30 - Ø, 92 2 11 2 2 (2 30 2 ) 2 921101 2 △ABC ˜ € +...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ø, 92 2 11 2 2 (2 30 2 ) 2 921101 2 △ABC ˜ € + ›Ä ª· ] ” ¨ B aB bB cB : △APBB △BPCB △CPAB aB bB c 2 ak + bk + ck = 1 ⇒ k = 2 VAPB = a,b,c B △ABC B PB △APBB △BPCB △CPA a b c VABC ,VBPC = VABC ,VCPA = VABC a+b+c a+b+c a+b+c CD, AE B CD, AE B D ', E ' B A 1 a+b+c 2 △ABC B AB, BC B DD ' = 2 D'B a b CD, EE ' = AE a+b+c a+b+c l1 // AB B E ' B P] Ø”ª · l2 // BC B D D' 2 l1 B l2 B · ] ” ت △PAB B △ABC B VPAB V 2 2 2 VABC V 2 a VPAB = VABC V a+b+c b VPBC = VABC V a+b+c a b VPAC = (1 − − )VABC a+b+c a+b+c c = VABC a+b+c a b c VAPB :VBPC :VCPA = : : = a :b:c a+b+c a+b+c a+b+c AB B a a+b+c l1 B P E' l2 E C V 1. • ¨]·ª* ]y Ÿ·ª* 2. + p~Eï)àPh©µª*ì 3. + p~Eï)àPh©µª*ì + p~Eï)àPh©µª*ì + p~Eï)àPh©µª*ì • ˜Ä›+°•]·ª*P W ì ì ì ì ∆APB = a b c ∆ABC , ∆BPC = ∆ABC , ∆CPA = ∆ABC a+b+c a+b+c a+b+c ∆ABC @+~EC 7n 2 921102 100Ä ˜›+ € 77 @+~E1 (A)3 (B)7 (C)9 (D)11 (E)17 2 * ª 2 ]N ¨ ·Ÿ N = x22 N+100 = y2 + 12 N+200 = z22 x2 y2 z 2 2 2 ∵ y –x + 1=1002 ∴ (y–x)(y + x) = 99 ∵x2 y 2 x 49 2 y 50 y+x 9 3 1 ⇒ N= x2 2401= 492 931 N+100= y2 + 1 2501=502+1 y–x 1 3 9 N+200= z2 2601=512 1¨·ª* Ÿ 100 2 15 18 225=152 325=182+1 425(2 ) 1 10 1 101=102+1 201(2 ) 2 N = 492 = 74 ª 2· ] 7 ¨ Ÿ (B) 2 1. E + ~› * @ 2.› ˜ Ä € + * ª µ © ì* h 3µ.©~ ì) @ hÐ1Ì•ô‘›+3 ªE E Q 1 77@ +@@ E ~ + @ 65.5 € + › ˜Ä 34.52 2 921103 5π Øc : B a = 1, b = 3 20π •Ä˜›+°”]·ª*W« π Øc 5, c = − 3 20 1 Q a 3 + b3 + c 3 − 3abc = ( a + b + c) ⋅ [(a − b) 2 + (b − c) 2 + (c − a ) 2 ] 2 3 3 3 ∴ a + b + c − 3abc a + b + c ⇒ 1 + 5 − 20 + 3 ⋅ 1 ⋅ 3 5 ⋅ 20 = −14 + 3 3 100 = − 3 2744 + 3 2700 < 0 ∴1 + 3 5 + (− 3 20) = a + b + c < 0 2 +˜ Ä € 3 20 > 1 + 3 5 · ] • ¨♠ ♠ ° ˜ Ä +˜ W • 1. • ∗ ] 3 3 3 1 : 3 5 : 3 20 2. Q a + b + c − 3abc = ( a + b + c ) ⋅ [( a − b) + (b − c) + (c − a ) ] V 2 2 2 1 2 a,b,c 2 2 (a − b) 2 + (b − c) 2 + (c − a ) 2 1 + 3 5 − 3 20 < 0 3 c ∴ a 3 + b3 + c 3 − 3abc a + b + π • ≅*˜+ Ä›ª ] 3. ≅ @ @ƒÞ-•›+p~E ( Ε ∼ + 2 + 20 > 1 + 3 5 2 1.π 2 •˜].≅*•+ ˜›ª 2Ä 2 π 3 2 · 17 ˆª ] ”* · 81 ˆª ] ”* · 23 ˆª ] ”* ·]” 716ˆª · ] 2 ˆª ”* 51Ä › € ˜+ 1+ 3 5 3 •20 ˜›ªW ˜ ]Ä ≅*•+« 3 2 ∼ + ›@+~ Ε ∗ 14 3 2 40Ä › € ˜+ 32 π 3 921104 d2 ª · ] n8 ˜ d @+~E1 77 *ª·]˜8 72 2 * ª · ] n 8 ˜ (a)18 (b)16 (c)15 (d)12 (e)11 2 82 2 :2 d B n n2 n-2 2 d + 8˜ ã·ª* n d ] (˜·ª* 2n d d (n − 2)( + 8) + 2 ⋅ = d + 72 n 2n d ∴ n 2 − 11n = n(n − 11) = , d , n ∈ N 8 n>11 2 n=12 2 12 ⋅ 1 = ]ª ( · 122 (d) d , d = 96 8 2 1. *~E@+ + › + 2.Ä • ˜”]·ª*W ›˜ * 3.¶ « W 5ò‘›+ á ª 49.2 € + › ˜Ä 50.82 2 921105 2 sol: B x+ B˜ y € ›Ä x + x 2 + x8 = y + y 2 + y 8 x= y x + x 2 + x8 = y + y 2 + y 8 x + x 2 + x8 − ( y + y 2 + y 8 ) = 0 2 ⇒ ( x − y ) + ( x 2 − y 2 ) + ( x8 − y 8 ) = 0 ⇒ ( x − y ) + ( x − y )( x + y ) + ( x 4 + y 4 )( x 2 + y 2 )( x + y )( x − y ) = 0 ⇒ ( x − y )[1 + ( x + y )[1 + ( x 4 + y 4 )( x 2 + y 2 )]] = 0 x 2 + y 2 = 0 x = y = 0· 8 ª] x2 + y2 ≠ 0 1 + ( x 4 + y 4 )( x 2 + y 2 ) > 1 ⇒ ( x + y )[1 + ( x 4 + y 4 )( x 2 + y 2 )] ≠ −1 ⇒ 1 + ( x + y )[1 + ( x 4 + y 4 )( x 2 + y 2 )] ≠ 0 ⇒ x− y =0⇒ x= y (1)2 (2)2 ♠] 8 • 1. • « x2 + y2 ≥ 0 B x2 + y2 ≥ 1B x2 + y2 ∈ N 2 (1) 2 x,y 8♠ · ]Ÿ ( x 2 + y 2 )( x 4 + y 4 ) + 1 > 1 (2) 2 (1)2 ( x 2 + y 2 )( x 4 + y 4 ) + 1 ≥ 1 B 2. 2 ( x 2 + y 2 )( x 4 + y 4 )( x + y ) + ( x + y ) + 1 = ˜ 0Ä € + 3. 94 2 ¶ + « › Wò ♠ ‘› 5 ∗ á 3+ 7- 1•p+~Eÿ)ð` Þ ƒ › x2 + y2 ...
View Full Document

This note was uploaded on 08/14/2010 for the course 數學 乙一 taught by Professor 鄭質明 during the Spring '07 term at Fudan University.

Ask a homework question - tutors are online