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!

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

I've rewritten your term to this:


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


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:


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.