[THOUGHT] Let me analyze this problem step by step.

1. First, I need to set up the geometry. We have the unit circle $O_0: x^2 + y^2 = 1$, and a point $P(x_0, y_0)$ on it with $x_0 > 0$.

2. The tangent line to $O_0$ at $P$ has equation $x \cdot x_0 + y \cdot y_0 = 1$.

3. Point $Q$ is where this tangent meets $y = -1$. So $x \cdot x_0 - y_0 = 1$, giving $x = \frac{1 + y_0}{x_0}$. Thus $Q = \left(\frac{1 + y_0}{x_0}, -1\right)$.

4. Point $P' = (0, -1)$.

5. The circles $O_1, O_2, \ldots$ are tangent to:

   • Segment $PQ$ (the tangent line)
   • The line $y = -1$
   • The previous circle $O_n$

Let me think about this more carefully. The circles form a chain (Pappus chain or similar) between the unit circle and the line $y = -1$.

Let me use the inversion approach or direct computation. Since each circle is tangent to the line $y = -1$ and to the tangent line $PQ$, and also to the previous circle, this forms a sequence of circles.

For a circle tangent to two lines, its center lies on the angle bisector. The two lines are $y = -1$ and $x \cdot x_0 + y \cdot y_0 = 1$.

Let me compute this using Python to find the relationship. [/THOUGHT] [PYTHON] import sympy as sp from sympy import symbols, sqrt, solve, simplify, pi, Rational, cos, sin

Let P = (x0, y0) be on the unit circle, so x0^2 + y0^2 = 1

x0, y0 = symbols('x0 y0', real=True, positive=True)

Tangent line at P: xx0 + yy0 = 1

Line y = -1

Q is intersection: x*x0 - y0 = 1, so x = (1+y0)/x0

Q = ((1+y0)/x0, -1)

P' = (0, -1)

The tangent line can be written as: x0x + y0y - 1 = 0

The line y = -1 can be written as: y + 1 = 0

For a circle tangent to both lines with center (h, k):

Distance to y = -1 is |k + 1|

Distance to tangent line is |x0h + y0k - 1| / sqrt(x0^2 + y0^2) = |x0h + y0k - 1| (since x0^2 + y0^2 = 1)

For the circle to be inside the triangle P'PQ and below the tangent line:

The center should have k > -1 (above y = -1)

And x0h + y0k < 1 (below the tangent line, assuming the origin is below)

So radius r = k + 1 = 1 - x0h - y0k

This gives: k + 1 = 1 - x0h - y0k

k + y0k = -x0h

k(1 + y0) = -x0*h

h = -k(1 + y0)/x0

So the center lies on the line h = -k(1+y0)/x0, which passes through P' = (0, -1)

When k = -1: h = -(-1)(1+y0)/x0 = (1+y0)/x0, which is Q!

So centers lie on line P'Q.

Let's parameterize: center = (h, k) where h = -k(1+y0)/x0

Radius r = k + 1

For O_1 to be tangent to O_0 (externally, since O_1 is inside the triangle):

Distance from center of O_1 to origin = 1 + r_1 (external tangency)

sqrt(h^2 + k^2) = 1 + r_1 = 1 + k + 1 = k + 2

h^2 + k^2 = (k + 2)^2

k^2 * (1+y0)^2 / x0^2 + k^2 = k^2 + 4k + 4

k^2 * [(1+y0)^2/x0^2 + 1 - 1] = 4k + 4

k^2 * (1+y0)^2/x0^2 = 4k + 4

Using x0^2 = 1 - y0^2 = (1-y0)(1+y0):

k^2 * (1+y0)^2 / [(1-y0)(1+y0)] = 4k + 4

k^2 * (1+y0) / (1-y0) = 4(k + 1)

Let's denote t = (1+y0)/(1-y0), then:

k^2 * t = 4(k + 1)

k^2 * t - 4k - 4 = 0

k = (4 ± sqrt(16 + 16t)) / (2t) = (4 ± 4*sqrt(1+t)) / (2t) = 2(1 ± sqrt(1+t)) / t

t = symbols('t', positive=True) # t = (1+y0)/(1-y0)

k = 2(1 - sqrt(1+t)) / t (taking minus for smaller circle)

Note: 1 + t = (1-y0+1+y0)/(1-y0) = 2/(1-y0)

sqrt(1+t) = sqrt(2/(1-y0)) = sqrt(2)/sqrt(1-y0)

k_1 = 2(1 - sqrt(2/(1-y0))) / ((1+y0)/(1-y0))

= 2(1-y0)(1 - sqrt(2/(1-y0))) / (1+y0)

= 2(1-y0 - sqrt(2(1-y0))) / (1+y0)

r_1 = k_1 + 1

Let me compute this more carefully

k1_expr = 2*(1 - sqrt(1+t)) / t r1_expr = k1_expr + 1 print("k1 in terms of t:", simplify(k1_expr)) print("r1 in terms of t:", simplify(r1_expr))

Substitute t = (1+y0)/(1-y0)

1 + t = 2/(1-y0)

sqrt(1+t) = sqrt(2)/sqrt(1-y0)

k1 = 2(1 - sqrt(2)/sqrt(1-y0)) * (1-y0)/(1+y0)

= 2(sqrt(1-y0) - sqrt(2)) * sqrt(1-y0) / (1+y0)

= 2(1-y0 - sqrt(2)*sqrt(1-y0)) / (1+y0)

r1 = k1 + 1 = [2(1-y0 - sqrt(2)*sqrt(1-y0)) + 1 + y0] / (1+y0)

= [2 - 2y0 - 2*sqrt(2)*sqrt(1-y0) + 1 + y0] / (1+y0)

= [3 - y0 - 2*sqrt(2)*sqrt(1-y0)] / (1+y0)

Let's verify with a specific value

y0_val = Rational(1, 2) x0_val = sqrt(1 - y0_val**2) t_val = (1 + y0_val)/(1 - y0_val) print(f"\nFor y0 = 1/2:") print(f"x0 = {x0_val}") print(f"t = {t_val}")

k1_val = 2*(1 - sqrt(1 + t_val)) / t_val r1_val = k1_val + 1 print(f"k1 = {simplify(k1_val)}") print(f"r1 = {simplify(r1_val)}") [/PYTHON]

[THOUGHT] Wait, I got k1 = -2/3 which means the center is at y = -2/3, which is above y = -1. And r1 = 1/3. Let me continue the analysis to find the pattern for the sequence of circles and compute the sum of areas.

Let me reconsider. The center has k > -1 for the circle to be above the line y = -1. With k1 = -2/3, we have r1 = k1 + 1 = 1/3. This seems correct.

Now I need to find the pattern for subsequent circles. Each O_{n+1} is tangent to O_n (internally or externally), the line y = -1, and the tangent line PQ.

For O_2 tangent to O_1 externally:

• Center on the same angle bisector line
• Distance between centers = r1 + r2
• r2 = k2 + 1

Let me work out the recurrence relation. If the n-th circle has radius r_n and center at height k_n = r_n - 1, then for the (n+1)-th circle:

The distance from center of O_{n+1} to center of O_n equals r_n + r_{n+1}.

Since both centers lie on the line h = -k(1+y0)/x0, I can parameterize by k.

Let me compute this more carefully with Python. [/THOUGHT]

Actually, I need to run more code to find the complete solution. Let me continue.

[PYTHON] import sympy as sp from sympy import symbols, sqrt, solve, simplify, pi, Rational, cos, sin, expand

For circles tangent to both lines (y=-1 and tangent line), centers lie on angle bisector

h = -k(1+y0)/x0, r = k + 1

For O_{n+1} tangent to O_n externally:

sqrt((h_{n+1} - h_n)^2 + (k_{n+1} - k_n)^2) = r_n + r_{n+1}

h_n = -k_n(1+y0)/x0

h_{n+1} - h_n = -(k_{n+1} - k_n)(1+y0)/x0

Let d = k_{n+1} - k_n (both negative, so d < 0 for decreasing sequence)

Distance = |d| * sqrt((1+y0)^2/x0^2 + 1) = |d| * sqrt((1+y0)^2 + x0^2)/x0

= |d| * sqrt((1+y0)^2 + 1 - y0^2)/x0

= |d| * sqrt(1 + 2y0 + y0^2 + 1 - y0^2)/x0

= |d| * sqrt(2 + 2y0)/x0

= |d| * sqrt(2(1+y0))/x0

r_n + r_{n+1} = (k_n + 1) + (k_{n+1} + 1) = k_n + k_{n+1} + 2

So: |k_{n+1} - k_n| * sqrt(2(1+y0))/x0 = k_n + k_{n+1} + 2

Since k_n < k_{n+1} < -1 + 1 = 0 (circles getting smaller, centers moving up toward y=0)

Wait, actually for smaller circles, k should be closer to -1.

r = k + 1, so smaller r means k closer to -1.

Let's use r_n directly. r_n = k_n + 1, so k_n = r_n - 1.

k_{n+1} - k_n = r_{n+1} - r_n

|r_{n+1} - r_n| * sqrt(2(1+y0))/x0 = (r_n - 1) + (r_{n+1} - 1) + 2 = r_n + r_{n+1}

Since r_{n+1} < r_n:

(r_n - r_{n+1}) * sqrt(2(1+y0))/x0 = r_n + r_{n+1}

Let s = sqrt(2(1+y0))/x0

(r_n - r_{n+1}) * s = r_n + r_{n+1}

sr_n - sr_{n+1} = r_n + r_{n+1}

r_n(s - 1) = r_{n+1}(s + 1)

r_{n+1}/r_n = (s - 1)/(s + 1)

So the radii form a geometric sequence with ratio q = (s-1)/(s+1)

y0 = symbols('y0', real=True) x0 = sqrt(1 - y0**2) s = sqrt(2*(1+y0))/x0 q = (s - 1)/(s + 1) q_simplified = simplify(q) print("Ratio q =", q_simplified)

Sum of areas = pi * (r1^2 + r2^2 + ...) = pi * r1^2 * (1 + q^2 + q^4 + ...) = pi * r1^2 / (1 - q^2)

Given sum = pi/8

r1^2 / (1 - q^2) = 1/8

Let's compute r1 in terms of y0

From earlier: r1 = (t - 2*sqrt(t+1) + 2)/t where t = (1+y0)/(1-y0)

t = (1 + y0)/(1 - y0) r1 = (t - 2*sqrt(t + 1) + 2)/t r1_simplified = simplify(r1) print("r1 =", r1_simplified)

Compute 1 - q^2

one_minus_q2 = simplify(1 - q**2) print("1 - q^2 =", one_minus_q2)

Sum of r_n^2 = r1^2 / (1 - q^2)

sum_r2 = simplify(r12 / (1 - q2)) print("Sum of r_n^2 =", sum_r2) [/PYTHON]

[THOUGHT] The expressions are getting complex. Let me run more code to simplify and solve for y0 when the sum equals 1/8. [/THOUGHT]

[PYTHON] import sympy as sp from sympy import symbols, sqrt, solve, simplify, pi, Rational, cos, sin, expand, nsimplify

y0 = symbols('y0', real=True, positive=True) x0 = sqrt(1 - y0**2)

Let me use a cleaner approach

s = sqrt(2(1+y0))/x0 = sqrt(2(1+y0))/sqrt(1-y0^2) = sqrt(2(1+y0))/sqrt((1-y0)(1+y0)) = sqrt(2/(1-y0))

s = sqrt(2/(1-y0)) q = (s - 1)/(s + 1)

For r1, from the tangency condition with O_0:

k^2 * (1+y0)/(1-y0) = 4(k+1)

Let u = 1+y0, v = 1-y0, so u + v = 2, u - v = 2y0

k^2 * u/v = 4(k+1)

k^2 * u = 4v(k+1)

k^2 * u - 4vk - 4v = 0

k = (4v ± sqrt(16v^2 + 16uv))/(2u) = (4v ± 4sqrt(v(u+v)))/(2u) = 2v(1 ± sqrt((u+v)/v))/(u)

Since u + v = 2: k = 2v(1 ± sqrt(2/v))/u

Taking minus for smaller k (smaller circle):

k1 = 2v(1 - sqrt(2/v))/u = 2(1-y0)(1 - sqrt(2/(1-y0)))/(1+y0)

v = 1 - y0 u = 1 + y0 k1 = 2v(1 - sqrt(2/v))/u r1 = k1 + 1 r1_simp = simplify(r1) print("r1 simplified:", r1_simp)

q^2

q2 = simplify(q**2) print("q^2:", q2)

1 - q^2

one_minus_q2 = simplify(1 - q2) print("1 - q^2:", one_minus_q2)

Sum = r1^2 / (1 - q^2) = 1/8

r1^2 = (1 - q^2)/8

sum_areas_coeff = simplify(r1**2 / (1 - q2)) print("Sum coefficient (should equal 1/8):", sum_areas_coeff)

Set equal to 1/8 and solve

equation = sp.Eq(sum_areas_coeff, Rational(1, 8)) solutions = solve(equation, y0) print("Solutions for y0:", solutions) [/PYTHON]

[THOUGHT] The solution gives y0 = 1/2. Now I need to compute 2024 · x0² · y0.

Since P(x0, y0) is on the unit circle with x0 > 0 and y0 = 1/2:

• x0² + y0² = 1
• x0² + 1/4 = 1
• x0² = 3/4

So: 2024 · x0² · y0 = 2024 · (3/4) · (1/2) = 2024 · 3/8 = 6072/8 = 759 [/THOUGHT]

\boxed{759}