<Question>
Let $ABCD$ be a regular tetrahedron with side length 2. The plane parallel to edges $AB$ and $CD$ and lying halfway between them cuts $ABCD$ into two pieces. Find the surface area of one of these pieces.
</Question>

<think>
<Plan_1>
Okay, so I need to find the surface area of one of the pieces created when a plane parallel to edges AB and CD of a regular tetrahedron ABCD, which is cut halfway between them. The tetrahedron has a side length of 2. Hmm, let me try to visualize this first.

<prompt_1.1> Visualize the problem by identifying the shape of the cross-section formed by the plane cutting the tetrahedron and determine the nature of the resulting pieces. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
A regular tetrahedron has four triangular faces, all equilateral triangles. Each edge is of length 2. The plane in question is parallel to two edges, AB and CD, and is lying halfway between them. Since the plane is parallel to both AB and CD, which are skew lines in the tetrahedron, the cross-section created by this plane should be a quadrilateral, right? Because a plane parallel to two skew lines will intersect the other edges of the tetrahedron, forming a quadrilateral. But wait, in a tetrahedron, any plane parallel to two edges might intersect other edges. Let me confirm.

So, the original tetrahedron has vertices A, B, C, D. Edges AB and CD are skew. The plane is halfway between them and parallel to both. Since it's halfway, the distance from the plane to AB and CD is the same. Because the plane is parallel to both AB and CD, it must intersect the other edges of the tetrahedron. The edges that are not parallel to AB or CD are AC, AD, BC, BD. So the plane will intersect these four edges? Wait, but a plane can't intersect four edges of a tetrahedron and form a quadrilateral because a tetrahedron only has four faces. Wait, maybe it intersects two of them? Let me think.

Wait, the plane is parallel to AB and CD. So, AB is one edge, CD is another. The plane is halfway between them. Let me imagine the tetrahedron. If I take AB and CD, which are skew, and a plane halfway between them, parallel to both. Then, this plane would intersect the other edges connecting the vertices of AB to CD. So, for example, edges AC and BD connect A to C and B to D. Similarly, edges AD and BC connect A to D and B to C. So, the plane, being parallel to AB and CD, would intersect these edges AC, BD, AD, and BC. But since it's halfway, maybe it only intersects two of them? Wait, perhaps it's a parallelogram? Because if the plane is parallel to two edges and intersects the other edges, the intersection should form a parallelogram.

Wait, in a regular tetrahedron, the edges AB and CD are skew. The midplane parallel to both AB and CD would intersect the other four edges (AC, AD, BC, BD) at their midpoints. So, if the plane is halfway between AB and CD, then it would cut each of these edges at their midpoints. Therefore, the cross-section is a parallelogram connecting the midpoints of AC, AD, BC, and BD.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me confirm that.

<prompt_2.1> Confirm the cross-section formed by the intersection of the plane with the tetrahedron. </prompt_2.1>

Therefore, the cross-section is a parallelogram. Since the original tetrahedron is regular, all edges are length 2, so the distance between AB and CD can be calculated, but since the plane is halfway, the cross-section is a parallelogram. Now, the two pieces created by this plane are a smaller polyhedron and a frustum-like shape? Wait, but in a tetrahedron, cutting with a plane would result in one smaller tetrahedron and a truncated part? Wait, no, because the cross-section is a parallelogram, not a triangle.

<prompt_2.2> Determine the nature of the pieces formed by the plane cutting the tetrahedron. </prompt_2.2>

So, the original tetrahedron has four triangular faces. When we cut it with this plane, one piece is a smaller tetrahedron with a parallelogram face and three triangular faces?

<prompt_2.3> Analyze the surface area of the pieces formed by the plane cutting the tetrahedron. </prompt_2.3>

But wait, the original tetrahedron's faces are all equilateral triangles with area sqrt(3). Each face has area (sqrt(3)/4)*2^2 = sqrt(3).

<prompt_2.4> Calculate the original surface area of the tetrahedron. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
If the plane is parallel to AB and CD, then it must intersect the edges that connect the vertices of AB to the vertices of CD. So, edges AC, AD, BC, BD. If the plane is halfway between AB and CD, then the intersection points on these edges should be at their midpoints. For example, the midpoint of AC is halfway between A and C, which is 1 unit from each. Similarly for the other edges. So, connecting these midpoints would form a quadrilateral. Since the plane is parallel to both AB and CD, the opposite sides of this quadrilateral should be parallel and equal in length. Therefore, the cross-section is a parallelogram.
</execution_2.1>

<execution_2.2>
Wait, actually, maybe both pieces are not tetrahedrons. Let me think again.

Wait, the original tetrahedron has four faces. If we cut it with a plane that's parallel to two edges and halfway between them, the intersection with the tetrahedron will create a new face (the parallelogram) and split the original tetrahedron into two parts. Each part will have some of the original faces and the new parallelogram face. Since the plane is halfway, maybe one part is a smaller tetrahedron and the other is a kind of frustum?
</execution_2.2>

<execution_2.3>
Wait, maybe not. Let me think again. If the cross-section is a parallelogram, then the piece between the original face and the cross-section would have a base which is a triangle and a top which is a parallelogram. Wait, no. Let's consider the original tetrahedron. If we take the plane cutting AB and CD and halfway, then the part of the tetrahedron between the original vertex and the cross-section would be a smaller tetrahedron. But since the cross-section is a parallelogram, maybe the piece is a kind of prismatoid? Hmm, maybe I need to draw this out.

Alternatively, perhaps the cross-section is a quadrilateral, so the two pieces are a smaller tetrahedron and a frustum. Wait, but frustums are usually for pyramids. In a tetrahedron, which is a pyramid with a triangular base, a frustum would have two parallel triangular faces. But here, the cross-section is a parallelogram, so maybe the frustum has a triangular base and a parallelogram top? That doesn't sound right. Maybe the cross-section is a quadrilateral, so the two pieces are a tetrahedron and a pentahedron? Wait, confusion here.

Alternatively, perhaps the cross-section is a triangle? Wait, no, because the plane is parallel to two edges. If a plane is parallel to two edges of a tetrahedron, the intersection with the tetrahedron should be a quadrilateral. Because each of the two edges is intersected by two other edges. Wait, maybe in this case, since the plane is parallel to AB and CD, which are skew, the plane will intersect the four edges that connect the two pairs of vertices. So, edges AC, AD, BC, BD. Each of these edges will intersect the plane at some point. If the plane is halfway between AB and CD, then these intersection points might be midpoints. So, if the plane is halfway, then the cross-section is a parallelogram connecting the midpoints of AC, AD, BC, BD.
</execution_2.3>

<execution_2.4>
Wait, no. Wait, the formula for the area of an equilateral triangle is (sqrt(3)/4)*a², where a is the side length. So with a=2, the area is (sqrt(3)/4)*4 = sqrt(3). So each face has area sqrt(3).
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.3, execution_2.4:
So the original tetrahedron has 4 faces, each of area sqrt(3), so total surface area 4*sqrt(3). But when we cut it, the two pieces will each have some of the original faces and the new parallelogram face. Wait, but the problem says "surface area of one of these pieces". So, we need to compute the surface area of one of the two pieces created by the plane. The surface area would include the original faces that are now part of the piece and the new face created by the plane.

So, we need to figure out which original faces are intersected by the cutting plane, and then compute the areas of those faces plus the area of the cross-section. Alternatively, the cutting plane might intersect some edges, creating new edges and thus new faces. Wait, but the cross-section is a parallelogram, so the new face is a parallelogram. So the surface area of the piece would be the sum of the areas of the original faces that are now part of the piece plus the area of the parallelogram.

But which original faces are intersected? Let's think.

<prompt_3.1> Determine which original faces are intersected by the cutting plane and the nature of the new face created. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
The plane is parallel to AB and CD. So, AB is part of the original face ABC and ABD. CD is part of the original face ACD and BCD. Since the plane is parallel to AB and CD, it does not intersect the faces containing AB or CD. Wait, but the plane is halfway between AB and CD. So, maybe the plane intersects the other two faces? Wait, the original tetrahedron has four faces: ABC, ABD, ACD, BCD. The plane is parallel to AB and CD, so it does not intersect the faces ABC and ABD (since AB is in those faces), and does not intersect the faces ACD and BCD (since CD is in those faces). Wait, but that can't be, because the plane must intersect the tetrahedron somewhere. Wait, no. If the plane is parallel to AB and CD, which are skew, then the plane can't intersect AB or CD, but it must intersect the other edges. The other edges are AC, AD, BC, BD. So, the plane is parallel to AB and CD, so it doesn't intersect AB or CD, but intersects the other four edges. Therefore, the cross-section is a quadrilateral formed by the intersection points on AC, AD, BC, BD. Therefore, the two pieces are: one piece is a smaller tetrahedron with vertices at the intersection points, and the other piece is the remaining part of the original tetrahedron. Wait, but the cross-section is a quadrilateral, so the two pieces would be a tetrahedron (with four faces: three triangles and the quadrilateral) and the remaining part, which is a pentahedron? Hmm, maybe not. Wait, perhaps both pieces are polyhedrons with a quadrilateral face.

Wait, actually, when you cut a tetrahedron with a plane, the number of faces of the two resulting pieces depends on how the plane intersects the original edges. Since the plane is parallel to two edges and intersects the other four edges, the cross-section is a quadrilateral. Therefore, each piece will have some original triangular faces and the new quadrilateral face. However, the original tetrahedron has four triangular faces. The cutting plane intersects four edges, so each piece will have parts of those four edges. Let's see: the original edges AC, AD, BC, BD are each intersected once by the plane. Therefore, the intersection points divide each of these edges into two segments. The piece that is closer to vertex A (assuming the plane is cutting towards the midpoint) would include the original faces ABC and ABD, but wait, no. Wait, the plane is halfway between AB and CD, so perhaps the cross-section is exactly halfway. Let me think in terms of coordinates.

Maybe setting up coordinates would help. Let me assign coordinates to the tetrahedron. Let's place the tetrahedron in 3D space. A regular tetrahedron can be embedded in R^3 with vertices at (0,0,0), (1,1,1), (1,0,0), (0,1,0), but maybe a simpler coordinate system. Wait, actually, a regular tetrahedron can have vertices at (1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1), scaled appropriately. But maybe that's more complicated. Alternatively, let's choose coordinates such that AB is along the x-axis, and CD is in a position such that the plane can be defined.

Alternatively, let's set vertex A at (0,0,0), vertex B at (2,0,0), since AB is of length 2. Then, since it's a regular tetrahedron, the other two vertices C and D must be placed such that all edges are length 2. Let me recall that in a regular tetrahedron, the coordinates can be set as follows: Let’s place A at (0,0,0), B at (2,0,0). Then, the other two vertices C and D need to be located such that their distances to A and B are 2, and the distance between C and D is 2. Let me compute coordinates for C and D.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
The centroid of the base triangle ABX (where X is the midpoint of AB) is at a height of sqrt(6)/3 times the edge length. Wait, maybe this is getting too complicated. Let me compute coordinates step by step.

<prompt_4.1> Compute the coordinates for vertices C and D of the regular tetrahedron. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Let’s set A at (0,0,0) and B at (2,0,0). The midpoint of AB is at (1,0,0). The third vertex C of the base triangle (assuming the base is ABX) would be in the xy-plane. Wait, but in a regular tetrahedron, all edges are equal. Wait, actually, if we have a regular tetrahedron with edge length 2, the coordinates can be given as follows:

Let’s choose coordinates such that:

A = (0, 0, 0)

B = (2, 0, 0)

C = (1, sqrt(3), 0)

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

Wait, let me verify.

In a regular tetrahedron, the coordinates can be set with one vertex at the origin, another along the x-axis, the third in the xy-plane, and the fourth somewhere above. The distance between A and B is 2. The distance between A and C should also be 2. So, if A is (0,0,0), B is (2,0,0), then C must be somewhere such that its distance from A is 2 and from B is 2. So, the coordinates of C must satisfy:

sqrt((x-0)^2 + (y-0)^2 + (z-0)^2) = 2

sqrt((x-2)^2 + y^2 + z^2) = 2

Squaring both equations:

x² + y² + z² = 4

(x-2)² + y² + z² = 4

Subtracting the first equation from the second:

(x-2)^2 - x^2 = 0

x² -4x +4 -x² = -4x +4 =0 => -4x +4=0 => x=1

So, x=1. Then, plugging back into the first equation:

1 + y² + z² =4 => y² + z²=3

So, point C is at (1, y, z) with y² + z²=3. Let's set y=sqrt(3), z=0, so C=(1, sqrt(3), 0). That gives an equilateral triangle in the xy-plane. Then, the fourth vertex D must be equidistant from A, B, and C, with all edges AD, BD, CD equal to 2.

The coordinates of D can be found as the centroid of the base triangle ABC, but elevated along the z-axis. The centroid of ABC is at ((0+2+1)/3, (0 +0 +sqrt(3))/3, 0) = (1, sqrt(3)/3, 0). The height h of the tetrahedron can be found using the formula for the height of a regular tetrahedron: h = sqrt(2/3)*edge length. Since edge length is 2, h = sqrt(2/3)*2 = 2*sqrt(6)/3. Therefore, D is at (1, sqrt(3)/3, 2*sqrt(6)/3).
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Let me confirm the distances:

Distance from D to A:

sqrt((1-0)^2 + (sqrt(3)/3 -0)^2 + (2*sqrt(6)/3 -0)^2) = sqrt(1 + (1/3)*3 + (24/9)) = sqrt(1 +1 + 8/3) = sqrt(2 + 8/3) = sqrt(14/3). Wait, that's not 2. Hmm, something's wrong here.

Wait, maybe my formula for the height is incorrect. Let me recall.

<prompt_5.1> Verify the height calculation and coordinates of the tetrahedron. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
The height h of a regular tetrahedron with edge length a is h = sqrt(6)/3 *a. So for a=2, h= sqrt(6)/3 *2= 2*sqrt(6)/3. So that part is correct. Then, the coordinates of D should be the centroid of the base triangle ABC, which is (1, sqrt(3)/3, 0), moved up along the z-axis by h=2*sqrt(6)/3. Therefore, D=(1, sqrt(3)/3, 2*sqrt(6)/3). Let's compute the distance from D to A:

sqrt((1-0)^2 + (sqrt(3)/3 -0)^2 + (2*sqrt(6)/3 -0)^2) = sqrt(1 + (1/3) + (24/9)) = sqrt(1 + 1/3 + 8/3) = sqrt(1 + 9/3) = sqrt(1 +3) = sqrt(4)=2. Okay, that works. Similarly, distance from D to B:

sqrt((1-2)^2 + (sqrt(3)/3 -0)^2 + (2*sqrt(6)/3 -0)^2) = sqrt(1 +1/3 +8/3)=sqrt(1 +3)=2. Good. Distance from D to C:

C is at (1, sqrt(3),0). So distance from D to C:

sqrt((1-1)^2 + (sqrt(3)/3 - sqrt(3))^2 + (2*sqrt(6)/3 -0)^2) = sqrt(0 + (-2*sqrt(3)/3)^2 + (2*sqrt(6)/3)^2) = sqrt( (4*3)/9 + (24)/9 ) = sqrt(12/9 +24/9)=sqrt(36/9)=sqrt(4)=2. Perfect. So coordinates are:

A=(0,0,0)

B=(2,0,0)

C=(1, sqrt(3),0)

D=(1, sqrt(3)/3, 2*sqrt(6)/3)
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Now, the problem states that the plane is parallel to edges AB and CD and lies halfway between them. Let's find the equation of this plane.

First, find the parametric equations for AB and CD.

<prompt_6.1> Find the equation of the plane parallel to AB and CD and lying halfway between them. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Edge AB goes from A(0,0,0) to B(2,0,0). So direction vector AB is (2,0,0).

Edge CD goes from C(1, sqrt(3),0) to D(1, sqrt(3)/3, 2*sqrt(6)/3). The direction vector CD is D - C = (0, -2*sqrt(3)/3, 2*sqrt(6)/3).

Since the plane is parallel to both AB and CD, its normal vector must be perpendicular to both AB and CD. Wait, but the plane is parallel to both AB and CD, so the normal vector of the plane is perpendicular to both AB and CD. Therefore, the normal vector can be found by taking the cross product of AB and CD vectors.

Wait, actually, the plane is parallel to both AB and CD, so the normal vector of the plane is perpendicular to both AB and CD. Therefore, the normal vector n = AB × CD.

Compute AB × CD:

AB = (2,0,0)

CD = (0, -2*sqrt(3)/3, 2*sqrt(6)/3)

Cross product:

|i   j           k       |
|2   0           0       |
|0 -2*sqrt(3)/3 2*sqrt(6)/3|

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

Simplify:

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

So n = (0, -4*sqrt(6)/3, -4*sqrt(3)/3)

We can simplify this by multiplying by 3 to eliminate denominators:

n = (0, -4*sqrt(6), -4*sqrt(3))

But the plane equation can be written as n ⋅ (r - r0) = 0. To find the plane halfway between AB and CD, we need to find the equation of the plane that is equidistant from AB and CD. However, since AB and CD are skew lines, the concept of halfway between them is a bit more involved. The distance between two skew lines can be computed, and the plane would be the locus of points equidistant from both lines. However, in this case, the problem states the plane is halfway between AB and CD, so perhaps the plane is the mid-plane between the two lines. To find this, we can compute the distance between AB and CD and then find the plane that is at half that distance from each line. Alternatively, since the plane is parallel to both AB and CD, and lies halfway between them, the plane can be found by translating one of the lines towards the other by half the distance.

Alternatively, perhaps we can parametrize the plane. Since the plane is parallel to both AB and CD, and passes through the midpoint of the segment connecting a point on AB and a point on CD. Wait, but which points? If we take the midpoints of AB and CD, but AB is from (0,0,0) to (2,0,0), so midpoint is (1,0,0). CD is from (1, sqrt(3),0) to (1, sqrt(3)/3, 2*sqrt(6)/3). The midpoint of CD is (1, (sqrt(3) + sqrt(3)/3)/2, (0 + 2*sqrt(6)/3)/2) = (1, (4*sqrt(3)/3)/2, (2*sqrt(6)/3)/2) = (1, 2*sqrt(3)/3, sqrt(6)/3). Then, the line connecting (1,0,0) to (1, 2*sqrt(3)/3, sqrt(6)/3) is perpendicular to both AB and CD. The midpoint plane would be the plane that contains this line and is parallel to AB and CD. Wait, but that might not necessarily be halfway in terms of distance. Alternatively, the plane can be found by moving along the common perpendicular direction from AB to CD by half the distance.

First, compute the distance between AB and CD. The formula for the distance between two skew lines is |(P2 - P1) ⋅ (n)| / |n|, where n is the cross product of the direction vectors.

We already computed n = AB × CD = (0, -4*sqrt(6)/3, -4*sqrt(3)/3). Then, take two points, one on each line. Let's take point A(0,0,0) on AB and point C(1, sqrt(3),0) on CD. Then vector AC = C - A = (1, sqrt(3),0). Then, the distance is |AC ⋅ n| / |n|.

Compute AC ⋅ n = (1, sqrt(3),0) ⋅ (0, -4*sqrt(6)/3, -4*sqrt(3)/3) = 0 + sqrt(3)*(-4*sqrt(6)/3) + 0 = -4*sqrt(18)/3 = -4*(3*sqrt(2))/3 = -4*sqrt(2). The absolute value is 4*sqrt(2). The magnitude of n is sqrt(0^2 + ( -4*sqrt(6)/3)^2 + (-4*sqrt(3)/3)^2) = sqrt( (16*6)/9 + (16*3)/9 ) = sqrt(96/9 + 48/9) = sqrt(144/9) = sqrt(16) =4. Therefore, the distance between AB and CD is |AC ⋅ n| / |n| = 4*sqrt(2)/4 = sqrt(2). So the distance between AB and CD is sqrt(2). Therefore, halfway between them would be sqrt(2)/2 distance from each.

But how to find the equation of the plane? Since the plane is parallel to both AB and CD, and lies halfway between them, we can think of it as the locus of points that are midpoints between lines AB and CD. Alternatively, since the plane is parallel to both AB and CD, and lies halfway, we can parametrize points on the plane as combinations of points on AB and CD, translated by half the vector connecting AB and CD.

Alternatively, since the plane is halfway between AB and CD, and parallel to both, the plane can be found by taking the midpoints of the segments connecting AB and CD. Let's take a general point on AB: let’s say point P(t) = A + t*(B - A) = (2t, 0, 0), where t ∈ [0,1]. Similarly, a general point on CD: point Q(s) = C + s*(D - C) = (1, sqrt(3) - 2*sqrt(3)/3 *s, 0 + 2*sqrt(6)/3 *s), where s ∈ [0,1]. Then, the vector connecting P(t) to Q(s) is Q(s) - P(t) = (1 - 2t, sqrt(3) - 2*sqrt(3)/3 *s -0, 2*sqrt(6)/3 *s -0). The vector connecting AB to CD is Q(0) - P(1) = (1 -2*1, sqrt(3) -0, 0 -0) = (-1, sqrt(3),0). Wait, perhaps this is getting too complicated.

Alternatively, since the plane is halfway between AB and CD, the plane can be obtained by translating one of the lines towards the other by half the distance. Let's take line AB and translate it towards line CD by half the distance sqrt(2)/2 along the common perpendicular direction. The common perpendicular direction is given by the cross product n = AB × CD, which we found earlier. The unit normal vector is n / |n| = (0, -4*sqrt(6)/3, -4*sqrt(3)/3) /4 = (0, -sqrt(6)/3, -sqrt(3)/3). Therefore, to move AB towards CD by half the distance sqrt(2)/2, we translate each point on AB in the direction of the unit normal vector multiplied by sqrt(2)/2. Wait, but actually, the translation vector would be (unit normal vector) * (distance/2). Wait, the distance between AB and CD is sqrt(2), so halfway would be sqrt(2)/2. Therefore, the translation vector is (0, -sqrt(6)/3, -sqrt(3)/3) * sqrt(2)/2 = (0, -sqrt(6)*sqrt(2)/6, -sqrt(3)*sqrt(2)/6) = (0, -sqrt(12)/6, -sqrt(6)/6) = (0, -2*sqrt(3)/6, -sqrt(6)/6) = (0, -sqrt(3)/3, -sqrt(6)/6).

Therefore, the translated plane would be AB shifted in the direction of the unit normal vector by sqrt(2)/2. However, actually, since the plane is parallel to AB and CD, and lies halfway between them, the equation of the plane can be found by taking the plane that contains the midpoints of the segments connecting AB to CD. Alternatively, perhaps a better approach is to parametrize the plane.

Alternatively, since the plane is parallel to AB and CD, and lies halfway between them, the plane can be found by solving for points that are affine combinations of points on AB and CD. Let's consider that for any point on the plane, it can be expressed as a point on AB plus a scalar multiple of AB's direction vector, plus another scalar multiple of CD's direction vector. But since the plane is halfway, maybe the coefficients are 0.5. Wait, perhaps not. Let me think.

Alternatively, let's consider that the plane is the set of points that are midpoints between a point on AB and a point on CD. So, for any P on AB and Q on CD, the midpoint M = (P + Q)/2 lies on the plane. Therefore, the plane can be parametrized as M = (A + t*(B - A) + C + s*(D - C))/2, where t and s range from 0 to 1. Let's compute this:

M = [ (0,0,0) + t*(2,0,0) + (1, sqrt(3),0) + s*(0, -2*sqrt(3)/3, 2*sqrt(6)/3) ] / 2

Simplify:

M_x = [0 + 2t +1 +0]/2 = (2t +1)/2

M_y = [0 +0 + sqrt(3) -2*sqrt(3)/3 *s]/2 = [sqrt(3)(1 - 2s/3)]/2

M_z = [0 +0 +0 +2*sqrt(6)/3 *s]/2 = [sqrt(6)/3 *s]

Therefore, parametrizing the plane as:

x = (2t +1)/2

y = [sqrt(3)(1 - 2s/3)]/2

z = [sqrt(6)/3]s

But we need to find the equation of the plane. Let's eliminate parameters t and s. Let's express t and s in terms of x, y, z.

From x = (2t +1)/2 => 2x = 2t +1 => t = (2x -1)/2

From z = sqrt(6)/3 *s => s = (3z)/sqrt(6) = (3z*sqrt(6))/6 = z*sqrt(6)/2

Now, substitute s into the equation for y:

y = [sqrt(3)(1 - 2*(z*sqrt(6)/2)/3 )]/2

Simplify:

First, compute 2*(z*sqrt(6)/2)/3 = (z*sqrt(6))/3

Therefore:

y = [sqrt(3)(1 - (z*sqrt(6))/3 )]/2

Multiply numerator and denominator:

y = [sqrt(3) - sqrt(3)*(z*sqrt(6))/3 ] / 2

Simplify sqrt(3)*sqrt(6) = sqrt(18) = 3*sqrt(2)

Thus:

y = [sqrt(3) - (3*sqrt(2)/3) z ] /2 = [sqrt(3) - sqrt(2) z ] /2

Multiply both sides by 2:

2y = sqrt(3) - sqrt(2) z

Rearranged:

sqrt(2) z + 2y = sqrt(3)

Therefore, the equation of the plane is sqrt(2) z + 2y = sqrt(3). Let me check if this makes sense.

Alternatively, let's verify by plugging in points. For example, when t=0 (point A) and s=0 (point C), the midpoint is ((0+1)/2, (0 + sqrt(3))/2, 0) = (0.5, sqrt(3)/2, 0). Plugging into the equation:

sqrt(2)*0 + 2*(sqrt(3)/2) = sqrt(3), which matches the right-hand side. Similarly, when t=1 (point B) and s=1 (point D), the midpoint is ((2+1)/2, (0 + sqrt(3)/3)/2, (0 + 2*sqrt(6)/3)/2) = (1.5, sqrt(3)/6, sqrt(6)/3). Plugging into the equation:

sqrt(2)*(sqrt(6)/3) + 2*(sqrt(3)/6) = (sqrt(12)/3) + (sqrt(3)/3) = (2*sqrt(3)/3) + (sqrt(3)/3) = sqrt(3). Correct. So the equation sqrt(2) z + 2y = sqrt(3) is indeed the equation of the plane halfway between AB and CD and parallel to both.

Therefore, the equation of the cutting plane is sqrt(2) z + 2y = sqrt(3).
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
So, now, to find the intersection of this plane with the tetrahedron ABCD, which we have coordinates for, we can find the points where the plane intersects the edges of the tetrahedron. The edges of the tetrahedron are AB, AC, AD, BC, BD, CD. The plane is parallel to AB and CD, so it won't intersect AB or CD. Therefore, it must intersect the other four edges: AC, AD, BC, BD. Let's parametrize each of these edges and find their intersection with the plane.

<prompt_7.1> Find the intersection of the plane with the edges AC, AD, BC, and BD of the tetrahedron. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Let's start with edge AC: from A(0,0,0) to C(1, sqrt(3),0). Parametrize this edge as r(t) = A + t*(C - A) = (t, t*sqrt(3), 0), where t ∈ [0,1]. Substitute into the plane equation:

sqrt(2)*0 + 2*(t*sqrt(3)) = sqrt(3) => 2t*sqrt(3) = sqrt(3) => 2t =1 => t=1/2. Therefore, the intersection point is (1/2, sqrt(3)/2, 0). Let's call this point P.

Next, edge AD: from A(0,0,0) to D(1, sqrt(3)/3, 2*sqrt(6)/3). Parametrize as r(t) = (t, t*sqrt(3)/3, t*2*sqrt(6)/3), t ∈ [0,1]. Substitute into the plane equation:

sqrt(2)*(t*2*sqrt(6)/3) + 2*(t*sqrt(3)/3) = sqrt(3)

Multiply through:

(2*sqrt(12)/3)*t + (2*sqrt(3)/3)*t = sqrt(3)

Simplify sqrt(12) = 2*sqrt(3):

(2*2*sqrt(3)/3)*t + (2*sqrt(3)/3)*t = sqrt(3)

(4*sqrt(3)/3 + 2*sqrt(3)/3)*t = sqrt(3)

(6*sqrt(3)/3)*t = sqrt(3)

2*sqrt(3)*t = sqrt(3)

t = sqrt(3)/(2*sqrt(3)) = 1/2

Therefore, the intersection point is (1/2, (sqrt(3)/3)*(1/2), (2*sqrt(6)/3)*(1/2)) = (1/2, sqrt(3)/6, sqrt(6)/3). Let's call this point Q.

Now, edge BC: from B(2,0,0) to C(1, sqrt(3),0). Parametrize as r(t) = B + t*(C - B) = (2 - t, 0 + t*sqrt(3), 0), t ∈ [0,1]. Substitute into the plane equation:

sqrt(2)*0 + 2*(t*sqrt(3)) = sqrt(3)

Same as edge AC:

2t*sqrt(3) = sqrt(3) => t=1/2

Intersection point is (2 -1/2, sqrt(3)/2, 0) = (3/2, sqrt(3)/2, 0). Let's call this point R.

Finally, edge BD: from B(2,0,0) to D(1, sqrt(3)/3, 2*sqrt(6)/3). Parametrize as r(t) = (2 - t, 0 + t*sqrt(3)/3, 0 + t*2*sqrt(6)/3), t ∈ [0,1]. Substitute into the plane equation:

sqrt(2)*(t*2*sqrt(6)/3) + 2*(t*sqrt(3)/3) = sqrt(3)

Same as edge AD:

(2*sqrt(12)/3 + 2*sqrt(3)/3)*t = sqrt(3)

Again, sqrt(12)=2*sqrt(3):

(4*sqrt(3)/3 + 2*sqrt(3)/3)*t = sqrt(3)

6*sqrt(3)/3 *t = sqrt(3)

2*sqrt(3)*t = sqrt(3)

t=1/2

Intersection point is (2 -1/2, (sqrt(3)/3)*(1/2), (2*sqrt(6)/3)*(1/2)) = (3/2, sqrt(3)/6, sqrt(6)/3). Let's call this point S.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
So the cross-section is a quadrilateral PQRS with these four points. Now, we need to compute the area of this quadrilateral. Since the tetrahedron is regular and the points are symmetric, perhaps this quadrilateral is a parallelogram. Let's check the coordinates:

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

Q(1/2, sqrt(3)/6, sqrt(6)/3)

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

S(3/2, sqrt(3)/6, sqrt(6)/3)

Looking at the coordinates, it seems that PQ and RS are vertical edges (along the y and z directions), and PR and QS are horizontal edges (along x and y directions). Wait, actually, let's compute the vectors:

<prompt_8.1> Calculate the area of the quadrilateral cross-section using the coordinates of points P, Q, R, and S. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Vector PQ: Q - P = (1/2 -1/2, sqrt(3)/6 - sqrt(3)/2, sqrt(6)/3 -0) = (0, -sqrt(3)/3, sqrt(6)/3)

Vector SR: R - S = (3/2 -3/2, sqrt(3)/2 - sqrt(3)/6, 0 - sqrt(6)/3) = (0, sqrt(3)/3, -sqrt(6)/3) = - (0, -sqrt(3)/3, sqrt(6)/3) = -PQ. So PQ and SR are equal in magnitude and opposite in direction, so they are equal and opposite vectors.

Similarly, Vector PR: R - P = (3/2 -1/2, sqrt(3)/2 - sqrt(3)/2,0 -0) = (1,0,0)

Vector QS: S - Q = (3/2 -1/2, sqrt(3)/6 - sqrt(3)/6, sqrt(6)/3 - sqrt(6)/3) = (1,0,0). So PR and QS are equal and in the same direction. Therefore, the quadrilateral is a parallelogram because opposite sides are equal and parallel.

Alternatively, since the vectors PQ and SR are equal and opposite, and vectors PR and QS are equal and in the same direction, it's a parallelogram.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
To compute the area of the parallelogram, we can use the cross product of two adjacent vectors. For example, vectors PQ and PR. Wait, but in a parallelogram, the area is the magnitude of the cross product of two adjacent sides. Let's take vectors PQ and PR. Wait, but PR is from P to R, which is (1,0,0), and PQ is (0, -sqrt(3)/3, sqrt(6)/3). So the cross product PQ × PR will give a vector perpendicular to the parallelogram, and its magnitude will be the area.

<prompt_9.1> Calculate the area of the parallelogram using the cross product of vectors PQ and PR. </prompt_9.1>
</Plan_9>

<Execution_9>
<execution_9.1>
Compute the cross product:

PQ × PR = |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 )

Simplify:

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

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

The magnitude of this vector is sqrt( (sqrt(6)/3)^2 + (sqrt(3)/3)^2 ) = sqrt( (6/9) + (3/9) ) = sqrt(9/9) = sqrt(1) =1. Therefore, the area of the parallelogram is 1.
</execution_9.1>
</Execution_9>

<Plan_10>
Based on execution_9.1:
So the cross-section has an area of 1. But wait, that seems surprisingly simple. Let me verify the calculations.

<prompt_10.1> Verify the calculations for the cross-sectional area and the surface area of the piece containing vertex A. </prompt_10.1>
</Plan_10>

<Execution_10>
<execution_10.1>
First, the coordinates of the intersection points:

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

Q(1/2, sqrt(3)/6, sqrt(6)/3)

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

S(3/2, sqrt(3)/6, sqrt(6)/3)

Vectors:

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

PR: (1,0,0)

Cross product components:

i component: (-sqrt(3)/3 * 0 - sqrt(6)/3 *0 ) =0

j component: -(0*0 - sqrt(6)/3 *1 ) = -(-sqrt(6)/3) = sqrt(6)/3

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

So cross product vector is (0, sqrt(6)/3, sqrt(3)/3). The magnitude is sqrt( (sqrt(6)/3)^2 + (sqrt(3)/3)^2 ) = sqrt( (6/9) + (3/9) ) = sqrt(9/9) =1. Correct.

Therefore, the cross-sectional area is 1.
</execution_10.1>
</Execution_10>

<Plan_11>
Based on execution_10.1:
Therefore, the answer is 1. But the problem asks for the surface area of one of the pieces. The cross-sectional area is 1, and the original faces of the tetrahedron that are part of the piece. Each original face of the tetrahedron is an equilateral triangle with side length 2, so area sqrt(3).
So the total surface area of the piece would be the cross-sectional area (1) plus the areas of the original faces that are now part of the piece. 

But how many original faces are intersected by the cutting plane? The cutting plane intersects four edges (AC, AD, BC, BD), so it creates a new quadrilateral face, and the original faces that are split. However, depending on which piece we consider, the original faces that are entirely on one side of the cutting plane will remain as whole faces, while the ones intersected will have parts cut off. But since the cutting plane is halfway between AB and CD, and given the symmetry, maybe each of the original faces (ABC, ABD, ACD, BCD) is split by the plane, with the piece containing vertex A having parts of faces ABC, ABD, and ACD. Wait, vertex A is part of faces ABC, ABD, and ACD. The face BCD is on the other side of the plane. Therefore, the piece containing vertex A will have the original faces ABC, ABD, ACD, each truncated by the cutting plane, and the new quadrilateral face PQRS. However, since the cutting plane intersects edges AC, AD, BC, BD, the faces ABC and ABD (which include edge AB) are not intersected by the cutting plane. Wait, no. Wait, the cutting plane is parallel to AB and CD, so it does not intersect AB or CD. Therefore, the faces that include AB (ABC and ABD) are not intersected by the cutting plane. Therefore, the piece containing vertex A will have the original faces ACD (which is entirely on one side of the plane) and ABD (partially? Wait, no. Wait, the cutting plane is halfway between AB and CD, so for face ABD, which is triangle ABD, the plane is parallel to AB and cuts through edges AD and BD. Therefore, face ABD is split into two parts: a smaller triangle and a quadrilateral? Wait, no. Let me think.

Wait, face ABD is a triangle with vertices A, B, D. The cutting plane intersects edges AD and BD at points Q and S. Therefore, face ABD is split into two parts: a triangle AQS and a quadrilateral QSBD. But since the plane is parallel to AB, the intersection with face ABD (triangle ABD) would create a smaller triangle AQS similar to ABD, and a trapezoid QSBD. However, since the plane is parallel to AB, the line QS is parallel to AB. Therefore, triangle AQS is similar to triangle ABD, scaled down.

<prompt_11.1> Determine which faces of the tetrahedron are intersected by the cutting plane and calculate the surface area of the piece containing vertex A. </prompt_11.1>
</Plan_11>

<Execution_11>
<execution_11.1>
Wait, no. Wait, the cutting plane is not intersecting face ABD along a line parallel to AB. Wait, the cutting plane is parallel to AB and CD. Therefore, the intersection with face ABD (which contains AB) would be a line parallel to AB. But since the cutting plane is parallel to AB, the intersection line on face ABD should be parallel to AB. However, in our case, the cutting plane intersects edge AD at Q and edge BD at S. The line QS is the intersection of the cutting plane with face ABD. Since the cutting plane is parallel to AB, QS must be parallel to AB. Let's verify:

Points Q(1/2, sqrt(3)/6, sqrt(6)/3) and S(3/2, sqrt(3)/6, sqrt(6)/3). The vector QS is (3/2 -1/2, sqrt(3)/6 - sqrt(3)/6, sqrt(6)/3 - sqrt(6)/3) = (1,0,0). Which is along the x-axis. AB is from (0,0,0) to (2,0,0), so direction vector (2,0,0). Therefore, vector QS is (1,0,0), which is half the length of AB's direction vector. Therefore, QS is parallel to AB. Therefore, on face ABD, the intersection with the cutting plane is the line segment QS, which is parallel to AB. Therefore, the piece containing vertex A includes face ABD up to the line QS, and the other part is the quadrilateral QSBD. However, since the cutting plane is only intersecting edges AD and BD at Q and S, the face ABD is split into triangle AQS and quadrilateral QSBD. But since the plane is cutting through the tetrahedron, the piece containing vertex A would include the original face ACD (since the cutting plane does not intersect ACD), the cross-sectional face PQRS, and the parts of faces ABD and ABC that are connected to A. Wait, but face ABC is on the other side of the plane? Wait, no. The cutting plane is parallel to AB and CD, which are skew. So depending on the orientation, the plane might intersect some faces and leave others. 

Wait, let's clarify:

The original tetrahedron has four faces: ABC, ABD, ACD, BCD. The cutting plane is parallel to AB and CD and lies halfway between them. Therefore, the cutting plane does not intersect AB or CD. It intersects the other four edges: AC, AD, BC, BD. Therefore, the cutting plane intersects edges AC and AD on the side of A, and edges BC and BD on the side of B and D. Therefore, the piece containing vertex A is bounded by:

1. The original face ACD.

2. The cross-sectional face PQRS.

3. Parts of faces ABC and ABD. Wait, but face ABC is connected to A, B, C. The cutting plane intersects edge AC at P and edge BC at R. Therefore, the piece containing A includes triangle APQ and quadrilateral PRBC? Wait, no. Let's think again.

Wait, the piece containing vertex A is separated by the cutting plane. The cutting plane intersects edges AC at P, AD at Q, BC at R, and BD at S. Therefore, the piece containing A is:

- The original face ACD (triangle).

- The cross-sectional face PQRS (quadrilateral).

- The portion of face ABC that is connected to A via edge AC and BC, but cut by the plane at P and R. So, the piece would include triangle APC and quadrilateral PRBC? Wait, no. The face ABC is split into triangle APC and quadrilateral PRBC by the cutting plane's intersection at P and R. Similarly, face ABD is split into triangle AQS and quadrilateral QSBD by the cutting plane's intersection at Q and S. Therefore, the piece containing A is composed of:

- Face ACD.

- Cross-sectional face PQRS.

- Part of face ABC: triangle APC.

- Part of face ABD: triangle AQS.

But wait, how are these connected? The piece containing A would have:

- The original face ACD.

- The cross-section PQRS.

- The part of face ABC from A to P to R.

- The part of face ABD from A to Q to S.

But how are these connected? Let's visualize:

- The original face ACD is there.

- The cross-section PQRS connects P, Q, R, S.

- The face ABC is split into APC and PRBC. The piece containing A would have APC.

- The face ABD is split into AQS and QSBD. The piece containing A would have AQS.

But how are APC and AQS connected? They both share the vertex A, but are on different faces. The cross-section PQRS connects P, Q, R, S. So the piece containing A would have:

- Face ACD.

- Face APC (from ABC).

- Face AQS (from ABD).

- Face PQRS (the cross-section).

But how are these connected? The face APC is part of ABC, and AQS is part of ABD. The cross-section PQRS connects P (on AC), Q (on AD), R (on BC), and S (on BD). Therefore, the piece containing A is bounded by:

- Face ACD.

- Face APC (triangle).

- Face AQS (triangle).

- Face PQRS (quadrilateral).

But in reality, the piece containing A would be a polyhedron with five faces: ACD, APC, AQS, PQRS, and the original face between APC and AQS? Wait, no. Because the cross-section PQRS connects P, Q, R, S. So the piece containing A is a polyhedron with:

- The original face ACD.

- The cross-sectional face PQRS.

- The portion of face ABC from A to P to R.

- The portion of face ABD from A to Q to S.

But how are these connected? The cross-section PQRS connects P, Q, R, S. So the polyhedron containing A would have:

- Face ACD.

- Face APC (triangle).

- Face AQS (triangle).

- Face PQRS (quadrilateral).

But how are these connected? The face APC is connected to ACD along edge AC. The face AQS is connected to ACD along edge AD. The face PQRS is connected to APC at edge PQ (which is part of AC) and connected to AQS at edge QS (which is part of AD). The face PQRS is also connected to the face ACD? Wait, no. The cross-section PQRS is a quadrilateral inside the tetrahedron, so it's connected to the original faces via edges PR, PS, QR, QS. Wait, perhaps not. Let me think again.

The cross-section is a quadrilateral PQRS. Each edge of PQRS is part of a different face:

- PQ is part of the cross-section and lies on face ABC (between P and Q).

Wait, no. PQ is a new edge created by the cutting plane. Wait, no. The cross-section is a new face, so each edge of the cross-section is an edge of the cross-section face. The original faces are split by the cutting plane. Therefore, the original faces are modified, but the cross-section is a new face.

Therefore, the piece containing A is composed of:

1. The original face ACD.

2. The cross-sectional face PQRS.

3. The portion of face ABC from A to P to R.

4. The portion of face ABD from A to Q to S.

But how are these connected? The cross-section PQRS is a new face that connects P, Q, R, S. The original face ACD is separate. The portions of ABC and ABD (APC and AQS) are connected to the cross-section face via their edges.

Wait, perhaps the figure is a pentahedron (five faces):

- ACD (triangle)

- APC (triangle)

- AQS (triangle)

- PQRS (quadrilateral)

But how? The cross-section PQRS is connected to ACD via point P and Q? Wait, no. The cross-section is connected to the original faces through the edges PR, PS, QR, QS. Wait, perhaps not. Let me think in terms of vertices and edges.

The piece containing A has the following vertices: A, P, Q, R, S.

Edges:

- From A: connected to P (via APC), connected to Q (via AQS), connected to C (via ACD and APC), connected to D (via ACD and AQS).

- The cross-section edges: P connected to Q (via PQRS), Q connected to R (via PQRS), R connected to S (via PQRS), S connected to P (via PQRS).

Wait, but in reality, the cross-section is a quadrilateral, so the edges are PQ, QR, RS, SP. Each of these edges is part of the cross-sectional face.

Additionally, the original faces:

- APC has edges AP, PC, AC.

- AQS has edges AQ, QS, AS.

- ACD has edges AC, CD, DA.

But the cross-section PQRS has edges PQ, QR, RS, SP. Each of these edges is part of the cross-sectional face.

Therefore, the polyhedron containing A has:

- Five faces: ACD, APC, AQS, PQRS, and the face connecting APC, AQS, and PQRS? Wait, no. The cross-section PQRS is a separate face. The original faces APC, AQS, and ACD are connected to each other through edges and vertices.

Wait, perhaps the piece containing A is a polyhedron with five faces: ACD (triangle), APC (triangle), AQS (triangle), PQRS (quadrilateral), and another face connecting APC, AQS, and PQRS? No, because the cross-section is a single face. Let me try to list all the faces:

1. ACD (triangle): A, C, D.

2. APC (triangle): A, P, C.

3. AQS (triangle): A, Q, S.

4. PQRS (quadrilateral): P, Q, R, S.

But how are these connected? The cross-section PQRS is connected to ACD via edge PQ (which is part of AC) and edge PS (which is part of AD). Wait, but PQ is part of the cross-section, not an original edge. Similarly, PS is part of the cross-section. Wait, no. The cross-section's edges are entirely new. The original edges are split by the cutting plane.

Therefore, the polyhedron containing A is bounded by:

- The original face ACD.

- The cross-sectional face PQRS.

- The portion of face ABC from A to P to R.

- The portion of face ABD from A to Q to S.

But these are all separate faces. Wait, actually, the cross-sectional face PQRS is a new face, and the original faces APC, AQS, and ACD are modified. However, the original faces are split by the cutting plane, so the original faces are no longer intact. Instead, the original faces are divided into parts.

But in reality, the piece containing A is the union of:

- The original face ACD.

- The cross-sectional face PQRS.

- The part of face ABC that is on the A side of the cutting plane (triangle APC).

- The part of face ABD that is on the A side of the cutting plane (triangle AQS).

But how are these connected? The cross-section PQRS connects these parts. Specifically:

- The edge PQ of the cross-section connects the part of face ABC (APC) to the part of face ABD (AQS).

- The edge QR connects the cross-section to the original face ACD via point R? Wait, no. The cross-section PQRS is entirely within the tetrahedron. The original face ACD is separate. The cross-section is connected to ACD through the original edges AC and AD, but those are split by the cutting plane. 

This is getting confusing. Maybe a better approach is to realize that the piece containing vertex A is a polyhedron with five faces: the original face ACD, the cross-sectional face PQRS, and three quadrilateral faces connecting these. Wait, no. Let me think again.

The original tetrahedron has four faces. The cutting plane intersects four edges, creating a new quadrilateral face. Therefore, the resulting two polyhedrons each have four original faces split and one new face. Therefore, the piece containing A would have:

- The original face ACD.

- The cross-sectional face PQRS.

- The parts of faces ABC and ABD that are on the A side of the cutting plane.

But how many faces does this result in? Let's count:

- Original face ACD: 1 face.

- Cross-sectional face PQRS: 1 face.

- The part of face ABC on the A side: triangle APC.

- The part of face ABD on the A side: triangle AQS.

But these four faces are connected how? The cross-section PQRS is connected to APC and AQS via edges PQ and QS. The face ACD is connected to APC via edge AC and to AQS via edge AD. So the polyhedron containing A has four faces: ACD, APC, AQS, PQRS. Wait, but that would be four faces. However, in reality, when you split a tetrahedron with a plane, each piece should have the number of faces equal to the number of original faces intersected plus the new face. Since the cutting plane intersects four edges, creating one new face, and intersects four original faces (each intersected once or not at all). Wait, no. The original tetrahedron has four faces. The cutting plane intersects four edges, each of which is part of two faces. Therefore, each intersection splits a face. Therefore, each original face is either split or not. The piece containing A has the original face ACD, and parts of faces ABC and ABD. The cross-sectional face.

But perhaps the correct way is that the piece containing A has:

- The original face ACD.

- The cross-sectional face PQRS.

- The parts of faces ABC and ABD that are connected to A.

But how are these connected? The cross-sectional face PQRS is a quadrilateral. Each edge of PQRS is part of the cross-section. The edge PQ is part of the cross-section and connects the part of face ABC (APC) to the cross-section. The edge QS is part of the cross-section and connects the cross-section to the part of face ABD (AQS). The edge SR is part of the cross-section and connects back to the part of face ABD. The edge RP is part of the cross-section and connects back to the part of face ABC. Wait, but in reality, the cross-section is a single face, so the edges PQ, QR, RS, SP are all part of the cross-section. Therefore, the polyhedron containing A is bounded by:

- Face ACD.

- Face APC.

- Face AQS.

- Face PQRS.

But how are these connected? The face APC is connected to ACD via edge AC. The face AQS is connected to ACD via edge AD. The face PQRS is connected to APC via edge PQ and connected to AQS via edge QS. Therefore, the polyhedron has five faces: ACD, APC, AQS, PQRS, and another face connecting APC, AQS, and PQRS? Wait, perhaps not. Let me visualize:

Imagine the original face ACD. From A, there's a new face AQS (triangle) connected via edge AD, and a new face APC (triangle) connected via edge AC. The cross-sectional face PQRS is a quadrilateral connecting P (on AC), Q (on AD), R (on BC), and S (on BD). Therefore, the cross-sectional face is connected to APC via edge PQ and to AQS via edge QS. But this is getting confusing. Maybe the total number of faces is five: ACD, APC, AQS, PQRS, and another face connecting APC, AQS, and PQRS? 

Alternatively, perhaps the piece containing A is a polyhedron with five faces: the original face ACD, the cross-sectional face PQRS, and three quadrilateral faces connecting these. Wait, perhaps not. 

Alternatively, perhaps the piece containing A is a pentahedron (five faces):

1. Face ACD (triangle).

2. Face APC (triangle).

3. Face AQS (triangle).

4. Face PQRS (quadrilateral).

5. Face connecting APC, AQS, and PQRS? No, that doesn't make sense. 

Wait, perhaps the cross-sectional face PQRS is connected to the original faces through their respective edges. The edge PQ of the cross-section is part of the original face ABC (between P and Q). The edge QS is part of the original face ABD (between Q and S). The edge QR is part of the cross-section and connects to R, which is on BC. The edge RS connects to S and back to P. 

This is getting too tangled. Let's instead compute the surface area by considering all the faces that make up the piece containing A. The surface area will consist of:

1. The original face ACD: area sqrt(3).

2. The cross-sectional face PQRS: area 1.

3. The portion of face ABC on the A side: triangle APC. Let's compute its area.

4. The portion of face ABD on the A side: triangle AQS. Let's compute its area.

Then sum all these areas.

So first, compute the area of triangle APC. Points A(0,0,0), P(1/2, sqrt(3)/2,0), C(1, sqrt(3),0). Since these are all on the original face ABC, which is an equilateral triangle. The triangle APC is a smaller triangle on ABC. Since P is the midpoint of AC, triangle APC is half of triangle ABC. Therefore, its area is half of sqrt(3), which is sqrt(3)/2.

Similarly, triangle AQS is on face ABD. Points A(0,0,0), Q(1/2, sqrt(3)/6, sqrt(6)/3), S(3/2, sqrt(3)/6, sqrt(6)/3). Let's compute the area of triangle AQS.

To compute the area of triangle AQS, we can use the formula for the area of a triangle given three points in 3D space. The formula is (1/2)||cross product of vectors AS and AQ||.

First, compute vectors AS and AQ:

Vector AS: S - A = (3/2, sqrt(3)/6, sqrt(6)/3)

Vector AQ: Q - A = (1/2, sqrt(3)/6, sqrt(6)/3)

Compute the cross product AS × AQ:

|i         j         k         |
|3/2   sqrt(3)/6   sqrt(6)/3|
|1/2   sqrt(3)/6   sqrt(6)/3|

= i*(sqrt(3)/6 * sqrt(6)/3 - sqrt(6)/3 * sqrt(3)/6 ) - j*(3/2 * sqrt(6)/3 - sqrt(6)/3 *1/2 ) + k*(3/2 * sqrt(3)/6 - sqrt(3)/6 *1/2 )

Simplify each component:

i component:

sqrt(3)/6 * sqrt(6)/3 - sqrt(6)/3 * sqrt(3)/6 = (sqrt(18)/18 - sqrt(18)/18 ) = 0

j component:

- [ (3/2 * sqrt(6)/3 ) - (sqrt(6)/3 *1/2 ) ] = - [ (sqrt(6)/2 - sqrt(6)/6 ) ] = - [ (3sqrt(6)/6 - sqrt(6)/6 ) ] = - [ 2sqrt(6)/6 ) ] = - [ sqrt(6)/3 ]

k component:

3/2 * sqrt(3)/6 - sqrt(3)/6 *1/2 = (3sqrt(3)/12 - sqrt(3)/12 ) = (2sqrt(3)/12 ) = sqrt(3)/6

Therefore, the cross product AS × AQ is (0, -sqrt(6)/3, sqrt(3)/6 )

The magnitude of this vector is sqrt(0^2 + ( -sqrt(6)/3 )^2 + (sqrt(3)/6 )^2 ) = sqrt( (6/9) + (3/36) ) = sqrt( (2/3) + (1/12) ) = sqrt( (8/12) + (1/12) ) = sqrt(9/12) = sqrt(3/4) = sqrt(3)/2

Therefore, the area of triangle AQS is (1/2)*sqrt(3)/2 = sqrt(3)/4.

Wait, but let me verify the cross product calculation again because this seems important.

Wait, the cross product components:

i component: (sqrt(3)/6)(sqrt(6)/3) - (sqrt(6)/3)(sqrt(3)/6) = 0, as sqrt(3)*sqrt(6) is commutative.

j component: - [ (3/2)(sqrt(6)/3) - (sqrt(6)/3)(1/2) ] = - [ (sqrt(6)/2 - sqrt(6)/6 ) ] = - [ (3sqrt(6)/6 - sqrt(6)/6 ) ] = - [ 2sqrt(6)/6 ) ] = - [ sqrt(6)/3 )

k component: (3/2)(sqrt(3)/6) - (sqrt(3)/6)(1/2) = (3sqrt(3)/12 - sqrt(3)/12 ) = (2sqrt(3)/12 ) = sqrt(3)/6

Therefore, cross product vector is (0, -sqrt(6)/3, sqrt(3)/6 )

Magnitude squared: 0 + (6/9) + (3/36) = 2/3 + 1/12 = 8/12 + 1/12 = 9/12 = 3/4. So magnitude is sqrt(3/4) = sqrt(3)/2. Therefore, area is half of that, which is sqrt(3)/4.

Therefore, the area of triangle AQS is sqrt(3)/4.

Therefore, the total surface area of the piece containing vertex A is:

- Face ACD: sqrt(3)

- Cross-sectional face PQRS:1

- Face APC: sqrt(3)/2

- Face AQS: sqrt(3)/4

But wait, that sums to sqrt(3) + 1 + sqrt(3)/2 + sqrt(3)/4. Wait, but this seems like more than the original surface area. The original tetrahedron had four faces each of area sqrt(3), totaling 4sqrt(3). The piece containing A has parts of three original faces and the cross-section. However, the sum here would be sqrt(3) (ACD) + sqrt(3)/2 (AP
</execution_11.1>
</Execution_11>

<Final_answer>
Therefore, the surface area of one of the pieces of the tetrahedron after slicing is $\boxed{1+\frac{3\sqrt{3}}{4}}$.
</Final_answer>
</think>