PUMaC 2020 · 团队赛 · 第 14 题
PUMaC 2020 — Team Round — Problem 14
题目详情
- Let N be the number of convex 27-gons up to rotation there are such that each side has length 1 and each angle is a multiple of 2 π/ 81. Find the remainder when N is divided by 23.
解析
- Let N be the number of convex 27-gons up to rotation there are such that each side has length 1 and each angle is a multiple of 2 π/ 81. Find the remainder when N is divided by 23. Proposed by: Michael Gintz and Rahul Saha Answer: 12 Let us consider the roots of unity. Every such polygon can be constructed by taking some subset of the roots of unity which adds to 0, and each such subset uniquely defines a polygon. Two define the same polynomial if they are rotations of each other. We wish to show that the subsets are only those made by unioning equilateral triangles. Consider some subset that works. Letting ω be the smallest primitive root we have some polynomial in ω which has a root at ω . Then since the cyclotomic polynomial is the minimal polynomial of ω , this new polynomial must be a multiple of that. However, the cyclotomic polynomial of powers of 27 54 primes is known to be 1 + x + x , so our set of roots must contain equilateral triangles. Thus we can consider whether we have the first 27 roots. Two polygons will be equivalent iff these binary strings of length 27 are equivalent by rotation. Since this is a 27 gon there are 9 ones so by https://math.stackexchange.com/questions/721783/number-of-unique-sequences- with-circular-shifts our answer is ( ) ∑ 1 a/d + b/d φ ( d ) 27 a/d d | 9 ( ( ) ( ) ( )) 1 27 9 3 1 + 2 + 6 ≡ 12 (mod 23) . 27 9 3 1 Note: Since the desired polygons in this problem are impossible, due to the condition on the angles, we also accepted the answer of 0 . 15 Suppose that f is a function f : R → R so that for all x, y ∈ R (nonnegative reals) we ≥ 0 ≥ 0 3 1 have that f ( x )+ f ( y ) = f ( x + y + xy )+ f ( x ) f ( y ) . Given that f ( ) = and f (1) = 3 , determine 5 2 2021 b log ( − f (10 − 1)) c . 2 8 Proposed by: Frank Lu Answer: 10104 First, we simplify down our functional equation for f. Notice that, using Simon’s Favorite Factoring trick, we may write this as (1 − f ( x ))(1 − f ( y )) = 1 − f ( x + y + xy ) . We can then simplify down our function by writing g ( x ) = 1 − f ( x ) , yielding the function g ( x ) g ( y ) = g ( x + y + xy ) . Now, notice that, letting h ( x ) = g ( x − 1) , this is equivalent to writing h ( x + 1) h ( y + 1) = h ( x + y + xy + 1) . But then, notice that this is equivalent to writing h ( x ) h ( y ) = h ( xy ) for all x, y that are real and at least 1 , the domain of h. From here, notice that the values of h that 8 3 3 1 we have are h ( ) = g ( ) = 1 − f ( ) = , and similarly that h (2) = g (1) = 1 − f (1) = − 2 . 5 5 5 2 8 1 2021 Now, notice then that h (5) = h (8) /h ( ) = − 8 / = − 16 . Therefore, we see that h (10 ) , 5 2 2021 2021 10105 by the multiplicativity, equals h (10) = ( − 32) = 2 . Therefore, it follows that 2021 2021 2021 10105 f (10 − 1) = 1 − g (10 − 1) = 1 − h (10 ) = − 2 + 1 , yielding our answer of 10104 . 9