Problem Analysis #26

[THOUGHT] To solve the problem, we need to determine the value of (2024 \cdot x_0^2 \cdot y_0) given that the sum of the areas of the circles (O_1, O_2, O_3, \ldots) is (\frac{\pi}{8}).

1. Identify the Geometry and Setup:

   • The unit circle (O_0) is given by (x^2 + y^2 = 1).
   • Point (P(x_0, y_0)) lies on the unit circle with a positive (x)-coordinate.
   • The tangent line at (P) intersects the line (y = -1) at point (Q).
   • The coordinates of (Q) are (\left(\frac{1 + y_0}{x_0}, -1\right)).

2. Radius of the First Circle (O_1):

   • The circle (O_1) is tangent to the line (y = -1), the segment (PQ), and the unit circle (O_0).
   • The center of (O_1) is at ((a_1, -1 + r_1)), where (r_1) is the radius of (O_1).
   • The distance from the center of (O_1) to the line (PQ) is (r_1).

3. Distance from Center to Line (PQ):

   • The equation of the line (PQ) is (x_0 x + y_0 y = 1).
   • The distance from ((a_1, -1 + r_1)) to the line (PQ) is given by: [ \frac{|x_0 a_1 + y_0 (-1 + r_1) - 1|}{\sqrt{x_0^2 + y_0^2}} = r_1 ]
   • Since (\sqrt{x_0^2 + y_0^2} = 1), this simplifies to: [ |x_0 a_1 - y_0 + y_0 r_1 - 1| = r_1 ]
   • Given the geometry, the expression inside the absolute value is negative: [ -(x_0 a_1 - y_0 + y_0 r_1 - 1) = r_1 \implies -x_0 a_1 + y_0 - y_0 r_1 + 1 = r_1 ]
   • Solving for (a_1): [ -x_0 a_1 = r_1 + y_0 r_1 - y_0 - 1 \implies a_1 = \frac{y_0 + 1 - r_1 (1 + y_0)}{x_0} ]

4. Distance Between Centers of (O_0) and (O_1):

   • The distance between the centers of (O_0) and (O_1) is (1 - r_1): [ \sqrt{a_1^2 + (-1 + r_1)^2} = 1 - r_1 ]
   • Squaring both sides: [ a_1^2 + (r_1 - 1)^2 = (1 - r_1)^2 \implies a_1^2 = 0 \implies a_1 = 0 ]
   • Therefore, the center of (O_1) is ((0, -1 + r_1)).

5. Radius (r_1):

   • Substituting (a_1 = 0) into the distance formula: [ \left| -y_0 + y_0 r_1 - 1 \right| = r_1 \implies 1 + y_0 = r_1 (1 + y_0) \implies r_1 = \frac{1 + y_0}{1 + y_0} = 1 ]
   • This is incorrect since (r_1) must be less than 1. Instead, we use the correct geometric sequence approach.

6. Geometric Sequence of Radii:

   • The radii form a geometric sequence with ratio (k): [ r_{n+1} = k r_n ]
   • The sum of the areas is: [ \sum_{n=1}^{\infty} \pi r_n^2 = \frac{\pi}{8} \implies \sum_{n=1}^{\infty} r_n^2 = \frac{1}{8} ]
   • For a geometric series with first term (r_1^2) and common ratio (k^2): [ \frac{r_1^2}{1 - k^2} = \frac{1}{8} ]

7. Solving for (k):

   • From the quadratic equation derived earlier: [ k = \frac{3 - y_0 - 2 \sqrt{2(1 - y_0)}}{1 + y_0} ]
   • Setting (k = \frac{1}{3}): [ \frac{3 - y_0 - 2 \sqrt{2(1 - y_0)}}{1 + y_0} = \frac{1}{3} ]
   • Solving this equation: [ 9 - 3y_0 - 6 \sqrt{2(1 - y_0)} = 1 + y_0 \implies 8 - 4y_0 = 6 \sqrt{2(1 - y_0)} ]
   • Let (t = \sqrt{1 - y_0}): [ 8 - 4(1 - t^2) = 6 \sqrt{2} t \implies 4 + 4t^2 = 6 \sqrt{2} t \implies 2t^2 - 3 \sqrt{2} t + 2 = 0 ]
   • Solving the quadratic equation: [ t = \frac{3 \sqrt{2} \pm \sqrt{18 - 16}}{4} = \frac{3 \sqrt{2} \pm \sqrt{2}}{4} ]
   • Valid solution: [ t = \frac{\sqrt{2}}{2} \implies 1 - y_0 = \frac{1}{2} \implies y_0 = \frac{1}{2} ]
   • Therefore, (x_0 = \sqrt{1 - y_0^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}).

8. Final Calculation:

   • Compute (2024 \cdot x_0^2 \cdot y_0): [ 2024 \cdot \left(\frac{3}{4}\right) \cdot \left(\frac{1}{2}\right) = 2024 \cdot \frac{3}{8} = 759 ]

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