HMMT 二月 2025 · ALGNT 赛 · 第 7 题
HMMT February 2025 — ALGNT Round — Problem 7
题目详情
- There exists a unique triple ( a, b, c ) of positive real numbers that satisfies the equations 2 2 2 2( a + 1) = 3( b + 1) = 4( c + 1) and ab + bc + ca = 1 . Compute a + b + c .
解析
- There exists a unique triple ( a, b, c ) of positive real numbers that satisfies the equations 2 2 2 2( a + 1) = 3( b + 1) = 4( c + 1) and ab + bc + ca = 1 . Compute a + b + c . Proposed by: David Wei √ 9 23 9 √ Answer: = 23 23 Solution 1: The crux of this problem is to apply the trigonometric substitutions a = cot α , b = cot β , and c = cot γ , with 0 < α, β, γ < π/ 2. Then, the given equations translate to 2 3 4 = = and cot α cot β + cot β cot γ + cot γ cot α = 1 . 2 2 2 sin α sin β sin γ From the second equation, we get 1 − cot α cot β cot γ = = − cot( α + β ) . cot α + cot β Since α , β , and γ all between 0 and π/ 2, we discover that α + β + γ = π. √ √ √ Let △ ABC be the (acute) triangle with side lengths BC = 2, CA = 3, and AB = 4. By Law of Sines, setting α = ∠ A , β = ∠ B , and γ = ∠ C will satisfy both equations. Thus, Law of Cosines gives 3 + 4 − 2 5 5 √ √ √ √ cos α = = = ⇒ a = cot α = 2 · 3 · 4 48 23 3 1 9 √ √ √ Similar calculations give b = and c = , so the answer is a + b + c = . 23 23 23 2 2 2 Solution 2: Let 2( a + 1) = 3( b + 1) = 4( c + 1) = x . Then, since ab + bc + ca = 1, we have the following system of equations: 2 2 ( a + b )( c + a ) = a + ab + bc + ca = a + 1 = x/ 2 2 2 ( b + c )( a + b ) = b + ab + bc + ca = b + 1 = x/ 3 2 2 ( c + a )( b + c ) = c + ab + bc + ca = c + 1 = x/ 4 . Taking advantage of symmetry, we discover that r r r 2 x x 3 x a + b = , b + c = , and c + a = . 3 6 8 To solve for x , notice that 2 = 2( ab + bc + ca ) 2 2 2 2 2 2 = ( a + b ) + ( b + c ) + ( c + a ) − 2( a + b + c ) 2 x x 3 x x x x = + + − 2 − 1 + − 1 + − 1 3 6 8 2 3 4 23 x = − + 6 , 24 96 so x = . Therefore, 23 ! r r r 1 2 x x 3 x a + b + c = + + 2 3 6 8 1 8 + 4 + 6 9 √ √ = = . 2 23 23