HMMT 二月 2025 · ALGNT 赛 · 第 3 题

HMMT February 2025 — ALGNT Round — Problem 3

Given that x , y , and z are positive real numbers such that log ( yz ) 8 4 log ( zx ) 9 6 log ( xy ) 5 10 2 2 2 x = 2 · 3 , y = 2 · 3 , and z = 2 · 3 , compute the smallest possible value of xyz.

解析

Given that x , y , and z are positive real numbers such that log ( yz ) 8 4 log ( zx ) 9 6 log ( xy ) 5 10 2 2 2 x = 2 · 3 , y = 2 · 3 , and z = 2 · 3 , compute the smallest possible value of xyz. Proposed by: Derek Liu 1 Answer: 576 Solution: Let k = log 3 for brevity. Taking the base-2 log of each equation gives 2 (log x )(log y + log z ) = 8 + 4 k, 2 2 2 (log y )(log z + log x ) = 9 + 6 k, 2 2 2 (log z )(log x + log y ) = 5 + 10 k. 2 2 2 Adding the first two equations and subtracting the third yields 2 log x log y = 12, so log x log y = 6. 2 2 2 2 Similarly, we get log x log y = 6 , 2 2 log y log z = 3 + 6 k, 2 2 log z log x = 2 + 4 k. 2 2 2 Multiplying the first two equations and dividing by the third yields (log y ) = 9, so log y = ± 3. 2 2 Then, the first and last equations tell us log x = ± 2 and log z = ± (1 + 2 k ), with all signs matching. 2 2 Thus log x + log y + log z = ± (3 + 2 + (1 + 2 k )) = ± (6 + 2 k ) , 2 2 2 so ± (6+2 k ) 6 2 − 6 − 2 xyz = 2 = 2 · 3 or 2 · 3 . 1 − 6 − 2 Clearly, the smallest solution is 2 · 3 = . 576