HMMT 二月 2015 · 几何 · 第 10 题

英文解析

10. 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 theappropriate interval of t —just take derivatives with respect to t to find slopes (the computationswould 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 →
    √
    ( x, 4 − 16 − y ). In general, Φ maps curves as follows: Φ( { ( x, y ) : f ( x, y ) = c } ) = { Φ( x, y ) : f ( x, y ) =2
    ′ ′ − 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 ≤
    √
    y ≤ 4 } and Φ( { 4 − y = mx : 0 ≤ y ≤ 4 } ) = { 16 − y = mx : 0 ≤ y ≤ 4 } . Since ℓ is unique, m must 2
    Geometryalso 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