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Let $\mathbb{S}^2$ be the unit sphere. Let $0<\lambda<1$ be fixed. What is the smallest number $0<\mu<1$ (depending on $\lambda$) such that for any three points $A,B,C\in \mathbb{S}^2$, with $d(C,A)\le \pi/2, d(C,B)\le \pi/2$, one has $d(A_{\lambda},B_{\lambda})\le \mu d(A,B)$. Here $A_{\lambda}$ and $B_{\lambda}$ are points on the geodesic paths $[A,C]$ and $[B,C]$ such that $d(A_{\lambda},C)=\lambda d(A,C)$ and $d(B_{\lambda},C)=\lambda d(B,C)$. ($d$ is the geodesic distance.)

My conjecture is $\sin(\lambda\pi/2)$. Is this correct?

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up vote 1 down vote accepted

Yes, this is correct.

Since you've found this result, I'll assume that you figured out the case where $A$ and $B$ are at equal distance from $C$, and it remains only to be shown that this is the worst case. Let me know if you'd like me to write more on the simpler case.

We can write $A$ and $B$ in spherical coordinates with $C$ at one pole and $A$ at $\phi=0$:

$$A=\pmatrix{\sin\theta_A\\0\\\cos\theta_A},\quad B=\pmatrix{\sin\theta_B\cos\phi\\\sin\theta_B\sin\phi\\\cos\theta_B}\;.$$


$$A_\lambda=\pmatrix{\sin\lambda\theta_A\\0\\\cos\lambda\theta_A},\quad B_\lambda=\pmatrix{\sin\lambda\theta_B\cos\phi\\\sin\lambda\theta_B\sin\phi\\\cos\lambda\theta_B}\;,$$

and the geodesic distances are the arccosines of the scalar products:

$$ \begin{eqnarray} d(A,B)&=&\arccos\left(\sin\theta_A\sin\theta_B\cos\phi+\cos\theta_A\cos\theta_B\right)\;,\\ d(A_\lambda,B_\lambda)&=&\arccos\left(\sin\lambda\theta_A\sin\lambda\theta_B\cos\phi+\cos\lambda\theta_A\cos\lambda\theta_B\right)\;. \end{eqnarray} $$

We can rewrite them in terms of $\Sigma=\theta_A+\theta_B$ and $\Delta=\theta_A-\theta_B$:

$$ \begin{eqnarray} d(A,B)&=&\arccos\frac12\left((1-\cos\phi)\cos\Sigma+(1+\cos\phi)\cos\Delta \right)\;,\\ d(A_\lambda,B_\lambda)&=& \arccos\frac12\left((1-\cos\phi)\cos\lambda\Sigma+(1+\cos\phi)\cos\lambda\Delta\right) \;. \end{eqnarray} $$

To see where the desired inequality is most strongly violated, we can differentiate the discrepancy with respect to $\Delta$:

$$\frac{\partial}{\partial\Delta}\left(d(A_\lambda,B_\lambda)-\mu d(A,B)\right)=\frac12(1+\cos\phi)\left(\frac{\lambda\sin\lambda\Delta}{\sin d(A_\lambda,B_\lambda)}-\mu\frac{\sin\Delta}{\sin d(A,B)}\right)\;.$$

Since we already know $\mu\ge\sin(\lambda\pi/2)\ge\lambda$, we must have $d(A_\lambda,B_\lambda)\gt\lambda d(A,B)$ for the inequality to be violated, and thus $\sin d(A_\lambda,B_\lambda)\gt\lambda \sin d(A,B)$. As all the quantities are non-negative (since $\theta\le\pi/2$ and $0\lt\lambda\lt1$), we can bound the derivative thus:

$$ \begin{eqnarray} \frac{\partial}{\partial\Delta}\left(d(A_\lambda,B_\lambda)-\mu d(A,B)\right) &\le& \frac12(1+\cos\phi)\left(\frac{\lambda\sin\lambda\Delta}{\lambda\sin d(A,B)}-\mu\frac{\sin\Delta}{\sin d(A,B)}\right)\\ &=& \frac12\frac{1+\cos\phi}{\sin d(A,B)}\left(\sin\lambda\Delta-\mu\sin\Delta\right)\\ &\le& \frac12\frac{1+\cos\phi}{\sin d(A,B)}\left(\sin\lambda\Delta-\sin\frac{\lambda\pi}2\sin\Delta\right)\\ &\le&0\;. \end{eqnarray} $$

Thus, the discrepancy is non-increasing wherever it is non-negative, and hence for it to be positive anywhere, it would have to be positive at $\Delta=0$. It therefore suffices to treat that case, which leads to the result you found.

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