tceic.com
简单学习网 让学习变简单
相关标签
当前位置:首页 >> 学科竞赛 >>

2004亚太数学奥林匹克研习营试题


? =?b?(APMO) _??t
2004 ?ù~ ?n

v???: lúüv (8:30 – 11:30) .)Uà??l?? ???} ???: I f ?*?b???bí?b/?—
f (x3 + y 3 ) = (x + y )((f (x))2 ? f (x)f (y ) + (f (y ))2 ),

w

2 x D y uL<?b t?: úL<?b x, f (2004x) = 2004f (x) ?p: (i) ? x = y = 0, ? f (0) = 0
(ii) ? y = 0 ? f (x3 ) = x((f (x))2 , ? f (x) = √ 3 √ x(f ( 3 x))2 ?¤?): x D f (x)

x ó°í?U I?? S = {k ∈ |f (kx) = kf (x)}, ¤T ?: 1 ∈ S Dcq k ∈ S , ?
√ √ √ 3 3 3 kx(f (x))2 = kf (x3 ) = f (kx3 ) = f (( kx)3 ) = kx(f ( kx))2 ,

[??bFAí?? ? S íì2

√ √ √ 3 kf (x) = f ( 3 kx), FJ 3 k ∈ S

?, ?) ( 3 kf (x))2 = (f ( 3 kx))2 ;W,H: x D f (x) x ó°í?U, ?? -?p : J h, k ∈ S ? h + k ∈ S 5?





√ √ 3 3 f ((h + k )x) = f (( hx)3 + ( kx)3 ) √ √ √ √ √ √ 3 3 3 3 3 3 = ( hx + kx)((f ( hx)2 ? f ( hx)f ( kx) + (f ( kx))2 ) √ √ √ √ √ √ √ 3 3 3 3 3 = ( h + k ) 3 x( h2 ? hk + k 2 )(f ( 3 x))2 = (h + k )f (x)

? 1 ∈ S , 1 + 1 = 2 ∈ S , FJ?? S ¨?7F í?cb ?¤??: 1996 ∈ S ] …?)?

16

??ù: ???í??, ,Hí?? ?? ?

24 _?/¤ 24 _????}A 24 _? ? 1 í?C

Dk*¤ 24 _??| 8 _?U)L<s_?FZAí?C?.? 3 C 8 t?: ??

j: ?¤ 24 _?J =?? 0 B 23 [?/駤 24 _ =??k 3 × 8 í? 2, à-F?:
0 3 8 11 16 19 6 9 12 15 18 21 14 17 20 23 2 5 22 1 4 7 10 13

,H? 2x s_4”: (i) ?ó°í 2, ó?s_j? ( ?_j?D|(?_j ??e?ó?s_j?) í?? 3; (ii) ?ó°íW2, ó?s_j? (|??íj? D| ?íj??e?ó?s_j?) í?? 8 ?¤, ????e??,Hí? 2 ?? 8 _?.ó?íj? I xn [?* 3 × n í? 2 ^??í_b, ¤T n ≥ 2, ? xn + xn?1 = 3 · 2n?1 , n ≥ 3 / x2 = 6 (??? ?W2
2 3

??, ?(íW2

?? ??,

??ê?? ?Wíj?D|(?Wíj?uó?í ?¤8$

ê?v, Bb.???|(?W, ?5?* 3 × (n ? 1) í? 2d ^í??) ?¤
x8 = (x8 + x7 ) ? (x7 + x6 ) + (x6 + x5 ) ? (x5 + x4 ) + (x4 + x3 ) ? (x3 + x2 ) + x2 = 3(27 ? 26 + 25 ? 24 + 23 ? 22 + 2) = 258

]…?

258

??

17

??ú: I a, b, c, b + c ? a, c + a ? b, a + b ? c D a + b + c ? 7 _ó?í”b/?—
a + b = 800 J M D m }?[?¤ 7 _bí|×MD|üM, ? M ? m 5M?S?

j: .^cq a < b ? a + b = 800 ?) a + b ≡ 2 (mod 3) n :

-Bb}ú 8$

(1) a ≡ 0 (mod 3) / b ≡ 2 (mod 3) ?¤?) a = 3 / c ≡ 0 (mod 3) (i) J c ≡ 1 (mod 3), ? a + b + c ≡ 0 (mod 3) ? a + b + c D a ??, FJ a + b + c u??Ab (pe) (ii) J c ≡ 2 (mod 3), ? a + b ? c ≡ 0 (mod 3) ? a + b ? c D a ??, FJ a + b ? c u??Ab (pe)

?? (i) D (ii), ?? (1) .??ê?
(2) a ≡ 2 (mod 3) / b ≡ 0 (mod 3) "? (1) ín , ?? (2) .??ê?

FJ, (3) a ≡ b ≡ 1 (mod 3) A ×b, FJ a + b + c = 3 (pe)

QOBbn ? (3) A 5-, c í

?”:

(i) J c ≡ 1 (mod 3), ? a + b + c ≡ 0 (mod 3) ? a + b + c u¤ 7 _”bí| (ii) J c ≡ 0 (mod 3), ? c = 3 / c + a ? b ≡ 0 (mod 3) ? c + a ? b D c ??,

FJ c + a ? b u??Ab (pe) ?? (i) D (ii), ?? c ≡ 2 (mod 3) ?¤, a + b ? c ≡ 0 (mod 3) O??< ?: a + b ? c u”b, ] a + b ? c = 3 / a + b ? c u¤ 7 _ó?”bí|üM ?¤ ?): c = 800 ? 3 = 797, M ? m = (a + b + c) ? (a + b ? c) = 2c = 1594 (?: a = 7 / b = 793 )?¤ 7 _ó?”b?: 3,7,11,793,797,1571,1597 )

18

???: ????D?$ ABCD ó~, Iw? AB , BC , CD D DA 5~?}??
E , F , G D H cq? M , N , P D Q }?? AB , BC , CD D DA ,í? / M N , P Q }?DC EF , C GH ó~ t?p: M Q D N P ?ó?W

?p: I O ??¤?í?-/ L ? M N DC EF ó~5~? ?
AOE = COF, EOM = M OL,

/
LON = F ON, 2 AOE + 2 EOM + 2 F ON = 180? .

FJ F ON = 90? ? AOE ? EOM = BOM ?¤?)
AM O = ABO + BOM = COF + F ON = CON.

°??): M AO = OCN , ?¤,

M AO ?

OCN ?)

AM CO = C AM · CN = AO · CO. AO CN

°???: AQ · CP = AO · CO, FJ
AM CP = . AQ CN

??j?, Bb???: M AQ = P CN , ?ó?W

M AQ ?

P CN ?)?: M Q D N P

19


推荐相关:
网站首页 | 网站地图
All rights reserved Powered by 简单学习网 www.tceic.com
copyright ©right 2010-2021。
文档资料库内容来自网络,如有侵犯请联系客服。zhit325@126.com