= Solution
Let $p$ be the midpoint of a geodesic $[y_1,y_2]\subseteq Y$ and put $L=d(y_1,y_2)$. Apply the <geodesic quadrilateral in a hyperbolic metric space> bound to the quadrilateral with consecutive vertices $y_1,z_1,z_2,y_2$. The point $p$ is within $2\delta$ of one of the other three sides. It cannot be within $2\delta$ of $[z_1,z_2]\subseteq Z$, because every point of $Y$ has distance greater than $2\delta$ from every point of $Z$.
By symmetry there is therefore a point $q\in[y_1,z_1]$ with $d(p,q)\le2\delta$. The <triangle inequality> gives
$$
d(y_1,q)\ge d(y_1,p)-d(p,q)\ge L/2-2\delta
$$
and hence
$$
d(z_1,p)
\le d(z_1,q)+2\delta
=d(z_1,y_1)-d(y_1,q)+2\delta
\le d(z_1,y_1)-L/2+4\delta.
$$
Since $p\in Y$ and $y_1$ is a closest point of $Y$ to $z_1$, we also have $d(z_1,y_1)\le d(z_1,p)$. Therefore $L\le8\delta$, as required.
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