Problem:
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 , where , , and are positive integers, and are relatively prime, and neither nor is divisible by the square of any prime. Find the remainder when the product is divided by .
Solution:
Without loss of generality, let the radius of the circle be . The radii to the endpoints of the chord, along with the chord, form an isosceles triangle with vertex angle . The area of the larger of the two regions is thus that of the circle plus the area of the isosceles triangle, and the area of the smaller of the two regions is thus that of the circle minus the area of the isosceles triangle. The requested ratio is therefore , so abcdef , and the requested remainder is .
The problems on this page are the property of the MAA's American Mathematics Competitions