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. There's an Asymptote code provided, which draws several lines connecting these points. Let me parse that.

From the Asymptote code:

- Points A, B, C, D are defined with coordinates:
  - A = (-3, 3)
  - B = (3, 0)
  - C = (0, -4)
  - D = (0, 0)

Then, the drawing commands are:

- D--B--C--cycle: So triangle DBC
- D--A--B--cycle: Triangle DAB
- D--A--C--cycle: Triangle DAC

So, connecting D to B to C to D, D to A to B to D, and D to A to C to D. So, the figure is a combination of triangles. Also, labels for distances:

- "3" on D--B (so DB is 3 units)
- "6" on A--B (AB is 6)
- "6" on A--C (AC is 6)
- "4" on D--C (DC is 4)
- "5" on A--D (AD is 5)
- "5" on C--B (CB is 5)

Wait, but according to coordinates, maybe the distances are different? Let me check if the coordinates match the given distances. Maybe the Asymptote code is just for drawing, and the labels are the actual distances. Let me confirm.

First, coordinates:

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

Let me compute the distances between these points to see if they match the labels.

First, distance between D (0,0) and B (3,0): that's sqrt[(3-0)^2 + (0-0)^2] = sqrt[9] = 3. Okay, matches the label "3" on D--B.

Distance between A (-3,3) and B (3,0): sqrt[(3 - (-3))^2 + (0 - 3)^2] = sqrt[(6)^2 + (-3)^2] = sqrt[36 + 9] = sqrt[45] = 3*sqrt(5) ≈ 6.708. But the label says "6" on A--B. Hmm, discrepancy here. Maybe the coordinates are not to scale? Or maybe the labels are the actual distances? The problem says "label" so maybe the coordinates are just for drawing, and the actual distances are as labeled. So perhaps the coordinates are not to scale? Maybe I should ignore coordinates and just use the given distances?

Wait, the problem says "start at any point, A, B, C, or D, and visit the other three points once." So it's a traveling salesman problem? But with only four points, and visiting all three others once. Wait, starting at one point, visiting the other three once. So total of four points, visiting all four? Wait, starting at one, then visiting the other three. So total of four points, visiting all four? Wait, but the question says "visit the other three points once." So starting at A, then visiting B, C, D once each. So the path is A -> B -> C -> D or some permutation. Similarly for starting at other points.

But the question is asking for the shortest distance that can be travelled. So the minimal total distance when starting at any of the four points and visiting the other three once. So it's equivalent to finding the shortest possible route that starts at any point and visits all three others, so effectively, the shortest Hamiltonian path in the complete graph of these four points, considering the distances between them.

But since the problem says "start at any point," maybe we need to consider all possible starting points and find the minimal total distance among all possible paths. However, since the graph is undirected, the minimal path should be the same regardless of starting point, but maybe not? Wait, no. For example, if the graph is asymmetric, but in this case, since it's a complete graph with symmetric distances, the minimal path should be the same regardless of starting point. Wait, but maybe not. Wait, the problem says "start at any point," so maybe we need to consider all possible starting points and find the minimal total distance. However, since the graph is undirected, the minimal path should be the same regardless of starting point. Wait, but maybe not. For example, if you have different distances between points, depending on the starting point, the path could be different. Wait, but since we can choose the starting point, maybe we can choose the one that gives the minimal total distance.

But actually, the problem says "start at any point, A, B, C, or D, and visit the other three points once." So the question is, what is the minimal total distance of such a path, considering that you can choose the starting point. So we need to find the minimal total distance over all possible starting points and all possible paths that start at that point and visit the other three once.

Alternatively, since the problem is about visiting all four points, but starting at one and visiting the other three. Wait, but if you start at a point and visit the other three, that's visiting all four points. So the problem is equivalent to finding the shortest Hamiltonian path in the complete graph of these four points, with the given distances between them. However, Hamiltonian path is a path that visits each vertex exactly once. So since there are four points, the path has three edges. So the total distance is the sum of three edges. However, since the problem says "start at any point," maybe we need to consider all possible starting points and all possible permutations of visiting the other three points, and find the minimal total distance.

But since the graph is undirected, the minimal path should be the same regardless of starting point. Wait, but maybe not. For example, if you have different distances between points, depending on the order. Let me think.

Alternatively, maybe the problem is equivalent to finding the minimal total distance of a path that starts at any point and visits all other three points once. So, for example, starting at A, then going to B, then to C, then to D. Or starting at A, then to C, then to B, then to D, etc. So for each starting point, there are 3! = 6 possible paths. But since the graph is undirected, some of these paths will be the same in reverse. However, since we are considering starting at a specific point, the direction matters.

But since the problem says "start at any point," we can choose the starting point that gives the minimal total distance. Therefore, we need to compute for each starting point, the minimal total distance of a path that starts there and visits the other three points once, and then take the minimum among those.

Alternatively, since the problem is symmetric, maybe the minimal total distance is the same regardless of starting point. But I need to check.

Alternatively, maybe the problem is asking for the minimal total distance of a path that starts at any point and visits all other three points once, so the minimal over all possible starting points and all possible paths. So we need to find the minimal total distance among all possible Hamiltonian paths in the complete graph of these four points.

But since there are four points, the number of Hamiltonian paths is 4 * 3! = 24, but since each path is counted twice (once in each direction), but since we are considering starting points, maybe not. Wait, if we fix the starting point, then for each starting point, there are 3! = 6 possible paths. So total of 4 * 6 = 24 paths. But since the graph is undirected, some of these paths are reverses of each other. However, since we are considering starting points, they are different.

But maybe there's a smarter way. Let me think. The minimal total distance would be the minimal sum of three edges that connect all four points. However, since it's a path, not a cycle, so it's a sequence of three edges connecting four points. So, for example, A-B-C-D, which is three edges: AB, BC, CD. Alternatively, A-B-D-C, which is AB, BD, DC. So different combinations.

But to find the minimal total distance, we need to find the combination of three edges that connect all four points with the minimal total length. However, since the problem allows starting at any point, maybe we can choose the starting point that allows the minimal total distance.

Alternatively, maybe the minimal total distance is the same regardless of starting point, but I need to check.

Alternatively, maybe the problem is equivalent to finding the minimal spanning tree, but no, because a spanning tree has three edges for four points, but a spanning tree is a tree, not a path. However, the minimal spanning tree would have the minimal total length, but the path is different. However, maybe the minimal path is related to the minimal spanning tree? Not sure.

Alternatively, maybe we can think of it as the Traveling Salesman Problem (TSP) where we need to visit all four points, but since we can start at any point, the minimal TSP tour would be the minimal total distance. However, in TSP, you have to return to the starting point, but here, we don't have to return. So it's a Hamiltonian path, not a cycle.

But since the problem says "visit the other three points once," starting at one point, so it's a Hamiltonian path. So the minimal Hamiltonian path.

But how do we compute that? For four points, it's manageable.

First, let me list all possible Hamiltonian paths and compute their total distances. However, since there are 4 points, the number of Hamiltonian paths is 4 * 3! = 24, but since each path is counted twice (once in each direction), but since we are considering starting points, maybe not. Wait, if we fix the starting point, then for each starting point, there are 3! = 6 paths. So total of 4 * 6 = 24 paths. However, since the graph is undirected, some of these paths are reverses of each other. But since we are considering starting points, they are different.

But maybe we can find the minimal total distance by considering all possible permutations.

Alternatively, since there are only four points, maybe we can find the minimal total distance by considering the minimal total of three edges that connect all four points. However, since it's a path, not a tree, the three edges must form a path, not a tree. So, for example, if we have points A, B, C, D, then a path could be A-B-C-D, which uses edges AB, BC, CD. Alternatively, A-B-D-C, which uses AB, BD, DC. Or A-D-B-C, which uses AD, DB, BC. Etc.

So, to find the minimal total distance, we need to find the path that has the minimal sum of three edges. However, since there are multiple possibilities, we need to check them all.

Alternatively, maybe we can use the fact that the minimal Hamiltonian path is the one that uses the three shortest edges. But since the edges are connected in a path, not just any three edges.

Alternatively, maybe we can think of it as follows: For each pair of points, we can compute the distances between them. Then, for each possible starting point, we can try to find the minimal path.

But first, let me list all the distances between the points. The problem gives some distances:

From the Asymptote code labels:

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

Wait, but according to coordinates, some distances are different. However, the problem says "label" so maybe the coordinates are just for drawing, and the actual distances are as labeled. Therefore, we should use the given distances. Let me confirm:

Given distances:

- DB = 3
- AB = 6
- AC = 6
- DC = 4
- AD = 5
- CB = 5

But what about other distances? For example, what is the distance between A and D? It's given as 5. What about between A and C? 6. Between A and B? 6. Between B and C? 5. Between D and C? 4. Between D and B? 3. What about between D and A? 5. What about between A and D? 5. What about between D and C? 4. What about between B and D? 3. What about between B and A? 6. What about between C and A? 6. What about between C and B? 5. What about between D and A? 5. What about between D and C? 4. What about between D and B? 3. What about between A and C? 6. What about between A and B? 6. What about between B and C? 5. What about between D and A? 5. What about between D and C? 4. What about between D and B? 3. What about between A and D? 5. So, the distances between all pairs:

Let me list all pairs:

AB: 6

AC: 6

AD: 5

BC: 5

BD: 3

CD: 4

What about the other pairs? Wait, there are four points: A, B, C, D. So total of 6 pairs:

AB, AC, AD, BC, BD, CD. So those are all. So the distances are:

AB = 6

AC = 6

AD = 5

BC = 5

BD = 3

CD = 4

So that's all. So we have all the distances between the points. Now, we need to find the minimal total distance of a path that starts at any point and visits the other three once. 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. Or starting at A, going to C, then to D, then to B: AC + CD + DB = 6 + 4 + 3 = 13. So starting at A, the minimal path is 13.

Similarly, let's check starting at B. Let's see. Starting at B, then visiting A, C, D. Let's try different permutations.

B -> A -> C -> D: BA + AC + CD = 6 + 6 + 4 = 16

B -> A -> D -> C: BA + AD + DC = 6 + 5 + 4 = 15

B -> C -> A -> D: BC + CA + AD = 5 + 6 + 5 = 16

B -> C -> D -> A: BC + CD + DA = 5 + 4 + 5 = 14

B -> D -> A -> C: BD + DA + AC = 3 + 5 + 6 = 14

B -> D -> C -> A: BD + DC + CA = 3 + 4 + 6 = 13

So starting at B, the minimal total distance is 13.

Starting at C. Let's check.

Starting at C, visiting A, B, D.

C -> A -> B -> D: CA + AB + BD = 6 + 6 + 3 = 15

C -> A -> D -> B: CA + AD + DB = 6 + 5 + 3 = 14

C -> B -> A -> D: CB + BA + AD = 5 + 6 + 5 = 16

C -> B -> D -> A: CB + BD + DA = 5 + 3 + 5 = 13

C -> D -> A -> B: CD + DA + AB = 4 + 5 + 6 = 15

C -> D -> B -> A: CD + DB + BA = 4 + 3 + 6 = 13

So starting at C, minimal total distance is 13.

Starting at D. Let's check.

Starting at D, visiting A, B, C.

D -> A -> B -> C: DA + AB + BC = 5 + 6 + 5 = 16

D -> A -> C -> B: DA + AC + CB = 5 + 6 + 5 = 16

D -> B -> A -> C: DB + BA + AC = 3 + 6 + 6 = 15

D -> B -> C -> A: DB + BC + CA = 3 + 5 + 6 = 14

D -> C -> A -> B: DC + CA + AB = 4 + 6 + 6 = 16

D -> C -> B -> A: DC + CB + BA = 4 + 5 + 6 = 15

So starting at D, the minimal total distance is 14.

Wait, so starting at A: 13, starting at B: 13, starting at C: 13, starting at D: 14. Therefore, the minimal total distance is 13. So the answer is 13? But let me check again.

Wait, when starting at D, the minimal path is 14? Let me check again:

Starting at D, the possible paths:

D -> B -> C -> A: DB + BC + CA = 3 + 5 + 6 = 14

D -> C -> B -> A: DC + CB + BA = 4 + 5 + 6 = 15

D -> B -> A -> C: DB + BA + AC = 3 + 6 + 6 = 15

D -> A -> B -> C: DA + AB + BC = 5 + 6 + 5 = 16

D -> A -> C -> B: DA + AC + CB = 5 + 6 + 5 = 16

D -> C -> A -> B: DC + CA + AB = 4 + 6 + 6 = 16

So yes, the minimal starting at D is 14. So the minimal total distance is 13, achieved when starting at A, B, or C. So the answer is 13? But let me check if there are other paths that I might have missed.

Wait, for example, starting at A, going to D, then to B, then to C: AD + DB + BC = 5 + 3 + 5 = 13. Similarly, starting at A, going to C, then to D, then to B: AC + CD + DB = 6 + 4 + 3 = 13. Starting at B, going to D, then to C, then to A: BD + DC + CA = 3 + 4 + 6 = 13. Starting at B, going to D, then to C, then to A: same as above. Starting at C, going to B, then to D, then to A: CB + BD + DA = 5 + 3 + 5 = 13. Starting at C, going to D, then to B, then to A: CD + DB + BA = 4 + 3 + 6 = 13. So all these paths give 13. So the minimal total distance is 13.

But let me check if there is a path with a shorter total distance. For example, is there a path with total distance less than 13? Let's see. Let's think of other combinations.

For example, starting at A, going to D (5), then to C (4), then to B (5): total 5 + 4 + 5 = 14. Not better. Starting at A, going to D (5), then to B (3), then to C (5): 5 + 3 + 5 = 13. Same as before.

Is there a way to have a path with total distance less than 13? Let's see. Let's think of the three edges. The three edges must connect all four points. So, for example, if we can find three edges that connect all four points with total distance less than 13. Let's see.

The three edges must form a path. So, for example, if we have edges BD (3), DC (4), and CA (6): total 3 + 4 + 6 = 13. Or BD (3), DA (5), and AC (6): 3 + 5 + 6 = 14. Or BD (3), BC (5), and CA (6): 3 + 5 + 6 = 14. Or DC (4), CB (5), BA (6): 4 + 5 + 6 = 15. Or DA (5), AB (6), BC (5): 5 + 6 + 5 = 16. Or DA (5), AC (6), CB (5): 5 + 6 + 5 = 16. Or DB (3), BA (6), AC (6): 3 + 6 + 6 = 15. Or DC (4), CA (6), AB (6): 4 + 6 + 6 = 16. Or DB (3), BC (5), CD (4): 3 + 5 + 4 = 12? Wait, hold on. Wait, if we have edges DB (3), BC (5), and CD (4). But that would form a cycle: D-B-C-D. But that's a cycle, not a path. Because a path must visit each point once. So if you have D-B-C-D, you are visiting D twice. So that's not allowed. So that's not a valid path. Similarly, if you have edges BD (3), DC (4), and CB (5), that would be D-B-C-D again, which is a cycle. So that's not a path. So the three edges must form a path, not a cycle. Therefore, the three edges must be connected in a sequence, visiting each point once. Therefore, the three edges cannot form a cycle. Therefore, the three edges must be such that they form a path. Therefore, the three edges must be connected in a way that they form a single path. Therefore, for example, edges AB, BC, CD form a path A-B-C-D. Similarly, edges AD, DB, BC form a path A-D-B-C. So, in that case, the total distance is AD + DB + BC = 5 + 3 + 5 = 13. Similarly, edges AC, CD, DB form a path A-C-D-B, which is AC + CD + DB = 6 + 4 + 3 = 13. Similarly, edges CB, BD, DA form a path C-B-D-A, which is CB + BD + DA = 5 + 3 + 5 = 13. So all these are valid paths with total distance 13. So there is no way to get a shorter total distance. Because the three edges that form a path must include three edges, and the minimal sum is 13. Let me check if there is a combination of three edges that form a path with a smaller total.

For example, if we take edges BD (3), DC (4), and CA (6): total 13. If we take edges BD (3), DA (5), and AC (6): total 14. If we take edges DC (4), CB (5), and BA (6): total 15. If we take edges AB (6), BC (5), CD (4): total 15. If we take edges AD (5), DC (4), CB (5): total 14. If we take edges AC (6), CB (5), BD (3): total 14. So no, there is no combination of three edges that form a path with total less than 13. Therefore, the minimal total distance is 13.

Therefore, the answer is 13. However, let me check once again if there is a path that I haven't considered. For example, starting at A, going to D (5), then to C (4), then to B (5): total 14. Starting at A, going to B (6), then to D (3), then to C (4): total 6 + 3 + 4 = 13. Wait, that's another path: A-B-D-C. Let me check the distances. AB is 6, BD is 3, DC is 4. Total 6 + 3 + 4 = 13. Yes, that's another path. So that's also 13. So that's another way to get 13. So yes, multiple paths give 13. Therefore, the minimal total distance is 13.

But let me check if there is a path with total distance less than 13. For example, if there is a path with total distance 12. Let's see. To get 12, we need three edges with total 12. The minimal edges are BD (3), DC (4), and... what's the next? If we have BD (3), DC (4), and then another edge. But the third edge must connect to the path. For example, if we have BD (3), DC (4), and then... if we have BD, DC, and CB (5), that would be 3 + 4 + 5 = 12, but that would form a cycle D-B-C-D, which is not a path. Similarly, BD (3), DA (5), and AC (6): 3 + 5 + 6 = 14. BD (3), DA (5), and AB (6): 3 + 5 + 6 = 14. DC (4), CB (5), BA (6): 4 + 5 + 6 = 15. So no, there is no way to get 12. Therefore, 13 is indeed the minimal.

Therefore, the answer is 13. However, let me check once again if I have considered all possible paths. For example, starting at A, going to D (5), then to B (3), then to C (5): total 5 + 3 + 5 = 13. Starting at A, going to C (6), then to D (4), then to B (3): total 6 + 4 + 3 = 13. Starting at B, going to D (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at B, going to D (3), then to A (5), then to C (6): 3 + 5 + 6 = 14. Wait, no, that's 14. Starting at B, going to D (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at C, going to B (5), then to D (3), then to A (5): 5 + 3 + 5 = 13. Starting at C, going to D (4), then to B (3), then to A (5): 4 + 3 + 5 = 12? Wait, hold on! Wait, if starting at C, going to D (4), then to B (3), then to A (5). Wait, is that a valid path? Let me check the distances. CD is 4, DB is 3, BA is 6. Wait, no, from B to A is 6, not 5. Wait, BA is AB, which is 6. So CD (4) + DB (3) + BA (6) = 4 + 3 + 6 = 13. So that's 13. Wait, but if I thought BA is 5, that was a mistake. So no, that's 13. So no, there is no path with total distance less than 13.

Wait, but what if there is a path that uses different edges? For example, starting at D, going to B (3), then to A (5), then to C (6): total 3 + 5 + 6 = 14. Starting at D, going to C (4), then to B (3), then to A (5): 4 + 3 + 5 = 12? Wait, no, from B to A is 6, not 5. Wait, BA is 6. So 4 + 3 + 6 = 13. So again, 13. So no, there is no way to get less than 13.

Therefore, the minimal total distance is 13. Therefore, the answer is 13. However, let me check once again if there is a path that I have not considered. For example, starting at A, going to D (5), then to C (4), then to B (5): total 5 + 4 + 5 = 14. Starting at A, going to B (6), then to D (3), then to C (4): 6 + 3 + 4 = 13. Starting at A, going to C (6), then to D (4), then to B (3): 6 + 4 + 3 = 13. Starting at B, going to D (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at B, going to C (5), then to D (3), then to A (5): 5 + 3 + 5 = 13. Starting at C, going to D (4), then to B (3), then to A (6): 4 + 3 + 6 = 13. Starting at C, going to B (5), then to D (3), then to A (5): 5 + 3 + 5 = 13. Starting at D, going to B (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at D, going to C (4), then to B (3), then to A (6): same as above. So all these paths give 13. Therefore, the minimal total distance is 13.

Therefore, the answer is 13. However, let me check if the problem says "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." So, the minimal total distance is 13. Therefore, the answer is \boxed{13}.

**Final Answer**
\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 **once**, we are essentially looking for the **shortest Hamiltonian path** among the four points. A Hamiltonian path is a path that visits each vertex exactly once, and since we are allowed to choose the starting point, we must consider all possible starting points and all possible permutations of visiting the other three points.

---

### Step 1: Understand the Given Distances

From the problem, we are given the following distances between the points:

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

These are the only distances between the four points, and we can use them to compute the total distance of any path that visits all four points exactly once.

---

### Step 2: Strategy

We need to find the **shortest total distance** of a path that starts at any one of the four points and visits the other three exactly once. Since the graph is undirected, the total distance of a path is the same regardless of the direction (i.e., the path $ A \to B \to C \to D $ has the same total distance as $ D \to C \to B \to A $).

Thus, we can consider all possible permutations of visiting the four points, starting from any one of them, and compute the total distance for each. Then, we select the **minimum** among all these.

However, since there are only four points, we can simplify the process by considering all possible **Hamiltonian paths** (i.e., all possible sequences of visiting the four points once), and compute the total distance for each.

---

### Step 3: Compute the Total Distance for All Valid Paths

We'll consider all permutations of the four points and compute the total distance for each. Since the graph is undirected, we can consider only the unique paths (i.e., avoid counting the reverse of each path as a separate one).

Let’s consider a few key examples:

#### Example 1: Path $ A \to D \to B \to C $
- $ AD = 5 $
- $ DB = 3 $
- $ BC = 5 $
- **Total = 5 + 3 + 5 = 13**

#### Example 2: Path $ A \to C \to D \to B $
- $ AC = 6 $
- $ CD = 4 $
- $ DB = 3 $
- **Total = 6 + 4 + 3 = 13**

#### Example 3: Path $ B \to D \to C \to A $
- $ BD = 3 $
- $ DC = 4 $
- $ CA = 6 $
- **Total = 3 + 4 + 6 = 13**

#### Example 4: Path $ C \to B \to D \to A $
- $ CB = 5 $
- $ BD = 3 $
- $ DA = 5 $
- **Total = 5 + 3 + 5 = 13**

#### Example 5: Path $ D \to B \to C \to A $
- $ DB = 3 $
- $ BC = 5 $
- $ CA = 6 $
- **Total = 3 + 5 + 6 = 14**

#### Example 6: Path $ D \to C \to B \to A $
- $ DC = 4 $
- $ CB = 5 $
- $ BA = 6 $
- **Total = 4 + 5 + 6 = 15**

From these examples, we see that the **minimum total distance** is **13**, and it is achieved by several different paths, such as:

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

---

### Step 4: Conclusion

After evaluating all possible paths that start at any of the four points and visit the other three exactly once, we find that the **shortest total distance** is:

$$
\boxed{13}
$$

[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. There's an Asymptote code provided, which draws several lines connecting these points. Let me parse that.

From the Asymptote code:

- Points A, B, C, D are defined with coordinates:
  - A = (-3, 3)
  - B = (3, 0)
  - C = (0, -4)
  - D = (0, 0)

Then, the drawing commands are:

- D--B--C--cycle: So triangle DBC
- D--A--B--cycle: Triangle DAB
- D--A--C--cycle: Triangle DAC

So, connecting D to B to C to D, D to A to B to D, and D to A to C to D. So, the figure is a combination of triangles. Also, labels for distances:

- "3" on D--B (so DB is 3 units)
- "6" on A--B (AB is 6)
- "6" on A--C (AC is 6)
- "4" on D--C (DC is 4)
- "5" on A--D (AD is 5)
- "5" on C--B (CB is 5)

Wait, but according to coordinates, maybe the distances are different? Let me check if the coordinates match the given distances. Maybe the Asymptote code is just for drawing, and the labels are the actual distances. Let me confirm.

First, coordinates:

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

Let me compute the distances between these points to see if they match the labels.

First, distance between D (0,0) and B (3,0): that's sqrt[(3-0)^2 + (0-0)^2] = sqrt[9] = 3. Okay, matches the label "3" on D--B.

Distance between A (-3,3) and B (3,0): sqrt[(3 - (-3))^2 + (0 - 3)^2] = sqrt[(6)^2 + (-3)^2] = sqrt[36 + 9] = sqrt[45] = 3*sqrt(5) ≈ 6.708. But the label says "6" on A--B. Hmm, discrepancy here. Maybe the coordinates are not to scale? Or maybe the labels are the actual distances? The problem says "label" so maybe the coordinates are just for drawing, and the actual distances are as labeled. So perhaps the coordinates are not to scale? Maybe I should ignore coordinates and just use the given distances?

Wait, the problem says "start at any point, A, B, C, or D, and visit the other three points once." So it's a traveling salesman problem? But with only four points, and visiting all three others once. Wait, starting at one point, visiting the other three once. So total of four points, visiting all four? Wait, starting at one, then visiting the other three. So total of four points, visiting all four? Wait, but the question says "visit the other three points once." So starting at A, then visiting B, C, D once each. So the path is A -> B -> C -> D or some permutation. Similarly for starting at other points.

But the question is asking for the shortest distance that can be travelled. So the minimal total distance when starting at any of the four points and visiting the other three once. So it's equivalent to finding the shortest possible route that starts at any point and visits all three others, so effectively, the shortest Hamiltonian path in the complete graph of these four points, considering the distances between them.

But since the problem says "start at any point," maybe we need to consider all possible starting points and find the minimal total distance among all possible paths. However, since the graph is undirected, the minimal path should be the same regardless of starting point, but maybe not? Wait, no. For example, if the graph is asymmetric, but in this case, since it's a complete graph with symmetric distances, the minimal path should be the same regardless of starting point. Wait, but maybe not. Wait, the problem says "start at any point," so maybe we need to consider all possible starting points and find the minimal total distance. However, since the graph is undirected, the minimal path should be the same regardless of starting point. Wait, but maybe not. For example, if you have different distances between points, depending on the starting point, the path could be different. Wait, but since we can choose the starting point, maybe we can choose the one that gives the minimal total distance.

But actually, the problem says "start at any point, A, B, C, or D, and visit the other three points once." So the question is, what is the minimal total distance of such a path, considering that you can choose the starting point. So we need to find the minimal total distance over all possible starting points and all possible paths that start at that point and visit the other three once.

Alternatively, since the problem is about visiting all four points, but starting at one and visiting the other three. Wait, but if you start at a point and visit the other three, that's visiting all four points. So the problem is equivalent to finding the shortest Hamiltonian path in the complete graph of these four points, with the given distances between them. However, Hamiltonian path is a path that visits each vertex exactly once. So since there are four points, the path has three edges. So the total distance is the sum of three edges. However, since the problem says "start at any point," maybe we need to consider all possible starting points and all possible permutations of visiting the other three points, and find the minimal total distance.

But since the graph is undirected, the minimal path should be the same regardless of starting point. Wait, but maybe not. For example, if you have different distances between points, depending on the order. Let me think.

Alternatively, maybe the problem is equivalent to finding the minimal total distance of a path that starts at any point and visits all other three points once. So, for example, starting at A, then going to B, then to C, then to D. Or starting at A, then to C, then to B, then to D, etc. So for each starting point, there are 3! = 6 possible paths. But since the graph is undirected, some of these paths will be the same in reverse. However, since we are considering starting at a specific point, the direction matters.

But since the problem says "start at any point," we can choose the starting point that gives the minimal total distance. Therefore, we need to compute for each starting point, the minimal total distance of a path that starts there and visits the other three points once, and then take the minimum among those.

Alternatively, since the problem is symmetric, maybe the minimal total distance is the same regardless of starting point. But I need to check.

Alternatively, maybe the problem is asking for the minimal total distance of a path that starts at any point and visits all other three points once, so the minimal over all possible starting points and all possible paths. So we need to find the minimal total distance among all possible Hamiltonian paths in the complete graph of these four points.

But since there are four points, the number of Hamiltonian paths is 4 * 3! = 24, but since each path is counted twice (once in each direction), but since we are considering starting points, maybe not. Wait, if we fix the starting point, then for each starting point, there are 3! = 6 possible paths. So total of 4 * 6 = 24 paths. But since the graph is undirected, some of these paths are reverses of each other. However, since we are considering starting points, they are different.

But maybe there's a smarter way. Let me think. The minimal total distance would be the minimal sum of three edges that connect all four points. However, since it's a path, not a cycle, so it's a sequence of three edges connecting four points. So, for example, A-B-C-D, which is three edges: AB, BC, CD. Alternatively, A-B-D-C, which is AB, BD, DC. So different combinations.

But to find the minimal total distance, we need to find the combination of three edges that connect all four points with the minimal total length. However, since the problem allows starting at any point, maybe we can choose the starting point that allows the minimal total distance.

Alternatively, maybe the minimal total distance is the same regardless of starting point, but I need to check.

Alternatively, maybe the problem is equivalent to finding the minimal spanning tree, but no, because a spanning tree has three edges for four points, but a spanning tree is a tree, not a path. However, the minimal spanning tree would have the minimal total length, but the path is different. However, maybe the minimal path is related to the minimal spanning tree? Not sure.

Alternatively, maybe we can think of it as the Traveling Salesman Problem (TSP) where we need to visit all four points, but since we can start at any point, the minimal TSP tour would be the minimal total distance. However, in TSP, you have to return to the starting point, but here, we don't have to return. So it's a Hamiltonian path, not a cycle.

But since the problem says "visit the other three points once," starting at one point, so it's a Hamiltonian path. So the minimal Hamiltonian path.

But how do we compute that? For four points, it's manageable.

First, let me list all possible Hamiltonian paths and compute their total distances. However, since there are 4 points, the number of Hamiltonian paths is 4 * 3! = 24, but since each path is counted twice (once in each direction), but since we are considering starting points, maybe not. Wait, if we fix the starting point, then for each starting point, there are 3! = 6 paths. So total of 4 * 6 = 24 paths. However, since the graph is undirected, some of these paths are reverses of each other. But since we are considering starting points, they are different.

But maybe we can find the minimal total distance by considering all possible permutations.

Alternatively, since there are only four points, maybe we can find the minimal total distance by considering the minimal total of three edges that connect all four points. However, since it's a path, not a tree, the three edges must form a path, not a tree. So, for example, if we have points A, B, C, D, then a path could be A-B-C-D, which uses edges AB, BC, CD. Alternatively, A-B-D-C, which uses AB, BD, DC. Or A-D-B-C, which uses AD, DB, BC. Etc.

So, to find the minimal total distance, we need to find the path that has the minimal sum of three edges. However, since there are multiple possibilities, we need to check them all.

Alternatively, maybe we can use the fact that the minimal Hamiltonian path is the one that uses the three shortest edges. But since the edges are connected in a path, not just any three edges.

Alternatively, maybe we can think of it as follows: For each pair of points, we can compute the distances between them. Then, for each possible starting point, we can try to find the minimal path.

But first, let me list all the distances between the points. The problem gives some distances:

From the Asymptote code labels:

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

Wait, but according to coordinates, some distances are different. However, the problem says "label" so maybe the coordinates are just for drawing, and the actual distances are as labeled. Therefore, we should use the given distances. Let me confirm:

Given distances:

- DB = 3
- AB = 6
- AC = 6
- DC = 4
- AD = 5
- CB = 5

But what about other distances? For example, what is the distance between A and D? It's given as 5. What about between A and C? 6. Between A and B? 6. Between B and C? 5. Between D and C? 4. Between D and B? 3. What about between D and A? 5. What about between A and D? 5. What about between D and C? 4. What about between B and D? 3. What about between B and A? 6. What about between C and A? 6. What about between C and B? 5. What about between D and A? 5. What about between D and C? 4. What about between D and B? 3. What about between A and C? 6. What about between A and B? 6. What about between B and C? 5. What about between D and A? 5. What about between D and C? 4. What about between D and B? 3. What about between A and D? 5. So, the distances between all pairs:

Let me list all pairs:

AB: 6

AC: 6

AD: 5

BC: 5

BD: 3

CD: 4

What about the other pairs? Wait, there are four points: A, B, C, D. So total of 6 pairs:

AB, AC, AD, BC, BD, CD. So those are all. So the distances are:

AB = 6

AC = 6

AD = 5

BC = 5

BD = 3

CD = 4

So that's all. So we have all the distances between the points. Now, we need to find the minimal total distance of a path that starts at any point and visits the other three once. 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. Or starting at A, going to C, then to D, then to B: AC + CD + DB = 6 + 4 + 3 = 13. So starting at A, the minimal path is 13.

Similarly, let's check starting at B. Let's see. Starting at B, then visiting A, C, D. Let's try different permutations.

B -> A -> C -> D: BA + AC + CD = 6 + 6 + 4 = 16

B -> A -> D -> C: BA + AD + DC = 6 + 5 + 4 = 15

B -> C -> A -> D: BC + CA + AD = 5 + 6 + 5 = 16

B -> C -> D -> A: BC + CD + DA = 5 + 4 + 5 = 14

B -> D -> A -> C: BD + DA + AC = 3 + 5 + 6 = 14

B -> D -> C -> A: BD + DC + CA = 3 + 4 + 6 = 13

So starting at B, the minimal total distance is 13.

Starting at C. Let's check.

Starting at C, visiting A, B, D.

C -> A -> B -> D: CA + AB + BD = 6 + 6 + 3 = 15

C -> A -> D -> B: CA + AD + DB = 6 + 5 + 3 = 14

C -> B -> A -> D: CB + BA + AD = 5 + 6 + 5 = 16

C -> B -> D -> A: CB + BD + DA = 5 + 3 + 5 = 13

C -> D -> A -> B: CD + DA + AB = 4 + 5 + 6 = 15

C -> D -> B -> A: CD + DB + BA = 4 + 3 + 6 = 13

So starting at C, minimal total distance is 13.

Starting at D. Let's check.

Starting at D, visiting A, B, C.

D -> A -> B -> C: DA + AB + BC = 5 + 6 + 5 = 16

D -> A -> C -> B: DA + AC + CB = 5 + 6 + 5 = 16

D -> B -> A -> C: DB + BA + AC = 3 + 6 + 6 = 15

D -> B -> C -> A: DB + BC + CA = 3 + 5 + 6 = 14

D -> C -> A -> B: DC + CA + AB = 4 + 6 + 6 = 16

D -> C -> B -> A: DC + CB + BA = 4 + 5 + 6 = 15

So starting at D, the minimal total distance is 14.

Wait, so starting at A: 13, starting at B: 13, starting at C: 13, starting at D: 14. Therefore, the minimal total distance is 13. So the answer is 13? But let me check again.

Wait, when starting at D, the minimal path is 14? Let me check again:

Starting at D, the possible paths:

D -> B -> C -> A: DB + BC + CA = 3 + 5 + 6 = 14

D -> C -> B -> A: DC + CB + BA = 4 + 5 + 6 = 15

D -> B -> A -> C: DB + BA + AC = 3 + 6 + 6 = 15

D -> A -> B -> C: DA + AB + BC = 5 + 6 + 5 = 16

D -> A -> C -> B: DA + AC + CB = 5 + 6 + 5 = 16

D -> C -> A -> B: DC + CA + AB = 4 + 6 + 6 = 16

So yes, the minimal starting at D is 14. So the minimal total distance is 13, achieved when starting at A, B, or C. So the answer is 13? But let me check if there are other paths that I might have missed.

Wait, for example, starting at A, going to D, then to B, then to C: AD + DB + BC = 5 + 3 + 5 = 13. Similarly, starting at A, going to C, then to D, then to B: AC + CD + DB = 6 + 4 + 3 = 13. Starting at B, going to D, then to C, then to A: BD + DC + CA = 3 + 4 + 6 = 13. Starting at B, going to D, then to C, then to A: same as above. Starting at C, going to B, then to D, then to A: CB + BD + DA = 5 + 3 + 5 = 13. Starting at C, going to D, then to B, then to A: CD + DB + BA = 4 + 3 + 6 = 13. So all these paths give 13. So the minimal total distance is 13.

But let me check if there is a path with a shorter total distance. For example, is there a path with total distance less than 13? Let's see. Let's think of other combinations.

For example, starting at A, going to D (5), then to C (4), then to B (5): total 5 + 4 + 5 = 14. Not better. Starting at A, going to D (5), then to B (3), then to C (5): 5 + 3 + 5 = 13. Same as before.

Is there a way to have a path with total distance less than 13? Let's see. Let's think of the three edges. The three edges must connect all four points. So, for example, if we can find three edges that connect all four points with total distance less than 13. Let's see.

The three edges must form a path. So, for example, if we have edges BD (3), DC (4), and CA (6): total 3 + 4 + 6 = 13. Or BD (3), DA (5), and AC (6): 3 + 5 + 6 = 14. Or BD (3), BC (5), and CA (6): 3 + 5 + 6 = 14. Or DC (4), CB (5), BA (6): 4 + 5 + 6 = 15. Or DA (5), AB (6), BC (5): 5 + 6 + 5 = 16. Or DA (5), AC (6), CB (5): 5 + 6 + 5 = 16. Or DB (3), BA (6), AC (6): 3 + 6 + 6 = 15. Or DC (4), CA (6), AB (6): 4 + 6 + 6 = 16. Or DB (3), BC (5), CD (4): 3 + 5 + 4 = 12? Wait, hold on. Wait, if we have edges DB (3), BC (5), and CD (4). But that would form a cycle: D-B-C-D. But that's a cycle, not a path. Because a path must visit each point once. So if you have D-B-C-D, you are visiting D twice. So that's not allowed. So that's not a valid path. Similarly, if you have edges BD (3), DC (4), and CB (5), that would be D-B-C-D again, which is a cycle. So that's not a path. So the three edges must form a path, not a cycle. Therefore, the three edges must be connected in a sequence, visiting each point once. Therefore, the three edges cannot form a cycle. Therefore, the three edges must be such that they form a path. Therefore, the three edges must be connected in a way that they form a single path. Therefore, for example, edges AB, BC, CD form a path A-B-C-D. Similarly, edges AD, DB, BC form a path A-D-B-C. So, in that case, the total distance is AD + DB + BC = 5 + 3 + 5 = 13. Similarly, edges AC, CD, DB form a path A-C-D-B, which is AC + CD + DB = 6 + 4 + 3 = 13. Similarly, edges CB, BD, DA form a path C-B-D-A, which is CB + BD + DA = 5 + 3 + 5 = 13. So all these are valid paths with total distance 13. So there is no way to get a shorter total distance. Because the three edges that form a path must include three edges, and the minimal sum is 13. Let me check if there is a combination of three edges that form a path with a smaller total.

For example, if we take edges BD (3), DC (4), and CA (6): total 13. If we take edges BD (3), DA (5), and AC (6): total 14. If we take edges DC (4), CB (5), and BA (6): total 15. If we take edges AB (6), BC (5), CD (4): total 15. If we take edges AD (5), DC (4), CB (5): total 14. If we take edges AC (6), CB (5), BD (3): total 14. So no, there is no combination of three edges that form a path with total less than 13. Therefore, the minimal total distance is 13.

Therefore, the answer is 13. However, let me check once again if there is a path that I haven't considered. For example, starting at A, going to D (5), then to C (4), then to B (5): total 14. Starting at A, going to B (6), then to D (3), then to C (4): total 6 + 3 + 4 = 13. Wait, that's another path: A-B-D-C. Let me check the distances. AB is 6, BD is 3, DC is 4. Total 6 + 3 + 4 = 13. Yes, that's another path. So that's also 13. So that's another way to get 13. So yes, multiple paths give 13. Therefore, the minimal total distance is 13.

But let me check if there is a path with total distance less than 13. For example, if there is a path with total distance 12. Let's see. To get 12, we need three edges with total 12. The minimal edges are BD (3), DC (4), and... what's the next? If we have BD (3), DC (4), and then another edge. But the third edge must connect to the path. For example, if we have BD (3), DC (4), and then... if we have BD, DC, and CB (5), that would be 3 + 4 + 5 = 12, but that would form a cycle D-B-C-D, which is not a path. Similarly, BD (3), DA (5), and AC (6): 3 + 5 + 6 = 14. BD (3), DA (5), and AB (6): 3 + 5 + 6 = 14. DC (4), CB (5), BA (6): 4 + 5 + 6 = 15. So no, there is no way to get 12. Therefore, 13 is indeed the minimal.

Therefore, the answer is 13. However, let me check once again if I have considered all possible paths. For example, starting at A, going to D (5), then to B (3), then to C (5): total 5 + 3 + 5 = 13. Starting at A, going to C (6), then to D (4), then to B (3): total 6 + 4 + 3 = 13. Starting at B, going to D (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at B, going to D (3), then to A (5), then to C (6): 3 + 5 + 6 = 14. Wait, no, that's 14. Starting at B, going to D (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at C, going to B (5), then to D (3), then to A (5): 5 + 3 + 5 = 13. Starting at C, going to D (4), then to B (3), then to A (5): 4 + 3 + 5 = 12? Wait, hold on! Wait, if starting at C, going to D (4), then to B (3), then to A (5). Wait, is that a valid path? Let me check the distances. CD is 4, DB is 3, BA is 6. Wait, no, from B to A is 6, not 5. Wait, BA is AB, which is 6. So CD (4) + DB (3) + BA (6) = 4 + 3 + 6 = 13. So that's 13. Wait, but if I thought BA is 5, that was a mistake. So no, that's 13. So no, there is no path with total distance less than 13.

Wait, but what if there is a path that uses different edges? For example, starting at D, going to B (3), then to A (5), then to C (6): total 3 + 5 + 6 = 14. Starting at D, going to C (4), then to B (3), then to A (5): 4 + 3 + 5 = 12? Wait, no, from B to A is 6, not 5. Wait, BA is 6. So 4 + 3 + 6 = 13. So again, 13. So no, there is no way to get less than 13.

Therefore, the minimal total distance is 13. Therefore, the answer is 13. However, let me check once again if there is a path that I have not considered. For example, starting at A, going to D (5), then to C (4), then to B (5): total 5 + 4 + 5 = 14. Starting at A, going to B (6), then to D (3), then to C (4): 6 + 3 + 4 = 13. Starting at A, going to C (6), then to D (4), then to B (3): 6 + 4 + 3 = 13. Starting at B, going to D (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at B, going to C (5), then to D (3), then to A (5): 5 + 3 + 5 = 13. Starting at C, going to D (4), then to B (3), then to A (6): 4 + 3 + 6 = 13. Starting at C, going to B (5), then to D (3), then to A (5): 5 + 3 + 5 = 13. Starting at D, going to B (3), then to C (4), then to A (6): 3 + 4 + 6 = 13. Starting at D, going to C (4), then to B (3), then to A (6): same as above. So all these paths give 13. Therefore, the minimal total distance is 13.

Therefore, the answer is 13. However, let me check if the problem says "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." So, the minimal total distance is 13. Therefore, the answer is \boxed{13}.

**Final Answer**
\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 **once**, we are essentially looking for the **shortest Hamiltonian path** among the four points. A Hamiltonian path is a path that visits each vertex exactly once, and since we are allowed to choose the starting point, we must consider all possible starting points and all possible permutations of visiting the other three points.

---

### Step 1: Understand the Given Distances

From the problem, we are given the following distances between the points:

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

These are the only distances between the four points, and we can use them to compute the total distance of any path that visits all four points exactly once.

---

### Step 2: Strategy

We need to find the **shortest total distance** of a path that starts at any one of the four points and visits the other three exactly once. Since the graph is undirected, the total distance of a path is the same regardless of the direction (i.e., the path $ A \to B \to C \to D $ has the same total distance as $ D \to C \to B \to A $).

Thus, we can consider all possible permutations of visiting the four points, starting from any one of them, and compute the total distance for each. Then, we select the **minimum** among all these.

However, since there are only four points, we can simplify the process by considering all possible **Hamiltonian paths** (i.e., all possible sequences of visiting the four points once), and compute the total distance for each.

---

### Step 3: Compute the Total Distance for All Valid Paths

We'll consider all permutations of the four points and compute the total distance for each. Since the graph is undirected, we can consider only the unique paths (i.e., avoid counting the reverse of each path as a separate one).

Let’s consider a few key examples:

#### Example 1: Path $ A \to D \to B \to C $
- $ AD = 5 $
- $ DB = 3 $
- $ BC = 5 $
- **Total = 5 + 3 + 5 = 13**

#### Example 2: Path $ A \to C \to D \to B $
- $ AC = 6 $
- $ CD = 4 $
- $ DB = 3 $
- **Total = 6 + 4 + 3 = 13**

#### Example 3: Path $ B \to D \to C \to A $
- $ BD = 3 $
- $ DC = 4 $
- $ CA = 6 $
- **Total = 3 + 4 + 6 = 13**

#### Example 4: Path $ C \to B \to D \to A $
- $ CB = 5 $
- $ BD = 3 $
- $ DA = 5 $
- **Total = 5 + 3 + 5 = 13**

#### Example 5: Path $ D \to B \to C \to A $
- $ DB = 3 $
- $ BC = 5 $
- $ CA = 6 $
- **Total = 3 + 5 + 6 = 14**

#### Example 6: Path $ D \to C \to B \to A $
- $ DC = 4 $
- $ CB = 5 $
- $ BA = 6 $
- **Total = 4 + 5 + 6 = 15**

From these examples, we see that the **minimum total distance** is **13**, and it is achieved by several different paths, such as:

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

---

### Step 4: Conclusion

After evaluating all possible paths that start at any of the four points and visit the other three exactly once, we find that the **shortest total distance** is:

$$
\boxed{13}
$$

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 $\boxed{13}$.