HMMT 二月 2022 · 冲刺赛 · 第 31 题

HMMT February 2022 — Guts Round — Problem 31

[16] For a point P = ( x, y ) in the Cartesian plane, let f ( P ) = ( x − y , 2 xy − y ). If S is the set of all P so that the sequence P, f ( P ) , f ( f ( P )) , f ( f ( f ( P ))) , . . . approaches (0 , 0), then the area of S can be √ expressed as π r for some positive real number r . Compute ⌊ 100 r ⌋ .

解析

[16] For a point P = ( x, y ) in the Cartesian plane, let f ( P ) = ( x − y , 2 xy − y ). If S is the set of all P so that the sequence P, f ( P ) , f ( f ( P )) , f ( f ( f ( P ))) , . . . approaches (0 , 0), then the area of S can be √ expressed as π r for some positive real number r . Compute ⌊ 100 r ⌋ . Proposed by: Daniel Zhu Answer: 133 Solution: For a point P = ( x, y ), let z ( P ) = x + yω , where ω is a nontrivial third root of unity. Then 2 2 2 2 2 z ( f ( P )) = ( x − y ) + (2 xy − y ) ω = x + 2 xyω + y ( − 1 − ω ) 2 2 2 2 2 = x + 2 xyω + y ω = ( x + yω ) = z ( P ) . n n n − 1 2 n − 2 4 2 Applying this recursively gives us z ( f ( P )) = z ( f ( P )) = z ( f ( P )) = · · · = z ( P ) . Thus the n condition f ( P ) → (0 , 0) is equivalent to | z ( P ) | < 1. The region of such points is the preimage of the √ 1 3 unit disk (area π ) upon the “shear” sending (0 , 1) to ( − , ). This shear multiplies areas by a factor 2 2 q √ 3 2 π 4 √ of , so the original area was = π . 2 3 3