ĐỀ THI CHỌN HỌC SINH GIỎI GIẢI TOÁN TRÊN MÁY TÍNH CASIO NĂM HỌC 2009 – 2010 -Lớp 12 THPT Qui định: !""#$%&'()&*+&,-%.&* /012 %34567 Bài 1 / 8 / / = + ( ) ( ) ( ) ( ) 9 8 8 8 8= + + +×××+ ! "#$% ! Bài 2.&'( )**+,-.*/-! 0 / 8 / / / + = + + ! "#$% ! Bài 3.&',-123,-42.*! ( ) 8 / / / /= + − + + ,0:;56 / / = + ! "#$% ! Bài 4.789* 3 :$%! ( ) ( ) 0 + − − = 3 ( ) ( ) ; ; 0 + − − = ∈N 3 ≥ <=789 < = + ∈N 3 ≥ * ,->.* 0 0 ? ? ? @ ? ? ? A BCDEF,%9/ + G + 3 @ + G + 3 HEF,%9/,I?3#$%9,JA2DIK ? + + 3 < + G ? ? ? + + ? ? ? = L:,->.*! M N ? ? ? ?< < < < < ! "#$% ! Bài 5.*F M M / / / /= − + + * JO ? ?5 .*ACA* 8 / / 5/ / = = + + + ?A#,P(** F 8 / *F / JQR*F7Q M 0 / / /= + + A S1,- ? ?5 3H*JQR?,-&'O.*#D%9#.*/ - 8 /= $%*T@ ! "#$% ! Bài 6. B8%2.*UV#(O.*WXY*3H*$%*IK*9T: X%VUA*&%,O%/318%2?;ZQ*&9W[?J8%2[ I?Z,V*[#DG3A:X%#DKV@*%V*[8%2 >%\?NZ?A:X%#DKVIW,\)*?(,'UA:X% QR 3]8;0;0;M?N/Q*,\^4A:X%8VU#(O, A*I%_`I%aQR$%9,JA2D,I9 ! "#$% ! Bài 7. * J / A# M 0;N0 / / / / $ = > / / + − + − − − = 31 > 3-.*D&V? = bRDCD(.*D&V? $ RDCD(.*D&V A JO.*:F* ; M ? ?/ / / ,(*,-F`%ca.* / / + ÷ ! "#$% ! Bài 8. *J 55 * ( ) ( ) ( ) ( ) 55 5 5 = + + × − − `I%$%9,JA2D QR(#$% AJ+I 42*(CDDQa*QR+I) %U%);3)&%dU%);! ;;;;;; = `I%aQR ! "#$% ! Bài 9. QYe ! @ ! @ ! / / / − + = − − = + − = ^*QYe 3 *%:f@*QYe 3 *%:T@*QYe 3 *%: * Jg*W.*f?T?3#7Q17:DX A &'O.*QYeF**DX,f.**fT 3g*W*h.**DX31:T &'7OD&JDe)*QY,\:#D3QY,\W#D *fT"#$% ,\#)iCDDX ! "#$% ! Bài 10.JDdU%:95?@A1?:AI?B11 * 7O>%$%*3.*JD A &'W?D'?X9.*RDAjk6AI369.*JD JD&j)*J&%W#D3J&%:#DJDU%8 ! "#$% ! HẾT Đáp án và biểu điểm Bài 1! / 8 / / = + l^mnlofl^mnloTfBp^fffBp^fqfBp^ffrfBp^f! fBp^fTfBp^fqfBp^fTrs fBp^ff÷fBp^f f÷r T2I#Dqqq#(fC,-J7H?g(# $% jA#T! N;9 ≈ − 9C!,-DE6..; FG&+. − H&'; IG&+. Bài 2!&'( )**+,-.*/- 0 / 8 / / / + = + + r:2DJ+:3+%.*!<5! R ( ) ( ) t 0 / / 8 / / / + − = + + @ t @ 8 / / / − − − + = ⇔ = = !^+,- / 3 / huF[fB,-+,-! fBp^f<> r÷fBp^f<> rfBp^f<r0fBCD ,- − ql^mnlof ;N ≈ ?fBCD#D + ql^mnloT M;000 ≈ " )**+,-.*/-! ( ) ( ) / / = − + − T29! rfBp^fT−fBp^ff> q(#$% ! M ≈ Bài 3:&',-123,-42.*! ( ) 8 / / / /= + − + + 56 ( ) 0 ? @ / / / = + = − ∈ − @ / = − ( ) ( ) ( ) ( ) 8 / / / / = + − + + = − − − + + = 0 0 ? @ = − + + + ∈ − t M M = − + ? t N00@ M0N@ M0M = ⇔ ≈ − ≈ ≈ ? ? @ ∈ − fBp^f<s0−0 fBp^f<> rfBp^f<rr fBCD3c q*QR ( ) 0MN00 − ≈ fBCD3 q*QR ( ) MN;0N0N ≈ Qa+?*! M;NMNM;;@ ;@ 0M; ≈ − ≈ ≈ SC9! MN;0;@ M;NMJ5/ 8 / J 8 /≈ ≈ − Bài 4: 0 ? ? M;@ ;0 = = = = 0 ? 0? ;? N = = = = . EF,%9/.*% r 7:! 5 + + + = + *ODQa,J! 0 M; @ M; ;0 5 5 5 5 5 = + + = ⇔ ⇔ = = − = + + = h! + + = − Qa+! 0 N + + = − v%9,JA2D! l^mnlofl^mnloTl^mnlo0l^mnlohl^mnlo< T## fBp^f<fBp^fqfBp^f<rfBp^f!fBp^fwfBp^fqfBp^fT− fBp^fffBp^f!fBp^fffBp^fqfBp^fTfBp^f!fBp^fTfBp^fq fBp^fwfBp^f!fBp^fnfBp^fq0fBp^fh − NfBp^ffBp^f! fBp^ffBp^fqfBp^fhfBp^f!fBp^fhfBp^fqfBp^fnfBp^f! fBp^fxfBp^fqfBp^fwrfBp^fnqqq,-.*wF31% r ?.* nF313 r ?.*xF31y r L:,-Q*%! M N ;? ;N? qM0N? y MMNN? y ;MM;;; < < <= = = = Bài 5: *O.**F>! @ @ 0 / / /= − = = G #*! M 0 8 / / / /= + + + ?%9,*! 0 M 0 0 0 M N ; 0 0 0 0 8 5 8 5 5 8 − = − = − + = + ÷ ÷ = = ⇔ + + = − + + = − = = ÷ ÷ L ODQa,J*QR! @ @ 0 M 0 5 = = = h! 0 M 0 8 / / / /= + + + ALg/- 0 M 0 8 / / / /= = + + + #D%9#.*$%*T@QYe ! /= + O ^ODQa,JW#D3O.*#D%9#.*$%*T! 0 M 0 0 0 t M M / / / / / / / / 8 / / / + + + = + + − = ⇔ = + + = = + + L DQa,J*QROW.*#DF31#D%9#.* $%*T@! M0@ M;0M@ ;0/ / /≈ − ≈ − ≈ huF[fBO.*#D%9#QaF.*! M;@ ;@ N ≈ − ≈ − ≈ Bài 6: Lg*V318%2?;Z?>V318%2?NZ?J V#(O!*rr>"?UV 3]8! ; N ;0;0;MN 5 / × × × = v%9,JA2D! × ;sfBp^ff× s× NsfBp^f< − ;0;0;MN fBp^fq l^mnloBSw`CD,-.*fq`CD,-&%<ql^mnloBSw (#$% <(E%9I B6D:$%9,J31fCD3&QR??0??#(CQR,-%9I.*<q0 (fq SC9A:X%V#(O!rr0q Bài 7: M 0;N0 / / / / $ = > / / + − + − − − = 0;N0l^mnlofl^mnlo<fBp^f<fBp^fqfBp^f<rfBp^f! ,fBp^f<rfBp^f<rp,fBp^f<−fBp^f<− l^mn>z−fBp^f<sM−fBp^f<s−fBp^ffqqq#(A%F AP?F31 NK = A ( ) ( ) / $ / / $ / $ / / − − + − − − = = = + = = = ÷ ÷ ÷ ÷ ∑ ∑ ∑ S1 M − = ⇔ = l%9,*O.* M / N;;$ = S1 ; N − = ⇔ = l%9,*O.* ; / N 0;$ = S1 − = ⇔ = l%9,*O.* / M0N$ = Bài 8: *l&J!MM ! ( ) 55 5 5 5 5 = + + + = + = + ( ) ( ) ( ) ( ) ( ) ( ) 5 5 5 + + × − − = + − h! ( ) ( ) ( ) ( ) ( ) ( ) 55 5 5 5 5 = + + × − − ⇔ + = + − `#% 5 = ⇒ = ?U%9(E> 9,* Qa+?#% 5= ⇒ + = ?U%9(E> 9,* v%9,JA29! fBp^ffrfBp^f<−fBp^ffrfBp^f<−fBp^f q l^mnloBSw`CD,-fq`CD#D,-&%<q(#$% <i CDDX l^mnloBSw`CD,-fq`CD#D,-&%<q(#$% <i CDDX l^mnloBSw`CD,-fq`CD#D,-&%<q(#$% <qM@ #DK$%9,J#(fqN *bJQR!MM A^a3-b ;= )%;S1* 5 b 0M;;= ) %U%; S1) ( ) 5 b; )%U%; * ! ;;; N////≈ @ ;;; NM ////≈ ? ;;; 0? @///× ≈ ; ;;; NN? @ ;;; NM? /// ///× ≈ × ≈ @ M ;;; 0;? @///× ≈ `Q3C9?CDDQa.*%E);D A&%Aj!N@NM@ 0@N>@NM>@0>@>q????N V! N; ;;0@ NM; ;MN@ 0; ;;;N0= = = SC9&J!q0;3 0; ;;;N00M0N;;;= Bài 9: * ( ) N @ 0 ? @ @ @ M 0 = L $ − − − − ÷ ÷ A µ * * = − − = − ÷ L)**DXf3o>! µ * * * = − − − + = + ÷ ÷ ÷ l%9,*!^O.*f! * * * 5 − − = + ÷ ÷ T29! *l^mn* c rl^mn* c * A{ l^mnlof(# $% ! N5 ≈ r5QYeF**DXf/-.*! 5/ = + ?f$%* @ 0= − − I 0 5 = − rg*W*h.*f3TO.*ODQa,J! 0 / 5/ 5 + = − = − + L OD APA29QCDO* 7ufBp^ff3CDO 7u−fBp^f fr0?*QR(#$% ! ?NM0@ ?0M 0 M 0 =L = + + − ÷ ÷ 3A#f N M 0 L$ = + + + ÷ ÷ 3A#T N 0 $= = − + + ÷ ÷ 3A# fBp^ffrfBp^fTrfBp^f÷l^mnloh`V*%3D hO.**fT! fBp^fhfBp^fh−fBp^fffBp^fh−fBp^fT fBp^fhl^mnlow T(QY,\:#D*fT! 0 5 N 9 = ! fBp^fffBp^fTfBp^f÷0÷fBp^fwl^mnlon T(QY,\W#D*fT! 9 = hOD&JDe)*QY,\W#D3QY,\:#D*fT! ( ) 9 N N π π π = − = − l^mn π fBp^fw> −fBp^fw÷fBp^fh> q(#$% 0?00 9 .≈ Bài 10: *A(QY,:#D93,%: .*JD! r ;M00M 5 N O== = ≈ ;0l^mnlof÷÷l^mnloT (#$% A(QY,\:#D9.*J D! ;M00MN ≈ rU%*.*JD! 9O N= = − N00> −fBp^fT> l^mnlo (#$% ;0NN0M ≈ r,%:.*JD! A B C D E S I O M J K [...]...2 a a - Tính OI: OI = Bấm máy: ⇒ d = SI = h 2 + OI 2 = h 2 + 0 0 ÷ 2 tan 36 2 tan 36 ( ALPHA C x2 + ( ALPHA A ÷ 2 ÷ tan 36 ) x2 ) SHIFT STO D cho kết quả trung đoạn hình chóp: d ≈ 8.817975958(cm) 1 2 2 2.5 ALPHA... Phân giác góc SIO cắt SO tại K là tâm mặt cầu nội tiếp hình chóp đều có bán kính r 1 = KO: 1 h r1 = KO = OI tan sin −1 ÷÷ d 2 ( ALPHA A ÷ 2 ÷ tan 36 ) tan ( 0.5 SHIFT sin -1 ( ÷ ALPHA C ÷ ALPHA D ) ) SHIFT STO E cho kết quả: r1 = KO ≈ 2,5851(cm) Trung trực đoạn SA trong mặt phẳng SAO cắt SO tại J Mặt cầu ngoại tiếp hình chóp đều có tâm J, bán kính SJ SM SO SA2 b 2 = ⇒ r = SJ = = SJ SA . ĐỀ THI CHỌN HỌC SINH GIỎI GIẢI TOÁN TRÊN MÁY TÍNH CASIO NĂM HỌC 2009 – 2010 -Lớp 12 THPT Qui định: