Explanation:
Okay, so I need to figure out the shortest distance someone can travel if they start at any of the points A, B, C, or D and visit the other three points once. Let me try to visualize the diagram first.

From the Asymptote code, the coordinates are given as A(-3,3), B(3,0), C(0,-4), D(0,0). There are several lines drawn: D connected to B, B to C, C to D; D to A, A to B, and D to A to C. The labels on the sides are: 3 units between D and B, 6 units between A and B, 6 units between A and C, 4 units between D and C, 5 units between A and D, and 5 units between C and B.

First, I need to confirm the distances between all pairs of points. Let me list them out:

- AB: 6 units
- AC: 6 units
- AD: 5 units
- BC: 5 units
- BD: 3 units
- CD: 4 units

Wait, but maybe I should check if these distances are consistent with the coordinates? Let me verify that.

Coordinates:

A(-3,3), B(3,0), C(0,-4), D(0,0)

Calculating AB distance: distance between (-3,3) and (3,0). Using distance formula sqrt[(3 - (-3))² + (0 - 3)²] = sqrt[(6)² + (-3)²] = sqrt[36 + 9] = sqrt[45] ≈ 6.708. But the label says AB is 6 units. Hmm, that's conflicting. Maybe the coordinates are not to scale? Or maybe the labels are different? Wait, the Asymptote code labels the sides with specific lengths. Let me check again.

The Asymptote code labels:

- D--B: 3
- A--B: 6
- A--C: 6
- D--C: 4
- A--D: 5
- C--B: 5

So maybe the coordinates are not exact? Or maybe the diagram is schematic? Maybe I shouldn't rely on coordinates but just on the given distances? Wait, but the coordinates are given, so maybe the distances are calculated from coordinates? Let me check again.

Wait, if AB is supposed to be 6 units, but according to coordinates, it's sqrt(45) ≈ 6.708. So there's a discrepancy. Maybe the coordinates are not accurate? Or maybe the labels are different? Wait, the Asymptote code draws D--B--C--cycle, so triangle DBC? Then D--A--B--cycle, triangle DAB? And D--A--C--cycle, triangle DAC?

But maybe the labels are not the lengths of the sides of the triangles but something else? Wait, the labels are on the sides. For example, D--B is labeled 3, so that's the length of segment DB. Similarly, A--B is labeled 6, so AB is 6. So maybe the coordinates are just for drawing purposes, but the actual distances are given? That would make sense. So perhaps I should ignore the coordinates and just use the given distances? But then how do I know the positions? Wait, but maybe the coordinates are correct, and the labels are wrong? Or maybe the labels are correct, and the coordinates are just approximate?

This is confusing. Maybe I need to check if the given distances are consistent with coordinates? Let me check:

First, distance from D(0,0) to B(3,0) is 3 units. That matches the label "3" on D--B. Good.

Distance from A(-3,3) to B(3,0): sqrt[(3 - (-3))² + (0 - 3)²] = sqrt[6² + (-3)²] = sqrt[36 + 9] = sqrt[45] ≈ 6.708. But the label says 6. So that's inconsistent. Similarly, distance from A(-3,3) to C(0,-4): sqrt[(0 - (-3))² + (-4 - 3)²] = sqrt[3² + (-7)²] = sqrt[9 + 49] = sqrt[58] ≈ 7.616, but the label says 6. Hmm. So the coordinates don't match the given distances. Therefore, maybe the coordinates are not to scale, and I should just use the given distances?

Alternatively, maybe the coordinates are correct, and the labels are different? Wait, the Asymptote code labels "3" on D--B, which is from (0,0) to (3,0), which is indeed 3 units. Then "6" on A--B, which is from A(-3,3) to B(3,0). But according to coordinates, that's sqrt(45). Maybe the diagram is not to scale? Maybe the labels are the actual distances? So perhaps the coordinates are just for drawing, but the actual distances are as labeled?

In that case, maybe I should consider the distances given in the labels as the actual distances between the points. So:

AB = 6, AC = 6, AD = 5, BC = 5, BD = 3, CD = 4.

So, with that, I can model the points as a graph with these distances. Now, the problem is: starting at any point A, B, C, or D, visit the other three points once. So, the question is asking for the shortest possible path that starts at one of the points, visits the other three, and ends at the last one. So, it's a traveling salesman problem (TSP) where we need to find the shortest possible route that visits all four points, but since we can start at any point, we need to find the minimal tour.

Wait, but the problem says "start at any point, A, B, C, or D, and visit the other three points once." So, it's a path that starts at one point, visits the other three, so it's a path that goes through all four points, but since it's a path, not necessarily a cycle. So, the total distance would be the sum of the distances between consecutive points. So, for example, starting at A, going to B, then to C, then to D. Or starting at A, going to D, then to C, then to B. Etc. So, the problem is to find the minimal such path.

But since we can start at any point, we need to find the minimal path that starts at any of the four points and visits the other three. So, the minimal path would be the minimal Hamiltonian path in the complete graph with these distances.

Alternatively, since the problem says "visit the other three points once," so starting at one, then visiting the other three, so the path has four points, with each point visited once. So, the total number of possible paths is 4 * 3! = 24, since starting at any of the four points, and then permuting the other three. But since the graph is undirected, some paths are the same in reverse. However, since we are looking for the minimal distance, we can consider all possible permutations.

But maybe there's a smarter way than enumerating all 24 possibilities. Let me think.

Alternatively, since the problem is small, maybe I can find the minimal path by considering the distances.

First, let me note the distances between all pairs:

AB = 6

AC = 6

AD = 5

BC = 5

BD = 3

CD = 4

So, the distances are:

Between A and B: 6

A and C: 6

A and D: 5

B and C: 5

B and D: 3

C and D: 4

So, the distances are as above.

Now, the problem is to find the shortest path that starts at any point, visits the other three. So, for example, starting at A, then going to B, then to C, then to D. The total distance would be AB + BC + CD = 6 + 5 + 4 = 15. Alternatively, starting at A, going to D, then to B, then to C: AD + DB + BC = 5 + 3 + 5 = 13. That's shorter. Or starting at A, going to D, then to C, then to B: AD + DC + CB = 5 + 4 + 5 = 14. So, 13 is better.

Alternatively, starting at A, going to D, then to B, then to C: 5 + 3 + 5 = 13.

Alternatively, starting at A, going to D, then to C, then to B: 5 + 4 + 5 = 14.

Alternatively, starting at A, going to B, then to D, then to C: AB + BD + DC = 6 + 3 + 4 = 13.

Same as before. So, 13.

Alternatively, starting at A, going to C, then to D, then to B: AC + CD + DB = 6 + 4 + 3 = 13.

Same.

So, starting at A, there are multiple paths that give 13.

Now, let's check starting at B.

Starting at B, then going to D (distance 3), then to C (distance 4), then to A (distance 6): total 3 + 4 + 6 = 13.

Or starting at B, going to D, then to A, then to C: BD + DA + AC = 3 + 5 + 6 = 14.

Or starting at B, going to C (distance 5), then to D (distance 4), then to A: BC + CD + DA = 5 + 4 + 5 = 14.

Or starting at B, going to C, then to A, then to D: BC + CA + AD = 5 + 6 + 5 = 16.

Alternatively, starting at B, going to A (distance 6), then to D (distance 5?), wait, distance from A to D is 5, so BD is 3. Wait, starting at B, going to A (distance 6), then to D (distance 5?), but distance from A to D is 5. So, BA + AD + DC = 6 + 5 + 4 = 15.

Alternatively, starting at B, going to A, then to C, then to D: BA + AC + CD = 6 + 6 + 4 = 16.

So, the minimal starting at B is 13.

Now starting at C.

Starting at C, going to D (distance 4), then to B (distance 3), then to A: CD + DB + BA = 4 + 3 + 6 = 13.

Or starting at C, going to D, then to A, then to B: CD + DA + AB = 4 + 5 + 6 = 15.

Or starting at C, going to B (distance 5), then to D (distance 4), then to A: CB + BD + DA = 5 + 3 + 5 = 13.

Or starting at C, going to B, then to A, then to D: CB + BA + AD = 5 + 6 + 5 = 16.

Or starting at C, going to A (distance 6), then to D (distance 4), then to B: CA + AD + DB = 6 + 5 + 3 = 14.

So, minimal starting at C is 13.

Now starting at D.

Starting at D, going to B (distance 3), then to C (distance 4), then to A: DB + BC + CA = 3 + 5 + 6 = 14.

Or starting at D, going to B, then to A, then to C: DB + BA + AC = 3 + 6 + 6 = 15.

Or starting at D, going to A (distance 5), then to B (distance 6?), wait, distance from A to B is 6. So DA + AB + BC = 5 + 6 + 5 = 16.

Or starting at D, going to A, then to C, then to B: DA + AC + CB = 5 + 6 + 5 = 16.

Or starting at D, going to C (distance 4), then to B (distance 3), then to A: DC + CB + BA = 4 + 5 + 6 = 15.

Alternatively, starting at D, going to C, then to A, then to B: DC + CA + AB = 4 + 6 + 6 = 16.

Alternatively, starting at D, going to A (distance 5), then to C (distance 6?), wait, distance from A to C is 6. So DA + AC + CB = 5 + 6 + 5 = 16.

Alternatively, starting at D, going to A, then to B, then to C: DA + AB + BC = 5 + 6 + 5 = 16.

So, the minimal starting at D is 14.

So, summarizing:

Starting at A: minimal path is 13

Starting at B: minimal path is 13

Starting at C: minimal path is 13

Starting at D: minimal path is 14

Therefore, the overall minimal distance is 13.

But let me check if there are any other paths that I might have missed. For example, starting at A, going to D, then to B, then to C: 5 + 3 + 5 = 13. Starting at A, going to D, then to C, then to B: 5 + 4 + 5 = 14. Starting at A, going to B, then to D, then to C: 6 + 3 + 4 = 13. Starting at A, going to C, then to D, then to B: 6 + 4 + 3 = 13.

Similarly, starting at B, going to D, then to C, then to A: 3 + 4 + 6 = 13. Starting at B, going to C, then to D, then to A: 5 + 4 + 5 = 14. Starting at B, going to D, then to A, then to C: 3 + 5 + 6 = 14.

Starting at C, going to D, then to B, then to A: 4 + 3 + 6 = 13. Starting at C, going to B, then to D, then to A: 5 + 3 + 5 = 13.

Starting at D, going to B, then to C, then to A: 3 + 5 + 6 = 14.

So, all the minimal paths are 13.

Therefore, the answer is 13.

But let me check if there's a shorter path. For example, is there a way to have a path with total distance less than 13? Let's think.

Suppose we try to find a path that uses the shortest edges. The shortest edges are BD = 3, CD = 4, AD = 5, BC = 5, AB = 6, AC = 6.

So, the shortest edge is BD = 3. If we can include that in the path.

For example, if we have a path that includes BD, which is 3, then maybe other edges.

But since we need to visit all four points, we need three edges.

So, if we have BD (3), then we need two more edges. Let's see.

For example, starting at B, going to D (3), then from D, we can go to A (5) or to C (4). If we go to C (4), then from C, we need to go to A (6). So total is 3 + 4 + 6 = 13.

Alternatively, starting at D, going to B (3), then to C (4), then to A (6): same as above.

Alternatively, starting at D, going to B (3), then to A (5), then to C (6): 3 + 5 + 6 = 14.

Alternatively, starting at B, going to D (3), then to A (5), then to C (6): same as above.

Alternatively, if we can find a path that uses BD (3) and CD (4) and something else. Wait, but if we have BD and CD, that would be two edges, but we need three edges.

Wait, if we have BD (3), CD (4), and then another edge. But if we go from D to B to C, then we need to go from C to A, which is 6.

Alternatively, if we have BD (3), then from B to C (5), then from C to A (6): total 3 + 5 + 6 = 14.

Alternatively, BD (3), then from D to A (5), then from A to C (6): 3 + 5 + 6 = 14.

So, the minimal seems to be 13.

Is there a way to have two of the shorter edges? For example, BD (3) and CD (4). But that would require going through D, B, C, but then need to go to A.

Alternatively, BD (3) and AD (5). Then from A, go to C (6).

But that's 3 + 5 + 6 = 14.

Alternatively, BD (3) and BC (5). Then from C to A (6). 3 + 5 + 6 = 14.

Alternatively, CD (4) and AD (5). Then from D to B (3). Wait, but if you go from A to D (5), then D to C (4), then C to B (5). That's 5 + 4 + 5 = 14.

Alternatively, CD (4) and BC (5). Then from B to A (6). 4 + 5 + 6 = 15.

Alternatively, AD (5) and BD (3). Then from D to C (4). 5 + 3 + 4 = 12? Wait, wait. Wait, if you start at A, go to D (5), then to B (3), then to C (5). Wait, that would be AD + DB + BC = 5 + 3 + 5 = 13.

Wait, that's the same as before.

So, no, there's no way to get lower than 13.

Alternatively, is there a path that uses BD (3) and CD (4) and something else? For example, starting at C, going to D (4), then to B (3), then to A (6). Total 4 + 3 + 6 = 13.

Same as before.

Alternatively, starting at C, going to B (5), then to D (4), then to A (5). 5 + 4 + 5 = 14.

So, no.

Alternatively, starting at D, going to C (4), then to B (3), then to A (6). 4 + 3 + 6 = 13.

Same.

So, all minimal paths are 13.

Therefore, the answer is 13.

But let me check once again if there's a different path. For example, starting at A, going to D (5), then to C (4), then to B (5). That's 5 + 4 + 5 = 14.

Starting at A, going to B (6), then to D (3), then to C (4). 6 + 3 + 4 = 13.

Yes, same as before.

Alternatively, starting at A, going to C (6), then to D (4), then to B (3). 6 + 4 + 3 = 13.

Same.

So, all these paths give 13.

Therefore, the minimal distance is 13.

But let me check if there's a way to have a shorter path by using some other combination. For example, if there's a way to have two short edges.

But since we have to visit all four points, we need three edges. The shortest edges are BD (3), CD (4), AD (5), BC (5).

So, if we can have BD (3) and CD (4) and something else. But BD and CD share point D. So, if we have BD and CD, that would be going from B to D to C. Then we need to go from C to A (6). So total 3 + 4 + 6 = 13.

Alternatively, BD (3) and BC (5) and something else. If we go from B to D to C to A, that's 3 + 4 + 6 = 13.

Alternatively, BD (3) and AD (5) and something else. If we go from A to D to B to C, that's 5 + 3 + 5 = 13.

So, same.

Alternatively, CD (4) and AD (5) and something else. From A to D to C to B: 5 + 4 + 5 = 14.

So, no.

Therefore, 13 is the minimal.

Therefore, the answer is 13.

But let me check once again if there's a path that uses BD (3), AD (5), and BC (5). Wait, but how? If you go from B to D (3), then D to A (5), then A to C (6). That's 3 + 5 + 6 = 14.

Alternatively, from A to D (5), D to B (3), then B to C (5). 5 + 3 + 5 = 13.

Yes, that's the same as before.

So, no shorter path.

Therefore, I think the minimal distance is 13.

**Final Answer**
The shortest distance is \boxed{13}.

To determine the **shortest distance** that can be traveled when starting at any of the points $ A, B, C, $ or $ D $ and visiting the other three points exactly once, we are essentially looking for the **shortest Hamiltonian path** in a complete graph of four nodes, where the edge weights are the given distances between the points.

---

### Step 1: Understand the Given Distances

From the problem, the distances between the points are:

- $ AB = 6 $
- $ AC = 6 $
- $ AD = 5 $
- $ BC = 5 $
- $ BD = 3 $
- $ CD = 4 $

These are the only distances we need to consider, as the problem provides them explicitly. The coordinates provided in the Asymptote code are likely for drawing purposes and not to scale.

---

### Step 2: Define the Problem

We are to find the **shortest path** that starts at any one of the four points and visits the other three exactly once. This is equivalent to finding the **shortest Hamiltonian path** in this graph.

A Hamiltonian path visits all four points exactly once, and since the graph is complete, any permutation of the four points is a valid path. However, since the graph is undirected, we can consider only half of the permutations (i.e., the path and its reverse are the same in terms of total distance).

---

### Step 3: Evaluate Possible Paths

We can evaluate the total distance for several paths that start at each point and visit the other three in different orders. Here are some key examples:

#### Starting at A:
- $ A \to D \to B \to C $: $ AD + DB + BC = 5 + 3 + 5 = 13 $
- $ A \to D \to C \to B $: $ AD + DC + CB = 5 + 4 + 5 = 14 $
- $ A \to B \to D \to C $: $ AB + BD + DC = 6 + 3 + 4 = 13 $
- $ A \to C \to D \to B $: $ AC + CD + DB = 6 + 4 + 3 = 13 $

#### Starting at B:
- $ B \to D \to C \to A $: $ BD + DC + CA = 3 + 4 + 6 = 13 $
- $ B \to D \to A \to C $: $ BD + DA + AC = 3 + 5 + 6 = 14 $
- $ B \to C \to D \to A $: $ BC + CD + DA = 5 + 4 + 5 = 14 $

#### Starting at C:
- $ C \to D \to B \to A $: $ CD + DB + BA = 4 + 3 + 6 = 13 $
- $ C \to B \to D \to A $: $ CB + BD + DA = 5 + 3 + 5 = 13 $
- $ C \to A \to D \to B $: $ CA + AD + DB = 6 + 5 + 3 = 14 $

#### Starting at D:
- $ D \to B \to C \to A $: $ DB + BC + CA = 3 + 5 + 6 = 14 $
- $ D \to A \to B \to C $: $ DA + AB + BC = 5 + 6 + 5 = 16 $
- $ D \to A \to C \to B $: $ DA + AC + CB = 5 + 6 + 5 = 16 $

---

### Step 4: Determine the Minimum

From the above evaluations, the **shortest path** that visits all four points exactly once is **13 units**, and this can be achieved from multiple starting points:

- $ A \to D \to B \to C $
- $ A \to B \to D \to C $
- $ A \to C \to D \to B $
- $ B \to D \to C \to A $
- $ B \to C \to D \to A $
- $ C \to D \to B \to A $
- $ C \to B \to D \to A $

All of these paths have a total distance of **13**.

---

### Final Answer

$$
\boxed{13}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the shortest distance someone can travel if they start at any of the points A, B, C, or D and visit the other three points once. Let me try to visualize the diagram first.

From the Asymptote code, the coordinates are given as A(-3,3), B(3,0), C(0,-4), D(0,0). There are several lines drawn: D connected to B, B to C, C to D; D to A, A to B, and D to A to C. The labels on the sides are: 3 units between D and B, 6 units between A and B, 6 units between A and C, 4 units between D and C, 5 units between A and D, and 5 units between C and B.

First, I need to confirm the distances between all pairs of points. Let me list them out:

- AB: 6 units
- AC: 6 units
- AD: 5 units
- BC: 5 units
- BD: 3 units
- CD: 4 units

Wait, but maybe I should check if these distances are consistent with the coordinates? Let me verify that.

Coordinates:

A(-3,3), B(3,0), C(0,-4), D(0,0)

Calculating AB distance: distance between (-3,3) and (3,0). Using distance formula sqrt[(3 - (-3))² + (0 - 3)²] = sqrt[(6)² + (-3)²] = sqrt[36 + 9] = sqrt[45] ≈ 6.708. But the label says AB is 6 units. Hmm, that's conflicting. Maybe the coordinates are not to scale? Or maybe the labels are different? Wait, the Asymptote code labels the sides with specific lengths. Let me check again.

The Asymptote code labels:

- D--B: 3
- A--B: 6
- A--C: 6
- D--C: 4
- A--D: 5
- C--B: 5

So maybe the coordinates are not exact? Or maybe the diagram is schematic? Maybe I shouldn't rely on coordinates but just on the given distances? Wait, but the coordinates are given, so maybe the distances are calculated from coordinates? Let me check again.

Wait, if AB is supposed to be 6 units, but according to coordinates, it's sqrt(45) ≈ 6.708. So there's a discrepancy. Maybe the coordinates are not accurate? Or maybe the labels are different? Wait, the Asymptote code draws D--B--C--cycle, so triangle DBC? Then D--A--B--cycle, triangle DAB? And D--A--C--cycle, triangle DAC?

But maybe the labels are not the lengths of the sides of the triangles but something else? Wait, the labels are on the sides. For example, D--B is labeled 3, so that's the length of segment DB. Similarly, A--B is labeled 6, so AB is 6. So maybe the coordinates are just for drawing purposes, but the actual distances are given? That would make sense. So perhaps I should ignore the coordinates and just use the given distances? But then how do I know the positions? Wait, but maybe the coordinates are correct, and the labels are wrong? Or maybe the labels are correct, and the coordinates are just approximate?

This is confusing. Maybe I need to check if the given distances are consistent with coordinates? Let me check:

First, distance from D(0,0) to B(3,0) is 3 units. That matches the label "3" on D--B. Good.

Distance from A(-3,3) to B(3,0): sqrt[(3 - (-3))² + (0 - 3)²] = sqrt[6² + (-3)²] = sqrt[36 + 9] = sqrt[45] ≈ 6.708. But the label says 6. So that's inconsistent. Similarly, distance from A(-3,3) to C(0,-4): sqrt[(0 - (-3))² + (-4 - 3)²] = sqrt[3² + (-7)²] = sqrt[9 + 49] = sqrt[58] ≈ 7.616, but the label says 6. Hmm. So the coordinates don't match the given distances. Therefore, maybe the coordinates are not to scale, and I should just use the given distances?

Alternatively, maybe the coordinates are correct, and the labels are different? Wait, the Asymptote code labels "3" on D--B, which is from (0,0) to (3,0), which is indeed 3 units. Then "6" on A--B, which is from A(-3,3) to B(3,0). But according to coordinates, that's sqrt(45). Maybe the diagram is not to scale? Maybe the labels are the actual distances? So perhaps the coordinates are just for drawing, but the actual distances are as labeled?

In that case, maybe I should consider the distances given in the labels as the actual distances between the points. So:

AB = 6, AC = 6, AD = 5, BC = 5, BD = 3, CD = 4.

So, with that, I can model the points as a graph with these distances. Now, the problem is: starting at any point A, B, C, or D, visit the other three points once. So, the question is asking for the shortest possible path that starts at one of the points, visits the other three, and ends at the last one. So, it's a traveling salesman problem (TSP) where we need to find the shortest possible route that visits all four points, but since we can start at any point, we need to find the minimal tour.

Wait, but the problem says "start at any point, A, B, C, or D, and visit the other three points once." So, it's a path that starts at one point, visits the other three, so it's a path that goes through all four points, but since it's a path, not necessarily a cycle. So, the total distance would be the sum of the distances between consecutive points. So, for example, starting at A, going to B, then to C, then to D. Or starting at A, going to D, then to C, then to B. Etc. So, the problem is to find the minimal such path.

But since we can start at any point, we need to find the minimal path that starts at any of the four points and visits the other three. So, the minimal path would be the minimal Hamiltonian path in the complete graph with these distances.

Alternatively, since the problem says "visit the other three points once," so starting at one, then visiting the other three, so the path has four points, with each point visited once. So, the total number of possible paths is 4 * 3! = 24, since starting at any of the four points, and then permuting the other three. But since the graph is undirected, some paths are the same in reverse. However, since we are looking for the minimal distance, we can consider all possible permutations.

But maybe there's a smarter way than enumerating all 24 possibilities. Let me think.

Alternatively, since the problem is small, maybe I can find the minimal path by considering the distances.

First, let me note the distances between all pairs:

AB = 6

AC = 6

AD = 5

BC = 5

BD = 3

CD = 4

So, the distances are:

Between A and B: 6

A and C: 6

A and D: 5

B and C: 5

B and D: 3

C and D: 4

So, the distances are as above.

Now, the problem is to find the shortest path that starts at any point, visits the other three. So, for example, starting at A, then going to B, then to C, then to D. The total distance would be AB + BC + CD = 6 + 5 + 4 = 15. Alternatively, starting at A, going to D, then to B, then to C: AD + DB + BC = 5 + 3 + 5 = 13. That's shorter. Or starting at A, going to D, then to C, then to B: AD + DC + CB = 5 + 4 + 5 = 14. So, 13 is better.

Alternatively, starting at A, going to D, then to B, then to C: 5 + 3 + 5 = 13.

Alternatively, starting at A, going to D, then to C, then to B: 5 + 4 + 5 = 14.

Alternatively, starting at A, going to B, then to D, then to C: AB + BD + DC = 6 + 3 + 4 = 13.

Same as before. So, 13.

Alternatively, starting at A, going to C, then to D, then to B: AC + CD + DB = 6 + 4 + 3 = 13.

Same.

So, starting at A, there are multiple paths that give 13.

Now, let's check starting at B.

Starting at B, then going to D (distance 3), then to C (distance 4), then to A (distance 6): total 3 + 4 + 6 = 13.

Or starting at B, going to D, then to A, then to C: BD + DA + AC = 3 + 5 + 6 = 14.

Or starting at B, going to C (distance 5), then to D (distance 4), then to A: BC + CD + DA = 5 + 4 + 5 = 14.

Or starting at B, going to C, then to A, then to D: BC + CA + AD = 5 + 6 + 5 = 16.

Alternatively, starting at B, going to A (distance 6), then to D (distance 5?), wait, distance from A to D is 5, so BD is 3. Wait, starting at B, going to A (distance 6), then to D (distance 5?), but distance from A to D is 5. So, BA + AD + DC = 6 + 5 + 4 = 15.

Alternatively, starting at B, going to A, then to C, then to D: BA + AC + CD = 6 + 6 + 4 = 16.

So, the minimal starting at B is 13.

Now starting at C.

Starting at C, going to D (distance 4), then to B (distance 3), then to A: CD + DB + BA = 4 + 3 + 6 = 13.

Or starting at C, going to D, then to A, then to B: CD + DA + AB = 4 + 5 + 6 = 15.

Or starting at C, going to B (distance 5), then to D (distance 4), then to A: CB + BD + DA = 5 + 3 + 5 = 13.

Or starting at C, going to B, then to A, then to D: CB + BA + AD = 5 + 6 + 5 = 16.

Or starting at C, going to A (distance 6), then to D (distance 4), then to B: CA + AD + DB = 6 + 5 + 3 = 14.

So, minimal starting at C is 13.

Now starting at D.

Starting at D, going to B (distance 3), then to C (distance 4), then to A: DB + BC + CA = 3 + 5 + 6 = 14.

Or starting at D, going to B, then to A, then to C: DB + BA + AC = 3 + 6 + 6 = 15.

Or starting at D, going to A (distance 5), then to B (distance 6?), wait, distance from A to B is 6. So DA + AB + BC = 5 + 6 + 5 = 16.

Or starting at D, going to A, then to C, then to B: DA + AC + CB = 5 + 6 + 5 = 16.

Or starting at D, going to C (distance 4), then to B (distance 3), then to A: DC + CB + BA = 4 + 5 + 6 = 15.

Alternatively, starting at D, going to C, then to A, then to B: DC + CA + AB = 4 + 6 + 6 = 16.

Alternatively, starting at D, going to A (distance 5), then to C (distance 6?), wait, distance from A to C is 6. So DA + AC + CB = 5 + 6 + 5 = 16.

Alternatively, starting at D, going to A, then to B, then to C: DA + AB + BC = 5 + 6 + 5 = 16.

So, the minimal starting at D is 14.

So, summarizing:

Starting at A: minimal path is 13

Starting at B: minimal path is 13

Starting at C: minimal path is 13

Starting at D: minimal path is 14

Therefore, the overall minimal distance is 13.

But let me check if there are any other paths that I might have missed. For example, starting at A, going to D, then to B, then to C: 5 + 3 + 5 = 13. Starting at A, going to D, then to C, then to B: 5 + 4 + 5 = 14. Starting at A, going to B, then to D, then to C: 6 + 3 + 4 = 13. Starting at A, going to C, then to D, then to B: 6 + 4 + 3 = 13.

Similarly, starting at B, going to D, then to C, then to A: 3 + 4 + 6 = 13. Starting at B, going to C, then to D, then to A: 5 + 4 + 5 = 14. Starting at B, going to D, then to A, then to C: 3 + 5 + 6 = 14.

Starting at C, going to D, then to B, then to A: 4 + 3 + 6 = 13. Starting at C, going to B, then to D, then to A: 5 + 3 + 5 = 13.

Starting at D, going to B, then to C, then to A: 3 + 5 + 6 = 14.

So, all the minimal paths are 13.

Therefore, the answer is 13.

But let me check if there's a shorter path. For example, is there a way to have a path with total distance less than 13? Let's think.

Suppose we try to find a path that uses the shortest edges. The shortest edges are BD = 3, CD = 4, AD = 5, BC = 5, AB = 6, AC = 6.

So, the shortest edge is BD = 3. If we can include that in the path.

For example, if we have a path that includes BD, which is 3, then maybe other edges.

But since we need to visit all four points, we need three edges.

So, if we have BD (3), then we need two more edges. Let's see.

For example, starting at B, going to D (3), then from D, we can go to A (5) or to C (4). If we go to C (4), then from C, we need to go to A (6). So total is 3 + 4 + 6 = 13.

Alternatively, starting at D, going to B (3), then to C (4), then to A (6): same as above.

Alternatively, starting at D, going to B (3), then to A (5), then to C (6): 3 + 5 + 6 = 14.

Alternatively, starting at B, going to D (3), then to A (5), then to C (6): same as above.

Alternatively, if we can find a path that uses BD (3) and CD (4) and something else. Wait, but if we have BD and CD, that would be two edges, but we need three edges.

Wait, if we have BD (3), CD (4), and then another edge. But if we go from D to B to C, then we need to go from C to A, which is 6.

Alternatively, if we have BD (3), then from B to C (5), then from C to A (6): total 3 + 5 + 6 = 14.

Alternatively, BD (3), then from D to A (5), then from A to C (6): 3 + 5 + 6 = 14.

So, the minimal seems to be 13.

Is there a way to have two of the shorter edges? For example, BD (3) and CD (4). But that would require going through D, B, C, but then need to go to A.

Alternatively, BD (3) and AD (5). Then from A, go to C (6).

But that's 3 + 5 + 6 = 14.

Alternatively, BD (3) and BC (5). Then from C to A (6). 3 + 5 + 6 = 14.

Alternatively, CD (4) and AD (5). Then from D to B (3). Wait, but if you go from A to D (5), then D to C (4), then C to B (5). That's 5 + 4 + 5 = 14.

Alternatively, CD (4) and BC (5). Then from B to A (6). 4 + 5 + 6 = 15.

Alternatively, AD (5) and BD (3). Then from D to C (4). 5 + 3 + 4 = 12? Wait, wait. Wait, if you start at A, go to D (5), then to B (3), then to C (5). Wait, that would be AD + DB + BC = 5 + 3 + 5 = 13.

Wait, that's the same as before.

So, no, there's no way to get lower than 13.

Alternatively, is there a path that uses BD (3) and CD (4) and something else? For example, starting at C, going to D (4), then to B (3), then to A (6). Total 4 + 3 + 6 = 13.

Same as before.

Alternatively, starting at C, going to B (5), then to D (4), then to A (5). 5 + 4 + 5 = 14.

So, no.

Alternatively, starting at D, going to C (4), then to B (3), then to A (6). 4 + 3 + 6 = 13.

Same.

So, all minimal paths are 13.

Therefore, the answer is 13.

But let me check once again if there's a different path. For example, starting at A, going to D (5), then to C (4), then to B (5). That's 5 + 4 + 5 = 14.

Starting at A, going to B (6), then to D (3), then to C (4). 6 + 3 + 4 = 13.

Yes, same as before.

Alternatively, starting at A, going to C (6), then to D (4), then to B (3). 6 + 4 + 3 = 13.

Same.

So, all these paths give 13.

Therefore, the minimal distance is 13.

But let me check if there's a way to have a shorter path by using some other combination. For example, if there's a way to have two short edges.

But since we have to visit all four points, we need three edges. The shortest edges are BD (3), CD (4), AD (5), BC (5).

So, if we can have BD (3) and CD (4) and something else. But BD and CD share point D. So, if we have BD and CD, that would be going from B to D to C. Then we need to go from C to A (6). So total 3 + 4 + 6 = 13.

Alternatively, BD (3) and BC (5) and something else. If we go from B to D to C to A, that's 3 + 4 + 6 = 13.

Alternatively, BD (3) and AD (5) and something else. If we go from A to D to B to C, that's 5 + 3 + 5 = 13.

So, same.

Alternatively, CD (4) and AD (5) and something else. From A to D to C to B: 5 + 4 + 5 = 14.

So, no.

Therefore, 13 is the minimal.

Therefore, the answer is 13.

But let me check once again if there's a path that uses BD (3), AD (5), and BC (5). Wait, but how? If you go from B to D (3), then D to A (5), then A to C (6). That's 3 + 5 + 6 = 14.

Alternatively, from A to D (5), D to B (3), then B to C (5). 5 + 3 + 5 = 13.

Yes, that's the same as before.

So, no shorter path.

Therefore, I think the minimal distance is 13.

**Final Answer**
The shortest distance is \boxed{13}.

To determine the **shortest distance** that can be traveled when starting at any of the points $ A, B, C, $ or $ D $ and visiting the other three points exactly once, we are essentially looking for the **shortest Hamiltonian path** in a complete graph of four nodes, where the edge weights are the given distances between the points.

---

### Step 1: Understand the Given Distances

From the problem, the distances between the points are:

- $ AB = 6 $
- $ AC = 6 $
- $ AD = 5 $
- $ BC = 5 $
- $ BD = 3 $
- $ CD = 4 $

These are the only distances we need to consider, as the problem provides them explicitly. The coordinates provided in the Asymptote code are likely for drawing purposes and not to scale.

---

### Step 2: Define the Problem

We are to find the **shortest path** that starts at any one of the four points and visits the other three exactly once. This is equivalent to finding the **shortest Hamiltonian path** in this graph.

A Hamiltonian path visits all four points exactly once, and since the graph is complete, any permutation of the four points is a valid path. However, since the graph is undirected, we can consider only half of the permutations (i.e., the path and its reverse are the same in terms of total distance).

---

### Step 3: Evaluate Possible Paths

We can evaluate the total distance for several paths that start at each point and visit the other three in different orders. Here are some key examples:

#### Starting at A:
- $ A \to D \to B \to C $: $ AD + DB + BC = 5 + 3 + 5 = 13 $
- $ A \to D \to C \to B $: $ AD + DC + CB = 5 + 4 + 5 = 14 $
- $ A \to B \to D \to C $: $ AB + BD + DC = 6 + 3 + 4 = 13 $
- $ A \to C \to D \to B $: $ AC + CD + DB = 6 + 4 + 3 = 13 $

#### Starting at B:
- $ B \to D \to C \to A $: $ BD + DC + CA = 3 + 4 + 6 = 13 $
- $ B \to D \to A \to C $: $ BD + DA + AC = 3 + 5 + 6 = 14 $
- $ B \to C \to D \to A $: $ BC + CD + DA = 5 + 4 + 5 = 14 $

#### Starting at C:
- $ C \to D \to B \to A $: $ CD + DB + BA = 4 + 3 + 6 = 13 $
- $ C \to B \to D \to A $: $ CB + BD + DA = 5 + 3 + 5 = 13 $
- $ C \to A \to D \to B $: $ CA + AD + DB = 6 + 5 + 3 = 14 $

#### Starting at D:
- $ D \to B \to C \to A $: $ DB + BC + CA = 3 + 5 + 6 = 14 $
- $ D \to A \to B \to C $: $ DA + AB + BC = 5 + 6 + 5 = 16 $
- $ D \to A \to C \to B $: $ DA + AC + CB = 5 + 6 + 5 = 16 $

---

### Step 4: Determine the Minimum

From the above evaluations, the **shortest path** that visits all four points exactly once is **13 units**, and this can be achieved from multiple starting points:

- $ A \to D \to B \to C $
- $ A \to B \to D \to C $
- $ A \to C \to D \to B $
- $ B \to D \to C \to A $
- $ B \to C \to D \to A $
- $ C \to D \to B \to A $
- $ C \to B \to D \to A $

All of these paths have a total distance of **13**.

---

### Final Answer

$$
\boxed{13}
$$

Solution: 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 $\boxed{13}$.