# If this relation holds, then is the triangle equilateral?

Let $ABC$ be a triangle. If $$\sum_{cyc}\frac{BC}{4AC\cos^2({\frac{\angle BAC}{2})}+BC}=\frac{3}{4}$$ then the triangle is equilateral? We can check if we set $\widehat{BAC}=\pi/3$ and $AB=BC=CA$ that the relation holds. If yes, how to prove this? Thank you!

• What is the source of the question? – Sawarnik May 2 '14 at 20:02
• It's from a problem and I got it trying to solve the problem, but this seems to take me nowhere. – user146371 May 2 '14 at 20:09
• @user146371 Can you post the original problem? – evil999man May 7 '14 at 3:25
• I just set a similar question (math.stackexchange.com/questions/783808/…) by a simple convexity argument, maybe the same works here, too. – Jack D'Aurizio Jul 12 '14 at 1:25

I've rewritten your term to this:

$$t(a,b,c)=\frac{a}{4b\left(\frac12+\frac{b^2+c^2-a^2}{4bc}\right)+a}$$

It's based on the cosine law $\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$ and the half angle formula $\cos^2\frac{\alpha}2=\frac{1+\cos\alpha}{2}$. Now you are essentially asking whether

$$p(a,b,c)=t(a,b,c)+t(b,c,a)+t(c,a,b)-\tfrac34$$

has any zeros for positive $a,b,c$ which also satisfy the triangle inequalities. First off, your formula is scale invariant, so w.l.o.g. we may assume $a=1$. You can also ignore the denominator of the whole expression, so you are left with a sixth degree polynomial in two variables:

$$3a^6+5a^5b-5a^4b^2-12a^3b^3-a^2b^4+7ab^5+3b^6+7a^5c+12a^4bc-5a^3b^2c-3a^2b^3c+12ab^4c+5b^5c-a^4c^2-3a^3bc^2-3a^2b^2c^2-5ab^3c^2-5b^4c^2-12a^3c^3-5a^2bc^3-3ab^2c^3-12b^3c^3-5a^2c^4+12abc^4-b^2c^4+5ac^5+7bc^5+3c^6$$

Ugly. But one can do math on that. For example, one can look for special values of $b$ where the number of associated values $c$ will change. There are four of them, computed as roots of the discriminant of the above polynomial:

$$b_1\approx0.0121788129\qquad b_2=1\qquad b_3\approx89.092238572\qquad b_4\approx1201.1620407$$

Theoretically you can compute the corresponding $c$ values for every position in between. By looking at $p(1,b,b-1), p(1,b,b+1), p(1,b,1-b)$ you can see that there is no position in between these $b_i$ where the resulting triangle would become degenerate, i.e. have one triangle inequality satisfied with equality. There are solutions at $b\le0$ and $b=1$, but the first is outside our scope and the second is one of the special points, which we'll treat in a moment. Also look at $p(1,b,0)$ to find sign changes, i.e. places where the positivity constraint might start being satisfied. You'll find solutions at $b\le0$ and $b=1$, but also one at $b_5=\frac{\sqrt{37}-1}{6}\approx0.8471270884$.

Next you can take each range (i.e. $b\in(0,b_1), (b_1,b_5), (b_5,b_2), (b_2,b_3), (b_3,b_4), (b_4,\infty)$ resp.) and compute one set of $c$ values anywhere in that range. You will find that they all violate one triangle inequality or another, or the positivity condition. Since the $c$ values form continuous algebraic curves between the special $b_i$, and since we have shown that none of these conditions will change inside the areas between the special $b_i$ values, we know that there can be no solution in the different areas.

So all that remains is computing valid $c$ values at each of the special $b_i$ itself. You'd have to do so using exact arithmetic, i.e. algebraic numbers. Then you have a finite set of possible combinations, and can verify that $a=b=c=1$ is the only one which satisfies all triangle inequalities and positivity constraints.

This is a pretty ugly solution, but it gets the job done. If anyone has a better solution, I'd like to hear about that.