HMMT 二月 2024 · 冲刺赛 · 第 5 题

HMMT February 2024 — Guts Round — Problem 5

[6] Let a , b , and c be real numbers such that a + b + c = 100 , ab + bc + ca = 20 , and ( a + b )( a + c ) = 24 . Compute all possible values of bc .

解析

[6] Let a , b , and c be real numbers such that a + b + c = 100 , ab + bc + ca = 20 , and ( a + b )( a + c ) = 24 . Compute all possible values of bc . Proposed by: Pitchayut Saengrungkongka Answer: 224 , − 176 2 Solution: We first expand the left-hand-side of the third equation to get ( a + b )( a + c ) = a + ac + 2 ab + bc = 24 . From this, we subtract the second equation to obtain a = 4 , so a = ± 2 . If a = 2 , plugging into the first equation gives us b + c = 98 and plugging into the second equation gives us 2( b + c ) + bc = 20 ⇒ 2(98) + bc = 20 ⇒ bc = − 176 . Then, if a = − 2 , plugging into the first equation gives us b + c = 102 , and plugging into the second equation gives us − 2( b + c ) + bc = 20 ⇒ − 2(102) + bc = 20 ⇒ bc = 224 . Therefore, the possible values of bc are 224 , − 176 .