HMMT 二月 2010 · 冲刺赛 · 第 18 题

HMMT February 2010 — Guts Round — Problem 18

[ 9 ] Find two lines of symmetry of the graph of the function y = x + . Express your answer as two x equations of the form y = ax + b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND

解析

[ 9 ] Find two lines of symmetry of the graph of the function y = x + . Express your answer as two x equations of the form y = ax + b . √ √ 1 Answer: y = (1 + 2) x and y = (1 − 2) x The graph of the function y = x + is a hyperbola. We x ( ) ( ) 2 2 y 1 2 can see this more clearly by writing it out in the standard form x − xy +1 = 0 or − x − y = 1. 2 2 The hyperbola has asymptotes given by x = 0 and y = x , so the lines of symmetry will be the (interior ◦ and exterior) angle bisectors of these two lines. This means that they will be y = tan(67 . 5 ) x and √ ( ) 1+cos( x ) ◦ x y = − cot(67 . 5 ) x , which, using the tangent half-angle formula tan = , gives the two 2 1 − cos( x ) √ √ lines y = (1 + 2) x and y = (1 − 2) x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND