) from being regular, and need to call the number $$q$$ somewhow (for example, quadrofactror), but wonder if $$q$$ already has some standard name. This motivates my

Question. Has the number $$q$$ some standard name in Plane Geometry?

• This is not an answer but just a comment that there is another function of the side lengths which does what you want--its minimum (zero) is taken at the degenerate cases and its maximum at the equilateral one. It does have a name (area) and its expression as a function of $a$, $b$ and $c$ is Heron's formula. Apr 8, 2021 at 6:57
• @bathalf15320 Thank you for the comment. The area is a good function but it is not invariant under similarity transformations. So, it does not evaluate the form (and the regularity) of the triangle. Apr 8, 2021 at 7:11
• Yes. I was tacitly normalising so that the longest side has length $1$--should have made that explicit. Apr 8, 2021 at 13:37
• MSE is a right forum for such type questions. Apr 8, 2021 at 16:50
• @user64494 Probably you are right concerning MSE, but I am not a member of MSE and would not like to register there only in sake of this single question. Apr 8, 2021 at 17:04

Added to my comment above, this time taking care of my carelessness in not normalising: one has the formula $$\frac{16 A^2}{(a^2+b^2+c^2)^2}=1-\frac{2(a^4+b^4+c^4)}{(a^2+b^2+c^2)^2}$$ which shows, at least in my book, that a normalised version of the area $$A$$ (more precisely of its square) does the trick.
• Note that $\dfrac{4A}{a^2+b^2+c^2}$ is the tangent of the Brocard angle $\omega$. Thus, $q = \left(1-\tan^2\omega\right)/2$. Apr 8, 2021 at 18:34
• @bathhalf15320 Thank you very much for your answer. It was very helpful. In fact, my formulas involved $q$ in the subformula $\sqrt{3−6q}$ and now I understand what this subformula actually means: it is the normalized area of the triangle! Great! I would like to write acknowledgement to your help. Should I write your nick bathhalf15320 or some real name will be better? Thank you. Apr 9, 2021 at 4:39
• Indeed, in the mentioned paper on Wikipedia there is a similar inequality with the same lower and upper bounds: $\frac13\le \frac{a^2+b^2+c^2}{(a+b+c)^2}<\frac12$. But the upper bound $\frac12$ is attainable only at triangles with two coinciding vertices. On the other hand, the upper bound $\frac12$ at my inequality is attainable at each triangle on a line. Apr 8, 2021 at 17:19