Distance to an apartment of the affine building of GL(N) Here $F$ is a locally compact non-archimedean non-discrete field.
Let $X$ be the reduced  (affine) Bruhat-Tits building of ${\rm GL}(n,F)$. Fix a maximal split torus $T$. Let $B$ be  a Borel subgroup  containing $T$ and write $U$ for the unipotent radical of $B$. Let $A$ be the unique apartment of $X$ stabilized by the normalizer $N_G (T)$ of $T$ in $G$. Finally fix a vertex $x$ of $X$.
It is not difficult to see that there exists a unique point $x_U$ of  $A$  such that $x_U =u.x\in A$ for some $u\in U$ (use Iwasawa decomposition). Making $B$ vary among the Borel subgroups containing $T$, one gets $n!$ points $x_U$ in $A$. 
My first question is: 

What is the link between the points $x_U$ and the projection $x_A$ of $x$ onto the apartment $A$ ($A$ is a closed convex subset of the CAT$(0)$ space $X$ and this projection is well defined) ? (For $n=2$, $x_A$ is the isobarycenter.)

Now fix $B$ and consider $x$ and $u$ as above. Assume that $x\not\in A$. Then $u$ fixes a sector $C$ of $A$. Let $c$ be a point of $C$. Consider the geodesic segments $[c,x]$ and $[c, x_U ]$ (so that $[c,x_U ]=u.[c,x]$). Let $c_0$ be the unique point of $[c,x]$ such that $[c,c_0 ]\subset A$ and $(c_0 ,x]\cap A =\emptyset$. 
My second question is: 

Is the point $c_0$ close to $x_A$ ? Can one choose $c$ such that $c_0 =x_A$ ?

My third question is : 

Do we have $d(x,x_A )=\frac{1}{2}d(x,x_U )$ ? 

 A: I think that the answers to your three questions are negative. Here's an example for $n=3$.
Choose first $x_A$ to be a vertex of $A$. In the link of $x_A$, it is possible to choose a chamber $d$ which is at distance $2$ from $2$ chambers in $A$, and at distance $3$ from the $4$ others. Choose $x$ in the alcove $d$ so that it  projects on $x_A$. To fix the ideas, you can take the middle of the side opposite to $x_A$.
Then the geodesic ray from $x$ to $x_A$ can be continuated  in only four ways, namely via one of the four alcoves opposite $d$ in the link of $x_A$. This answers negatively to your second question. The retraction from the 4 points towards which these geodesics are going are easily calculated: you just have to retract the geodesic segment from $x$ to $x_A$ from somewhere further on your geodesic ray; the four points you get are the four points in the alcove opposite to the four alcoves which were opposite to $d$ (and, of course, on the middle of the side opposite to $x_A$).
From this, it is possible also to answer negatively to the third question: there is some $x_U$ which is on in one of the two alcoves which are at distance $2$ from $d$. Then the distance from $x$ to this $x_U$ is the distance between the middle of two sides of a regular hexagon which are at distance $2$, and the distance from $x$ to $x_A$ is the distance to the center of this hexagon, so we do not have $d(x,x_A)=\frac 1 2 d(x,x_U)$. 
The calculation of the retractions centered at the two other points is not so easy. Let $\xi_1$ be one of them. The sector from $x_A$ to $\xi_1$ starts with some alcove $c$ which is at distance $2$ from $d$. Let $d'$ be the alcove adjacent to both $c$ and $d$, and let $H$ be the wall of $A$ which separates $c$ and $d'$. $H$ separates $A$ in two half-planes, say $\alpha$ and $-\alpha$, with $\xi_1$ in $\alpha$. Let $\beta$ be another half-plane bounded by $H$, such that $d'\in \beta$. Then $\beta\cup\alpha$ is another apartment $A_1$.
The retraction on $A_1$ centered at $\xi_1$ sends $d$ to a chamber which is at distance $2$ from $c$ and adjacent to $d'$, so it is the alcove in $\beta$ which is adjacent to $d'$. The retraction of $d'$ on $A$ is then the retraction of this alcove is the alcove which is at distance 2 from $c$ and in $-\alpha$. Since this alcove is already opposite to an alcove which is opposite to $d'$, we get the same point as some other $x_U$.
Of course, the retraction centered at the last point is treated in a similar way. So, in conclusion, there are only 4 different retractions of $x$. The two "double" points form a segment whose middle is $x_A$. The two other are symmetric with respect to $H$, but are not in the same sector. So the barycenter of our 6 points is some point of $H$ which is not $x_A$.
