HMMT 二月 2010 · 冲刺赛 · 第 18 题
HMMT February 2010 — Guts Round — Problem 18
题目详情
- [ 9 ] Find two lines of symmetry of the graph of the function y = x + . Express your answer as twoxequations of the form y = ax + b .
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth
英文原题
[ 9 ] Find two lines of symmetry of the graph of the function y = x + 1
x . Express your answer as two
equations of the form y = ax + b .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13 th ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND
解析
英文解析
- [ 9 ] Find two lines of symmetry of the graph of the function y = x + . Express your answer as twoxequations of the form y = ax + b .
√ √
Answer: y = (1 + 2) x and y = (1 − 2) x The graph of the function y = x + is a hyperbola. We 1
( ) ( )x
2 2
1 ycan see this more clearly by writing it out in the standard form x − xy +1 = 0 or − x − y = 1.2
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 )
° xy = − 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 .
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth