HMMT 二月 2024 · 冲刺赛 · 第 32 题
HMMT February 2024 — Guts Round — Problem 32
题目详情
- [16] Over all pairs of complex numbers ( x, y ) satisfying the equations 2 4 2 4 x + 2 y = x and y + 2 x = y , compute the minimum possible real part of x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2024, February 17, 2024 — GUTS ROUND Organization Team Team ID#
解析
- [16] Over all pairs of complex numbers ( x, y ) satisfying the equations 2 4 2 4 x + 2 y = x and y + 2 x = y , compute the minimum possible real part of x . Proposed by: Jaedon Whyte √ √ 3 1 − 33 Answer: 2 Solution 1: Note the following observations: 2 3 (a) if ( x, y ) is a solution then ( ωx, ω y ) is also a solution if ω = 1 and ω ̸ = 1 . 4 2 (b) we have some solutions ( x, x ) where x is a solution of x − 2 x − x = 0 . These are really the only necessary observations and the first does not need to be noticed immediately. 1 2 4 Indeed, we can try to solve this directly as follows: first, from the first equation, we get y = ( x − x ) , 2 so inserting this into the second equation gives 1 4 2 2 ( x − x ) − 2 x = y 4 ( ) 2 ( ) 2 4 2 4 x − x − 8 x − 8 x + 8 x = 0 ( ) 2 8 5 2 4 x − 2 x − 7 x − 8 x + 8 x = 0 16 4 x + · · · + 41 x + 8 x = 0 ︸ ︷︷ ︸ P ( x ) 3 By the second observation, we have that x ( x − 2 x − 1) should be a factor of P . The first observation 3 3 2 3 3 gives that ( x − 2 ωx − 1)( x − 2 ω x − 1) should therefore also be factor. Now ( x − 2 ωx − 1)( x − 2 6 4 3 2 2 2 2 ω x − 1) = x + 2 x − 2 x + 4 x − 2 x + 1 since ω and ω are roots of x + x + 1 . So now we see 4 that the last two terms of the product of all of these is − 5 x − x . Hence the last two terms of the 3 polynomial we get after dividing out should be − x − 8 , and given what we know about the degree 6 3 and the fact that everything is monic, the quotient must be exactly x − x − 8 which has roots being √ √ 3 1 ± 33 2 3 the cube roots of the roots to x − x − 8 , which are . Now x − 2 x − 1 is further factorable as 2 √ 1 ± 5 2 ( x − 1)( x − x − 1) with roots 1 , so it is not difficult to compare the real parts of all roots of P , 2 1 especially since 5 are real and non-zero, and we have that Re( ωx ) = − x if x ∈ R . We conclude that 2 √ √ 3 1 − 33 the smallest is . 2 Solution 2: Subtracting the second equation from the first, we get: 2 2 4 4 ( y + 2 x ) − ( x + 2 y ) = y − x = ⇒ 2 2 2 2 2 2 ( x − y ) + 2( x − y ) = ( x − y )( x + y ) = ⇒ 2 2 ( x − y )(1 − ( x + y )( x + y + 2)) = 0 Subtracting y times the first equation from x times the second, we get: 3 3 4 4 ( xy + 2 y ) − ( xy + 2 x ) = x y − xy = ⇒ 3 3 3 3 2( y − x ) = xy ( x − y ) = ⇒ 3 3 ( x − y )(2 + xy ) = 0 2 2 Subtracting y times the second equation from x times the first, we get: 3 2 2 3 2 2 6 6 ( x + 2 x y ) − ( y + 2 x y ) = x − y = ⇒ 3 3 3 3 3 3 x − y = ( x + y )( x − y ) = ⇒ 3 3 3 3 ( x − y )(1 − x − y ) = 0 We have three cases. Case 0. x = 0 Thus, ( x, y ) = (0 , 0) is the only valid solution. 2 4 2 2 Case 1. x = ωy for some third root of unity ω . Thus, y = ω x = ωx 2 4 x + 2 y = x = ⇒ 2 4 x + 2 ωx = x = ⇒ 2 x (1 + ω )(2 − ωx ) = 1 Note that x = − ω is always a solution to the above, and so we can factor as: 3 x + 2(1 + ω ) x − 1 = 0 2 2 ( x + ω )( x − ωx − ω ) = 0 and so the other solutions are of the form: √ 1 ± 5 x = · ω 2 √ 1+ 5 for the third root of unity. The minimum real part in this case is − when ω = 1 . 2 3 3 3 3 Case 2. Since x − y ̸ = 0 , we have xy = − 2 and x + y = 1 . ( ) √ √ √ 3 1 ± 33 3 3 3 3 2 2 Thus, x − y = ( x + y ) − 4( xy ) = ± 33 = ⇒ x = 2 ( ) √ 1 / 3 1 − 33 This yields the minimum solution of x = as desired. This is satisfied by letting 2 ( ) √ 1 / 3 1+ 33 y = . 2