Krdytkyu

Characterize all triples $(a,b,c)$ of positive real numbers such that
$$ a^2-ab+bc = b^2-bc+ca = c^2-ca+ab. $$

This is the equality case of the so-called Vasc inequality. I think the answer is that $a=b=c$ or $a:b:c = sin^2(4pi/7) : sin^2(2pi/7) : sin^2(pi/7)$ and cyclic equality cases. I'm mostly interested if there is a way to derive this reasonably, other than just magically guessing the solutions and then doing degree-counting. This was left as a "good exercise" in Mildorf that has bothered me for years.

Too long for comment

Using $sin(pi-A)=+sin A,cos(pi-B)=-cos B,sin2C=2sin Ccos C$

$$dfrac{sin^2dfrac{2pi}7}{sin^2dfrac{pi}7}=4cos^2dfracpi7=2+2cosdfrac{2pi}7$$

$$dfrac{sin^2dfrac{pi}7}{sin^2dfrac{3pi}7}=dfrac{sin^2dfrac{6pi}7}{sin^2dfrac{3pi}7}=4cos^2dfrac{3pi}7=2+2cosdfrac{6pi}7$$

$$dfrac{sin^2dfrac{3pi}7}{sin^2dfrac{2pi}7}=dfrac{sin^2dfrac{4pi}7}{sin^2dfrac{2pi}7}=4cos^2dfrac{2pi}7=2+2cosdfrac{4pi}7$$

From How to solve $8t^3-4t^2-4t+1=0$ and factor $z^7-1$ into linear and quadratic factors and prove that $ cos(pi/7) cdotcos(2pi/7) cdotcos(3pi/7)=1/8$,

the roots of $$x^3+x^2-2x-1=0$$ are $2cosdfrac{2pi}7,2cosdfrac{4pi}7,2cosdfrac{6pi}7$

Replace $x$ with $y-2$ to find $$y^3-5y^2+6y-1=0$$

If $c=0$ then $a^2-ab=b^2=ab$, which gives $a=b=c=0.$

Let $abcneq0$ and $a=xb$.

Thus, from the first equation we obtain:
$$a^2-ab-b^2=(a-2b)c.$$
If $a=2b$ so $a=b=c=0$, which is impossible here.

Thus, $c=frac{a^2-ab-b^2}{a-2b}$ and from the second equation we obtain:
$$b^2+frac{(a-b)(a^2-ab-b^2)}{a-2b}=frac{(a^2-ab-b^2)^2}{(a-2b)^2}-frac{(a^2-ab-b^2)a}{a-2b}+ab$$ or
$$1+frac{(x-1)(x^2-x-1)}{x-2}=frac{(x^2-x-1)^2}{(x-2)^2}-frac{(x^2-x-1)x}{x-2}+x$$ or
$$(x-1)(x^3-5x^2+6x-1)=0,$$ which gives $x=1$ and $a=b=c$ or
$$x^3-5x^2+6x-1=0.$$
Now, easy to show that $frac{sin^2frac{2pi}{7}}{sin^2frac{pi}{7}}$, $frac{sin^2frac{pi}{7}}{sin^2frac{3pi}{7}}$ and $frac{sin^2frac{3pi}{7}}{sin^2frac{2pi}{7}}$ they are roots of the last equation.

For example:
$$left(frac{sin^2frac{2pi}{7}}{sin^2frac{pi}{7}}right)^3-5left(frac{sin^2frac{2pi}{7}}{sin^2frac{pi}{7}}right)^2+6cdotfrac{sin^2frac{2pi}{7}}{sin^2frac{pi}{7}}-1=$$
$$=left(4cos^2frac{pi}{7}right)^3-5left(4cos^2frac{pi}{7}right)^2+6left(4cos^2frac{pi}{7}right)-1=$$
$$=left(2+2cosfrac{2pi}{7}right)^3-5left(2+2cosfrac{2pi}{7}right)^2+6left(2+2cosfrac{2pi}{7}right)-1=$$
$$=8cos^3frac{2pi}{7}+4cos^2frac{2pi}{7}-4cosfrac{2pi}{7}-1=$$
$$=2left(4cos^3frac{2pi}{7}-3cosfrac{2pi}{7}right)+6cosfrac{2pi}{7}+2+2cosfrac{4pi}{7}-4cosfrac{2pi}{7}-1=$$
$$=2cosfrac{2pi}{7}+2cosfrac{4pi}{7}+2cosfrac{6pi}{7}+1=$$
$$=frac{2sinfrac{pi}{7}cosfrac{2pi}{7}+2sinfrac{pi}{7}cosfrac{4pi}{7}+2sinfrac{pi}{7}cosfrac{6pi}{7}}{sinfrac{pi}{7}}+1=$$
$$=frac{sinfrac{3pi}{7}-sinfrac{pi}{7}+sinfrac{5pi}{7}-sinfrac{3pi}{7}+sinfrac{7pi}{7}-sinfrac{5pi}{7}}{sinfrac{pi}{7}}+1=0.$$
Since we have no another roots, we are done!

Suppose $a=b$, then $bc=b^2=c^2-cb+b^2$, those $a=b=c$. Which includes cyclic equality cases as well.

Now lets assume WLOG $a>b>c$. Show that in this case above equations don't hold simultaneously.