Let be the unit circle, and let be a point on with positive -coordinate.Let be the intersection of the line tangent to at and the line .Let . Among all circles tangent simultaneously to segment , the line, and the circle , choose the one lying inside triangle and havingradius smaller than that of . Call this circle .Similarly, for each circle , define to be the circle tangent tosegment , the line , and the circle , lying inside triangle ,and having radius smaller than that of .If the sum of the areas of the circlesiscompute the value ofoindent
[THOUGHT] The circles (O_n) (including (O_0)) are all tangent to the two intersecting lines (y=-1) and (PQ). In an angle of half-angle (\alpha), consecutive circles tangent to both sides and to each other have radii ratio [ \frac{r_{n+1}}{r_n}=t=\frac{1-\sin\alpha}{1+\sin\alpha}. ] Given (\sum_{n\ge1}\pi r_n^2=\frac{\pi}{8}) and (r_0=1), we have (r_n=t^n), so [ \sum_{n\ge1}\pi t^{2n}=\pi\frac{t^2}{1-t^2}=\frac{\pi}{8}\Rightarrow t^2=\frac19\Rightarrow t=\frac13. ] Thus (\frac{1-\sin\alpha}{1+\sin\alpha}=\frac13\Rightarrow \sin\alpha=\frac12\Rightarrow \alpha=\frac{\pi}{6}), so the angle between the lines is (2\alpha=\frac{\pi}{3}) (60°). Therefore the tangent line (PQ) makes a 60° angle with the horizontal (y=-1), so its slope has magnitude (\tan60=\sqrt3). The tangent line at (P(x_0,y_0)) has slope (-x_0/y_0), hence (|x_0/y_0|=\sqrt3). The configuration inside triangle (P'PQ) forces the interior angle at (Q) to be 60°, giving (y_0>0), so (x_0/y_0=\sqrt3\Rightarrow y_0=x_0/\sqrt3). With (x_0^2+y_0^2=1), we get (x_0^2+x_0^2/3=1\Rightarrow x_0^2=3/4) and (y_0=1/2). Compute (2024\cdot x_0^2\cdot y_0). [/THOUGHT] [PYTHON] from fractions import Fraction x0_sq = Fraction(3,4) y0 = Fraction(1,2) value = 2024 * x0_sq * y0 print(value) print(int(value)) [/PYTHON]
[THOUGHT] \boxed{759}