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HMMT 二月 2025 · ALGNT 赛 · 第 10 题

HMMT February 2025 — ALGNT Round — Problem 10

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Let a , b , and c be pairwise distinct complex numbers such that 2 2 2 a = b + 6 , b = c + 6 , and c = a + 6 . Compute the two possible values of a + b + c .
解析
  1. Let a , b , and c be pairwise distinct complex numbers such that 2 2 2 a = b + 6 , b = c + 6 , and c = a + 6 . Compute the two possible values of a + b + c . Proposed by: Vasawat Rawangwong √ √ − 1+ 17 − 1 − 17 Answer: , 2 2 Solution 1: Notice that any of a , b , or c being 3 or − 2 implies a = b = c , which is invalid. Thus, 2 2 2 ( a − 9)( b − 9)( c − 9) = ( b − 3)( c − 3)( a − 3) = ⇒ ( a + 3)( b + 3)( c + 3) = 1 , 2 2 2 ( a − 4)( b − 4)( c − 4) = ( b + 2)( c + 2)( a + 2) = ⇒ ( a − 2)( b − 2)( c − 2) = 1 . Therefore, 2 and − 3 are roots of the polynomial ( x − a )( x − b )( x − c ) + 1, and so there exists some t such that ( x − t )( x − 2)( x + 3) = ( x − a )( x − b )( x − c ) + 1 . Comparing coefficients gives a + b + c = t − 1 and ab + bc + ca = − ( t + 6). We can then solve for t by 2 2 2 noting a + b + c = ( b + 6) + ( c + 6) + ( a + 6) = a + b + c + 18, so 1 1 2 2 2 2 2 ab + bc + ca = (( a + b + c ) − ( a + b + c )) = (( a + b + c ) − ( a + b + c + 18)) . 2 2 Hence, √ 1 1 ± 17 2 2 − ( t + 6) = (( t − 1) − ( t + 17)) = ⇒ t − t − 4 = 0 = ⇒ t = . 2 2 √ − 1 ± 17 Therefore a + b + c = are the two possible values of a + b + c . 2 2 2 Solution 2: Let s = a + b + c . Subtracting two adjacent equations gives a − b = b − c , or ( a − b )( a + b ) = ( b − c ). Multiplying this and its cyclic variants gives ( a + b )( b + c )( c + a ) = 1 . Now, we recall the identity 3 3 3 3 ( a + b + c ) = a + b + c + 3( a + b )( b + c )( c + a ) 3 3 3 3 = ⇒ s = a + b + c + 3 . 3 3 3 To simplify a + b + c , we add a times the first equation, b times the second, and c times the third to obtain 3 3 3 a + b + c = a ( b + 6) + b ( c + 6) + c ( a + 6) = ( ab + bc + ca ) + 6 s 2 2 2 2 1 = ( a + b + c ) − ( a + b + c ) + 6 s 2 2 1 1 = s − ( b + 6) + ( c + 6) + ( a + 6) + 6 s 2 2 2 1 11 = s + s − 9 . 2 2 Therefore, 3 2 2 1 11 3 s = s + s − 6 = ⇒ s − s + s − 4 = 0 . 2 2 2 3 At this point, the only reasonable guess is that s = is an extra solution, and the remaining two roots 2 √ − 1 ± 17 s = are the possible answers. We now justify this guess. Assume for sake of contradiction 2 3 that s = . Then, 2 2 2 2 39 a + b + c = ( b + 6) + ( c + 6) + ( a + 6) = 2 1 9 39 69 ab + bc + ca = − = − . 2 4 2 8 Then, observe abc = ( a + b + c )( ab + bc + ca ) − ( a + b )( b + c )( c + a ) 207 223 = − − 1 = − . 16 16 On the other hand, ( a + 6)( b + 6)( c + 6) = 216 + 36( a + b + c ) + 6( ab + bc + ca ) + abc 3 69 223 = 216 + 36 · − 6 · − , 2 8 16 2 2 2 2 223 which is a rational number of denominator 16. But ( a + 6)( b + 6)( c + 6) = b c a = − has 16 2 3 1 denominator 16 = 256, a contradiction. Thus s = is impossible. (It arises from a = b = c = , 2 2 which satisfies ( a + b )( b + c )( c + a ) = 1 but not the given conditions.) 2 2 Solution 3: Subtracting any two adjacent equations gives a − b = b − c , which is equivalent to both ( a − b )( a + b ) = ( b − c ) and ( a − b )( a + b + 1) = ( a − c ). Multiplying each of these with its respective cyclic variants and canceling the ( a − b )( b − c )( c − a ) factor (which is given to be nonzero), we get ( a + b )( b + c )( c + a ) = 1 and ( a + b + 1)( b + c + 1)( c + a + 1) = − 1 . Expanding the latter equation and using the given equations gives the following result. 2 2 2 ( a + b )( b + c )( c + a ) + ( a + b + c ) + 3( ab + bc + ca ) + 2( a + b + c ) + 1 = − 1 1 + ( b + 6 + c + 6 + a + 6) + 3( ab + bc + ca ) + 2( a + b + c ) + 1 = − 1 3( a + b + c ) + 3( ab + bc + ca ) = − 21 a + b + c + ab + bc + ca = − 7 . Let s = a + b + c . We can then solve for s by considering the following: 2 2 2 2 s = ( a + b + c ) + 2( ab + bc + ca ) = ( b + 6 + c + 6 + a + 6) + 2( − 7 − a − b − c ) = − s + 4 , √ − 1 ± 17 so s = . 2 Solution 4: Let s = a + b + c and consider the polynomial 2 2 2 4 2 x + ( x − 6) + (( x − 6) − 6) − s = x − 11 x + x + 24 − s. This polynomial has roots a , b , and c . By Vieta’s, the sum of all four roots is 0, so its fourth root must be − s . Using Vieta’s again, we have ab + bc + ca − sa − sb − sc = − 11. We can now solve for s . 2 ab + bc + ca − ( a + b + c ) = − 11 2 2 2 a + b + c + ab + bc + ca = 11 1 2 2 2 2 (( a + b + c ) + ( a + b + c )) = 11 2 2 ( a + b + c ) + ( b + 6 + c + 6 + a + 6) = 22 √ − 1 ± 17 2 s + s − 4 = 0 = ⇒ s = . 2 4 2 Remark. Another way to finish using this approach is to substitute − s directly into x − 11 x + x + 2 24 − s = 0 to get ( s − 3)( s + 2)( x + x − 4) = 0, then discard the solutions s = 3 and s = − 2, which arise from the invalid values a = b = c = 3 and a = b = c = − 2. (In the invalid cases, s ̸ = a + b + c because a = b = c is only a single root to the polynomial.)