HMMT 二月 2020 · 团队赛 · 第 7 题

HMMT February 2020 — Team Round — Problem 7

[50] Positive real numbers x and y satisfy ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ · · · | x | − y − x · · · − y − x = · · · | y | − x − y · · · − x − y ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ where there are 2019 absolute value signs | · | on each side. Determine, with proof, all possible values x of . y

解析

[50] Positive real numbers x and y satisfy ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ · · · | x | − y − x · · · − y − x = · · · | y | − x − y · · · − x − y ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ where there are 2019 absolute value signs | · | on each side. Determine, with proof, all possible values x of . y Proposed by: Krit Boonsiriseth 1 Answer: , 1 , 3 3 ∣ ∣ ∣ ∣ Solution: Clearly x = y works. Else WLOG x < y , define d = y − x , and define f ( z ) := | z − y | − x so our expression reduces to ∣ ∣ 1009 1009 ∣ ∣ f ( x ) = f (0) − y . Now note that for z ∈ [0 , y ], f ( z ) can be written as { d − z, 0 ≤ z ≤ d f ( z ) = z − d, d < z ≤ y Hence f ( f ( z )) = f ( d − z ) = z for all z ∈ [0 , d ]. Therefore ∣ ∣ 1009 ∣ ∣ f (0) − y = | f (0) − y | = x. 1009 1009 If x > d then f ( x ) < x which is impossible (if f ( x ) ≤ d then the conclusion trivially holds, and 1009 1009 1009 if f ( x ) > d we must have f ( x ) = x − 1009 d < x ). Therefore x ≤ d , so f ( x ) = f ( x ) = d − x and we must have d − x = x . Hence y = 3 x which is easily seen to work. To summarize, the possible x 1 values of are , 1 , 3. y 3