HMMT 二月 2005 · 冲刺赛 · 第 39 题

HMMT February 2005 — Guts Round — Problem 39

[15] How many regions of the plane are bounded by the graph of 6 5 4 2 3 2 2 4 4 6 x − x + 3 x y + 10 x y + 3 x y − 5 xy + y = 0? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HARVARD-MIT MATHEMATICS TOURNAMENT, FEBRUARY 19, 2005 — GUTS ROUND

解析

How many regions of the plane are bounded by the graph of 6 5 4 2 3 2 2 4 4 6 x − x + 3 x y + 10 x y + 3 x y − 5 xy + y = 0? Solution: 5 The left-hand side decomposes as 6 4 2 2 4 6 5 3 2 4 2 2 3 5 3 2 4 ( x + 3 x y + 3 x y + y ) − ( x − 10 x y + 5 xy ) = ( x + y ) − ( x − 10 x y + 5 xy ) . Now, note that 5 5 4 3 2 2 3 4 5 ( x + iy ) = x + 5 ix y − 10 x y − 10 ix y + 5 xy + iy , 2 2 3 5 so that our function is just ( x + y ) − < (( x + iy ) ). Switching to polar coordinates, 6 5 5 6 5 this is r − < ( r (cos θ + i sin θ ) ) = r − r cos 5 θ by de Moivre’s rule. The graph of our 6 5 function is then the graph of r − r cos 5 θ = 0, or, more suitably, of r = cos 5 θ . This is a five-petal rose, so the answer is 5. 15