[THOUGHT] We need to understand the geometry. O_0 is the unit circle x^2+y^2=1. P(x0,y0) is on O_0 with x0>0. The tangent line at P to O_0: equation is x0x + y0y = 1 (since for unit circle, tangent at (x0,y0) is x0 x + y0 y = 1). Q is intersection of this tangent line with y=-1. So plug y=-1: x0x + y0(-1)=1 => x0*x = 1+y0 => x = (1+y0)/x0. So Q = ((1+y0)/x0, -1). P' = (0,-1). Triangle P'PQ: vertices P'(0,-1), P(x0,y0), Q((1+y0)/x0, -1). The triangle is inside? Let's visualize: P' and Q are on y=-1 line, P above on unit circle. The segment PQ is part of tangent line from P to Q. The line y=-1 is base. The circles O_n are tangent to segment PQ, line y=-1, and circle O_{n-1} (with O_0 being the unit circle). They lie inside triangle P'PQ and have radius smaller than previous. This is reminiscent of a chain of circles inscribed in a curvilinear triangle formed by two lines (PQ and y=-1) and a circle (O_0). Actually, O_1 is tangent to segment PQ (a line), line y=-1 (a line), and circle O_0. So O_1 is a circle inscribed in a triangular region bounded by two lines and a circle. Then O_2 is tangent to same two lines and O_1, etc. This is a classic chain of circles (Soddy circles) but with two lines and a circle.

We need to find radii r_n of circles O_n. The sum of areas sum_{n=1}^∞ π r_n^2 = π/8 => sum r_n^2 = 1/8. We need to find x0 and y0 given that condition. Then compute 2024 * x0^2 * y0.

We need to derive recurrence for radii. Let's set up coordinate system. Let’s denote the two lines: L1: line PQ (tangent line), L2: line y=-1. The circle O_0: unit circle centered at (0,0) radius R0=1. The circles O_n are all tangent to L1, L2, and O_{n-1}. They lie inside triangle P'PQ, which is the region between L1, L2, and the vertical line? Actually triangle P'PQ has vertices P'(0,-1), P(x0,y0), Q((1+y0)/x0,-1). Its sides are: P'P (line from (0,-1) to (x0,y0)), PQ (tangent line), and QP' (horizontal line y=-1 from Q to P'). So the triangle is bounded by three lines: y=-1 (L2), line through P and Q (L1), and line through P' and P (call L3). Wait, but O_n are tangent to segment PQ (i.e., L1) and line y=-1 (L2), but also to circle O_{n-1}. They do NOT mention being tangent to L3 (line P'P). So the circles are inscribed in region bounded by L1, L2, and O_0? Actually O_1 is tangent to L1, L2, and O_0. That region is like a curvilinear triangle: two lines and a circle. O_2 is tangent to L1, L2, and O_1, etc. So they are all tangent to same two lines and previous circle. So they form a chain of circles along the "corner" between L1 and L2, touching the previous circle.

We can use inversion or analytic geometry to find recurrence. Let's set up coordinates conveniently.

Let’s denote the angle between L1 and L2. L2 is horizontal y=-1. L1 is tangent to unit circle at P. The slope of L1? Derivative: dy/dx = -x0/y0 (since x^2+y^2=1 => 2x+2y y'=0 => y'=-x/y). So slope m = -x0/y0. But note: y0 = sqrt(1-x0^2) positive? Not necessarily positive y-coordinate? It says P(x0,y0) on O_0 with positive x-coordinate. It doesn't specify y0 sign. Could be positive or negative? Usually on unit circle, y0 could be positive or negative. But triangle P'PQ: P'=(0,-1), Q on y=-1, P on circle. For triangle to contain circles inside, likely P is above y=-1? Actually if y0 is negative, triangle might be degenerate. Let's assume y0>0 for interior region. We'll see.

Let’s find angle between L1 and L2. L2 horizontal, slope 0. L1 slope m = -x0/y0. The angle θ between lines: tan θ = |m| = x0/y0. Actually angle from horizontal to L1 (since L1 goes downward if x0>0,y0>0? slope negative). But magnitude.

Better: Let’s find distance from center of O_n to lines L1 and L2. Let circle O_n have radius r, center (h,k). It is tangent to y=-1 => distance from center to line y=-1 is |k - (-1)| = |k+1| = r (since circle lies above y=-1? Actually inside triangle, likely circles are above y=-1, so k > -1? Actually circle tangent to y=-1 from above, so distance = k+1 = r => k = r-1.

It is tangent to L1: line equation x0x + y0y = 1. Distance from (h,k) to L1: |x0h + y0k - 1| / sqrt(x0^2+y0^2) = r (since radius). sqrt(x0^2+y0^2)=1 because (x0,y0) on unit circle. So |x0 h + y0 k - 1| = r.

Also tangent to circle O_{n-1} with radius R_{n-1} and center (0,0) for n=1? Actually O_0 centered at (0,0) radius 1. For n>=2, O_{n-1} has center somewhere? Wait, O_1 is tangent to O_0, but O_2 is tangent to O_1, not O_0. So each circle O_n is tangent to previous circle O_{n-1}, not necessarily to O_0 for n>1. So centers change.

We need recurrence. Perhaps the chain of circles is such that their radii form a geometric progression. Let's try to compute r1 explicitly.

Let’s denote d_n = distance between centers of O_n and O_{n-1}. For tangency externally? Since both circles inside triangle and tangent to same two lines, they likely are externally tangent (they don't overlap). So d_n = r_n + r_{n-1}.

Also, centers lie on a line? Since they are tangent to two fixed lines, their centers lie on the angle bisector of those lines. Indeed, for a circle tangent to two lines, its center lies on the angle bisector. So all centers O_n lie on the same angle bisector of L1 and L2.

Let’s find that angle bisector. Lines L1: x0 x + y0 y -1 =0, L2: y+1=0. The angle bisectors are given by (x0 x + y0 y -1)/√(x0^2+y0^2) = ± (y+1). Since √(x0^2+y0^2)=1, we have x0 x + y0 y -1 = ± (y+1). We need the bisector that lies inside triangle P'PQ. Which sign? Let's test point inside triangle. Possibly the one with +? Let's compute later.

Alternatively, we can find the incenter of triangle P'PQ? But circles are not necessarily inscribed in triangle, they are tangent to two sides and the previous circle.

Maybe easier: Use coordinate geometry with specific P. Since the sum condition is independent of P? Possibly x0 and y0 are determined uniquely. Let's set up variables.

Let’s denote angle between L1 and L2 as 2α. Then the distance from center of circle to each line is r. If center lies on bisector, distance to each line is r. Actually if circle tangent to both lines, distance from center to each line equals radius r. So center lies on line parallel to both lines at distance r? Actually if two lines intersect, the locus of points equidistant from both lines is the angle bisector. But distance to each line equals r means center lies on a line parallel to each line at distance r? No, the condition is: distance to L1 = r and distance to L2 = r. So center must be at intersection of two lines parallel to L1 and L2 at distance r. That gives two points symmetric about bisector. But also must lie on bisector? Actually if distances to both lines are equal, then point lies on angle bisector. So indeed center lies on bisector and distance to each line = r. So we can find center coordinates along bisector.

Let’s find equation of bisector. Let’s compute with symbolic x0,y0.

We'll use Python to derive formulas.

First, define x0,y0 with relation x0^2+y0^2=1. Let y0 = sqrt(1-x0^2) positive? We'll keep symbolic.

We'll compute r1 from conditions: circle radius r, center (h,k). Conditions:

1. k = r-1 (distance to y=-1).
2. Distance to L1: |x0 h + y0 k - 1| = r.
3. Distance to O_0 center (0,0): distance between (h,k) and (0,0) = |1 - r|? Since O_0 radius 1, and O_1 tangent to it. Are they externally tangent? O_1 lies inside triangle, likely inside region between O_0 and lines. O_0 is unit circle, O_1 is smaller and inside triangle. They might be externally tangent? Actually O_1 touches O_0 from inside? Wait, O_0 is the unit circle, O_1 is inside triangle P'PQ. The triangle P'PQ includes region outside O_0? Actually O_0 is the unit circle, triangle vertices: P' (0,-1), P on circle, Q on tangent line. The triangle likely contains region outside O_0? Not sure. But O_1 is tangent to O_0, and lies inside triangle. Probably O_1 is externally tangent to O_0, meaning distance between centers = 1 + r. But could be internally tangent: distance = 1 - r. Which one? Since O_1 is inside triangle and O_0 is the unit circle, if O_1 is inside O_0, then distance <1. But O_1 is tangent to segment PQ and line y=-1, which are outside O_0? Actually line y=-1 is below circle (circle center at origin, radius 1, y=-1 is below). Tangent line at P touches circle at P. So region between lines and circle is outside circle? Possibly O_1 is outside O_0 but inside triangle. Let's determine.

Take a concrete example: choose P at (1,0) (x0=1,y0=0). But y0=0 gives tangent line vertical? Actually tangent at (1,0) is x=1. Then Q intersection with y=-1 is (1,-1). Triangle P'PQ: P'(0,-1), P(1,0), Q(1,-1). That's a right triangle. O_1 tangent to line x=1, y=-1, and unit circle. That's a circle in corner. Its radius? Let's compute numerically to guess.

Let's code to compute r1 for general x0,y0.

We'll solve equations for r1.

[/THOUGHT] [PYTHON] import sympy as sp

variables

x0, y0, r = sp.symbols('x0 y0 r', real=True)

relation x0^2 + y0^2 = 1

eq_circle = x02 + y02 - 1

center (h,k)

h, k = sp.symbols('h k')

conditions:

1) distance to y=-1 is r: |k+1| = r, assuming k > -1 so k+1 = r

eq1 = k + 1 - r

2) distance to line L1: x0x + y0y = 1 is r

|x0h + y0k - 1| = r. Assume sign? Since center likely on same side of L1 as origin? Actually origin (0,0) gives x00+y00-1 = -1 <0. So distance =1. For circle O_1, radius r<1, and center likely also on same side as origin? Let's assume x0h + y0k - 1 <0 so that distance = -(x0h + y0k -1) = r => 1 - x0h - y0k = r.

eq2 = 1 - x0h - y0k - r

3) distance to O_0 center (0,0) = 1 + r (external tangency) or |1 - r| (internal). Let's try external first: distance = 1 + r

eq3_ext = h2 + k2 - (1 + r)**2

internal: distance = |1 - r|, assuming r<1, distance = 1 - r

eq3_int = h2 + k2 - (1 - r)**2

Solve with external tangency

solutions_ext = sp.solve([eq1, eq2, eq3_ext, eq_circle], [h, k, r, x0], dict=True) print("External tangency solutions:") print(solutions_ext)

This gives many solutions. Let's solve for r in terms of x0,y0 eliminating h,k.

Actually we can solve eq1, eq2 for h,k in terms of r, then plug into eq3.

from eq1: k = r - 1. from eq2: 1 - x0h - y0(r-1) = r => 1 - x0h - y0r + y0 = r => 1 + y0 - x0h = r + y0r => x0h = 1 + y0 - r(1 + y0) = (1+y0)(1 - r) => h = (1+y0)(1 - r)/x0. Now plug into eq3_ext: h^2 + (r-1)^2 = (1+r)^2. Compute symbolically. h_expr = (1+y0)(1 - r)/x0 k_expr = r - 1 eq = h_expr2 + k_expr2 - (1+r)**2

simplify with x0^2+y0^2=1

Use sympy to simplify eq with relation.

eq_simplified = sp.simplify(eq.subs(x02, 1 - y02)) print("Equation for r (external):") print(eq_simplified) sol_r_ext = sp.solve(eq_simplified, r) print("r solutions (external):", sol_r_ext)

Now internal tangency:

eq3_int = h_expr2 + k_expr2 - (1 - r)2 eq_int_simplified = sp.simplify(eq3_int.subs(x02, 1 - y0**2)) print("Equation for r (internal):") print(eq_int_simplified) sol_r_int = sp.solve(eq_int_simplified, r) print("r solutions (internal):", sol_r_int) [/PYTHON]

[THOUGHT] There was a syntax error in the code. Likely due to missing import or incorrect substitution. Let's rewrite more carefully.

We have derived expressions: k = r - 1 h = (1+y0)*(1 - r)/x0

Now plug into distance condition to O_0 center.

We need to decide whether tangency is external or internal. Let's test with a numeric example to see which yields positive r < 1.

Take P at (1,0): x0=1, y0=0. Then h = (1+0)(1-r)/1 = 1 - r. k = r - 1. Distance to origin: sqrt((1-r)^2 + (r-1)^2) = sqrt(2(1-r)^2) = sqrt(2)(1-r). If external tangency: distance = 1 + r. So sqrt(2)(1-r) = 1 + r => solve for r. If internal tangency: distance = 1 - r. So sqrt(2)*(1-r) = 1 - r => either 1-r=0 or sqrt(2)=1, impossible. So internal tangency gives r=1 (trivial) or no solution. So external tangency likely.

Let's compute r for (1,0) using external condition.

We'll compute numerically.

[/THOUGHT] [PYTHON] import sympy as sp

numeric test for x0=1, y0=0

x0_val = 1 y0_val = 0 r = sp.symbols('r')

from external tangency condition: distance between centers = 1 + r

center (h,k):

h_expr = (1 + y0_val)*(1 - r)/x0_val k_expr = r - 1 eq = h_expr2 + k_expr2 - (1 + r)**2 sol_r = sp.solve(eq, r) print("Solutions for r (x0=1,y0=0):", sol_r)

evaluate numerically

for sol in sol_r: print(float(sol)) [/PYTHON]

[THOUGHT] For P(1,0), r1 = 3 - 2√2 ≈ 0.1716. That seems plausible: a small circle tangent to vertical line x=1, horizontal line y=-1, and unit circle. Now we need to find the recurrence for subsequent circles. Since all circles are tangent to the same two lines and the previous circle, their radii likely follow a geometric progression. In many problems of inscribed circles in a curvilinear triangle formed by two lines and a circle, the radii form a geometric progression. Let's test: compute r2 if possible.

But we have sum of areas = π/8 => sum_{n=1}^∞ r_n^2 = 1/8. If r_n form a geometric progression r_n = r1 * k^{n-1} for some ratio k (0<k<1), then sum r_n^2 = r1^2 / (1 - k^2) = 1/8. So we need to find k in terms of x0,y0.

We need to derive recurrence relation for r_n. Let’s denote the distance from center of O_n to the vertex where lines L1 and L2 intersect. Actually, the two lines intersect at some point V. Let's find V: solve L1: x0 x + y0 y = 1 and L2: y = -1. Then x0 x + y0*(-1) = 1 => x0 x = 1 + y0 => x = (1+y0)/x0. That's Q! Wait, Q is intersection of L1 and L2? Actually Q is intersection of tangent line and y=-1, yes that's exactly V. So V = Q. So lines intersect at Q.

Now, consider the chain of circles tangent to both lines. Their centers lie on the angle bisector of lines L1 and L2. Let angle between lines be 2α. Then distance from center of circle of radius r to vertex Q is r / sin α (since distance from center to each line is r, and center lies on bisector). Actually, if circle tangent to both lines, distance from center to vertex Q along bisector is r / sin α. Because if we drop perpendiculars to both lines, each perpendicular length r, and the angle between radius to line and line to vertex is α. So center distance d = r / sin α.

Thus for circle O_n with radius r_n, distance from its center to Q is d_n = r_n / sin α.

Now, also O_n is tangent to O_{n-1}. Their centers are on same bisector, so distance between centers = d_n - d_{n-1} (if they are on same side of Q). Since both circles are inside the angle, their centers lie on same ray from Q. So distance between centers = |d_n - d_{n-1}| = r_n + r_{n-1} (external tangency). So we have:

r_n / sin α - r_{n-1} / sin α = r_n + r_{n-1} => (r_n - r_{n-1}) / sin α = r_n + r_{n-1} => r_n - r_{n-1} = sin α (r_n + r_{n-1}) => r_n - r_{n-1} = sin α r_n + sin α r_{n-1} => r_n - sin α r_n = r_{n-1} + sin α r_{n-1} => r_n (1 - sin α) = r_{n-1} (1 + sin α) => r_n / r_{n-1} = (1 + sin α) / (1 - sin α)

Let k = (1 + sin α) / (1 - sin α) > 1? Wait, r_n is smaller than r_{n-1}? The problem says radius smaller than that of O_n? Actually O_{n+1} has radius smaller than O_n. So r_{n+1} < r_n. So ratio r_{n+1}/r_n = (1 - sin α)/(1 + sin α) maybe? Let's check indices.

We have O_n tangent to O_{n-1}. For n>=2, O_n is smaller than O_{n-1}. So r_n < r_{n-1}. So ratio r_n / r_{n-1} = (1 - sin α)/(1 + sin α) < 1. That matches.

Thus r_n = r1 * ρ^{n-1} where ρ = (1 - sin α)/(1 + sin α).

Now sum of squares: sum_{n=1}^∞ r_n^2 = r1^2 / (1 - ρ^2) = 1/8.

We need sin α. α is half the angle between L1 and L2. Let's compute sin α.

Lines L1 and L2. L2 horizontal, slope 0. L1 slope m = -x0/y0. Angle between lines: tan(2α) = |m| = x0/|y0|. Actually angle between lines is 2α, where α is angle from horizontal to L1? Let's compute sin α.

We can compute sin α using formula: sin α = ??? Let's compute cos(2α) = dot product of unit direction vectors. Or use formula: tan(2α) = slope difference? Actually if line L1 makes angle θ with horizontal, then tan θ = slope = -x0/y0. Since slope negative, θ is negative angle? But magnitude: tan|θ| = x0/|y0|. Then angle between L1 (θ) and L2 (0) is |θ|. So 2α = |θ|? Wait, α is half the angle between lines, so 2α = |θ|. So α = |θ|/2.

Thus sin α = sin(|θ|/2). We can compute sin(θ/2) from tan θ = x0/y0? Actually tan θ = -x0/y0, but sign doesn't matter for sin(θ/2). Use identity: sin(θ/2) = sqrt((1 - cos θ)/2). And cos θ = 1/√(1+tan^2 θ) = |y0|/√(x0^2+y0^2) = |y0|. Since x0^2+y0^2=1. So cos θ = |y0|. But θ is angle of line L1 relative to horizontal, so cos θ = x0/√(x0^2+y0^2)? Wait, direction vector of line L1 is ( -y0, x0 )? Actually line L1: x0 x + y0 y = 1. Normal vector is (x0, y0). Direction vector perpendicular to normal: ( -y0, x0 ). So slope = x0 / (-y0) = -x0/y0. So tan θ = slope = -x0/y0. So cos θ = 1/√(1+tan^2 θ) = 1/√(1+(x0^2/y0^2)) = |y0|/√(x0^2+y0^2) = |y0|. So cos θ = |y0|. Then sin(θ/2) = sqrt((1 - |y0|)/2). But α = |θ|/2, so sin α = sqrt((1 - |y0|)/2). However, we need sin α positive. Since y0 could be positive or negative? But triangle P'PQ likely requires y0 positive? Let's assume y0>0.

Thus sin α = sqrt((1 - y0)/2).

Now ρ = (1 - sin α)/(1 + sin α).

Now r1 we can compute from earlier expression. For external tangency, we derived equation for r1. Let's compute r1 symbolically.

From earlier: h = (1+y0)(1 - r)/x0, k = r-1. Distance to origin squared = h^2 + (r-1)^2 = (1+r)^2 (external tangency). So equation: ((1+y0)(1 - r)/x0)^2 + (r-1)^2 = (1+r)^2. Simplify: (1+y0)^2 (1 - r)^2 / x0^2 + (1 - r)^2 = (1+r)^2. Factor (1 - r)^2: (1 - r)^2 [ (1+y0)^2 / x0^2 + 1 ] = (1+r)^2. But x0^2 = 1 - y0^2 = (1-y0)(1+y0). So (1+y0)^2 / x0^2 = (1+y0)^2 / ((1-y0)(1+y0)) = (1+y0)/(1-y0). Thus (1 - r)^2 [ (1+y0)/(1-y0) + 1 ] = (1+r)^2. Compute bracket: (1+y0)/(1-y0) + 1 = (1+y0 + 1 - y0)/(1-y0) = 2/(1-y0). So equation: (1 - r)^2 * 2/(1-y0) = (1+r)^2. => 2(1 - r)^2 = (1-y0)(1+r)^2. Take square roots? Solve for r.

Let's solve symbolically.

[/THOUGHT] [PYTHON] import sympy as sp

y0, r = sp.symbols('y0 r', real=True)

equation: 2*(1 - r)**2 = (1 - y0)*(1 + r)**2

eq = sp.Eq(2*(1 - r)**2, (1 - y0)*(1 + r)**2) sol_r = sp.solve(eq, r) print("r solutions:", sol_r)

Simplify

for sol in sol_r: print(sp.simplify(sol)) [/PYTHON]

[THOUGHT] We have two solutions for r1. Which one is positive and less than 1? For y0 between -1 and 1. Since y0>0 likely, let's test y0=0.5. Compute numerically:

First solution: (-y0 - 2sqrt(2-2y0) + 3)/(y0+1) Second solution: (-y0 + 2sqrt(2-2y0) + 3)/(y0+1)

For y0=0.5: sqrt(2-1)=1. So first: (-0.5 -21 +3)/(1.5)= (-0.5-2+3)/1.5 = (0.5)/1.5=0.3333. Second: (-0.5+2+3)/1.5 = (4.5)/1.5=3. So first is smaller. So r1 = (-y0 - 2sqrt(2-2y0) + 3)/(y0+1). But note sqrt(2-2y0)=sqrt(2(1-y0))=√2 * √(1-y0). And sin α = √((1-y0)/2). So √(1-y0)=√2 sin α. Thus sqrt(2-2y0)=√2 * √(2) sin α = 2 sin α? Wait: sqrt(2-2y0)=√(2(1-y0))=√2 * √(1-y0)=√2 * √2 sin α = 2 sin α? Because √(1-y0)=√2 sin α. Indeed sin α = √((1-y0)/2) => √(1-y0)=√2 sin α. Then sqrt(2-2*y0)=√2 * √2 sin α = 2 sin α. Good.

Thus r1 = (-y0 - 2*(2 sin α) + 3)/(y0+1) = (-y0 -4 sin α +3)/(y0+1). But y0 = cos θ? Actually cos θ = y0 (since earlier cos θ = y0). And sin α = √((1-y0)/2). We can express r1 in terms of sin α.

Also y0 = cos(2α)? Because θ = 2α? Actually earlier we had angle between lines is 2α, and cos θ = y0? Wait, we had cos θ = |y0|. But θ is angle of line L1 relative to horizontal. Since slope = -x0/y0, tan θ = -x0/y0. Then cos θ = y0/√(x0^2+y0^2)= y0 (since √(x0^2+y0^2)=1). So y0 = cos θ. And θ = 2α? Actually angle between L1 and horizontal is θ, and angle between L1 and L2 (horizontal) is |θ|, which is 2α. So θ = ±2α. So y0 = cos(2α). Good.

Thus y0 = cos(2α) = 1 - 2 sin^2 α. So 1 - y0 = 2 sin^2 α. Indeed sin α = √((1-y0)/2) matches.

Now express r1 in terms of sin α.

r1 = (-y0 -4 sin α +3)/(y0+1). Substitute y0 = 1 - 2 sin^2 α.

Then numerator: -(1 - 2 sin^2 α) -4 sin α +3 = -1 + 2 sin^2 α -4 sin α +3 = 2 sin^2 α -4 sin α +2 = 2(sin^2 α -2 sin α +1) = 2 (sin α -1)^2. Denominator: y0+1 = (1 - 2 sin^2 α) +1 = 2 - 2 sin^2 α = 2(1 - sin^2 α) = 2 cos^2 α.

Thus r1 = [2 (sin α -1)^2] / [2 cos^2 α] = (sin α -1)^2 / cos^2 α = (1 - sin α)^2 / cos^2 α = ((1 - sin α)/cos α)^2.

But note cos α positive (α acute). So r1 = ((1 - sin α)/cos α)^2 = ( (1 - sin α)^2 / (1 - sin^2 α) ) = (1 - sin α)^2 / ((1 - sin α)(1 + sin α)) = (1 - sin α)/(1 + sin α).

That's exactly ρ! So r1 = ρ.

Thus r1 = ρ = (1 - sin α)/(1 + sin α). Then r2 = ρ * ρ = ρ^2, etc. So r_n = ρ^n? Wait, r1 = ρ, r2 = ρ * ρ = ρ^2, so r_n = ρ^n.

Check: For n=1, r1 = ρ. For n=2, r2 = ρ^2, etc.

Thus sum of squares: sum_{n=1}^∞ r_n^2 = sum_{n=1}^∞ ρ^{2n} = ρ^2 / (1 - ρ^2) = 1/8.

Let t = ρ^2. Then t/(1 - t) = 1/8 => t = (1 - t)/8 => 8t = 1 - t => 9t = 1 => t = 1/9. So ρ^2 = 1/9 => ρ = 1/3 (since ρ positive).

Thus ρ = (1 - sin α)/(1 + sin α) = 1/3 => cross multiply: 3(1 - sin α) = 1 + sin α => 3 - 3 sin α = 1 + sin α => 3 - 1 = sin α + 3 sin α => 2 = 4 sin α => sin α = 1/2.

Thus α = 30° (since α acute). Then y0 = cos(2α) = cos(60°) = 1/2.

Also x0^2 = 1 - y0^2 = 1 - 1/4 = 3/4 => x0 = √(3)/2 (positive x0).

Now compute 2024 * x0^2 * y0 = 2024 * (3/4) * (1/2) = 2024 * (3/8) = 2024 * 0.375 = 759.

Compute precisely: 2024 * 3 / 8 = (2024/8)*3 = 253 * 3 = 759.

Thus answer is 759.

[/THOUGHT] \boxed{759}