PUMaC 2020 · 团队赛 · 第 8 题
PUMaC 2020 — Team Round — Problem 8
题目详情
- Let there be a tiger, William, at the origin. William leaps 1 unit in a random direction, then leaps 2 units in a random direction, and so forth until he leaps 15 units in a random direction to celebrate PUMaC’s 15th year. There exists a circle centered at the origin such that the probability that William is contained 1 in the circle (assume William is a point) is exactly after the 15 leaps. The area of that circle 2 can be written as Aπ . What is A ?
解析
- Let there be a tiger, William, at the origin. William leaps 1 unit in a random direction, then leaps 2 units in a random direction, and so forth until he leaps 15 units in a random direction to celebrate PUMaC’s 15th year. There exists a circle centered at the origin such that the probability that William is contained 1 in the circle (assume William is a point) is exactly after the 15 leaps. The area of that circle 2 can be written as Aπ . What is A ? Proposed by: Aditya Gollapudi Answer: 1240 Let D = { θ , θ , ..., θ } represent the random directions that William has selected. Then the 1 2 15 ∑ ∑ 15 15 point that William is at can be represented by ( i cos( θ ) , i sin( θ )). Thus the area i i i =1 i =1 ∑ ∑ 15 15 2 2 of the smallest circle containing is π (( i cos( θ )) + ( i sin( θ )) ) and we need only i i i =1 i =1 ∑ ∑ ∑ 15 15 15 2 2 2 2 solve for ( i cos( θ )) + ( i sin( θ )) Expanding this out we get i cos( θ ) + i i i i =1 i =1 i =1 ∑ ∑ ∑ ∑ ∑ 15 15 15 15 15 2 2 i sin( θ ) + ij cos( θ + θ ) + ij sin( θ + θ ) which can be i i j i j i =1 i =1 j =1 ,j 6 = i i =1 j =1 ,j 6 = i ∑ ∑ 15 15 2 simplified to i + cos( θ + θ ) By the symmetry of the distribution and the i j i =1 j =1 ,j 6 = i 1 1 symmetry of cos, the second term is less than zero of the time and greater than zero half 2 2 1 of the time. Thus the area of the circle in which William is contained of the time is simply 2 ∑ 15 2 15 ∗ 16 ∗ 31 π i which is well known to be π = 1240 π i =1 6