Explanation:
Okay, so I need to find the median of 27 consecutive positive integers whose sum is 3^7. Let me start by recalling some concepts.

First, consecutive integers. If there are 27 consecutive numbers, the median would be the 14th number because there are an odd number of terms. Wait, let me check that. For 27 numbers, the median is the middle one, which is the (27 + 1)/2 = 14th term. Yeah, that's right. So if I can find the 14th term, that's the median.

Now, the sum of these numbers is given as 3^7. Let me compute 3^7 first. 3^1 is 3, 3^2 is 9, 3^3 is 27, 3^4 is 81, 3^5 is 243, 3^6 is 729, and 3^7 is 2187. So the sum is 2187.

Now, how do I relate the sum of consecutive numbers to their median? Let me think. If I have consecutive integers, they can be represented as starting from some integer 'a' and then increasing by 1 each time. So the sequence would be a, a+1, a+2, ..., a+26. Since there are 27 terms.

The sum of an arithmetic sequence is (number of terms)/2 times (first term + last term). So here, the sum S = 27/2 * (a + (a + 26)). Let me compute that.

S = (27/2)*(2a + 26) = (27/2)*2*(a + 13) = 27*(a + 13).

Wait, that simplifies nicely. So the sum is 27*(a + 13). And we know the sum is 2187. Therefore:

27*(a + 13) = 2187

Divide both sides by 27:

a + 13 = 2187 / 27

Let me compute that. 27*80 = 2160. Then 2187 - 2160 = 27. So 2187 / 27 = 80 + 1 = 81. Therefore, a + 13 = 81. Therefore, a = 81 - 13 = 68.

So the first term is 68. Then the median is the 14th term. Let me check: the first term is a = 68, so the 14th term is a + 13. Wait, because the first term is a, the second is a + 1, ..., the 14th term is a + 13. Which is exactly the same as the expression we had earlier. So since a + 13 = 81, the median is 81.

Wait, so is the answer 81? Let me verify.

Alternatively, since the sum is 27*(median), because in an arithmetic sequence with an odd number of terms, the sum is equal to the number of terms multiplied by the median. Wait, is that true? Let me think.

Yes, because the median is the middle term, and the sum is the number of terms times the median. For example, if you have numbers symmetrically distributed around the median, each pair equidistant from the median adds up to twice the median. So for 27 terms, there are 13 pairs that add up to 2*median, and then the median itself. So total sum is 13*2*median + median = 27*median.

Therefore, sum = 27*median. Therefore, median = sum / 27.

Given that the sum is 3^7 = 2187, then median = 2187 / 27 = 81.

So that's another way to see it. Therefore, regardless of the starting number, the median is just the total sum divided by the number of terms.

Therefore, the median is 81.

But let me check with the previous calculation. Earlier, I found that a = 68, so the sequence is from 68 to 68 + 26 = 94. Let me check the median. The median is the 14th term, which is 68 + 13 = 81. So that's consistent.

Alternatively, if I compute the average of the numbers, which is the same as the median here because the numbers are consecutive. The average is sum / number of terms = 2187 / 27 = 81. So that's the average, which is equal to the median.

Therefore, the answer is 81.

But let me just make sure there isn't a trick here. The problem says "positive integers," so we need to ensure that the starting number is positive. In this case, a = 68, which is positive, so that's okay.

Alternatively, if there was a different way to represent the numbers? For example, sometimes with consecutive numbers, you can represent them around the median. Let me think. If the median is m, then the numbers are m - 13, m - 12, ..., m - 1, m, m + 1, ..., m + 13. Because there are 27 numbers, so 13 numbers below the median, the median itself, and 13 numbers above.

So the sum would be 27*m. Which again gives the same result. Therefore, m = 2187 / 27 = 81.

Therefore, regardless of the approach, the median is 81.

So I think that's solid. Therefore, the answer is 81.

**Final Answer**
The median of the 27 consecutive positive integers is \boxed{81}.

To solve the problem, we are given that the sum of 27 consecutive positive integers is $ 3^7 = 2187 $, and we are asked to find their **median**.

---

### Step 1: Understanding the Median in a Sequence of Consecutive Integers

For a sequence of **odd** number of terms, the **median** is the **middle term**. Since there are 27 terms, the median is the **14th term** in the sequence.

---

### Step 2: Representing the Sequence

Let the first term of the sequence be $ a $. Then the sequence is:

$$
a, a+1, a+2, \ldots, a+26
$$

This is an arithmetic sequence with:

- First term: $ a $
- Last term: $ a + 26 $
- Number of terms: 27

The sum of an arithmetic sequence is given by:

$$
\text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term})
$$

Substituting the known values:

$$
\text{Sum} = \frac{27}{2} \times (a + (a + 26)) = \frac{27}{2} \times (2a + 26) = 27(a + 13)
$$

We are told that the sum is $ 2187 $, so:

$$
27(a + 13) = 2187
$$

---

### Step 3: Solving for the Median

Divide both sides by 27:

$$
a + 13 = \frac{2187}{27} = 81
$$

So:

$$
a = 81 - 13 = 68
$$

Now, the **median** is the 14th term, which is:

$$
a + 13 = 68 + 13 = 81
$$

---

### Step 4: Alternative Confirmation

Another way to think about this is that in a sequence of **odd** number of terms, the **median** is also the **average** of the sequence. Since the sum is 2187 and there are 27 terms:

$$
\text{Median} = \frac{\text{Sum}}{\text{Number of terms}} = \frac{2187}{27} = 81
$$

This confirms our earlier result.

---

### Final Answer

$$
\boxed{81}
$$

[Assistant]
Explanation:
Okay, so I need to find the median of 27 consecutive positive integers whose sum is 3^7. Let me start by recalling some concepts.

First, consecutive integers. If there are 27 consecutive numbers, the median would be the 14th number because there are an odd number of terms. Wait, let me check that. For 27 numbers, the median is the middle one, which is the (27 + 1)/2 = 14th term. Yeah, that's right. So if I can find the 14th term, that's the median.

Now, the sum of these numbers is given as 3^7. Let me compute 3^7 first. 3^1 is 3, 3^2 is 9, 3^3 is 27, 3^4 is 81, 3^5 is 243, 3^6 is 729, and 3^7 is 2187. So the sum is 2187.

Now, how do I relate the sum of consecutive numbers to their median? Let me think. If I have consecutive integers, they can be represented as starting from some integer 'a' and then increasing by 1 each time. So the sequence would be a, a+1, a+2, ..., a+26. Since there are 27 terms.

The sum of an arithmetic sequence is (number of terms)/2 times (first term + last term). So here, the sum S = 27/2 * (a + (a + 26)). Let me compute that.

S = (27/2)*(2a + 26) = (27/2)*2*(a + 13) = 27*(a + 13).

Wait, that simplifies nicely. So the sum is 27*(a + 13). And we know the sum is 2187. Therefore:

27*(a + 13) = 2187

Divide both sides by 27:

a + 13 = 2187 / 27

Let me compute that. 27*80 = 2160. Then 2187 - 2160 = 27. So 2187 / 27 = 80 + 1 = 81. Therefore, a + 13 = 81. Therefore, a = 81 - 13 = 68.

So the first term is 68. Then the median is the 14th term. Let me check: the first term is a = 68, so the 14th term is a + 13. Wait, because the first term is a, the second is a + 1, ..., the 14th term is a + 13. Which is exactly the same as the expression we had earlier. So since a + 13 = 81, the median is 81.

Wait, so is the answer 81? Let me verify.

Alternatively, since the sum is 27*(median), because in an arithmetic sequence with an odd number of terms, the sum is equal to the number of terms multiplied by the median. Wait, is that true? Let me think.

Yes, because the median is the middle term, and the sum is the number of terms times the median. For example, if you have numbers symmetrically distributed around the median, each pair equidistant from the median adds up to twice the median. So for 27 terms, there are 13 pairs that add up to 2*median, and then the median itself. So total sum is 13*2*median + median = 27*median.

Therefore, sum = 27*median. Therefore, median = sum / 27.

Given that the sum is 3^7 = 2187, then median = 2187 / 27 = 81.

So that's another way to see it. Therefore, regardless of the starting number, the median is just the total sum divided by the number of terms.

Therefore, the median is 81.

But let me check with the previous calculation. Earlier, I found that a = 68, so the sequence is from 68 to 68 + 26 = 94. Let me check the median. The median is the 14th term, which is 68 + 13 = 81. So that's consistent.

Alternatively, if I compute the average of the numbers, which is the same as the median here because the numbers are consecutive. The average is sum / number of terms = 2187 / 27 = 81. So that's the average, which is equal to the median.

Therefore, the answer is 81.

But let me just make sure there isn't a trick here. The problem says "positive integers," so we need to ensure that the starting number is positive. In this case, a = 68, which is positive, so that's okay.

Alternatively, if there was a different way to represent the numbers? For example, sometimes with consecutive numbers, you can represent them around the median. Let me think. If the median is m, then the numbers are m - 13, m - 12, ..., m - 1, m, m + 1, ..., m + 13. Because there are 27 numbers, so 13 numbers below the median, the median itself, and 13 numbers above.

So the sum would be 27*m. Which again gives the same result. Therefore, m = 2187 / 27 = 81.

Therefore, regardless of the approach, the median is 81.

So I think that's solid. Therefore, the answer is 81.

**Final Answer**
The median of the 27 consecutive positive integers is \boxed{81}.

To solve the problem, we are given that the sum of 27 consecutive positive integers is $ 3^7 = 2187 $, and we are asked to find their **median**.

---

### Step 1: Understanding the Median in a Sequence of Consecutive Integers

For a sequence of **odd** number of terms, the **median** is the **middle term**. Since there are 27 terms, the median is the **14th term** in the sequence.

---

### Step 2: Representing the Sequence

Let the first term of the sequence be $ a $. Then the sequence is:

$$
a, a+1, a+2, \ldots, a+26
$$

This is an arithmetic sequence with:

- First term: $ a $
- Last term: $ a + 26 $
- Number of terms: 27

The sum of an arithmetic sequence is given by:

$$
\text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term})
$$

Substituting the known values:

$$
\text{Sum} = \frac{27}{2} \times (a + (a + 26)) = \frac{27}{2} \times (2a + 26) = 27(a + 13)
$$

We are told that the sum is $ 2187 $, so:

$$
27(a + 13) = 2187
$$

---

### Step 3: Solving for the Median

Divide both sides by 27:

$$
a + 13 = \frac{2187}{27} = 81
$$

So:

$$
a = 81 - 13 = 68
$$

Now, the **median** is the 14th term, which is:

$$
a + 13 = 68 + 13 = 81
$$

---

### Step 4: Alternative Confirmation

Another way to think about this is that in a sequence of **odd** number of terms, the **median** is also the **average** of the sequence. Since the sum is 2187 and there are 27 terms:

$$
\text{Median} = \frac{\text{Sum}}{\text{Number of terms}} = \frac{2187}{27} = 81
$$

This confirms our earlier result.

---

### Final Answer

$$
\boxed{81}
$$

Solution: The median of a set of consecutive positive integers is equal to the mean of the set of integers.  Therefore, we can find the median by dividing the sum by the number of integers: $3^7/3^3=3^4=\boxed{81}$.


Explanation:
Okay, so I need to find the value of $ xy + yz + zx $ given these three equations:

1. $ xyz = 4 $
2. $ x^3 + y^3 + z^3 = 4 $
3. $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $

Hmm, let me think. I remember that there are some identities related to symmetric sums. Maybe I can use those? Let me recall.

First, the equation $ x^3 + y^3 + z^3 $ can be related to $ (x + y + z)^3 $, but I need to remember the exact formula. Let me write it down:

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x) $. Wait, no, that doesn't seem right. Let me check again.

Actually, the correct expansion is:

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x) $? Wait, no, that's not correct. Let me recall the standard identity:

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x) $? Hmm, no, that's not. Let me think again.

Wait, the standard identity is:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Yes, that's right. So maybe I can use that. Also, there's another identity for $ x^3 + y^3 + z^3 $. But I also have the third equation, which is $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $. Let me see if I can relate that to other symmetric sums.

Let me denote:

Let me recall that $ (x + y + z)(xy + yz + zx) = x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 3xyz $. Wait, yes! Let me check:

Expanding $ (x + y + z)(xy + yz + zx) $:

Multiply term by term:

First, x*(xy) = x^2 y

x*(yz) = xyz

x*(zx) = x^2 z

Similarly for y*(xy) = xy^2, y*(yz) = y^2 z, y*(zx) = xyz

Similarly for z*(xy) = xyz, z*(yz) = y z^2, z*(zx) = z^2 x

So adding all these up:

x^2 y + x^2 z + xy^2 + y^2 z + xz^2 + y z^2 + 3xyz

Therefore, indeed:

$ (x + y + z)(xy + yz + zx) = x^2 y + x^2 z + xy^2 + y^2 z + xz^2 + y z^2 + 3xyz $

Which means that:

$ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = (x + y + z)(xy + yz + zx) - 3xyz $

So according to the third equation, this sum is 12. Therefore:

$ (x + y + z)(xy + yz + zx) - 3xyz = 12 $

But we know that $ xyz = 4 $, so substituting:

$ (x + y + z)(xy + yz + zx) - 3*4 = 12 $

Which simplifies to:

$ (x + y + z)(xy + yz + zx) = 12 + 12 = 24 $

So that gives me $ (x + y + z)(xy + yz + zx) = 24 $. Let me note that.

Now, let me think about the other equations. The second equation is $ x^3 + y^3 + z^3 = 4 $. And from the identity I mentioned earlier:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

We know $ x^3 + y^3 + z^3 = 4 $, and $ xyz = 4 $, so:

$ 4 - 3*4 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Calculating left side: 4 - 12 = -8

Therefore:

$ -8 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Hmm. Now, I need to relate this to other terms. Let me recall that $ x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) $. Let me denote $ S = x + y + z $, $ P = xy + yz + zx $, and $ Q = xyz = 4 $. Then, the above equation becomes:

$ -8 = S( (S^2 - 2P) - P ) = S(S^2 - 3P) $

So:

$ -8 = S^3 - 3 S P $

But earlier, from the third equation, we had $ S * P = 24 $. Therefore, substituting $ S * P = 24 $ into the above equation:

$ -8 = S^3 - 3 * 24 $

Wait, no. Wait, the equation is:

$ -8 = S^3 - 3 S P $

But since $ S P = 24 $, then:

$ -8 = S^3 - 3 * 24 $

Wait, that would be:

$ -8 = S^3 - 72 $

Therefore:

$ S^3 = -8 + 72 = 64 $

Therefore, $ S^3 = 64 $, so $ S = \sqrt[3]{64} = 4 $. Since 4 cubed is 64. So S = 4.

Therefore, $ x + y + z = 4 $.

Then, since $ S * P = 24 $, and S = 4, then:

$ 4 * P = 24 \implies P = 6 $.

Therefore, $ xy + yz + zx = 6 $. Wait, so is that the answer? Let me check if this is consistent with all equations.

Wait, let me verify. Let me check if with S = 4, P = 6, Q = 4, whether the equations hold.

First, check the third equation:

We had $ (x + y + z)(xy + yz + zx) - 3xyz = 4*6 - 12 = 24 - 12 = 12 $, which matches the third equation. Good.

Then, check the second equation:

We had $ x^3 + y^3 + z^3 = 4 $. From the identity:

$ x^3 + y^3 + z^3 = 3xyz + (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Wait, but earlier we had:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

So, substituting:

Left side: 4 - 12 = -8

Right side: S*(x^2 + y^2 + z^2 - P). Since S = 4, and x^2 + y^2 + z^2 = S^2 - 2P = 16 - 12 = 4. Therefore, x^2 + y^2 + z^2 - P = 4 - 6 = -2. Therefore, right side is 4*(-2) = -8. Which matches. So that's correct.

So, seems like all equations are satisfied with S = 4, P = 6, Q = 4. Therefore, the value of $ xy + yz + zx = P = 6 $. Therefore, the answer is 6?

Wait, but let me check if there are other possibilities. For example, when we solved $ S^3 = 64 $, we took S = 4, but technically, in real numbers, cube roots are unique. So S = 4 is the only real solution. If we consider complex numbers, there are other roots, but since the problem doesn't specify, and given that xyz = 4 is positive, and the other equations, probably real numbers. So S = 4 is the only real solution. Therefore, the answer is 6.

But let me check if there's another way to approach this, just to confirm.

Alternatively, maybe we can think of variables. Suppose that x, y, z are roots of a cubic equation. Let me try that approach.

Let me assume that x, y, z are roots of the equation $ t^3 - S t^2 + P t - Q = 0 $, where S = x + y + z, P = xy + yz + zx, Q = xyz. So, in our case, Q = 4, S = 4, P = 6. Therefore, the equation would be $ t^3 - 4 t^2 + 6 t - 4 = 0 $. Let me check if this cubic factors.

Let me try rational roots. Possible rational roots are factors of 4 over factors of 1: ±1, ±2, ±4.

Testing t=1: 1 - 4 + 6 - 4 = -1 ≠ 0

t=2: 8 - 16 + 12 - 4 = 0. Yes! So t=2 is a root. Then, we can factor (t - 2) out.

Using polynomial division or synthetic division:

Divide $ t^3 - 4 t^2 + 6 t - 4 $ by (t - 2):

Coefficients: 1 | -4 | 6 | -4

Bring down 1. Multiply by 2: 2. Add to next coefficient: -4 + 2 = -2

Multiply by 2: -4. Add to next coefficient: 6 + (-4) = 2

Multiply by 2: 4. Add to last coefficient: -4 + 4 = 0. Perfect.

So the cubic factors as (t - 2)(t^2 - 2t + 2). Then, the quadratic equation t^2 - 2t + 2 = 0 has discriminant 4 - 8 = -4, so complex roots. Therefore, the roots are 2, 1 + i, 1 - i.

So, if x, y, z are 2, 1 + i, 1 - i, then let's check the equations.

First, xyz: 2*(1 + i)*(1 - i) = 2*(1 - i^2) = 2*(1 - (-1)) = 2*2 = 4. Correct.

x^3 + y^3 + z^3: Let's compute each term.

First, 2^3 = 8.

(1 + i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = (-2 + 2i)

Similarly, (1 - i)^3 = 1 - 3i + 3i^2 - i^3 = 1 - 3i - 3 + i = (-2 - 2i)

Adding them up: 8 + (-2 + 2i) + (-2 - 2i) = 8 - 4 = 4. Correct.

Third equation: xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z.

Let me compute this. Let me note that since two of the variables are complex conjugates, maybe the expression will be real.

Alternatively, since we already derived that with S = 4, P = 6, and the equations are satisfied, maybe we don't need to check. But just to be thorough.

Let me compute each term:

First, take x = 2, y = 1 + i, z = 1 - i.

Compute xy^2:

x*y^2 = 2*(1 + i)^2 = 2*(1 + 2i + i^2) = 2*(1 + 2i -1) = 2*(2i) = 4i

Similarly, x^2 y = (2)^2*(1 + i) = 4*(1 + i) = 4 + 4i

xz^2 = 2*(1 - i)^2 = 2*(1 - 2i + i^2) = 2*(1 - 2i -1) = 2*(-2i) = -4i

x^2 z = 4*(1 - i) = 4 - 4i

yz^2: y*z^2 = (1 + i)*(1 - i)^2. First compute z^2 = (1 - i)^2 = -2i, so y*z^2 = (1 + i)*(-2i) = -2i - 2i^2 = -2i + 2 = 2 - 2i

Similarly, y^2 z: (1 + i)^2 * z = (2i)*(1 - i) = 2i - 2i^2 = 2i + 2 = 2 + 2i

Now, adding all these terms:

xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z =

4i + (4 + 4i) + (-4i) + (4 - 4i) + (2 - 2i) + (2 + 2i)

Let me compute term by term:

Start with 4i + 4 + 4i = 4 + 8i

Then add -4i: 4 + 8i - 4i = 4 + 4i

Then add 4 - 4i: 4 + 4i + 4 - 4i = 8

Then add 2 - 2i: 8 + 2 - 2i = 10 - 2i

Then add 2 + 2i: 10 - 2i + 2 + 2i = 12. Perfect! So that's 12, which matches the third equation. So all equations are satisfied. Therefore, even though two variables are complex, the equations hold. Therefore, the value of $ xy + yz + zx = 6 $ is correct.

Therefore, the answer is 6. Wait, but the problem didn't specify that x, y, z are real numbers. But even if they are complex, the value of $ xy + yz + zx $ is still 6. So regardless of the nature of the variables, the answer is 6. Therefore, I think that's the correct answer.

But let me check if there's another approach, maybe using substitution or something else, just to confirm.

Alternatively, suppose we let $ a = x + y + z $, $ b = xy + yz + zx $, $ c = xyz $. Then, we have:

Given:

1. $ c = 4 $

2. $ x^3 + y^3 + z^3 = 4 $

3. $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $

From equation 3, as we did before, we have:

$ (x + y + z)(xy + yz + zx) - 3xyz = 12 \implies a b - 3c = 12 \implies a b - 12 = 12 \implies a b = 24 $

From equation 2, using the identity:

$ x^3 + y^3 + z^3 = a^3 - 3a b + 3c $. Wait, let me check that identity again.

Wait, earlier we had:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

But $ x^2 + y^2 + z^2 = a^2 - 2b $, so substituting:

$ x^3 + y^3 + z^3 - 3c = a( (a^2 - 2b) - b ) = a(a^2 - 3b) $

Therefore:

$ x^3 + y^3 + z^3 = a(a^2 - 3b) + 3c $

So given that $ x^3 + y^3 + z^3 = 4 $, then:

$ 4 = a(a^2 - 3b) + 3c $

But since c = 4, then:

$ 4 = a^3 - 3ab + 12 \implies a^3 - 3ab = 4 - 12 = -8 $

But from equation 3, we have $ ab = 24 $, so substituting:

$ a^3 - 3*(24) = -8 \implies a^3 - 72 = -8 \implies a^3 = 64 \implies a = 4 $

Therefore, $ a = 4 $, then $ b = 24 / a = 24 / 4 = 6 $. Therefore, $ b = 6 $. Therefore, $ xy + yz + zx = 6 $. So same result. Therefore, regardless of the approach, the answer is 6. Therefore, I think that's solid.

Therefore, the answer is 6. So I can write that as \boxed{6}.

**Final Answer**
\boxed{6}

To solve the problem, we are given the following system of equations involving three variables $ x, y, z $:

$$
\begin{align*}
(1) \quad & xyz = 4 \\
(2) \quad & x^3 + y^3 + z^3 = 4 \\
(3) \quad & xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12
\end{align*}
$$

We are asked to find the value of $ xy + yz + zx $.

---

### Step 1: Use Symmetric Identities

Let us denote the symmetric sums as follows:

- $ S = x + y + z $
- $ P = xy + yz + zx $
- $ Q = xyz $

We are given:

- $ Q = 4 $
- $ x^3 + y^3 + z^3 = 4 $
- $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $

We can use the identity:

$$
(x + y + z)(xy + yz + zx) = x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 3xyz
$$

This implies:

$$
xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = (x + y + z)(xy + yz + zx) - 3xyz
$$

Substituting the known values:

$$
12 = S \cdot P - 3 \cdot 4 = S \cdot P - 12
$$

So:

$$
S \cdot P = 24 \tag{1}
$$

---

### Step 2: Use the Identity for $ x^3 + y^3 + z^3 $

We use the identity:

$$
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
$$

We also know that:

$$
x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) = S^2 - 2P
$$

So:

$$
x^3 + y^3 + z^3 - 3xyz = S(S^2 - 3P)
$$

Substituting the known values:

$$
4 - 3 \cdot 4 = S(S^2 - 3P) \Rightarrow -8 = S(S^2 - 3P)
$$

From equation (1), we know $ S \cdot P = 24 $, so $ S^2 - 3P = \frac{S^3 - 3SP}{S} = \frac{S^3 - 72}{S} $

Substitute into the equation:

$$
-8 = S \cdot \left( \frac{S^3 - 72}{S} \right) = S^3 - 72
$$

So:

$$
S^3 = 64 \Rightarrow S = 4
$$

Now, using $ S \cdot P = 24 $, we find:

$$
4 \cdot P = 24 \Rightarrow P = 6
$$

---

### Final Answer

Thus, the value of $ xy + yz + zx $ is:

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ xy + yz + zx $ given these three equations:

1. $ xyz = 4 $
2. $ x^3 + y^3 + z^3 = 4 $
3. $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $

Hmm, let me think. I remember that there are some identities related to symmetric sums. Maybe I can use those? Let me recall.

First, the equation $ x^3 + y^3 + z^3 $ can be related to $ (x + y + z)^3 $, but I need to remember the exact formula. Let me write it down:

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x) $. Wait, no, that doesn't seem right. Let me check again.

Actually, the correct expansion is:

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x) $? Wait, no, that's not correct. Let me recall the standard identity:

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3(x + y)(y + z)(z + x) $? Hmm, no, that's not. Let me think again.

Wait, the standard identity is:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Yes, that's right. So maybe I can use that. Also, there's another identity for $ x^3 + y^3 + z^3 $. But I also have the third equation, which is $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $. Let me see if I can relate that to other symmetric sums.

Let me denote:

Let me recall that $ (x + y + z)(xy + yz + zx) = x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 3xyz $. Wait, yes! Let me check:

Expanding $ (x + y + z)(xy + yz + zx) $:

Multiply term by term:

First, x*(xy) = x^2 y

x*(yz) = xyz

x*(zx) = x^2 z

Similarly for y*(xy) = xy^2, y*(yz) = y^2 z, y*(zx) = xyz

Similarly for z*(xy) = xyz, z*(yz) = y z^2, z*(zx) = z^2 x

So adding all these up:

x^2 y + x^2 z + xy^2 + y^2 z + xz^2 + y z^2 + 3xyz

Therefore, indeed:

$ (x + y + z)(xy + yz + zx) = x^2 y + x^2 z + xy^2 + y^2 z + xz^2 + y z^2 + 3xyz $

Which means that:

$ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = (x + y + z)(xy + yz + zx) - 3xyz $

So according to the third equation, this sum is 12. Therefore:

$ (x + y + z)(xy + yz + zx) - 3xyz = 12 $

But we know that $ xyz = 4 $, so substituting:

$ (x + y + z)(xy + yz + zx) - 3*4 = 12 $

Which simplifies to:

$ (x + y + z)(xy + yz + zx) = 12 + 12 = 24 $

So that gives me $ (x + y + z)(xy + yz + zx) = 24 $. Let me note that.

Now, let me think about the other equations. The second equation is $ x^3 + y^3 + z^3 = 4 $. And from the identity I mentioned earlier:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

We know $ x^3 + y^3 + z^3 = 4 $, and $ xyz = 4 $, so:

$ 4 - 3*4 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Calculating left side: 4 - 12 = -8

Therefore:

$ -8 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Hmm. Now, I need to relate this to other terms. Let me recall that $ x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) $. Let me denote $ S = x + y + z $, $ P = xy + yz + zx $, and $ Q = xyz = 4 $. Then, the above equation becomes:

$ -8 = S( (S^2 - 2P) - P ) = S(S^2 - 3P) $

So:

$ -8 = S^3 - 3 S P $

But earlier, from the third equation, we had $ S * P = 24 $. Therefore, substituting $ S * P = 24 $ into the above equation:

$ -8 = S^3 - 3 * 24 $

Wait, no. Wait, the equation is:

$ -8 = S^3 - 3 S P $

But since $ S P = 24 $, then:

$ -8 = S^3 - 3 * 24 $

Wait, that would be:

$ -8 = S^3 - 72 $

Therefore:

$ S^3 = -8 + 72 = 64 $

Therefore, $ S^3 = 64 $, so $ S = \sqrt[3]{64} = 4 $. Since 4 cubed is 64. So S = 4.

Therefore, $ x + y + z = 4 $.

Then, since $ S * P = 24 $, and S = 4, then:

$ 4 * P = 24 \implies P = 6 $.

Therefore, $ xy + yz + zx = 6 $. Wait, so is that the answer? Let me check if this is consistent with all equations.

Wait, let me verify. Let me check if with S = 4, P = 6, Q = 4, whether the equations hold.

First, check the third equation:

We had $ (x + y + z)(xy + yz + zx) - 3xyz = 4*6 - 12 = 24 - 12 = 12 $, which matches the third equation. Good.

Then, check the second equation:

We had $ x^3 + y^3 + z^3 = 4 $. From the identity:

$ x^3 + y^3 + z^3 = 3xyz + (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Wait, but earlier we had:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

So, substituting:

Left side: 4 - 12 = -8

Right side: S*(x^2 + y^2 + z^2 - P). Since S = 4, and x^2 + y^2 + z^2 = S^2 - 2P = 16 - 12 = 4. Therefore, x^2 + y^2 + z^2 - P = 4 - 6 = -2. Therefore, right side is 4*(-2) = -8. Which matches. So that's correct.

So, seems like all equations are satisfied with S = 4, P = 6, Q = 4. Therefore, the value of $ xy + yz + zx = P = 6 $. Therefore, the answer is 6?

Wait, but let me check if there are other possibilities. For example, when we solved $ S^3 = 64 $, we took S = 4, but technically, in real numbers, cube roots are unique. So S = 4 is the only real solution. If we consider complex numbers, there are other roots, but since the problem doesn't specify, and given that xyz = 4 is positive, and the other equations, probably real numbers. So S = 4 is the only real solution. Therefore, the answer is 6.

But let me check if there's another way to approach this, just to confirm.

Alternatively, maybe we can think of variables. Suppose that x, y, z are roots of a cubic equation. Let me try that approach.

Let me assume that x, y, z are roots of the equation $ t^3 - S t^2 + P t - Q = 0 $, where S = x + y + z, P = xy + yz + zx, Q = xyz. So, in our case, Q = 4, S = 4, P = 6. Therefore, the equation would be $ t^3 - 4 t^2 + 6 t - 4 = 0 $. Let me check if this cubic factors.

Let me try rational roots. Possible rational roots are factors of 4 over factors of 1: ±1, ±2, ±4.

Testing t=1: 1 - 4 + 6 - 4 = -1 ≠ 0

t=2: 8 - 16 + 12 - 4 = 0. Yes! So t=2 is a root. Then, we can factor (t - 2) out.

Using polynomial division or synthetic division:

Divide $ t^3 - 4 t^2 + 6 t - 4 $ by (t - 2):

Coefficients: 1 | -4 | 6 | -4

Bring down 1. Multiply by 2: 2. Add to next coefficient: -4 + 2 = -2

Multiply by 2: -4. Add to next coefficient: 6 + (-4) = 2

Multiply by 2: 4. Add to last coefficient: -4 + 4 = 0. Perfect.

So the cubic factors as (t - 2)(t^2 - 2t + 2). Then, the quadratic equation t^2 - 2t + 2 = 0 has discriminant 4 - 8 = -4, so complex roots. Therefore, the roots are 2, 1 + i, 1 - i.

So, if x, y, z are 2, 1 + i, 1 - i, then let's check the equations.

First, xyz: 2*(1 + i)*(1 - i) = 2*(1 - i^2) = 2*(1 - (-1)) = 2*2 = 4. Correct.

x^3 + y^3 + z^3: Let's compute each term.

First, 2^3 = 8.

(1 + i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = (-2 + 2i)

Similarly, (1 - i)^3 = 1 - 3i + 3i^2 - i^3 = 1 - 3i - 3 + i = (-2 - 2i)

Adding them up: 8 + (-2 + 2i) + (-2 - 2i) = 8 - 4 = 4. Correct.

Third equation: xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z.

Let me compute this. Let me note that since two of the variables are complex conjugates, maybe the expression will be real.

Alternatively, since we already derived that with S = 4, P = 6, and the equations are satisfied, maybe we don't need to check. But just to be thorough.

Let me compute each term:

First, take x = 2, y = 1 + i, z = 1 - i.

Compute xy^2:

x*y^2 = 2*(1 + i)^2 = 2*(1 + 2i + i^2) = 2*(1 + 2i -1) = 2*(2i) = 4i

Similarly, x^2 y = (2)^2*(1 + i) = 4*(1 + i) = 4 + 4i

xz^2 = 2*(1 - i)^2 = 2*(1 - 2i + i^2) = 2*(1 - 2i -1) = 2*(-2i) = -4i

x^2 z = 4*(1 - i) = 4 - 4i

yz^2: y*z^2 = (1 + i)*(1 - i)^2. First compute z^2 = (1 - i)^2 = -2i, so y*z^2 = (1 + i)*(-2i) = -2i - 2i^2 = -2i + 2 = 2 - 2i

Similarly, y^2 z: (1 + i)^2 * z = (2i)*(1 - i) = 2i - 2i^2 = 2i + 2 = 2 + 2i

Now, adding all these terms:

xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z =

4i + (4 + 4i) + (-4i) + (4 - 4i) + (2 - 2i) + (2 + 2i)

Let me compute term by term:

Start with 4i + 4 + 4i = 4 + 8i

Then add -4i: 4 + 8i - 4i = 4 + 4i

Then add 4 - 4i: 4 + 4i + 4 - 4i = 8

Then add 2 - 2i: 8 + 2 - 2i = 10 - 2i

Then add 2 + 2i: 10 - 2i + 2 + 2i = 12. Perfect! So that's 12, which matches the third equation. So all equations are satisfied. Therefore, even though two variables are complex, the equations hold. Therefore, the value of $ xy + yz + zx = 6 $ is correct.

Therefore, the answer is 6. Wait, but the problem didn't specify that x, y, z are real numbers. But even if they are complex, the value of $ xy + yz + zx $ is still 6. So regardless of the nature of the variables, the answer is 6. Therefore, I think that's the correct answer.

But let me check if there's another approach, maybe using substitution or something else, just to confirm.

Alternatively, suppose we let $ a = x + y + z $, $ b = xy + yz + zx $, $ c = xyz $. Then, we have:

Given:

1. $ c = 4 $

2. $ x^3 + y^3 + z^3 = 4 $

3. $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $

From equation 3, as we did before, we have:

$ (x + y + z)(xy + yz + zx) - 3xyz = 12 \implies a b - 3c = 12 \implies a b - 12 = 12 \implies a b = 24 $

From equation 2, using the identity:

$ x^3 + y^3 + z^3 = a^3 - 3a b + 3c $. Wait, let me check that identity again.

Wait, earlier we had:

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

But $ x^2 + y^2 + z^2 = a^2 - 2b $, so substituting:

$ x^3 + y^3 + z^3 - 3c = a( (a^2 - 2b) - b ) = a(a^2 - 3b) $

Therefore:

$ x^3 + y^3 + z^3 = a(a^2 - 3b) + 3c $

So given that $ x^3 + y^3 + z^3 = 4 $, then:

$ 4 = a(a^2 - 3b) + 3c $

But since c = 4, then:

$ 4 = a^3 - 3ab + 12 \implies a^3 - 3ab = 4 - 12 = -8 $

But from equation 3, we have $ ab = 24 $, so substituting:

$ a^3 - 3*(24) = -8 \implies a^3 - 72 = -8 \implies a^3 = 64 \implies a = 4 $

Therefore, $ a = 4 $, then $ b = 24 / a = 24 / 4 = 6 $. Therefore, $ b = 6 $. Therefore, $ xy + yz + zx = 6 $. So same result. Therefore, regardless of the approach, the answer is 6. Therefore, I think that's solid.

Therefore, the answer is 6. So I can write that as \boxed{6}.

**Final Answer**
\boxed{6}

To solve the problem, we are given the following system of equations involving three variables $ x, y, z $:

$$
\begin{align*}
(1) \quad & xyz = 4 \\
(2) \quad & x^3 + y^3 + z^3 = 4 \\
(3) \quad & xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12
\end{align*}
$$

We are asked to find the value of $ xy + yz + zx $.

---

### Step 1: Use Symmetric Identities

Let us denote the symmetric sums as follows:

- $ S = x + y + z $
- $ P = xy + yz + zx $
- $ Q = xyz $

We are given:

- $ Q = 4 $
- $ x^3 + y^3 + z^3 = 4 $
- $ xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = 12 $

We can use the identity:

$$
(x + y + z)(xy + yz + zx) = x^2 y + x^2 z + y^2 x + y^2 z + z^2 x + z^2 y + 3xyz
$$

This implies:

$$
xy^2 + x^2 y + xz^2 + x^2 z + yz^2 + y^2 z = (x + y + z)(xy + yz + zx) - 3xyz
$$

Substituting the known values:

$$
12 = S \cdot P - 3 \cdot 4 = S \cdot P - 12
$$

So:

$$
S \cdot P = 24 \tag{1}
$$

---

### Step 2: Use the Identity for $ x^3 + y^3 + z^3 $

We use the identity:

$$
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
$$

We also know that:

$$
x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) = S^2 - 2P
$$

So:

$$
x^3 + y^3 + z^3 - 3xyz = S(S^2 - 3P)
$$

Substituting the known values:

$$
4 - 3 \cdot 4 = S(S^2 - 3P) \Rightarrow -8 = S(S^2 - 3P)
$$

From equation (1), we know $ S \cdot P = 24 $, so $ S^2 - 3P = \frac{S^3 - 3SP}{S} = \frac{S^3 - 72}{S} $

Substitute into the equation:

$$
-8 = S \cdot \left( \frac{S^3 - 72}{S} \right) = S^3 - 72
$$

So:

$$
S^3 = 64 \Rightarrow S = 4
$$

Now, using $ S \cdot P = 24 $, we find:

$$
4 \cdot P = 24 \Rightarrow P = 6
$$

---

### Final Answer

Thus, the value of $ xy + yz + zx $ is:

$$
\boxed{6}
$$

Solution:
Let $s_1 = x + y + z$ and $s_2 = xy + xz + yz.$  Then
\begin{align*}
s_1 s_2 &= (x + y + z)(xy + xz + yz) \\
&= x^2 y + xy^2 + x^2 z + xz^2 + y^2 z + yz^2 + 3xyz \\
&= 12 + 3 \cdot 4 = 24.
\end{align*}Also,
\begin{align*}
s_1^3 &= (x + y + z)^3 \\
&= (x^3 + y^3 + z^3) + 3(x^2 y + xy^2 + x^2 z + xz^2 + y^2 z + yz^2) + 6xyz \\
&= 4 + 3 \cdot 12 + 6 \cdot 4 = 64,
\end{align*}so $s_1 = 4.$  Hence, $s_2 = \frac{24}{s_1} = \boxed{6}.$


Explanation:
Okay, so I need to find the side length of an equilateral triangle inscribed in the parabola $ x^2 = 8y $, with one vertex at the vertex of the parabola. Let me start by recalling some properties of parabolas and equilateral triangles.

First, the parabola $ x^2 = 8y $ is an upward-opening parabola with vertex at the origin (0,0). The standard form of such a parabola is $ x^2 = 4py $, where p is the focal length. Comparing, 4p = 8, so p = 2. That means the focus is at (0, 2), but maybe that's not directly relevant here.

Now, the problem says an equilateral triangle is inscribed in the parabola, with one vertex at the vertex of the parabola, which is (0,0). Let me denote the three vertices of the triangle as A, B, and C. Let's say point A is at (0,0). Then points B and C are somewhere on the parabola.

Since it's an equilateral triangle, all sides must be equal. So AB = BC = CA. But since A is at (0,0), I need to find coordinates for B and C on the parabola such that the distances AB, BC, and CA are equal.

Let me denote the coordinates of B and C. Let me assume that points B and C are symmetric with respect to the y-axis. Because if the triangle is equilateral and one vertex is at the origin, maybe the other two points are symmetric. Let me check if that's a valid assumption.

If I assume symmetry, then coordinates of B could be (a, b) and C would be (-a, b), since they are symmetric across the y-axis. Then, since they lie on the parabola $ x^2 = 8y $, substituting x = a, we get $ a^2 = 8b $, so b = $ a^2 / 8 $.

So coordinates of B: (a, $ a^2 / 8 $), coordinates of C: (-a, $ a^2 / 8 $).

Now, let's compute the distances AB, AC, and BC.

First, AB: distance between (0,0) and (a, $ a^2 / 8 $):

AB = sqrt[(a - 0)^2 + (a^2 / 8 - 0)^2] = sqrt[a^2 + (a^4)/64]

Similarly, AC is the same as AB because of symmetry, so AC = sqrt[a^2 + (a^4)/64]

Now, BC: distance between (a, $ a^2 / 8 $) and (-a, $ a^2 / 8 $):

BC = sqrt[(-a - a)^2 + (a^2 / 8 - a^2 / 8)^2] = sqrt[(-2a)^2 + 0] = sqrt[4a^2] = 2|a|

Since the triangle is equilateral, AB = BC. Therefore:

sqrt[a^2 + (a^4)/64] = 2|a|

Since a can be positive or negative, but since we are squaring, let's assume a > 0 for simplicity.

So:

sqrt[a^2 + (a^4)/64] = 2a

Let me square both sides to eliminate the square root:

a^2 + (a^4)/64 = 4a^2

Subtract a^2 from both sides:

(a^4)/64 = 3a^2

Multiply both sides by 64:

a^4 = 192a^2

Divide both sides by a^2 (assuming a ≠ 0, which it can't be because then points B and C would coincide with A):

a^2 = 192

Therefore, a = sqrt(192) or a = -sqrt(192). Since we assumed a > 0, a = sqrt(192). Simplify sqrt(192):

sqrt(192) = sqrt(64 * 3) = 8*sqrt(3)

So a = 8√3. Then, b = a² / 8 = (64 * 3) / 8 = 8 * 3 = 24. Therefore, coordinates of B and C are (8√3, 24) and (-8√3, 24).

Now, let's check if AB equals BC. Let me compute AB:

AB = sqrt[(8√3)^2 + (24)^2] = sqrt[ 64*3 + 576 ] = sqrt[192 + 576] = sqrt[768] = sqrt[256 * 3] = 16√3

BC is 2a = 2*(8√3) = 16√3. So yes, AB = BC. But wait, the triangle is supposed to be equilateral, so we also need to check that AC = AB and that angle at A is 60 degrees? Wait, but since we already have AB = AC = BC, then it's equilateral. Wait, but hold on, in this case, points A(0,0), B(8√3,24), C(-8√3,24). Let me check the distance AC as well. Since AC is same as AB, so yes, it's 16√3. So all sides are equal. Therefore, this seems to satisfy the condition. Therefore, the side length is 16√3?

Wait, but let me think again. Is this triangle really equilateral? Because if you have points at (0,0), (a, b), (-a, b), then the triangle is isoceles with AB=AC, but BC is the base. But for it to be equilateral, AB must equal BC. Which we have done. So according to this calculation, the side length is 16√3. But let me check if this is correct.

But wait, let me think again. Maybe there is another triangle where the points are not symmetric? But the problem says "an equilateral triangle is inscribed in the parabola... with one vertex at the vertex of the parabola." Maybe there are multiple such triangles, but perhaps the symmetric one is the only one? Or maybe not?

Alternatively, maybe my assumption of symmetry is wrong? Let me check.

Suppose that the triangle is not symmetric. Let me try to think. Let me suppose that the three points are A(0,0), B(x1, y1), C(x2, y2), with both B and C on the parabola. Then, AB = AC = BC. But solving this system might be complicated. However, maybe the symmetric case is the only possible one? Or maybe there are two triangles, one symmetric and one not?

But given that the problem says "the" equilateral triangle, implying that there is only one, so maybe the symmetric one is the answer. However, let me check if my answer is correct.

Wait, but let me check with coordinates. Let me compute the distance between B(8√3, 24) and C(-8√3, 24). That's 16√3, as before. The distance from A(0,0) to B is sqrt( (8√3)^2 + 24^2 ) = sqrt( 192 + 576 ) = sqrt(768) = 16√3. So yes, all sides are equal. Therefore, this seems correct.

But wait, let me check if the triangle is actually equilateral. Let me compute the angles. For example, the angle at point A. Let me compute vectors AB and AC. Vector AB is (8√3, 24), vector AC is (-8√3, 24). The dot product is (8√3)(-8√3) + (24)(24) = - 64*3 + 576 = -192 + 576 = 384. The magnitude of AB and AC is 16√3, so the dot product is |AB||AC|cos(theta) = (16√3)^2 cos(theta) = 768 cos(theta). Therefore:

768 cos(theta) = 384 => cos(theta) = 384 / 768 = 0.5 => theta = 60 degrees. So yes, the angle at A is 60 degrees, which is consistent with an equilateral triangle. Therefore, this triangle is indeed equilateral. Therefore, the side length is 16√3. Therefore, is this the answer?

Wait, but let me check again. The problem says "an equilateral triangle is inscribed in the parabola". But inscribed usually means that all three vertices lie on the parabola. In this case, yes, A(0,0), B(8√3, 24), C(-8√3, 24) are all on the parabola. So that's correct.

But let me check if there are other possibilities. For example, maybe the triangle is not symmetric with respect to the y-axis. Suppose that the triangle has vertex at (0,0), and the other two vertices are not symmetric. Let me try to think if such a triangle can exist.

Let me suppose that points B and C are not symmetric. Let me denote coordinates of B as (x1, y1) and C as (x2, y2), both lying on the parabola, so x1² = 8y1 and x2² = 8y2. Then, the distances AB, AC, BC must be equal.

So AB² = x1² + y1² = x1² + (x1² / 8)² = x1² + x1^4 / 64

Similarly, AC² = x2² + y2² = x2² + x2^4 / 64

And BC² = (x1 - x2)^2 + (y1 - y2)^2 = (x1 - x2)^2 + (x1² / 8 - x2² / 8)^2

Since AB = AC = BC, so AB² = AC², which implies that x1² + x1^4 / 64 = x2² + x2^4 / 64. Which would imply that either x1 = x2 or some other relation. If x1 ≠ x2, then we have a quartic equation. However, solving this seems complicated. However, if we assume symmetry, then x2 = -x1, which satisfies the equation. Therefore, the symmetric case is a solution. But are there other solutions?

Alternatively, maybe there are other solutions where AB = AC = BC without symmetry. However, given that the problem says "the" equilateral triangle, maybe there is only one such triangle? Or maybe there are two, but the side length is same?

Alternatively, maybe my answer is correct, and the side length is 16√3. However, let me check with another approach.

Alternatively, maybe parametrize the parabola. Let me parametrize the parabola $ x^2 = 8y $. A standard parametrization for parabola $ x^2 = 4py $ is (2pt, pt²). In our case, 4p = 8 => p = 2, so parametrization is (4t, 2t²). Wait, let me check:

If x = 4t, then x² = 16t², and 8y = 16t² => y = 2t². Yes, so parametric equations are x = 4t, y = 2t². So any point on the parabola can be written as (4t, 2t²). Therefore, let me use this parametrization.

Let me denote the three points as follows: A(0,0), B(4t, 2t²), C(4s, 2s²). Since A is the vertex.

Now, since the triangle is equilateral, AB = AC = BC.

First, compute AB²: distance from A(0,0) to B(4t, 2t²):

AB² = (4t - 0)^2 + (2t² - 0)^2 = 16t² + 4t^4

Similarly, AC² = 16s² + 4s^4

Since AB = AC, so AB² = AC²:

16t² + 4t^4 = 16s² + 4s^4 => 4t^4 + 16t² = 4s^4 + 16s² => t^4 + 4t² = s^4 + 4s²

Let me rearrange:

t^4 - s^4 + 4t² - 4s² = 0 => (t² - s²)(t² + s²) + 4(t² - s²) = 0 => (t² - s²)(t² + s² + 4) = 0

Therefore, either t² - s² = 0 => t = ±s, or t² + s² + 4 = 0. But since t and s are real numbers, t² + s² + 4 = 0 is impossible. Therefore, only possibility is t = ±s.

Case 1: t = s. Then points B and C coincide, which is not allowed for a triangle. Therefore, discard.

Case 2: t = -s. Then, points B(4t, 2t²) and C(4(-t), 2(-t)^2) = (-4t, 2t²). Therefore, symmetric with respect to y-axis, which is the case I considered earlier. Therefore, this shows that the only possible case is symmetric points. Therefore, the triangle must be symmetric with respect to y-axis. Therefore, my initial assumption was correct, and there are no other triangles. Therefore, the side length is 16√3. Wait, but let me check with this parametrization.

So, if t = -s, then let me compute AB² = 16t² + 4t^4. Then BC² is distance between B(4t, 2t²) and C(-4t, 2t²):

BC² = ( -4t - 4t )² + (2t² - 2t²)^2 = (-8t)^2 + 0 = 64t²

Therefore, AB² = 16t² + 4t^4 = BC² = 64t² => 16t² + 4t^4 = 64t² => 4t^4 - 48t² = 0 => 4t²(t² - 12) = 0 => t² = 0 or t² = 12. t² = 0 gives t = 0, which would make points B and C coincide with A, so invalid. Therefore, t² = 12 => t = ±2√3. Therefore, t = 2√3, so coordinates of B: (4*(2√3), 2*(12)) = (8√3, 24), same as before. Therefore, AB² = 16*(12) + 4*(144) = 192 + 576 = 768, so AB = sqrt(768) = 16√3. Therefore, same result. Therefore, this seems correct.

But let me check if there is another case where maybe the triangle is not symmetric? But according to the parametrization approach, the only possibility is symmetric. Therefore, the answer is 16√3. However, let me check with another method.

Alternatively, maybe using calculus or geometry. Let me think.

Alternatively, since the triangle is equilateral, the slopes of AB and AC should satisfy certain conditions. But since we already have the answer through parametrization and symmetry, maybe that's sufficient.

But let me check with coordinates. Suppose we have points A(0,0), B(8√3, 24), C(-8√3, 24). Let me check if the triangle is equilateral. As before, AB = 16√3, BC = 16√3, and AC = 16√3. So yes. Therefore, the side length is 16√3. Therefore, the answer is 16√3. However, let me check if this is correct with another approach.

Alternatively, maybe using complex numbers or rotation. Since in an equilateral triangle, rotating one vertex around another by 60 degrees gives the third vertex. Let me try that.

Let me consider point A(0,0), point B(8√3, 24). If we rotate point B around point A by 60 degrees, we should get point C. Let me check.

Rotation of point (x, y) around origin by 60 degrees is given by:

x' = x cos θ - y sin θ

y' = x sin θ + y cos θ

For θ = 60°, cos θ = 0.5, sin θ = (√3)/2.

So, rotating point B(8√3, 24) by 60 degrees:

x' = 8√3 * 0.5 - 24 * (√3)/2 = 4√3 - 12√3 = -8√3

y' = 8√3 * (√3)/2 + 24 * 0.5 = (8*3)/2 + 12 = 12 + 12 = 24

Therefore, the rotated point is (-8√3, 24), which is point C. Therefore, this confirms that rotating point B around A by 60 degrees gives point C, so triangle ABC is indeed equilateral. Therefore, this is correct.

Therefore, the side length is 16√3. Therefore, the answer is \boxed{16\sqrt{3}}.

Wait, but let me check once more. Maybe I made a mistake in calculation. Let me compute AB again. Coordinates of B: (8√3, 24). So x-coordinate squared is (8√3)^2 = 64 * 3 = 192. y-coordinate squared is 24^2 = 576. Therefore, AB squared is 192 + 576 = 768. sqrt(768) is sqrt(256 * 3) = 16√3. Correct. So yes, that's correct.

Alternatively, maybe the problem expects a different answer? Let me check with another approach.

Alternatively, suppose we use coordinates for general points. Let me suppose that the triangle has vertex at (0,0), and other two points (a, a²/8) and (b, b²/8). Then, the distances between (0,0) and (a, a²/8) is sqrt(a² + (a^4)/64). Similarly for (0,0) and (b, b²/8). The distance between (a, a²/8) and (b, b²/8) is sqrt[(a - b)^2 + (a²/8 - b²/8)^2]. For the triangle to be equilateral, all three distances must be equal. Therefore:

sqrt(a² + (a^4)/64) = sqrt(b² + (b^4)/64) = sqrt[(a - b)^2 + ((a² - b²)/8)^2]

From the first equality, sqrt(a² + (a^4)/64) = sqrt(b² + (b^4)/64). Squaring both sides:

a² + (a^4)/64 = b² + (b^4)/64 => same as before, leading to a^4 - b^4 + 4a² - 4b² = 0 => (a² - b²)(a² + b² + 4) = 0. Therefore, same conclusion as before, so either a = ±b. If a = b, then points coincide. If a = -b, then symmetric case. Therefore, same result. Therefore, the only solution is symmetric case, leading to side length 16√3.

Therefore, I think this is correct. Therefore, the answer is 16√3.

But let me check with another resource or example. Wait, I recall that in some parabola problems, the side length of an equilateral triangle can be found using specific formulas, but I don't remember exactly. Alternatively, maybe I can check with specific values.

Alternatively, let me check if the coordinates (8√3, 24) lie on the parabola x² = 8y. Let me compute x²: (8√3)^2 = 64 * 3 = 192. 8y = 8*24 = 192. Yes, correct. Therefore, correct.

Therefore, I think my answer is correct. Therefore, the side length is 16√3.

**Final Answer**
The side length of the equilateral triangle is \boxed{16\sqrt{3}}.

To determine the side length of an equilateral triangle inscribed in the parabola $ x^2 = 8y $, with one vertex at the vertex of the parabola (which is at the origin, $ (0, 0) $), we proceed as follows:

---

### **Step 1: Understand the Geometry**

The parabola $ x^2 = 8y $ is an upward-opening parabola with vertex at the origin. We are to inscribe an equilateral triangle such that one of its vertices is at the origin, and the other two lie on the parabola.

Let the three vertices of the triangle be:
- $ A = (0, 0) $ (the vertex of the parabola),
- $ B = (a, b) $,
- $ C = (c, d) $.

Since the triangle is equilateral, all sides must be equal:
$$
AB = AC = BC
$$

---

### **Step 2: Use Symmetry to Simplify**

To simplify the problem, we assume that the triangle is symmetric with respect to the y-axis. This is a reasonable assumption because the parabola is symmetric about the y-axis, and the triangle is equilateral, which often implies symmetry in such geometric configurations.

Let:
- $ B = (a, b) $,
- $ C = (-a, b) $.

Since both $ B $ and $ C $ lie on the parabola $ x^2 = 8y $, we have:
$$
a^2 = 8b \quad \Rightarrow \quad b = \frac{a^2}{8}
$$

So the coordinates of the points are:
- $ A = (0, 0) $,
- $ B = (a, \frac{a^2}{8}) $,
- $ C = (-a, \frac{a^2}{8}) $

---

### **Step 3: Compute the Side Lengths**

We now compute the distances between the points:

#### **Distance AB:**
$$
AB = \sqrt{(a - 0)^2 + \left(\frac{a^2}{8} - 0\right)^2} = \sqrt{a^2 + \frac{a^4}{64}}
$$

#### **Distance BC:**
$$
BC = \sqrt{(-a - a)^2 + \left(\frac{a^2}{8} - \frac{a^2}{8}\right)^2} = \sqrt{(-2a)^2} = 2|a|
$$

Since the triangle is equilateral, we set $ AB = BC $:
$$
\sqrt{a^2 + \frac{a^4}{64}} = 2a
$$

---

### **Step 4: Solve the Equation**

Square both sides:
$$
a^2 + \frac{a^4}{64} = 4a^2
$$

Subtract $ a^2 $ from both sides:
$$
\frac{a^4}{64} = 3a^2
$$

Multiply both sides by 64:
$$
a^4 = 192a^2
$$

Divide both sides by $ a^2 $ (assuming $ a \ne 0 $):
$$
a^2 = 192 \quad \Rightarrow \quad a = \sqrt{192} = 8\sqrt{3}
$$

---

### **Step 5: Compute the Side Length**

Now that we know $ a = 8\sqrt{3} $, we can compute the side length of the triangle. Since $ AB = BC $, and we already found:
$$
AB = \sqrt{a^2 + \frac{a^4}{64}} = \sqrt{192 + \frac{(8\sqrt{3})^4}{64}} = \sqrt{192 + \frac{192^2}{64}} = \sqrt{192 + 576} = \sqrt{768}
$$

$$
\sqrt{768} = \sqrt{256 \cdot 3} = 16\sqrt{3}
$$

---

### **Conclusion**

All sides of the triangle are equal to $ 16\sqrt{3} $, confirming that the triangle is equilateral and inscribed in the parabola as required.

---

### **Final Answer**

$$
\boxed{16\sqrt{3}}
$$

[Assistant]
Explanation:
Okay, so I need to find the side length of an equilateral triangle inscribed in the parabola $ x^2 = 8y $, with one vertex at the vertex of the parabola. Let me start by recalling some properties of parabolas and equilateral triangles.

First, the parabola $ x^2 = 8y $ is an upward-opening parabola with vertex at the origin (0,0). The standard form of such a parabola is $ x^2 = 4py $, where p is the focal length. Comparing, 4p = 8, so p = 2. That means the focus is at (0, 2), but maybe that's not directly relevant here.

Now, the problem says an equilateral triangle is inscribed in the parabola, with one vertex at the vertex of the parabola, which is (0,0). Let me denote the three vertices of the triangle as A, B, and C. Let's say point A is at (0,0). Then points B and C are somewhere on the parabola.

Since it's an equilateral triangle, all sides must be equal. So AB = BC = CA. But since A is at (0,0), I need to find coordinates for B and C on the parabola such that the distances AB, BC, and CA are equal.

Let me denote the coordinates of B and C. Let me assume that points B and C are symmetric with respect to the y-axis. Because if the triangle is equilateral and one vertex is at the origin, maybe the other two points are symmetric. Let me check if that's a valid assumption.

If I assume symmetry, then coordinates of B could be (a, b) and C would be (-a, b), since they are symmetric across the y-axis. Then, since they lie on the parabola $ x^2 = 8y $, substituting x = a, we get $ a^2 = 8b $, so b = $ a^2 / 8 $.

So coordinates of B: (a, $ a^2 / 8 $), coordinates of C: (-a, $ a^2 / 8 $).

Now, let's compute the distances AB, AC, and BC.

First, AB: distance between (0,0) and (a, $ a^2 / 8 $):

AB = sqrt[(a - 0)^2 + (a^2 / 8 - 0)^2] = sqrt[a^2 + (a^4)/64]

Similarly, AC is the same as AB because of symmetry, so AC = sqrt[a^2 + (a^4)/64]

Now, BC: distance between (a, $ a^2 / 8 $) and (-a, $ a^2 / 8 $):

BC = sqrt[(-a - a)^2 + (a^2 / 8 - a^2 / 8)^2] = sqrt[(-2a)^2 + 0] = sqrt[4a^2] = 2|a|

Since the triangle is equilateral, AB = BC. Therefore:

sqrt[a^2 + (a^4)/64] = 2|a|

Since a can be positive or negative, but since we are squaring, let's assume a > 0 for simplicity.

So:

sqrt[a^2 + (a^4)/64] = 2a

Let me square both sides to eliminate the square root:

a^2 + (a^4)/64 = 4a^2

Subtract a^2 from both sides:

(a^4)/64 = 3a^2

Multiply both sides by 64:

a^4 = 192a^2

Divide both sides by a^2 (assuming a ≠ 0, which it can't be because then points B and C would coincide with A):

a^2 = 192

Therefore, a = sqrt(192) or a = -sqrt(192). Since we assumed a > 0, a = sqrt(192). Simplify sqrt(192):

sqrt(192) = sqrt(64 * 3) = 8*sqrt(3)

So a = 8√3. Then, b = a² / 8 = (64 * 3) / 8 = 8 * 3 = 24. Therefore, coordinates of B and C are (8√3, 24) and (-8√3, 24).

Now, let's check if AB equals BC. Let me compute AB:

AB = sqrt[(8√3)^2 + (24)^2] = sqrt[ 64*3 + 576 ] = sqrt[192 + 576] = sqrt[768] = sqrt[256 * 3] = 16√3

BC is 2a = 2*(8√3) = 16√3. So yes, AB = BC. But wait, the triangle is supposed to be equilateral, so we also need to check that AC = AB and that angle at A is 60 degrees? Wait, but since we already have AB = AC = BC, then it's equilateral. Wait, but hold on, in this case, points A(0,0), B(8√3,24), C(-8√3,24). Let me check the distance AC as well. Since AC is same as AB, so yes, it's 16√3. So all sides are equal. Therefore, this seems to satisfy the condition. Therefore, the side length is 16√3?

Wait, but let me think again. Is this triangle really equilateral? Because if you have points at (0,0), (a, b), (-a, b), then the triangle is isoceles with AB=AC, but BC is the base. But for it to be equilateral, AB must equal BC. Which we have done. So according to this calculation, the side length is 16√3. But let me check if this is correct.

But wait, let me think again. Maybe there is another triangle where the points are not symmetric? But the problem says "an equilateral triangle is inscribed in the parabola... with one vertex at the vertex of the parabola." Maybe there are multiple such triangles, but perhaps the symmetric one is the only one? Or maybe not?

Alternatively, maybe my assumption of symmetry is wrong? Let me check.

Suppose that the triangle is not symmetric. Let me try to think. Let me suppose that the three points are A(0,0), B(x1, y1), C(x2, y2), with both B and C on the parabola. Then, AB = AC = BC. But solving this system might be complicated. However, maybe the symmetric case is the only possible one? Or maybe there are two triangles, one symmetric and one not?

But given that the problem says "the" equilateral triangle, implying that there is only one, so maybe the symmetric one is the answer. However, let me check if my answer is correct.

Wait, but let me check with coordinates. Let me compute the distance between B(8√3, 24) and C(-8√3, 24). That's 16√3, as before. The distance from A(0,0) to B is sqrt( (8√3)^2 + 24^2 ) = sqrt( 192 + 576 ) = sqrt(768) = 16√3. So yes, all sides are equal. Therefore, this seems correct.

But wait, let me check if the triangle is actually equilateral. Let me compute the angles. For example, the angle at point A. Let me compute vectors AB and AC. Vector AB is (8√3, 24), vector AC is (-8√3, 24). The dot product is (8√3)(-8√3) + (24)(24) = - 64*3 + 576 = -192 + 576 = 384. The magnitude of AB and AC is 16√3, so the dot product is |AB||AC|cos(theta) = (16√3)^2 cos(theta) = 768 cos(theta). Therefore:

768 cos(theta) = 384 => cos(theta) = 384 / 768 = 0.5 => theta = 60 degrees. So yes, the angle at A is 60 degrees, which is consistent with an equilateral triangle. Therefore, this triangle is indeed equilateral. Therefore, the side length is 16√3. Therefore, is this the answer?

Wait, but let me check again. The problem says "an equilateral triangle is inscribed in the parabola". But inscribed usually means that all three vertices lie on the parabola. In this case, yes, A(0,0), B(8√3, 24), C(-8√3, 24) are all on the parabola. So that's correct.

But let me check if there are other possibilities. For example, maybe the triangle is not symmetric with respect to the y-axis. Suppose that the triangle has vertex at (0,0), and the other two vertices are not symmetric. Let me try to think if such a triangle can exist.

Let me suppose that points B and C are not symmetric. Let me denote coordinates of B as (x1, y1) and C as (x2, y2), both lying on the parabola, so x1² = 8y1 and x2² = 8y2. Then, the distances AB, AC, BC must be equal.

So AB² = x1² + y1² = x1² + (x1² / 8)² = x1² + x1^4 / 64

Similarly, AC² = x2² + y2² = x2² + x2^4 / 64

And BC² = (x1 - x2)^2 + (y1 - y2)^2 = (x1 - x2)^2 + (x1² / 8 - x2² / 8)^2

Since AB = AC = BC, so AB² = AC², which implies that x1² + x1^4 / 64 = x2² + x2^4 / 64. Which would imply that either x1 = x2 or some other relation. If x1 ≠ x2, then we have a quartic equation. However, solving this seems complicated. However, if we assume symmetry, then x2 = -x1, which satisfies the equation. Therefore, the symmetric case is a solution. But are there other solutions?

Alternatively, maybe there are other solutions where AB = AC = BC without symmetry. However, given that the problem says "the" equilateral triangle, maybe there is only one such triangle? Or maybe there are two, but the side length is same?

Alternatively, maybe my answer is correct, and the side length is 16√3. However, let me check with another approach.

Alternatively, maybe parametrize the parabola. Let me parametrize the parabola $ x^2 = 8y $. A standard parametrization for parabola $ x^2 = 4py $ is (2pt, pt²). In our case, 4p = 8 => p = 2, so parametrization is (4t, 2t²). Wait, let me check:

If x = 4t, then x² = 16t², and 8y = 16t² => y = 2t². Yes, so parametric equations are x = 4t, y = 2t². So any point on the parabola can be written as (4t, 2t²). Therefore, let me use this parametrization.

Let me denote the three points as follows: A(0,0), B(4t, 2t²), C(4s, 2s²). Since A is the vertex.

Now, since the triangle is equilateral, AB = AC = BC.

First, compute AB²: distance from A(0,0) to B(4t, 2t²):

AB² = (4t - 0)^2 + (2t² - 0)^2 = 16t² + 4t^4

Similarly, AC² = 16s² + 4s^4

Since AB = AC, so AB² = AC²:

16t² + 4t^4 = 16s² + 4s^4 => 4t^4 + 16t² = 4s^4 + 16s² => t^4 + 4t² = s^4 + 4s²

Let me rearrange:

t^4 - s^4 + 4t² - 4s² = 0 => (t² - s²)(t² + s²) + 4(t² - s²) = 0 => (t² - s²)(t² + s² + 4) = 0

Therefore, either t² - s² = 0 => t = ±s, or t² + s² + 4 = 0. But since t and s are real numbers, t² + s² + 4 = 0 is impossible. Therefore, only possibility is t = ±s.

Case 1: t = s. Then points B and C coincide, which is not allowed for a triangle. Therefore, discard.

Case 2: t = -s. Then, points B(4t, 2t²) and C(4(-t), 2(-t)^2) = (-4t, 2t²). Therefore, symmetric with respect to y-axis, which is the case I considered earlier. Therefore, this shows that the only possible case is symmetric points. Therefore, the triangle must be symmetric with respect to y-axis. Therefore, my initial assumption was correct, and there are no other triangles. Therefore, the side length is 16√3. Wait, but let me check with this parametrization.

So, if t = -s, then let me compute AB² = 16t² + 4t^4. Then BC² is distance between B(4t, 2t²) and C(-4t, 2t²):

BC² = ( -4t - 4t )² + (2t² - 2t²)^2 = (-8t)^2 + 0 = 64t²

Therefore, AB² = 16t² + 4t^4 = BC² = 64t² => 16t² + 4t^4 = 64t² => 4t^4 - 48t² = 0 => 4t²(t² - 12) = 0 => t² = 0 or t² = 12. t² = 0 gives t = 0, which would make points B and C coincide with A, so invalid. Therefore, t² = 12 => t = ±2√3. Therefore, t = 2√3, so coordinates of B: (4*(2√3), 2*(12)) = (8√3, 24), same as before. Therefore, AB² = 16*(12) + 4*(144) = 192 + 576 = 768, so AB = sqrt(768) = 16√3. Therefore, same result. Therefore, this seems correct.

But let me check if there is another case where maybe the triangle is not symmetric? But according to the parametrization approach, the only possibility is symmetric. Therefore, the answer is 16√3. However, let me check with another method.

Alternatively, maybe using calculus or geometry. Let me think.

Alternatively, since the triangle is equilateral, the slopes of AB and AC should satisfy certain conditions. But since we already have the answer through parametrization and symmetry, maybe that's sufficient.

But let me check with coordinates. Suppose we have points A(0,0), B(8√3, 24), C(-8√3, 24). Let me check if the triangle is equilateral. As before, AB = 16√3, BC = 16√3, and AC = 16√3. So yes. Therefore, the side length is 16√3. Therefore, the answer is 16√3. However, let me check if this is correct with another approach.

Alternatively, maybe using complex numbers or rotation. Since in an equilateral triangle, rotating one vertex around another by 60 degrees gives the third vertex. Let me try that.

Let me consider point A(0,0), point B(8√3, 24). If we rotate point B around point A by 60 degrees, we should get point C. Let me check.

Rotation of point (x, y) around origin by 60 degrees is given by:

x' = x cos θ - y sin θ

y' = x sin θ + y cos θ

For θ = 60°, cos θ = 0.5, sin θ = (√3)/2.

So, rotating point B(8√3, 24) by 60 degrees:

x' = 8√3 * 0.5 - 24 * (√3)/2 = 4√3 - 12√3 = -8√3

y' = 8√3 * (√3)/2 + 24 * 0.5 = (8*3)/2 + 12 = 12 + 12 = 24

Therefore, the rotated point is (-8√3, 24), which is point C. Therefore, this confirms that rotating point B around A by 60 degrees gives point C, so triangle ABC is indeed equilateral. Therefore, this is correct.

Therefore, the side length is 16√3. Therefore, the answer is \boxed{16\sqrt{3}}.

Wait, but let me check once more. Maybe I made a mistake in calculation. Let me compute AB again. Coordinates of B: (8√3, 24). So x-coordinate squared is (8√3)^2 = 64 * 3 = 192. y-coordinate squared is 24^2 = 576. Therefore, AB squared is 192 + 576 = 768. sqrt(768) is sqrt(256 * 3) = 16√3. Correct. So yes, that's correct.

Alternatively, maybe the problem expects a different answer? Let me check with another approach.

Alternatively, suppose we use coordinates for general points. Let me suppose that the triangle has vertex at (0,0), and other two points (a, a²/8) and (b, b²/8). Then, the distances between (0,0) and (a, a²/8) is sqrt(a² + (a^4)/64). Similarly for (0,0) and (b, b²/8). The distance between (a, a²/8) and (b, b²/8) is sqrt[(a - b)^2 + (a²/8 - b²/8)^2]. For the triangle to be equilateral, all three distances must be equal. Therefore:

sqrt(a² + (a^4)/64) = sqrt(b² + (b^4)/64) = sqrt[(a - b)^2 + ((a² - b²)/8)^2]

From the first equality, sqrt(a² + (a^4)/64) = sqrt(b² + (b^4)/64). Squaring both sides:

a² + (a^4)/64 = b² + (b^4)/64 => same as before, leading to a^4 - b^4 + 4a² - 4b² = 0 => (a² - b²)(a² + b² + 4) = 0. Therefore, same conclusion as before, so either a = ±b. If a = b, then points coincide. If a = -b, then symmetric case. Therefore, same result. Therefore, the only solution is symmetric case, leading to side length 16√3.

Therefore, I think this is correct. Therefore, the answer is 16√3.

But let me check with another resource or example. Wait, I recall that in some parabola problems, the side length of an equilateral triangle can be found using specific formulas, but I don't remember exactly. Alternatively, maybe I can check with specific values.

Alternatively, let me check if the coordinates (8√3, 24) lie on the parabola x² = 8y. Let me compute x²: (8√3)^2 = 64 * 3 = 192. 8y = 8*24 = 192. Yes, correct. Therefore, correct.

Therefore, I think my answer is correct. Therefore, the side length is 16√3.

**Final Answer**
The side length of the equilateral triangle is \boxed{16\sqrt{3}}.

To determine the side length of an equilateral triangle inscribed in the parabola $ x^2 = 8y $, with one vertex at the vertex of the parabola (which is at the origin, $ (0, 0) $), we proceed as follows:

---

### **Step 1: Understand the Geometry**

The parabola $ x^2 = 8y $ is an upward-opening parabola with vertex at the origin. We are to inscribe an equilateral triangle such that one of its vertices is at the origin, and the other two lie on the parabola.

Let the three vertices of the triangle be:
- $ A = (0, 0) $ (the vertex of the parabola),
- $ B = (a, b) $,
- $ C = (c, d) $.

Since the triangle is equilateral, all sides must be equal:
$$
AB = AC = BC
$$

---

### **Step 2: Use Symmetry to Simplify**

To simplify the problem, we assume that the triangle is symmetric with respect to the y-axis. This is a reasonable assumption because the parabola is symmetric about the y-axis, and the triangle is equilateral, which often implies symmetry in such geometric configurations.

Let:
- $ B = (a, b) $,
- $ C = (-a, b) $.

Since both $ B $ and $ C $ lie on the parabola $ x^2 = 8y $, we have:
$$
a^2 = 8b \quad \Rightarrow \quad b = \frac{a^2}{8}
$$

So the coordinates of the points are:
- $ A = (0, 0) $,
- $ B = (a, \frac{a^2}{8}) $,
- $ C = (-a, \frac{a^2}{8}) $

---

### **Step 3: Compute the Side Lengths**

We now compute the distances between the points:

#### **Distance AB:**
$$
AB = \sqrt{(a - 0)^2 + \left(\frac{a^2}{8} - 0\right)^2} = \sqrt{a^2 + \frac{a^4}{64}}
$$

#### **Distance BC:**
$$
BC = \sqrt{(-a - a)^2 + \left(\frac{a^2}{8} - \frac{a^2}{8}\right)^2} = \sqrt{(-2a)^2} = 2|a|
$$

Since the triangle is equilateral, we set $ AB = BC $:
$$
\sqrt{a^2 + \frac{a^4}{64}} = 2a
$$

---

### **Step 4: Solve the Equation**

Square both sides:
$$
a^2 + \frac{a^4}{64} = 4a^2
$$

Subtract $ a^2 $ from both sides:
$$
\frac{a^4}{64} = 3a^2
$$

Multiply both sides by 64:
$$
a^4 = 192a^2
$$

Divide both sides by $ a^2 $ (assuming $ a \ne 0 $):
$$
a^2 = 192 \quad \Rightarrow \quad a = \sqrt{192} = 8\sqrt{3}
$$

---

### **Step 5: Compute the Side Length**

Now that we know $ a = 8\sqrt{3} $, we can compute the side length of the triangle. Since $ AB = BC $, and we already found:
$$
AB = \sqrt{a^2 + \frac{a^4}{64}} = \sqrt{192 + \frac{(8\sqrt{3})^4}{64}} = \sqrt{192 + \frac{192^2}{64}} = \sqrt{192 + 576} = \sqrt{768}
$$

$$
\sqrt{768} = \sqrt{256 \cdot 3} = 16\sqrt{3}
$$

---

### **Conclusion**

All sides of the triangle are equal to $ 16\sqrt{3} $, confirming that the triangle is equilateral and inscribed in the parabola as required.

---

### **Final Answer**

$$
\boxed{16\sqrt{3}}
$$

Solution:
The vertex of the parabola is the origin.  Let $A = \left( a, \frac{a^2}{8} \right)$ be one vertex of the triangle.  Then by symmetry, $B = \left( -a, \frac{a^2}{8} \right)$ is another vertex of the triangle.

[asy]
unitsize(0.2 cm);

real parab (real x) {
  return(x^2/8);
}

pair A, B, O;

A = (8*sqrt(3),24);
B = (-8*sqrt(3),24);
O = (0,0);

draw(O--A--B--cycle);
draw(graph(parab,-15,15));

dot("$A = (a,\frac{a^2}{8})$", A, E);
dot("$B = (-a,\frac{a^2}{8})$", B, W);
dot("$O = (0,0)$", O, S);
[/asy]

Then $AB^2 = (a + a)^2 = 4a^2,$ and
\[OA^2 = a^2 + \frac{a^4}{64}.\]Hence, $4a^2 = a^2 + \frac{a^4}{64}.$  Then $3a^2 = \frac{a^4}{64},$ so $a^2 = 3 \cdot 64,$ which means $a = 8 \sqrt{3}.$

Therefore, the side length of the triangle is $2a = \boxed{16 \sqrt{3}}.$


Explanation:
Okay, so I need to figure out the equation of the set of image points after reflecting each point on the circle x² + y² = 25 over the point (4,1). The final equation is supposed to be in the form x² + a y² + b x + c y + d = 0, and I need to find the ordered quadruple (a, b, c, d). Let me start by recalling what reflecting a point over another point means.

Reflecting a point P over a point Q results in a point P' such that Q is the midpoint of the segment PP'. So, if I have a point (x, y) on the original circle, its reflection (let's call it (x', y')) over (4,1) should satisfy the midpoint formula. That is, the midpoint between (x, y) and (x', y') is (4,1).

So, the midpoint formula says that:

( (x + x') / 2 , (y + y') / 2 ) = (4, 1)

Therefore, solving for x' and y', we get:

x + x' = 8 => x' = 8 - x

y + y' = 2 => y' = 2 - y

So, the reflection of any point (x, y) over (4,1) is (8 - x, 2 - y).

Now, since the original points lie on the circle x² + y² = 25, their reflections will satisfy some equation. Let me denote the reflected point as (X, Y). So, X = 8 - x and Y = 2 - y. Therefore, we can express x and y in terms of X and Y:

x = 8 - X

y = 2 - Y

Since (x, y) lies on the original circle, substituting these into the original equation:

(8 - X)² + (2 - Y)² = 25

Now, I need to expand this equation and rearrange it into the form x² + a y² + b x + c y + d = 0. Wait, but here the variables are X and Y, which are the coordinates of the image points. However, the problem says the set of image points satisfies the equation x² + a y² + b x + c y + d = 0. So, I need to write the equation in terms of X and Y, but since X and Y are just variables, we can just replace them with x and y for the equation. Let me check that.

So, expanding (8 - X)² + (2 - Y)² = 25:

First, expand (8 - X)²:

= 64 - 16X + X²

Then, expand (2 - Y)²:

= 4 - 4Y + Y²

Adding them together:

64 - 16X + X² + 4 - 4Y + Y² = 25

Combine like terms:

(64 + 4) + (-16X - 4Y) + (X² + Y²) = 25

Which is:

68 - 16X - 4Y + X² + Y² = 25

Now, subtract 25 from both sides to set the equation to zero:

X² + Y² - 16X - 4Y + 68 - 25 = 0

Calculating 68 - 25:

68 - 25 = 43

So, the equation becomes:

X² + Y² - 16X - 4Y + 43 = 0

But the problem states that the equation is in the form x² + a y² + b x + c y + d = 0. Comparing this with our equation, we can see that:

x² + y² - 16x - 4y + 43 = 0

Therefore, the coefficients are:

a = 1 (since the coefficient of y² is 1)

b = -16

c = -4

d = 43

Wait, but the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0". So, in our case, the equation is x² + y² -16x -4y +43 = 0, so comparing term by term:

x² + (1)y² + (-16)x + (-4)y + 43 = 0. Therefore, the ordered quadruple is (a, b, c, d) = (1, -16, -4, 43).

But let me double-check my steps to make sure I didn't make any mistakes.

First, reflection over a point: yes, midpoint formula. So if original point is (x, y), then reflection is (8 - x, 2 - y). Then substituting into original equation.

Original equation: x² + y² = 25. So replacing x with 8 - X and y with 2 - Y. Then expanding:

(8 - X)^2 + (2 - Y)^2 = 25.

Expanding:

(64 - 16X + X²) + (4 - 4Y + Y²) = 25.

Adding 64 + 4 = 68. Then -16X -4Y. Then X² + Y².

So 68 -16X -4Y + X² + Y² = 25.

Subtract 25: X² + Y² -16X -4Y + 43 = 0.

Yes, that seems correct.

But let me check if there's a possibility that the problem expects the equation in terms of the original variables? Wait, no. The image points are (X, Y), so the equation they satisfy is in terms of X and Y. But since the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0", so they just use x and y as variables for the image points. So replacing X and Y with x and y, the equation is x² + y² -16x -4y +43 = 0. Therefore, the coefficients are a=1, b=-16, c=-4, d=43.

But let me check if there's a different approach. Maybe I can think about the reflection of the circle. Reflecting a circle over a point... Well, reflecting a circle over a point results in another circle. The original circle has center at (0,0) and radius 5. Reflecting over the point (4,1) would result in a circle whose center is the reflection of the original center over (4,1).

Let me try this approach. The original center is (0,0). Reflecting (0,0) over (4,1) gives a point (8, 2), since midpoint between (0,0) and (8,2) is (4,1). Therefore, the reflected circle has center (8,2) and same radius 5. Therefore, the equation is (x - 8)^2 + (y - 2)^2 = 25.

Expanding this equation:

x² - 16x + 64 + y² -4y +4 = 25

Combine terms:

x² + y² -16x -4y + 64 + 4 -25 = 0

Which is x² + y² -16x -4y + 43 = 0.

Same result as before! So that confirms that the equation is indeed x² + y² -16x -4y +43 = 0. Therefore, the ordered quadruple is (1, -16, -4, 43).

But let me check once more if the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0". So, in our case, the equation is x² + y² -16x -4y +43 = 0. Therefore, comparing term by term:

x² + (1)y² + (-16)x + (-4)y + 43 = 0. Therefore, a=1, b=-16, c=-4, d=43.

So the ordered quadruple is (1, -16, -4, 43).

But let me check if there's any possible mistake in the reflection. For example, if I had messed up the reflection formula. Let me think again. If you reflect a point (x, y) over the point (h, k), then the reflection point (x', y') is such that (h, k) is the midpoint. Therefore:

h = (x + x')/2 => x' = 2h - x

Similarly, y' = 2k - y.

So in this case, h = 4, k = 1. Therefore, x' = 8 - x, y' = 2 - y. Which is what I had before. So that's correct.

Therefore, substituting into original equation, we get the equation of the reflected circle.

Alternatively, if I think of the reflected circle as having center (8, 2) and radius 5, then equation is (x - 8)^2 + (y - 2)^2 = 25. Expanding that gives x² -16x +64 + y² -4y +4 = 25 => x² + y² -16x -4y + 68 -25 = 0 => x² + y² -16x -4y +43 = 0.

Same result. So that's correct.

Therefore, the answer is (a, b, c, d) = (1, -16, -4, 43).

But let me check if the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0". So, in our case, the equation is x² + y² -16x -4y +43 = 0. Therefore, the coefficients are:

a = 1 (coefficient of y²)

b = -16 (coefficient of x)

c = -4 (coefficient of y)

d = 43 (constant term)

Yes, so ordered quadruple is (1, -16, -4, 43).

Therefore, I think that's the answer.

But just to be thorough, let me check with a specific point. Let's take a point on the original circle, reflect it, and check if it satisfies the equation.

Take the point (5, 0) on the original circle. Reflecting over (4,1):

x' = 8 - 5 = 3

y' = 2 - 0 = 2

So the image point is (3, 2). Let's plug into the equation x² + y² -16x -4y +43:

3² + 2² -16*3 -4*2 +43 = 9 + 4 -48 -8 +43 = (13) - (56) +43 = 13 -56 +43 = (13 +43) -56 = 56 -56 = 0. Correct.

Another point: (0,5). Reflecting over (4,1):

x' = 8 - 0 = 8

y' = 2 - 5 = -3

So image point (8, -3). Plug into equation:

8² + (-3)^2 -16*8 -4*(-3) +43 = 64 + 9 -128 +12 +43 = (73) -128 +12 +43 = 73 -128 is -55, then -55 +12 is -43, then -43 +43 = 0. Correct.

Another point: (0,0). Wait, (0,0) is on the original circle? Wait, 0² + 0² = 0 ≠ 25. So not on the circle. Let me pick (3,4). Since 3² +4² = 25. Reflecting over (4,1):

x' = 8 - 3 = 5

y' = 2 - 4 = -2

So image point (5, -2). Plug into equation:

5² + (-2)^2 -16*5 -4*(-2) +43 = 25 +4 -80 +8 +43 = 29 -80 +8 +43 = (29 +8 +43) -80 = 80 -80 = 0. Correct.

So all these points check out. Therefore, the equation is correct.

Therefore, the ordered quadruple is (1, -16, -4, 43).

**Final Answer**
The ordered quadruple is \boxed{(1, -16, -4, 43)}.

To solve the problem, we are given a circle defined by the equation:

$$
x^2 + y^2 = 25
$$

We are to reflect **each point** on this circle over the point $(4, 1)$, and then determine the equation of the set of image points in the form:

$$
x^2 + a y^2 + b x + c y + d = 0
$$

---

### Step 1: Understanding the Reflection

Reflecting a point $(x, y)$ over a point $(h, k)$ results in a new point $(x', y')$ such that $(h, k)$ is the **midpoint** of the segment joining $(x, y)$ and $(x', y')$. Using the midpoint formula:

$$
\left( \frac{x + x'}{2}, \frac{y + y'}{2} \right) = (4, 1)
$$

Solving for $x'$ and $y'$:

$$
x' = 8 - x, \quad y' = 2 - y
$$

So, the image of a point $(x, y)$ on the original circle is $(8 - x, 2 - y)$.

---

### Step 2: Substituting into the Original Equation

Let the image point be $(X, Y)$. Then:

$$
X = 8 - x \Rightarrow x = 8 - X \\
Y = 2 - y \Rightarrow y = 2 - Y
$$

Substitute into the original circle equation:

$$
x^2 + y^2 = 25 \Rightarrow (8 - X)^2 + (2 - Y)^2 = 25
$$

---

### Step 3: Expanding the Equation

$$
(8 - X)^2 = 64 - 16X + X^2 \\
(2 - Y)^2 = 4 - 4Y + Y^2
$$

Adding both:

$$
64 - 16X + X^2 + 4 - 4Y + Y^2 = 25
$$

Simplify:

$$
X^2 + Y^2 - 16X - 4Y + 68 = 25
$$

Subtract 25 from both sides:

$$
X^2 + Y^2 - 16X - 4Y + 43 = 0
$$

---

### Step 4: Matching the Form

The problem asks for the equation in the form:

$$
x^2 + a y^2 + b x + c y + d = 0
$$

Comparing with our derived equation:

$$
x^2 + y^2 - 16x - 4y + 43 = 0
$$

We identify the coefficients:

- $a = 1$ (coefficient of $y^2$)
- $b = -16$ (coefficient of $x$)
- $c = -4$ (coefficient of $y$)
- $d = 43$ (constant term)

---

### Final Answer

$$
\boxed{(1, -16, -4, 43)}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the equation of the set of image points after reflecting each point on the circle x² + y² = 25 over the point (4,1). The final equation is supposed to be in the form x² + a y² + b x + c y + d = 0, and I need to find the ordered quadruple (a, b, c, d). Let me start by recalling what reflecting a point over another point means.

Reflecting a point P over a point Q results in a point P' such that Q is the midpoint of the segment PP'. So, if I have a point (x, y) on the original circle, its reflection (let's call it (x', y')) over (4,1) should satisfy the midpoint formula. That is, the midpoint between (x, y) and (x', y') is (4,1).

So, the midpoint formula says that:

( (x + x') / 2 , (y + y') / 2 ) = (4, 1)

Therefore, solving for x' and y', we get:

x + x' = 8 => x' = 8 - x

y + y' = 2 => y' = 2 - y

So, the reflection of any point (x, y) over (4,1) is (8 - x, 2 - y).

Now, since the original points lie on the circle x² + y² = 25, their reflections will satisfy some equation. Let me denote the reflected point as (X, Y). So, X = 8 - x and Y = 2 - y. Therefore, we can express x and y in terms of X and Y:

x = 8 - X

y = 2 - Y

Since (x, y) lies on the original circle, substituting these into the original equation:

(8 - X)² + (2 - Y)² = 25

Now, I need to expand this equation and rearrange it into the form x² + a y² + b x + c y + d = 0. Wait, but here the variables are X and Y, which are the coordinates of the image points. However, the problem says the set of image points satisfies the equation x² + a y² + b x + c y + d = 0. So, I need to write the equation in terms of X and Y, but since X and Y are just variables, we can just replace them with x and y for the equation. Let me check that.

So, expanding (8 - X)² + (2 - Y)² = 25:

First, expand (8 - X)²:

= 64 - 16X + X²

Then, expand (2 - Y)²:

= 4 - 4Y + Y²

Adding them together:

64 - 16X + X² + 4 - 4Y + Y² = 25

Combine like terms:

(64 + 4) + (-16X - 4Y) + (X² + Y²) = 25

Which is:

68 - 16X - 4Y + X² + Y² = 25

Now, subtract 25 from both sides to set the equation to zero:

X² + Y² - 16X - 4Y + 68 - 25 = 0

Calculating 68 - 25:

68 - 25 = 43

So, the equation becomes:

X² + Y² - 16X - 4Y + 43 = 0

But the problem states that the equation is in the form x² + a y² + b x + c y + d = 0. Comparing this with our equation, we can see that:

x² + y² - 16x - 4y + 43 = 0

Therefore, the coefficients are:

a = 1 (since the coefficient of y² is 1)

b = -16

c = -4

d = 43

Wait, but the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0". So, in our case, the equation is x² + y² -16x -4y +43 = 0, so comparing term by term:

x² + (1)y² + (-16)x + (-4)y + 43 = 0. Therefore, the ordered quadruple is (a, b, c, d) = (1, -16, -4, 43).

But let me double-check my steps to make sure I didn't make any mistakes.

First, reflection over a point: yes, midpoint formula. So if original point is (x, y), then reflection is (8 - x, 2 - y). Then substituting into original equation.

Original equation: x² + y² = 25. So replacing x with 8 - X and y with 2 - Y. Then expanding:

(8 - X)^2 + (2 - Y)^2 = 25.

Expanding:

(64 - 16X + X²) + (4 - 4Y + Y²) = 25.

Adding 64 + 4 = 68. Then -16X -4Y. Then X² + Y².

So 68 -16X -4Y + X² + Y² = 25.

Subtract 25: X² + Y² -16X -4Y + 43 = 0.

Yes, that seems correct.

But let me check if there's a possibility that the problem expects the equation in terms of the original variables? Wait, no. The image points are (X, Y), so the equation they satisfy is in terms of X and Y. But since the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0", so they just use x and y as variables for the image points. So replacing X and Y with x and y, the equation is x² + y² -16x -4y +43 = 0. Therefore, the coefficients are a=1, b=-16, c=-4, d=43.

But let me check if there's a different approach. Maybe I can think about the reflection of the circle. Reflecting a circle over a point... Well, reflecting a circle over a point results in another circle. The original circle has center at (0,0) and radius 5. Reflecting over the point (4,1) would result in a circle whose center is the reflection of the original center over (4,1).

Let me try this approach. The original center is (0,0). Reflecting (0,0) over (4,1) gives a point (8, 2), since midpoint between (0,0) and (8,2) is (4,1). Therefore, the reflected circle has center (8,2) and same radius 5. Therefore, the equation is (x - 8)^2 + (y - 2)^2 = 25.

Expanding this equation:

x² - 16x + 64 + y² -4y +4 = 25

Combine terms:

x² + y² -16x -4y + 64 + 4 -25 = 0

Which is x² + y² -16x -4y + 43 = 0.

Same result as before! So that confirms that the equation is indeed x² + y² -16x -4y +43 = 0. Therefore, the ordered quadruple is (1, -16, -4, 43).

But let me check once more if the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0". So, in our case, the equation is x² + y² -16x -4y +43 = 0. Therefore, comparing term by term:

x² + (1)y² + (-16)x + (-4)y + 43 = 0. Therefore, a=1, b=-16, c=-4, d=43.

So the ordered quadruple is (1, -16, -4, 43).

But let me check if there's any possible mistake in the reflection. For example, if I had messed up the reflection formula. Let me think again. If you reflect a point (x, y) over the point (h, k), then the reflection point (x', y') is such that (h, k) is the midpoint. Therefore:

h = (x + x')/2 => x' = 2h - x

Similarly, y' = 2k - y.

So in this case, h = 4, k = 1. Therefore, x' = 8 - x, y' = 2 - y. Which is what I had before. So that's correct.

Therefore, substituting into original equation, we get the equation of the reflected circle.

Alternatively, if I think of the reflected circle as having center (8, 2) and radius 5, then equation is (x - 8)^2 + (y - 2)^2 = 25. Expanding that gives x² -16x +64 + y² -4y +4 = 25 => x² + y² -16x -4y + 68 -25 = 0 => x² + y² -16x -4y +43 = 0.

Same result. So that's correct.

Therefore, the answer is (a, b, c, d) = (1, -16, -4, 43).

But let me check if the problem says "the set of image points satisfies the equation x² + a y² + b x + c y + d = 0". So, in our case, the equation is x² + y² -16x -4y +43 = 0. Therefore, the coefficients are:

a = 1 (coefficient of y²)

b = -16 (coefficient of x)

c = -4 (coefficient of y)

d = 43 (constant term)

Yes, so ordered quadruple is (1, -16, -4, 43).

Therefore, I think that's the answer.

But just to be thorough, let me check with a specific point. Let's take a point on the original circle, reflect it, and check if it satisfies the equation.

Take the point (5, 0) on the original circle. Reflecting over (4,1):

x' = 8 - 5 = 3

y' = 2 - 0 = 2

So the image point is (3, 2). Let's plug into the equation x² + y² -16x -4y +43:

3² + 2² -16*3 -4*2 +43 = 9 + 4 -48 -8 +43 = (13) - (56) +43 = 13 -56 +43 = (13 +43) -56 = 56 -56 = 0. Correct.

Another point: (0,5). Reflecting over (4,1):

x' = 8 - 0 = 8

y' = 2 - 5 = -3

So image point (8, -3). Plug into equation:

8² + (-3)^2 -16*8 -4*(-3) +43 = 64 + 9 -128 +12 +43 = (73) -128 +12 +43 = 73 -128 is -55, then -55 +12 is -43, then -43 +43 = 0. Correct.

Another point: (0,0). Wait, (0,0) is on the original circle? Wait, 0² + 0² = 0 ≠ 25. So not on the circle. Let me pick (3,4). Since 3² +4² = 25. Reflecting over (4,1):

x' = 8 - 3 = 5

y' = 2 - 4 = -2

So image point (5, -2). Plug into equation:

5² + (-2)^2 -16*5 -4*(-2) +43 = 25 +4 -80 +8 +43 = 29 -80 +8 +43 = (29 +8 +43) -80 = 80 -80 = 0. Correct.

So all these points check out. Therefore, the equation is correct.

Therefore, the ordered quadruple is (1, -16, -4, 43).

**Final Answer**
The ordered quadruple is \boxed{(1, -16, -4, 43)}.

To solve the problem, we are given a circle defined by the equation:

$$
x^2 + y^2 = 25
$$

We are to reflect **each point** on this circle over the point $(4, 1)$, and then determine the equation of the set of image points in the form:

$$
x^2 + a y^2 + b x + c y + d = 0
$$

---

### Step 1: Understanding the Reflection

Reflecting a point $(x, y)$ over a point $(h, k)$ results in a new point $(x', y')$ such that $(h, k)$ is the **midpoint** of the segment joining $(x, y)$ and $(x', y')$. Using the midpoint formula:

$$
\left( \frac{x + x'}{2}, \frac{y + y'}{2} \right) = (4, 1)
$$

Solving for $x'$ and $y'$:

$$
x' = 8 - x, \quad y' = 2 - y
$$

So, the image of a point $(x, y)$ on the original circle is $(8 - x, 2 - y)$.

---

### Step 2: Substituting into the Original Equation

Let the image point be $(X, Y)$. Then:

$$
X = 8 - x \Rightarrow x = 8 - X \\
Y = 2 - y \Rightarrow y = 2 - Y
$$

Substitute into the original circle equation:

$$
x^2 + y^2 = 25 \Rightarrow (8 - X)^2 + (2 - Y)^2 = 25
$$

---

### Step 3: Expanding the Equation

$$
(8 - X)^2 = 64 - 16X + X^2 \\
(2 - Y)^2 = 4 - 4Y + Y^2
$$

Adding both:

$$
64 - 16X + X^2 + 4 - 4Y + Y^2 = 25
$$

Simplify:

$$
X^2 + Y^2 - 16X - 4Y + 68 = 25
$$

Subtract 25 from both sides:

$$
X^2 + Y^2 - 16X - 4Y + 43 = 0
$$

---

### Step 4: Matching the Form

The problem asks for the equation in the form:

$$
x^2 + a y^2 + b x + c y + d = 0
$$

Comparing with our derived equation:

$$
x^2 + y^2 - 16x - 4y + 43 = 0
$$

We identify the coefficients:

- $a = 1$ (coefficient of $y^2$)
- $b = -16$ (coefficient of $x$)
- $c = -4$ (coefficient of $y$)
- $d = 43$ (constant term)

---

### Final Answer

$$
\boxed{(1, -16, -4, 43)}
$$

Solution:
The center of the original circle is $(0,0).$  The reflection of the point $(0,0)$ in the point $(4,1)$ is $(8,2),$ so the equation of the new circle is
\[(x - 8)^2 + (y - 2)^2 = 25.\]This simplifies to $x^2 + y^2 - 16x - 4y + 43 = 0.$  Hence, $(a,b,c,d) = \boxed{(1,-16,-4,43)}.$