To visit all four points, we notice that we must travel along at least three different segments. The sum of the shortest three segments is $3+4+5=12$, but we quickly notice that it is impossible to start at one point and visit the other three points by traveling on a path with length $12$ ($DB$, $BC$, and $CD$ do not let us visit point $A$, and it is not possible to travel on $AD$, $CD$, and $BD$ in a continuous path). We now look for a path with length $13$ and notice that traveling from point $A$ to $D$ to $B$ to $C$ works. Alternatively, $B$ to $D$ to $C$ to $A$ also works. Both paths have length $\boxed{13}$.
Question:
What is the shortest distance that can be travelled if you start at any point, $A$, $B$, $C$, or $D$, and visit the other three points once?

[asy]
pair A,B,C, D;
A = (-3,3); B = (3,0); C = (0, -4); D = (0,0);
draw(D--B--C--cycle); draw(D--A--B--cycle);draw(D--A--C--cycle);
label("$A$", A, NW);label("$B$", B, E); label("$C$", C, S);label("$D$", D, NE);
label("3", D--B, S); label("6", A--B, NE); label("6", A--C, SW); label("4", D--C, NW+N);
label("5", A--D, SE+NE); label("5", C--B, E);
[/asy]
Answer:
Let's think step by step.

user
13
What is the shortest distance that can be travelled if you start at any point, $A$, $B$, $C$, or $D$, and visit the other three points once?

[asy]
pair A,B,C, D;
A = (-3,3); B = (3,0); C = (0, -4); D = (0,0);
draw(D--B--C--cycle); draw(D--A--B--cycle);draw(D--A--C--cycle);
label("$A$", A, NW);label("$B$", B, E); label("$C$", C, S);label("$D$", D, NE);
label("3", D--B, S); label("6", A--B, NE); label("6", A--C, SW); label("4", D--C, NW+N);
label("5", A--D, SE+NE); label("5", C--B, E);
[/asy]
test