[THOUGHT] We need to parse the geometry problem carefully. Let's restate:

Tetrahedron ABCD with:

• AB = CD = 4
• BC = BD = 2√5
• H = foot of perpendicular from A to line CD (so AH ⟂ CD)
• AH = 4
• Planes ABH and BCD are perpendicular.
• G = centroid of triangle ABH
• S = sphere with center G tangent to plane ACD
• T = locus of points P on S such that ∠APG = π/2 (i.e., right angle at P)
• Find area of orthogonal projection of T onto plane ABC.
• If area = π/k, compute k (integer).

We need to find k.

We'll need to set up coordinates, compute positions of points A, B, C, D, H, G, sphere S, then find T, project onto plane ABC, compute area.

Plan:

1. Choose coordinate system conveniently. Since ABH and BCD are perpendicular planes, maybe we can align CD along some axis, etc. Let’s set D at origin? Or maybe C and D on x-axis.

Given CD = 4, let’s put C and D on x-axis: D = (0,0,0), C = (4,0,0) (since CD = 4). But careful: CD is length 4, but orientation arbitrary. We can set CD along x-axis.

H is foot from A to line CD, so H lies on CD. AH ⟂ CD, so AH is perpendicular to CD (x-axis). So AH lies in a plane perpendicular to x-axis. Since AH = 4, we can set H somewhere on CD, and A at distance 4 perpendicular to CD.

Let H = (h,0,0) for some h between 0 and 4. Then AH is vector from A to H perpendicular to CD, so AH lies in yz-plane (since CD along x-axis). Let’s set AH along y-axis for simplicity: A = (h, 4, 0). But careful: AH is perpendicular from A to line CD, so AH is perpendicular to CD (x-axis). So indeed AH can be along y-axis. So A = (h, 4, 0).

Now we have AB = 4. B unknown. Also BC = 2√5, BD = 2√5. So B is equidistant from C and D with distance 2√5. So B lies on perpendicular bisector of CD. Since C=(4,0,0), D=(0,0,0), midpoint of CD is (2,0,0). Perpendicular bisector plane is x=2. So B has x-coordinate 2.

Let B = (2, yb, zb). Distances: BD^2 = (2-0)^2 + yb^2 + zb^2 = 4 + yb^2 + zb^2 = (2√5)^2 = 20. So yb^2 + zb^2 = 16.

BC^2 = (2-4)^2 + yb^2 + zb^2 = 4 + yb^2 + zb^2 = 20, same condition. So yb^2+zb^2=16.

Also AB = 4: A = (h,4,0). So AB^2 = (2-h)^2 + (yb-4)^2 + zb^2 = 16.

We also have plane ABH perpendicular to plane BCD. Plane ABH contains points A, B, H. Plane BCD contains B, C, D. Need condition: normal vectors perpendicular.

Compute normal vectors: Plane ABH: vectors AH = H - A = (h - h, 0-4, 0-0) = (0, -4, 0). AB = B - A = (2-h, yb-4, zb). Cross product n1 = AH × AB.

AH = (0, -4, 0) AB = (2-h, yb-4, zb) n1 = AH × AB = determinant i j k; 0 -4 0; 2-h yb-4 zb = i*(-4zb - 0(yb-4)) - j*(0zb - 0(2-h)) + k*(0*(yb-4) - (-4)(2-h)) = i(-4zb) - j(0) + k*(4*(2-h)). So n1 = (-4zb, 0, 4(2-h)).

Plane BCD: vectors BD = D - B = (-2, -yb, -zb), BC = C - B = (2, -yb, -zb). Cross product n2 = BD × BC.

Compute: BD = (-2, -yb, -zb) BC = (2, -yb, -zb) n2 = determinant i j k; -2 -yb -zb; 2 -yb -zb = i*((-yb)(-zb) - (-zb)(-yb)) - j*((-2)(-zb) - (-zb)2) + k((-2)(-yb) - (-yb)*2).

Simplify: i: (-yb)(-zb) = ybzb, (-zb)(-yb) = zbyb, so difference = 0. j: (-2)*(-zb) = 2zb, (-zb)2 = -2zb, so 2zb - (-2zb) = 4zb? Wait careful: j component is -[ (-2)(-zb) - (-zb)2 ] = -[2zb - (-2zb)] = -[2zb + 2zb] = -4zb. k: (-2)(-yb) = 2yb, (-yb)*2 = -2yb, so 2yb - (-2yb) = 4yb.

Thus n2 = (0, -4zb, 4yb).

Now planes are perpendicular if n1·n2 = 0.

Compute dot product: (-4zb)0 + 0(-4zb) + 4(2-h)*4yb = 16(2-h)yb = 0.

So either yb = 0 or h = 2.

But yb^2 + zb^2 = 16. If yb = 0, then zb = ±4. That's possible.

If h = 2, then H is midpoint of CD. But we don't know yet.

We also have AH = 4 already satisfied. No other constraints? We also have AB = 4 equation.

Let's consider both cases.

Case 1: yb = 0. Then zb = ±4. Then B = (2, 0, ±4). Then AB^2 = (2-h)^2 + (0-4)^2 + (±4)^2 = (2-h)^2 + 16 + 16 = (2-h)^2 + 32 = 16? Wait AB = 4 => AB^2 = 16. So (2-h)^2 + 32 = 16 => (2-h)^2 = -16 impossible. So yb cannot be 0.

Thus h = 2.

So h = 2. Then H = (2,0,0) (midpoint of CD). Good.

Now with h=2, A = (2,4,0). Then AB^2 = (2-2)^2 + (yb-4)^2 + zb^2 = (yb-4)^2 + zb^2 = 16.

Also yb^2 + zb^2 = 16.

Subtract: (yb-4)^2 + zb^2 - (yb^2+zb^2) = 0 => (yb^2 -8yb +16) - yb^2 = -8yb +16 = 0 => yb = 2.

Then yb^2 = 4, so zb^2 = 16 - 4 = 12 => zb = ±√12 = ±2√3.

So B = (2, 2, ±2√3). Two possibilities; choose one, say positive for simplicity: B = (2, 2, 2√3).

Now we have points: D = (0,0,0) C = (4,0,0) H = (2,0,0) A = (2,4,0) B = (2,2,2√3)

Check distances: AB: (0, -2, 2√3) length sqrt(4 + 12) = sqrt16 = 4 ok. BC: (2-4,2-0,2√3-0)=(-2,2,2√3) length sqrt(4+4+12)=sqrt20=2√5 ok. BD: (-2, -2, -2√3) length sqrt(4+4+12)=2√5 ok. CD: (4,0,0) length 4 ok. AH: (0, -4, 0) length 4 ok.

Now planes ABH and BCD perpendicular: we already enforced condition giving h=2, which satisfied.

Now compute centroid G of triangle ABH: points A(2,4,0), B(2,2,2√3), H(2,0,0). All have x=2. So triangle ABH lies in plane x=2.

Centroid: average of coordinates: G = ((2+2+2)/3, (4+2+0)/3, (0+2√3+0)/3) = (6/3, 6/3, (2√3)/3) = (2, 2, (2√3)/3).

So G = (2, 2, 2√3/3).

Now sphere S: center G, tangent to plane ACD. Need distance from G to plane ACD = radius R.

Plane ACD: points A(2,4,0), C(4,0,0), D(0,0,0). Let's compute plane equation.

Vectors: AC = (2, -4, 0), AD = (-2, -4, 0). Cross product AC × AD = determinant i j k; 2 -4 0; -2 -4 0 = i*((-4)0 - 0(-4)) - j*(20 - 0(-2)) + k*(2*(-4) - (-4)(-2)) = i(0) - j*(0) + k*(-8 - 8) = (0,0,-16). So normal vector n = (0,0,-16) or simply (0,0,1) direction. So plane ACD is z=0? Wait check: points A, C, D all have z=0. Indeed A(2,4,0), C(4,0,0), D(0,0,0) all have z=0. So plane ACD is the xy-plane (z=0). Good.

Thus distance from G to plane z=0 is |z-coordinate of G| = |2√3/3| = 2√3/3.

So radius R = 2√3/3.

Thus sphere S: center G(2,2,2√3/3), radius R = 2√3/3.

Now T = locus of points P on sphere S such that ∠APG = π/2, i.e., right angle at P in triangle APG. So AP ⟂ PG.

That is (A-P)·(G-P) = 0.

Let P = (x,y,z) on sphere: (x-2)^2 + (y-2)^2 + (z - 2√3/3)^2 = R^2 = (2√3/3)^2 = 4*3/9 = 12/9 = 4/3.

Condition: (A-P)·(G-P) = 0.

Compute A-P = (2-x, 4-y, 0-z) = (2-x, 4-y, -z). G-P = (2-x, 2-y, 2√3/3 - z).

Dot product: (2-x)(2-x) + (4-y)(2-y) + (-z)(2√3/3 - z) = 0.

Simplify: (2-x)^2 + (4-y)(2-y) - z(2√3/3 - z) = 0.

Expand (4-y)(2-y) = 8 -4y -2y + y^2 = 8 -6y + y^2.

So equation: (2-x)^2 + y^2 -6y + 8 - z*(2√3/3) + z^2 = 0.

But (2-x)^2 = (x-2)^2.

Thus: (x-2)^2 + y^2 -6y + 8 + z^2 - (2√3/3)z = 0.

Now sphere equation: (x-2)^2 + (y-2)^2 + (z - 2√3/3)^2 = 4/3.

Expand sphere: (x-2)^2 + (y-2)^2 + z^2 - (4√3/3)z + (43/9) = 4/3? Wait compute: (z - 2√3/3)^2 = z^2 - (4√3/3)z + (43/9) = z^2 - (4√3/3)z + 12/9 = z^2 - (4√3/3)z + 4/3.

Thus sphere: (x-2)^2 + (y-2)^2 + z^2 - (4√3/3)z + 4/3 = 4/3.

Cancel 4/3: (x-2)^2 + (y-2)^2 + z^2 - (4√3/3)z = 0.

So sphere equation simplifies to: (x-2)^2 + (y-2)^2 + z^2 - (4√3/3)z = 0. (Eq1)

Now condition from right angle: (x-2)^2 + y^2 -6y + 8 + z^2 - (2√3/3)z = 0. (Eq2)

Subtract Eq2 from Eq1? Let's subtract Eq2 from Eq1 to eliminate (x-2)^2 and z^2.

Eq1 - Eq2: [(x-2)^2 + (y-2)^2 + z^2 - (4√3/3)z] - [(x-2)^2 + y^2 -6y + 8 + z^2 - (2√3/3)z] = 0 - 0? Actually both equal 0, so their difference is 0.

Compute:

Cancel (x-2)^2, z^2.

Remaining: (y-2)^2 - [y^2 -6y + 8] + [ - (4√3/3)z - ( - (2√3/3)z ) ] = 0.

Simplify: (y^2 -4y +4) - y^2 +6y -8 + [ - (4√3/3)z + (2√3/3)z ] = 0.

Combine y terms: (-4y +6y) = 2y, constants: 4 -8 = -4. So: 2y -4 + [ - (2√3/3)z ] = 0.

Thus: 2y -4 - (2√3/3)z = 0 => divide 2: y -2 - (√3/3)z = 0 => y = 2 + (√3/3)z. (Eq3)

So the locus T is intersection of sphere S with plane Eq3.

Thus T is a circle (intersection of sphere with plane). We need orthogonal projection of T onto plane ABC.

Now we need plane ABC: points A(2,4,0), B(2,2,2√3), C(4,0,0). Compute its equation.

Vectors: AB = (0, -2, 2√3), AC = (2, -4, 0). Normal vector n_ABC = AB × AC.

Compute cross product:

AB = (0, -2, 2√3) AC = (2, -4, 0)

Determinant i j k; 0 -2 2√3; 2 -4 0 = i*((-2)0 - (2√3)(-4)) - j*(00 - (2√3)2) + k(0(-4) - (-2)2) = i(0 + 8√3) - j*(0 - 4√3) + k*(0 + 4) = (8√3, 4√3, 4).

Simplify divide by 4: n_ABC = (2√3, √3, 1). So plane ABC equation: 2√3(x - 2) + √3(y - 4) + 1(z - 0) = 0.

Check: at A(2,4,0): 0. At B(2,2,2√3): 2√3(0) + √3(-2) + 2√3 = -2√3 + 2√3 = 0. Good.

Thus plane ABC: 2√3 x + √3 y + z - (4√3 + 4√3)?? Let's compute constant: using point A: 2√32 + √34 + 0 = 4√3 + 4√3 = 8√3. So plane equation: 2√3 x + √3 y + z = 8√3.

Now we need orthogonal projection of circle T onto plane ABC. Projection of a circle onto a plane is generally an ellipse (unless plane parallel to circle's plane). But we need area of that ellipse.

We have circle T: intersection of sphere S with plane Eq3: y = 2 + (√3/3)z.

Let’s find center and radius of this circle.

Sphere center G = (2,2,2√3/3). Plane Eq3: y = 2 + (√3/3)z, or rewrite: y - 2 - (√3/3)z = 0.

Distance from G to plane: compute d = |2 - 2 - (√3/3)*(2√3/3)| / sqrt(1^2 + ( -√3/3 )^2 )? Wait plane normal vector: coefficients: 0 for x, 1 for y, -√3/3 for z. So normal vector n_plane = (0,1, -√3/3). Magnitude = sqrt(1 + ( (√3/3)^2 )) = sqrt(1 + 3/9) = sqrt(1 + 1/3) = sqrt(4/3) = 2/√3.

Distance from G(2,2,2√3/3) to plane: |12 - 12 - (√3/3)(2√3/3)|? Wait compute: plane equation: y - 2 - (√3/3)z = 0. So plug G: y_G - 2 - (√3/3) z_G = 2 - 2 - (√3/3)(2√3/3) = 0 - (2*3)/9 = -6/9 = -2/3. Absolute value = 2/3.

Thus distance d = | -2/3 | = 2/3.

But careful: distance formula uses normal vector magnitude. Actually we computed numerator = 2/3, but denominator = |n| = 2/√3. So distance = (2/3) / (2/√3) = (2/3)*(√3/2) = √3/3.

Wait: let's recalc properly.

Plane: y - 2 - (√3/3)z = 0. Normal vector = (0,1, -√3/3). Magnitude = sqrt(1 + (√3/3)^2) = sqrt(1 + 3/9) = sqrt(1 + 1/3) = sqrt(4/3) = 2/√3.

Distance from point (x0,y0,z0) = |0x0 + 1y0 + (-√3/3)*z0 - 2| / |n|. But plane equation is y - (√3/3)z - 2 = 0. So constant term -2. So distance = |y0 - (√3/3)z0 - 2| / (2/√3).

For G: y0=2, z0=2√3/3. Compute y0 - (√3/3)z0 - 2 = 2 - (√3/3)(2√3/3) - 2 = 2 - (23)/9 - 2 = -6/9 = -2/3. Absolute value = 2/3.

Thus distance d = (2/3) / (2/√3) = (2/3)*(√3/2) = √3/3.

Yes.

Now sphere radius R = 2√3/3.

Thus circle radius r = sqrt(R^2 - d^2) = sqrt( (2√3/3)^2 - (√3/3)^2 ) = sqrt( (4*3/9) - (3/9) ) = sqrt(12/9 - 3/9) = sqrt(9/9) = 1.

So circle T has radius 1.

Now we need area of orthogonal projection of this circle onto plane ABC.

Projection of a circle onto a plane: area = π * (radius of circle) * (cos θ), where θ is angle between plane of circle and projection plane? Actually orthogonal projection of a circle of radius r onto a plane: the projected shape is an ellipse with semi-major axis r, semi-minor axis r cos φ, where φ is angle between the plane of the circle and the projection plane. Wait careful: If we have a circle of radius r lying in a plane with unit normal n1, and we project orthogonally onto a plane with unit normal n2, the projection is an ellipse with area = π r^2 * |cos φ|, where φ is the angle between the two planes (i.e., angle between normals). Actually area of projection = area of original shape times |cos θ|, where θ is angle between the plane of the shape and the projection plane. For a circle, area = π r^2, projected area = π r^2 * |cos α|, where α is angle between the two planes.

But is that correct? Let's think: Orthogonal projection onto a plane: the scaling factor for area when projecting from plane1 to plane2 is |n1·n2|? Actually if you project a planar region of area A onto a plane, the projected area is A * |cos φ|, where φ is angle between the two planes (i.e., between their normals). Yes, because projection compresses area by factor cos φ.

Thus we need angle between plane of T and plane ABC.

Plane T is defined by Eq3: y - 2 - (√3/3)z = 0, i.e., normal vector n_T = (0, 1, -√3/3).

Plane ABC normal vector n_ABC = (2√3, √3, 1). We can simplify n_ABC by dividing by something? Let's keep as is.

Compute dot product n_T · n_ABC = 0*(2√3) + 1*√3 + (-√3/3)*1 = √3 - √3/3 = (3√3/3 - √3/3) = (2√3/3).

Magnitudes: |n_T| = 2/√3 (as computed). |n_ABC| = sqrt( (2√3)^2 + (√3)^2 + 1^2 ) = sqrt(12 + 3 + 1) = sqrt(16) = 4.

Thus cos φ = |n_T·n_ABC| / (|n_T| |n_ABC|) = (2√3/3) / ( (2/√3)4 ) = (2√3/3) / (8/√3) = (2√3/3) * (√3/8) = (23)/(3*8) = 6/24 = 1/4.

So cos φ = 1/4.

Thus projected area = area of circle T * |cos φ| = π * r^2 * cos φ = π * 1^2 * (1/4) = π/4.

Thus area = π/4, so k = 4.

But wait: Is the projected shape an ellipse? Yes, orthogonal projection of a circle onto a plane is an ellipse with area = π r^2 |cos φ|. That seems correct.

Thus k = 4.

We should verify with coordinates maybe compute projection explicitly to be sure.

Let's compute projection of circle onto plane ABC. But maybe we can double-check with Python.

We'll compute circle T as intersection of sphere and plane, find parametric form, project onto plane ABC, compute area of ellipse.

Plan for verification:

1. Define sphere center G, radius R.
2. Define plane T: y = 2 + (√3/3)z.
3. Find circle center C_T = projection of G onto plane T? Actually center of circle is foot of perpendicular from G to plane T. That's not necessarily the center of sphere. Wait: The intersection circle of a sphere and a plane has center at the projection of the sphere center onto the plane. Yes.

We already computed distance from G to plane T = d = √3/3. So center of circle T is point M = G + d * (unit normal of plane T). But careful: unit normal of plane T: n_T = (0,1,-√3/3). Magnitude = 2/√3. Unit normal = (0, 1/(2/√3), -√3/3/(2/√3)) = (0, √3/2, -√3/3 * √3/2) = (0, √3/2, -1/2). Check: magnitude sqrt( (3/4) + (1/4) ) = sqrt(1) = 1. Good.

But note direction: distance d = √3/3 positive? We computed signed distance = -2/3 divided by magnitude 2/√3 gave -√3/3. Actually signed distance = -2/3 / (2/√3) = -√3/3. So point M = G + (-√3/3) * unit normal? Wait: if signed distance is -√3/3, then moving from G towards plane in direction of unit normal yields plane. So M = G + (signed distance) * unit normal = G + (-√3/3) * (0, √3/2, -1/2) = (2,2,2√3/3) + (0, - (√3/3)(√3/2), - (√3/3)(-1/2)) = (2,2,2√3/3) + (0, - (3/6), (√3/6)) = (2,2,2√3/3) + (0, -1/2, √3/6).

Compute: y: 2 - 1/2 = 3/2. z: 2√3/3 + √3/6 = (4√3/6 + √3/6) = 5√3/6.

Thus M = (2, 3/2, 5√3/6).

Check if M lies on plane T: y = 3/2, z = 5√3/6. Then 2 + (√3/3)z = 2 + (√3/3)(5√3/6) = 2 + (53)/(18) = 2 + 15/18 = 2 + 5/6 = 17/6 ≈ 2.8333, but y=1.5. That's not equal. Something wrong: I must have messed sign.

Let's recompute plane T equation: y = 2 + (√3/3)z. So plane: y - (√3/3)z - 2 = 0. Normal vector n = (0,1,-√3/3). Unit normal u = (0, 1/√(1+ (√3/3)^2), -√3/3/√(1+ (√3/3)^2)) = (0, 1/(2/√3), -√3/3/(2/√3)) = (0, √3/2, -√3/3 * √3/2) = (0, √3/2, -1/2). Good.

Now signed distance from G to plane: d_signed = (y_G - (√3/3)z_G - 2) / |n|? Actually formula: d_signed = (a x0 + b y0 + c z0 + d) / √(a^2+b^2+c^2). For plane: 0x + 1y + (-√3/3)z - 2 = 0. So a=0,b=1,c=-√3/3,d=-2. Then d_signed = (02 + 12 + (-√3/3)(2√3/3) - 2) / (2/√3) = (2 - (23)/9 - 2) / (2/√3) = (2 - 6/9 - 2) / (2/√3) = (-6/9) / (2/√3) = (-2/3) * (√3/2) = -√3/3.

Thus signed distance = -√3/3. So moving from G in direction of unit normal u (pointing from plane to G? Actually unit normal direction: (0, √3/2, -1/2). If we move from G by signed distance d_signed along u, we get foot: M = G + d_signed * u = G + (-√3/3)(0, √3/2, -1/2) = (2,2,2√3/3) + (0, - (√3/3)(√3/2), - (√3/3)*(-1/2)) = (2,2,2√3/3) + (0, - (3/6), (√3/6)) = (2,2,2√3/3) + (0, -0.5, √3/6).

Compute z: 2√3/3 = 4√3/6. So z = 4√3/6 + √3/6 = 5√3/6 ≈ 1.443. y = 2 - 0.5 = 1.5.

Now check plane: y - (√3/3)z - 2 = 1.5 - (√3/3)(5√3/6) - 2 = 1.5 - (53)/(18) - 2 = 1.5 - 15/18 - 2 = 1.5 - 0.8333 - 2 = -1.3333, not zero. Something wrong.

Let's compute exactly: (√3/3)(5√3/6) = (53)/(18) = 15/18 = 5/6 ≈ 0.8333. So y - (√3/3)z - 2 = 1.5 - 0.8333 - 2 = -1.3333. That's not zero. So M is not on plane. Something inconsistent.

Maybe I messed unit normal direction. The unit normal we computed points in some direction. The signed distance is negative, meaning G is on the opposite side of plane relative to the direction of unit normal? Actually signed distance formula uses normal vector orientation. If we use normal vector n = (0,1,-√3/3), then unit normal u = n/|n| = (0, √3/2, -1/2). The signed distance is (n·(G - any point on plane)) / |n|. Since plane passes through origin? No, plane equation is y - (√3/3)z - 2 = 0. Take point on plane: say (0,2,0). Then G - (0,2,0) = (2,0,2√3/3). Dot with n: (0,1,-√3/3)·(2,0,2√3/3) = 0 + 0 + (-√3/3)*(2√3/3) = -2/3. Divide by |n| = 2/√3 gives -√3/3. So signed distance is -√3/3. That means G is on the side of plane opposite to the direction of n. So foot M = G - d_signed * u? Actually foot = G - d_signed * u? Wait: If we move from G towards the plane along direction of normal n, we need to move by distance |d_signed| in direction of n if d_signed positive? Actually standard: foot = G - (signed distance) * unit normal? Let's derive: Let plane: n·(X - X0) = 0. Signed distance from point P to plane is (n·(P - X0))/|n|. Foot Q = P - (signed distance) * (n/|n|). Because moving from P by - (signed distance) * unit normal brings us to plane. Indeed if signed distance positive, P is on side of normal, so moving opposite to normal reduces distance. So Q = P - (signed distance) * u.

Thus M = G - d_signed * u = G - (-√3/3) * u = G + (√3/3) * u.

Compute: u = (0, √3/2, -1/2). So (√3/3)u = (0, (√3/3)(√3/2), (√3/3)*(-1/2)) = (0, 1/2, -√3/6).

Then M = (2,2,2√3/3) + (0, 1/2, -√3/6) = (2, 2.5, 2√3/3 - √3/6) = (2, 5/2, (4√3/6 - √3/6)) = (2, 5/2, 3√3/6) = (2, 5/2, √3/2).

Check plane: y - (√3/3)z - 2 = 2.5 - (√3/3)*(√3/2) - 2 = 2.5 - (3/6) - 2 = 2.5 - 0.5 - 2 = 0. Good.

Thus circle center M = (2, 5/2, √3/2). Radius r = 1.

Now we need to project this circle onto plane ABC. Projection is orthogonal projection onto plane ABC.

We can compute projection of circle as ellipse. The ellipse's area = area of circle * |cos φ|, where φ is angle between plane of circle and projection plane.

We already computed cos φ = 1/4. So area = π * 1^2 * 1/4 = π/4.

Thus k = 4.

But let's verify with Python by computing projection explicitly: find ellipse axes.

We'll compute parametric equations of circle T, project onto plane ABC, compute area of ellipse.

Steps:

• Circle center M.
• Radius r=1.
• Need orthonormal basis vectors for circle plane (plane T).
• Project these vectors onto plane ABC.
• Compute semi-axes lengths from projected basis vectors.
• Area = π * a * b.

Let's do that in Python.

We'll also verify that the projection is indeed an ellipse.

We'll compute using sympy for precision.

We'll compute unit normal of plane T: n_T = (0,1,-√3/3). Normalize to u_T = (0, √3/2, -1/2) as before.

Find two orthonormal vectors in plane T: choose v1 = (1,0,0) maybe? But need perpendicular to u_T. Actually plane T is vertical plane? Since x can be anything? Plane T equation: y = 2 + (√3/3)z, independent of x. So plane T is parallel to x-axis. Indeed, normal has no x component, so x-axis lies in plane. So we can take v1 = (1,0,0) as a unit vector in plane T.

Then v2 = u_T × v1 (cross product) gives another unit vector in plane T, perpendicular to both.

Compute v2 = u_T × (1,0,0) = determinant i j k; 0 √3/2 -1/2; 1 0 0 = i*( (√3/2)0 - (-1/2)0 ) - j(00 - (-1/2)1) + k(00 - (√3/2)1) = i(0) - j(0 - (-1/2)) + k*(0 - √3/2) = (0, -1/2, -√3/2). Wait compute: j component: -[0*0 - (-1/2)*1] = -[0 + 1/2] = -1/2. So v2 = (0, -1/2, -√3/2). Magnitude: sqrt(0 + 1/4 + 3/4) = 1. Good.

Thus circle parametric: M + r*(cosθ * v1 + sinθ * v2).

Now project onto plane ABC. Orthogonal projection onto plane: we can find projection operator: for any vector, subtract component along normal of plane ABC.

Let n_ABC = (2√3, √3, 1). Unit normal u_ABC = n_ABC / |n_ABC| = (2√3/4, √3/4, 1/4) = (√3/2, √3/4, 1/4). Check magnitude: sqrt(3/4 + 3/16 + 1/16) = sqrt(12/16 + 3/16 + 1/16) = sqrt(16/16)=1. Good.

Projection of a vector v onto plane ABC: v_proj = v - (v·u_ABC) * u_ABC.

Thus we can compute projected ellipse as image of circle under linear map P (projection). The ellipse is centered at projection of M onto plane ABC? Actually orthogonal projection of circle onto plane: center projects to center of ellipse. So we can compute center C_proj = projection of M onto plane ABC.

Then ellipse axes are projections of circle's axes vectors (scaled by radius). The semi-axes lengths are lengths of projected basis vectors.

We can compute area as π * (length of projected v1) * (length of projected v2) if v1 and v2 are orthogonal in plane T and projection preserves orthogonality? Not necessarily, but we can compute ellipse area via determinant of projection matrix restricted to plane T.

Alternatively, compute area = π * r^2 * |cos φ|, which we already have.

But to be safe, compute explicitly.

Let's write Python code to compute area.

We'll use sympy for exact values.

We'll compute:

• Plane ABC normal n_ABC = (2√3, √3, 1)
• Unit normal u_ABC.
• Projection operator P = I - u_ABC ⊗ u_ABC (outer product).
• Compute projected vectors p1 = P(v1), p2 = P(v2).
• Compute area of ellipse: area = π * sqrt( det( [p1·p1, p1·p2; p2·p1, p2·p2] ) )? Actually area of parallelogram spanned by p1 and p2 is sqrt( det( Gram matrix ) ). But ellipse is image of unit circle under linear map that takes (cosθ, sinθ) to r*(cosθ p1 + sinθ p2). The ellipse area is π * r^2 * |det( matrix with columns p1, p2 )|? Wait: The ellipse is { r*(cosθ p1 + sinθ p2) }. The area of ellipse = π * r^2 * |det([p1, p2])|, where p1 and p2 are column vectors. Because linear map from unit circle to ellipse has matrix [p1, p2] times r. The area scaling factor is absolute determinant of that matrix. So ellipse area = π * r^2 * |det([p1, p2])|.

But careful: p1 and p2 are 3D vectors, but ellipse lies in plane ABC. The determinant of 3x2 matrix? Actually we need area of ellipse in 3D, but since it's planar, we can compute area as sqrt( (p1·p1)*(p2·p2) - (p1·p2)^2 ) * π * r^2. That's the area of parallelogram spanned by p1 and p2, which is |p1 × p2| (magnitude). Since p1 and p2 are perpendicular in plane T? v1 and v2 are orthogonal unit vectors. But projection may not preserve orthogonality. So we need to compute area as π * r^2 * |det( matrix of transformation from circle plane to projection plane )|. The determinant of the linear map from the plane T (with orthonormal basis) to the projected plane (with standard metric in 3D) is the absolute value of the determinant of the 2x2 matrix of inner products? Actually simpler: area of ellipse = π * (product of semi-axes lengths). We can compute semi-axes lengths by singular values of projection matrix restricted to plane T.

But maybe easier: compute area of parallelogram spanned by p1 and p2 = |p1 × p2|. Since p1 and p2 are projections of orthogonal unit vectors, the area of ellipse is π * r^2 * |p1 × p2|? Wait: The ellipse is image of unit circle under linear map L = [p1, p2] (columns). The ellipse area = π * |det(L)| where L is 2x2 matrix representing L in orthonormal basis of the plane of circle? Actually careful: The circle is in plane T, with orthonormal basis {v1, v2}. The map from that plane to plane ABC via orthogonal projection is linear. In coordinates (c1, c2) where point = c1 v1 + c2 v2, the projected point is c1 p1 + c2 p2, where p1 = projection of v1, p2 = projection of v2. The ellipse is image of unit circle c1^2 + c2^2 = 1. The area of this ellipse is π * |det( matrix whose columns are p1, p2 in orthonormal basis of the target plane )|? Actually the ellipse lies in plane ABC. To compute area, we can compute the area of parallelogram spanned by p1 and p2, which is |p1 × p2| (since they are 3D vectors, cross product magnitude gives area of parallelogram). The ellipse area = π * (area of parallelogram) / (area of unit circle?) Wait: The unit circle maps to ellipse via linear map L. The area of ellipse = area of unit circle * |det(L)|. The unit circle area = π. So ellipse area = π * |det(L)|. But L is a linear map from R^2 (circle coordinates) to R^3 (projected vectors). However, the image is 2D in plane ABC. The determinant of L (as map from R^2 to R^3) is not defined as usual. Instead, the area of ellipse = π * sqrt( det( (L^T L) ) ), where L is 3x2 matrix with columns p1, p2. Because L^T L is 2x2 matrix of inner products. The singular values of L are square roots of eigenvalues of L^T L. The area scaling factor from unit circle to ellipse is product of singular values, i.e., sqrt( det(L^T L) ). Indeed, ellipse area = π * σ1 * σ2 = π * sqrt( det(L^T L) ).

Thus area = π * sqrt( (p1·p1)*(p2·p2) - (p1·p2)^2 ).

Since r=1, ellipse area = π * sqrt( det( Gram matrix ) ).

Thus we can compute that.

Let's compute with Python.

We'll use sympy for exact symbolic computation.

We'll compute k = 4 if area = π/4.

Let's do it. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

sqrt3 = sp.sqrt(3)

Points

G = (2,2,2*sqrt3/3)

M = (2,5/2, sqrt3/2)

v1 = (1,0,0)

v2 = (0, -1/2, -sqrt3/2)

r = 1

Plane ABC normal n_ABC = (2*sqrt3, sqrt3, 1)

Unit normal u_ABC = n_ABC / |n_ABC| = (2*sqrt3/4, sqrt3/4, 1/4) = (sqrt3/2, sqrt3/4, 1/4)

Projection operator P(v) = v - (v·u_ABC)*u_ABC

Compute u_ABC

n_ABC = sp.Matrix([2*sqrt3, sqrt3, 1]) norm_n = sp.sqrt(n_ABC.dot(n_ABC)) # = 4 u_ABC = n_ABC / norm_n # (sqrt3/2, sqrt3/4, 1/4)

v1 and v2

v1 = sp.Matrix([1, 0, 0]) v2 = sp.Matrix([0, -sp.Rational(1,2), -sqrt3/2])

Projected vectors

def project(v): return v - (v.dot(u_ABC)) * u_ABC

p1 = project(v1) p2 = project(v2)

Compute area of ellipse: area = pi * sqrt(det(Gram))

Gram = sp.Matrix([[p1.dot(p1), p1.dot(p2)], [p2.dot(p1), p2.dot(p2)]]) det_gram = Gram.det() area = sp.pi * sp.sqrt(det_gram)

print("Projected vectors p1, p2:") print("p1 =", p1) print("p2 =", p2) print("Gram matrix:") print(Gram) print("det(Gram) =", det_gram) print("Area =", area) print("Area as rational times pi:", area / sp.pi) [/PYTHON]

[THOUGHT] The Python output shows that the area of the orthogonal projection of T onto plane ABC is π/4. Therefore, if the area is given as π/k, then k = 4.

\boxed{4}