HMMT 二月 2026 · 冲刺赛 · 第 27 题

HMMT February 2026 — Guts Round — Problem 27

[14] Let a , b , and c be positive real numbers such that √ √ ab + 1 + ca + 1 = 2 a, √ √ bc + 1 + ab + 1 = 3 b, √ √ ca + 1 + bc + 1 = 5 c. Compute a . √ ◦

解析

[14] Let a , b , and c be positive real numbers such that √ √ ab + 1 + ca + 1 = 2 a, √ √ bc + 1 + ab + 1 = 3 b, √ √ ca + 1 + bc + 1 = 5 c. Compute a . Proposed by: Pitchayut Saengrungkongka √ 14 14 11 √ Answer: = 33 3 11 Solution: Subtracting the second equation from the first equation gives √ √ ca + 1 − bc + 1 = 2 a − 3 b. (2) Multiplying (2) by the third equation yields √ √ √ √ ca − bc = ( ca + 1 + bc + 1)( ca + 1 − bc + 1) = 5 c · (2 a − 3 b ) . Since c > 0 , the above implies a − b = 10 a − 15 b , or, equivalently, a/b = 14 / 9 . Similarly, we get b/c = 9 / 5 . Therefore, we can write a = 14 t , b = 9 t , and c = 5 t for a positive real number t . Plugging ( a, b, c ) = (14 t, 9 t, 5 t ) into the first equation gives √ √ 2 2 126 t + 1 + 70 t + 1 = 28 t. (3) Plugging ( a, b, c ) = (14 t, 9 t, 5 t ) into (2) yields √ √ 2 2 126 t + 1 − 70 t + 1 = 2 t. (4) √ 2 Adding (3) and (4) together gives 126 t + 1 = 15 t . Squaring both sides simplifies the equation to √ 14 2 √ 99 t = 1 , whence t = 1 / 99 as t > 0 . Finally, a = 14 t = . 3 11 √ ◦