Explanation:
Okay, let me try to figure out this problem. So, Jim and Martha are at the corner of a rectangular field. The field is 300 feet wide and 400 feet long. Jim walks diagonally across the field, while Martha walks along the length and width to get to the same location. The question is asking how many feet less Jim walks compared to Martha.

First, I need to visualize the field. It's a rectangle, right? So, if Jim and Martha are at one corner, let's say the bottom-left corner, then Martha would walk along the length and width to reach the opposite corner. Wait, actually, the problem says Martha walks along the length and width to get to the same location. So maybe Martha is walking along both the length and the width? Wait, but if they're starting at the same corner, and the field is rectangular, then the opposite corner would be 300 feet in one direction and 400 feet in the other.

So Martha is walking along the length and the width. Let me think. If Martha is going from one corner to the opposite corner, she would walk along the length and then the width, or the width and then the length. Either way, the total distance she walks would be the sum of the length and the width.

So Martha's distance is 300 + 400. Let me check that again. The field is 300 feet wide and 400 feet long. So if you walk along the length, that's 400 feet, and then along the width, that's 300 feet. So total distance Martha walks is 400 + 300, which is 700 feet.

Now Jim is walking diagonally across the field. So the distance Jim walks is the diagonal of the rectangle. To find the length of the diagonal, I can use the Pythagorean theorem. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (which is the diagonal here) is equal to the sum of the squares of the other two sides.

So the formula is: diagonal^2 = length^2 + width^2.

Plugging in the numbers: diagonal^2 = 400^2 + 300^2.

Let me calculate that.

First, 400 squared. 400 times 400. Well, 4 times 4 is 16, and then adding the zeros: 400 squared is 160,000.

Then 300 squared is 90,000.

Adding those together: 160,000 + 90,000 = 250,000.

So the diagonal squared is 250,000. Therefore, the diagonal is the square root of 250,000.

What's the square root of 250,000? Let me think.

Well, sqrt(250,000) can be broken down as sqrt(25 * 10,000). Because 25 times 10,000 is 250,000.

sqrt(25) is 5, and sqrt(10,000) is 100.

So multiplying those together, 5 * 100 = 500.

Therefore, the diagonal is 500 feet.

So Jim walks 500 feet, and Martha walks 700 feet.

Therefore, the difference is 700 - 500 = 200 feet.

So Jim walks 200 feet less than Martha.

Wait, let me check if I did everything correctly.

First, the field's dimensions: 300 feet wide and 400 feet long. So if we consider the rectangle, the sides are 300 and 400. Martha walks along both sides, so 300 + 400 = 700.

Jim walks the diagonal, so Pythagorean theorem. 300 squared is 90,000, 400 squared is 160,000, sum is 250,000. Square root of that is 500.

So 700 - 500 is indeed 200.

Therefore, Jim walks 200 feet less than Martha.

I don't see any mistakes here. Let me think again.

Alternatively, maybe there's a different interpretation of the problem? Let me check again.

"Jim and Martha are standing together at the corner of a rectangular field. Jim walks diagonally across the field. Martha gets to the same location by walking along its length and width."

So, "the same location" meaning the opposite corner? Because if they start at the same corner, then walking along the length and width would get them to the opposite corner. So Martha walks along the length and width, which are 400 and 300. So total distance 700.

Jim walks diagonally, which is 500.

Difference is 200.

Yes. That seems correct.

Alternatively, maybe Martha walks along the length and then the width? So if the field is 400 feet long and 300 feet wide, then from the corner, Martha walks along the length (400 ft) and then along the width (300 ft) to reach the opposite corner. So total distance is 700.

So Jim's path is shorter by 200 feet.

Alternatively, if Martha walked along the length and then the width, but maybe she's walking along the length and width in some other way? But no, since they are at the corner, the only way to get to the opposite corner is to go along both sides.

Alternatively, maybe the problem is that the field is 300 feet wide and 400 feet long, so if they are at the corner, maybe Martha walks along the width and then the length? But regardless, the total distance is still 300 + 400 = 700.

So I think the answer is 200.

But just to be thorough, let me check the Pythagorean theorem.

For a 3-4-5 triangle, if sides are 3 and 4, hypotenuse is 5.

So in this case, the sides are 300 and 400. Which is 3*100 and 4*100, so hypotenuse would be 5*100=500.

Yes, so that's exactly the case here.

Therefore, the diagonal is 500.

So Martha walks 700, Jim walks 500, difference is 200.

Therefore, the answer is 200.

**Final Answer**
Jim walks \boxed{200} feet less than Martha.

To solve this problem, we are given a rectangle that measures **300 feet in width** and **400 feet in length**. Jim and Martha begin from the same corner and head to the opposite corner, but their paths differ:

- **Martha** walks along the **length** and **width** of the rectangle.
- **Jim** walks **diagonally** across the field.

We are asked: **How many feet less does Jim walk than Martha?**

---

### Step 1: Distance Martha Walks

Martha walks along the length and the width of the rectangle to reach the opposite corner. Thus, her total distance is:

$$
\text{Martha's distance} = \text{length} + \text{width} = 400 + 300 = 700 \text{ feet}
$$

---

### Step 2: Distance Jim Walks

Jim takes the shortest path across the field, which is the **diagonal**. To find the length of the diagonal, we use the **Pythagorean Theorem**:

$$
\text{Diagonal}^2 = \text{length}^2 + \text{width}^2
$$

$$
\text{Diagonal}^2 = 400^2 + 300^2 = 160000 + 90000 = 250000
$$

$$
\text{Diagonal} = \sqrt{250000} = 500 \text{ feet}
$$

So, **Jim walks 500 feet**.

---

### Step 3: Difference in Distances

Now, we subtract Jim's distance from Martha's distance:

$$
\text{Difference} = \text{Martha's distance} - \text{Jim's distance} = 700 - 500 = 200 \text{ feet}
$$

---

### Final Answer

$$
\boxed{200}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, Jim and Martha are at the corner of a rectangular field. The field is 300 feet wide and 400 feet long. Jim walks diagonally across the field, while Martha walks along the length and width to get to the same location. The question is asking how many feet less Jim walks compared to Martha.

First, I need to visualize the field. It's a rectangle, right? So, if Jim and Martha are at one corner, let's say the bottom-left corner, then Martha would walk along the length and width to reach the opposite corner. Wait, actually, the problem says Martha walks along the length and width to get to the same location. So maybe Martha is walking along both the length and the width? Wait, but if they're starting at the same corner, and the field is rectangular, then the opposite corner would be 300 feet in one direction and 400 feet in the other.

So Martha is walking along the length and the width. Let me think. If Martha is going from one corner to the opposite corner, she would walk along the length and then the width, or the width and then the length. Either way, the total distance she walks would be the sum of the length and the width.

So Martha's distance is 300 + 400. Let me check that again. The field is 300 feet wide and 400 feet long. So if you walk along the length, that's 400 feet, and then along the width, that's 300 feet. So total distance Martha walks is 400 + 300, which is 700 feet.

Now Jim is walking diagonally across the field. So the distance Jim walks is the diagonal of the rectangle. To find the length of the diagonal, I can use the Pythagorean theorem. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (which is the diagonal here) is equal to the sum of the squares of the other two sides.

So the formula is: diagonal^2 = length^2 + width^2.

Plugging in the numbers: diagonal^2 = 400^2 + 300^2.

Let me calculate that.

First, 400 squared. 400 times 400. Well, 4 times 4 is 16, and then adding the zeros: 400 squared is 160,000.

Then 300 squared is 90,000.

Adding those together: 160,000 + 90,000 = 250,000.

So the diagonal squared is 250,000. Therefore, the diagonal is the square root of 250,000.

What's the square root of 250,000? Let me think.

Well, sqrt(250,000) can be broken down as sqrt(25 * 10,000). Because 25 times 10,000 is 250,000.

sqrt(25) is 5, and sqrt(10,000) is 100.

So multiplying those together, 5 * 100 = 500.

Therefore, the diagonal is 500 feet.

So Jim walks 500 feet, and Martha walks 700 feet.

Therefore, the difference is 700 - 500 = 200 feet.

So Jim walks 200 feet less than Martha.

Wait, let me check if I did everything correctly.

First, the field's dimensions: 300 feet wide and 400 feet long. So if we consider the rectangle, the sides are 300 and 400. Martha walks along both sides, so 300 + 400 = 700.

Jim walks the diagonal, so Pythagorean theorem. 300 squared is 90,000, 400 squared is 160,000, sum is 250,000. Square root of that is 500.

So 700 - 500 is indeed 200.

Therefore, Jim walks 200 feet less than Martha.

I don't see any mistakes here. Let me think again.

Alternatively, maybe there's a different interpretation of the problem? Let me check again.

"Jim and Martha are standing together at the corner of a rectangular field. Jim walks diagonally across the field. Martha gets to the same location by walking along its length and width."

So, "the same location" meaning the opposite corner? Because if they start at the same corner, then walking along the length and width would get them to the opposite corner. So Martha walks along the length and width, which are 400 and 300. So total distance 700.

Jim walks diagonally, which is 500.

Difference is 200.

Yes. That seems correct.

Alternatively, maybe Martha walks along the length and then the width? So if the field is 400 feet long and 300 feet wide, then from the corner, Martha walks along the length (400 ft) and then along the width (300 ft) to reach the opposite corner. So total distance is 700.

So Jim's path is shorter by 200 feet.

Alternatively, if Martha walked along the length and then the width, but maybe she's walking along the length and width in some other way? But no, since they are at the corner, the only way to get to the opposite corner is to go along both sides.

Alternatively, maybe the problem is that the field is 300 feet wide and 400 feet long, so if they are at the corner, maybe Martha walks along the width and then the length? But regardless, the total distance is still 300 + 400 = 700.

So I think the answer is 200.

But just to be thorough, let me check the Pythagorean theorem.

For a 3-4-5 triangle, if sides are 3 and 4, hypotenuse is 5.

So in this case, the sides are 300 and 400. Which is 3*100 and 4*100, so hypotenuse would be 5*100=500.

Yes, so that's exactly the case here.

Therefore, the diagonal is 500.

So Martha walks 700, Jim walks 500, difference is 200.

Therefore, the answer is 200.

**Final Answer**
Jim walks \boxed{200} feet less than Martha.

To solve this problem, we are given a rectangle that measures **300 feet in width** and **400 feet in length**. Jim and Martha begin from the same corner and head to the opposite corner, but their paths differ:

- **Martha** walks along the **length** and **width** of the rectangle.
- **Jim** walks **diagonally** across the field.

We are asked: **How many feet less does Jim walk than Martha?**

---

### Step 1: Distance Martha Walks

Martha walks along the length and the width of the rectangle to reach the opposite corner. Thus, her total distance is:

$$
\text{Martha's distance} = \text{length} + \text{width} = 400 + 300 = 700 \text{ feet}
$$

---

### Step 2: Distance Jim Walks

Jim takes the shortest path across the field, which is the **diagonal**. To find the length of the diagonal, we use the **Pythagorean Theorem**:

$$
\text{Diagonal}^2 = \text{length}^2 + \text{width}^2
$$

$$
\text{Diagonal}^2 = 400^2 + 300^2 = 160000 + 90000 = 250000
$$

$$
\text{Diagonal} = \sqrt{250000} = 500 \text{ feet}
$$

So, **Jim walks 500 feet**.

---

### Step 3: Difference in Distances

Now, we subtract Jim's distance from Martha's distance:

$$
\text{Difference} = \text{Martha's distance} - \text{Jim's distance} = 700 - 500 = 200 \text{ feet}
$$

---

### Final Answer

$$
\boxed{200}
$$

Solution: The diagonal of the field is $\sqrt{300^2+400^2}=500$ feet long so Jim walks 500 feet. Two adjacent sides of the field are $300+400=700$ feet long so Martha walks 700 feet. Jim walks $700-500=\boxed{200}$ feet less than Martha.


Explanation:
Okay, so I need to find all possible values of the expression $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$ where $\omega$ is a complex number such that $\omega^3 = 1$. Let me think about how to approach this step by step.

First, I remember that the equation $\omega^3 = 1$ has three solutions, which are the cube roots of unity. These are $1$, $\omega$, and $\omega^2$, where $\omega$ is a primitive cube root of unity. So, the possible values of $\omega$ are $1$, $e^{2\pi i /3}$, and $e^{4\pi i /3}$. However, the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$." So, $\omega$ could be any of these three roots. But the question asks for all possible values of the given expression. Therefore, I need to compute the expression for each of these roots and see what values I get.

But maybe there's a smarter way than just plugging in each root. Let me recall some properties of roots of unity. The cube roots of unity satisfy the equation $x^3 - 1 = 0$, which factors as $(x - 1)(x^2 + x + 1) = 0$. Therefore, the non-1 roots satisfy $x^2 + x + 1 = 0$. So, for $\omega \neq 1$, we have $\omega^2 + \omega + 1 = 0$. That seems useful. Let me note that down.

So, if $\omega^3 = 1$, then $\omega \neq 1$ implies $\omega^2 + \omega + 1 = 0$. Therefore, maybe I can use this relation to simplify the given expression.

Let me write down the expression again:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

I need to compute this sum. Let me first try to combine these two fractions. To add them, I need a common denominator, which would be $(1 + \omega)(1 + \omega^2)$. Let me compute the numerator:

$$
(1 + \omega^2) + (1 + \omega) = 1 + \omega^2 + 1 + \omega = 2 + \omega + \omega^2
$$

Therefore, the expression becomes:

$$
\frac{2 + \omega + \omega^2}{(1 + \omega)(1 + \omega^2)}
$$

Now, from the equation $\omega^2 + \omega + 1 = 0$ (for $\omega \neq 1$), we know that $\omega + \omega^2 = -1$. Therefore, substituting into the numerator:

$$
2 + (-1) = 1
$$

So the numerator simplifies to 1. Now, I need to compute the denominator: $(1 + \omega)(1 + \omega^2)$.

Let me expand this:

$$
(1 + \omega)(1 + \omega^2) = 1 \cdot 1 + 1 \cdot \omega^2 + \omega \cdot 1 + \omega \cdot \omega^2
$$

Calculating each term:

- $1 \cdot 1 = 1$
- $1 \cdot \omega^2 = \omega^2$
- $\omega \cdot 1 = \omega$
- $\omega \cdot \omega^2 = \omega^3$

But $\omega^3 = 1$, so that term is 1. Therefore, the entire expression becomes:

$$
1 + \omega^2 + \omega + 1 = 2 + \omega + \omega^2
$$

Wait, that's the same as the numerator! Wait, but earlier, we found that $\omega + \omega^2 = -1$, so substituting again:

$$
2 + (-1) = 1
$$

Wait, so the denominator is also 1? That would make the entire expression $\frac{1}{1} = 1$. But that seems strange. Let me check my calculations again because if both numerator and denominator are 1, then the expression is 1. However, if $\omega = 1$, then the original expression would have denominators 1 + 1 = 2 and 1 + 1 = 2, so the expression would be 1/2 + 1/2 = 1 as well. Wait, so does that mean regardless of which root we take, the value is 1? But the question says "all possible values", implying that there might be more than one. But according to this, it's always 1? Let me verify.

Wait, but if $\omega = 1$, then $\omega^3 = 1$, so that's allowed. However, when we derived the numerator and denominator, we used the equation $\omega^2 + \omega + 1 = 0$, which only holds when $\omega \neq 1$. Therefore, maybe we need to consider separately the case when $\omega = 1$ and when $\omega \neq 1$?

Let me check. If $\omega = 1$, then the original expression becomes:

$$
\frac{1}{1 + 1} + \frac{1}{1 + 1^2} = \frac{1}{2} + \frac{1}{2} = 1
$$

So even when $\omega = 1$, the value is 1. If $\omega$ is a primitive root, then as above, we have the expression is 1. Therefore, regardless of which root we take, the expression is 1? Therefore, there's only one possible value, which is 1?

But the problem says "all possible values", so maybe I made a mistake here. Let me check again with specific roots.

Let me take $\omega = e^{2\pi i /3}$. Then $\omega^2 = e^{4\pi i /3}$, and $\omega^3 = 1$. Let me compute the expression:

First, compute $1 + \omega$. Let me recall that $1 + \omega + \omega^2 = 0$, so $1 + \omega = -\omega^2$, and $1 + \omega^2 = -\omega$. Therefore:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} = \frac{1}{-\omega^2} + \frac{1}{-\omega} = -\frac{1}{\omega^2} - \frac{1}{\omega}
$$

But since $\omega^3 = 1$, we have that $\omega^2 = \omega^{-1}$, so $1/\omega^2 = \omega$, and $1/\omega = \omega^2$. Therefore, substituting:

$$
- \omega - \omega^2
$$

But from the equation $\omega^2 + \omega + 1 = 0$, we have $\omega + \omega^2 = -1$, so substituting:

$$
-(-1) = 1
$$

Same result. Therefore, even with this specific root, we get 1. Therefore, regardless of which root we take, the expression is 1. Therefore, the only possible value is 1. But then why does the problem say "all possible values"? Maybe there's a misunderstanding here?

Wait, let me check once more with $\omega = 1$. If $\omega = 1$, then the original expression is $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. So again, same result. Therefore, no matter which root we take, the value is 1. Therefore, there is only one possible value. Therefore, the answer is just 1.

But let me think again. Could there be another way where maybe the expression evaluates to something else? For example, if there was some mistake in my algebra?

Wait, let me check the denominator again. When I expanded $(1 + \omega)(1 + \omega^2)$, I got:

$$
1 + \omega^2 + \omega + \omega^3 = 1 + \omega + \omega^2 + 1 = 2 + \omega + \omega^2
$$

Which is equal to 2 + (-1) = 1. Therefore, the denominator is 1. Numerator is also 1, so the expression is 1. Therefore, regardless of $\omega$, as long as $\omega^3 = 1$, the expression evaluates to 1. Therefore, the only possible value is 1.

Wait, so then the answer is just 1? But maybe I need to check if there's a case where denominators are zero? For example, if $1 + \omega = 0$ or $1 + \omega^2 = 0$. Let me check that. If $1 + \omega = 0$, then $\omega = -1$. Then $\omega^3 = (-1)^3 = -1 \neq 1$, so $\omega = -1$ is not a cube root of unity. Therefore, none of the roots $\omega$ satisfy $1 + \omega = 0$ or $1 + \omega^2 = 0$, so denominators are never zero. Hence, the expression is always defined and equal to 1.

Therefore, the only possible value is 1. So, I think the answer is \boxed{1}.

Wait, but maybe I need to check with another approach. Let me consider using algebraic manipulation without splitting into cases.

Let me write the original expression:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

Let me note that $\omega^3 = 1$, so $\omega^2 = \omega^{-1}$, and $\omega = \omega^{-2}$. Also, since $\omega^3 = 1$, we can write everything in terms of $\omega$ or $\omega^2$. Let me try to compute:

Let me compute $ \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} $. Let me also note that since $\omega^2 + \omega + 1 = 0$, we can replace $\omega^2$ with $-1 - \omega$, and vice versa. Let me try substituting $\omega^2 = -1 - \omega$ into the expression.

First, compute $1 + \omega^2 = 1 + (-1 - \omega) = -\omega$. Similarly, $1 + \omega$ is just $1 + \omega$. Therefore, the expression becomes:

$$
\frac{1}{1 + \omega} + \frac{1}{ - \omega} = \frac{1}{1 + \omega} - \frac{1}{\omega}
$$

Now, let me combine these terms. To do that, I need a common denominator, which would be $\omega(1 + \omega)$. Let me write:

$$
\frac{\omega}{\omega(1 + \omega)} - \frac{1 + \omega}{\omega(1 + \omega)} = \frac{\omega - (1 + \omega)}{\omega(1 + \omega)} = \frac{\omega - 1 - \omega}{\omega(1 + \omega)} = \frac{ -1 }{\omega(1 + \omega)}
$$

Hmm, so that simplifies to $-1 / [\omega(1 + \omega)]$. Wait, but how does that relate to previous results? Let me compute this denominator: $\omega(1 + \omega)$. Let me expand that:

$$
\omega(1 + \omega) = \omega + \omega^2
$$

Again, from $\omega^2 + \omega + 1 = 0$, we have $\omega + \omega^2 = -1$, so the denominator is $-1$. Therefore, the expression is $-1 / (-1) = 1$. There we go! So again, we get 1. Therefore, regardless of the approach, we get 1. So that confirms that the value is always 1.

Alternatively, another approach: Let me consider multiplying numerator and denominator by something. For example, since we know that $\omega^3 = 1$, maybe multiplying numerator and denominator by $\omega - 1$ or something else?

Wait, but maybe I can consider the original expression:

Let me denote $S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$

Let me compute $S$:

First, note that $1 + \omega + \omega^2 = 0$, so $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$. Therefore:

$$
S = \frac{1}{- \omega^2} + \frac{1}{- \omega} = - \left( \frac{1}{\omega^2} + \frac{1}{\omega} \right)
$$

But $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$, since $\omega^3 = 1$. Therefore:

$$
S = - (\omega + \omega^2 ) = -(-1) = 1
$$

Again, same result. So all approaches lead to 1. Therefore, it seems that regardless of the root, the value is 1. Hence, the only possible value is 1.

But then why does the problem say "Find all possible values"? Maybe there's something wrong here? Wait, but the problem states "Let $\omega$ be a complex number such that $\omega^3 = 1$." So $\omega$ can be 1, $\omega$, or $\omega^2$. But in all cases, the expression evaluates to 1. Therefore, only one possible value. So maybe the answer is just 1.

Alternatively, perhaps there's a mistake in assuming that all roots give the same value? Let me check with specific numbers.

Let me take $\omega = e^{2\pi i /3} = -1/2 + i \sqrt{3}/2$. Then compute $1/(1 + \omega) + 1/(1 + \omega^2)$.

First compute $1 + \omega = 1 + (-1/2 + i \sqrt{3}/2) = 1/2 + i \sqrt{3}/2$. The reciprocal of this is $2/(1 + i \sqrt{3})$. Multiply numerator and denominator by $1 - i \sqrt{3}$:

$$
\frac{2(1 - i \sqrt{3})}{(1)^2 + (\sqrt{3})^2} = \frac{2(1 - i \sqrt{3})}{1 + 3} = \frac{2(1 - i \sqrt{3})}{4} = \frac{1 - i \sqrt{3}}{2}
$$

Similarly, compute $1 + \omega^2$. Since $\omega^2 = e^{4\pi i /3} = -1/2 - i \sqrt{3}/2$, so $1 + \omega^2 = 1 + (-1/2 - i \sqrt{3}/2) = 1/2 - i \sqrt{3}/2$. Reciprocal is $2/(1 - i \sqrt{3})$. Multiply numerator and denominator by $1 + i \sqrt{3}$:

$$
\frac{2(1 + i \sqrt{3})}{(1)^2 + (\sqrt{3})^2} = \frac{2(1 + i \sqrt{3})}{4} = \frac{1 + i \sqrt{3}}{2}
$$

Therefore, adding the two terms:

$$
\frac{1 - i \sqrt{3}}{2} + \frac{1 + i \sqrt{3}}{2} = \frac{1 - i \sqrt{3} + 1 + i \sqrt{3}}{2} = \frac{2}{2} = 1
$$

So that's 1. Therefore, with actual numbers, it checks out. Similarly, if I take $\omega = e^{4\pi i /3}$, same thing. And for $\omega = 1$, also 1. Therefore, conclusion is that the expression is always 1, so the only possible value is 1. Therefore, answer is \boxed{1}.

But maybe the question is trying to trick us? Let me check once more. Suppose someone thought that $\omega$ can be any of the roots, but maybe for different roots, even though they are different, the expression gives same value. Which is the case here. So the answer is only 1.

But the problem says "all possible values", so maybe there is another value? Wait, let me think again. Suppose someone tried to compute it as follows:

Let me assume $\omega$ is a root of $x^3 - 1 = 0$, so maybe I can use polynomial division or something else?

Alternatively, let me consider that:

Let me call the expression S:

$$
S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

Multiply numerator and denominator by something. Let me multiply numerator and denominator by $(1 + \omega)(1 + \omega^2)$, but that's what I did before. Alternatively, note that $\omega^3 = 1$, so $ \omega^2 = \omega^{-1} $, so $1 + \omega^2 = 1 + \omega^{-1}$. Let me write S as:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^{-1}} = \frac{1}{1 + \omega} + \frac{\omega}{\omega + 1} = \frac{1 + \omega}{1 + \omega} = 1
$$

Wait, that seems surprising. Let me check:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^{-1}} = \frac{1}{1 + \omega} + \frac{\omega}{\omega + 1} = \frac{1 + \omega}{1 + \omega} = 1
$$

Yes! Because $1 + \omega^{-1} = 1 + \omega^2$ since $\omega^{-1} = \omega^2$ (since $\omega^3 = 1$). Therefore, the second term is $\frac{1}{1 + \omega^2} = \frac{\omega}{\omega + 1}$. Therefore, adding them gives $\frac{1 + \omega}{\omega + 1} = 1$. So that's another way to see it. Therefore, regardless of $\omega$, as long as $\omega \ne -1$ (which it isn't, because otherwise $\omega^3 = -1 \ne 1$), the expression is 1. Hence, the answer is 1.

Therefore, I think confidently that the only possible value is 1, so the answer is $\boxed{1}$.

**Final Answer**
\boxed{1}

To evaluate the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
where $\omega$ is a complex number satisfying $\omega^3 = 1$, we begin by analyzing the nature of $\omega$.

---

### Step 1: Understand the roots of unity

The equation $\omega^3 = 1$ has three solutions, known as the **cube roots of unity**. These are:

- $\omega = 1$
- $\omega = e^{2\pi i / 3}$
- $\omega = e^{4\pi i / 3}$

These roots satisfy the identity:
$$
\omega^2 + \omega + 1 = 0 \quad \text{for } \omega \ne 1
$$
This identity will be key for simplification.

---

### Step 2: Combine the terms

We aim to simplify:
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

To combine, we use a common denominator:
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} = \frac{(1 + \omega^2) + (1 + \omega)}{(1 + \omega)(1 + \omega^2)}
$$

Simplify numerator and denominator:

- **Numerator**:
  $$
  1 + \omega^2 + 1 + \omega = 2 + \omega + \omega^2
  $$
- **Denominator**:
  $$
  (1 + \omega)(1 + \omega^2) = 1 + \omega + \omega^2 + \omega^3
  $$
  Using $\omega^3 = 1$, we simplify:
  $$
  1 + \omega + \omega^2 + 1 = 2 + \omega + \omega^2
  $$

So both numerator and denominator simplify to:
$$
2 + \omega + \omega^2
$$

---

### Step 3: Use the cubic identity

Recall:
$$
\omega^2 + \omega + 1 = 0 \quad \text{(for } \omega \ne 1 \text{)}
$$
This implies:
$$
\omega + \omega^2 = -1
$$

So:
$$
2 + \omega + \omega^2 = 2 - 1 = 1
$$

Thus, the entire expression simplifies to:
$$
\frac{1}{1} = 1
$$

---

### Step 4: Check the case $\omega = 1$

Now consider $\omega = 1$. Then:
$$
\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1
$$

Therefore, even in this case, the value of the expression is $1$.

---

### Step 5: Conclusion

For all possible values of $\omega$ such that $\omega^3 = 1$, the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
is always equal to $1$. Thus, the **only possible value** is:

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to find all possible values of the expression $\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$ where $\omega$ is a complex number such that $\omega^3 = 1$. Let me think about how to approach this step by step.

First, I remember that the equation $\omega^3 = 1$ has three solutions, which are the cube roots of unity. These are $1$, $\omega$, and $\omega^2$, where $\omega$ is a primitive cube root of unity. So, the possible values of $\omega$ are $1$, $e^{2\pi i /3}$, and $e^{4\pi i /3}$. However, the problem says "Let $\omega$ be a complex number such that $\omega^3 = 1$." So, $\omega$ could be any of these three roots. But the question asks for all possible values of the given expression. Therefore, I need to compute the expression for each of these roots and see what values I get.

But maybe there's a smarter way than just plugging in each root. Let me recall some properties of roots of unity. The cube roots of unity satisfy the equation $x^3 - 1 = 0$, which factors as $(x - 1)(x^2 + x + 1) = 0$. Therefore, the non-1 roots satisfy $x^2 + x + 1 = 0$. So, for $\omega \neq 1$, we have $\omega^2 + \omega + 1 = 0$. That seems useful. Let me note that down.

So, if $\omega^3 = 1$, then $\omega \neq 1$ implies $\omega^2 + \omega + 1 = 0$. Therefore, maybe I can use this relation to simplify the given expression.

Let me write down the expression again:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

I need to compute this sum. Let me first try to combine these two fractions. To add them, I need a common denominator, which would be $(1 + \omega)(1 + \omega^2)$. Let me compute the numerator:

$$
(1 + \omega^2) + (1 + \omega) = 1 + \omega^2 + 1 + \omega = 2 + \omega + \omega^2
$$

Therefore, the expression becomes:

$$
\frac{2 + \omega + \omega^2}{(1 + \omega)(1 + \omega^2)}
$$

Now, from the equation $\omega^2 + \omega + 1 = 0$ (for $\omega \neq 1$), we know that $\omega + \omega^2 = -1$. Therefore, substituting into the numerator:

$$
2 + (-1) = 1
$$

So the numerator simplifies to 1. Now, I need to compute the denominator: $(1 + \omega)(1 + \omega^2)$.

Let me expand this:

$$
(1 + \omega)(1 + \omega^2) = 1 \cdot 1 + 1 \cdot \omega^2 + \omega \cdot 1 + \omega \cdot \omega^2
$$

Calculating each term:

- $1 \cdot 1 = 1$
- $1 \cdot \omega^2 = \omega^2$
- $\omega \cdot 1 = \omega$
- $\omega \cdot \omega^2 = \omega^3$

But $\omega^3 = 1$, so that term is 1. Therefore, the entire expression becomes:

$$
1 + \omega^2 + \omega + 1 = 2 + \omega + \omega^2
$$

Wait, that's the same as the numerator! Wait, but earlier, we found that $\omega + \omega^2 = -1$, so substituting again:

$$
2 + (-1) = 1
$$

Wait, so the denominator is also 1? That would make the entire expression $\frac{1}{1} = 1$. But that seems strange. Let me check my calculations again because if both numerator and denominator are 1, then the expression is 1. However, if $\omega = 1$, then the original expression would have denominators 1 + 1 = 2 and 1 + 1 = 2, so the expression would be 1/2 + 1/2 = 1 as well. Wait, so does that mean regardless of which root we take, the value is 1? But the question says "all possible values", implying that there might be more than one. But according to this, it's always 1? Let me verify.

Wait, but if $\omega = 1$, then $\omega^3 = 1$, so that's allowed. However, when we derived the numerator and denominator, we used the equation $\omega^2 + \omega + 1 = 0$, which only holds when $\omega \neq 1$. Therefore, maybe we need to consider separately the case when $\omega = 1$ and when $\omega \neq 1$?

Let me check. If $\omega = 1$, then the original expression becomes:

$$
\frac{1}{1 + 1} + \frac{1}{1 + 1^2} = \frac{1}{2} + \frac{1}{2} = 1
$$

So even when $\omega = 1$, the value is 1. If $\omega$ is a primitive root, then as above, we have the expression is 1. Therefore, regardless of which root we take, the expression is 1? Therefore, there's only one possible value, which is 1?

But the problem says "all possible values", so maybe I made a mistake here. Let me check again with specific roots.

Let me take $\omega = e^{2\pi i /3}$. Then $\omega^2 = e^{4\pi i /3}$, and $\omega^3 = 1$. Let me compute the expression:

First, compute $1 + \omega$. Let me recall that $1 + \omega + \omega^2 = 0$, so $1 + \omega = -\omega^2$, and $1 + \omega^2 = -\omega$. Therefore:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} = \frac{1}{-\omega^2} + \frac{1}{-\omega} = -\frac{1}{\omega^2} - \frac{1}{\omega}
$$

But since $\omega^3 = 1$, we have that $\omega^2 = \omega^{-1}$, so $1/\omega^2 = \omega$, and $1/\omega = \omega^2$. Therefore, substituting:

$$
- \omega - \omega^2
$$

But from the equation $\omega^2 + \omega + 1 = 0$, we have $\omega + \omega^2 = -1$, so substituting:

$$
-(-1) = 1
$$

Same result. Therefore, even with this specific root, we get 1. Therefore, regardless of which root we take, the expression is 1. Therefore, the only possible value is 1. But then why does the problem say "all possible values"? Maybe there's a misunderstanding here?

Wait, let me check once more with $\omega = 1$. If $\omega = 1$, then the original expression is $\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1$. So again, same result. Therefore, no matter which root we take, the value is 1. Therefore, there is only one possible value. Therefore, the answer is just 1.

But let me think again. Could there be another way where maybe the expression evaluates to something else? For example, if there was some mistake in my algebra?

Wait, let me check the denominator again. When I expanded $(1 + \omega)(1 + \omega^2)$, I got:

$$
1 + \omega^2 + \omega + \omega^3 = 1 + \omega + \omega^2 + 1 = 2 + \omega + \omega^2
$$

Which is equal to 2 + (-1) = 1. Therefore, the denominator is 1. Numerator is also 1, so the expression is 1. Therefore, regardless of $\omega$, as long as $\omega^3 = 1$, the expression evaluates to 1. Therefore, the only possible value is 1.

Wait, so then the answer is just 1? But maybe I need to check if there's a case where denominators are zero? For example, if $1 + \omega = 0$ or $1 + \omega^2 = 0$. Let me check that. If $1 + \omega = 0$, then $\omega = -1$. Then $\omega^3 = (-1)^3 = -1 \neq 1$, so $\omega = -1$ is not a cube root of unity. Therefore, none of the roots $\omega$ satisfy $1 + \omega = 0$ or $1 + \omega^2 = 0$, so denominators are never zero. Hence, the expression is always defined and equal to 1.

Therefore, the only possible value is 1. So, I think the answer is \boxed{1}.

Wait, but maybe I need to check with another approach. Let me consider using algebraic manipulation without splitting into cases.

Let me write the original expression:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

Let me note that $\omega^3 = 1$, so $\omega^2 = \omega^{-1}$, and $\omega = \omega^{-2}$. Also, since $\omega^3 = 1$, we can write everything in terms of $\omega$ or $\omega^2$. Let me try to compute:

Let me compute $ \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} $. Let me also note that since $\omega^2 + \omega + 1 = 0$, we can replace $\omega^2$ with $-1 - \omega$, and vice versa. Let me try substituting $\omega^2 = -1 - \omega$ into the expression.

First, compute $1 + \omega^2 = 1 + (-1 - \omega) = -\omega$. Similarly, $1 + \omega$ is just $1 + \omega$. Therefore, the expression becomes:

$$
\frac{1}{1 + \omega} + \frac{1}{ - \omega} = \frac{1}{1 + \omega} - \frac{1}{\omega}
$$

Now, let me combine these terms. To do that, I need a common denominator, which would be $\omega(1 + \omega)$. Let me write:

$$
\frac{\omega}{\omega(1 + \omega)} - \frac{1 + \omega}{\omega(1 + \omega)} = \frac{\omega - (1 + \omega)}{\omega(1 + \omega)} = \frac{\omega - 1 - \omega}{\omega(1 + \omega)} = \frac{ -1 }{\omega(1 + \omega)}
$$

Hmm, so that simplifies to $-1 / [\omega(1 + \omega)]$. Wait, but how does that relate to previous results? Let me compute this denominator: $\omega(1 + \omega)$. Let me expand that:

$$
\omega(1 + \omega) = \omega + \omega^2
$$

Again, from $\omega^2 + \omega + 1 = 0$, we have $\omega + \omega^2 = -1$, so the denominator is $-1$. Therefore, the expression is $-1 / (-1) = 1$. There we go! So again, we get 1. Therefore, regardless of the approach, we get 1. So that confirms that the value is always 1.

Alternatively, another approach: Let me consider multiplying numerator and denominator by something. For example, since we know that $\omega^3 = 1$, maybe multiplying numerator and denominator by $\omega - 1$ or something else?

Wait, but maybe I can consider the original expression:

Let me denote $S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}$

Let me compute $S$:

First, note that $1 + \omega + \omega^2 = 0$, so $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$. Therefore:

$$
S = \frac{1}{- \omega^2} + \frac{1}{- \omega} = - \left( \frac{1}{\omega^2} + \frac{1}{\omega} \right)
$$

But $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$, since $\omega^3 = 1$. Therefore:

$$
S = - (\omega + \omega^2 ) = -(-1) = 1
$$

Again, same result. So all approaches lead to 1. Therefore, it seems that regardless of the root, the value is 1. Hence, the only possible value is 1.

But then why does the problem say "Find all possible values"? Maybe there's something wrong here? Wait, but the problem states "Let $\omega$ be a complex number such that $\omega^3 = 1$." So $\omega$ can be 1, $\omega$, or $\omega^2$. But in all cases, the expression evaluates to 1. Therefore, only one possible value. So maybe the answer is just 1.

Alternatively, perhaps there's a mistake in assuming that all roots give the same value? Let me check with specific numbers.

Let me take $\omega = e^{2\pi i /3} = -1/2 + i \sqrt{3}/2$. Then compute $1/(1 + \omega) + 1/(1 + \omega^2)$.

First compute $1 + \omega = 1 + (-1/2 + i \sqrt{3}/2) = 1/2 + i \sqrt{3}/2$. The reciprocal of this is $2/(1 + i \sqrt{3})$. Multiply numerator and denominator by $1 - i \sqrt{3}$:

$$
\frac{2(1 - i \sqrt{3})}{(1)^2 + (\sqrt{3})^2} = \frac{2(1 - i \sqrt{3})}{1 + 3} = \frac{2(1 - i \sqrt{3})}{4} = \frac{1 - i \sqrt{3}}{2}
$$

Similarly, compute $1 + \omega^2$. Since $\omega^2 = e^{4\pi i /3} = -1/2 - i \sqrt{3}/2$, so $1 + \omega^2 = 1 + (-1/2 - i \sqrt{3}/2) = 1/2 - i \sqrt{3}/2$. Reciprocal is $2/(1 - i \sqrt{3})$. Multiply numerator and denominator by $1 + i \sqrt{3}$:

$$
\frac{2(1 + i \sqrt{3})}{(1)^2 + (\sqrt{3})^2} = \frac{2(1 + i \sqrt{3})}{4} = \frac{1 + i \sqrt{3}}{2}
$$

Therefore, adding the two terms:

$$
\frac{1 - i \sqrt{3}}{2} + \frac{1 + i \sqrt{3}}{2} = \frac{1 - i \sqrt{3} + 1 + i \sqrt{3}}{2} = \frac{2}{2} = 1
$$

So that's 1. Therefore, with actual numbers, it checks out. Similarly, if I take $\omega = e^{4\pi i /3}$, same thing. And for $\omega = 1$, also 1. Therefore, conclusion is that the expression is always 1, so the only possible value is 1. Therefore, answer is \boxed{1}.

But maybe the question is trying to trick us? Let me check once more. Suppose someone thought that $\omega$ can be any of the roots, but maybe for different roots, even though they are different, the expression gives same value. Which is the case here. So the answer is only 1.

But the problem says "all possible values", so maybe there is another value? Wait, let me think again. Suppose someone tried to compute it as follows:

Let me assume $\omega$ is a root of $x^3 - 1 = 0$, so maybe I can use polynomial division or something else?

Alternatively, let me consider that:

Let me call the expression S:

$$
S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

Multiply numerator and denominator by something. Let me multiply numerator and denominator by $(1 + \omega)(1 + \omega^2)$, but that's what I did before. Alternatively, note that $\omega^3 = 1$, so $ \omega^2 = \omega^{-1} $, so $1 + \omega^2 = 1 + \omega^{-1}$. Let me write S as:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^{-1}} = \frac{1}{1 + \omega} + \frac{\omega}{\omega + 1} = \frac{1 + \omega}{1 + \omega} = 1
$$

Wait, that seems surprising. Let me check:

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^{-1}} = \frac{1}{1 + \omega} + \frac{\omega}{\omega + 1} = \frac{1 + \omega}{1 + \omega} = 1
$$

Yes! Because $1 + \omega^{-1} = 1 + \omega^2$ since $\omega^{-1} = \omega^2$ (since $\omega^3 = 1$). Therefore, the second term is $\frac{1}{1 + \omega^2} = \frac{\omega}{\omega + 1}$. Therefore, adding them gives $\frac{1 + \omega}{\omega + 1} = 1$. So that's another way to see it. Therefore, regardless of $\omega$, as long as $\omega \ne -1$ (which it isn't, because otherwise $\omega^3 = -1 \ne 1$), the expression is 1. Hence, the answer is 1.

Therefore, I think confidently that the only possible value is 1, so the answer is $\boxed{1}$.

**Final Answer**
\boxed{1}

To evaluate the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
where $\omega$ is a complex number satisfying $\omega^3 = 1$, we begin by analyzing the nature of $\omega$.

---

### Step 1: Understand the roots of unity

The equation $\omega^3 = 1$ has three solutions, known as the **cube roots of unity**. These are:

- $\omega = 1$
- $\omega = e^{2\pi i / 3}$
- $\omega = e^{4\pi i / 3}$

These roots satisfy the identity:
$$
\omega^2 + \omega + 1 = 0 \quad \text{for } \omega \ne 1
$$
This identity will be key for simplification.

---

### Step 2: Combine the terms

We aim to simplify:
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$

To combine, we use a common denominator:
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} = \frac{(1 + \omega^2) + (1 + \omega)}{(1 + \omega)(1 + \omega^2)}
$$

Simplify numerator and denominator:

- **Numerator**:
  $$
  1 + \omega^2 + 1 + \omega = 2 + \omega + \omega^2
  $$
- **Denominator**:
  $$
  (1 + \omega)(1 + \omega^2) = 1 + \omega + \omega^2 + \omega^3
  $$
  Using $\omega^3 = 1$, we simplify:
  $$
  1 + \omega + \omega^2 + 1 = 2 + \omega + \omega^2
  $$

So both numerator and denominator simplify to:
$$
2 + \omega + \omega^2
$$

---

### Step 3: Use the cubic identity

Recall:
$$
\omega^2 + \omega + 1 = 0 \quad \text{(for } \omega \ne 1 \text{)}
$$
This implies:
$$
\omega + \omega^2 = -1
$$

So:
$$
2 + \omega + \omega^2 = 2 - 1 = 1
$$

Thus, the entire expression simplifies to:
$$
\frac{1}{1} = 1
$$

---

### Step 4: Check the case $\omega = 1$

Now consider $\omega = 1$. Then:
$$
\frac{1}{1 + 1} + \frac{1}{1 + 1} = \frac{1}{2} + \frac{1}{2} = 1
$$

Therefore, even in this case, the value of the expression is $1$.

---

### Step 5: Conclusion

For all possible values of $\omega$ such that $\omega^3 = 1$, the expression
$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2}
$$
is always equal to $1$. Thus, the **only possible value** is:

$$
\boxed{1}
$$

Solution:
We can write
\begin{align*}
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} &= \frac{1 + \omega^2 + 1 + \omega}{(1 + \omega)(1 + \omega^2)} \\
&= \frac{2 + \omega + \omega^2}{1 + \omega + \omega^2 + \omega^3} \\
&= \frac{2 + \omega + \omega^2}{2 + \omega + \omega^2} \\
&= \boxed{1}.
\end{align*}


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). Let me sketch that mentally. Point D is at the origin, B is at (3,0), which is to the right on the x-axis, C is straight down at (0,-4), and A is at (-3,3), which is to the left and up.

The diagram also has some labels for distances: between D and B is labeled 3, A and B is 6, A and C is 6, D and C is 4, A and D is 5, and C and B is 5. Let me confirm that with coordinates.

First, distance between D(0,0) and B(3,0) should be sqrt[(3-0)^2 + (0-0)^2] = sqrt[9] = 3. That matches. Then A to B: A(-3,3) to B(3,0). The distance is sqrt[(3 - (-3))^2 + (0 - 3)^2] = sqrt[(6)^2 + (-3)^2] = sqrt[36 + 9] = sqrt[45] = 3√5 ≈ 6.708. Wait, but the label says 6. Hmm, that's a discrepancy. Did I do something wrong?

Wait, maybe the diagram isn't to scale? Or perhaps the labels are approximate? Wait, the problem says "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);" So according to the diagram, the distances are 3, 6, 6, 4, 5, 5 as per the labels. So maybe the coordinates are not exact? Or maybe they are? Maybe I need to verify.

Wait, if A is at (-3,3) and D is at (0,0), then distance between A and D is sqrt[(-3 - 0)^2 + (3 - 0)^2] = sqrt[9 + 9] = sqrt[18] = 3√2 ≈ 4.24, but the label says 5. Hmm, that's conflicting. Similarly, distance between D and C is (0,-4) to (0,0): sqrt[0 + 16] = 4, which matches the label. So maybe the coordinates are not exact? Or perhaps there's a mistake in my assumption.

Alternatively, maybe the coordinates are given as per the diagram's labels, but the coordinates are not to scale? That is, maybe the coordinates are just for drawing purposes, but the actual distances are given by the labels? But then, how can we compute the distances? Or maybe the coordinates are correct, and the labels are approximate?

This is confusing. Wait, the problem says "visit the other three points once"—so it's a traveling salesman problem? But with only four points? Wait, the problem is to find the shortest path that starts at any of the four points, visits the other three once. So it's like a path starting at A, then going to B, C, D in some order, and then ending? Or is it a closed loop? Wait, the problem says "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 it's a path that starts at one point, visits all the other three once, and then ends? So it's not a cycle. So the total distance is the sum of the distances between consecutive points. So we need to find the minimal such path starting at any point.

But since the points are connected with given distances, perhaps we can model this as a graph where the nodes are A, B, C, D, and the edges have given weights. Then, for each starting point, find the shortest path that goes through all other points once. Since there are only four points, starting at one point and visiting the other three, so three edges. So the total path would be four points with three edges. So the minimal total distance.

But how do we compute this? Since there are four points, starting at any one, and visiting the other three. So for each starting node, there are different permutations of the other three nodes. However, since the graph is fully connected? Wait, are all the edges given? Let me check. The diagram shows several connections:

From the Asymptote code: draw(D--B--C--cycle); draw(D--A--B--cycle); draw(D--A--C--cycle);

So that means:

- D connected to B, B connected to C, C connected to D? Wait, D--B--C--cycle would be D connected to B, B connected to C, C connected back to D? So triangle D-B-C?

Similarly, D--A--B--cycle: D connected to A, A connected to B, B connected back to D?

And D--A--C--cycle: D connected to A, A connected to C, C connected back to D?

So, in terms of edges:

Edges are:

D-B, B-C, C-D, D-A, A-B, A-D, A-C, C-D. Wait, so:

From D to B, B to C, C to D, D to A, A to B, A to D, A to C, and also C to D again? Wait, maybe I need to list all edges.

But the problem labels several distances:

- D-B: 3

- A-B: 6

- A-C: 6

- D-C: 4

- A-D: 5

- C-B: 5

So, that's those edges. So the edges are D-B (3), A-B (6), A-C (6), D-C (4), A-D (5), and C-B (5). So what about other edges? For example, is there an edge between D and A? Yes, D-A is 5. Between A and C, yes, 6. Between B and C, yes, 5. Between D and C, 4. Between D and B, 3. Between A and B, 6. What about between D and A? 5. So the edges given 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 that's all edges? Wait, so between every pair? Wait, there are four points, so six possible edges. Let me check.

AB: 6

AC: 6

AD: 5

BC: 5

BD: 3

CD: 4

Yes, that's all. So the graph is complete with those edge weights.

So now, the problem is to find the minimal path that starts at any of the four points, visits all other three exactly once, so it's a path of three edges. So for each starting point, there are 3! = 6 permutations of the other points, but since the path is directional, maybe? Wait, no. For example, starting at A, then going to B, C, D in some order. So for each starting point, there are 3! possible paths. However, since the graph is undirected, the path from A to B to C to D is the same as D to C to B to A, but since we are starting at a specific point, maybe we need to consider all permutations.

But since the problem says "visit the other three points once", so the path must go through all three other points. So the total number of possible paths starting at each point is 3! = 6. So for each starting point, we can have 6 different paths, and then we need to find the minimal one among all starting points.

Alternatively, since the graph is small, maybe we can calculate all possibilities.

Alternatively, since it's a traveling salesman problem with visiting all nodes once, but since it's just four nodes, we can compute the minimal path.

But maybe there's a better way. Let me think.

First, let me confirm the distances between all pairs again:

AB: 6

AC: 6

AD: 5

BC: 5

BD: 3

CD: 4

So, for example, the distance between D and B is 3, shortest.

Now, if I start at D, then I need to go to A, B, C. Let me think.

But maybe I need to find the shortest possible path that goes through all four points? Wait, but the problem says starting at any point, visiting the other three once. So the path is from start to end, visiting three other points. So the total number of points visited is four? Wait, starting at one point, then visiting three others. So total of four points. So the path is a sequence of four points, with the first being the starting point, and the next three being the other three in some order.

So for example, starting at A, then going to B, then to C, then to D. So the distance would be AB + BC + CD. Alternatively, starting at A, going to D, then to C, then to B: AD + DC + CB. Etc.

Therefore, for each starting point, we need to consider all permutations of the other three points and calculate the total distance, then take the minimal one.

But since there are four starting points, each with 3! = 6 permutations, that would be 4 * 6 = 24 possibilities. However, since some paths may be symmetric, we can find the minimal ones more efficiently.

Alternatively, since the problem is small, we can compute the minimal path for each starting point.

Let me start with each starting point.

First, starting at A:

We need to visit B, C, D. The possible permutations are B-C-D, B-D-C, C-B-D, C-D-B, D-B-C, D-C-B.

Let me compute the total distance for each:

1. A-B-C-D: AB + BC + CD = 6 + 5 + 4 = 15

2. A-B-D-C: AB + BD + DC = 6 + 3 + 4 = 13

3. A-C-B-D: AC + CB + BD = 6 + 5 + 3 = 14

4. A-C-D-B: AC + CD + DB = 6 + 4 + 3 = 13

5. A-D-B-C: AD + DB + BC = 5 + 3 + 5 = 13

6. A-D-C-B: AD + DC + CB = 5 + 4 + 5 = 14

So the minimal starting at A is 13.

Now starting at B:

We need to visit A, C, D. Permutations: A-C-D, A-D-C, C-A-D, C-D-A, D-A-C, D-C-A.

Calculating each:

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

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

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

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

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

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

So the minimal starting at B is 13.

Now starting at C:

Need to visit A, B, D. Permutations: A-B-D, A-D-B, B-A-D, B-D-A, D-A-B, D-B-A.

Calculating:

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

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

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

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

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

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

So minimal starting at C is 13.

Starting at D:

Need to visit A, B, C. Permutations: A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, C-B-A.

Calculating:

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

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

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

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

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

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

So the minimal starting at D is 14.

Therefore, the minimal of all the starting points is 13. It occurs starting at A (for example, A-B-D-C or A-C-D-B or A-D-B-C), starting at B (B-D-C-A or B-D-A-C), and starting at C (C-B-D-A or C-D-B-A). So the minimal total distance is 13.

Wait, but let me check again. For example, when starting at B, the path B-D-C-A had total distance BD + DC + CA = 3 + 4 + 6 = 13. Similarly, when starting at C, path C-B-D-A: CB + BD + DA = 5 + 3 + 5 = 13. Also, starting at A, path A-D-B-C: 5 + 3 + 5 = 13. Similarly, A-B-D-C: 6 + 3 + 4 = 13. And starting at B, path B-D-A-C: BD + DA + AC = 3 + 5 + 6 = 14? Wait no, earlier I thought B-D-C-A was 3+4+6=13? Wait, let me check again:

Starting at B:

Permutation 5: B-D-A-C: BD + DA + AC = 3 + 5 + 6 = 14. Wait, but earlier I wrote:

Starting at B, permutations:

1. B-A-C-D: 16

2. B-A-D-C: 15

3. B-C-A-D:16

4. B-C-D-A: 5 + 4 + 5 = 14

Wait, BC + CD + DA: BC is 5, CD is 4, DA is 5? Wait CD is 4? Wait, CD is between C and D, which is 4. DA is between D and A, which is 5. Yes. So that's 5 + 4 + 5 = 14. Then permutation 5: B-D-A-C is BD (3) + DA (5) + AC (6): 3 + 5 + 6 = 14. Then permutation 6: B-D-C-A is BD (3) + DC (4) + CA (6): 3 + 4 + 6 = 13. So that's correct. So starting at B, the minimal is 13.

Similarly, starting at C, permutation 4: C-B-D-A: CB (5) + BD (3) + DA (5): 5 + 3 + 5 = 13. Yes. So that's correct. So the minimal total distance is 13, and that can be achieved starting at A, B, or C.

Therefore, the answer should be 13. Wait, but let me check if there's a shorter path. For example, maybe some overlapping or something? Wait, no. Since we have to visit all three points once each, so three edges. So for example, the path from D to A to D to something... but no, you can't visit a point twice. So it's strictly visiting each of the other three once.

But let me check for any miscalculations. For example, when starting at A, the path A-B-D-C: AB is 6, BD is 3, DC is 4. Total 13. Correct. Starting at D, is there a way to get less than 13? The minimal starting at D was 14. Starting at D, the minimal path is 14. So no, 13 is the minimal.

Wait, but 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 answer is 13. But let me check once again if there's a different path that I haven't considered.

Alternatively, maybe there's a way to have overlapping distances? For example, using some diagonal path? Wait, but the problem must be using the given edge distances. Since the Asymptote code labels the distances between the points. So, for example, you cannot go from A to C directly unless that's an edge. Wait, but the problem says "visit the other three points once". So you can only move along the edges given. So the distances given are between the points. So you can't take other paths, like straight lines. So the distances are fixed as the labels.

So for instance, you can't go from A to C directly via some other route, you have to go through the given edges. Therefore, the distances between points are as given. Therefore, the answer is 13.

But just to confirm once again, let me check the minimal path starting at B: B-D-C-A, which is BD (3) + DC (4) + CA (6) = 13. Yes. Similarly, starting at C: C-B-D-A, CB (5) + BD (3) + DA (5) = 13. Starting at A: A-D-B-C, AD (5) + DB (3) + BC (5) = 13. All three of these are 13. So that's the minimal.

Alternatively, if there's a path that uses a different combination of edges, but I don't think so. Since we have to visit three points, three edges. So the minimal total distance is 13. Therefore, the answer is 13.

But wait, just to be thorough, let me check if there's a path that goes through some other order that I might have missed. For example, starting at D, can we have a shorter path? Let's see:

Starting at D, visiting A, B, C. The minimal was 14. For instance: D-A-B-C (5 + 6 + 5 = 16), D-A-C-B (5 + 6 + 5 = 16), D-B-A-C (3 + 6 + 6 = 15), D-B-C-A (3 + 5 + 6 = 14), D-C-A-B (4 + 6 + 6 = 16), D-C-B-A (4 + 5 + 6 = 15). So the minimal is 14, which is higher than 13. So no.

Similarly, starting at D, the minimal is 14. So 13 is better. Similarly, starting at A, B, or C, we can get 13. Therefore, the minimal total distance is 13.

Therefore, the answer is 13.

**Final Answer**
\boxed{13}

To determine the **shortest distance** that can be travelled starting at any one of the points $ A $, $ B $, $ C $, or $ D $ and visiting the other three points exactly once, we treat this as a **path optimization problem**. Given the points and the distances between them, we are essentially looking for the shortest path that starts at any of the four nodes, visits the other three exactly once, and ends at the last node.

---

### Step 1: Understand the Problem

We are told the distance between each pair of the points (this information is derived from the diagram's edge labels). These are:

| Pair  | Distance |
|-------|----------|
| AB    | 6        |
| AC    | 6        |
| AD    | 5        |
| BC    | 5        |
| BD    | 3        |
| CD    | 4        |

We can model this as a **complete undirected graph** with the above edge weights.

---

### Step 2: Strategy

We need to find the shortest path that:

- Starts at **any** one of the four nodes.
- Visits the **other three nodes** **exactly once**.
- Uses the given distances between the points.

This is equivalent to finding a **shortest Hamiltonian path starting at any node**, with the length being the **sum of the distances** of three consecutive edges.

Each node has **three other nodes** that need to be visited in **some order** — that is, **3! = 6 permutations** per starting node — so we evaluate all such permutations to find the **minimal total distance** across all possible paths.

---

### Step 3: Minimal Paths from Each Starting Point

Let's compute the total distance for each starting point:

#### **Starting at A**
We need to visit B, C, D.

Possible paths and distances:

1. **A → B → C → D**: $6 + 5 + 4 = 15$
2. **A → B → D → C**: $6 + 3 + 4 = 13$
3. **A → C → B → D**: $6 + 5 + 3 = 14$
4. **A → C → D → B**: $6 + 4 + 5 = 15$
5. **A → D → B → C**: $5 + 3 + 5 = 13$
6. **A → D → C → B**: $5 + 4 + 5 = 14$

**Minimum:** 13

#### **Starting at B**
We need to visit A, C, D.

1. **B → A → C → D**: $6 + 6 + 4 = 16$
2. **B → A → D → C**: $6 + 5 + 4 = 15$
3. **B → C → A → D**: $5 + 6 + 5 = 16$
4. **B → C → D → A**: $5 + 4 + 5 = 14$
5. **B → D → A → C**: $3 + 5 + 6 = 14$
6. **B → D → C → A**: $3 + 4 + 6 = 13$

**Minimum:** 13

#### **Starting at C**
We need to visit A, B, D.

1. **C → A → B → D**: $6 + 6 + 3 = 15$
2. **C → A → D → B**: $6 + 5 + 3 = 14$
3. **C → B → A → D**: $5 + 6 + 5 = 16$
4. **C → B → D → A**: $5 + 3 + 5 = 13$
5. **C → D → A → B**: $4 + 5 + 6 = 15$
6. **C → D → B → A**: $4 + 3 + 6 = 13$

**Minimum:** 13

#### **Starting at D**
We need to visit A, B, C.

1. **D → A → B → C**: $5 + 6 + 5 = 16$
2. **D → A → C → B**: $5 + 6 + 5 = 16$
3. **D → B → A → C**: $3 + 6 + 6 = 15$
4. **D → B → C → A**: $3 + 5 + 6 = 14$
5. **D → C → A → B**: $4 + 6 + 6 = 16$
6. **D → C → B → A**: $4 + 5 + 6 = 15$

**Minimum:** 14

---

### Step 4: Global Minimal Path

From the above evaluations, the **shortest distance** is achieved when starting at **A**, **B**, or **C** with a path length of **13**. Examples of such optimal paths include:

- **A → D → B → C**
- **B → D → C → A**
- **C → B → D → A**

All of these have total length **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). Let me sketch that mentally. Point D is at the origin, B is at (3,0), which is to the right on the x-axis, C is straight down at (0,-4), and A is at (-3,3), which is to the left and up.

The diagram also has some labels for distances: between D and B is labeled 3, A and B is 6, A and C is 6, D and C is 4, A and D is 5, and C and B is 5. Let me confirm that with coordinates.

First, distance between D(0,0) and B(3,0) should be sqrt[(3-0)^2 + (0-0)^2] = sqrt[9] = 3. That matches. Then A to B: A(-3,3) to B(3,0). The distance is sqrt[(3 - (-3))^2 + (0 - 3)^2] = sqrt[(6)^2 + (-3)^2] = sqrt[36 + 9] = sqrt[45] = 3√5 ≈ 6.708. Wait, but the label says 6. Hmm, that's a discrepancy. Did I do something wrong?

Wait, maybe the diagram isn't to scale? Or perhaps the labels are approximate? Wait, the problem says "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);" So according to the diagram, the distances are 3, 6, 6, 4, 5, 5 as per the labels. So maybe the coordinates are not exact? Or maybe they are? Maybe I need to verify.

Wait, if A is at (-3,3) and D is at (0,0), then distance between A and D is sqrt[(-3 - 0)^2 + (3 - 0)^2] = sqrt[9 + 9] = sqrt[18] = 3√2 ≈ 4.24, but the label says 5. Hmm, that's conflicting. Similarly, distance between D and C is (0,-4) to (0,0): sqrt[0 + 16] = 4, which matches the label. So maybe the coordinates are not exact? Or perhaps there's a mistake in my assumption.

Alternatively, maybe the coordinates are given as per the diagram's labels, but the coordinates are not to scale? That is, maybe the coordinates are just for drawing purposes, but the actual distances are given by the labels? But then, how can we compute the distances? Or maybe the coordinates are correct, and the labels are approximate?

This is confusing. Wait, the problem says "visit the other three points once"—so it's a traveling salesman problem? But with only four points? Wait, the problem is to find the shortest path that starts at any of the four points, visits the other three once. So it's like a path starting at A, then going to B, C, D in some order, and then ending? Or is it a closed loop? Wait, the problem says "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 it's a path that starts at one point, visits all the other three once, and then ends? So it's not a cycle. So the total distance is the sum of the distances between consecutive points. So we need to find the minimal such path starting at any point.

But since the points are connected with given distances, perhaps we can model this as a graph where the nodes are A, B, C, D, and the edges have given weights. Then, for each starting point, find the shortest path that goes through all other points once. Since there are only four points, starting at one point and visiting the other three, so three edges. So the total path would be four points with three edges. So the minimal total distance.

But how do we compute this? Since there are four points, starting at any one, and visiting the other three. So for each starting node, there are different permutations of the other three nodes. However, since the graph is fully connected? Wait, are all the edges given? Let me check. The diagram shows several connections:

From the Asymptote code: draw(D--B--C--cycle); draw(D--A--B--cycle); draw(D--A--C--cycle);

So that means:

- D connected to B, B connected to C, C connected to D? Wait, D--B--C--cycle would be D connected to B, B connected to C, C connected back to D? So triangle D-B-C?

Similarly, D--A--B--cycle: D connected to A, A connected to B, B connected back to D?

And D--A--C--cycle: D connected to A, A connected to C, C connected back to D?

So, in terms of edges:

Edges are:

D-B, B-C, C-D, D-A, A-B, A-D, A-C, C-D. Wait, so:

From D to B, B to C, C to D, D to A, A to B, A to D, A to C, and also C to D again? Wait, maybe I need to list all edges.

But the problem labels several distances:

- D-B: 3

- A-B: 6

- A-C: 6

- D-C: 4

- A-D: 5

- C-B: 5

So, that's those edges. So the edges are D-B (3), A-B (6), A-C (6), D-C (4), A-D (5), and C-B (5). So what about other edges? For example, is there an edge between D and A? Yes, D-A is 5. Between A and C, yes, 6. Between B and C, yes, 5. Between D and C, 4. Between D and B, 3. Between A and B, 6. What about between D and A? 5. So the edges given 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 that's all edges? Wait, so between every pair? Wait, there are four points, so six possible edges. Let me check.

AB: 6

AC: 6

AD: 5

BC: 5

BD: 3

CD: 4

Yes, that's all. So the graph is complete with those edge weights.

So now, the problem is to find the minimal path that starts at any of the four points, visits all other three exactly once, so it's a path of three edges. So for each starting point, there are 3! = 6 permutations of the other points, but since the path is directional, maybe? Wait, no. For example, starting at A, then going to B, C, D in some order. So for each starting point, there are 3! possible paths. However, since the graph is undirected, the path from A to B to C to D is the same as D to C to B to A, but since we are starting at a specific point, maybe we need to consider all permutations.

But since the problem says "visit the other three points once", so the path must go through all three other points. So the total number of possible paths starting at each point is 3! = 6. So for each starting point, we can have 6 different paths, and then we need to find the minimal one among all starting points.

Alternatively, since the graph is small, maybe we can calculate all possibilities.

Alternatively, since it's a traveling salesman problem with visiting all nodes once, but since it's just four nodes, we can compute the minimal path.

But maybe there's a better way. Let me think.

First, let me confirm the distances between all pairs again:

AB: 6

AC: 6

AD: 5

BC: 5

BD: 3

CD: 4

So, for example, the distance between D and B is 3, shortest.

Now, if I start at D, then I need to go to A, B, C. Let me think.

But maybe I need to find the shortest possible path that goes through all four points? Wait, but the problem says starting at any point, visiting the other three once. So the path is from start to end, visiting three other points. So the total number of points visited is four? Wait, starting at one point, then visiting three others. So total of four points. So the path is a sequence of four points, with the first being the starting point, and the next three being the other three in some order.

So for example, starting at A, then going to B, then to C, then to D. So the distance would be AB + BC + CD. Alternatively, starting at A, going to D, then to C, then to B: AD + DC + CB. Etc.

Therefore, for each starting point, we need to consider all permutations of the other three points and calculate the total distance, then take the minimal one.

But since there are four starting points, each with 3! = 6 permutations, that would be 4 * 6 = 24 possibilities. However, since some paths may be symmetric, we can find the minimal ones more efficiently.

Alternatively, since the problem is small, we can compute the minimal path for each starting point.

Let me start with each starting point.

First, starting at A:

We need to visit B, C, D. The possible permutations are B-C-D, B-D-C, C-B-D, C-D-B, D-B-C, D-C-B.

Let me compute the total distance for each:

1. A-B-C-D: AB + BC + CD = 6 + 5 + 4 = 15

2. A-B-D-C: AB + BD + DC = 6 + 3 + 4 = 13

3. A-C-B-D: AC + CB + BD = 6 + 5 + 3 = 14

4. A-C-D-B: AC + CD + DB = 6 + 4 + 3 = 13

5. A-D-B-C: AD + DB + BC = 5 + 3 + 5 = 13

6. A-D-C-B: AD + DC + CB = 5 + 4 + 5 = 14

So the minimal starting at A is 13.

Now starting at B:

We need to visit A, C, D. Permutations: A-C-D, A-D-C, C-A-D, C-D-A, D-A-C, D-C-A.

Calculating each:

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

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

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

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

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

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

So the minimal starting at B is 13.

Now starting at C:

Need to visit A, B, D. Permutations: A-B-D, A-D-B, B-A-D, B-D-A, D-A-B, D-B-A.

Calculating:

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

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

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

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

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

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

So minimal starting at C is 13.

Starting at D:

Need to visit A, B, C. Permutations: A-B-C, A-C-B, B-A-C, B-C-A, C-A-B, C-B-A.

Calculating:

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

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

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

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

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

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

So the minimal starting at D is 14.

Therefore, the minimal of all the starting points is 13. It occurs starting at A (for example, A-B-D-C or A-C-D-B or A-D-B-C), starting at B (B-D-C-A or B-D-A-C), and starting at C (C-B-D-A or C-D-B-A). So the minimal total distance is 13.

Wait, but let me check again. For example, when starting at B, the path B-D-C-A had total distance BD + DC + CA = 3 + 4 + 6 = 13. Similarly, when starting at C, path C-B-D-A: CB + BD + DA = 5 + 3 + 5 = 13. Also, starting at A, path A-D-B-C: 5 + 3 + 5 = 13. Similarly, A-B-D-C: 6 + 3 + 4 = 13. And starting at B, path B-D-A-C: BD + DA + AC = 3 + 5 + 6 = 14? Wait no, earlier I thought B-D-C-A was 3+4+6=13? Wait, let me check again:

Starting at B:

Permutation 5: B-D-A-C: BD + DA + AC = 3 + 5 + 6 = 14. Wait, but earlier I wrote:

Starting at B, permutations:

1. B-A-C-D: 16

2. B-A-D-C: 15

3. B-C-A-D:16

4. B-C-D-A: 5 + 4 + 5 = 14

Wait, BC + CD + DA: BC is 5, CD is 4, DA is 5? Wait CD is 4? Wait, CD is between C and D, which is 4. DA is between D and A, which is 5. Yes. So that's 5 + 4 + 5 = 14. Then permutation 5: B-D-A-C is BD (3) + DA (5) + AC (6): 3 + 5 + 6 = 14. Then permutation 6: B-D-C-A is BD (3) + DC (4) + CA (6): 3 + 4 + 6 = 13. So that's correct. So starting at B, the minimal is 13.

Similarly, starting at C, permutation 4: C-B-D-A: CB (5) + BD (3) + DA (5): 5 + 3 + 5 = 13. Yes. So that's correct. So the minimal total distance is 13, and that can be achieved starting at A, B, or C.

Therefore, the answer should be 13. Wait, but let me check if there's a shorter path. For example, maybe some overlapping or something? Wait, no. Since we have to visit all three points once each, so three edges. So for example, the path from D to A to D to something... but no, you can't visit a point twice. So it's strictly visiting each of the other three once.

But let me check for any miscalculations. For example, when starting at A, the path A-B-D-C: AB is 6, BD is 3, DC is 4. Total 13. Correct. Starting at D, is there a way to get less than 13? The minimal starting at D was 14. Starting at D, the minimal path is 14. So no, 13 is the minimal.

Wait, but 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 answer is 13. But let me check once again if there's a different path that I haven't considered.

Alternatively, maybe there's a way to have overlapping distances? For example, using some diagonal path? Wait, but the problem must be using the given edge distances. Since the Asymptote code labels the distances between the points. So, for example, you cannot go from A to C directly unless that's an edge. Wait, but the problem says "visit the other three points once". So you can only move along the edges given. So the distances given are between the points. So you can't take other paths, like straight lines. So the distances are fixed as the labels.

So for instance, you can't go from A to C directly via some other route, you have to go through the given edges. Therefore, the distances between points are as given. Therefore, the answer is 13.

But just to confirm once again, let me check the minimal path starting at B: B-D-C-A, which is BD (3) + DC (4) + CA (6) = 13. Yes. Similarly, starting at C: C-B-D-A, CB (5) + BD (3) + DA (5) = 13. Starting at A: A-D-B-C, AD (5) + DB (3) + BC (5) = 13. All three of these are 13. So that's the minimal.

Alternatively, if there's a path that uses a different combination of edges, but I don't think so. Since we have to visit three points, three edges. So the minimal total distance is 13. Therefore, the answer is 13.

But wait, just to be thorough, let me check if there's a path that goes through some other order that I might have missed. For example, starting at D, can we have a shorter path? Let's see:

Starting at D, visiting A, B, C. The minimal was 14. For instance: D-A-B-C (5 + 6 + 5 = 16), D-A-C-B (5 + 6 + 5 = 16), D-B-A-C (3 + 6 + 6 = 15), D-B-C-A (3 + 5 + 6 = 14), D-C-A-B (4 + 6 + 6 = 16), D-C-B-A (4 + 5 + 6 = 15). So the minimal is 14, which is higher than 13. So no.

Similarly, starting at D, the minimal is 14. So 13 is better. Similarly, starting at A, B, or C, we can get 13. Therefore, the minimal total distance is 13.

Therefore, the answer is 13.

**Final Answer**
\boxed{13}

To determine the **shortest distance** that can be travelled starting at any one of the points $ A $, $ B $, $ C $, or $ D $ and visiting the other three points exactly once, we treat this as a **path optimization problem**. Given the points and the distances between them, we are essentially looking for the shortest path that starts at any of the four nodes, visits the other three exactly once, and ends at the last node.

---

### Step 1: Understand the Problem

We are told the distance between each pair of the points (this information is derived from the diagram's edge labels). These are:

| Pair  | Distance |
|-------|----------|
| AB    | 6        |
| AC    | 6        |
| AD    | 5        |
| BC    | 5        |
| BD    | 3        |
| CD    | 4        |

We can model this as a **complete undirected graph** with the above edge weights.

---

### Step 2: Strategy

We need to find the shortest path that:

- Starts at **any** one of the four nodes.
- Visits the **other three nodes** **exactly once**.
- Uses the given distances between the points.

This is equivalent to finding a **shortest Hamiltonian path starting at any node**, with the length being the **sum of the distances** of three consecutive edges.

Each node has **three other nodes** that need to be visited in **some order** — that is, **3! = 6 permutations** per starting node — so we evaluate all such permutations to find the **minimal total distance** across all possible paths.

---

### Step 3: Minimal Paths from Each Starting Point

Let's compute the total distance for each starting point:

#### **Starting at A**
We need to visit B, C, D.

Possible paths and distances:

1. **A → B → C → D**: $6 + 5 + 4 = 15$
2. **A → B → D → C**: $6 + 3 + 4 = 13$
3. **A → C → B → D**: $6 + 5 + 3 = 14$
4. **A → C → D → B**: $6 + 4 + 5 = 15$
5. **A → D → B → C**: $5 + 3 + 5 = 13$
6. **A → D → C → B**: $5 + 4 + 5 = 14$

**Minimum:** 13

#### **Starting at B**
We need to visit A, C, D.

1. **B → A → C → D**: $6 + 6 + 4 = 16$
2. **B → A → D → C**: $6 + 5 + 4 = 15$
3. **B → C → A → D**: $5 + 6 + 5 = 16$
4. **B → C → D → A**: $5 + 4 + 5 = 14$
5. **B → D → A → C**: $3 + 5 + 6 = 14$
6. **B → D → C → A**: $3 + 4 + 6 = 13$

**Minimum:** 13

#### **Starting at C**
We need to visit A, B, D.

1. **C → A → B → D**: $6 + 6 + 3 = 15$
2. **C → A → D → B**: $6 + 5 + 3 = 14$
3. **C → B → A → D**: $5 + 6 + 5 = 16$
4. **C → B → D → A**: $5 + 3 + 5 = 13$
5. **C → D → A → B**: $4 + 5 + 6 = 15$
6. **C → D → B → A**: $4 + 3 + 6 = 13$

**Minimum:** 13

#### **Starting at D**
We need to visit A, B, C.

1. **D → A → B → C**: $5 + 6 + 5 = 16$
2. **D → A → C → B**: $5 + 6 + 5 = 16$
3. **D → B → A → C**: $3 + 6 + 6 = 15$
4. **D → B → C → A**: $3 + 5 + 6 = 14$
5. **D → C → A → B**: $4 + 6 + 6 = 16$
6. **D → C → B → A**: $4 + 5 + 6 = 15$

**Minimum:** 14

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### Step 4: Global Minimal Path

From the above evaluations, the **shortest distance** is achieved when starting at **A**, **B**, or **C** with a path length of **13**. Examples of such optimal paths include:

- **A → D → B → C**
- **B → D → C → A**
- **C → B → D → A**

All of these have total length **13**.

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### ✅ 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 $\boxed{13}$.