HMMT 二月 2015 · 几何 · 第 10 题
HMMT February 2015 — Geometry — Problem 10
题目详情
- Let G be the set of all points ( x, y ) in the Cartesian plane such that 0 ≤ y ≤ 8 and √ 2 2 ( x − 3) + 31 = ( y − 4) + 8 y (8 − y ) . There exists a unique line ℓ of negative slope tangent to G and passing through the point (0 , 4). Suppose ℓ is tangent to G at a unique point P . Find the coordinates ( α, β ) of P .
解析
- Let G be the set of all points ( x, y ) in the Cartesian plane such that 0 ≤ y ≤ 8 and √ 2 2 ( x − 3) + 31 = ( y − 4) + 8 y (8 − y ) . There exists a unique line ℓ of negative slope tangent to G and passing through the point (0 , 4). Suppose ℓ is tangent to G at a unique point P . Find the coordinates ( α, β ) of P . 12 8 Answer: ( , ) Let G be G restricted to the strip of plane 0 ≤ y ≤ 4 (we only care about this 5 5 region since ℓ has negative slope going down from (0 , 4)). By completing the square, the original √ 2 2 equation rearranges to ( x − 3) + ( y (8 − y ) − 4) = 1. One could finish the problem in a completely √ standard way via the single-variable parameterization ( x, y (8 − y )) = (3 + cos t, 4 + sin t ) on the appropriate interval of t —just take derivatives with respect to t to find slopes (the computations would probably not be too bad)—but we will present a slightly cleaner solution. √ − 1 Consider the bijective plane transformation Φ : ( x, y ) 7 → ( x, y (8 − y )), with inverse Φ : ( x, y ) 7 → √ 2 ( x, 4 − 16 − y ). In general, Φ maps curves as follows: Φ( { ( x, y ) : f ( x, y ) = c } ) = { Φ( x, y ) : f ( x, y ) = ′ ′ − 1 ′ ′ c } = { ( x , y ) : f (Φ ( x , y )) = c } . 2 2 Our line ℓ has the form y − 4 = − mx for some m > 0. We have Φ( G ) = { ( x − 3) + ( y − 4) = 1 : 0 ≤ √ 2 y ≤ 4 } and Φ( { 4 − y = mx : 0 ≤ y ≤ 4 } ) = { 16 − y = mx : 0 ≤ y ≤ 4 } . Since ℓ is unique, m must Geometry also be. But it’s easy to see that m = 1 gives a tangency point, so if our original tangency point was √ ( ) ( ) 4 12 16 12 8 ( u, v ), then our new tangency point is ( u, v (8 − v )) = (3 , 4) = , , and so ( u, v ) = , . 5 5 5 5 5 Remark. To see what G looks like, see Wolfram Alpha using the plotting/graphing commands. Geometry