HMMT 二月 2020 · 团队赛 · 第 4 题
HMMT February 2020 — Team Round — Problem 4
题目详情
- [35] Alan draws a convex 2020-gon A = A A · · · A with vertices in clockwise order and chooses 1 2 2020 2020 angles θ , θ , . . . , θ ∈ (0 , π ) in radians with sum 1010 π . He then constructs isosceles triangles 1 2 2020 4 A B A on the exterior of A with B A = B A and ∠ A B A = θ . (Here, A = A .) i i i +1 i i i i +1 i i i +1 i 2021 1 Finally, he erases A and the point B . He then tells Jason the angles θ , θ , . . . , θ he chose. Show 1 1 2 2020 that Jason can determine where B was from the remaining 2019 points, i.e. show that B is uniquely 1 1 determined by the information Jason has.
解析
- [35] Alan draws a convex 2020-gon A = A A · · · A with vertices in clockwise order and chooses 1 2 2020 2020 angles θ , θ , . . . , θ ∈ (0 , π ) in radians with sum 1010 π . He then constructs isosceles triangles 1 2 2020 4 A B A on the exterior of A with B A = B A and ∠ A B A = θ . (Here, A = A .) i i i +1 i i i i +1 i i i +1 i 2021 1 Finally, he erases A and the point B . He then tells Jason the angles θ , θ , . . . , θ he chose. Show 1 1 2 2020 that Jason can determine where B was from the remaining 2019 points, i.e. show that B is uniquely 1 1 determined by the information Jason has. Proposed by: Andrew Gu, Colin Tang Solution 1: For each i , let τ be the transformation of the plane which is rotation by θ counterclockwise i i about B . Recall that a composition of rotations is a rotation or translation, and that the angles of i rotation add. Consider the composition τ ◦ τ ◦· · ·◦ τ , with total rotation angle 1010 π . This must 2020 2019 1 be a translation because 1010 π = 505(2 π ). Also note that the composition sends A to itself because 1 − 1 − 1 − 1 τ ( A ) = A . Therefore it is the identity. Now Jason can identify the map τ as τ ◦ τ ◦ · · · ◦ τ , i i i +1 1 2 3 2020 and B is the unique fixed point of this map. 1 Solution 2: Fix an arbitrary coordinate system. For 1 ≤ k ≤ 2020, let a , b be the complex numbers k k corresponding to A , B . The given condition translates to k k iθ k e ( b − a ) = ( b − a ) . (1 ≤ k ≤ 2020) k k k k +1 In other words iθ iθ k k ( e − 1) b = e a − a , k k k +1 or − i ( θ + ··· + θ ) − i ( θ + ··· + θ ) − i ( θ + ··· + θ ) − i ( θ + ··· + θ ) k − 1 1 k 1 k − 1 1 k 1 ( e − e ) b = e a − e a . k k k +1 Summing over all k , and using the fact that − i ( θ + ··· + θ ) 1 2020 e = 1 , we see that the right hand side cancels to 0, thus 2020 ∑ − i ( θ + ··· + θ ) − i ( θ + ··· + θ ) k − 1 1 k 1 ( e − e ) b = 0 . k k =1 Jason knows b , . . . , b and all the θ , so the equation above is a linear equation in b . We finish by 2 2020 i 1 − iθ 1 noting that the coefficient of b is 1 − e which is non-zero, as θ ∈ (0 , π ). Thus Jason can solve for 1 1 b uniquely. 1 ˜ ˜ ˜ Solution 3: Let A A · · · A and A A · · · A be two 2020-gons that satisfy the conditions in 1 2 2020 1 2 2020 ˜ the problem statement, and let B , B be the points Alan would construct with respect to these two k k ˜ ˜ polygons. It suffices to show that if B = B for k = 2, 3, . . . , 2020, then B = B . k k 1 1 For 2 ≤ k ≤ 2020, we note that ˜ ˜ A B = A B , A B = A B k k k +1 k k k k +1 k ˜ ˜ Furthermore, we have the equality of directed angles ∠ A B A = ∠ A B A = θ , therefore k k k +1 k k k +1 k ˜ ˜ ˜ ∼ ˜ ∠ A B A = ∠ A B A . This implies the congruence 4 A B A 4 A B A . = k k k k +1 k k +1 k k k k +1 k k +1 ˜ ˜ The congruence shows that A A = A A ; furthermore, the angle from the directed segment k k k +1 k +1 − − − → − − − − − − − → ˜ ˜ A A to A A is θ counterclockwise. This holds for k = 2, 3, . . . , 2020; we conclude that k k k +1 k +1 k − − − → − − − → ˜ ˜ ˜ ˜ A A = A A , and the angle from the directed segments A A to A A is 1 1 2 2 1 1 2 2 2020 ∑ − θ = θ − 1010 π = θ k 1 1 k =2 counterclockwise. − − − → − − − → Finally we observe that A B = A B , and the angle from the directed segment A B to A B is 1 1 2 1 1 1 2 1 ˜ ˜ ˜ ∼ ˜ ˜ θ counterclockwise. This implies ∠ B A A = ∠ B A A , so 4 A B A 4 A B A . Thus A B = = 1 1 1 1 1 2 2 1 1 1 2 1 2 1 1 − − − → − − − → ˜ ˜ ˜ ˜ A B , and the angle from A B to A B is θ counterclockwise. We conclude that B = B . 2 1 1 1 2 1 1 1 1