n********y
发帖数: 187 | 1 Y2K以前是TG科学的冬天,生物科学的核冬天,能活下来的不多,基数自然小.
现在有钱了,还是游学期,积累需要很长时间,没有20年自主的CNS不会多的.
Nature
破。 |
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a*****g
发帖数: 19398 | 2 美国数学教育发展简史(1950～)zz
Teaching Math in 1950's — A logger sells a truckload of lumber for $100. Hi
s cost of production is 4/5 of the price. What is his profit ?
Teaching Math in 1960's — A logger sells a truckload of lumber for $100. Hi
s cost of production is 4/5 of the price, or $80. What is his profit?
Teaching Math in 1970's — A logger exchanges a set "L" of lumber for a set
"M" of money. The cardinality of set "M" is 100. Each element is worth one d
ollar. Make 100 dots representing the eleme... 阅读全帖 |
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l*****o
发帖数: 82 | 3 哪一年的文章？
我估计审稿的的是Physics/EE方面的吧。他们比较倾向于阅读
APPLIED PHYSICS LITTER/JAP/SCIENCE/NATURE,连ADV MATER.
和Syn. Metal都很少看就更别提Macromolecules了。而且共聚物
QW毕竟还是qw，请物理／电子器件领域的专家审稿也比较理所
当然。（发光学报本身也是偏物理的刊物）
（老实说我干这行的时候也只读物理／半导体领域的文章。y2k出
来的，也没看过你们说的那篇文章。）
国内有很多人不是胆子大，而是脸皮厚。记得当年读过好几篇
吉林大学的OLED文章，相同内容，发表在国内外至少４种学术刊物
上，只有极少差别。不过也许科学界就是这样，通病．即便是PRINCETON
Forrest组也这么干，只不过文章间的差别稍稍多一点而已。 |
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b*k
发帖数: 27 | 4 Let a, b, c be positive real numbers such that abc = 1.
Prove that
(a-1+1/b)(b-1+1/c)(c-1+1/a) <= 1 |
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u**x
发帖数: 45 | 5
Consider LHS(Left Hand Side) of above inequality:
LHS=c/b*(1/c-b+1)*(b-1+1/c)*(1-1/c+b)
Let x=(1/c-b+1), y=(b-1+1/c), z=(1-1/c+b).
Since the condition of x,y,z being negative/zero is one of 1/c, b, 1 is larger
than/equal to the sum of the other two, there can be only 2 cases:
1) only one of x, y, z is negative or 0 , obviously LHS<=0<1
2) x>0, y>0, z>0, we have
1/c=(x+y)/2, b=(y+z)/2, 1=(x+z)/2
8xyz
LHS=-------------- (*)
(x+y)(y+z)(z+x)
since x+y>=2(xy)^0.5, y+z>=2(yz)^0.5, z |
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b*k
发帖数: 27 | 6 Determine whether or not therer exists a positive integer n such that
n is divisible exactly 2000 different prime numbers, and
2^n+1 is divisible by n |
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l**i
发帖数: 5 | 7 I think k could be 3^n or 2*3^n.
It's easy to check that 9*19|2^{9}+1
Use induction to prove that p1^k*p2*...*pk|2^{3^k}+1 where p1=3,p2=19
with all the pi's distinct
we already have the case k=2,assume k>2 is correct,prove case k+1
because 2^{3^{k+1}}+1=(2^{3^k}+1){(2^{3^k})^2-2^{3^k}+1}
let A=(2^{3^k})^2-2^{3^k}+1,we have 3|A,A=3(mod pi)(i>1),so pi!|A(i>1)
assume 2^{3^k}=3m-1,then A=9m^2-9m+3=3*{3m*m-3m+1),let B=3m*m-3m+1
we see pi!|B for all i,so B must have a different prime factor p{k+1}
an |
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b*k
发帖数: 27 | 8 A magician has one hundred cards numbered 1 to 100. He puts them into three
boxes, a red one, a white one and a blue one, so that each box contains
at least one card.
A member of the audience selects two of the three boxes, chose one card
from each and announces teh sum of the numbers on the chosen cards. Given
this sum, the magician identifies the box from which no card has been chosen.
How many ways are there to put all the cards into the boxes so that this
trick always works?(Tow ways are con |
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C******a
发帖数: 115 | 9 答案是12个解，实质上只有两种：B={1}, C={100}, A=其它；
A={1,4,7,10,...}, B={2,5,8,11...}, C={3,6,9,12...}。
由条件可以推出若任意一个集合中有两个数相差为n，则其他两个集合各取
一数得到的两个数相差不可能是n（因为会出现魔术师无法判断的情况）。
不可能有两个集合，比如说A和B中各有相邻的一对数，因为如果那样的话，
由上面的结果，C中的数既不会和B中的数相邻，也不会和A中的数相邻，矛盾。
于是最多有一个集合，比如说A，中有一对相邻的数。于是B和C中都没有相邻
的数对，B和C各选一数也不相邻。令D＝B并C，则D中的任两个数都是不相邻的。
假设n=min(|b-c|:b\in B, c\in C)，且n=b_0-c_0, b_0\in B, c_0\in C。
那么n>1，在b_0和c_0之间有n-1个A中的数。如果在b_0或c_0不是1或100，
则在b_0或c_0的外侧有一个A中的数，这样就能找到两个A中的数相差为n，矛盾。
这其实也得到了一个解：B={1}, C={100}, A=其它。颠倒顺序即得6个解。 |
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b*k
发帖数: 27 | 10 Let n >= 2 be a positive integer. Initially, there are n fleas on a horizontal
line, not all at the same point.
For a positive real number r, define a moves as follows:
choose any two fleas, at points A and B, with A to the left of B;
let the flea ar A jump to the point C on the line to the right of B
with BC/AB = r.
Determine all values of r such that, for any point M on the line and any
initial positions of the n fleas, there is a finite sequence of moves that
will take all the fleas to the po |
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C******a
发帖数: 115 | 11 对于一个n元实数集合X，定义f(X)=n*max(X)-sum(X)。
下面证明如果r>=1/(n-1)，则满足条件。
在这种情况下，对开始的任意n个数，进行这样的变化：
以最大的数为跳板，把最小的数跳到最右面，即成为新的最大。
初始时的集合记为X_0，跳过以后的集合记作X_1。
则max(X_1)-max(X_0)=r*l_0, sum(X_1)-sum(X_0)=(1+r)*l_0。
其中l_0=max(X_0)-min(X_0)>=f(X_0)/n>0。
那么f(X_1)-f(X_0)=n*r*l_0-(1+r)*l_0=((n-1)*r-1)*l_0>=0。
继续类似地做下去，则f(X_i)是不减正数列，
max(X_{i+1})-max(X_i)=r*l_i>=r*f(X_i)/n>=f(X_0)/(n*(n-1))。
于是max(X_i)是以不小于一个给定正数的步伐递增。
经过有限步后，必然有max(X)>M，然后再经过n-1步，则都大于了M。
下面证明如果r<1/(n-1)，则不满足条件。
考察从初始情况开始的任意一次跳动。要证明存在C(n,r)>0，使得
max |
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b*k
发帖数: 27 | 12 Let AH1, BH2, CH3 be the altitudes of an acute-angled triangle ABC.
The incircle of the triangle ABC touches the sides BC, CA, AB at T1, T2, T3
respectively. Let the lines L1, L2, L3 be the reflections of the lines
H2H3, H3H1, H1H2 in the lines T2T3, T3T1, T1T2 repectively.
Prove that L1, L2, L3 determine a triangle whose vertices lin on the
incircle of the triangle ABC. |
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l**i
发帖数: 5 | 13 it's not hard to prove that L1//BC etc,
the next thing is to prove the distance
from the center of the incircle to L1
is rcosA,where r is the radius of the
incircle,etc,and we are done.
To prove this,using trigonometrical
relations to find |H3T3| and suppose L1
intersects AB at P3,calculate |T3P3|
we can get the propotion of |AP3| to |AB|
the remaning thing is just checking it
using r and |AH1|.
I believe there should be simpler plane
geometrical method to do this. |
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u**x
发帖数: 45 | 14 Here is a geometric proof
1) easy to show that L1//BC, L2//AC, L3//AB. omitted
2) Let O be the incenter of ABC, let the incenter of triangle AH2H3 be X.
O, and X are on the bisector of angle A. AO cross T1T2 perpendicularly at
Y.
Since AH2H3 ~= ABC and AH2=AB*cos(A), ----(*)
thus AX=AO*cos(A)
Then XY=AY-AX=AO*cos(A/2)^2-AO*cos(A)=AO*sin(A/2)^2
And since YO=AO*sin(A/2)^2
X is the reflection of O in line T2T3.
So the distance of O to L1 is equal to the distance of X to H2H3, i.e.,
the in-radius of |
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b*k
发帖数: 27 | 15 Two circles T1 and T2 intersect at M and N.
Let L be the common tangent to T1 and T2 so that M is closer to
L than N is. Let L touch T1 at A and T2 at B. Let the line through M parallel
to L meet the circle T1 again at C and circle T2 at D.
Lines CA and DB meet at E; lines AN and CD meet at P; lines BN and CD meet at
Q.
show that EP = EQ |
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l**n
发帖数: 67 | 16 1.EM is perpendicular to AB (in triangle ECD, CA = AM, DB = BM)
2.PM = MQ (MN 平分AB,PQ)
=>EP = EQ |
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u**x
发帖数: 45 | 17 设原心 O1, O2,
1)
由O1A, O2B 垂直平分CM, MD
则AB=1/2CD
CA=E, DB=BE,
有EM//OA1//OA2, EM垂直CD.
2) 延MN交AB于G. GA^2=GB^2=GM*GN
GA=GB
又PM/GA=QM/GB
PM=QM
由1)2) EP=EQ |
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h*l
发帖数: 19 | 18 美丽几何
嗨! 其实有一个漂亮的解法.当然本身是最简单的
一道,并不一定值得这样漂亮. ----
容易化归到证 EM 垂直 CD. 以 M 为中心做反演,
你就会发现: T1, T2 变成两直线,与仍是直线的
CD形成三角形. AB 变成它的切园, 而 E ---,
正好变成这个切园的圆心! 故而 EM 垂直 CD !
由于 反演 是保角的, 自然在变化前也是对的!
(哈,有点象我在新语丝提到的那只蝴蝶的样子.)
可能受了当年蒋声的"几何变换"的影响,但肯定
不是严镇军的那本讲复数的小册子影响,呵呵,开
个玩笑.
----两月前看到这里的Y2KIMO,呵呵,这多年了,
还是不能立马全部搞定,毕竟比高手差一截的说.不
过这年居然有两个平几题目,而且自己感觉是6题由
难到易,也算独特了.(Y2K嘛!)咱立马搞定的就是这
两道平几题,"虫跳"也搞定了,前三道象很技巧的样
子,就没想啦.前一道这里的平几解法太妙了,我还真
没想到.我就是按前一方案用三角硬算得到的.用了
正弦定理和万能代换.后一道就是上面的了.现在有
空闲贴了出来,呵呵.
...美丽的几何,就向刘翎在"初恋人生"里写到的: |
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k**********g
发帖数: 989 | 19 The printf thing is getting funny. Let's see if Microsoft will create a
Windows-only Y2K Problem by adding C99 compliance into VCRT14. |
|
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e***e
发帖数: 3872 | 21 这也是个陈包子，看ZST改名前的水：
发信人: xexe (x.exe-y2k virus), 信区: TOBE
标 题: Re: 灌水
发信站: The unknown SPACE (Sat May 20 08:51:03 2000), 站内信件
无名氏【梧叶儿○失题】
青铜镜，不敢磨，磨著后照人多。一尺水，一尺波，信人唆，那一个心肠似
我。 |
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x**e
发帖数: 315 | 22 【 以下文字转载自 test 讨论区,原文如下 】
发信人: xexe (x.exe-y2k virus), 信区: test
标 题: 让艺术板更加绚丽多彩
发信站: The unknown SPACE (Wed Apr 26 21:17:14 2000), 站内信
件
【 以下文字转载自 Arts 讨论区 】
【 原文由 weidong 所发表 】
change to blue |
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e***e
发帖数: 3872 | 23 这个有没有用？帮主们？
发信人: xexe (x.exe-y2k virus), 信区: TOBE
标 题: 版主手册
发信站: The unknown SPACE (Sat May 20 22:55:55 2000), 站内信件
发信人: space (排骨教主), 信区: Science
标 题: [转载] 转贴--版主手册
发信站: The unknown SPACE (Tue May 18 21:27:16 1999), 转信
【 以下文字转载自 Whisper 讨论区 】
【 原文由 whg 所发表 】
1)
按 B ,就可以把和当前信件主题相同的信件一起处理了.
将有如下功能
相同主题 (0)取消 (1)删除 (2)保留 (3)文摘 (4)放入精华区 (5)放入暂存档 ? [0]:
如果你选择 5 ,然后再到你的精华区中按 i .随便打个文件名
再打出标题,嗯.OK 了.你的合集做好了.
你可以在精华区中用^P再把它转回你的版来.
2)
按 i ,可以把很多东西放到一起来的哦 :-) |
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