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A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form \(\frac{a \pi+b \sqrt{c}}{d \pi-e \sqrt{f}}\), where \(a, b, c, d, e\), and \(f\) are positive integers, \(a\) and \(e\) are relatively prime, and neither \(c\) nor \(f\) is divisible by the square of any prime. Find the remainder when the product \(a \cdot b \cdot c \cdot d \cdot e \cdot f\) is divided by 1000 .
in Mathematics by Diamond (66,769 points) | 97 views

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Without loss of generality, let the radius of the circle be 2 . The radii to the endpoints of the chord, along with the chord, form an isosceles triangle with vertex angle \(120^{\circ}\). The area of the larger of the two regions is thus \(2 / 3\) that of the circle plus the area of the isosceles triangle, and the area of the smaller of the two regions is thus \(1 / 3\) that of the circle minus the area of the isosceles triangle. The requested ratio is therefore \(\frac{\frac{2}{3} \cdot 4 \pi+\sqrt{3}}{\frac{1}{3} \cdot 4 \pi-\sqrt{3}}=\frac{8 \pi+3 \sqrt{3}}{4 \pi-3 \sqrt{3}}\), so abcd \(f=8 \cdot 3 \cdot 3 \cdot 4 \cdot 3 \cdot 3=2592\), and the requested remainder is 592 .

Final Answer: The final answer is 592
by Diamond (66,769 points)

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