Explanation:
Okay, so I have this problem about a regular tetrahedron with side length 2. The question says that there's a plane parallel to edges AB and CD, and it's lying halfway between them. This plane cuts the tetrahedron into two pieces, and I need to find the surface area of one of these pieces. Alright, let me try to visualize this first.

First, let me recall what a regular tetrahedron is. It's a three-dimensional shape with four triangular faces, each of which is an equilateral triangle. All edges are equal, so every edge here is length 2. The vertices are A, B, C, D, and each pair of vertices is connected by an edge of length 2.

Now, the plane in question is parallel to edges AB and CD. Hmm. So edges AB and CD are skew lines in a tetrahedron, right? Because in a tetrahedron, any two edges that don't share a common vertex are skew. Let me confirm that. AB connects A and B, and CD connects C and D. Since A, B, C, D are all vertices of the tetrahedron, AB and CD don't share a common vertex. So they are indeed skew lines.

So the plane is parallel to both AB and CD. Since AB and CD are skew, the plane can't be parallel to both unless it's somehow in a specific orientation. Wait, but planes can be parallel to lines even if the lines are skew. For example, a plane can be parallel to two skew lines if it's parallel to the direction of both lines. So, if the plane is parallel to both AB and CD, then it must be parallel to the direction of both edges AB and CD. Therefore, the plane's normal vector is perpendicular to the direction vectors of AB and CD. Maybe I need to think about coordinates here.

Maybe assigning coordinates to the tetrahedron will help. Let me try that. Let me recall that coordinates for a regular tetrahedron can be set up in 3D space. One common coordinate system for a regular tetrahedron is to have vertices at (1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1), scaled appropriately. But maybe there's a simpler way. Alternatively, maybe I can place the tetrahedron with base ABC and apex D. Let me try that.

Let me consider the base triangle ABC. Let me put point A at (0, 0, 0), point B at (2, 0, 0), since the side length is 2. Now, since ABC is an equilateral triangle, point C will be somewhere in the xy-plane. Let me calculate its coordinates. The coordinates of C can be found by knowing that the distance from A to C is 2, and from B to C is 2. Let me denote C as (x, y, 0). Then, the distance from A to C is sqrt(x² + y²) = 2, and from B to C is sqrt((x - 2)² + y²) = 2. So:

x² + y² = 4,

and

(x - 2)² + y² = 4.

Subtracting the first equation from the second:

(x - 2)² + y² - x² - y² = 4 - 4 => (x² -4x +4) -x² = 0 => -4x +4 = 0 => x = 1.

So x is 1. Then from x² + y² = 4, we get 1 + y² = 4 => y² = 3 => y = sqrt(3) or -sqrt(3). Let me take y = sqrt(3) to place the base triangle in the xy-plane. So point C is at (1, sqrt(3), 0).

Now, the apex D of the tetrahedron. Since it's a regular tetrahedron, all edges are length 2. So the coordinates of D must satisfy the distances from A, B, and C. Let me denote D as (p, q, r). Then the distance from D to A is sqrt((p - 0)^2 + (q - 0)^2 + (r - 0)^2) = sqrt(p² + q² + r²) = 2. Similarly, distance from D to B is sqrt((p - 2)^2 + q² + r²) = 2, and distance from D to C is sqrt((p - 1)^2 + (q - sqrt(3))² + r²) = 2.

So let me write down these equations:

1. p² + q² + r² = 4.

2. (p - 2)^2 + q² + r² = 4.

3. (p - 1)^2 + (q - sqrt(3))² + r² = 4.

Subtract equation 1 from equation 2:

(p - 2)^2 - p² = 0 => p² -4p +4 - p² = -4p +4 = 0 => p = 1.

So p = 1. Then, substituting p = 1 into equation 1:

1² + q² + r² = 4 => q² + r² = 3.

Now equation 3 becomes:

(1 - 1)^2 + (q - sqrt(3))² + r² = 4 => 0 + (q - sqrt(3))² + r² = 4.

But since q² + r² = 3, substitute that into equation 3:

(q - sqrt(3))² + r² = q² - 2 q sqrt(3) + 3 + r² = (q² + r²) - 2 q sqrt(3) + 3 = 3 - 2 q sqrt(3) + 3 = 6 - 2 q sqrt(3) = 4.

Therefore, 6 - 2 q sqrt(3) = 4 => -2 q sqrt(3) = -2 => q sqrt(3) = 1 => q = 1 / sqrt(3) = sqrt(3)/3.

Then, from q² + r² = 3:

( (sqrt(3)/3)^2 ) + r² = 3 => (3/9) + r² = 3 => 1/3 + r² = 3 => r² = 3 - 1/3 = 8/3 => r = ± 2*sqrt(6)/3. Since a tetrahedron can be oriented either way, but let's take r positive, so r = 2√6 / 3.

Therefore, coordinates of D are (1, sqrt(3)/3, 2√6/3). Let me confirm that this makes sense. Let me check the distance from D to A: sqrt( (1)^2 + (sqrt(3)/3)^2 + (2√6/3)^2 ). Calculating:

1 + (3/9) + (24/9) = 1 + 1/3 + 8/3 = 1 + 9/3 = 1 + 3 = 4. Yes, sqrt(4) = 2. Good. Similarly, distance from D to B would be same. Let me check distance from D to C:

Coordinates of C: (1, sqrt(3), 0). Coordinates of D: (1, sqrt(3)/3, 2√6/3). So difference in x: 0, difference in y: sqrt(3)/3 - sqrt(3) = -2 sqrt(3)/3, difference in z: 2√6/3 - 0 = 2√6/3.

So squared distance: 0 + ( -2 sqrt(3)/3 )² + ( 2√6 / 3 )² = (4 * 3)/9 + (4 * 6)/9 = 12/9 + 24/9 = 36/9 = 4. So sqrt(4) = 2. Perfect. So coordinates are correct.

So now, the tetrahedron has coordinates:

A(0, 0, 0)

B(2, 0, 0)

C(1, sqrt(3), 0)

D(1, sqrt(3)/3, 2√6/3)

Alright, so now we need to find a plane that's parallel to edges AB and CD, and halfway between them. Let me first figure out what it means for a plane to be parallel to edges AB and CD. Since AB is the edge from A(0,0,0) to B(2,0,0), its direction vector is (2,0,0) - (0,0,0) = (2,0,0). Similarly, edge CD goes from C(1, sqrt(3), 0) to D(1, sqrt(3)/3, 2√6/3). So direction vector is D - C = (1 - 1, sqrt(3)/3 - sqrt(3), 2√6/3 - 0) = (0, -2 sqrt(3)/3, 2√6/3).

So direction vectors AB is (2, 0, 0) and CD is (0, -2√3/3, 2√6/3). Now, the plane is parallel to both AB and CD. So the plane must contain direction vectors AB and CD. Wait, no. Wait, a plane that's parallel to two lines can be parallel to their direction vectors. So the plane's direction vectors are the same as AB and CD's direction vectors. However, since AB and CD are skew, their direction vectors are not in the same plane, so the plane must be parallel to both direction vectors. Therefore, the normal vector of the plane is perpendicular to both direction vectors of AB and CD.

Alternatively, since the plane is parallel to AB and CD, then the plane can be defined by containing lines parallel to AB and CD. However, since AB and CD are skew, there is no plane that contains both AB and CD. But there can be planes that are parallel to both AB and CD. So the plane in question is parallel to both AB and CD, meaning that the direction vectors of AB and CD are direction vectors of the plane.

So, given that, the plane has a normal vector that is perpendicular to both direction vectors of AB and CD. Let me compute that. Let me denote direction vector AB as vector u = (2, 0, 0), and direction vector CD as vector v = (0, -2√3/3, 2√6/3). Then, normal vector n = u × v.

Compute cross product u × v:

u × v = |i   j   k|
         |2   0   0|
         |0 -2√3/3 2√6/3|

Calculating determinant:

i*(0*2√6/3 - 0*(-2√3/3)) - j*(2*2√6/3 - 0*0) + k*(2*(-2√3/3) - 0*0)

Simplify:

i*(0 - 0) - j*(4√6/3 - 0) + k*(-4√3/3 - 0)

So n = (0, -4√6/3, -4√3/3)

Alternatively, we can write it as n = (0, -4√6, -4√3) divided by 3, but for the purposes of direction, it's okay. So the normal vector of the plane is (0, -4√6/3, -4√3/3). However, for the plane equation, we can use this normal vector.

But maybe I can simplify or find a direction vector. However, maybe I need to find the equation of the plane. But we also know that the plane is halfway between AB and CD. So how do we determine "halfway between" AB and CD?

Since AB and CD are skew lines, the distance between them is some value, and the plane is halfway between them. But how do we define distance between skew lines? The distance between AB and CD can be calculated, then the plane is at halfway between them. Alternatively, since the plane is parallel to both, maybe it's halfway in terms of some parameter along the common perpendicular?

Alternatively, since the plane is parallel to both lines, it might be that the plane intersects the tetrahedron in a certain way, and the "halfway between AB and CD" refers to the plane being equidistant to AB and CD?

But how do we define the distance from a line to a plane? Wait, if the plane is parallel to AB and CD, then the distance from AB to the plane should be equal to the distance from CD to the plane? Because the plane is halfway between AB and CD. So the plane is equidistant to both AB and CD, and is parallel to both.

Therefore, if I can find the plane that is equidistant to AB and CD and parallel to both. So perhaps I can parametrize the plane.

Alternatively, since the plane is parallel to AB and CD, then it must be parallel to the direction vectors AB and CD. Therefore, as earlier, the plane can be parametrized as follows: Let me think.

Alternatively, maybe I can parametrize the plane by finding a point on it. Since it is halfway between AB and CD, perhaps we can take the line segment connecting AB and CD? Wait, but AB and CD are skew, so there is no line segment connecting them. However, there is a common perpendicular between AB and CD. The plane is parallel to both AB and CD, so it must be parallel to the common perpendicular? Wait, perhaps not. Maybe the plane is somewhere in between, such that the distance from AB to the plane equals the distance from CD to the plane.

Alternatively, since AB and CD are skew, there is a unique line that is perpendicular to both, and the distance between AB and CD is the distance between these two lines, which is the distance between the plane and the line AB and CD. Wait, maybe I need to compute the distance between AB and CD first. Let me try that.

The distance between skew lines AB and CD can be found by taking the absolute value of the scalar triple product of vectors AB, AC, and AD divided by the magnitude of the cross product of AB and CD? Wait, let me recall the formula.

The formula for the distance between two skew lines is |(b - a) ⋅ (n)| / |n|, where n is the cross product of the direction vectors of the two lines, and a and b are points on each line.

Let me confirm. Let me denote line AB and line CD. Let me take point A on AB and point C on CD. Then vector AC is from A to C. The distance between AB and CD is |AC ⋅ (u × v)| / |u × v|, where u and v are direction vectors of AB and CD. Wait, yes, I think that's correct.

So let me compute that. Let me denote:

u = direction vector of AB = (2, 0, 0)

v = direction vector of CD = (0, -2√3/3, 2√6/3)

Then, n = u × v = (0, -4√6/3, -4√3/3) as before.

Then, vector AC is from A to C: C - A = (1, sqrt(3), 0) - (0, 0, 0) = (1, sqrt(3), 0).

Then, the distance between AB and CD is |AC ⋅ n| / |n|.

First compute AC ⋅ n:

AC is (1, sqrt(3), 0). n is (0, -4√6/3, -4√3/3).

Dot product = 1*0 + sqrt(3)*(-4√6/3) + 0*(-4√3/3) = 0 - (4 sqrt(18))/3 + 0 = - (4 * 3 * sqrt(2))/3 = -4 sqrt(2).

So absolute value is 4 sqrt(2).

Now compute |n|:

n = (0, -4√6/3, -4√3/3)

So |n| = sqrt( (0)^2 + (-4√6/3)^2 + (-4√3/3)^2 ) = sqrt( (16*6)/9 + (16*3)/9 ) = sqrt( (96 + 48)/9 ) = sqrt(144/9) = sqrt(16) = 4.

Therefore, the distance between AB and CD is |AC ⋅ n| / |n| = 4√2 / 4 = √2. Therefore, the distance between AB and CD is √2. So the plane halfway between them would be at distance √2 / 2 from each of AB and CD.

Therefore, the plane is located at a distance of √2 / 2 from AB and CD, and is parallel to both. Now, how do I find the equation of this plane?

Since the plane is parallel to AB and CD, as we determined earlier, its normal vector is n = (0, -4√6/3, -4√3/3). However, we can use a simpler normal vector by dividing by the scalar. Since the direction vector of the plane is along n, but perhaps we can use a simplified version. Alternatively, since the plane is parallel to AB and CD, and we need to find the plane at distance √2 / 2 from both AB and CD. Maybe we can find the equation of such a plane.

Alternatively, since the plane is parallel to AB and CD, then it can be represented parametrically. However, since it's halfway between AB and CD, perhaps we can find a point that lies on the plane. Let me think. Since the plane is halfway between AB and CD, maybe it intersects some edges of the tetrahedron? Let me think about how the plane cuts the tetrahedron.

Alternatively, since the plane is parallel to AB and CD, which are skew lines, then the intersection of the plane with the tetrahedron must form a quadrilateral? Because cutting a tetrahedron with a plane parallel to two edges... Wait, a regular tetrahedron has four triangular faces, so cutting with a plane that's parallel to two edges... Wait, but maybe the intersection is a line? Or maybe a quadrilateral?

Wait, if the plane is parallel to two edges AB and CD, then it can't intersect those edges, since they are parallel to the plane. Therefore, the plane would intersect the other edges of the tetrahedron. Let me think about the edges of the tetrahedron. The tetrahedron has edges AB, AC, AD, BC, BD, CD. So since the plane is parallel to AB and CD, it doesn't intersect AB or CD, so it must intersect the other edges: AC, AD, BC, BD. Wait, but maybe? Or maybe not?

Wait, if the plane is parallel to AB and CD, then it can't intersect AB or CD. But depending on where the plane is, it can intersect other edges. Let me think. For example, if we imagine moving a plane parallel to AB and CD from one side to the other, it would intersect some edges. Since AB is one edge, and CD is another, which are skew. So the plane would intersect the edges connecting AB to CD. These edges are AC, BD, and also maybe others? Wait, the tetrahedron's edges are AB, AC, AD, BC, BD, CD. So edges adjacent to AB are AC and AD, and edges adjacent to CD are BC and BD. So maybe the plane intersects AC, BD, and also maybe other edges?

Alternatively, since the plane is parallel to AB and CD, which are skew. Maybe the intersection is a quadrilateral? For instance, if the plane intersects edges AC, BD, and also edges AD and BC? Wait, maybe not. Let me try to think.

Alternatively, let me try to parametrize the plane. Let me consider parametric equations.

Since the plane is parallel to AB and CD, it has direction vectors AB and CD. However, as AB and CD are skew, we need another point to define the plane. But since the plane is halfway between AB and CD, perhaps we can find a point on the plane by moving halfway from AB to CD along the common perpendicular.

Alternatively, since the distance between AB and CD is √2, and the plane is halfway, which is √2 / 2 away from each. Therefore, perhaps we can find a point that is on the common perpendicular line midway between AB and CD, and then define the plane through that point with normal vector n.

Alternatively, since we have coordinates for AB and CD, maybe we can find such a plane.

Alternatively, maybe I can parametrize the plane as follows: since it's parallel to AB and CD, its equation can be written in terms of direction vectors. Let me think.

Let me suppose the plane has equation ax + by + cz = d. Since it's parallel to AB, which has direction vector (2, 0, 0), then the normal vector (a, b, c) must be perpendicular to AB. Therefore, (a, b, c) ⋅ (2, 0, 0) = 2a = 0 ⇒ a = 0. Similarly, since the plane is parallel to CD, whose direction vector is (0, -2√3/3, 2√6/3), then the normal vector must also be perpendicular to this direction vector. Therefore, (a, b, c) ⋅ (0, -2√3/3, 2√6/3) = 0 * a + (-2√3/3)*b + (2√6/3)*c = 0. Since we already have a = 0, this reduces to (-2√3/3)*b + (2√6/3)*c = 0 ⇒ -√3 b + √6 c = 0 ⇒ √6 c = √3 b ⇒ c = (√3 / √6) b = (1/√2) b. Therefore, c = b / √2. Therefore, the normal vector is (0, b, b / √2). We can choose b = √2 to make c = 1. Therefore, normal vector is (0, √2, 1). So the plane equation is 0x + √2 y + z = d. Therefore, √2 y + z = d.

Now, we need to find d such that the plane is halfway between AB and CD. So how do we find d?

Earlier, we computed that the distance between AB and CD is √2, so the plane is at distance √2 / 2 from each. Let me compute the distance from AB to the plane √2 y + z = d, and set it equal to √2 / 2.

But since AB is the line from A(0,0,0) to B(2,0,0). We can use the formula for the distance from a line to a plane. Wait, but actually, since the plane is parallel to AB, the distance from AB to the plane is constant along AB. So we can compute the distance from any point on AB to the plane.

Let me pick a point on AB, say point A(0,0,0). The distance from A to the plane √2 y + z = d is |√2 * 0 + 0 - d| / sqrt( (√2)^2 + 1^2 ) = | -d | / sqrt(2 + 1) = |d| / sqrt(3). Similarly, we can compute the distance from the plane to line CD. Let me pick a point on CD, say point C(1, sqrt(3), 0). The distance from point C to the plane is |√2 * sqrt(3) + 0 - d| / sqrt(3) = |√6 - d| / sqrt(3). Since the plane is midway, these distances should be equal to √2 / 2. Wait, but the plane is halfway between AB and CD, so the distance from AB to the plane is equal to distance from CD to the plane. But since AB and CD are on opposite sides of the plane? Wait, actually, depending on the orientation, but since the distance between AB and CD is √2, then the distance from AB to the plane plus the distance from the plane to CD is √2. Since the plane is halfway, each distance is √2 / 2. Therefore:

From AB's distance: |d| / sqrt(3) = √2 / 2.

From CD's distance: |√6 - d| / sqrt(3) = √2 / 2.

Therefore, we have two equations:

1) |d| = (√2 / 2) * sqrt(3) = √6 / 2.

2) |√6 - d| = (√2 / 2) * sqrt(3) = √6 / 2.

So solving these equations. Let me consider equation 1: |d| = √6 / 2 ⇒ d = ±√6 / 2.

Now, equation 2: |√6 - d| = √6 / 2. Let me consider both possibilities for d.

Case 1: d = √6 / 2.

Then |√6 - √6 / 2| = |√6 / 2| = √6 / 2, which matches the equation. Good.

Case 2: d = -√6 / 2.

Then |√6 - (-√6 / 2)| = |√6 + √6 / 2| = |(3√6)/2| = 3√6 / 2, which is not equal to √6 / 2. Therefore, case 2 is invalid. Therefore, d = √6 / 2. Therefore, the equation of the plane is √2 y + z = √6 / 2.

Therefore, the plane is √2 y + z = √6 / 2.

Now, let me check if this makes sense. Let me check the distance from point A(0,0,0) to the plane: |√2*0 + 0 - √6 / 2| / sqrt( (√2)^2 + 1^2 ) = | -√6 / 2 | / sqrt(3) = (√6 / 2) / sqrt(3) ) = (√6 / 2) / (√3) ) = (√2 / 2), which is indeed √2 / 2. Similarly, distance from point C(1, sqrt(3), 0): |√2 * sqrt(3) + 0 - √6 / 2| / sqrt(3) = |√6 - √6 / 2| / sqrt(3) = |√6 / 2| / sqrt(3) = same as above, √2 / 2. Perfect. So this plane is indeed halfway between AB and CD.

Therefore, now I have the plane equation: √2 y + z = √6 / 2.

Now, the next step is to find how this plane cuts the tetrahedron into two pieces, and find the surface area of one of these pieces.

First, I need to figure out how the plane intersects the tetrahedron. Since the plane is parallel to AB and CD, and doesn't intersect them, as established earlier, it must intersect the other edges. Let me list all edges of the tetrahedron again:

AB, AC, AD, BC, BD, CD.

Plane doesn't intersect AB or CD, so it must intersect AC, AD, BC, BD. Wait, let me check each edge.

Edge AC: from A(0,0,0) to C(1, sqrt(3), 0). Let me find if the plane intersects this edge.

Parametrize AC: points on AC can be written as A + t(C - A) = (0 + t*1, 0 + t*sqrt(3), 0 + t*0) = (t, t sqrt(3), 0), where t ∈ [0,1].

Plug into plane equation: √2 y + z = √6 / 2 ⇒ √2*(t sqrt(3)) + 0 = √6 / 2 ⇒ t * sqrt(6) = √6 / 2 ⇒ t = 1/2. Therefore, the plane intersects AC at t = 1/2, which is point (1/2, (sqrt(3))/2, 0). Let me denote this as point P.

Similarly, check edge AD: from A(0,0,0) to D(1, sqrt(3)/3, 2√6/3). Parametrize as A + s(D - A) = (s*1, s*sqrt(3)/3, s*2√6/3), s ∈ [0,1].

Plug into plane equation: √2 y + z = √6 / 2.

Compute:

√2*(s*sqrt(3)/3) + (s*2√6/3) = √6 / 2.

Let me compute left side:

s [ √2 * sqrt(3)/3 + 2√6 / 3 ] = s [ (sqrt(6)/3 + 2 sqrt(6)/3 ) ] = s [ 3 sqrt(6)/3 ] = s sqrt(6).

Therefore, s sqrt(6) = √6 / 2 ⇒ s = 1/2. Therefore, the plane intersects AD at s = 1/2. The coordinates are:

x = 1*(1/2) = 1/2,

y = (sqrt(3)/3)*(1/2) = sqrt(3)/6,

z = (2√6 / 3)*(1/2) = √6 / 3.

So point Q(1/2, sqrt(3)/6, √6 / 3).

Next, check edge BC: from B(2, 0, 0) to C(1, sqrt(3), 0). Parametrize as B + r(C - B) = (2 - r, 0 + r*sqrt(3), 0 + r*0) = (2 - r, r sqrt(3), 0), r ∈ [0,1].

Plug into plane equation: √2 y + z = √6 / 2 ⇒ √2*(r sqrt(3)) + 0 = √6 / 2 ⇒ r sqrt(6) = √6 / 2 ⇒ r = 1/2. Therefore, intersection at r = 1/2. Coordinates: (2 - 1/2, (sqrt(3)/2), 0) = (3/2, sqrt(3)/2, 0). Let me call this point R.

Similarly, check edge BD: from B(2, 0, 0) to D(1, sqrt(3)/3, 2√6/3). Parametrize as B + u(D - B) = (2 - u, 0 + u*(sqrt(3)/3 - 0), 0 + u*(2√6/3 - 0)) = (2 - u, u*sqrt(3)/3, 2√6 u / 3), u ∈ [0,1].

Plug into plane equation √2 y + z = √6 / 2:

√2*(u sqrt(3)/3) + (2√6 u / 3) = √6 / 2.

Left side: u [ √2 * sqrt(3)/3 + 2√6 / 3 ] = same as before: u [ sqrt(6)/3 + 2 sqrt(6)/3 ] = u [ sqrt(6) ]. Therefore:

u sqrt(6) = √6 / 2 ⇒ u = 1/2.

Therefore, the plane intersects BD at u = 1/2. Coordinates:

x = 2 - 1/2 = 3/2,

y = (sqrt(3)/3)*(1/2) = sqrt(3)/6,

z = (2√6 / 3)*(1/2) = √6 / 3.

So point S(3/2, sqrt(3)/6, √6 / 3).

Therefore, the plane intersects the tetrahedron in four points: P(1/2, sqrt(3)/2, 0), Q(1/2, sqrt(3)/6, √6 / 3), R(3/2, sqrt(3)/2, 0), and S(3/2, sqrt(3)/6, √6 / 3). Therefore, the intersection is a quadrilateral PQRS.

Wait, let me check if PQRS is a parallelogram? Since the plane is parallel to AB and CD, maybe the intersection is a parallelogram. Let me check the coordinates.

Let me compute vectors PQ and RS.

Point P(1/2, sqrt(3)/2, 0), Q(1/2, sqrt(3)/6, √6 / 3), R(3/2, sqrt(3)/2, 0), S(3/2, sqrt(3)/6, √6 / 3).

Vector PQ: Q - P = (0, sqrt(3)/6 - sqrt(3)/2, √6/3 - 0) = (0, -sqrt(3)/3, √6 / 3).

Vector RS: S - R = (3/2 - 3/2, sqrt(3)/6 - sqrt(3)/2, √6 / 3 - 0) = (0, -sqrt(3)/3, √6 / 3). So PQ and RS are equal vectors. Similarly, vector QR: R - Q = (3/2 - 1/2, sqrt(3)/2 - sqrt(3)/6, 0 - √6 / 3) = (1, sqrt(3)/3, -√6 / 3).

Vector SP: P - S = (1/2 - 3/2, sqrt(3)/2 - sqrt(3)/6, 0 - √6 / 3) = (-1, sqrt(3)/3, -√6 / 3). So QR and SP are negatives of each other. Therefore, PQRS is a parallelogram. So the intersection is a parallelogram.

Therefore, when the plane cuts the tetrahedron, it creates a smaller piece (probably a smaller tetrahedron and a frustum-like shape?) Wait, no. Since the plane cuts through four edges, creating a quadrilateral cross-section, the two pieces are a smaller polyhedron with a quadrilateral face and the rest of the original tetrahedron. But since the original tetrahedron is cut by a plane, the two pieces will be: one piece is the part containing the vertex D (since the plane is halfway between AB and CD, so maybe not?), and the other piece is the part containing AB. Wait, need to visualize.

Alternatively, perhaps the plane divides the tetrahedron into a smaller tetrahedron and a pentahedron? Wait, since the intersection is a quadrilateral, the upper piece (assuming the plane is cutting through) would have a quadrilateral face and three original faces? Wait, maybe not. Let me think.

Wait, the original tetrahedron has four triangular faces. The cutting plane intersects four edges, so the lower piece (assuming the plane is between AB and CD) would have the original base face ABC? Wait, no, the base is ABC, but the plane cuts through AC and BC, so the lower piece might have a quadrilateral face PQRS and parts of the original faces. Alternatively, perhaps it's easier to think of the two pieces: one is a smaller tetrahedron and the other is a kind of prism or something. But since the plane is parallel to AB and CD, which are skew lines, and halfway between them, perhaps the intersection is a parallelogram, and the two pieces are a prism with the parallelogram and another piece?

Alternatively, since the plane cuts through four edges, maybe each piece has a quadrilateral and three triangular faces? Wait, let me count the faces. The original tetrahedron has four triangular faces. After cutting, each piece will have some original faces and the new quadrilateral face. For example, the piece containing vertex D would have the original face ABD, original face ACD, and parts of the other faces? Wait, no, since the cutting plane cuts through edges AD and BD, so the piece containing D would have the quadrilateral PQRS and the original face ADC? Wait, maybe not. Let me try to figure out.

Alternatively, maybe it's better to think about which vertices are on which side of the plane. Let me check the position of each vertex relative to the plane √2 y + z = √6 / 2.

Compute for each vertex:

A(0,0,0): √2*0 + 0 = 0 < √6 / 2 ≈ 1.2247. So A is on one side.

B(2,0,0): √2*0 + 0 = 0 < √6 / 2. So B is on the same side as A.

C(1, sqrt(3), 0): √2*sqrt(3) + 0 ≈ 2.449 > √6 / 2 ≈ 1.2247. So C is on the other side.

D(1, sqrt(3)/3, 2√6/3): √2*(sqrt(3)/3) + 2√6 / 3.

Compute:

√2 * sqrt(3)/3 = sqrt(6)/3 ≈ 0.816,

2√6 / 3 ≈ 1.632,

Total ≈ 0.816 + 1.632 ≈ 2.448 ≈ √6 ≈ 2.449, which is greater than √6 / 2. Therefore, D is also on the same side as C.

Therefore, the plane divides the tetrahedron into two parts: one part contains A and B (and the other part contains C and D). Therefore, the piece containing A and B would have the original faces AB, and parts of the other faces, and the quadrilateral cross-section PQRS. Similarly, the piece containing C and D would have the original faces CD and parts of other faces, and the quadrilateral cross-section.

But since the problem says "the plane... cuts ABCD into two pieces", and asks for the surface area of one of these pieces. So we need to calculate the surface area of one of them. Since surface area is additive, but we need to be careful which piece we choose.

But depending on which piece we take, the surface area will include different faces. However, given that the plane is halfway between AB and CD, it's possible that the piece containing A and B has a certain surface area, and the piece containing C and D has another. However, since the original tetrahedron has all edges equal, maybe the two pieces are congruent? But since the plane is between AB and CD, which are skew, maybe not. However, given the regular tetrahedron is symmetric, perhaps the surface areas are the same? Wait, but AB and CD are not symmetric in the tetrahedron. Wait, AB is an edge, CD is another edge, but the tetrahedron is regular, so all edges are the same. Therefore, the two pieces should have congruent surface areas, so the answer is the same for either piece. So maybe I can compute it for one and it will be the same.

But to be safe, I need to figure out exactly which piece's surface area they are asking for. However, since it's just asking for one of the pieces, we can compute for either. Let me assume we are to compute the surface area of the piece that contains A and B, which is the lower piece (since the plane is halfway between AB and CD, so AB is on one side, and CD on the other). But maybe not. Anyway, let's proceed.

First, the surface area of one piece will consist of:

- The original faces that are entirely on one side of the plane.

- The new face created by the plane, which is the quadrilateral PQRS.

But we have to check which original faces are entirely on one side. Let me think again. The piece containing A and B (assuming the plane is cutting between AB and CD). The original faces:

Face ABC: points A, B, C. Since A and B are on one side, and C is on the other side, so the face ABC is divided by the plane. Similarly, face ABD: points A, B, D. Both A and B on one side, D on the other side. So the face ABD is also divided by the plane. Face ACD: points A, C, D. A is on one side, C and D on the other. So divided. Face BCD: points B, C, D. B on one side, C and D on the other. Divided. Therefore, all original faces are intersected by the plane. Therefore, the piece containing A and B will have:

- A portion of each original face.

Plus the new quadrilateral face PQRS.

But that complicates things because calculating the area of portions of the original faces. However, maybe there is a smarter way. Wait, but the problem says the plane is parallel to edges AB and CD. So, maybe the surface area of one piece includes the original face AB (but AB is entirely on one side?), but AB is on the same side as A and B. Wait, but AB is an edge. However, since the plane is parallel to AB, the edge AB is not intersected by the plane. Therefore, the original face AB is actually not part of the surface area of the piece? Wait, no. The piece containing A and B: its surface area includes the original face AB? Wait, no. Wait, the original face AB is part of the tetrahedron. If the plane is cutting off a part, then the original face AB is entirely on one side of the plane. Since AB is on the same side as A and B, which are part of the piece containing A and B. However, the plane is parallel to AB, so the original face AB is not intersected by the plane. Therefore, the original face AB is entirely part of the surface of the piece containing A and B. Similarly, the original face CD is entirely part of the surface of the other piece. Wait, but AB is an edge, not a face. The original face containing AB is triangle ABC. Wait, triangle ABC is intersected by the plane, as point C is on the other side. Therefore, triangle ABC is split into two parts. Therefore, the piece containing A and B will have a portion of triangle ABC, and the original face ABD is split into two parts as well. Similarly for the other faces. So actually, the surface area of the piece containing A and B will consist of:

- The portion of face ABC that's on the A/B side.

- The portion of face ABD that's on the A/B side.

- The portion of face ACD that's on the A/B side.

- The portion of face BCD that's on the A/B side.

Plus the new face PQRS.

But calculating these portions might be complex. However, since the plane is parallel to AB and CD and cuts the edges AC, AD, BC, BD at their midpoints (since t = 1/2, u = 1/2, etc.), maybe these portions are triangles or something with known areas.

Wait, when the plane intersects AC at t = 1/2 and BC at t = 1/2, then the portion of triangle ABC on the A/B side is the triangle ABP'? Wait, no. Wait, point P is on AC, and point R is on BC. So in triangle ABC, the plane cuts AC at P and BC at R, so the portion of triangle ABC on the A/B side is the triangle APR? Wait, but AP is from A to P (midpoint of AC), and BR is from B to R (midpoint of BC). Wait, actually, connecting P and R. The line PR divides triangle ABC into two regions: the smaller triangle APR and the quadrilateral PRBC. But since the plane is cutting through the tetrahedron, the piece containing A and B would have the area which is the part of triangle ABC below the plane? Since we're assuming the plane is cutting through AC and BC at their midpoints. However, actually, the plane cuts AC and BC at their midpoints, so PR is a line connecting midpoints of AC and BC, which is a midline of triangle ABC, parallel to AB and half its length. Wait, since AB is length 2, then PR would be length 1. Therefore, the area of triangle APR would be 1/4 of the area of ABC? Wait, the midline divides the triangle into two parts: the smaller triangle APR has area 1/4 of ABC, and the trapezoid PRBC has area 3/4? Wait, no. Wait, if you connect midpoints of AC and BC, then the line PR is parallel to AB and half its length, so the area of APR is (1/2 * base * height). Since PR is half of AB, and the height is half of the height of ABC, so area is 1/4 of original. Therefore, the area of APR is 1/4 of ABC. Similarly for other faces?

But since the plane is cutting through other edges as well, maybe similar logic applies.

Alternatively, maybe it's easier to note that since all the cuts are at midpoints (since t = 1/2, etc.), the intersection is a midline parallelogram, and the resulting piece might have some proportional areas.

But let me think again. For the piece containing A and B, which has the face AB (original) and the quadrilateral PQRS. However, the original face AB is part of the surface area, but also parts of other original faces. But perhaps when you take the piece containing AB, its surface area is the area of AB (which is 2, since AB is length 2, but wait, the surface area would include the area of the face AB. Wait, no. Wait, the surface area of a piece is the sum of all its faces. The piece containing AB would have:

- The original face AB (which is a triangle? No, AB is an edge. Wait, AB is an edge of the original tetrahedron. The original face containing AB is triangle ABC and triangle ABD.

Wait, this is getting confusing. Maybe I need to approach this more systematically.

First, let me note that the plane cuts the tetrahedron into two pieces. Each piece has:

- The original faces that are entirely on one side of the plane.

- The portions of the original faces that are cut by the plane.

- The new face, which is the quadrilateral PQRS.

But since all original faces are intersected by the plane, there are no original faces entirely on one side. Therefore, the surface area of each piece will consist of:

- The new face PQRS.

- The portions of each original face that are on that piece.

Therefore, to compute the surface area of one piece, we need to calculate the area of PQRS, plus the areas of the portions of the original faces.

But maybe there is a smarter way. Since the plane is parallel to AB and CD, and cuts the tetrahedron symmetrically? Wait, but AB and CD are skew lines, so maybe not symmetric. However, since the plane is halfway between AB and CD, and cuts each of the edges AC, AD, BC, BD at their midpoints, maybe the piece containing A and B is similar to the original tetrahedron scaled down? But not sure.

Alternatively, since the plane is cutting the edges AC, AD, BC, BD at midpoints, which are halfway along each edge. Therefore, the piece containing A and B is a kind of pyramid with base PQRS and apexes at A and B? Wait, no, since the piece is a polyhedron with two parts? Wait, perhaps the piece containing A and B is a kind of frustum-like shape with a quadrilateral base and a triangular top? No, maybe not. Alternatively, maybe it's a pentahedron with the quadrilateral face PQRS and three original triangular faces?

Wait, let me think about the vertices of the piece containing A and B. It includes the original vertices A and B, and the points P, Q, R, S. Therefore, the polyhedron has vertices A, B, P, Q, R, S. Let me try to see how the faces are structured.

Original face ABC is split by the plane into two parts: triangle APR (connecting A, P, R) and quadrilateral PRBC. But since we are taking the piece containing A and B, we have triangle APR and the quadrilateral APR... Wait, no. The piece containing A and B would have the triangle APR and the triangle ABP? No, since P is on AC and R is on BC.

Wait, maybe the piece containing A and B has the following faces:

- The original face ABD: split into two parts by the plane. The portion on the A/B side is triangle AQS and something else? Similarly, the original face ACD is split into triangle APQ and something else. The original face ABC is split into APR and another part. And the original face BCD is split into BR something and something else. This seems too vague.

Alternatively, since the plane cuts each of the edges AC, AD, BC, BD at their midpoints, we can think of the piece containing A and B as follows:

- It includes triangle ABD, but with parts removed. However, the intersection with the plane cuts AD at Q and BD at S. So in triangle ABD, the plane cuts it along QS. So the piece containing A and B would have triangle ABQ and triangle ABS? Wait, no. If you have triangle ABD, and you cut it with a line QS, which connects Q (midpoint of AD) to S (midpoint of BD). Then, the portion of triangle ABD on the A/B side is the triangle ABQ? No, because Q is on AD and S is on BD. The line QS divides triangle ABD into two parts: a smaller triangle ASQ and a quadrilateral QBS? Wait, maybe not. Let me visualize.

Alternatively, since Q and S are midpoints, QS is a midline of triangle ABD, parallel to AB. Therefore, the portion of triangle ABD on the A/B side is a smaller triangle similar to ABD with half the side lengths, so area 1/4 of ABD. Similarly, the other portion is a trapezoid.

But if the piece containing A and B includes triangle ABD, but only up to QS, then the area would be 1/4 of the area of ABD? But I need to figure out exactly which parts are included.

Alternatively, since the plane cuts the edges AC, AD, BC, BD at their midpoints, the piece containing A and B is a polyhedron whose surface areas consist of:

- The original face AB (which is an edge, but the face containing AB is ABC and ABD). However, since the plane is parallel to AB, the face AB is not intersected, but the faces ABC and ABD are intersected. Therefore, the piece containing A and B has:

- For face ABC: the trapezoid APRB? Wait, AP is from A to midpoint of AC, and BR is from B to midpoint of BC. Connecting P and R, which is parallel to AB and half its length. So the area of APRB would be the area of triangle ABC minus the area of triangle PRC? Wait, since PR divides ABC into a smaller triangle (ARPR?) Wait, confusing again.

Alternatively, since PR is the midline of ABC, parallel to AB, then APRB is a trapezoid with bases AB and PR, and height half of the height of ABC. The area of this trapezoid is (AB + PR) * height / 2. Since AB = 2, PR = 1 (since it's half the length of AB), and the height is half of the original height. The original area of ABC is (base * height)/2 = (2 * h)/2 = h, where h is the height. Wait, but in a regular tetrahedron, all faces are equilateral triangles. So the area of ABC is (sqrt(3)/4) * (2)^2 = sqrt(3).

So area of ABC is sqrt(3). The trapezoid APRB (assuming that's correct) would have area (AB + PR)/2 * height_between_them. Since PR is halfway up, the height between AB and PR is half the height of ABC. Therefore, area is (2 + 1)/2 * (h / 2) = (3/2) * (h / 2) = 3h / 4. Earlier, area of ABC is h, so this would be 3/4 of the area? Wait, but that seems conflicting with the earlier thought that the area of APR is 1/4 of ABC. Maybe not. Let me calculate with actual numbers.

The area of ABC is sqrt(3). Let me compute the area of APRB (assuming it's a trapezoid). The length of AB is 2, length of PR is 1, and the distance between AB and PR is half the height of ABC.

The height of ABC is the height of an equilateral triangle with side 2: h = sqrt(2^2 - 1^2) = sqrt(3). Therefore, the distance between AB and PR is sqrt(3)/2. Therefore, the area of the trapezoid APRB is (AB + PR)/2 * distance_between = (2 + 1)/2 * sqrt(3)/2 = 3/2 * sqrt(3)/2 = (3 sqrt(3))/4. Which is 3/4 of the area of ABC, since the area of ABC is sqrt(3). Therefore, the portion of face ABC on the A/B side is (3 sqrt(3))/4. Similarly, for face ABD, since it's also an equilateral triangle with area sqrt(3), the portion on the A/B side would also be (3 sqrt(3))/4. Wait, but then what about the other faces?

Similarly, for face ACD: this face is triangle ACD, which is split by the plane. The plane intersects AC at P and AD at Q. Therefore, the portion of ACD on the A/B side is the triangle APQ. Since P and Q are midpoints, triangle APQ is similar to ACD with scaling factor 1/2, so its area is (1/2)^2 * area of ACD = 1/4 * sqrt(3). Similarly, for face BCD, which is triangle BCD, the portion on the A/B side is triangle BR S? Wait, the plane intersects BC at R and BD at S. Therefore, triangle BRS is similar to BCD with area 1/4 * sqrt(3). But wait, triangle BRS is not on the A/B side, since R and S are on the opposite side? Wait, no. Wait, the piece containing A and B would have the portion of BCD that's on the A/B side. However, since point B is on the A/B side and points C and D are on the other side, the portion of BCD on the A/B side is triangle BR... wait, point R is on BC, and point S is on BD. So the intersection of the plane with BCD is line RS. Therefore, the portion of triangle BCD on the A/B side is the triangle BRS. Since B is on the A/B side, and R and S are on the plane. Since R and S are midpoints, triangle BRS is similar to BCD with scaling factor 1/2, so area is 1/4 of sqrt(3). Therefore, the area of BRS is sqrt(3)/4.

Similarly, for face ACD, the area of APQ is sqrt(3)/4.

Therefore, putting it all together, the surface area of the piece containing A and B would be:

- The area of APRB (trapezoid): 3 sqrt(3)/4

- The area of ABQ... Wait, no. Wait, what about the other portions? Wait, the piece containing A and B includes:

- The portion of ABC: 3 sqrt(3)/4

- The portion of ABD: 3 sqrt(3)/4

- The portion of ACD: sqrt(3)/4

- The portion of BCD: sqrt(3)/4

Plus the area of the new quadrilateral PQRS.

But wait, if I add all these, I would be overcounting? Wait, no. Because the piece contains parts of each original face. For example, the trapezoid APRB is part of ABC, the trapezoid ... something is part of ABD, and so on? Wait, no, earlier I said for ABC, the area is 3 sqrt(3)/4, and for ABD, similarly 3 sqrt(3)/4. However, the areas for ACD and BCD are each sqrt(3)/4. But I need to check if those portions are indeed part of the piece containing A and B.

Wait, for ACD: the piece containing A and B includes the portion of ACD that is on the A/B side. Since ACD has vertices A, C, D. A is on the A/B side, while C and D are on the other side. The plane cuts ACD at P (on AC) and Q (on AD). Therefore, the portion of ACD on the A/B side is triangle APQ, which has area sqrt(3)/4. Similarly, the portion of BCD on the A/B side is triangle BRS, area sqrt(3)/4. Therefore, adding up:

- ABC: 3 sqrt(3)/4

- ABD: 3 sqrt(3)/4

- ACD: sqrt(3)/4

- BCD: sqrt(3)/4

Plus the area of PQRS.

But wait, but the original tetrahedron has four faces, and we're adding up all four portions? However, each of these portions is part of the surface area of the piece. However, when you add them, you have to make sure that you are not double-counting or missing something. But seems okay.

However, actually, the surface area of the piece should include:

- All the areas of the parts of the original faces that are on the piece.

- The area of the new face PQRS.

So that's the total.

So let's calculate:

First, sum of the original portions:

For ABC: 3√3 / 4

For ABD: 3√3 / 4

For ACD: √3 / 4

For BCD: √3 / 4

Total original portions: 3√3 / 4 + 3√3 / 4 + √3 / 4 + √3 / 4 = (3 + 3 + 1 + 1) * √3 / 4 = 8 * √3 / 4 = 2√3.

Then, add the area of PQRS.

So now we need to calculate the area of PQRS, the quadrilateral.

Given that PQRS is a parallelogram, as established earlier. To find its area, we can compute the magnitude of the cross product of vectors PQ and PS or something similar. But since it's a parallelogram, area is base * height or |PQ × PR| or something.

Alternatively, since PQRS is a parallelogram, and we have coordinates for all four points, we can compute its area using coordinates.

Points:

P(1/2, sqrt(3)/2, 0)

Q(1/2, sqrt(3)/6, √6 / 3)

R(3/2, sqrt(3)/2, 0)

S(3/2, sqrt(3)/6, √6 / 3)

Let me compute vectors PQ and PS.

Vector PQ = Q - P = (0, sqrt(3)/6 - sqrt(3)/2, √6 / 3 - 0) = (0, -sqrt(3)/3, √6 / 3)

Vector PS = S - P = (3/2 - 1/2, sqrt(3)/6 - sqrt(3)/2, √6 / 3 - 0) = (1, -sqrt(3)/3, √6 / 3)

Then, the area of the parallelogram is |PQ × PS|.

Compute cross product:

PQ = (0, -sqrt(3)/3, √6 / 3)

PS = (1, -sqrt(3)/3, √6 / 3)

Cross product PQ × PS:

|i       j         k         |
|0   -sqrt(3)/3   √6/3|
|1   -sqrt(3)/3   √6/3|

= i [ (-sqrt(3)/3 * √6/3 - √6/3 * (-sqrt(3)/3) ) ] - j [ 0 * √6/3 - √6/3 * 1 ] + k [ 0 * (-sqrt(3)/3) - (-sqrt(3)/3) * 1 ]

Let me compute each component:

i component:

(-sqrt(3)/3)(√6/3) - (√6/3)(-sqrt(3)/3) = (-sqrt(18)/9) + sqrt(18)/9 = 0.

j component:

- [ 0 - √6/3 * 1 ] = - [ -√6 / 3 ] = √6 / 3.

k component:

0 - (-sqrt(3)/3 * 1) = sqrt(3)/3.

Therefore, the cross product is (0, √6/3, sqrt(3)/3).

Therefore, the magnitude of this vector is sqrt( (0)^2 + ( √6 / 3 )^2 + ( √3 / 3 )^2 ) = sqrt( (6/9) + (3/9) ) = sqrt(9/9) = sqrt(1) = 1.

So the area of the parallelogram PQRS is 1.

Wait, that's interesting. So the area of PQRS is 1.

Therefore, the total surface area of the piece containing A and B is 2√3 + 1.

But wait, let me check this calculation again. Because the cross product result magnitude was 1? Let me recheck.

Calculating the cross product components:

i component: (-sqrt(3)/3 * √6/3 - √6/3 * (-sqrt(3)/3))

First term: (-sqrt(3)*sqrt(6))/9 = -sqrt(18)/9 = - (3*sqrt(2))/9 = -sqrt(2)/3.

Second term: - √6/3 * (-sqrt(3)/3) = + sqrt(18)/9 = same as before, sqrt(2)/3.

Therefore, total i component: -sqrt(2)/3 + sqrt(2)/3 = 0. Correct.

j component: - [ (0 * √6/3 - √6/3 * 1) ] = - [ -√6 / 3 ] = √6 / 3.

Wait, but earlier I had j component as √6 / 3.

k component: 0 * (-sqrt(3)/3) - (-sqrt(3)/3)*1 = 0 + sqrt(3)/3 = sqrt(3)/3.

Therefore, vector is (0, √6/3, sqrt(3)/3). Then magnitude squared:

(√6/3)^2 + (sqrt(3)/3)^2 = (6/9) + (3/9) = 9/9 = 1. So magnitude is 1. Yes. Therefore, area of PQRS is 1. Interesting.

So the total surface area of the piece containing A and B is 2√3 + 1? Wait, earlier we calculated the sum of original portions as 2√3, and then adding 1 from PQRS.

But let me check again: the original portions sum to 2√3? Earlier steps:

Sum of original portions: ABC (3√3/4), ABD (3√3/4), ACD (√3/4), BCD (√3/4). Adding up:

3√3/4 + 3√3/4 = 6√3/4 = 3√3/2.

Then √3/4 + √3/4 = √3/2.

So total original portions: 3√3/2 + √3/2 = 4√3/2 = 2√3. Yes. Then adding the area of PQRS, which is 1. So total surface area = 2√3 + 1.

But let me check if this answer makes sense. The original surface area of the tetrahedron is 4*(sqrt(3)/4)*2^2 = 4*(sqrt(3)/4)*4 = 4*sqrt(3). Wait, original surface area is 4*(area of one face). Each face has area sqrt(3), as calculated earlier. Therefore, original surface area is 4*sqrt(3). If the two pieces have surface areas 2√3 + 1 and something else, then their sum should be equal to 4√3 + (area of PQRS). Wait, no. Wait, when you cut the tetrahedron with the plane, the total surface area of the two pieces is equal to the original surface area plus twice the area of the new face PQRS. Because each piece has the new face PQRS as a part of its surface area. Therefore, total surface area of both pieces is 4√3 + 2*1 = 4√3 + 2. But if one of the pieces has surface area 2√3 + 1, then the other piece would have surface area (4√3 + 2) - (2√3 + 1) = 2√3 + 1. Therefore, both pieces have the same surface area, which makes sense, as we thought earlier. Therefore, the answer should be 2√3 + 1. But wait, let me check this with another approach.

Alternatively, maybe there is a miscalculation. Let me think again about the original portions. The piece containing A and B has parts of all four original faces. But when we add those parts, do they cover all the original faces? For example, the original face ABC is split into the trapezoid APRB (area 3√3 /4) and the triangle PRC (area sqrt(3)/4). Similarly, original face ABD is split into trapezoid ABQS (area 3√3 /4) and triangle QSD (area sqrt(3)/4). Original face ACD is split into triangle APQ (area sqrt(3)/4) and quadrilateral PQDC (area 3√3 /4). Original face BCD is split into triangle BRS (area sqrt(3)/4) and quadrilateral RSPC (area 3√3 /4). Therefore, the piece containing A and B includes:

- From ABC: 3√3 /4

- From ABD: 3√3 /4

- From ACD: sqrt(3)/4

- From BCD: sqrt(3)/4

Total: 3√3 /4 + 3√3 /4 + sqrt(3)/4 + sqrt(3)/4 = 8√3 /4 = 2√3, as before.

Plus the new face area 1. So total 2√3 + 1. So this seems to check out.

But wait, what about the original face AB? Wait, AB is part of both faces ABC and ABD. But we already considered portions of ABC and ABD. Since AB is an edge, not a face, it's part of both ABC and ABD. However, when we calculated the area of APRB (from ABC) and ABQS (from ABD), the area of AB is counted twice? Wait, no. Because APRB is a trapezoid that includes AB. Wait, APRB is part of ABC, and ABQS is part of ABD. However, AB is common to both faces. But when we add the areas of APRB and ABQS, we are counting the area of AB twice? No, because APRB is part of ABC, and ABQS is part of ABD. Since APRB includes AB and some other area, and ABQS includes AB and some other area. But the line AB is shared between these two trapezoids. However, in reality, AB is an edge, and the area calculation for the trapezoids would have AB as a base. Therefore, when we add 3√3 /4 and 3√3 /4, we might be overcounting the area of AB? Wait, no. Because each trapezoid is part of a different face. The trapezoid APRB is in ABC, and ABQS is in ABD. The overlap between them is just the edge AB, which has no area. Therefore, adding them is okay.

So total surface area is indeed 2√3 + 1.

But let me check with numerical values. The original surface area is 4√3 ≈ 6.928. The total surface area of both pieces is 4√3 + 2 ≈ 6.928 + 2 = 8.928. The individual piece surface area is 2√3 + 1 ≈ 3.464 + 1 = 4.464. So each piece has surface area ≈ 4.464. Seems reasonable.

But let me check if the area of PQRS is indeed 1. Given that the cross product magnitude is 1, which is correct. Let me verify with coordinates.

Alternatively, since PQRS is a parallelogram, we can compute its area by calculating the length of PQ and the height between the sides. Let me calculate the length of PQ.

Vector PQ is (0, -sqrt(3)/3, sqrt(6)/3). Its magnitude is sqrt(0 + (sqrt(3)/3)^2 + (sqrt(6)/3)^2 ) = sqrt( (3/9) + (6/9) ) = sqrt(9/9) = 1. So length of PQ is 1.

Similarly, the length of PR is sqrt( (3/2 - 1/2)^2 + (sqrt(3)/2 - sqrt(3)/2)^2 + (0 - 0)^2 ) = sqrt( (1)^2 + 0 + 0 ) = 1. Wait, PR is from P(1/2, sqrt(3)/2, 0) to R(3/2, sqrt(3)/2, 0). So the vector PR is (1, 0, 0), so length 1. Wait, then if PQ and PR are both length 1, and the angle between them is such that the area is |PQ × PR|, which we already calculated as 1. Wait, but PQ and PR would not be adjacent sides of the parallelogram. Wait, PQ and PS are adjacent sides. Wait, earlier we computed PQ × PS = (0, √6/3, √3/3), magnitude 1. Therefore, the area is indeed 1. So PQRS has area 1.

Therefore, the total surface area of the piece is 2√3 + 1.

But the problem says "the plane... cuts ABCD into two pieces. Find the surface area of one of these pieces."

Therefore, the answer is 2√3 + 1. However, let me check if there is an alternative way to get this or if I made a mistake.

Alternatively, since the original surface area is 4√3, and the new surface area is 2*1 = 2, then total surface area of both pieces is 4√3 + 2. If one piece has surface area 2√3 + 1, then the other is 4√3 + 2 - (2√3 + 1) = 2√3 + 1, which matches. So that seems consistent.

But let me check with another approach. If I consider that the piece containing A and B is a kind of prism or frustum. However, given the complexity, and that cross-checks seem consistent, maybe 2√3 + 1 is correct.

But let me check with vector areas.

Alternatively, think about the fact that the piece containing A and B has the following faces:

- The trapezoid APRB (area 3√3 /4)

- The trapezoid ABQS (area 3√3 /4)

- The triangle APQ (area sqrt(3)/4)

- The triangle BRS (area sqrt(3)/4)

- The parallelogram PQRS (area 1)

Wait, but adding those would be 3√3/4 + 3√3/4 + sqrt(3)/4 + sqrt(3)/4 + 1 = same as before, 2√3 + 1.

But actually, these faces are different? Wait, APQ is part of ACD, BRS is part of BCD, APRB is part of ABC, ABQS is part of ABD, and PQRS is the new face. Therefore, yes, the total surface area is as calculated.

Therefore, I think this is correct. However, let me verify with approximate numbers. Let me compute 2√3 + 1. Since √3 ≈ 1.732, so 2*1.732 ≈ 3.464 + 1 ≈ 4.464.

Original surface area is 4*√3 ≈ 6.928. The total surface area after cutting is 6.928 + 2 ≈ 8.928. Divided by two, each piece has ≈ 4.464. Which seems okay.

But let me check if there is an alternative interpretation. For example, maybe the problem considers only the surface area of the piece that is "above" the plane, but since we had to compute for one piece, and the answer is 2√3 + 1, maybe that is the answer.

Alternatively, the question may expect a different form. Let me check if 2√3 + 1 can be simplified or expressed differently. However, 2√3 + 1 is already simplified.

But let me check again the calculation of the area of PQRS. Because earlier I found that the cross product magnitude was 1. Given that PQRS is a parallelogram with sides of length 1 (as calculated) and area 1. Which is interesting. Given that the original tetrahedron has edge length 2, and the area of PQRS is 1, which is less than the area of the original faces.

Alternatively, perhaps there's a mistake in calculating the area of PQRS. Let me try another approach.

Since PQRS is a parallelogram, its area can also be calculated as base * height. Let me take PQ as a base, which has length 1. What is the height corresponding to this base?

The height would be the distance between the lines PQ and RS. Since PQ and RS are opposite sides of the parallelogram. But since the plane is parallel to AB and CD, and PQ is parallel to AB? Wait, earlier we found that PQ is parallel to AB? Wait, vector PQ is (0, -sqrt(3)/3, sqrt(6)/3). AB is (2, 0, 0). Are they parallel? No, because the direction vectors are not scalar multiples. Wait, but earlier we said the plane is parallel to AB and CD. So the plane contains lines parallel to AB and CD. However, PQ is a vector in the plane, so it should be a combination of directions parallel to AB and CD. But vector PQ was computed as (0, -sqrt(3)/3, sqrt(6)/3), which is not a scalar multiple of AB or CD. However, since PQ is in the plane, which is parallel to AB and CD, then PQ can be expressed as a combination of AB and CD.

But regardless, since we already calculated the area of the parallelogram as 1 using cross product, and verified with coordinates, I think it's safe.

Therefore, I think the answer is 2√3 + 1. However, let me check if the question asks for the surface area of one of the pieces, and if there's an alternative way that the answer is different.

Wait, another approach: the surface area of one piece includes the area of the new face plus the areas of the parts of the original faces. However, what about the other piece? If the other piece also has surface area 2√3 + 1, then the total is 4√3 + 2, which matches with original total surface area plus twice the area of the new face. Therefore, seems correct.

Alternatively, maybe the question expects the answer as 2√3 + 1, but I need to check if there's any error in assuming that the original portions sum to 2√3. Let me verify with another method.

Each of the original faces is divided into a portion on the piece and a portion on the other piece. For example, face ABC is split into 3√3 /4 (on A/B side) and sqrt(3)/4 (on C/D side). Similarly for other faces. Therefore, total original surface area is 4 * sqrt(3). The total of the portions on the A/B side is 2√3, and similarly on the other side is 2√3. Therefore, adding the new face area (1) for each piece, the total surface area for each piece is 2√3 + 1. So yes.

Therefore, I think this is correct.

But to make sure, let me check the area of PQRS again. Given that PQRS is a parallelogram with area 1, and the original tetrahedron has volume, maybe we can compute volume of the piece and check some consistency?

The volume of the original tetrahedron with side length 2 is V = (sqrt(2)/12) * (2)^3 = (sqrt(2)/12)*8 = (2 sqrt(2))/3 ≈ 0.9428.

If we compute the volume of the piece containing A and B, which includes the original portions of the faces. However, since the plane cuts the tetrahedron halfway between AB and CD, and since AB and CD are skew, it's not straightforward. However, since the plane cuts the edges AC, AD, BC, BD at their midpoints, the piece containing A and B is a kind of "half" of the tetrahedron. However, in a regular tetrahedron, cutting through midpoints of edges... Maybe the volume of this piece is 1/2 of the original? But let me check.

Actually, if you connect midpoints of edges, the resulting figure might be a smaller tetrahedron or something else. Wait, but in this case, the intersection is a parallelogram, so the piece containing A and B is a kind of frustum-like shape. However, its volume can be calculated using determinants or other methods. However, since this is getting complex, and we already have surface area as 2√3 + 1, and all the checks seem consistent, I think it's safe to go with this answer.

But just to be thorough, let me calculate the volume of the piece containing A and B. Since the plane cuts the edges AC, AD, BC, BD at their midpoints, we can use the formula for the volume of a pyramid or something else.

Alternatively, consider that the piece containing A and B can be considered as the union of pyramids or prisms. However, given the time constraints, maybe not. Alternatively, note that the plane divides the tetrahedron into two congruent pieces? Wait, no, since AB and CD are not symmetric in the tetrahedron. However, since the plane is halfway between them, perhaps the volumes are equal? If so, then volume of each piece would be half of original, which is (2√2)/6 = √2 / 3 ≈ 0.4714. But how does this relate to the surface area? Not directly, but maybe the surface area is connected.

However, given that the surface area calculation seems consistent, and cross-checked, and the total surface area makes sense, I think I can go with the answer 2√3 + 1.

But let me check with another perspective. The problem says the plane is parallel to AB and CD, and halfway between them. In a regular tetrahedron, AB and CD are skew lines, and the distance between them is √2 as calculated before. The midpoint plane would be at half that distance. The intersection is a parallelogram, and the surface area includes that quadrilateral. Given that, and the calculations above, I think the answer is 2√3 + 1.

However, to cross-verify with an example, suppose the edge length was different. Suppose edge length is 1, then calculations change, but since we are given edge length 2, I think calculations are correct.

Alternatively, check if my value for the area of PQRS being 1 is correct. Suppose the edge length is 2, and PQRS area is 1. Let me think about units. If the tetrahedron has edge length 2, then the area of a face is sqrt(3). The area of PQRS is 1, which is less than that, which is plausible. Alternatively, if I consider the coordinates:

Points P(1/2, sqrt(3)/2, 0), Q(1/2, sqrt(3)/6, √6 / 3), R(3/2, sqrt(3)/2, 0), S(3/2, sqrt(3)/6, √6 / 3).

Let me calculate the area of PQRS using another method. For example, divide the parallelogram into two triangles. Take triangle PQR and triangle PRS.

Compute area of triangle PQR and double it.

Coordinates:

P(1/2, sqrt(3)/2, 0)

Q(1/2, sqrt(3)/6, √6 / 3)

R(3/2, sqrt(3)/2, 0)

Compute vectors PQ and PR.

Vector PQ = Q - P = (0, -sqrt(3)/3, sqrt(6)/3)

Vector PR = R - P = (1, 0, 0)

Area of triangle PQR is (1/2)|PQ × PR|.

Compute cross product PQ × PR:

PQ = (0, -sqrt(3)/3, sqrt(6)/3)

PR = (1, 0, 0)

Cross product:

|i       j         k         |
|0   -sqrt(3)/3   sqrt(6)/3|
|1     0           0         |

= i [ (-sqrt(3)/3 * 0 - sqrt(6)/3 * 0) ] - j [ 0*0 - sqrt(6)/3 * 1 ] + k [ 0*0 - (-sqrt(3)/3 * 1) ]

= i (0) - j (-sqrt(6)/3) + k (sqrt(3)/3)

= (0, sqrt(6)/3, sqrt(3)/3)

Which has magnitude sqrt( (sqrt(6)/3)^2 + (sqrt(3)/3)^2 ) = sqrt(6/9 + 3/9) = sqrt(9/9) = 1. Therefore, the area of triangle PQR is 1/2 * 1 = 1/2. Then the area of the parallelogram PQRS is 2*(1/2) = 1. So yes, correct.

Therefore, the area of PQRS is indeed 1.

Therefore, the surface area of the piece is 2√3 + 1.

Wait, but the problem says "surface area of one of these pieces". However, the answer choices are not given, but the problem says "put your final answer within \boxed{}". So likely the answer is 2√3 + 1. However, let me check with another way. Maybe the answer is 2√3? Or something else?

Wait, but according to all calculations, it's 2√3 + 1. Maybe I made a mistake in the original portions. Let me check again:

Original portions:

- From ABC: 3√3 /4

- From ABD: 3√3 /4

- From ACD: √3 /4

- From BCD: √3 /4

Total: 3√3 /4 + 3√3 /4 = 6√3 /4 = 3√3 /2

√3 /4 + √3 /4 = √3 / 2

Total: 3√3 /2 + √3 /2 = 4√3 /2 = 2√3. Yes. Then adding 1 gives 2√3 + 1. So this seems correct.

Alternatively, let me think about the problem again. Maybe the surface area of one piece includes only the new face and some of the original faces, but not all. However, according to the earlier analysis, it includes portions of all four original faces. However, since each original face is intersected by the plane, and the piece contains parts of each.

Wait, but for example, the face ACD is split into APQ and PDC. The piece containing A and B contains APQ, and the other part is PDC. Similarly for BCD. Therefore, the original face areas partially belong to each piece. Therefore, the total of the original portions is 2√3, and adding the new face gives 2√3 + 1.

Yes, I think this is correct. Therefore, the answer should be 2√3 + 1. Therefore, I will go with that.

**Final Answer**
\boxed{1 + 2\sqrt{3}}

To solve the problem, we analyze the geometry of a regular tetrahedron $ABCD$ with edge length 2, and the effect of cutting it with a plane parallel to edges $AB$ and $CD$, lying halfway between them.

---

### Step 1: Coordinate Setup

We assign coordinates to the vertices to better analyze the structure. A common coordinate setup for a regular tetrahedron is:

- $A = (0, 0, 0)$
- $B = (2, 0, 0)$
- $C = (1, \sqrt{3}, 0)$
- $D = (1, \sqrt{3}/3, 2\sqrt{6}/3)$

These coordinates ensure all edges are of length 2 and the tetrahedron is regular.

---

### Step 2: Direction Vectors

The direction vector of edge $AB$ is $ \vec{u} = (2, 0, 0) $, and for edge $CD$, it is:

$$
\vec{v} = D - C = (1 - 1, \sqrt{3}/3 - \sqrt{3}, 2\sqrt{6}/3 - 0) = (0, -2\sqrt{3}/3, 2\sqrt{6}/3)
$$

A plane **parallel** to both direction vectors and **halfway between** edges $AB$ and $CD$ has a normal vector perpendicular to both $ \vec{u} $ and $ \vec{v} $, i.e., the cross product $ \vec{n} = \vec{u} \times \vec{v} $.

Computing $ \vec{n} $:

$$
\vec{n} = \vec{u} \times \vec{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 0 & 0 \\
0 & -2\sqrt{3}/3 & 2\sqrt{6}/3
\end{vmatrix}
= (0, -4\sqrt{6}/3, -4\sqrt{3}/3)
$$

This can be normalized or simplified for the plane equation.

Using the standard form $ \vec{n} \cdot \vec{r} = d $, and knowing the plane is equidistant from both lines, we compute the point where this plane lies **midway** between $AB$ and $CD$, leading to the plane equation:

$$
\sqrt{2}y + z = \frac{\sqrt{6}}{2}
$$

---

### Step 3: Intersections with Edges

The plane intersects the following edges of the tetrahedron at their midpoints:

- **AC** at point $P = (1/2, \sqrt{3}/2, 0)$
- **AD** at point $Q = (1/2, \sqrt{3}/6, \sqrt{6}/3)$
- **BC** at point $R = (3/2, \sqrt{3}/2, 0)$
- **BD** at point $S = (3/2, \sqrt{3}/6, \sqrt{6}/3)$

These four points form a **parallelogram $PQRS$** (since the direction vectors of the plane are parallel to the directions of the original edge vectors, and the plane is not intersecting those edges).

---

### Step 4: Surface Area of the Cut Piece

We compute the **surface area** of one of the two resulting pieces. Let's consider the piece containing vertices $A$ and $B$.

This piece has:

- **Portions** of the original faces $ABC$, $ABD$, $ACD$, and $BCD$ that are on the same side of the plane.
- **A new face**, the parallelogram $PQRS$, formed by the cut.

We compute each contribution:

---

#### **Area of PQRS (New Face)**

Using the cross product of adjacent vectors:

- $ \vec{PQ} = (0, -\sqrt{3}/3, \sqrt{6}/3) $
- $ \vec{PS} = (1, -\sqrt{3}/3, \sqrt{6}/3) $

$$
\vec{PQ} \times \vec{PS} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & -\sqrt{3}/3 & \sqrt{6}/3 \\
1 & -\sqrt{3}/3 & \sqrt{6}/3
\end{vmatrix}
= (0, \sqrt{6}/3, \sqrt{3}/3)
$$

Magnitude of this vector:

$$
|\vec{PQ} \times \vec{PS}| = \sqrt{0^2 + (\sqrt{6}/3)^2 + (\sqrt{3}/3)^2} = \sqrt{(6/9 + 3/9)} = \sqrt{9/9} = 1
$$

So, the area of quadrilateral $PQRS$ is:

$$
\text{Area}_{PQRS} = 1
$$

---

#### **Area of Original Face Portions**

Each original face is cut by the plane into a portion lying on the piece containing $A$ and $B$:

- **Face $ABC$** → portion $APRB$ (trapezoid): $ \frac{3\sqrt{3}}{4} $
- **Face $ABD$** → portion $ABQS$ (trapezoid): $ \frac{3\sqrt{3}}{4} $
- **Face $ACD$** → portion $APQ$ (triangle): $ \frac{\sqrt{3}}{4} $
- **Face $BCD$** → portion $BRS$ (triangle): $ \frac{\sqrt{3}}{4} $

Sum of these:

$$
\frac{3\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{8\sqrt{3}}{4} = 2\sqrt{3}
$$

---

### Step 5: Total Surface Area

Final step:

$$
\text{Surface Area} = \text{Original face portions} + \text{New face area} = 2\sqrt{3} + 1
$$

---

### ✅ Final Answer

$$
\boxed{1 + 2\sqrt{3}}
$$

[Assistant]
Explanation:
Okay, so I have this problem about a regular tetrahedron with side length 2. The question says that there's a plane parallel to edges AB and CD, and it's lying halfway between them. This plane cuts the tetrahedron into two pieces, and I need to find the surface area of one of these pieces. Alright, let me try to visualize this first.

First, let me recall what a regular tetrahedron is. It's a three-dimensional shape with four triangular faces, each of which is an equilateral triangle. All edges are equal, so every edge here is length 2. The vertices are A, B, C, D, and each pair of vertices is connected by an edge of length 2.

Now, the plane in question is parallel to edges AB and CD. Hmm. So edges AB and CD are skew lines in a tetrahedron, right? Because in a tetrahedron, any two edges that don't share a common vertex are skew. Let me confirm that. AB connects A and B, and CD connects C and D. Since A, B, C, D are all vertices of the tetrahedron, AB and CD don't share a common vertex. So they are indeed skew lines.

So the plane is parallel to both AB and CD. Since AB and CD are skew, the plane can't be parallel to both unless it's somehow in a specific orientation. Wait, but planes can be parallel to lines even if the lines are skew. For example, a plane can be parallel to two skew lines if it's parallel to the direction of both lines. So, if the plane is parallel to both AB and CD, then it must be parallel to the direction of both edges AB and CD. Therefore, the plane's normal vector is perpendicular to the direction vectors of AB and CD. Maybe I need to think about coordinates here.

Maybe assigning coordinates to the tetrahedron will help. Let me try that. Let me recall that coordinates for a regular tetrahedron can be set up in 3D space. One common coordinate system for a regular tetrahedron is to have vertices at (1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1), scaled appropriately. But maybe there's a simpler way. Alternatively, maybe I can place the tetrahedron with base ABC and apex D. Let me try that.

Let me consider the base triangle ABC. Let me put point A at (0, 0, 0), point B at (2, 0, 0), since the side length is 2. Now, since ABC is an equilateral triangle, point C will be somewhere in the xy-plane. Let me calculate its coordinates. The coordinates of C can be found by knowing that the distance from A to C is 2, and from B to C is 2. Let me denote C as (x, y, 0). Then, the distance from A to C is sqrt(x² + y²) = 2, and from B to C is sqrt((x - 2)² + y²) = 2. So:

x² + y² = 4,

and

(x - 2)² + y² = 4.

Subtracting the first equation from the second:

(x - 2)² + y² - x² - y² = 4 - 4 => (x² -4x +4) -x² = 0 => -4x +4 = 0 => x = 1.

So x is 1. Then from x² + y² = 4, we get 1 + y² = 4 => y² = 3 => y = sqrt(3) or -sqrt(3). Let me take y = sqrt(3) to place the base triangle in the xy-plane. So point C is at (1, sqrt(3), 0).

Now, the apex D of the tetrahedron. Since it's a regular tetrahedron, all edges are length 2. So the coordinates of D must satisfy the distances from A, B, and C. Let me denote D as (p, q, r). Then the distance from D to A is sqrt((p - 0)^2 + (q - 0)^2 + (r - 0)^2) = sqrt(p² + q² + r²) = 2. Similarly, distance from D to B is sqrt((p - 2)^2 + q² + r²) = 2, and distance from D to C is sqrt((p - 1)^2 + (q - sqrt(3))² + r²) = 2.

So let me write down these equations:

1. p² + q² + r² = 4.

2. (p - 2)^2 + q² + r² = 4.

3. (p - 1)^2 + (q - sqrt(3))² + r² = 4.

Subtract equation 1 from equation 2:

(p - 2)^2 - p² = 0 => p² -4p +4 - p² = -4p +4 = 0 => p = 1.

So p = 1. Then, substituting p = 1 into equation 1:

1² + q² + r² = 4 => q² + r² = 3.

Now equation 3 becomes:

(1 - 1)^2 + (q - sqrt(3))² + r² = 4 => 0 + (q - sqrt(3))² + r² = 4.

But since q² + r² = 3, substitute that into equation 3:

(q - sqrt(3))² + r² = q² - 2 q sqrt(3) + 3 + r² = (q² + r²) - 2 q sqrt(3) + 3 = 3 - 2 q sqrt(3) + 3 = 6 - 2 q sqrt(3) = 4.

Therefore, 6 - 2 q sqrt(3) = 4 => -2 q sqrt(3) = -2 => q sqrt(3) = 1 => q = 1 / sqrt(3) = sqrt(3)/3.

Then, from q² + r² = 3:

( (sqrt(3)/3)^2 ) + r² = 3 => (3/9) + r² = 3 => 1/3 + r² = 3 => r² = 3 - 1/3 = 8/3 => r = ± 2*sqrt(6)/3. Since a tetrahedron can be oriented either way, but let's take r positive, so r = 2√6 / 3.

Therefore, coordinates of D are (1, sqrt(3)/3, 2√6/3). Let me confirm that this makes sense. Let me check the distance from D to A: sqrt( (1)^2 + (sqrt(3)/3)^2 + (2√6/3)^2 ). Calculating:

1 + (3/9) + (24/9) = 1 + 1/3 + 8/3 = 1 + 9/3 = 1 + 3 = 4. Yes, sqrt(4) = 2. Good. Similarly, distance from D to B would be same. Let me check distance from D to C:

Coordinates of C: (1, sqrt(3), 0). Coordinates of D: (1, sqrt(3)/3, 2√6/3). So difference in x: 0, difference in y: sqrt(3)/3 - sqrt(3) = -2 sqrt(3)/3, difference in z: 2√6/3 - 0 = 2√6/3.

So squared distance: 0 + ( -2 sqrt(3)/3 )² + ( 2√6 / 3 )² = (4 * 3)/9 + (4 * 6)/9 = 12/9 + 24/9 = 36/9 = 4. So sqrt(4) = 2. Perfect. So coordinates are correct.

So now, the tetrahedron has coordinates:

A(0, 0, 0)

B(2, 0, 0)

C(1, sqrt(3), 0)

D(1, sqrt(3)/3, 2√6/3)

Alright, so now we need to find a plane that's parallel to edges AB and CD, and halfway between them. Let me first figure out what it means for a plane to be parallel to edges AB and CD. Since AB is the edge from A(0,0,0) to B(2,0,0), its direction vector is (2,0,0) - (0,0,0) = (2,0,0). Similarly, edge CD goes from C(1, sqrt(3), 0) to D(1, sqrt(3)/3, 2√6/3). So direction vector is D - C = (1 - 1, sqrt(3)/3 - sqrt(3), 2√6/3 - 0) = (0, -2 sqrt(3)/3, 2√6/3).

So direction vectors AB is (2, 0, 0) and CD is (0, -2√3/3, 2√6/3). Now, the plane is parallel to both AB and CD. So the plane must contain direction vectors AB and CD. Wait, no. Wait, a plane that's parallel to two lines can be parallel to their direction vectors. So the plane's direction vectors are the same as AB and CD's direction vectors. However, since AB and CD are skew, their direction vectors are not in the same plane, so the plane must be parallel to both direction vectors. Therefore, the normal vector of the plane is perpendicular to both direction vectors of AB and CD.

Alternatively, since the plane is parallel to AB and CD, then the plane can be defined by containing lines parallel to AB and CD. However, since AB and CD are skew, there is no plane that contains both AB and CD. But there can be planes that are parallel to both AB and CD. So the plane in question is parallel to both AB and CD, meaning that the direction vectors of AB and CD are direction vectors of the plane.

So, given that, the plane has a normal vector that is perpendicular to both direction vectors of AB and CD. Let me compute that. Let me denote direction vector AB as vector u = (2, 0, 0), and direction vector CD as vector v = (0, -2√3/3, 2√6/3). Then, normal vector n = u × v.

Compute cross product u × v:

u × v = |i   j   k|
         |2   0   0|
         |0 -2√3/3 2√6/3|

Calculating determinant:

i*(0*2√6/3 - 0*(-2√3/3)) - j*(2*2√6/3 - 0*0) + k*(2*(-2√3/3) - 0*0)

Simplify:

i*(0 - 0) - j*(4√6/3 - 0) + k*(-4√3/3 - 0)

So n = (0, -4√6/3, -4√3/3)

Alternatively, we can write it as n = (0, -4√6, -4√3) divided by 3, but for the purposes of direction, it's okay. So the normal vector of the plane is (0, -4√6/3, -4√3/3). However, for the plane equation, we can use this normal vector.

But maybe I can simplify or find a direction vector. However, maybe I need to find the equation of the plane. But we also know that the plane is halfway between AB and CD. So how do we determine "halfway between" AB and CD?

Since AB and CD are skew lines, the distance between them is some value, and the plane is halfway between them. But how do we define distance between skew lines? The distance between AB and CD can be calculated, then the plane is at halfway between them. Alternatively, since the plane is parallel to both, maybe it's halfway in terms of some parameter along the common perpendicular?

Alternatively, since the plane is parallel to both lines, it might be that the plane intersects the tetrahedron in a certain way, and the "halfway between AB and CD" refers to the plane being equidistant to AB and CD?

But how do we define the distance from a line to a plane? Wait, if the plane is parallel to AB and CD, then the distance from AB to the plane should be equal to the distance from CD to the plane? Because the plane is halfway between AB and CD. So the plane is equidistant to both AB and CD, and is parallel to both.

Therefore, if I can find the plane that is equidistant to AB and CD and parallel to both. So perhaps I can parametrize the plane.

Alternatively, since the plane is parallel to AB and CD, then it must be parallel to the direction vectors AB and CD. Therefore, as earlier, the plane can be parametrized as follows: Let me think.

Alternatively, maybe I can parametrize the plane by finding a point on it. Since it is halfway between AB and CD, perhaps we can take the line segment connecting AB and CD? Wait, but AB and CD are skew, so there is no line segment connecting them. However, there is a common perpendicular between AB and CD. The plane is parallel to both AB and CD, so it must be parallel to the common perpendicular? Wait, perhaps not. Maybe the plane is somewhere in between, such that the distance from AB to the plane equals the distance from CD to the plane.

Alternatively, since AB and CD are skew, there is a unique line that is perpendicular to both, and the distance between AB and CD is the distance between these two lines, which is the distance between the plane and the line AB and CD. Wait, maybe I need to compute the distance between AB and CD first. Let me try that.

The distance between skew lines AB and CD can be found by taking the absolute value of the scalar triple product of vectors AB, AC, and AD divided by the magnitude of the cross product of AB and CD? Wait, let me recall the formula.

The formula for the distance between two skew lines is |(b - a) ⋅ (n)| / |n|, where n is the cross product of the direction vectors of the two lines, and a and b are points on each line.

Let me confirm. Let me denote line AB and line CD. Let me take point A on AB and point C on CD. Then vector AC is from A to C. The distance between AB and CD is |AC ⋅ (u × v)| / |u × v|, where u and v are direction vectors of AB and CD. Wait, yes, I think that's correct.

So let me compute that. Let me denote:

u = direction vector of AB = (2, 0, 0)

v = direction vector of CD = (0, -2√3/3, 2√6/3)

Then, n = u × v = (0, -4√6/3, -4√3/3) as before.

Then, vector AC is from A to C: C - A = (1, sqrt(3), 0) - (0, 0, 0) = (1, sqrt(3), 0).

Then, the distance between AB and CD is |AC ⋅ n| / |n|.

First compute AC ⋅ n:

AC is (1, sqrt(3), 0). n is (0, -4√6/3, -4√3/3).

Dot product = 1*0 + sqrt(3)*(-4√6/3) + 0*(-4√3/3) = 0 - (4 sqrt(18))/3 + 0 = - (4 * 3 * sqrt(2))/3 = -4 sqrt(2).

So absolute value is 4 sqrt(2).

Now compute |n|:

n = (0, -4√6/3, -4√3/3)

So |n| = sqrt( (0)^2 + (-4√6/3)^2 + (-4√3/3)^2 ) = sqrt( (16*6)/9 + (16*3)/9 ) = sqrt( (96 + 48)/9 ) = sqrt(144/9) = sqrt(16) = 4.

Therefore, the distance between AB and CD is |AC ⋅ n| / |n| = 4√2 / 4 = √2. Therefore, the distance between AB and CD is √2. So the plane halfway between them would be at distance √2 / 2 from each of AB and CD.

Therefore, the plane is located at a distance of √2 / 2 from AB and CD, and is parallel to both. Now, how do I find the equation of this plane?

Since the plane is parallel to AB and CD, as we determined earlier, its normal vector is n = (0, -4√6/3, -4√3/3). However, we can use a simpler normal vector by dividing by the scalar. Since the direction vector of the plane is along n, but perhaps we can use a simplified version. Alternatively, since the plane is parallel to AB and CD, and we need to find the plane at distance √2 / 2 from both AB and CD. Maybe we can find the equation of such a plane.

Alternatively, since the plane is parallel to AB and CD, then it can be represented parametrically. However, since it's halfway between AB and CD, perhaps we can find a point that lies on the plane. Let me think. Since the plane is halfway between AB and CD, maybe it intersects some edges of the tetrahedron? Let me think about how the plane cuts the tetrahedron.

Alternatively, since the plane is parallel to AB and CD, which are skew lines, then the intersection of the plane with the tetrahedron must form a quadrilateral? Because cutting a tetrahedron with a plane parallel to two edges... Wait, a regular tetrahedron has four triangular faces, so cutting with a plane that's parallel to two edges... Wait, but maybe the intersection is a line? Or maybe a quadrilateral?

Wait, if the plane is parallel to two edges AB and CD, then it can't intersect those edges, since they are parallel to the plane. Therefore, the plane would intersect the other edges of the tetrahedron. Let me think about the edges of the tetrahedron. The tetrahedron has edges AB, AC, AD, BC, BD, CD. So since the plane is parallel to AB and CD, it doesn't intersect AB or CD, so it must intersect the other edges: AC, AD, BC, BD. Wait, but maybe? Or maybe not?

Wait, if the plane is parallel to AB and CD, then it can't intersect AB or CD. But depending on where the plane is, it can intersect other edges. Let me think. For example, if we imagine moving a plane parallel to AB and CD from one side to the other, it would intersect some edges. Since AB is one edge, and CD is another, which are skew. So the plane would intersect the edges connecting AB to CD. These edges are AC, BD, and also maybe others? Wait, the tetrahedron's edges are AB, AC, AD, BC, BD, CD. So edges adjacent to AB are AC and AD, and edges adjacent to CD are BC and BD. So maybe the plane intersects AC, BD, and also maybe other edges?

Alternatively, since the plane is parallel to AB and CD, which are skew. Maybe the intersection is a quadrilateral? For instance, if the plane intersects edges AC, BD, and also edges AD and BC? Wait, maybe not. Let me try to think.

Alternatively, let me try to parametrize the plane. Let me consider parametric equations.

Since the plane is parallel to AB and CD, it has direction vectors AB and CD. However, as AB and CD are skew, we need another point to define the plane. But since the plane is halfway between AB and CD, perhaps we can find a point on the plane by moving halfway from AB to CD along the common perpendicular.

Alternatively, since the distance between AB and CD is √2, and the plane is halfway, which is √2 / 2 away from each. Therefore, perhaps we can find a point that is on the common perpendicular line midway between AB and CD, and then define the plane through that point with normal vector n.

Alternatively, since we have coordinates for AB and CD, maybe we can find such a plane.

Alternatively, maybe I can parametrize the plane as follows: since it's parallel to AB and CD, its equation can be written in terms of direction vectors. Let me think.

Let me suppose the plane has equation ax + by + cz = d. Since it's parallel to AB, which has direction vector (2, 0, 0), then the normal vector (a, b, c) must be perpendicular to AB. Therefore, (a, b, c) ⋅ (2, 0, 0) = 2a = 0 ⇒ a = 0. Similarly, since the plane is parallel to CD, whose direction vector is (0, -2√3/3, 2√6/3), then the normal vector must also be perpendicular to this direction vector. Therefore, (a, b, c) ⋅ (0, -2√3/3, 2√6/3) = 0 * a + (-2√3/3)*b + (2√6/3)*c = 0. Since we already have a = 0, this reduces to (-2√3/3)*b + (2√6/3)*c = 0 ⇒ -√3 b + √6 c = 0 ⇒ √6 c = √3 b ⇒ c = (√3 / √6) b = (1/√2) b. Therefore, c = b / √2. Therefore, the normal vector is (0, b, b / √2). We can choose b = √2 to make c = 1. Therefore, normal vector is (0, √2, 1). So the plane equation is 0x + √2 y + z = d. Therefore, √2 y + z = d.

Now, we need to find d such that the plane is halfway between AB and CD. So how do we find d?

Earlier, we computed that the distance between AB and CD is √2, so the plane is at distance √2 / 2 from each. Let me compute the distance from AB to the plane √2 y + z = d, and set it equal to √2 / 2.

But since AB is the line from A(0,0,0) to B(2,0,0). We can use the formula for the distance from a line to a plane. Wait, but actually, since the plane is parallel to AB, the distance from AB to the plane is constant along AB. So we can compute the distance from any point on AB to the plane.

Let me pick a point on AB, say point A(0,0,0). The distance from A to the plane √2 y + z = d is |√2 * 0 + 0 - d| / sqrt( (√2)^2 + 1^2 ) = | -d | / sqrt(2 + 1) = |d| / sqrt(3). Similarly, we can compute the distance from the plane to line CD. Let me pick a point on CD, say point C(1, sqrt(3), 0). The distance from point C to the plane is |√2 * sqrt(3) + 0 - d| / sqrt(3) = |√6 - d| / sqrt(3). Since the plane is midway, these distances should be equal to √2 / 2. Wait, but the plane is halfway between AB and CD, so the distance from AB to the plane is equal to distance from CD to the plane. But since AB and CD are on opposite sides of the plane? Wait, actually, depending on the orientation, but since the distance between AB and CD is √2, then the distance from AB to the plane plus the distance from the plane to CD is √2. Since the plane is halfway, each distance is √2 / 2. Therefore:

From AB's distance: |d| / sqrt(3) = √2 / 2.

From CD's distance: |√6 - d| / sqrt(3) = √2 / 2.

Therefore, we have two equations:

1) |d| = (√2 / 2) * sqrt(3) = √6 / 2.

2) |√6 - d| = (√2 / 2) * sqrt(3) = √6 / 2.

So solving these equations. Let me consider equation 1: |d| = √6 / 2 ⇒ d = ±√6 / 2.

Now, equation 2: |√6 - d| = √6 / 2. Let me consider both possibilities for d.

Case 1: d = √6 / 2.

Then |√6 - √6 / 2| = |√6 / 2| = √6 / 2, which matches the equation. Good.

Case 2: d = -√6 / 2.

Then |√6 - (-√6 / 2)| = |√6 + √6 / 2| = |(3√6)/2| = 3√6 / 2, which is not equal to √6 / 2. Therefore, case 2 is invalid. Therefore, d = √6 / 2. Therefore, the equation of the plane is √2 y + z = √6 / 2.

Therefore, the plane is √2 y + z = √6 / 2.

Now, let me check if this makes sense. Let me check the distance from point A(0,0,0) to the plane: |√2*0 + 0 - √6 / 2| / sqrt( (√2)^2 + 1^2 ) = | -√6 / 2 | / sqrt(3) = (√6 / 2) / sqrt(3) ) = (√6 / 2) / (√3) ) = (√2 / 2), which is indeed √2 / 2. Similarly, distance from point C(1, sqrt(3), 0): |√2 * sqrt(3) + 0 - √6 / 2| / sqrt(3) = |√6 - √6 / 2| / sqrt(3) = |√6 / 2| / sqrt(3) = same as above, √2 / 2. Perfect. So this plane is indeed halfway between AB and CD.

Therefore, now I have the plane equation: √2 y + z = √6 / 2.

Now, the next step is to find how this plane cuts the tetrahedron into two pieces, and find the surface area of one of these pieces.

First, I need to figure out how the plane intersects the tetrahedron. Since the plane is parallel to AB and CD, and doesn't intersect them, as established earlier, it must intersect the other edges. Let me list all edges of the tetrahedron again:

AB, AC, AD, BC, BD, CD.

Plane doesn't intersect AB or CD, so it must intersect AC, AD, BC, BD. Wait, let me check each edge.

Edge AC: from A(0,0,0) to C(1, sqrt(3), 0). Let me find if the plane intersects this edge.

Parametrize AC: points on AC can be written as A + t(C - A) = (0 + t*1, 0 + t*sqrt(3), 0 + t*0) = (t, t sqrt(3), 0), where t ∈ [0,1].

Plug into plane equation: √2 y + z = √6 / 2 ⇒ √2*(t sqrt(3)) + 0 = √6 / 2 ⇒ t * sqrt(6) = √6 / 2 ⇒ t = 1/2. Therefore, the plane intersects AC at t = 1/2, which is point (1/2, (sqrt(3))/2, 0). Let me denote this as point P.

Similarly, check edge AD: from A(0,0,0) to D(1, sqrt(3)/3, 2√6/3). Parametrize as A + s(D - A) = (s*1, s*sqrt(3)/3, s*2√6/3), s ∈ [0,1].

Plug into plane equation: √2 y + z = √6 / 2.

Compute:

√2*(s*sqrt(3)/3) + (s*2√6/3) = √6 / 2.

Let me compute left side:

s [ √2 * sqrt(3)/3 + 2√6 / 3 ] = s [ (sqrt(6)/3 + 2 sqrt(6)/3 ) ] = s [ 3 sqrt(6)/3 ] = s sqrt(6).

Therefore, s sqrt(6) = √6 / 2 ⇒ s = 1/2. Therefore, the plane intersects AD at s = 1/2. The coordinates are:

x = 1*(1/2) = 1/2,

y = (sqrt(3)/3)*(1/2) = sqrt(3)/6,

z = (2√6 / 3)*(1/2) = √6 / 3.

So point Q(1/2, sqrt(3)/6, √6 / 3).

Next, check edge BC: from B(2, 0, 0) to C(1, sqrt(3), 0). Parametrize as B + r(C - B) = (2 - r, 0 + r*sqrt(3), 0 + r*0) = (2 - r, r sqrt(3), 0), r ∈ [0,1].

Plug into plane equation: √2 y + z = √6 / 2 ⇒ √2*(r sqrt(3)) + 0 = √6 / 2 ⇒ r sqrt(6) = √6 / 2 ⇒ r = 1/2. Therefore, intersection at r = 1/2. Coordinates: (2 - 1/2, (sqrt(3)/2), 0) = (3/2, sqrt(3)/2, 0). Let me call this point R.

Similarly, check edge BD: from B(2, 0, 0) to D(1, sqrt(3)/3, 2√6/3). Parametrize as B + u(D - B) = (2 - u, 0 + u*(sqrt(3)/3 - 0), 0 + u*(2√6/3 - 0)) = (2 - u, u*sqrt(3)/3, 2√6 u / 3), u ∈ [0,1].

Plug into plane equation √2 y + z = √6 / 2:

√2*(u sqrt(3)/3) + (2√6 u / 3) = √6 / 2.

Left side: u [ √2 * sqrt(3)/3 + 2√6 / 3 ] = same as before: u [ sqrt(6)/3 + 2 sqrt(6)/3 ] = u [ sqrt(6) ]. Therefore:

u sqrt(6) = √6 / 2 ⇒ u = 1/2.

Therefore, the plane intersects BD at u = 1/2. Coordinates:

x = 2 - 1/2 = 3/2,

y = (sqrt(3)/3)*(1/2) = sqrt(3)/6,

z = (2√6 / 3)*(1/2) = √6 / 3.

So point S(3/2, sqrt(3)/6, √6 / 3).

Therefore, the plane intersects the tetrahedron in four points: P(1/2, sqrt(3)/2, 0), Q(1/2, sqrt(3)/6, √6 / 3), R(3/2, sqrt(3)/2, 0), and S(3/2, sqrt(3)/6, √6 / 3). Therefore, the intersection is a quadrilateral PQRS.

Wait, let me check if PQRS is a parallelogram? Since the plane is parallel to AB and CD, maybe the intersection is a parallelogram. Let me check the coordinates.

Let me compute vectors PQ and RS.

Point P(1/2, sqrt(3)/2, 0), Q(1/2, sqrt(3)/6, √6 / 3), R(3/2, sqrt(3)/2, 0), S(3/2, sqrt(3)/6, √6 / 3).

Vector PQ: Q - P = (0, sqrt(3)/6 - sqrt(3)/2, √6/3 - 0) = (0, -sqrt(3)/3, √6 / 3).

Vector RS: S - R = (3/2 - 3/2, sqrt(3)/6 - sqrt(3)/2, √6 / 3 - 0) = (0, -sqrt(3)/3, √6 / 3). So PQ and RS are equal vectors. Similarly, vector QR: R - Q = (3/2 - 1/2, sqrt(3)/2 - sqrt(3)/6, 0 - √6 / 3) = (1, sqrt(3)/3, -√6 / 3).

Vector SP: P - S = (1/2 - 3/2, sqrt(3)/2 - sqrt(3)/6, 0 - √6 / 3) = (-1, sqrt(3)/3, -√6 / 3). So QR and SP are negatives of each other. Therefore, PQRS is a parallelogram. So the intersection is a parallelogram.

Therefore, when the plane cuts the tetrahedron, it creates a smaller piece (probably a smaller tetrahedron and a frustum-like shape?) Wait, no. Since the plane cuts through four edges, creating a quadrilateral cross-section, the two pieces are a smaller polyhedron with a quadrilateral face and the rest of the original tetrahedron. But since the original tetrahedron is cut by a plane, the two pieces will be: one piece is the part containing the vertex D (since the plane is halfway between AB and CD, so maybe not?), and the other piece is the part containing AB. Wait, need to visualize.

Alternatively, perhaps the plane divides the tetrahedron into a smaller tetrahedron and a pentahedron? Wait, since the intersection is a quadrilateral, the upper piece (assuming the plane is cutting through) would have a quadrilateral face and three original faces? Wait, maybe not. Let me think.

Wait, the original tetrahedron has four triangular faces. The cutting plane intersects four edges, so the lower piece (assuming the plane is between AB and CD) would have the original base face ABC? Wait, no, the base is ABC, but the plane cuts through AC and BC, so the lower piece might have a quadrilateral face PQRS and parts of the original faces. Alternatively, perhaps it's easier to think of the two pieces: one is a smaller tetrahedron and the other is a kind of prism or something. But since the plane is parallel to AB and CD, which are skew lines, and halfway between them, perhaps the intersection is a parallelogram, and the two pieces are a prism with the parallelogram and another piece?

Alternatively, since the plane cuts through four edges, maybe each piece has a quadrilateral and three triangular faces? Wait, let me count the faces. The original tetrahedron has four triangular faces. After cutting, each piece will have some original faces and the new quadrilateral face. For example, the piece containing vertex D would have the original face ABD, original face ACD, and parts of the other faces? Wait, no, since the cutting plane cuts through edges AD and BD, so the piece containing D would have the quadrilateral PQRS and the original face ADC? Wait, maybe not. Let me try to figure out.

Alternatively, maybe it's better to think about which vertices are on which side of the plane. Let me check the position of each vertex relative to the plane √2 y + z = √6 / 2.

Compute for each vertex:

A(0,0,0): √2*0 + 0 = 0 < √6 / 2 ≈ 1.2247. So A is on one side.

B(2,0,0): √2*0 + 0 = 0 < √6 / 2. So B is on the same side as A.

C(1, sqrt(3), 0): √2*sqrt(3) + 0 ≈ 2.449 > √6 / 2 ≈ 1.2247. So C is on the other side.

D(1, sqrt(3)/3, 2√6/3): √2*(sqrt(3)/3) + 2√6 / 3.

Compute:

√2 * sqrt(3)/3 = sqrt(6)/3 ≈ 0.816,

2√6 / 3 ≈ 1.632,

Total ≈ 0.816 + 1.632 ≈ 2.448 ≈ √6 ≈ 2.449, which is greater than √6 / 2. Therefore, D is also on the same side as C.

Therefore, the plane divides the tetrahedron into two parts: one part contains A and B (and the other part contains C and D). Therefore, the piece containing A and B would have the original faces AB, and parts of the other faces, and the quadrilateral cross-section PQRS. Similarly, the piece containing C and D would have the original faces CD and parts of other faces, and the quadrilateral cross-section.

But since the problem says "the plane... cuts ABCD into two pieces", and asks for the surface area of one of these pieces. So we need to calculate the surface area of one of them. Since surface area is additive, but we need to be careful which piece we choose.

But depending on which piece we take, the surface area will include different faces. However, given that the plane is halfway between AB and CD, it's possible that the piece containing A and B has a certain surface area, and the piece containing C and D has another. However, since the original tetrahedron has all edges equal, maybe the two pieces are congruent? But since the plane is between AB and CD, which are skew, maybe not. However, given the regular tetrahedron is symmetric, perhaps the surface areas are the same? Wait, but AB and CD are not symmetric in the tetrahedron. Wait, AB is an edge, CD is another edge, but the tetrahedron is regular, so all edges are the same. Therefore, the two pieces should have congruent surface areas, so the answer is the same for either piece. So maybe I can compute it for one and it will be the same.

But to be safe, I need to figure out exactly which piece's surface area they are asking for. However, since it's just asking for one of the pieces, we can compute for either. Let me assume we are to compute the surface area of the piece that contains A and B, which is the lower piece (since the plane is halfway between AB and CD, so AB is on one side, and CD on the other). But maybe not. Anyway, let's proceed.

First, the surface area of one piece will consist of:

- The original faces that are entirely on one side of the plane.

- The new face created by the plane, which is the quadrilateral PQRS.

But we have to check which original faces are entirely on one side. Let me think again. The piece containing A and B (assuming the plane is cutting between AB and CD). The original faces:

Face ABC: points A, B, C. Since A and B are on one side, and C is on the other side, so the face ABC is divided by the plane. Similarly, face ABD: points A, B, D. Both A and B on one side, D on the other side. So the face ABD is also divided by the plane. Face ACD: points A, C, D. A is on one side, C and D on the other. So divided. Face BCD: points B, C, D. B on one side, C and D on the other. Divided. Therefore, all original faces are intersected by the plane. Therefore, the piece containing A and B will have:

- A portion of each original face.

Plus the new quadrilateral face PQRS.

But that complicates things because calculating the area of portions of the original faces. However, maybe there is a smarter way. Wait, but the problem says the plane is parallel to edges AB and CD. So, maybe the surface area of one piece includes the original face AB (but AB is entirely on one side?), but AB is on the same side as A and B. Wait, but AB is an edge. However, since the plane is parallel to AB, the edge AB is not intersected by the plane. Therefore, the original face AB is actually not part of the surface area of the piece? Wait, no. The piece containing A and B: its surface area includes the original face AB? Wait, no. Wait, the original face AB is part of the tetrahedron. If the plane is cutting off a part, then the original face AB is entirely on one side of the plane. Since AB is on the same side as A and B, which are part of the piece containing A and B. However, the plane is parallel to AB, so the original face AB is not intersected by the plane. Therefore, the original face AB is entirely part of the surface of the piece containing A and B. Similarly, the original face CD is entirely part of the surface of the other piece. Wait, but AB is an edge, not a face. The original face containing AB is triangle ABC. Wait, triangle ABC is intersected by the plane, as point C is on the other side. Therefore, triangle ABC is split into two parts. Therefore, the piece containing A and B will have a portion of triangle ABC, and the original face ABD is split into two parts as well. Similarly for the other faces. So actually, the surface area of the piece containing A and B will consist of:

- The portion of face ABC that's on the A/B side.

- The portion of face ABD that's on the A/B side.

- The portion of face ACD that's on the A/B side.

- The portion of face BCD that's on the A/B side.

Plus the new face PQRS.

But calculating these portions might be complex. However, since the plane is parallel to AB and CD and cuts the edges AC, AD, BC, BD at their midpoints (since t = 1/2, u = 1/2, etc.), maybe these portions are triangles or something with known areas.

Wait, when the plane intersects AC at t = 1/2 and BC at t = 1/2, then the portion of triangle ABC on the A/B side is the triangle ABP'? Wait, no. Wait, point P is on AC, and point R is on BC. So in triangle ABC, the plane cuts AC at P and BC at R, so the portion of triangle ABC on the A/B side is the triangle APR? Wait, but AP is from A to P (midpoint of AC), and BR is from B to R (midpoint of BC). Wait, actually, connecting P and R. The line PR divides triangle ABC into two regions: the smaller triangle APR and the quadrilateral PRBC. But since the plane is cutting through the tetrahedron, the piece containing A and B would have the area which is the part of triangle ABC below the plane? Since we're assuming the plane is cutting through AC and BC at their midpoints. However, actually, the plane cuts AC and BC at their midpoints, so PR is a line connecting midpoints of AC and BC, which is a midline of triangle ABC, parallel to AB and half its length. Wait, since AB is length 2, then PR would be length 1. Therefore, the area of triangle APR would be 1/4 of the area of ABC? Wait, the midline divides the triangle into two parts: the smaller triangle APR has area 1/4 of ABC, and the trapezoid PRBC has area 3/4? Wait, no. Wait, if you connect midpoints of AC and BC, then the line PR is parallel to AB and half its length, so the area of APR is (1/2 * base * height). Since PR is half of AB, and the height is half of the height of ABC, so area is 1/4 of original. Therefore, the area of APR is 1/4 of ABC. Similarly for other faces?

But since the plane is cutting through other edges as well, maybe similar logic applies.

Alternatively, maybe it's easier to note that since all the cuts are at midpoints (since t = 1/2, etc.), the intersection is a midline parallelogram, and the resulting piece might have some proportional areas.

But let me think again. For the piece containing A and B, which has the face AB (original) and the quadrilateral PQRS. However, the original face AB is part of the surface area, but also parts of other original faces. But perhaps when you take the piece containing AB, its surface area is the area of AB (which is 2, since AB is length 2, but wait, the surface area would include the area of the face AB. Wait, no. Wait, the surface area of a piece is the sum of all its faces. The piece containing AB would have:

- The original face AB (which is a triangle? No, AB is an edge. Wait, AB is an edge of the original tetrahedron. The original face containing AB is triangle ABC and triangle ABD.

Wait, this is getting confusing. Maybe I need to approach this more systematically.

First, let me note that the plane cuts the tetrahedron into two pieces. Each piece has:

- The original faces that are entirely on one side of the plane.

- The portions of the original faces that are cut by the plane.

- The new face, which is the quadrilateral PQRS.

But since all original faces are intersected by the plane, there are no original faces entirely on one side. Therefore, the surface area of each piece will consist of:

- The new face PQRS.

- The portions of each original face that are on that piece.

Therefore, to compute the surface area of one piece, we need to calculate the area of PQRS, plus the areas of the portions of the original faces.

But maybe there is a smarter way. Since the plane is parallel to AB and CD, and cuts the tetrahedron symmetrically? Wait, but AB and CD are skew lines, so maybe not symmetric. However, since the plane is halfway between AB and CD, and cuts each of the edges AC, AD, BC, BD at their midpoints, maybe the piece containing A and B is similar to the original tetrahedron scaled down? But not sure.

Alternatively, since the plane is cutting the edges AC, AD, BC, BD at midpoints, which are halfway along each edge. Therefore, the piece containing A and B is a kind of pyramid with base PQRS and apexes at A and B? Wait, no, since the piece is a polyhedron with two parts? Wait, perhaps the piece containing A and B is a kind of frustum-like shape with a quadrilateral base and a triangular top? No, maybe not. Alternatively, maybe it's a pentahedron with the quadrilateral face PQRS and three original triangular faces?

Wait, let me think about the vertices of the piece containing A and B. It includes the original vertices A and B, and the points P, Q, R, S. Therefore, the polyhedron has vertices A, B, P, Q, R, S. Let me try to see how the faces are structured.

Original face ABC is split by the plane into two parts: triangle APR (connecting A, P, R) and quadrilateral PRBC. But since we are taking the piece containing A and B, we have triangle APR and the quadrilateral APR... Wait, no. The piece containing A and B would have the triangle APR and the triangle ABP? No, since P is on AC and R is on BC.

Wait, maybe the piece containing A and B has the following faces:

- The original face ABD: split into two parts by the plane. The portion on the A/B side is triangle AQS and something else? Similarly, the original face ACD is split into triangle APQ and something else. The original face ABC is split into APR and another part. And the original face BCD is split into BR something and something else. This seems too vague.

Alternatively, since the plane cuts each of the edges AC, AD, BC, BD at their midpoints, we can think of the piece containing A and B as follows:

- It includes triangle ABD, but with parts removed. However, the intersection with the plane cuts AD at Q and BD at S. So in triangle ABD, the plane cuts it along QS. So the piece containing A and B would have triangle ABQ and triangle ABS? Wait, no. If you have triangle ABD, and you cut it with a line QS, which connects Q (midpoint of AD) to S (midpoint of BD). Then, the portion of triangle ABD on the A/B side is the triangle ABQ? No, because Q is on AD and S is on BD. The line QS divides triangle ABD into two parts: a smaller triangle ASQ and a quadrilateral QBS? Wait, maybe not. Let me visualize.

Alternatively, since Q and S are midpoints, QS is a midline of triangle ABD, parallel to AB. Therefore, the portion of triangle ABD on the A/B side is a smaller triangle similar to ABD with half the side lengths, so area 1/4 of ABD. Similarly, the other portion is a trapezoid.

But if the piece containing A and B includes triangle ABD, but only up to QS, then the area would be 1/4 of the area of ABD? But I need to figure out exactly which parts are included.

Alternatively, since the plane cuts the edges AC, AD, BC, BD at their midpoints, the piece containing A and B is a polyhedron whose surface areas consist of:

- The original face AB (which is an edge, but the face containing AB is ABC and ABD). However, since the plane is parallel to AB, the face AB is not intersected, but the faces ABC and ABD are intersected. Therefore, the piece containing A and B has:

- For face ABC: the trapezoid APRB? Wait, AP is from A to midpoint of AC, and BR is from B to midpoint of BC. Connecting P and R, which is parallel to AB and half its length. So the area of APRB would be the area of triangle ABC minus the area of triangle PRC? Wait, since PR divides ABC into a smaller triangle (ARPR?) Wait, confusing again.

Alternatively, since PR is the midline of ABC, parallel to AB, then APRB is a trapezoid with bases AB and PR, and height half of the height of ABC. The area of this trapezoid is (AB + PR) * height / 2. Since AB = 2, PR = 1 (since it's half the length of AB), and the height is half of the original height. The original area of ABC is (base * height)/2 = (2 * h)/2 = h, where h is the height. Wait, but in a regular tetrahedron, all faces are equilateral triangles. So the area of ABC is (sqrt(3)/4) * (2)^2 = sqrt(3).

So area of ABC is sqrt(3). The trapezoid APRB (assuming that's correct) would have area (AB + PR)/2 * height_between_them. Since PR is halfway up, the height between AB and PR is half the height of ABC. Therefore, area is (2 + 1)/2 * (h / 2) = (3/2) * (h / 2) = 3h / 4. Earlier, area of ABC is h, so this would be 3/4 of the area? Wait, but that seems conflicting with the earlier thought that the area of APR is 1/4 of ABC. Maybe not. Let me calculate with actual numbers.

The area of ABC is sqrt(3). Let me compute the area of APRB (assuming it's a trapezoid). The length of AB is 2, length of PR is 1, and the distance between AB and PR is half the height of ABC.

The height of ABC is the height of an equilateral triangle with side 2: h = sqrt(2^2 - 1^2) = sqrt(3). Therefore, the distance between AB and PR is sqrt(3)/2. Therefore, the area of the trapezoid APRB is (AB + PR)/2 * distance_between = (2 + 1)/2 * sqrt(3)/2 = 3/2 * sqrt(3)/2 = (3 sqrt(3))/4. Which is 3/4 of the area of ABC, since the area of ABC is sqrt(3). Therefore, the portion of face ABC on the A/B side is (3 sqrt(3))/4. Similarly, for face ABD, since it's also an equilateral triangle with area sqrt(3), the portion on the A/B side would also be (3 sqrt(3))/4. Wait, but then what about the other faces?

Similarly, for face ACD: this face is triangle ACD, which is split by the plane. The plane intersects AC at P and AD at Q. Therefore, the portion of ACD on the A/B side is the triangle APQ. Since P and Q are midpoints, triangle APQ is similar to ACD with scaling factor 1/2, so its area is (1/2)^2 * area of ACD = 1/4 * sqrt(3). Similarly, for face BCD, which is triangle BCD, the portion on the A/B side is triangle BR S? Wait, the plane intersects BC at R and BD at S. Therefore, triangle BRS is similar to BCD with area 1/4 * sqrt(3). But wait, triangle BRS is not on the A/B side, since R and S are on the opposite side? Wait, no. Wait, the piece containing A and B would have the portion of BCD that's on the A/B side. However, since point B is on the A/B side and points C and D are on the other side, the portion of BCD on the A/B side is triangle BR... wait, point R is on BC, and point S is on BD. So the intersection of the plane with BCD is line RS. Therefore, the portion of triangle BCD on the A/B side is the triangle BRS. Since B is on the A/B side, and R and S are on the plane. Since R and S are midpoints, triangle BRS is similar to BCD with scaling factor 1/2, so area is 1/4 of sqrt(3). Therefore, the area of BRS is sqrt(3)/4.

Similarly, for face ACD, the area of APQ is sqrt(3)/4.

Therefore, putting it all together, the surface area of the piece containing A and B would be:

- The area of APRB (trapezoid): 3 sqrt(3)/4

- The area of ABQ... Wait, no. Wait, what about the other portions? Wait, the piece containing A and B includes:

- The portion of ABC: 3 sqrt(3)/4

- The portion of ABD: 3 sqrt(3)/4

- The portion of ACD: sqrt(3)/4

- The portion of BCD: sqrt(3)/4

Plus the area of the new quadrilateral PQRS.

But wait, if I add all these, I would be overcounting? Wait, no. Because the piece contains parts of each original face. For example, the trapezoid APRB is part of ABC, the trapezoid ... something is part of ABD, and so on? Wait, no, earlier I said for ABC, the area is 3 sqrt(3)/4, and for ABD, similarly 3 sqrt(3)/4. However, the areas for ACD and BCD are each sqrt(3)/4. But I need to check if those portions are indeed part of the piece containing A and B.

Wait, for ACD: the piece containing A and B includes the portion of ACD that is on the A/B side. Since ACD has vertices A, C, D. A is on the A/B side, while C and D are on the other side. The plane cuts ACD at P (on AC) and Q (on AD). Therefore, the portion of ACD on the A/B side is triangle APQ, which has area sqrt(3)/4. Similarly, the portion of BCD on the A/B side is triangle BRS, area sqrt(3)/4. Therefore, adding up:

- ABC: 3 sqrt(3)/4

- ABD: 3 sqrt(3)/4

- ACD: sqrt(3)/4

- BCD: sqrt(3)/4

Plus the area of PQRS.

But wait, but the original tetrahedron has four faces, and we're adding up all four portions? However, each of these portions is part of the surface area of the piece. However, when you add them, you have to make sure that you are not double-counting or missing something. But seems okay.

However, actually, the surface area of the piece should include:

- All the areas of the parts of the original faces that are on the piece.

- The area of the new face PQRS.

So that's the total.

So let's calculate:

First, sum of the original portions:

For ABC: 3√3 / 4

For ABD: 3√3 / 4

For ACD: √3 / 4

For BCD: √3 / 4

Total original portions: 3√3 / 4 + 3√3 / 4 + √3 / 4 + √3 / 4 = (3 + 3 + 1 + 1) * √3 / 4 = 8 * √3 / 4 = 2√3.

Then, add the area of PQRS.

So now we need to calculate the area of PQRS, the quadrilateral.

Given that PQRS is a parallelogram, as established earlier. To find its area, we can compute the magnitude of the cross product of vectors PQ and PS or something similar. But since it's a parallelogram, area is base * height or |PQ × PR| or something.

Alternatively, since PQRS is a parallelogram, and we have coordinates for all four points, we can compute its area using coordinates.

Points:

P(1/2, sqrt(3)/2, 0)

Q(1/2, sqrt(3)/6, √6 / 3)

R(3/2, sqrt(3)/2, 0)

S(3/2, sqrt(3)/6, √6 / 3)

Let me compute vectors PQ and PS.

Vector PQ = Q - P = (0, sqrt(3)/6 - sqrt(3)/2, √6 / 3 - 0) = (0, -sqrt(3)/3, √6 / 3)

Vector PS = S - P = (3/2 - 1/2, sqrt(3)/6 - sqrt(3)/2, √6 / 3 - 0) = (1, -sqrt(3)/3, √6 / 3)

Then, the area of the parallelogram is |PQ × PS|.

Compute cross product:

PQ = (0, -sqrt(3)/3, √6 / 3)

PS = (1, -sqrt(3)/3, √6 / 3)

Cross product PQ × PS:

|i       j         k         |
|0   -sqrt(3)/3   √6/3|
|1   -sqrt(3)/3   √6/3|

= i [ (-sqrt(3)/3 * √6/3 - √6/3 * (-sqrt(3)/3) ) ] - j [ 0 * √6/3 - √6/3 * 1 ] + k [ 0 * (-sqrt(3)/3) - (-sqrt(3)/3) * 1 ]

Let me compute each component:

i component:

(-sqrt(3)/3)(√6/3) - (√6/3)(-sqrt(3)/3) = (-sqrt(18)/9) + sqrt(18)/9 = 0.

j component:

- [ 0 - √6/3 * 1 ] = - [ -√6 / 3 ] = √6 / 3.

k component:

0 - (-sqrt(3)/3 * 1) = sqrt(3)/3.

Therefore, the cross product is (0, √6/3, sqrt(3)/3).

Therefore, the magnitude of this vector is sqrt( (0)^2 + ( √6 / 3 )^2 + ( √3 / 3 )^2 ) = sqrt( (6/9) + (3/9) ) = sqrt(9/9) = sqrt(1) = 1.

So the area of the parallelogram PQRS is 1.

Wait, that's interesting. So the area of PQRS is 1.

Therefore, the total surface area of the piece containing A and B is 2√3 + 1.

But wait, let me check this calculation again. Because the cross product result magnitude was 1? Let me recheck.

Calculating the cross product components:

i component: (-sqrt(3)/3 * √6/3 - √6/3 * (-sqrt(3)/3))

First term: (-sqrt(3)*sqrt(6))/9 = -sqrt(18)/9 = - (3*sqrt(2))/9 = -sqrt(2)/3.

Second term: - √6/3 * (-sqrt(3)/3) = + sqrt(18)/9 = same as before, sqrt(2)/3.

Therefore, total i component: -sqrt(2)/3 + sqrt(2)/3 = 0. Correct.

j component: - [ (0 * √6/3 - √6/3 * 1) ] = - [ -√6 / 3 ] = √6 / 3.

Wait, but earlier I had j component as √6 / 3.

k component: 0 * (-sqrt(3)/3) - (-sqrt(3)/3)*1 = 0 + sqrt(3)/3 = sqrt(3)/3.

Therefore, vector is (0, √6/3, sqrt(3)/3). Then magnitude squared:

(√6/3)^2 + (sqrt(3)/3)^2 = (6/9) + (3/9) = 9/9 = 1. So magnitude is 1. Yes. Therefore, area of PQRS is 1. Interesting.

So the total surface area of the piece containing A and B is 2√3 + 1? Wait, earlier we calculated the sum of original portions as 2√3, and then adding 1 from PQRS.

But let me check again: the original portions sum to 2√3? Earlier steps:

Sum of original portions: ABC (3√3/4), ABD (3√3/4), ACD (√3/4), BCD (√3/4). Adding up:

3√3/4 + 3√3/4 = 6√3/4 = 3√3/2.

Then √3/4 + √3/4 = √3/2.

So total original portions: 3√3/2 + √3/2 = 4√3/2 = 2√3. Yes. Then adding the area of PQRS, which is 1. So total surface area = 2√3 + 1.

But let me check if this answer makes sense. The original surface area of the tetrahedron is 4*(sqrt(3)/4)*2^2 = 4*(sqrt(3)/4)*4 = 4*sqrt(3). Wait, original surface area is 4*(area of one face). Each face has area sqrt(3), as calculated earlier. Therefore, original surface area is 4*sqrt(3). If the two pieces have surface areas 2√3 + 1 and something else, then their sum should be equal to 4√3 + (area of PQRS). Wait, no. Wait, when you cut the tetrahedron with the plane, the total surface area of the two pieces is equal to the original surface area plus twice the area of the new face PQRS. Because each piece has the new face PQRS as a part of its surface area. Therefore, total surface area of both pieces is 4√3 + 2*1 = 4√3 + 2. But if one of the pieces has surface area 2√3 + 1, then the other piece would have surface area (4√3 + 2) - (2√3 + 1) = 2√3 + 1. Therefore, both pieces have the same surface area, which makes sense, as we thought earlier. Therefore, the answer should be 2√3 + 1. But wait, let me check this with another approach.

Alternatively, maybe there is a miscalculation. Let me think again about the original portions. The piece containing A and B has parts of all four original faces. But when we add those parts, do they cover all the original faces? For example, the original face ABC is split into the trapezoid APRB (area 3√3 /4) and the triangle PRC (area sqrt(3)/4). Similarly, original face ABD is split into trapezoid ABQS (area 3√3 /4) and triangle QSD (area sqrt(3)/4). Original face ACD is split into triangle APQ (area sqrt(3)/4) and quadrilateral PQDC (area 3√3 /4). Original face BCD is split into triangle BRS (area sqrt(3)/4) and quadrilateral RSPC (area 3√3 /4). Therefore, the piece containing A and B includes:

- From ABC: 3√3 /4

- From ABD: 3√3 /4

- From ACD: sqrt(3)/4

- From BCD: sqrt(3)/4

Total: 3√3 /4 + 3√3 /4 + sqrt(3)/4 + sqrt(3)/4 = 8√3 /4 = 2√3, as before.

Plus the new face area 1. So total 2√3 + 1. So this seems to check out.

But wait, what about the original face AB? Wait, AB is part of both faces ABC and ABD. But we already considered portions of ABC and ABD. Since AB is an edge, not a face, it's part of both ABC and ABD. However, when we calculated the area of APRB (from ABC) and ABQS (from ABD), the area of AB is counted twice? Wait, no. Because APRB is a trapezoid that includes AB. Wait, APRB is part of ABC, and ABQS is part of ABD. However, AB is common to both faces. But when we add the areas of APRB and ABQS, we are counting the area of AB twice? No, because APRB is part of ABC, and ABQS is part of ABD. Since APRB includes AB and some other area, and ABQS includes AB and some other area. But the line AB is shared between these two trapezoids. However, in reality, AB is an edge, and the area calculation for the trapezoids would have AB as a base. Therefore, when we add 3√3 /4 and 3√3 /4, we might be overcounting the area of AB? Wait, no. Because each trapezoid is part of a different face. The trapezoid APRB is in ABC, and ABQS is in ABD. The overlap between them is just the edge AB, which has no area. Therefore, adding them is okay.

So total surface area is indeed 2√3 + 1.

But let me check with numerical values. The original surface area is 4√3 ≈ 6.928. The total surface area of both pieces is 4√3 + 2 ≈ 6.928 + 2 = 8.928. The individual piece surface area is 2√3 + 1 ≈ 3.464 + 1 = 4.464. So each piece has surface area ≈ 4.464. Seems reasonable.

But let me check if the area of PQRS is indeed 1. Given that the cross product magnitude is 1, which is correct. Let me verify with coordinates.

Alternatively, since PQRS is a parallelogram, we can compute its area by calculating the length of PQ and the height between the sides. Let me calculate the length of PQ.

Vector PQ is (0, -sqrt(3)/3, sqrt(6)/3). Its magnitude is sqrt(0 + (sqrt(3)/3)^2 + (sqrt(6)/3)^2 ) = sqrt( (3/9) + (6/9) ) = sqrt(9/9) = 1. So length of PQ is 1.

Similarly, the length of PR is sqrt( (3/2 - 1/2)^2 + (sqrt(3)/2 - sqrt(3)/2)^2 + (0 - 0)^2 ) = sqrt( (1)^2 + 0 + 0 ) = 1. Wait, PR is from P(1/2, sqrt(3)/2, 0) to R(3/2, sqrt(3)/2, 0). So the vector PR is (1, 0, 0), so length 1. Wait, then if PQ and PR are both length 1, and the angle between them is such that the area is |PQ × PR|, which we already calculated as 1. Wait, but PQ and PR would not be adjacent sides of the parallelogram. Wait, PQ and PS are adjacent sides. Wait, earlier we computed PQ × PS = (0, √6/3, √3/3), magnitude 1. Therefore, the area is indeed 1. So PQRS has area 1.

Therefore, the total surface area of the piece is 2√3 + 1.

But the problem says "the plane... cuts ABCD into two pieces. Find the surface area of one of these pieces."

Therefore, the answer is 2√3 + 1. However, let me check if there is an alternative way to get this or if I made a mistake.

Alternatively, since the original surface area is 4√3, and the new surface area is 2*1 = 2, then total surface area of both pieces is 4√3 + 2. If one piece has surface area 2√3 + 1, then the other is 4√3 + 2 - (2√3 + 1) = 2√3 + 1, which matches. So that seems consistent.

But let me check with another approach. If I consider that the piece containing A and B is a kind of prism or frustum. However, given the complexity, and that cross-checks seem consistent, maybe 2√3 + 1 is correct.

But let me check with vector areas.

Alternatively, think about the fact that the piece containing A and B has the following faces:

- The trapezoid APRB (area 3√3 /4)

- The trapezoid ABQS (area 3√3 /4)

- The triangle APQ (area sqrt(3)/4)

- The triangle BRS (area sqrt(3)/4)

- The parallelogram PQRS (area 1)

Wait, but adding those would be 3√3/4 + 3√3/4 + sqrt(3)/4 + sqrt(3)/4 + 1 = same as before, 2√3 + 1.

But actually, these faces are different? Wait, APQ is part of ACD, BRS is part of BCD, APRB is part of ABC, ABQS is part of ABD, and PQRS is the new face. Therefore, yes, the total surface area is as calculated.

Therefore, I think this is correct. However, let me verify with approximate numbers. Let me compute 2√3 + 1. Since √3 ≈ 1.732, so 2*1.732 ≈ 3.464 + 1 ≈ 4.464.

Original surface area is 4*√3 ≈ 6.928. The total surface area after cutting is 6.928 + 2 ≈ 8.928. Divided by two, each piece has ≈ 4.464. Which seems okay.

But let me check if there is an alternative interpretation. For example, maybe the problem considers only the surface area of the piece that is "above" the plane, but since we had to compute for one piece, and the answer is 2√3 + 1, maybe that is the answer.

Alternatively, the question may expect a different form. Let me check if 2√3 + 1 can be simplified or expressed differently. However, 2√3 + 1 is already simplified.

But let me check again the calculation of the area of PQRS. Because earlier I found that the cross product magnitude was 1. Given that PQRS is a parallelogram with sides of length 1 (as calculated) and area 1. Which is interesting. Given that the original tetrahedron has edge length 2, and the area of PQRS is 1, which is less than the area of the original faces.

Alternatively, perhaps there's a mistake in calculating the area of PQRS. Let me try another approach.

Since PQRS is a parallelogram, its area can also be calculated as base * height. Let me take PQ as a base, which has length 1. What is the height corresponding to this base?

The height would be the distance between the lines PQ and RS. Since PQ and RS are opposite sides of the parallelogram. But since the plane is parallel to AB and CD, and PQ is parallel to AB? Wait, earlier we found that PQ is parallel to AB? Wait, vector PQ is (0, -sqrt(3)/3, sqrt(6)/3). AB is (2, 0, 0). Are they parallel? No, because the direction vectors are not scalar multiples. Wait, but earlier we said the plane is parallel to AB and CD. So the plane contains lines parallel to AB and CD. However, PQ is a vector in the plane, so it should be a combination of directions parallel to AB and CD. But vector PQ was computed as (0, -sqrt(3)/3, sqrt(6)/3), which is not a scalar multiple of AB or CD. However, since PQ is in the plane, which is parallel to AB and CD, then PQ can be expressed as a combination of AB and CD.

But regardless, since we already calculated the area of the parallelogram as 1 using cross product, and verified with coordinates, I think it's safe.

Therefore, I think the answer is 2√3 + 1. However, let me check if the question asks for the surface area of one of the pieces, and if there's an alternative way that the answer is different.

Wait, another approach: the surface area of one piece includes the area of the new face plus the areas of the parts of the original faces. However, what about the other piece? If the other piece also has surface area 2√3 + 1, then the total is 4√3 + 2, which matches with original total surface area plus twice the area of the new face. Therefore, seems correct.

Alternatively, maybe the question expects the answer as 2√3 + 1, but I need to check if there's any error in assuming that the original portions sum to 2√3. Let me verify with another method.

Each of the original faces is divided into a portion on the piece and a portion on the other piece. For example, face ABC is split into 3√3 /4 (on A/B side) and sqrt(3)/4 (on C/D side). Similarly for other faces. Therefore, total original surface area is 4 * sqrt(3). The total of the portions on the A/B side is 2√3, and similarly on the other side is 2√3. Therefore, adding the new face area (1) for each piece, the total surface area for each piece is 2√3 + 1. So yes.

Therefore, I think this is correct.

But to make sure, let me check the area of PQRS again. Given that PQRS is a parallelogram with area 1, and the original tetrahedron has volume, maybe we can compute volume of the piece and check some consistency?

The volume of the original tetrahedron with side length 2 is V = (sqrt(2)/12) * (2)^3 = (sqrt(2)/12)*8 = (2 sqrt(2))/3 ≈ 0.9428.

If we compute the volume of the piece containing A and B, which includes the original portions of the faces. However, since the plane cuts the tetrahedron halfway between AB and CD, and since AB and CD are skew, it's not straightforward. However, since the plane cuts the edges AC, AD, BC, BD at their midpoints, the piece containing A and B is a kind of "half" of the tetrahedron. However, in a regular tetrahedron, cutting through midpoints of edges... Maybe the volume of this piece is 1/2 of the original? But let me check.

Actually, if you connect midpoints of edges, the resulting figure might be a smaller tetrahedron or something else. Wait, but in this case, the intersection is a parallelogram, so the piece containing A and B is a kind of frustum-like shape. However, its volume can be calculated using determinants or other methods. However, since this is getting complex, and we already have surface area as 2√3 + 1, and all the checks seem consistent, I think it's safe to go with this answer.

But just to be thorough, let me calculate the volume of the piece containing A and B. Since the plane cuts the edges AC, AD, BC, BD at their midpoints, we can use the formula for the volume of a pyramid or something else.

Alternatively, consider that the piece containing A and B can be considered as the union of pyramids or prisms. However, given the time constraints, maybe not. Alternatively, note that the plane divides the tetrahedron into two congruent pieces? Wait, no, since AB and CD are not symmetric in the tetrahedron. However, since the plane is halfway between them, perhaps the volumes are equal? If so, then volume of each piece would be half of original, which is (2√2)/6 = √2 / 3 ≈ 0.4714. But how does this relate to the surface area? Not directly, but maybe the surface area is connected.

However, given that the surface area calculation seems consistent, and cross-checked, and the total surface area makes sense, I think I can go with the answer 2√3 + 1.

But let me check with another perspective. The problem says the plane is parallel to AB and CD, and halfway between them. In a regular tetrahedron, AB and CD are skew lines, and the distance between them is √2 as calculated before. The midpoint plane would be at half that distance. The intersection is a parallelogram, and the surface area includes that quadrilateral. Given that, and the calculations above, I think the answer is 2√3 + 1.

However, to cross-verify with an example, suppose the edge length was different. Suppose edge length is 1, then calculations change, but since we are given edge length 2, I think calculations are correct.

Alternatively, check if my value for the area of PQRS being 1 is correct. Suppose the edge length is 2, and PQRS area is 1. Let me think about units. If the tetrahedron has edge length 2, then the area of a face is sqrt(3). The area of PQRS is 1, which is less than that, which is plausible. Alternatively, if I consider the coordinates:

Points P(1/2, sqrt(3)/2, 0), Q(1/2, sqrt(3)/6, √6 / 3), R(3/2, sqrt(3)/2, 0), S(3/2, sqrt(3)/6, √6 / 3).

Let me calculate the area of PQRS using another method. For example, divide the parallelogram into two triangles. Take triangle PQR and triangle PRS.

Compute area of triangle PQR and double it.

Coordinates:

P(1/2, sqrt(3)/2, 0)

Q(1/2, sqrt(3)/6, √6 / 3)

R(3/2, sqrt(3)/2, 0)

Compute vectors PQ and PR.

Vector PQ = Q - P = (0, -sqrt(3)/3, sqrt(6)/3)

Vector PR = R - P = (1, 0, 0)

Area of triangle PQR is (1/2)|PQ × PR|.

Compute cross product PQ × PR:

PQ = (0, -sqrt(3)/3, sqrt(6)/3)

PR = (1, 0, 0)

Cross product:

|i       j         k         |
|0   -sqrt(3)/3   sqrt(6)/3|
|1     0           0         |

= i [ (-sqrt(3)/3 * 0 - sqrt(6)/3 * 0) ] - j [ 0*0 - sqrt(6)/3 * 1 ] + k [ 0*0 - (-sqrt(3)/3 * 1) ]

= i (0) - j (-sqrt(6)/3) + k (sqrt(3)/3)

= (0, sqrt(6)/3, sqrt(3)/3)

Which has magnitude sqrt( (sqrt(6)/3)^2 + (sqrt(3)/3)^2 ) = sqrt(6/9 + 3/9) = sqrt(9/9) = 1. Therefore, the area of triangle PQR is 1/2 * 1 = 1/2. Then the area of the parallelogram PQRS is 2*(1/2) = 1. So yes, correct.

Therefore, the area of PQRS is indeed 1.

Therefore, the surface area of the piece is 2√3 + 1.

Wait, but the problem says "surface area of one of these pieces". However, the answer choices are not given, but the problem says "put your final answer within \boxed{}". So likely the answer is 2√3 + 1. However, let me check with another way. Maybe the answer is 2√3? Or something else?

Wait, but according to all calculations, it's 2√3 + 1. Maybe I made a mistake in the original portions. Let me check again:

Original portions:

- From ABC: 3√3 /4

- From ABD: 3√3 /4

- From ACD: √3 /4

- From BCD: √3 /4

Total: 3√3 /4 + 3√3 /4 = 6√3 /4 = 3√3 /2

√3 /4 + √3 /4 = √3 / 2

Total: 3√3 /2 + √3 /2 = 4√3 /2 = 2√3. Yes. Then adding 1 gives 2√3 + 1. So this seems correct.

Alternatively, let me think about the problem again. Maybe the surface area of one piece includes only the new face and some of the original faces, but not all. However, according to the earlier analysis, it includes portions of all four original faces. However, since each original face is intersected by the plane, and the piece contains parts of each.

Wait, but for example, the face ACD is split into APQ and PDC. The piece containing A and B contains APQ, and the other part is PDC. Similarly for BCD. Therefore, the original face areas partially belong to each piece. Therefore, the total of the original portions is 2√3, and adding the new face gives 2√3 + 1.

Yes, I think this is correct. Therefore, the answer should be 2√3 + 1. Therefore, I will go with that.

**Final Answer**
\boxed{1 + 2\sqrt{3}}

To solve the problem, we analyze the geometry of a regular tetrahedron $ABCD$ with edge length 2, and the effect of cutting it with a plane parallel to edges $AB$ and $CD$, lying halfway between them.

---

### Step 1: Coordinate Setup

We assign coordinates to the vertices to better analyze the structure. A common coordinate setup for a regular tetrahedron is:

- $A = (0, 0, 0)$
- $B = (2, 0, 0)$
- $C = (1, \sqrt{3}, 0)$
- $D = (1, \sqrt{3}/3, 2\sqrt{6}/3)$

These coordinates ensure all edges are of length 2 and the tetrahedron is regular.

---

### Step 2: Direction Vectors

The direction vector of edge $AB$ is $ \vec{u} = (2, 0, 0) $, and for edge $CD$, it is:

$$
\vec{v} = D - C = (1 - 1, \sqrt{3}/3 - \sqrt{3}, 2\sqrt{6}/3 - 0) = (0, -2\sqrt{3}/3, 2\sqrt{6}/3)
$$

A plane **parallel** to both direction vectors and **halfway between** edges $AB$ and $CD$ has a normal vector perpendicular to both $ \vec{u} $ and $ \vec{v} $, i.e., the cross product $ \vec{n} = \vec{u} \times \vec{v} $.

Computing $ \vec{n} $:

$$
\vec{n} = \vec{u} \times \vec{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 0 & 0 \\
0 & -2\sqrt{3}/3 & 2\sqrt{6}/3
\end{vmatrix}
= (0, -4\sqrt{6}/3, -4\sqrt{3}/3)
$$

This can be normalized or simplified for the plane equation.

Using the standard form $ \vec{n} \cdot \vec{r} = d $, and knowing the plane is equidistant from both lines, we compute the point where this plane lies **midway** between $AB$ and $CD$, leading to the plane equation:

$$
\sqrt{2}y + z = \frac{\sqrt{6}}{2}
$$

---

### Step 3: Intersections with Edges

The plane intersects the following edges of the tetrahedron at their midpoints:

- **AC** at point $P = (1/2, \sqrt{3}/2, 0)$
- **AD** at point $Q = (1/2, \sqrt{3}/6, \sqrt{6}/3)$
- **BC** at point $R = (3/2, \sqrt{3}/2, 0)$
- **BD** at point $S = (3/2, \sqrt{3}/6, \sqrt{6}/3)$

These four points form a **parallelogram $PQRS$** (since the direction vectors of the plane are parallel to the directions of the original edge vectors, and the plane is not intersecting those edges).

---

### Step 4: Surface Area of the Cut Piece

We compute the **surface area** of one of the two resulting pieces. Let's consider the piece containing vertices $A$ and $B$.

This piece has:

- **Portions** of the original faces $ABC$, $ABD$, $ACD$, and $BCD$ that are on the same side of the plane.
- **A new face**, the parallelogram $PQRS$, formed by the cut.

We compute each contribution:

---

#### **Area of PQRS (New Face)**

Using the cross product of adjacent vectors:

- $ \vec{PQ} = (0, -\sqrt{3}/3, \sqrt{6}/3) $
- $ \vec{PS} = (1, -\sqrt{3}/3, \sqrt{6}/3) $

$$
\vec{PQ} \times \vec{PS} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & -\sqrt{3}/3 & \sqrt{6}/3 \\
1 & -\sqrt{3}/3 & \sqrt{6}/3
\end{vmatrix}
= (0, \sqrt{6}/3, \sqrt{3}/3)
$$

Magnitude of this vector:

$$
|\vec{PQ} \times \vec{PS}| = \sqrt{0^2 + (\sqrt{6}/3)^2 + (\sqrt{3}/3)^2} = \sqrt{(6/9 + 3/9)} = \sqrt{9/9} = 1
$$

So, the area of quadrilateral $PQRS$ is:

$$
\text{Area}_{PQRS} = 1
$$

---

#### **Area of Original Face Portions**

Each original face is cut by the plane into a portion lying on the piece containing $A$ and $B$:

- **Face $ABC$** → portion $APRB$ (trapezoid): $ \frac{3\sqrt{3}}{4} $
- **Face $ABD$** → portion $ABQS$ (trapezoid): $ \frac{3\sqrt{3}}{4} $
- **Face $ACD$** → portion $APQ$ (triangle): $ \frac{\sqrt{3}}{4} $
- **Face $BCD$** → portion $BRS$ (triangle): $ \frac{\sqrt{3}}{4} $

Sum of these:

$$
\frac{3\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{8\sqrt{3}}{4} = 2\sqrt{3}
$$

---

### Step 5: Total Surface Area

Final step:

$$
\text{Surface Area} = \text{Original face portions} + \text{New face area} = 2\sqrt{3} + 1
$$

---

### ✅ Final Answer

$$
\boxed{1 + 2\sqrt{3}}
$$

Solution: The plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length 1. The surface area of each piece is half the total surface area of the tetrahedron plus the area of the square, that is, $\frac{1}{2}\cdot 4\cdot \frac{2^2 \sqrt{3}}{4}+1=\boxed{1+2\sqrt{3}}$.