PUMaC 2019 · 团队赛 · 第 5 题

PUMaC 2019 — Team Round — Problem 5

Let f ( x ) = x + 3 x + 1. There is a unique line of the form y = mx + b such that m > 0 and this line intersects f ( x ) at three points, A, B, C such that AB = BC = 2. Find b 100 m c .

解析

Let f ( x ) = x + 3 x + 1. There is a unique line of the form y = mx + b such that m > 0 and this line intersects f ( x ) at three points, A, B, C such that AB = BC = 2. Find b 100 m c . Proposed by: Frank Lu Answer: 41 Let g ( x ) = mx + b . We know that f ( x ) − g ( x ) = ( x − k )( x − k − d )( x − k + d ) for some real numbers 3 2 2 2 3 2 k, d . Expanding this out gives us that f ( x ) − g ( x ) = x − 3 kx +(3 k − d ) x +( − k + kd ), which 3 2 3 2 2 2 means that k = − 1. This then simplifies to x +3 x +1 − mx − b = x +3 x +(3 − d ) x − d +1. Observe that in order for this to hold, since the RHS has a root at x = − 1, the LHS needs to as well. Plugging in x = − 1 here yields that 3 + m − b = 0 = ⇒ b = m + 3, so we have 3 2 2 2 2 x +3 x − 2 − m ( x +1) = ( x +2 x − 2 − m )( x +1) = ( x +1)(( x +1) − d ). Factoring out ( x +1) 2 2 2 2 yields x + 2 x − 2 − m = ( x + 1) − d , which implies that d = ( m + 3). Now, applying simple √ √ 2 2 trigonometry, the distance AB = BC is equal to d m + 1 = ( m + 1)( m + 3) = 2. We now 3 2 solve for m . Squaring both sides and moving everything to one side yields m +3 m + m − 1 = 0. 2 Noting that m + 1 is a factor of this, we see that we this factors to ( m + 2 m − 1)( m + 1) = 0. √ We want positive m , so solving the quadratic and taking the positive solution yields 2 − 1. √ This yields the answer b 100( 2 − 1) c = 41