奧數比賽題目

oscar

以下的題目是剛剛1星期前比賽的題目，我只抽一些難做的。

1)
Suppose p, q are positive integers and 96/35 > p/q > 97/36, find the smallest possible value of q.

2)
Let n be a natural number, the area of the triangle bounded by the line nx + (n+1)y = √2 and the two coordinate axes is S(n). If K = S(1) + S(2) + S(3) + ... + S(2005), find the value of K.

3)
Given that 60^a = 3 and 60^b = 5. If R = 12^[(1-a-b)/2(1-b)], find the value of R.

4)
Given that p, q and r are distinct roots of the equation x^3 - x^2 + x - 2 = 0. If Q = p^3 + q^3 + r^3, find the value of Q.

5)
If B is an integer and B > (√2 + √3)^6, find the smallest possible value of B.

6)
Given that the perpendicular distances from the point O to three sides of a triangle ABC are all equal to 2 cm and the perimeter of triangle ABC is equal to 20 cm. If the area of triangle ABC is equal to k cm^2, find the value of k.

7)
In Figure 1, ten people are sitting in a round table with sitting numbers 1, 2, 3, ..., 10 respectively. Each of them chooses an integer A, B, C, ..., J respectively and tells the people on his left and right about his chosen number. Then each of them calculates the average number of the chosen numbers of his two neighbourhoods and announces this average number. If all the announced average numbers are the same as the corresponding sitting number, find the value of F.

抹茶可樂

第6條淺到無野講=.=

oscar

有冇人識做then？

suny

第7條最易~o個日我一睇就寫左ans~~係6



補:諗諗下都係比埋個解你~

設...a~f每人係佢坐位上ge number
即...a=1,b=2.......j=10
咁除左a同j外~b至f都係佢自己隔離兩個人ge平均數.....
即b=2,c=3....i=9   (a=6,j=5)
所以~f=6

就係咁~唔arm ge請指教!!!

[ Last edited by suny on 2005-2-7 at 04:13 PM ]

oscar

Originally posted by suny at 2005-2-7 15:56:
第7條最易~o個日我一睇就寫左ans~~係6



補:諗諗下都係比埋個解你~

設...a~f每人係佢坐位上ge number
即...a=1,b=2.......j=10
咁除左a同j外~b至f都係佢自己隔離兩個人ge平均數.....
即b=2,c=3....i= ...

我諗你唔係好明條題目問乜...
算啦...我計到啦...

解7)
From the questions above, we can get the following equations:
(J+B)/2=1---(1)
(A+C)/2=2---(2)
(B+D)/2=3---(3)
(C+E)/2=4---(4)
(D+F)/2=5---(5)
(E+G)/2=6---(6)
(F+H)/2=7---(7)
(G+I)/2=8---(8)
(H+J)/2=9---(9)
(I+A)/2=10---(10)

Simplify the above equations:
(1): J+B=2
(2): A+C=4
(3): B+D=6
(4): C+E=8
(5): D+F=10
(6): E+G=12
(7): F+H=14
(8): G+I=16
(9): H+J=18
(10): I+A=20

Since,
(3)-(1): D-J=4
(4)-(2): E-A=4
(5)-(3): F-B=4
(6)-(4): G-C=4
(7)-(5): H-D=4
(8)-(6): I-E=4
(9)-(7): J-F=4
(10)-(8): A-G=4
(1)-(9): B-H=4
(2)-(10): C-I=4

Therefore,
D-J=E-A=F-B=G-C=H-D=I-E=J-F=A-G=B-H=C-I ---(11)

From (11),
F-B=J-F
2F=J+B
From (1), J+B=2
So, 2F=2
F=1

細薯

第七題睇完answer又覺得ng係好難~~
只係煩左D~

oscar

Originally posted by smallpotato226 at 2005-2-13 16:05:
第七題睇完answer又覺得ng係好難~~
只係煩左D~

係呀...
不過比賽時間有限，當時諗唔到咁多野出黎...
其實我又已經諗到第4-6題點做...

細薯

Originally posted by oscar at 2005-2-13 08:35 PM:

係呀...
不過比賽時間有限，當時諗唔到咁多野出黎...
其實我又已經諗到第4-6題點做...

ng~
我D fd (參加奧數比賽果D)話佢一定輸
因為諗得慢

try:
5)
If B is an integer and B > (√2 + √3)^6, find the smallest possible value of B.

(√2 + √3)^6

=  √2^6    +  6  √2^5 √3    +  15  √2^4 √3^2  + 20 √2^3 √3^3   +  15  √2^2 √3^4   +  6  √2 √3^5   +   √3^6

= 8 +  24 √2√3  + 180 + 120 √2 √3   +  270  +  54  √2 √3   +   27

= 8 + 180 +  270  +   27 +  24 √2√3  + 120 √2 √3   +  54  √2 √3  

= 485 +  198 √2√3

之後ng識la~~XD
(o岩 ng o岩 ??)

[ Last edited by smallpotato226 on 2005-2-14 at 11:37 AM ]

oscar

我上年只參加group
今年只參加individual
下年應該group+individual...

細薯

幫我see see 7樓條數丫~~