HMMT 二月 2025 · 冲刺赛 · 第 30 题

HMMT February 2025 — Guts Round — Problem 30

[16] Let a , b , and c be real numbers satisfying the system of equations p p 3 2 2 a 1 + b + b 1 + a = , 4 p p 5 2 2 b 1 + c + c 1 + b = , and 12 p p 21 2 2 c 1 + a + a 1 + c = . 20 Compute a . 2

解析

[16] Let a , b , and c be real numbers satisfying the system of equations p p 3 2 2 a 1 + b + b 1 + a = , 4 p p 5 2 2 b 1 + c + c 1 + b = , and 12 p p 21 2 2 c 1 + a + a 1 + c = . 20 Compute a . Proposed by: Pitchayut Saengrungkongka 7 √ Answer: 2 30 x − x − x − x e − e e + e Solution 1: Recall that the functions sinh( x ) = and cosh( x ) = satisfy the relation 2 2 p p 2 2 sinh( x + y ) = sinh( x ) cosh( y ) + cosh( x ) sinh( y ) = sinh( x ) 1 + sinh( y ) + sinh( y ) 1 + sinh( x ) . Since sinh is surjective, we can perform the substitution a = sinh( x ), b = sinh( y ), and c = sinh( z ), which turns the equations into 1 2 − 2 sinh( x + y ) = , 2 3 2 − 2 3 sinh( y + z ) = , 2 5 2 − 2 5 sinh( z + x ) = . 2 Thus, x + y = log(2), y + z = log(3 / 2), and z + x = log(5 / 2). Solving these equations gives x = q q p 1 10 3 7 √ log( 10 / 3), so a = − = . 2 3 10 2 30 2 2 2 y − 1 x − 1 z − 1 Solution 2: We can find positive real numbers x , y , and z such that a = , b = , and c = . 2 x 2 y 2 z Then, the first equation becomes 2 2 2 2 x − 1 y + 1 y − 1 x + 1 3 · + · = , 2 x 2 y 2 y 2 x 4 which simplifies to 1 3 xy − = , xy 2 1 5 1 21 3 5 from which it follows that xy = 2. Similarly, yz − = and zx − = , so yz = and zx = . yz 6 zx 10 2 2 q q (10 / 3) − 1 5 3 10 7 √ √ Thus x = (2 · ) / ( ) = , and a = = . 2 2 3 2 30 2 10 / 3 2