HMMT 二月 2025 · 冲刺赛 · 第 34 题
HMMT February 2025 — Guts Round — Problem 34
题目详情
- [20] On the perimeter of a unit circle, 12 points are chosen uniformly and independently at random. Estimate the expected value of the area of the convex 12-gon formed by these points. Submit a positive number E written in decimal. If the correct answer is A , you will receive − 15 | E − A | round 20 e points.
解析
- [20] On the perimeter of a unit circle, 12 points are chosen uniformly and independently at random. Estimate the expected value of the area of the convex 12-gon formed by these points. Submit a positive number E written in decimal. If the correct answer is A , you will receive − 15 | E − A | round 20 e points. Proposed by: Isabella Zhu 467775 155925 10395 1485 33 Answer: − + − + ≈ 2 . 55915198 9 7 5 3 2 π 2 π π 2 π π Solution: We compute the exact answer as given above. Let n = 12. and θ , θ , . . . θ be uniformly 1 2 n randomly generated such that θ + · · · + θ = 2 π. We are trying to estimate 1 n " # n n X X 1 1 E sin( θ ) = E [sin( θ )] . i i 2 2 i =1 i =1 We do this by computing the marginal distribution of θ , which is proportional to the area of the n − 2 i dimensional cross section X θ = 2 π − θ . j i j ̸ = i n − 2 The volume of this cross section is proportional to (2 π − θ ) . Thus, the marginal probability i distribution of θ can be written as i n − 2 p ( θ ) = C (2 π − θ ) i i R 2 π for some constant C . We can solve for C because we know p ( θ ) dθ = 1. The result is that i i 0 n − 2 n − 1 θ i p ( θ ) = 1 − . i 2 π 2 π Thus, the quantity that we seek to estimate is " # n X 1 n E sin θ = E [sin θ ] i i 2 2 i =1 Z 2 π = n p ( θ ) sin θ dθ i i i 0 Z n − 2 2 π n ( n − 1) θ = 1 − sin θ dθ 4 π 2 π 0 Z 10 2 π 33 θ = 1 − sin θ dθ. π 2 π 0 We can integrate this using tabular integration by parts. Differentiate Integrate Sign 10 θ 1 − sin θ 2 π 9 5 θ − 1 − cos θ + π 2 π 8 45 θ 1 − − sin θ − 2 2 π 2 π 7 90 θ − 1 − − cos θ + 3 π 2 π 6 315 θ 1 − sin θ − 4 π 2 π 5 945 θ − 1 − cos θ + 5 π 2 π 4 4725 θ 1 − − sin θ − 6 2 π 2 π 3 4725 θ − 1 − − cos θ + 7 π 2 π 2 14175 θ 1 − sin θ − 8 2 π 2 π 1 14175 θ − 1 − cos θ + 9 2 π 2 π 14175 − sin θ − 10 4 π cos θ + n θ To compute the product, note that any term of the form C 1 − sin θ vanishes when taking definite 2 π integral from 0 to 2 π . Therefore, the desired value is 10 8 6 4 33 θ 45 θ 315 θ 4725 θ 1 − − 1 − + 1 − − 1 − 2 4 6 π 2 π 2 π 2 π π 2 π 2 π 2 π 2 π 2 14175 θ 14175
- 1 − + cos θ 8 10 2 π 2 π 4 π 0 33 45 315 4725 14175 = 1 − + − + 2 4 6 8 π 2 π π 2 π 2 π 33 1485 10395 155925 467775 = − + − + 3 5 7 9 π 2 π π π 2 π