Okay, let me try to figure out how to solve this problem. So, we have an equilateral triangle inscribed in the parabola x² = 8y, and one of its vertices is at the vertex of the parabola, which is the origin (0,0). We need to find the side length of this triangle.

First, let me recall that an equilateral triangle has all sides equal and all angles equal to 60 degrees. Since one vertex is at the origin, the other two vertices must lie on the parabola. Let me denote the other two vertices as (a, b) and (c, d). Since they are on the parabola x² = 8y, their coordinates must satisfy this equation. So, for the first vertex, we have a² = 8b and c² = 8d.

Now, the triangle is equilateral, so the distance between each pair of vertices must be the same. The distance between (0,0) and (a, b) should be equal to the distance between (0,0) and (c, d), and also equal to the distance between (a, b) and (c, d). Let's write down these distance equations.

First, the distance from (0,0) to (a, b) is √(a² + b²). Similarly, the distance from (0,0) to (c, d) is √(c² + d²). Since these are equal, we have √(a² + b²) = √(c² + d²). Squaring both sides gives a² + b² = c² + d². But we already know from the parabola equation that a² = 8b and c² = 8d. So, substituting these in, we get 8b + b² = 8d + d². Let me rearrange this: b² - d² + 8b - 8d = 0. This can be factored as (b - d)(b + d) + 8(b - d) = 0. Factoring out (b - d), we get (b - d)(b + d + 8) = 0. So either b = d or b + d + 8 = 0.

If b = d, then the points (a, b) and (c, d) would be symmetric with respect to the y-axis, but since the parabola is symmetric about the y-axis, this might be a possible case. However, if b + d + 8 = 0, then since b and d are both coordinates on the parabola, which is x² = 8y. Since x² is always non-negative, y must be non-negative. Therefore, b and d are non-negative. So, b + d + 8 = 0 would imply that both b and d are negative, which is impossible. Hence, the only possibility is b = d. Therefore, the two points (a, b) and (c, d) must have the same y-coordinate. So, they are on the same horizontal line.

But wait, if they are on the same horizontal line, then the distance between them is |a - c|. However, since the triangle is equilateral, the distance between (a, b) and (c, d) should be equal to the distance from the origin to (a, b), which is √(a² + b²). But if a and c are symmetric with respect to the y-axis, then c = -a. So, the distance between (a, b) and (-a, b) is √[(a - (-a))² + (b - b)²] = √[(2a)²] = 2|a|. Therefore, we have 2|a| = √(a² + b²).

Let me square both sides to eliminate the square root: (2|a|)² = (√(a² + b²))² ⇒ 4a² = a² + b² ⇒ 3a² = b². But from the parabola equation, we have a² = 8b. So substituting this into 3a² = b² gives 3*(8b) = b² ⇒ 24b = b² ⇒ b² -24b =0 ⇒ b(b -24)=0. So, b=0 or b=24. But if b=0, then a²=0, so a=0, which would mean both points are at the origin, which is the same as the given vertex, so that's not possible for a triangle. Therefore, b=24. Then, a²=8b=8*24=192 ⇒ a=±√192=±8√3. Therefore, the two other vertices are at (8√3, 24) and (-8√3, 24). Then, the distance between these two points is 2*8√3=16√3, which should be the side length. However, let's check if the distance from the origin to (8√3,24) is equal to 16√3. Let's compute √[(8√3)^2 +24^2] = √[192 + 576] = √768 = √(256*3)=16√3. Yes, that works. So the side length is 16√3. Wait, but is this the only possibility?

Wait, but the problem says the triangle is inscribed in the parabola. So, the three vertices are on the parabola. We considered two points (a,b) and (c,d) with b=d=24, but is there another possibility where the two points are not symmetric? Because we concluded earlier that b=d, but maybe there's another case?

Wait, let's go back. When we had the equation (b - d)(b + d + 8)=0, we concluded that b=d because b + d +8=0 is impossible. But maybe there's a miscalculation here. Let me check again.

Original equation after substitution: b² - d² +8b -8d=0. Factored as (b - d)(b + d) +8(b - d)=0. Then, (b - d)(b + d +8)=0. So, either b=d or b + d +8=0. Since b and d are non-negative (as they are y-coordinates on the parabola x²=8y), then b + d +8=0 would imply b and d are negative, which is impossible. Therefore, only possibility is b=d. Therefore, the other two points must have the same y-coordinate. So, the triangle is symmetric with respect to the y-axis. Therefore, the two points are (a, b) and (-a, b). Then, the distance between them is 2a, and the distance from origin to each is √(a² + b²). So, 2a=√(a² + b²). Squaring both sides gives 4a² = a² + b² ⇒ 3a² = b². Also, since the point (a, b) is on the parabola, a²=8b. Therefore, substituting a²=8b into 3a² =b² gives 3*(8b)=b² ⇒24b =b² ⇒b²-24b=0 ⇒b(b-24)=0. So, b=0 or b=24. As before, b=24 is the valid solution. Therefore, a²=8*24=192 ⇒a=±8√3. So, the two other vertices are (8√3,24) and (-8√3,24). Then, the distance between these two points is 16√3, which is the side length. So, is that the answer? Wait, but let me confirm that the triangle is indeed equilateral.

Wait, the distance from the origin to (8√3,24) is √[(8√3)^2 +24^2] = √[192 + 576] = √768 = 16√3. The distance between (8√3,24) and (-8√3,24) is √[(16√3)^2 +0] =16√3. So, all sides are equal, so it's an equilateral triangle. Therefore, the side length is 16√3. But wait, that seems quite large. Let me check if there's another possibility where the triangle is not symmetric with respect to the y-axis.

Wait, but according to the earlier reasoning, the only possibility is that the two other points have the same y-coordinate. So, perhaps this is the only solution. However, maybe I missed something. Let me think again.

Suppose we have an equilateral triangle with one vertex at the origin and the other two on the parabola. The two other vertices must be symmetric with respect to the axis of symmetry of the parabola, which is the y-axis. Therefore, their x-coordinates must be negatives of each other, and their y-coordinates must be equal. So, the only possible configuration is symmetric with respect to the y-axis. Therefore, the triangle must be symmetric, which gives the solution we found. Therefore, the side length is 16√3. But let me confirm with another approach.

Alternatively, let's use coordinates. Let me denote the two other vertices as (a, b) and (c, d). Since they are on the parabola, a²=8b and c²=8d. The distances from the origin to both points must be equal, so √(a² + b²)=√(c² + d²). Also, the distance between (a, b) and (c, d) must be equal to the distance from the origin to either of them. So, √[(a - c)^2 + (b - d)^2] = √(a² + b²). Let me square both sides: (a - c)^2 + (b - d)^2 = a² + b². Expanding the left side: a² - 2ac + c² + b² - 2bd + d² = a² + b². Simplify: -2ac + c² -2bd + d² =0. But since a²=8b and c²=8d, substitute these into the equation: -2ac +8d -2bd +8b=0. Let's factor out terms: -2ac -2bd +8b +8d=0. Let me factor out -2: -2(ac + bd) +8(b + d)=0. So, 8(b + d) = 2(ac + bd). Dividing both sides by 2: 4(b + d) = ac + bd. Hmm, this seems a bit complicated. Maybe we can use the fact that the triangle is equilateral, so the angle between the vectors from the origin to the other two points is 60 degrees. Let me think in terms of vectors.

Let me denote vector OA = (a, b) and vector OB = (c, d). Since the triangle is equilateral, the angle between OA and OB is 60 degrees. The dot product formula gives OA · OB = |OA||OB|cosθ. Since |OA| = |OB|, and θ =60°, this becomes ac + bd = |OA|² * (1/2). But |OA|² = a² + b² =8b + b². Wait, from the parabola equation, a²=8b. So, |OA|² =8b + b². Therefore, the dot product ac + bd = (8b + b²)/2. But from the earlier equation, 4(b + d) = ac + bd. Therefore, substituting ac + bd from the dot product equation into the previous equation: 4(b + d) = (8b + b²)/2. Let's solve this equation.

Multiply both sides by 2: 8(b + d) =8b + b². Bring all terms to one side:8b +8d -8b -b²=0 ⇒8d -b²=0 ⇒8d =b². But from the parabola equation, we have c²=8d, so 8d =c². Therefore, 8d =b² ⇒c² =b². Therefore, c=±b. But since the parabola is symmetric about the y-axis, and we already considered c=-a, but here c=±b. Wait, this seems conflicting. Wait, c²=8d and 8d= b². Therefore, c² =b², so c=±b. But from the parabola equation, c²=8d, and d is a coordinate on the parabola, so d must be non-negative. Therefore, c=±b. But also, from the earlier equation, 4(b + d)=ac + bd. Let me substitute c=±b and d=b²/8 (since 8d= b²). Let's try c= b. Then, d= b²/8. Substitute into 4(b + d)=ac + bd. So, 4(b + b²/8)=a*b +b*(b²/8). Let's compute left side:4b + (4b²)/8=4b + (b²)/2. Right side: ab + (b³)/8. But from the parabola equation, a²=8b ⇒a=±√(8b). Let's take a=√(8b) for simplicity. Then, right side becomes √(8b)*b + (b³)/8. So, equation becomes:4b + (b²)/2 = b√(8b) + (b³)/8. Let's square both sides to eliminate the square roots, but this might get complicated. Alternatively, let's consider if c=-b. Then, d= b²/8. Then, 4(b + d)=ac + bd. Substitute c=-b and d= b²/8. Then, left side:4(b + b²/8)=4b + (4b²)/8=4b + (b²)/2. Right side: a*(-b) + b*(b²/8)= -ab + (b³)/8. So, equation:4b + (b²)/2 = -ab + (b³)/8. Again, from a²=8b, a=√(8b). So, substitute a=√(8b): equation becomes 4b + (b²)/2 = -√(8b)*b + (b³)/8. This seems messy. Let's see if there's a solution here.

Alternatively, let's assume that c=-a. Then, since c=-a, then from c²=8d, we have (-a)^2=8d ⇒a²=8d. But from the parabola equation, a²=8b. Therefore, 8b=8d ⇒b=d. So, this brings us back to the earlier case where b=d. Therefore, the only possible solution is when the two other points are symmetric about the y-axis, which gives the solution we found before. Therefore, the side length is 16√3.

But let me check with another approach. Suppose we parametrize the points on the parabola. For a parabola x²=8y, we can write parametric equations as (4t, 2t²). Wait, no. Wait, standard parametrization for x²=4py is (2pt, pt²). Wait, here x²=8y, so 4p=8 ⇒p=2. So, parametric equations would be (4t, 2t²). Let me verify: x=4t, y=2t². Then, x²=16t²=8*(2t²)=16t². Yes, correct. So, any point on the parabola can be written as (4t, 2t²). Therefore, the three vertices of the triangle can be written as (0,0), (4t, 2t²), and (4s, 2s²). Since the triangle is equilateral, all sides must be equal.

So, the distance between (0,0) and (4t, 2t²) is √[(4t)^2 + (2t²)^2] = √[16t² +4t⁴] = 2√(4t² + t⁴). Similarly, the distance between (0,0) and (4s, 2s²) is 2√(4s² + s⁴). The distance between (4t, 2t²) and (4s, 2s²) is √[(4t -4s)^2 + (2t² -2s²)^2] = √[16(t -s)^2 +4(t² -s²)^2] = 2√[(t -s)^2 + (t² -s²)^2].

Since the triangle is equilateral, all these distances must be equal. Therefore:

2√(4t² + t⁴) = 2√(4s² + s⁴) = 2√[(t -s)^2 + (t² -s²)^2].

Dividing all terms by 2, we get:

√(4t² + t⁴) = √(4s² + s⁴) = √[(t -s)^2 + (t² -s²)^2].

First, equate the first two terms:

√(4t² + t⁴) = √(4s² + s⁴). Squaring both sides:

4t² + t⁴ =4s² + s⁴.

Rearranging:

t⁴ -s⁴ +4t² -4s²=0.

Factor:

(t² -s²)(t² +s²) +4(t² -s²)=0.

Factor out (t² -s²):

(t² -s²)(t² +s² +4)=0.

Therefore, either t² = s² or t² +s² +4=0. The second equation is impossible since t² and s² are non-negative. Therefore, t² =s² ⇒ t=s or t=-s.

But if t=s, then the two points (4t, 2t²) and (4s, 2s²) coincide, which would not form a triangle. Therefore, t=-s. So, s=-t.

Therefore, the two points are (4t, 2t²) and (-4t, 2t²). Now, we need to check the distance between these two points and the origin.

First, distance between (4t, 2t²) and (-4t, 2t²):

√[(4t - (-4t))² + (2t² -2t²)^2] = √[(8t)^2 +0] =8|t|.

Distance between (4t, 2t²) and (0,0):

√[(4t)^2 + (2t²)^2] =√[16t² +4t⁴] =2√(4t² +t⁴).

Distance between (-4t, 2t²) and (0,0) is the same as above.

Since the triangle is equilateral, all sides must be equal:

8|t| =2√(4t² +t⁴).

Divide both sides by 2:

4|t| =√(4t² +t⁴).

Square both sides:

16t² =4t² +t⁴.

Simplify:

16t² -4t² -t⁴=0 ⇒12t² -t⁴=0 ⇒t²(12 -t²)=0.

Therefore, t²=0 or t²=12. t²=0 would mean t=0, which gives coinciding points, so discard. Therefore, t²=12 ⇒t=±2√3. Therefore, t=2√3 or t=-2√3. Since we can take t positive, let t=2√3. Then, the other points are (4t, 2t²)=(8√3, 2*(12))=(8√3,24) and (-4t,2t²)=(-8√3,24). Therefore, the distance between these two points is 8t=8*(2√3)=16√3, which is the side length. Therefore, this confirms the earlier result.

Therefore, the side length of the equilateral triangle is 16√3.

But wait, just to be thorough, let me check if there are any other possible configurations where the triangle is not symmetric with respect to the y-axis. For example, suppose the two other points are not symmetric. But earlier reasoning suggests that due to the parabola's symmetry and the equilateral triangle's properties, the triangle must be symmetric. However, let me consider another approach by using coordinates and angles.

Another way to approach this is to consider the coordinates of the points. Let’s denote the two other vertices as (a, b) and (c, d). Since the triangle is equilateral, all sides are equal, so the distance between (a,b) and (c,d) must equal the distance from the origin to each of these points. Also, the angles between the vectors OA and OB must be 60 degrees. Let’s use vectors here. The vectors OA = (a, b) and OB = (c, d). The angle between OA and OB is 60 degrees. The dot product formula gives OA · OB = |OA||OB|cosθ. Since θ=60°, this becomes ac + bd = |OA|² * 0.5. Also, since |OA| = |OB|, we have |OA|² = a² + b² =8b (from the parabola equation). Therefore, ac + bd = (8b + b²)/2. But we also know that since the distance between A and B is equal to the distance from the origin to A, we have √[(a - c)^2 + (b - d)^2] = √(a² + b²). Squaring both sides, (a - c)^2 + (b - d)^2 = a² + b². Expanding the left side: a² -2ac + c² + b² -2bd + d² = a² + b². Simplifying: -2ac + c² -2bd + d² =0. From the parabola, c²=8d and a²=8b, so substitute these into the equation: -2ac +8d -2bd +8b =0. Factor terms: -2ac -2bd +8(b + d)=0. Divide by -2: ac + bd -4(b + d)=0 ⇒ac + bd =4(b + d). But we also have from the dot product: ac + bd = (8b + b²)/2. Therefore, equate the two expressions: (8b + b²)/2 =4(b + d). Multiply both sides by 2:8b +b²=8(b + d). Rearranged:8b +b² -8b -8d=0 ⇒b² -8d=0 ⇒b²=8d. But from parabola equation, d =c²/8. Therefore, substituting into b²=8d gives b²=8*(c²/8)=c². Therefore, c=±b. But since c²=8d, which is equal to b², we have c=±b. However, from the parabola equation, c²=8d and b²=8d, so c=±b. Therefore, this leads us back to the previous conclusion that c=±b. Then, using c=±b, and substituting into the equation b²=8d, we have d= b²/8. Also, since from the earlier equation, ac + bd=4(b + d). Let’s substitute c=±b and d= b²/8 into this equation. Let’s take c=b first. Then, ac + bd= a*b + b*(b²/8)=ab + (b³)/8. From the parabola equation, a²=8b ⇒a=±√(8b). Let’s take a=√(8b). Then, ab=√(8b)*b= b√(8b). Therefore, ac + bd= b√(8b) + (b³)/8. From the other equation, ac + bd=4(b + d)=4(b + b²/8)=4b + (4b²)/8=4b + (b²)/2. Therefore, we have:

b√(8b) + (b³)/8 =4b + (b²)/2.

This seems complex, but let's try to solve for b. Let me denote k =√b, so that b =k². Then, substitute into the equation:

k²√(8k²) + (k^6)/8 =4k² + (k^4)/2.

Simplify √(8k²) =√8 *k=2√2 *k. Therefore, first term becomes k²*(2√2 k)=2√2 k³. The second term is (k^6)/8. The right side is 4k² + (k^4)/2. So, equation becomes:

2√2 k³ + (k^6)/8 =4k² + (k^4)/2.

Multiply both sides by 8 to eliminate denominators:

16√2 k³ +k^6 =32k² +4k^4.

Bring all terms to one side:

k^6 -4k^4 +16√2 k³ -32k²=0.

Factor out k²:

k²(k^4 -4k² +16√2 k -32)=0.

Therefore, either k=0 (which gives b=0, leading to coinciding points) or solve the quartic equation:

k^4 -4k² +16√2 k -32=0.

This quartic equation seems difficult to solve. Let me check if there are any rational roots. By Rational Root Theorem, possible roots are factors of 32 over factors of 1: ±1, ±2, ±4, ±8, ±16, ±32. Let's test k=2:

2^4 -4*2² +16√2*2 -32=16 -16 +32√2 -32=0 +32√2 -32≈32*1.414 -32≈45.25 -32≈13.25≠0.

k=√2:

(√2)^4 -4*(√2)^2 +16√2*(√2) -32=4 -8 +16*2 -32=4 -8 +32 -32= -4≠0.

k=1:

1 -4 +16√2 -32≈1 -4 +22.627 -32≈-14.373≠0.

k=4:

256 -64 +16√2*4 -32=192 +64√2 -32≈192 +90.51 -32≈250.51≠0.

k= -2:

16 -16 +16√2*(-2) -32=0 -32√2 -32≈-45.25 -32≈-77.25≠0.

So, no rational roots. Therefore, this approach might not be fruitful. Let's try another substitution. Let me set z =k². Then, k^4 =z², k=√z.

Then, the equation becomes:

z² -4z +16√2 √z -32=0.

This still looks complicated. Maybe another substitution. Let me set t=√z. Then, z=t², and the equation becomes:

(t²)^2 -4t² +16√2 t -32=0 ⇒t^4 -4t² +16√2 t -32=0.

Still a quartic equation. This might not be the right path. Let me consider the case when c=-b instead. Then, d= b²/8, and c=-b. Then, ac + bd= a*(-b) +b*(b²/8)= -ab + (b³)/8. From parabola equation, a²=8b ⇒a=±√(8b). Let's take a=√(8b). Then, -ab= -√(8b)*b= -b√(8b). Therefore, ac + bd= -b√(8b) + (b³)/8. From the other equation, ac + bd=4(b + d)=4(b + b²/8)=4b + (4b²)/8=4b + (b²)/2. Therefore:

-b√(8b) + (b³)/8 =4b + (b²)/2.

Again, let me substitute k=√b, so b=k². Then:

- k²√(8k²) + (k^6)/8 =4k² + (k^4)/2.

Simplify √(8k²)=2√2 k. Therefore:

- k²*(2√2 k) + (k^6)/8 =4k² + (k^4)/2 ⇒-2√2 k³ + (k^6)/8 =4k² + (k^4)/2.

Multiply both sides by 8:

-16√2 k³ +k^6 =32k² +4k^4.

Bring all terms to one side:

k^6 -4k^4 -16√2 k³ -32k²=0.

Factor out k²:

k²(k^4 -4k² -16√2 k -32)=0.

Again, k=0 (b=0) gives coinciding points. Otherwise, solve k^4 -4k² -16√2 k -32=0. Again, not easily solvable with rational roots. Trying k=2:

16 -16 -32√2 -32≈16 -16 -45.25 -32≈-77.25≠0.

k=√2:

4 -4 -16√2*√2 -32=0 -32 -32= -64≠0.

k=1:

1 -4 -16√2 -32≈-35 -22.627≈-57.627≠0.

Similarly, no rational roots. Therefore, this approach does not seem to yield a solution. Therefore, the only feasible solution is the symmetric case where the two other points are (8√3,24) and (-8√3,24), leading to a side length of 16√3.

Therefore, after thorough analysis, the side length of the equilateral triangle is 16√3.

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