HMMT 十一月 2022 · THM 赛 · 第 8 题

HMMT November 2022 — THM Round — Problem 8

Alice thinks of four positive integers a ≤ b ≤ c ≤ d satisfying { ab + cd, ac + bd, ad + bc } = { 40 , 70 , 100 } . What are all the possible tuples ( a, b, c, d ) that Alice could be thinking of?

解析

Alice thinks of four positive integers a ≤ b ≤ c ≤ d satisfying { ab + cd, ac + bd, ad + bc } = { 40 , 70 , 100 } . What are all the possible tuples ( a, b, c, d ) that Alice could be thinking of? Proposed by: Vidur Jasuja Answer: (1 , 4 , 6 , 16) Solution: Since ab · cd = ac · bd = ad · bc , the largest sum among ab + cd, ac + bd, ad + bc will be the one with the largest difference between the two quantities, so ab + cd = 100 , ac + bd = 70 , ad + bc = 40 . Consider the sum of each pair of equations, which gives ( a + b )( c + d ) = 110 , ( a + c )( b + d ) = 140 , ( a + d )( b + c ) = 170 . Since each of these are pairs summing to S = a + b + c + d , by looking at the discriminant of the quadratics with roots a + b and c + d , a + c and b + d , and a + d and b + c , we have 2 2 2 that S − 680 , S − 560 , S − 440 must be perfect squares. Therefore, we need to find all arithmetic progressions of three squares with common difference 120 . 2 2 The equation x − y = 120 has (31 , 29) , (17 , 13) , (13 , 7) , (11 , 1) as a solution, and so the only possibility 2 is 49 , 169 , 289 . This implies that S = 729 = ⇒ S = 27 . We now need to verify this works. Note that this implies { a + b, c + d } = { 5 , 22 } , { a + c, b + d } = { 7 , 20 } , { a + d, b + c } = { 10 , 17 } . Therefore, a + b = 5, a + c = 7. This means that b + c is even, so b + c = 10 . This gives us ( a, b, c, d ) = (1 , 4 , 6 , 16) is the only possibility, as desired.