2009數學競賽題目

(1) 和 (3) 太易不用回答吧!

(4) 9997n = (10000 - 3)n
=n x10000 - 3 x n
如 3 x n < 10000 萬位數字必為偶數, 所以
3 x n > 10000
n > 3333.33
又n必為奇數,所以n最少為3335.

(2) 設A於坐標中心,B於(1,0)
向量AB為i+0j
向量BC為cos84i+sin84j
向量CD為cos156i+sin156j
向量DE為cos228i+sin228j
向量AE計算為-0.478i+0.251j
AE角度為152.27
所以內角E為540-96-108-108-152.27=75.73

(14)設共N人
第一名分數=n+2-2
第二名分數=n+2-2x2
第三名分數=n+2-2x3
...
第N名分數=n+2-2xN
總分為N(n+2)-2(1+2+3+...+N)
= N(n+2) - N(N+1)
= N (n - N + 1) = 2009 = 7 x 7 x 41
N = 1, 7, 41, 49, 287, 2009
相對n為2009, 293, 89, 89, 293, 2009
所以n最細值為89

(19) x+1/y = t
y=1/(t-x)...(1)
相似地:
z=1/(t-y)...(2)
w=1/(t-z)...(3)
x=1/(t-w)...(4)

(1)代入(2)
z=(t-x)/(t^2-xt-1)
代入(3)
w=(t^2-xt-1)/(t^3-xt^2-2t+x)
代入(4)
x=(t^3-xt^2-2t+x)/(t^4-xt^3-2t^2+xt-t^2+xt+1)
xt^4-x^2t^3-2xt^2+x^2t-xt^2+x^2t+x=t^3-xt^2-2t+x
xt^4-(1+x^2)t^3-2xt^2+2(1+x^2)t=0
t(t^2-2)[xt-(1+x^2)]=0
t=0 (rejected); t=+/- sqrt(2) ; t=x+1/x(rejected)

2009-06-07 13:47:26 補充：
(15) integrate sqrt(x) from 2009 to 6027
=2/3[sqrt(6027^3)-sqrt(2009^3)]
=251901.0057
using trapezoidal rule,

2009-06-07 13:47:48 補充：
Area A =1/2sqrt(2009)+sqrt(2010)+sqrt(2011)+...+sqrt(6026)+1/2sqrt(6027)
therefore A=sqrt(2010)+sqrt(2011)+...+sqrt(6027) + 1/2sqrt(2009) - 1/2sqrt(6027)
A=sqrt(2010)+sqrt(2011)+...+sqrt(6027) - 16.4059
sqrt(2010)+sqrt(2011)+...+sqrt(6027) = A + 16.4059 = 251917.4116

2009-06-07 13:48:06 補充：
Consider error term for integration from 2009 to 2010 is
2/3[sqrt(2010^3)-sqrt(2009^3)]-[(sqrt(2010)+sqrt(2009)]/2
=44.8274468337488-44.8274466024749
=0.0000002312739

2009-06-07 13:48:22 補充：
Consider error term for integration from 6026 to 6027 is
2/3[sqrt(6027^3)-sqrt(6026^3)]-[(sqrt(6026)+sqrt(6027)]/2
=77.6305352071145-77.6305351625837
=0.0000000445308

Obviously the 4018 error terms together is very much small than 1

Hence the required result is 251917

2009-06-07 13:58:06 補充：
(12) no. of coins must be odd even odd even in pattern. To maximum n, we can choose 0 and 1 for most of the boxes. As such n^n/2<2009. and max n is 63.
Consider coins are put
101010...0101
010101...1010
...
101010...0101
Total coins = 32+31+32+31+...+32 = 1985 (24 below 2009)

2009-06-07 14:03:37 補充：
Change the pattern at one corner as below will make up 2009 coins
1010...012343
0101...123232
1010...012121

2009-06-08 10:48:40 補充：
(12) Use M1,M2,M3,M3 to represents mathematicians, and X1,X2,X3,X4 otherwise
Groupings are (A,A),(A,A),(A,A),(A,A) in sequence with 8! possibilities
For no 2 mathematicians in any group, the grouping must be
(M,X),(M,X),(M,X),(M,X)
(X,M),(M,X),(M,X),(M,X), etc

2009-06-08 10:48:44 補充：
Total combination is 2 x 2 x 2 x 2 = 16
In each of these possibilities, the M can be arranged with 4! possibilities.
Likewise for X, it is als 4!
Total possibilities = 16 x 4! x 4! = 9216
probability = 9216/8! = 8/35

2009-06-08 13:09:45 補充：
(16) 4x+3y+2z=2009
y must be odd
max (x) = (2009-3-2)\4 = 501
max (y) = (2009-4-2)\3 = 667
max (z) = (2009-4-3)\2 = 1001
First solution when x=1 is (1,1,1001)
Second solution is (1,3,998)
Third solution is (1,5,995), etc

2009-06-08 13:09:58 補充：
So number of solutions for x=1 is 334 i.e. smallest integer greater than or equal to (1001/3)
I denote 334 as [1001/3] where [x] is smallest integer greater than or equal to x
First solution when x=2 is (2,1,1001-2)
Second solution is (1,3,998-2)

2009-06-08 13:10:02 補充：
Third solution is (1,5,995-2), etc
So number of solutions for x=2 is [(1001-2)/3]
Total number of solutions will be [1001/3]+[999/3]+[997/3]+[993/3]+...

2009-06-08 13:14:36 補充：
Correction:
Total number of solutions will be [1001/3]+[999/3]+[997/3]+[995/3]+[993/3]+...
4x+3y+2z=2000
y must be even
First solution for x=1 is (1,2,995)
Second solution is (1,4,992)

2009-06-08 13:14:41 補充：
third solution is (1,6,989), etc
So no. of solution for x=1 is [995/3]
Similarly for x=2, number of solution is [993/3]
and total number of all solution is [995/3]+[993/3]+...
Compare with the case when the sum is 2009, the difference is
[1001/3]+[999/3]+[997/3]
=334+333+333=1000

2009-06-08 13:28:37 補充：
(18) You should draw your own diagram.
Since R is the 外心 of APC, so R is the centre of the circle APC.
Since angle ACP = 45 (inscribed angle) so angle ARP = 90
Likewise since angle ABP =45, angle AQP=90
Since AR=PR so angle APR= angle PAR = 45

2009-06-08 13:29:04 補充：
Similarly QP=QA and
angle QPA= angle QAP = 45
angle QPR=angleQAR=90
AQPR is a square
Since QR=2; QP=sqrt(2) ;QB=sqrt(2); BP= sqrt(2) given

2009-06-08 13:29:08 補充：
So ABP is equilateral triangle
angle QBP=60
angle QBA=60-45-15
since QB=QA, angle QAB also = 15
AB=2xQBxcos15=2sqrt(2) (sqrt(2)+sqrt(6))/4=1+sqrt(3)
BC=sqrt(2)xAB = sqrt(2)+sqrt(6)
so PC=sqrt(6)

2009-06-08 14:11:10 補充：
(20) a,b,c and d are each with factor of 3^n where n can be 0,1,2,or3
Since LCM for any 3 number has a factor of 3^3, so at least 2 of the 4 numbers have the factor 3^3.
Consider the combination of n, it can be the following:

2009-06-08 14:11:13 補充：
2 is 3, the other 2 non-3 … total combination = 4C2 (consider the positioning) x 3 (0 to 2) x 3 (0 to 2) = 54
3 is 3, the other non-3 … total combination = 4C3 x 3 = 12
All 4 is 3 …only 1 possibility
Total possibilities = 54+12+1 = 67

2009-06-08 14:11:34 補充：
Use similar reasoning for 7^5:
2 is 5, the other 2 non-5 … total combination = 4C2 (consider the positioning) x 5 (0 to 4) x 5 (0 to 4) = 150
3 is 5, the other non-5 … total combination = 4C3 x 5 = 20
All 4 is 5 …only 1 possibility

2009-06-08 14:11:43 補充：
Total possibilities = 150+20+1=171
Combine the 2 results, total number of possibilities is 67 x 171 = 11457

2009-06-08 15:38:35 補充：
(6) Let HK cuts BC at L
angle ACB= angle ABC= X
cos x=5/13, sin x =12/13, tanx = 12/5
angle PKH =angle PHK = X
since HK=2, PL/(HK/2)=tanX
PL=12/5, CL=BL=5, so PC=12/5+5=7.4

2009-06-08 16:49:52 補充：
(8) Let A is the original of an x-y co-ordinate, and B is the point (1,0).
By similar triangle ADP and BXP
DA/AP=XB/BP
1/AP=(1-m)/(AP-1)
AP=1/m

2009-06-08 16:50:02 補充：
P is the point (1/m,0)
By symmetry, Q is (0,1/m)
Slope QP=-1
By similar triangle ADR and XCR
1/DR=m/CR
1/DR=m/(DR-1)
DR=1/(1-m)

2009-06-08 16:50:07 補充：
co-ordinate of R is [1/(1/m),1]
Slope RP = (1-0)/[1/(1-m)-1/m] =-1
m^2-3m+1=0
m=[3-sqrt(5)]/2 or [3+sqrt(5)]/2 (rejected)

2009-06-09 11:23:17 補充：
(17) Let's view the dice as a box build from a paper. Unfolding the dice will give something like this (I do not know how to post pciture here, sorry!)
A
BCDE
F or denote as A[BCDE]F
where A must be different from BCDE, and F must be different from BCDE as well.

2009-06-09 11:23:36 補充：
Let’s choose a colour for A, there are 6C1 = 6 possibilities. Different cases for BCDE:
<1> BCDE are all different.
The possible combinations will be:

2009-06-09 11:23:45 補充：
6 [for A] x 5C4 [choose 4 colours from the remaining 5] x 4! [the possible permutation of the 4 colours] x 2 [F can be the remaining colour or the same colour as A]
=6 x 5 x 24 x 2 = 6 x 240

2009-06-09 11:24:33 補充：
<2> 3 colours for BCDE where BD can be same or CE can be same
The possible combinations will be:
6 [for A] x 5C3 [choose 3 colours from the remaining 5] x 3C1 [choose one colour from the 3 so that 2 of the faces are same colour] x

2009-06-09 11:24:49 補充：
x 2 [BD same or CE same] x 2! [permutation of the other 2 colours] x 3 [F can be the remaining 2 colours or the same colour as A]
= 6 x 10 x 3 x 2 x 2 x 3 = 6 x 360

2009-06-09 11:25:05 補充：
<3> 2 colours for BCDE
The possible combinations will be:
6 [for A] x 5C2 x 2 [either 1212 or 2121] x 4 [the rest 3 colours + same colour as A]
= 6 x 80

2009-06-09 11:25:13 補充：
Total possibilities = 6 x (240 + 360 + 80) = 6 x 680
Since A[BCDE]F, B[ACEF]D, C[ABDF]E, D[ACEF]B, E[ABDF]C and F[BCDE]A generate the same colour combinations, the answer is reduced to 680

2009-06-09 11:56:03 補充：
(13) Let CD=AE=n, AB=m, AD=CE=p
Let G’ is point on AB such that GG’ // CE
Let F’ is point on AB such that FF’//CE
Let G’E=y
By similar triangles ACE and AGG’,
AE/CE=AG’/GG’ or n/p=(n-y)/GG’ … (1)

2009-06-09 11:56:23 補充：
By similar triangles ADB and G’GB,
AB/AD=BG’/GG’ or m/p=(m-n+y)/GG’ … (2)
Eliminate GG’ from (1) and (2),
n/m=(n-y)/(m-n+y) or y=n^2/(m+n) …(3)
By similar triangles ACE and AGG’,
FG:FC = F’G’:G’E = n/2-y:n/2 = (m-n)/(m+n):1

2009-06-09 11:56:32 補充：
Area ABCD = (m+n)h/2 where h is height of the figure.
Area AECD = nh
Area DFC = nh/4
Area DFG = nh/4 x (m-n)/(m+n)
Let AB/CD=r=m/n
R=DFG/ABCD =n(m-n)/[2(m+n)^2]=(r-1)/[2(r+1)^2]
Using differentiation, it can be shown that R is maximum when r=2.

2009-06-09 11:59:48 補充：
Correction : r=3

2009-06-09 16:39:55 補充：
(13) @(1/2)+@(1/2x2)+@(1/2x3)+…+@(1/2*2009)
Consider integral @(x/2)=x@(x/2)-ln(1+x^2/4)+C
Area from 0 to 2009 is 2009@(2009/2)-ln(1+2009^2/4) = 3139.90533
Use Trapezoidal rule Area = 1/2@(0) + [@(1/2)+@(1/2x2)+…+@(1/2x2008)] + 1/2@(1/2*2009)

2009-06-09 16:40:07 補充：
Therefore @(1/2)+@(1/2x2)+@(1/2x3)+…+@(1/2*2009) can be approximated by:
Area - 1/2@(0) + 1/2@(1/2*2009) = 3140.69023
tan(3140.69023)=-1.2664
But note that tan(x) is very sensitive to small error in x, for example,
tan(3140.59)=-1.567
tan(3140.69)=-1.267
tan(3140.79)=-1.035

2009-06-09 16:40:17 補充：
Therefore the approximated value may not give very accurate result.
Consider approximate area from x=5 to 2009.
Area is 2009@(2009/2)-ln(1+2009^2/4)-5@(5/2)+ln(1+5^2/4) = 3135.93488
Use Trapezoidal rule Area = 1/2@(5/2) + [@(1/2)+@(1/2x2)+…+@(1/2x2008)] + 1/2@(1/2*2009)

2009-06-09 16:40:27 補充：
Therefore @(1/2)+@(1/2x2)+@(1/2x3)+…+@(1/2*2009) can be approximated by:
Area + 1/2@(1/2*5) + 1/2@(1/2*2009) + @(1/2)+@(1/2x2) + @(1/2x3) + @(1/2x4) = 3140.65392
tan(3140.65392)=-1.3656

2009-06-09 16:40:43 補充：
Using similar approximation by integrate from 10 to 2009 yields tan(3140.64979 = -1.3775
Using similar approximation by integrate from 20 to 2009 yields tan(3140.64860 = -1.3810
Using similar approximation by integrate from 30 to 2009 yields tan(3140.64837 = -1.3816

2009-06-09 16:40:47 補充：
Using similar approximation by integrate from 40 to 2009 yields tan(3140.64829 = -1.3819
Using similar approximation by integrate from 50 to 2009 yields tan(3140.64825 = -1.3820
Therefore the required result is 1.38 to 2 decimal point.

2009-06-09 21:47:58 補充：
(2) correction
向量AE計算為-0.478i+0.658j
AE角度為126
所以內角E為540-96-108-108-126=102

2009-06-10 12:12:13 補充：
(13) Correction:
3 colours: There are 6C3 = 20 ways to choose the colours. One 1 way to colour the dice.

第11題唔值7星，咁EASY

是IMO Prelim嗎? 比我拿銅獎的那一年淺

第四題問咩??