3
$\begingroup$

Let ABC be a spherical triangle, where the spherical distance (or angle) AB is $\pi/2$ and $C\neq -A$. For $t\in[0,1]$, let $B(t)$ (resp. $C(t)$) be the only point on the segment $[AB]$ (resp. $[AC]$) such that $AB(t) = t AB$ (resp. $AC(t) = t AC$).

Question: Is it true that $B(t)C(t) \leq 2 BC$ for any $t\in [0,1]$?

The worst case seems to happen when $C$ is close to $-A$, so that $BC\simeq \pi/2$ but $C(t)$ goes in the direction opposite to $B$, so that at some point $B(t)C(t) = \pi$. (See the picture below, where D is that B(t) and E that C(t).)

Worst case


For those who care, the question is related to my paper arXiv:1507.05485 where Lemma 15 is basically the question I ask here but the proof is wrong, as pointed out by a very diligent referee. I can fix the proof, but only with $B(t)C(t) \leq 3 BC$...

$\endgroup$
2
  • 1
    $\begingroup$ Using spherical cosine law it would be left to prove that $$\cos(t\pi/2)\cos(tAC)+\sin(t\pi/2)\sin(tAC)\cos(\alpha)\geq 2\sin^2(AC)\cos^2(\alpha)-1$$ for all $t\in [0,1]$ where $\alpha$ is the angle at $A$. $\endgroup$ – user35593 Mar 9 '16 at 14:54
  • $\begingroup$ @user35593 In the case where $BC \leq \pi/2$, indeed... $\endgroup$ – Lierre Mar 9 '16 at 15:30
2
$\begingroup$

Unfortunately, the answer of user35593 is broken but I obtained something in the same line : $$\cos(B(t)C(t)) = \cos(tAB-tAC) - \sin(tAB)\sin(tAC)(1-\cos a),$$ where $a$ is the angle at $A$. Thus $$\cos(B(t)C(t)) \geq \cos(AB - AC) + \cos a - 1 = \sin(AC) + \cos a -1.$$ We assume that $BC < \frac\pi2$, so that $\cos a \sin(AC) = \cos(BC) > 0$. Since $0\leq AC \leq \pi$ we have $\sin(AC)\geq 0$ and hence $\sin(AC)>0$ and $\cos(a)>0$. This is enough to have $$\sin(AC) + \cos a -1 \geq 2 \cos(a)^2 \sin(AC)^2 - 1 = \cos(2 BC).$$ This gives the claim.

$\endgroup$

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.