2007 奥林匹克 OMK Bongsu Muda Sulong 全部问题+ 1germany ques.+USM Maths Quiz

91_kiat

1。首先，我希望在这里的每位的pro pro可以提供你们的答案和你们的解答方式。
2。每题的题目很有可能有多过一个解答放方式。
3。我将会发omk muda的先，
     然后看你们的反应，
     我才会放OMK BONGSU,
     最后才会曝光OMK Sulong 的。
4。除了你们回答2007 的问题，
      我建议你们也可以分享一下你们如何破解“奥林匹克的数学比赛”,
      因为我们不只是要会回答2007的问题，
      而是要会它的回答方式和解答的过程。
5。不要问我是怎样拿到的，我会回答我不知道。
6。我想问大家“奥林匹克数学”你们是如何训练的。。。
7。那题germany olypia 的题目是我的老师给的。。。。
8。我接下来就跟大家分享2007/2008Pertandingan Kuiz Maths Antara Sekolah2 Menengah(槟城理科大学的数理学会主办)

目录

OMK 2007 MUDA--------第二楼
解答-------------------------第三楼至第十三楼
OMK 2007 BONGSU---第四楼
解答-------------------------第五楼至第十三楼
OMK 2007 SULONG---第十四楼
解答-------------------------第十五楼起

[ 本帖最后由 91_kiat 于 26-1-2008 08:57 PM 编辑 ]

91_kiat

由于我最近忙一些东西，所以我不能帮你们翻译，所以请你们多多谅解。
我有几题不会解，3，4，5，8，9，所以希望你们可以帮帮我解答一下。

OMK 2007 MUDA

1.
GIVEN
x+y=2 AND 2xy-z^2＝１ .FIND THE INTEGER SOLUTION OF THE EQUATION .

2.
IN A SPORT’S TOURNAMENT LASTING FOR 4 DAYS, THERE ARE
“n” medals to be won. On the first day, 1/5 of the “n” medals are won . On the second day , 2/5 of the remainder from the first day are won . On the third day , 3/5 of the remainder from the second day are won . On the final day, 24 medal are won . What was the total number of medals?

3.

Let 2xyz7
be a five-digit number such that the product of the digits is zero and the sum of the digits is divisible by 9. Find how many such number.

4.
Let
ABCD be a rectangle . The line DP intersects the diagonal AC at Q and divides it in the ratio of 1:4 . If the area of the triangle APQ is one unit square , determine the area of the rectangle.

5.
Find the smallest integer such that if divided by2 leaves a remainder of 1, if divided by 3 leaves a remainder of 2 , if divided by 4 leaves a remainder of 3, and if divided by 5 leaves
a remainder of 4.

6.
Let file:///C:/DOCUME~1/5583/LOCALS~1/Temp/msohtmlclip1/08/clip_image004.gif
be a function defined on integers such that
f(2n)=-2f(n), f(2n+1)=f(n)-1, and f(0)=2

7.
Given 8(y^2 + 1/y^2) – 56(y + 1/y)+112 =0 , find all the values of y^2 + 1/y^2 .

8.
Let f,g be two functions defined on(0,2c) where c>0 . show that there exists x,yÎ(0,2c) such that
|xy-f(x)+ g(y)|³ c^2

9.
One side of the triangle is 4 cm . The
other two sides are in the ratio 1:3 . What is the largest area of the triangle.



[ 本帖最后由 91_kiat 于 4-1-2008 04:01 PM 编辑 ]

hamilan911

讨论过了
按一下连接
http://chinese.cari.com.my/myforum/viewthread.php?tid=903267&extra=page%3D2

91_kiat

OMK 2007 BONGSU
1. Let ABCD be a parallelogram where E is a point on AB such that 3BE=2DC. The lines CE and BD intersect at the point Q. If the area of triangle DQC is 36 , find the area of triangle BQE
2. A man was driving at 90km/h. Realizing that he was late , he increase his speed to 110km/h and completed his journey of 395km in 4hrs. For how long did he drive at 90km/h.
3.Solve 200520062007^2 -200520062006 x 200520062008 .
4. Find integers
x,y and z satisfying
xy+1/z = yz+1/x where x not equal z.
5. 20 persons dig a fish pond in 12 days if they work 6 hours per day. How many days are required to dig are required to dig the same pond if 4 persons work 5 hours per day ?
6. let 0<x<1 . If A=x , B=x^2 , C=1/x ,and D=1/x^2, arrange them from the smallest to the largest value .
7.Using only the numbers 1,2,3,6,7,9,0, find which number goes with each letter in the addition problem below to make it correct .

PAK

+
MAK

PIUK

8. Find the sum of angles a+b+c+d+e+f+g+h in the following diagram .(是关于angle 的问题，我懒惰画了)
9.
Given 8(y^2 + 1/y^2) – 56(y + 1/y)+112 =0 , find all the values of y^2 + 1/y^2 .




[ 本帖最后由 91_kiat 于 23-12-2007 05:15 PM 编辑 ]

flash

感觉上还不会很难，很多都是基本数学题。。。。。。例如第三题答案等于 1，道理就从 (a+1)(a-1) = a^2 - 1 来。

kenny56

1.16
那两个三角形是相似三角形

2.2.25
90x+110y=395
x+y=4

3.1
(a+1)^2-a(a+2)=1

4.这题无解是吗？
xy+1/z = yz+1/x
xy-yz=1/x-1/z
y(x-z)=1/x-1/y
欲令1/x-1/y是整数，只可以是x=y=1
x=2,y=-2或x=-2.y=2也可以，但左边就不配了。

5. 工作天数和人手及工作量成反比
12=k/(20*6)
   k=1440
1440/(5*4)=72
需72天

6。B<A<C<D

7.     160
   +   960
       1120
P=1 A=6 K=0   M=9  U=2
P=I=1?  还是题目是写1？

9.  Let y + 1/y=x
        y^2 + 1/y^2=x^2-2
8(x^2-2) – 56x+112 =0
8x^2-56x+96=0
x^2-7x+12=0
(x-3)(x-4）=0
x=3 or x=4
  y^2 + 1/y^2=7 or 14

请高手们帮忙检查答案。
还有这两题在上个贴没人讨论到，小弟也不会。请高手们帮忙解答。

4.
Let
ABCD be a rectangle . The line DP intersects the diagonal AC at Q and divides it in the ratio of 1:4 . If the area of the triangle APQ is one unit square , determine the area of the triangle.   
8.
Let f,g be two functions defined on(0,2c) where c>0 . show that there exists x,y&Icirc;(0,2c) such that
|xy-f(x)+ g(y)|&sup3; c^2

dunwan2tellu

Let f,g be two functions defined on [0,2c]  where c>0 . show that there exists x,y in[0,2c] such that
|xy-f(x)+ g(y)| >= c^2

先 define h(x,y) = xy - f(x) + g(y)
我们的目的是要 prove , 存在着 p,q in [0,2c] such that |h(p,q)| >= c^2
分2个 case

case 1 : 当 g(0)-g(2c) =< 2c^2 ; 那么
g(0) =< 2c^2 + g(2c)
g(0) + c^2 =< 2c*2c + g(2c) - c^2
2c*0 + g(0) + c^2 =< 2c*2c + g(2c) - c^2
LHS =< RHS
所以 f(2c) 可能在其中一个 range(当然他也有可能在 LHS 和 RHS 的中间) ,也就是
f(2c) >= LHS  or f(2c) =< RHS  
[ 这里你们就要想像一下为什么。给你一个 example 吧。如果 3 =< 8 那么任何一个号码 k 有可能是 k >= 3 或 k=< 8 。不信的话 test k = 2,5,9... 或其他号码看看]
如果 f(2c) >= LHS 也就是 f(2c) >= 2c*0 + g(0) + c^2
也就是 0*2c + g(0) - f(2c) =< - c^2  .拿modulus 就符合 |2c*0 + g(0) - f(2c)| >= c^2
或用h(x,y) 来看的话 |h(2c,0)| >= c^2

不然的话 f(2c) =< RHS ==> f(2c) =<  2c*2c + g(2c) - c^2
==> 2c*2c - f(2c) + g(2c) >= c^2 ==> |2c*2c - f(2c) + g(2c)| >= c^2
|h(2c,2c)| >= c^2
case 1 小总结 : 在case 1 里存在着至少一 pair of x,y satisfying |h(x,y)| >= c^2 , i.e either |h(2c,0)|>= c^2 or |h(2c,2c)|>=c^2

case 2 : 当 g(0) - g(2c) > 2c^2 时候，证明方法大致上相同。就当作练习啦。
得到的结论是 |h(0,0)| >= c^2 or |h(0,2c)|>= c^2

dunwan2tellu

Let
ABCD be a rectangle . The line DP intersects the diagonal AC at Q and divides it in the ratio of 1:4 . If the area of the triangle APQ is one unit square , determine the area of the triangle.   

先证明 AP : PB = 1 : 3 (如果不会，提示就是考虑 similar triangle APQ 和 CDQ)
那么你就能下结论 BPQ 的面积 = 3 x APQ 面积 。
又因为 AQ : QC = 1 : 4 所以 BQC 面积 = 4 x ABQ 面积。
那么你就能找到 ABC 的面积 = ABQ + BQC 。
之后 x2 就是 ABCD 的面积。

kenny56

g(0) - g(2c) > 2c^2
-g(2c)>2c^2-g(0)
g(2c)+c^2<0*0+g(0)-c^2
0*2c+g(2c)+c^2>0*0+g(0)-c^2

let f(0)>=0*2c+g(2c)+c^2
    -0*2c+f(0)-g(2c)>=c^2
     O*2c-f(0)+g(2c)=<-c^2
    IO*2c-f(0)+g(2c)I>=c^2
so  h(0,2c)>=c^2

let    f(0)=<0*0+g(0)-c^2
      -0*0-g(0)+f(0)=<-c^2
      0*0-f(0)+g(0)>=c^2
     I0*0-f(0)+g(0)I>=c^2
so h(0,0)>=c^2

我们这样证明  h(0,0)>=c^2 or h(0,2c)>=c^2 or h(2c,0)>=c^2 or h(2c,2c)>=c^2

但如果变数x,y在0到2c间呢？如  h(0,c)可以肯定大过c^2吗？
这点我搞不清楚，请版主教下

dunwan2tellu

我们的目的只是要证明 [0,2c] x [0,2c] 里会有会找到至少一 pair 的 (x,y) 使到 |h(x,y)|>= c^2 . 并不是 每一个 (x,y) 都可以得到那个 relationship.

举个例子 f(x) = 1 , g(y) = y^2 for all x,y in [0,2c]
那么 h(0,c) = 0*c - f(0) + g(c) = -1 + c^2 明显 < c^2
所以不一定 |h(0,c)| >= c^2
但如果用上面的 case 来看 g(0) - g(2c) = 0 - (2c)^2 = -4c^2 < 2c^2 所以是 case 1
那么我们知道一定是 either |h(2c,0)| >= c^2 or |h(2c,2c)| >= c^2

如果真的是 |h(2c,0)| >= c^2 的话就表示
|-f(2c) + g(0)| >= c^2 也就是说 1 >= c^2 ==> -1 =< c =< 1
否则的话就必须要有 |c|>1
那么 |h(2c,2c)| = |8c^2 - 1| = |c^2 + (7c^2-1)| >= c^2  ( 因为 7c^2 - 1 > 0 for |c|>1)

dunwan2tellu

我们的目的只是要证明 [0,2c] x [0,2c] 里会有会找到至少一 pair 的 (x,y) 使到 |h(x,y)|>= c^2 . 并不是 每一个 (x,y) 都可以得到那个 relationship.

举个例子 f(x) = 1 , g(y) = y^2 for all x,y in [0,2c]
那么 h(0,c) = 0*c - f(0) + g(c) = -1 + c^2 明显 < c^2
所以不一定 |h(0,c)| >= c^2
但如果用上面的 case 来看 g(0) - g(2c) = 0 - (2c)^2 = -4c^2 < 2c^2 所以是 case 1
那么我们知道一定是 either |h(2c,0)| >= c^2 or |h(2c,2c)| >= c^2

如果真的是 |h(2c,0)| >= c^2 的话就表示
|-f(2c) + g(0)| >= c^2 也就是说 1 >= c^2 ==> -1 =< c =< 1
否则的话就必须要有 |c|>1
那么 |h(2c,2c)| = |8c^2 - 1| = |c^2 + (7c^2-1)| >= c^2  ( 因为 7c^2 - 1 > 0 for |c|>1)

kenny56

谢谢你哦，版主。

刚才我忽略了"there exists"这两个字，才以为每个x,y只要是在0到2c间都符合。

91_kiat

OMK 2007 SULONG
1.
Let a1= 6,…..an =6^an-1 .Find the reminder when a100 is divided by 11.

2.
Let -1<y<0<0<x<1 . If A=(x^2) y, B= 1/(x^2) y ,C=(y^2) x and D=1/xy^2 , arrange them from the smallest to the greatest
value .

3.
One side of a triangle is 4 cm .The other two sides are in the ratio 1:3 . Find the largest area of the triangle .

4.
Simplify log24log46log68……log2n(2n+2)

5.
Write 580 as a sum of two squares .

6.
Let f be a function defined on non-negative integers satisfying the following conditions .
f(2n+1)=f(n), f(2n)=1-f(n)
Find f(2007)

7.
Let f, g be
two function defined on [0,2c ] where c>0 .Show that there exists x,y&Icirc;[0,2c] such that
|xy-f(x)+g(y)|&sup3; c^2

8.
Two circles of radius 1 and 2 respectively are tangential to one another externally . Another circles is drawn tangential to both circles such that their centres form a right angle triangle .Find the radius of the third circles .

9.
Determine the maximum value of m^2 + n^2 where m,n&Icirc;{1,2,3,…….2007} and (n^2 –mn – m^2)^2 =1



[ 本帖最后由 91_kiat 于 29-12-2007 09:59 AM 编辑 ]

Dexer

楼主，你给的muda组题目请修改一下，很多问题。谢谢。
例，
GIVEN
x+y=2 AND "2xy-z^2" .FIND THE INTEGER SOLUTION OF THE EQUATION

2xy-z^2=???????

Let
ABCD be a rectangle . The line DP intersects the diagonal AC at Q and divides it in the ratio of 1:4 . If the area of the triangle APQ is one unit square , determine the area of the "triangle"
should be rectangle right??

91_kiat

谢谢你。。。。真的对不起

双子座12

Let a1= 6,…..an =6^an-1 .Find the reminder when a100 is divided by 11.

这题不会，数目好像很大leh。谁可以教教我，我只会一些mod的皮毛。

是问        6^..............^(6^(6^(6^6)))除11的余数吗？

hamilan911

利用fermat's little theorem和6power n的特征
6^10 = 1(mod11)  ,   6^n = 6(mod10)

所以a100 = 6^(10X+6) = 6^6 = 5(mod11)

91_kiat

IF a,b,c are odd number , show ax^2+bx+c=0 , has no rational number

91_kiat

2007/2008 Pertandingan Kuiz Matematik
Antara Sekolah-sekolah Menangah

有些答案我听不清楚。。。。。对不起，答案有可能是错的。。。
Roud 2
1. Given that f^(-1)(x)=1/3 In x ,x>0 .Find the value of f(0). Answer: 1

2.Find the value of
t if a straight line y=2-tx is the tangent to the curve


y+4x^2 -6x +7 =0 . Answer : 68.18

3.For what real value of k will the equation below describe a circle a radius
equal to 3 ?
x^2 + y^2 +2x- 4y +k =0 Answer: -4

4.A circle is inscribed in a square S1 and a square S2is inscribed in the circle .

What is the ratio of the area of S2 to the area of S1 .Answer : &frac12;

5. A cube measuring 9 units on each is painted . It is then cut into 729 one
unit cubes. How many faces of the one-unit cubes are not painted .
Answer :3888.

6.6 points are equally spared round a circle of radius 1 unit. If all pairs of
points are rejoined by line. What is the total length of all lines .
Answer:22.39

7.If d(sin ax)/dx = a cos ax . Find the derivative of y=sin 3x sin 2x sin x , when x= p/4 . Answer: -1

8. Given the universal set such that
x x{:x is natural number less that 100}
P { x:x^2 &Icirc;x}
Q{ x:x^3 &Icirc;x}
R{ x:square root of x is a positive integers}
Find the value of n(p^(,) n Q) U (P U R) Answer = 16

9. No enough time to copy

10. A group of student sit for both physics and math paper in SPM .

75% pass math
7% pass physics
40% fall in react one subject
A student is selected randomly from the group . Find the probality that the student only pass only one of the two subject . Answer: 0.25

11. A candy maker mixes chocolate , milk and almonds to produce 3 kinds of candy x,y , and z with the following properties :
x: 7 kg chocolate , 5 litter of milk , 1g of almonds
y: 2 kg chocolate , 2litter of milk , 2g of almonds
z : 4 kg chocolate , 3 litter of milk , 3g of almonds

if 67 kg of chocolate , 48 litter of milk and 32 g of almonds are used , how many of each kind of candy can be produced .
Answer : x= 4, y=5, z=6

12. When 5 new classroom were built for wering……..be school .The average class size reduce by 6, when another 4 . If the number of students remained the same throughout the changes , how many students were there at the school . Answer :600

13. No enough time to copy

14.Find the solution for the equation

(x-99)(x-101)(x-103)(x-100)+16=0

X=100 plus minus square root of 4

15 Given that the composition function
F^2(x)=x^4 +18x^2+90

Find the value of f(3)
Answer : 18

16. Given that f:x&reg;(1+x)/(1-x), x not equal to 1. Find the value of f^466 (1/3)
Answer : -3

17. The function y=f(x) is a function such that f(f(x))=6x – 2005 for every real number
x. As integer t satisfies the equation f(t)=6t -2007 ,finf the value of t . Answer : 401

18. . No enough time to copy
19. . No enough time to copy

20. Find the value of (84871^3 + 123 ^ 6)/ (84871^2 – 84871
x 15129 + 123^4) .Answer : 100 000

dunwan2tellu

原帖由 91_kiat 于 26-1-2008 09:00 PM 发表
IF a,b,c are odd number , show ax^2+bx+c=0 , has no rational number

应该是 no rational roots ..

要有 rational roots 条件是 b^2 - 4ac = k^2 , where k = integer .

考虑 mod 8 , 发现到 4ac == 4 (mod 8) if a,c = odd .; b^2 == 1 (mod 8) if b = odd ==> b^2 - 4ac == 5 (mod 8) .

反观 k^2 只能是 0,1,4 (mod 8) , 无解。所以他没有 rational root