 Solution

Assume (i), and suppose a map $f$ as in (ii) existed. Thenare closed, cover $T$, and contain $I,J,K$, respectively. Butsince the three sides have no common point. This contradicts (i), so (i) implies (ii).

Conversely, suppose (i) fails. Let $A,B,C$ be a counterexample and putNo point lies in all three closed sets, so $D=d_A+d_B+d_C>0$. Identify $T$ affinely with the standard simplex so that side $I$ is the side on which the first barycentric coordinate is zero, and similarly for $J,K$. Define $f(x)$ to be the point with barycentric coordinatesBecause $A\cup B\cup C$ covers $T$, at least one distance is zero, so $f(x)\in \partial T$. If $x\in I\subset A$, then $d_A(x)=0$, so $f(x)\in I$; likewise for $J,K$. Thus $f$ is a forbidden map from (ii). This proves (ii) implies (i).