Problem:
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region C to the area of shaded region B is 11/5. Find the ratio of the area of shaded region D to the area of shaded region A.
Solution:
Without loss of generality, choose a unit of length equal to the distance between two adjacent parallel lines. Let x be the distance from the vertex of the angle to the closest of the parallel lines that intersect the angle. Denote the areas of the regions bounded by the parallel lines and the angle by K0​,K1​,K2​,…, as shown. Then for m>0 and n>0,
K0​Km​​Kn​Km​​​=K0​K0​+K1​+K2​+⋯+Km​​−K0​K0​+K1​+K2​+⋯+Km−1​​=(xx+m​)2−(xx+m−1​)2=x22x+2m−1​, so =Kn​/K0​Km​/K0​​=2x+2n−12x+2m−1​.​
Thus the ratio of the area of region C to the area of region B is K2​K4​​=2x+32x+7​, and so 2x+32x+7​=511​. Solve this equation to obtain x=1/6. Also, the ratio of the area of region D to the area of region A is K0​K6​​=x22x+11​. Substitute 1/6 for x to find that the requested ratio is 408​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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