HMMT 十一月 2013 · 冲刺赛 · 第 25 题
HMMT November 2013 — Guts Round — Problem 25
题目详情
- [ 13 ] Let a, b be positive reals with a > b > a . Place two squares of side lengths a, b next to each 2 other, such that the larger square has lower left corner at (0 , 0) and the smaller square has lower left corner at ( a, 0). Draw the line passing through (0 , a ) and ( a + b, 0). The region in the two squares a lying above the line has area 2013. If ( a, b ) is the unique pair maximizing a + b , compute . b 2 2 2
解析
- [ 13 ] Let a, b be positive reals with a > b > a . Place two squares of side lengths a, b next to each 2 other, such that the larger square has lower left corner at (0 , 0) and the smaller square has lower left corner at ( a, 0). Draw the line passing through (0 , a ) and ( a + b, 0). The region in the two squares a lying above the line has area 2013. If ( a, b ) is the unique pair maximizing a + b , compute . b 5 a Answer: Let t = ∈ (1 , 2); we will rewrite the sum a + b as a function of t . The area 3 b √ 2 2 a − ab +2 b 4026 2 2 condition easily translates to = 2013, or b ( t − t + 2) = 4026 ⇐⇒ b = . Thus 2 2 t − t +2 √ 4026 a + b is a function f ( t ) = (1 + t ) of t , and our answer is simply the value of t maximizing f , 2 t − t +2 2 2 (1+ t ) f or equivalently g ( t ) = = , over the interval (1 , 2). (In general, such maximizers/maximums 2 4026 t − t +2 need not exist, but we shall prove there’s a unique maximum here.) 2 (1+ t ) 16 We claim that λ = is the maximum of . Indeed, 2 7 t − t +2 2 ( λ − 1) t − ( λ + 2) t + (2 λ − 1) λ − g ( t ) = 2 t − t + 2 2 2 1 9 t − 30 t + 25 1 (3 t − 5) = = ≥ 0 1 7 2 2 7 t − t + 2 7 ( t − ) + 2 4 5 for all reals t (not just t ∈ (1 , 2)), with equality at t = ∈ (1 , 2). 3 Comment. To motivate the choice of λ , suppose λ were the maximum of f , attained at t ∈ (1 , 2); 0 2 2 then h ( t ) = λ ( t − t + 2) − ( t + 1) is quadratic and nonnegative on (1 , 2), but zero at t = t . If g is a 0 nontrivial quadratic (nonzero leading coefficient), then t must be a double root, so g has determinant 0
- Of course, g could also be constant or linear over (1 , 2), but we can easily rule out both of these possibilities. Alternatively, we can simply take a derivative of f to find critical points. 2 2 2