奥数问题大家讨论

我要开心

试题是选择题。另一个解法：

Let AD = 1 and DE = x

△ AGE ~ △ CGB

(EF+FG)/GB = (EF+FG)/EF = (1+x)/1 ---(1)

△ ABE ~ △ DEF

BF/EF = (EF+FG)/EF = 1/x ---(2)

Solving (1) & (2),

(1+x) = 1/x  

x2 + x – 1 = 0

x = (-1+√5)/2

DE/AE = x /(1+x) = (after simplifying)  (3 - √5)/2 (答案A)

教育版淘气包

我要开心 发表于 31-5-2012 17:00
试题是选择题。另一个解法：
Let AD = 1 and DE = x△ AGE ~ △ CGB(EF+FG)/GB = ...

David的思路，我也是一样能够得到开心女侠的答案的。

但是哇塞，你这个简单啊，我看到这个解题思路真是开心死~~~

回复下面静雯的点评：我儿子很一般，我还被老师叫去喝茶了的。。。但是我知道你家公子很厉害~~~

davidbin

朋友让我做的 RI open house question，有兴趣你可以试一下，据说五分钟答对，直接DSA
1000 个学生，每人一个LOCKER，编号1~1000，第一个学生打开所有LOCKER，第二个学生走到编号是2的倍数的LOCKER前，如果LOCKER是OPEN的，就CLOSE， 如果LOCKER是CLOSE的，就OPEN， 第三个学生走到编号是3的倍数的LOCKER前重复以上的动作，问1000 个学生走过以后，有多少个LOCKER是打开的？是那些编号的LOCKER？

frekiwang

davidbin 发表于 5-6-2012 12:09
朋友让我做的 RI open house question，有兴趣你可以试一下，据说五分钟答对，直接DSA
1000 个学生，每人一 ...

所有完全平方数（square numbers)是打开的，因为只有这些数有奇数个约数（odd number of factors)，所以打开的是1，4，9，16...961（1至31的平方共计31个）

frekiwang

题目：你有6个半径为1厘米的圆，要不重叠的放进一个边长为x厘米的正方形里，求x的最小值（精确至小数点后5位）

davidbin

楼上看来是专家，我先探个路
5.32820 CM
2r+2*2r*SinA = 2r+ 3*2r*CosA
SinA/CosA = 3/2
TanA = 3/2

鹏飞狮城

我要开心 发表于 31-5-2012 17:00
试题是选择题。另一个解法：
Let AD = 1 and DE = x△ AGE ~ △ CGB(EF+FG)/GB = ...

自认为读书时几何学得还不错，汗颜的是花了近一个小时才看懂开心的解答。看来辅导奥数是对我的一个巨大考验啊！

鹏飞狮城

frekiwang 发表于 5-6-2012 22:53
题目：你有6个半径为1厘米的圆，要不重叠的放进一个边长为x厘米的正方形里，求x的最小值（精确至小数点后5 ...

我投降。。。

静雯

2011 SMO Junior Round 2

Q5. Initially, the number 10 is written on the board. In each subsequent moves, you can either
(i) erase the number 1 and replace it with a 10,
or
(ii) erase the number 10 and replace it with a 1 and a 25,
or
(iii) erase a 25 and replace it with two 10.
After sometime, you notice that there are exactly one hundred copies of 1 on the board. What is the least possible sum of all the numbers on the board at that moment?

(我这儿没有解答。有解答的版亲请将见解或解答回复在下面。先谢谢了！）

仙都峰居士

先声明这是书本上抄的
Suppose the number of times that operations (i), (ii) and (iii) have been performed are x, y and z respectively.
Then the number of 1, 10 and 25 are y-x, 1+x-y+2z and y-z, respectively, with y-x=100.
Thus the sum is S=y-x+10(1+x-y+2z)+25(y-z)=-890+5(5y-z)
Since we want the minimum values of S, y has to be as small as possible, and z as large as possible.
Since y-x=100, 1+x-y+2z>=0, y-z>=0
We get, from the first equation, y>=100, from the second inequation, 2z>=99 or z>=50 and y>=z from the third.
Thus the minimum is achieved when y=100, x=0 and z=100.
Thus the sum S=1110