HMMT 十一月 2023 · 冲刺赛 · 第 18 题
HMMT November 2023 — Guts Round — Problem 18
题目详情
- [10] Over all real numbers x and y such that 3 3 x = 3 x + y and y = 3 y + x, 2 2 compute the sum of all possible values of x + y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2023, November 11, 2023 — GUTS ROUND Organization Team Team ID#
解析
- [10] Over all real numbers x and y such that 3 3 x = 3 x + y and y = 3 y + x, 2 2 compute the sum of all possible values of x + y . Proposed by: Rishabh Das Answer: 15 Solution 1: First, we eliminate easy cases. √ √ √ √ √ √ 3 • if x = − y , then x = 3 x − x = 2 x , so x ∈ { 0 , 2 , − 2 } . Therefore, we get ( 2 , − 2) , ( − 2 , 2), and (0 , 0). 3 • if x = y ̸ = 0, then x = 3 x + x = 4 x , so x ∈ { 2 , − 2 } . Therefore, we get (2 , 2) and ( − 2 , − 2). Otherwise, adding two equations gives 3 3 x + y = 4 x + 4 y 2 2 ( x + y )( x − xy + y ) = 4( x + y ) 2 2 x − xy + y = 4 , and subtracting the two equations gives 3 3 x − y = 2 x − 2 y 2 2 ( x − y )( x + xy + y ) = 2( x − y ) 2 2 x + xy + y = 2 . 2 2 2 2 2 2 We have x − xy + y = 4 and x + xy + y = 2, so adding these gives x + y = 3. One can also see that xy = − 1, so the solution obtained will be real. The final answer is 4 + 8 + 0 + 3 = 15. 1 1 1 3 3 Solution 2: Let x = a + and y = b + for nonzero a and b . Then x − 3 x = a + and 3 a b a 1 3 3 y − 3 y = b + , so 3 b 1 1 1 1 3 3 a + = b + and b + = a + . 3 3 a b b a 3 3 3 3 9 9 1 8 10 These imply b ∈ { a , 1 /a } and a ∈ { b , 1 /b } . These mean a = a or a = , so a = 1 or a = 1. a 3 We can WLOG b = a for simplicity. √ √ 8 First suppose a = 1. Now a = 1 gives (2 , 2), a = − 1 gives ( − 2 , − 2), a = cis( ± π/ 4) gives ( 2 , − 2), √ √ and a = cis( ± 3 π/ 4) gives ( − 2 , 2). Finally a = ± i gives (0 , 0). This gives 8 + 4 + 0 = 12 to our sum. 10 Now suppose a = 1; we can assume a ̸ = 1 , − 1 since these were covered by the previous case. It can be seen that for all other values of a , the pairs ( a, b ) that work can be get by either swapping a 2 2 and b or negating one of the variables, all of which give the same value of x + y , so we only need √ 1+ 5 to work on one of them. Suppose a = cis( π/ 5) and b = cis(3 π/ 5), so x = 2 cos( π/ 5) = and 2 √ 1 − 5 2 2 y = 2 cos(3 π/ 5) = . Then x + y = 3. 2 The final answer is 12 + 3 = 15.