modular arithmetic
quadratics
number theory
relatively prime
Divisibility Theory
pen

[quote="PDatK40SP"]We can also prove that $m \equiv 2 \ (mod \ 12)$ ( and replace $m$ by $12t+2$ in the formula of $A$, we will have $A$ is odd immediately ). Use this lemma: ( I think it's well-known ) Let $n$ is an odd integer, and $k$ is an odd divisor of $3^{n}+1$. Then, $k \equiv 1 \ (mod \ 6)$
$m|(m+3)^{n}+1 \Rightarrow m|3^{n}+1$
Of course $3 \not | m$, so if $n$ is even, then $(m+3)^{n}+1 \equiv 2 \ (mod \ 3)$, which is impossible because $3m|(m+3)^{n}+1$.
So $n$ is even, then, from the lemma, we have $m = 2^{k}(6t+1) \ \ (k,t \in \mathbb{N})$
If $k$ is even, we have $(m+3)^{n}+1 \equiv 2 \ (mod \ 3)$, which is impossible. Moreovers, easy to see that $8 \not | 3^{n}+1 \ \forall n \in \mathbb{N}$ so $k \geq 3$ is impossible.
So, $k=1 \Rightarrow m=12t+2$[/quote]

[quote="Студент"][quote="Peter"][quote="PDatK40SP"] Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n}\plus{}1$. Then, $ k\equiv 1\ (mod\ 6)$[/quote][/quote]
Why?[/quote]

It's equivalent to showing that none of its prime divisors can be 5 modulo 6.

Let $ a \equal{} 3^{(n\plus{}1)/2}$ and suppose $ 3^{n}\plus{}1$ is divisible by a prime $ p\equiv 5\pmod{6}$. Then $ a^{2}\equiv\minus{}3\pmod{p}$. At this point, Quadratic Reciprocity would work. You could also show that
\[ x\equal{}2^{\minus{}1}(a\minus{}1)\Rightarrow x^{2}\plus{}x\plus{}1\equiv 0\pmod{p}\]
So $ x^{3}\equiv 1\pmod{p}$ and $ x^{p\minus{}1}\equiv 1\pmod{p}$. We have $ p\minus{}1$ and 3 relatively prime, so $ x\equiv 1\pmod{p}$. But as $ x^{2}\plus{}x\plus{}1\equiv 0\not\equiv 3\pmod{p}$, this is a contradiction.

[quote="Peter"][quote="PDatK40SP"] Use this lemma: ( I think it's well-known ) Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n} \plus{} 1$. Then, $ k \equiv 1 \ (mod \ 6)$[/quote][/quote]

what's well known lemma?

[quote=TTsphn]More :
1) There exist infinite $ (m,n)\in N$ satisfy the condition.
2)$ m\equiv 2(\mod 12)$ (take from my teacher).[/quote]

I agree with your Teacher. Except of some ifs and buts.  According to the mysterious Lemma the solution of the problem A17 is:

m = 2 + 12t  for ANY natural t, including t = 0. Those t's may also be perfect squares like t = 4, 9, 16, 25,... so that m = 50, 110, 194, 302,..

All those m's make A an odd number, at least for certain n's.  Right?   If you agree, let me offer you my CONJECTURE A17. :

                                        If  t > 1  is a perfect square, then A  is not  an integer, for ANY n.

Unable to prove or disprove my conjecture.  If I'm right, then maybe Peter's Lemma has to be modified.(?)  Please help. Thanks.

Peter,
The first sentence of your post.  You “…will have A odd immediately”(?)
But what if t = 4, 9, 16, 25, 36 , 49…or  12, 5, 19 and n is any number. Any!
Then A will be proper or improper fraction - not an odd as you claim.
    If  t = 0  AND  n = 1, 3, 5, 7,...    or   t = 7 AND  n = 21, 63, 105, 147, …. or   t = 22 AND  n = 9, 27, 45, 63, 81….
then in each case we get infinite number of odd A’s  “immediately” and  "m = 2 + 12t" works.  Nothing is always
   You ignore(?) that A is not A(t), but  A(t, n).   t is parameter, n is a variable (see below)
Trivial: A can be either  a fraction (for any n) or an odd  (for certain t’s and certain unlimited  n’s).

Let’s subdivide the set of 26 consecutive t’s (from 0 to 25)  by two subsets:  ‘bad’  t’s  and  ‘good’  t’s.

Proposition 1.         ‘Bad’ t’s do not make A an odd. For any n.

Here they are:   t =    4,  9,  12,  14,  15,  16,  18,  19,  20,  21,  23,  24,  25.

Proposition 2.       ‘Good’ t’s do make A an odd.  But not for any n.

Good  t’s:            t =     0,  1,  2,  3,  5,  6,  7,  8,  10,  11,  13,  17,  22

Bad t’s do not solve the problem A17.  We’ll ignore them from now on.

Now we’ll find all unlimited odd A’s (all of them) for first 13 good t’s and and predefined n’s
We ‘ll show that A can not be just any odd  but unlimited strictly defined ones.
In the table below k runs through all odds:   k = 1, 3, 5, 7, ….

In A(t, n):  t is parameter,  n is variable (a linear function of odd k’s for each ‘good’ t). Some examples:
t = 0     A(0, k) is  an  odd  for any n  =   k  =  1, 3, 5, 7, …….             then A(0,  k) = 1,  21,  521,  13 021,   325 521, …
t = 1     A(1, 3k) is an odd   for any n = 3k   =   3, 9, 15, 21…..           then A(1, 3k) = 117,   2 823 520 869,  ………
t = 7     A(7, 21k) is an odd  for any  n = 21k = 21,  63,  105,  147,… then A(7,21k) =…..infinite number of huge odds
t = 13   A(13, 39k) is an odd  for any n = 39k =  39,  117,  195,…      then A(13, 39k) = …..          ‘’

‘good’ t___m=2+12t_____n_______A(t,n)  =  ((5+12t)^n + 1)/3(2+12t)_____A(t, n),  all odds.
============ ==================================================================

0_________2__________k________A (0, k) = (5^k + 1)/6_____________A(0, k)  = 1,  21,  521,  13021,  325521,.

1________14_________ 3k_______A (1, 3k) = (17^3k + 1)/42_________ A(1, 3k) =  117,  2823520869, …..

2________26__________3k_______A(2, 3k) = (29^3k + 1)/78_________ A(2, 3k) = Infinite number of  huge odds

3________38__________9k_______A(3, 9k) = (41^9k + 1)/114________ A(3, 9k) = _______ same

5________62_________15k_______A(5,15k) = (65^15k + 1)/186_______A(5, 15k)________ same

6________74__________9k_______A (6, 9k) =  (77^9k + 1)/222_______ A(6, 9k)_________ same

7________86_________ 21k______ A(7, 21k)  =  (89^21k + 1)/258_____ A(7, 21k)________ same

8________98__________21k______A(8, 21k)  = (101^21k + 1)/294_____A(8, 21k)________ same

10______122___________5k______A(10, 5k) = (125^5k + 1)/366______ A(10, 5k)__________"

11______134__________11k______A(11, 11k) = (137^11k + 1)/402_____A(11, 11k)__________"

13______158__________39k______A(13, 39k) = (161^39k + 1)/474_____A(13, 39k)_________‘’

17______206__________17k______A(17, 17k) = (209^17k + 1)/618_____A(17, 17k)_________‘’

22______266___________9k_____ A(22, 9k ) = (269^9k  + 1)/798______A(22, 9k)___________‘’

There are 13 good and 13 bad t's. Just coincidence.
That’s all I can say about that. As an exercise: for  25 < t < 51 find all ‘good’ t’s and corresponding n’s.

all odd prime factors of m are $\equiv 1 \pmod{3}$
write $m=2^k \cdot j$, $k \geq 0$ and $j\equiv 1 \pmod{3}$
taking modulo 3, $m\equiv 2\pmod{3}$ and n is odd
$2^k \cdot j \equiv 2 (mod 3) \implies$ k is odd
but $k\geq 3 \iff 8|m \implies 8|3^n+1$ creates contradiction
so $k=1 \iff 2||m$
so $m\equiv 2\pmod {12}$

$(m+3)^n+1\equiv 2\pmod4$ , so A is odd
one example is $(m,n,A)=(2,1,1)$

[quote=Peter]Let $m$ and $n$ be natural numbers such that \[A=\frac{(m+3)^{n}+1}{3m}\] is an integer. Prove that $A$ is odd.[/quote]

When $m$ is odd ,$ A$ is always odd because both numerator and denominator are odd.

 [b]Claim:[/b]  When $  m,n$ are both even ,or  when $ m$ is divisible by $3$ then $ A$ is never an integer.
[b]Proof[/b] If $m \equiv 0 \mod {3}$ then $$(m+3)^n+1 \equiv m^n+1 \equiv 1 \mod 3 $$ wich contradicts the fact that $ A$ is integer .
When $m  \not\equiv 0 \mod 3$ and $n $ even we have that$$(m+3)^{2k}+1 \equiv m^{2k}+1 \equiv 2 \mod 3 $$wich again contradicts that $A$ is integer

[b]Claim:[/b] When $m$ is even , then we need $m=2k $ or $m= 4k$ where $k$ is odd integer for $A $ to be integer.
[b]Proof:[/b] We must have $$ (m+3)^n+1 \equiv 3^n+1\equiv 0 \mod m$$ or $$ m| 3^n+1$$ since $n$ is odd

By LTE we can say that $$v_2(m) \leq v_2 (3^n+1)=v_2(4)+v_2(n)=2 $$

[b]Claim: [/b] When $m=2k$ or $m=4k$ then $v_2(A)=0$
[b]Proof[/b] $$v_2(A)=v_2 ((m+3)^n+1)-v_2(3m)=v_2(2k+4)-v_2(6k)=0$$ for $ m=2k$ and
$$v_2(A)=v_2(4k+4)-v_2(12k)=0$$ for $m=4k$

After checking all cases we can say that $A$ is always odd. $\blacksquare$

[quote=XbenX]
[b]Claim: [/b] When $m=2k$ or $m=4k$ then $v_2(A)=0$
[b]Proof[/b] $$v_2(A)=v_2 ((m+3)^n+1)-v_2(3m)=v_2(2k+4)-v_2(6k)=0$$ for $ m=2k$ and
$$v_2(A)=v_2(4k+4)-v_2(12k)=0$$ for $m=4k$
[/quote]

This part is wrong. Suppose we have $m \equiv 4 \text{ (mod }8 \text{)}$ and $n$ an odd integer, then $(m+3)^n$ is congruent to $7$ (mod $8$). then $v_2(A)$ might be positive, and in that case we need to prove the divisibility relation can't hold, instead of computing $v_2(A)$ more.


Five students $ A, B, C, D, E$ took part in a contest. One prediction was that the contestants would finish in the order $ ABCDE$. This prediction was very poor. In fact, no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order $ DAECB$. This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.

[hide]
We use casework on the first place student.

It is impossible for $ A$ to be first, as this would agree with Prediction 1.

If $ B$ is first, then $ (CB)$ could not have finished consecutively.  So the possible pairs from Prediction 2 are $ (DA)$, $ (AE)$, and $ (EC)$.  Only $ (DA)$ and $ (EC)$ are disjoint.  We cannot start with $ BEC$, as this would place $ C$ as predicted in Prediction 1.  So our order is $ BDAEC$, which has no agreement with Prediction 2.  So $ B$ cannot be first.

If $ C$ is first, then the possible pairs from Prediction 2 are $ (DA)$, $ (AE)$, and $ (CB)$.  We cannot use $ (CB)$ because it would place $ B$ as predicted by Prediction 1.  This leaves only $ (DA)$ and $ (AE)$, which are not disjoint, so it is impossible for $ C$ to be first.

If $ D$ is first, we can either use $ (DA)$ or not.  If we do, then we must choose either $ (EC)$ or $ (CB)$ as our second pair.  If we use $ (EC)$, we cannot start $ DAEC$, as this agrees in four places with Prediction 2.  So we have $ DABEC$, which shares $ (AB)$ with Prediction 1, so it is impossible.  If we use $ (CB)$, we cannot start $ DACB$, as $ C$ agrees with Prediction 1.  So we are left with $ DAECB$, which agrees in five places with Prediction 2.

So we cannot use $ (DA)$.  The possible pairs from Prediction 2 are $ (AE)$, $ (EC)$, and $ (CB)$.  Only $ (AE)$ and $ (CB)$ are disjoint.  We cannot place $ (AE)$ right after $ D$, as $ (DA)$ is forbidden.  So we must have $ DCBAE$, in which $ E$ agrees with Prediction 1.  So $ D$ cannot be first.

Thus, $ E$ is first.  The possible pairs from Prediction 2 are $ (DA)$, $ (EC)$, and $ (CB)$. If we use $ (EC)$, then we must also use $ (DA)$.  But $ ECBDA$ and $ ECDAB$ have only zero and one correspondences with Prediction 2, respectively.  So we cannot use $ (EC)$

If we don't use $ (EC)$, then we must have $ EDACB$ (preventing $ (EC)$ by not placing $ (CB)$ right after $ E$).  This clearly satisfies all requirements, so it is the desired order.
[/hide]

Wouldn't that order be wrong, though? B and C finish consecutively, as do D and E.

Or is it that they must finish consecutively in the order that was predicted? So A can't be right before B, but B can be right before A?

[quote=sunjae62700]Why there are no pure combinatorics questions in the early IMOs?[/quote]

Do you want to mean counting??


1,Suppose $ A,B\in F^{n\times n},AB\equal{}BA\equal{}\mathbf{0},rank(A^2)\equal{}rank(A)$,show that $ rank(A\plus{}B)\equal{}rank(A)\plus{}rank(B)$.
2,Suppose $ A,B\in F^{n\times n},AB\equal{}BA\equal{}\mathbf{0}$.Prove that,there exists a positive integer $ k$ such that $ rank(A^k\plus{}B^k)\equal{}rank(A^k)\plus{}rank(B^k)$.
3,Suppose $ A\in F^{m\times n},B\in F^{n\times m}$.Prove: $ rank(AB)\equal{}rank(A)$ if and only if there exists $ C\in F^{m\times n}$ such that $ A\equal{}ABC$. Furthermore ,Prove that if $ (AB)^2\equal{}AB$ then $ (BA)^2\equal{}BA$.

I found these problems difficult for me. Any help will be appreciated.

I found on CMJ,  David Calan proved the following theorem:

If $ A$ and $ B$ are two matrices. Then
\[ \text{Rank}(A) \plus{} \text{Rank} (B) \equal{} \text{Rank}(A \plus{} B)
\]
if and only if $ C(A)\cap C(B) \equal{} \{0\}$ and $ R(A)\cap R(B) \equal{} \{0\}$, where $ C(X)$ and $ R(X)$ mean the column and row space of $ X$ respectively.

The proof is quite elementary (I'll post the proof if anybody need it. :) ).

This theorem solves the first problem since form $ AB \equal{} 0$ implies $ C(B) \subset \text{Ker}(A)$ and $ \text{Rank}(A^2) \equal{} \text{Rank}(A)$ implies $ C(A) \cap \text{Ker}(A) \equal{} \{0\}$ thus  $ C(A)\cap C(B) \equal{} \{0\}$. Similarly $ R(A)\cap R(B) \equal{} \{0\}$.

Wich issue of  CMJ did you find the result of David Calan ? thank

For those who have access to the jstor: [url=http://www.jstor.org/pss/2687710]When is rank additive?[/url]

For 2/

There exist $ q$ positive integer such that

$ Im(A^k)\equal{}Im(A^q)$ for any $ k\geq q$

We have also $ Im(A^q) \cap ker(A^q) \equal{} 0$

Since A,B commute we have $ A^qB^q\equal{}0$

$ Im(B^q)$ is included in $ ker(A^q)$

We obtain

$ Im(A^q) \cap Im(B^q)$ is included in $ Im(A^q) \cap ker(A^q) \equal{} 0$

So $ C(A^q) \cap C(B^q) \equal{} 0$

to finish apply the result of beginner

$ rank(A^q\plus{}B^q)\equal{}rank(A^q)  \plus{} rank(B^q)$

can some one post a proof of beginner theorem


Greetings! I was the fourth deadly SIN at the Glasgow IMO.
I am now a conscript. *sob*
Sadly, this happens to all male singaporean citizens who aren't exempted on medical grounds. It is a sad fate. So while my foreign peers are out there in university, I'm languishing in an undisclosed location, somewhere within 20km of the coast. If you look at the map of Singapore, you will realise that this is not saying much.

The reason why I am online the whole of this sunday is really because I'm on manning duty, which means that I'm sitting at my office computer waiting for a phone call that I hope will never come. (This is ******. ****** is activated. ****** will be at ******)

My path to the IMO:
2001: Lose tie-breaker, get sent to SEAMEO-MO instead, first (but vietnam wasn't there)

2002: Win tiebreaker (notice the pattern?). Get silver. Become extremely depressed a few months later upon getting conscripted. (and upon the realisation of the fact that I'll probably never have so much fun)

IMO2002

Zachary Leung (no longer a citizen - who'd want to be at our age? Ans. Females)
21 bronze
Calvin Lin (now a conscript)
10
Meng Dazhe (not a citizen)
20 bronze
Kiat Chuan (conscript)
24 silver
Julius Poh (conscript)
25 silver
Colin Tan (medically exempted.)
12 HM

Note that 3/6 are exempt from conscription. Clearly I have been less than cunning.

We were 30th. I would like to point out that there was only 1 team <20 pts below us, whereas there were 6 teams within 10 pts (more) of us, and 12 within 20.

The only thing I can say to this is - darn!

But this year our team's 18th! Makes an old fogey like me feel dull.
*wave flag*


From:   http://www.xxhh.net

我是一个贼传统的男人（当时是传统男孩），上大学之前我都不曾离开过农村，即使上了大学，第一个学期还是象一个乡村少年，不识得任何都市风情！记得当时大部分同学对某些人开学不到两个星期就开始同居见怪不怪，但这事叫我惊奇了好长一段时间，因为我和gf拍了半年的时候还没有牵过手。
开始进入正题，在上大学之前，我当然从未听说过卫生巾，因为我们高中是没人用那玩意的，再说了，就是真的有人用，我也不知道啊，不过当时的确很少人用。上了大学，这个词我听的多了起来，但我并不了解它的真正用途，所以对它也没有什么特别的；在我的脑袋里头，卫生巾就是高级的卫生纸，是有钱人用的卫生纸，我没钱，当然不去买啦，但是，说实话，我当时还是很向往有一天可以用上卫生巾地。

　在这种向往持续了大约一个学期之后，我终于等到机会了。大一的寒假，我在广州读大学的gf留在学校赚钱，当时拍拖不久，有不经常见面，当然十分想念，于是，我决定到广州“探亲“；在借到学生证、买好车票后，万分激动的我突然记起一件事：卫生巾；当时我暗地里想：广州消费水平多高啊，那里的人应该人人都是用卫生巾的吧，要是我带了一卷卫生纸去那也太没自尊了，而且，也给我们家乡的形象抹灰啊，不行，不管它有多贵，我一定要买一卷（那时我认为卫生巾也是一卷一卷的），让广州的小子看看，我那地方一样都用卫生巾。主意拿定，我立即就去了学校的小卖部，笔蹦抢锉咧挥?、5个女生买零食，正在那里叽叽喳喳，我有点失望，我可是第一次买这么高档的生活用品啊，当然想多欣赏一下别人羡慕的目光，不过算了，反正主要是展示给那些广州人看的，于是我就指着包装最华丽的卫生巾对售货的小姐说：给我拿一卷那种卫生巾（为了耍酷，我这次买东西没问价钱），旁边叽叽喳喳的女生立马静了下来，齐刷刷的把头转向我，售货的小姐也做了同样的动作，我心里那个得意啊，心里想：原来耍酷这么简单的！为了酷到底，我故意抬头挺胸不看她们，只盯着那包卫生巾；大约5秒钟以后，那小姐终于缓过劲来，稍有些脸红的说：13块6，我吓了一跳：这卫生纸果然够高级，竟然如此之贵，心里有点后悔，但是，为了一个字：酷，我一狠心就把钱拿出来了,然后我自豪的迈这大步走了出去；路上人纷纷侧目，我就更自豪了。快到寝室的时候，我把那宝贝藏了起来，怕寝室的那帮兄弟给我瓜分了。
　　火车上我和三个在我这城市上学的广州学生坐在一起，两男一女，其中有一对情侣；大家在路上玩的很开心；本来，我是不准备用那宝贵的卫生巾地，但是，火车里暖气太足了，感觉脸上有点粘粘的，挺不好受地；这时，同坐的几个拿出了纸巾，我想，反正那包上面写的有12片嘛，用一片无妨，不过想到一片一块多，的确心疼啊，但是，当时顾不上那么多了，于是，我装做很随意地样子拿出了那包用来撑门面地卫生巾；立时，那两男一女都瞪大眼睛看着我，我心里想：你丫傻了吧你，我用的纸比你的高级；然后，我开始优雅地想撕开口，但遭遇小小尴尬--撕不开，这当然难不倒我，我拿出指甲剪，剪了一个小口出来，然后用力顺那小口一撕，大功告成了！于是我拿了一个出来，不由的暗自赞叹：这么厚，这么软，怪不得这么贵啊；由于之前我没用过这玩意，但又不能细心研究（那样就证明我是第一次用这种高级玩意的，其实当时很奇怪为什么一面有双面胶贴地），我用没有胶贴的那面擦了擦脸和嘴;然后，我想，有胶贴的那面是让使用者方便存放吧，我很为自己的聪明骄傲，因此，我就大方的把它贴在车壁上 ，心想，正好靠窗，爽。
　在他们惊奇和羡慕的眼光里，我第一次来到了广州，按一向独立，这次也不例外，不用gf接，我自己就来到她寝室楼门口，不过这寝室男生免进，好在看寝室的大姐是我老乡，她好心地把我gf叫了下来，然后，gf就叫了许多好朋友下来分享我带的家乡特产，然后，就在看寝室大姐的小房间里，我开始大包小包的往外拿东西，当我拿到那包卫生巾时，我故意把它放到一个显眼的地方，gf大惊：你买这个干什么？我答到，拿来用啊，众人皆爆笑……


topology
function
advanced fields
advanced fields unsolved

1. Prove that in $2^{nd}$ countable space a non countable set has an accumulation point.
2. Prove that a space with a closed basis is completely regular.

Thanks!

[b]edit[/b]: Corrected the statement of the 1st problem.

What's the first question really like? :) As it is, it's clearly false: take $\mathbb R$ with the usual topology. It's second countable, and there are lots of uncountable closed sets.

I'm really sorry grobber.  :blush:

And now I see it's pretty easy..  :oops:
[hide]
Let [tex]<X,\tau>[/tex] be [tex]2^{nd}[/tex] countable topological space. Let [tex]A \subseteq X[/tex] s.t. [tex]|A| > \aleph_{0}[/tex]. Let's assume there is no accumulation point of [tex]A[/tex]. Then for any [tex]p \in A[/tex] we have an open set [tex]G_{p}[/tex] about [tex]p[/tex] s.t. it does not contain any other points from [tex]A[/tex]. Let's define [tex]f: P(A) \rightarrow \tau[/tex] like this:

$f(I)= \bigcup_{\mu \in I} G_{\mu}$

Let $I_1 \neq I_2$ be subsets of $A$. Then WLOG, there is $\mu \in I_1$ s.t. $\mu \notin I_2$. This clearly means that $\mu \in f(I_1)$ but $\mu \notin f(I_2)$. Therefore, $f(I_1) \neq f(I_2)$. So $f$ is an injective map, therefore $|\tau| \geq 2^{|A|} > 2^{\aleph_0}$. But $2^{nd}$ countable space can't have topology of cardinality above $2^{\aleph_0}$! Therefore we got our contradiction.
$\blacksquare$
[/hide]

You can prove a lot more: every isolated point of the set is the only point of the set in some properly chosen basis element. This means that we can establish an injection from the set of isolated points of our set to the basis. If the basis is countable, this implies that every set has countably many isolated points, and the first question is weaker than this.

As for the second one, what do you mean by closed basis? :? Usually, bases of open sets are considered. Do you mean here that you have some open sets which form a basis for the topology and are also closed?

I mean that all the members of the basis are closed and open at once.

grobber, I don't understand how are you building your injection from the set of isolated points to the members of a countable basis.. Can you explain please?

For every isolated point you consider a basis element which contains only that point (there must be such a basis element, by the definition of an isolated point).

As for the other question, if $x$ is a point and $F\supset x$ is a closed set not containing $x$, then take any basis element $B$ around $x$ and disjoint from $F$, and define the function you're looking for to be $0$ in $B$ and $1$ in the complementary of $B$. It's easy to check that's it's continuous.


geometry parallelogram trigonometry quadratics trapezoid

[hide][quote="FieryHydra"][quote="TRAN THAI HUNG"]Let ABCD is a quarilateral , $\ S_{ABCD}=8$, $\ AB+AC+CD=8$.Prove that ABCD is a parallelogram and find its diagonal :lol:[/quote]
In the solution below, we use the idea that $ax^{2}+bx+c \geq 0$ for all $x \in \mathbb{R}$ iff $a \geq 0$ and $b^{2}-4ac \leq 0 .$

Let $\angle BAC = \alpha$ and $\angle ACD = \beta.$ Then we have the relation $\frac12 (AB)(AC) (\sin \alpha)+\frac12 (CD)(AC)(\sin \beta) = S_{ABCD}= 8$

$\Rightarrow (\sin \beta ){AC}^{2}+[\sin \beta (AB-8)-AB \sin \alpha] AC+16 =0$ ... (I) (quadratic in $AC$ with real roots.)

$\Rightarrow{[(\sin \beta (AB-8)-AB \sin \alpha)]}^{2}-4 \cdot 16 \cdot \sin \beta \geq 0$

$\Rightarrow{(\sin \beta-\sin \alpha)}^{2}{AB}^{2}-16 \sin \beta (\sin \beta-\sin \alpha) AB+64({\sin}^{2}\beta-\sin \beta ) \geq 0$

Now using the result noted in the beginning, we obtain

${16}^{2}{\sin}^{2}\beta{(\sin \beta-\sin \alpha)}^{2}-4{(\sin \beta-\sin \alpha)}^{2}\cdot 64 ({\sin}^{2}\beta-\sin \beta) \leq 0$

$\Rightarrow{(\sin \beta-\sin \alpha)}^{2}\sin \beta \leq 0$

$\because \sin \beta \neq 0,$ we must have $\sin \beta = \sin \alpha$ ... (II)

Plugging this back into equation (I), we get $AC(8-AC) \sin \beta =16.$ But note that $AC$ and $AC-8$ are both positive, and so applying $AM-GM,$ we get $AC(8-AC) \leq 16 \Rightarrow \sin \beta \geq 1 \Rightarrow \beta = 90^{\circ}= \alpha.$

Thus, we have $AC(8-AC) = 16 \Rightarrow{(AC-4)}^{2}= 0 \Rightarrow AC=4$ (the length of the diagonal.)

Can someone finish this off by proving that $ABCD$ is really a parallelogram? :)[/quote][/hide]
You can use the another way for the easier.I will post the solution later :wink:

[quote="delta"][quote="FieryHydra"]
Let $\angle BAC = \alpha$ and $\angle ACD = \beta.$ Then we have the relation $\frac12 (AB)(AC) (\sin \alpha)+\frac12 (CD)(AC)(\sin \beta) = S_{ABCD}= 8$
[/quote]
From this $8=\frac{AB \sin \alpha+CD \sin \beta}{2}AC \le \frac{AB+CD}{2}AC = \frac{(8-AC)AC}{2}\le 8$ and the equality holds only when $AC=4$ and $\alpha = \beta =90^\circ$
so [b]ABCD is trapezoid[/b]. Consider the case $AB=3$, $CD=1$. Hope I'm not wrong but it satisfies all conditions, so its impossible to prove that ABCD is paralelogram.  :maybe:[/quote]
Ahh! I should have seen that! :) Thanks!

Yeah, it seems, $ABCD$ need not be a parallelogram after all.

Sorry,that my mistake it's a trapezoid :blush:  :wallbash_red:


Let $n$ be a positive integer which has more than $k$ different prime factors. Prove that there exists a convex $n$-gon in the plane such that

[b]1.[/b] all angles of the $n$-gon are equal;
[b]2.[/b] the lengths of the sides of the $n$-gon are the numbers $1^{k}$, $2^{k}$, ..., $n^{k}$ in some order.

  Darij

As I remember there is a generalisation with $P(x)=\sum_{i=1}^{k}a_{i}x^{i}$
and the polygon with sides $P(1),P(2),...,P(n)$
but the proof is almost the same :wink:


For a positive integer $n$, let $r(n)$ denote the sum of the remainders when $n$ is divided by $1, 2, \ldots , n$ respectively. Prove that $r(k) = r(k-1)$ for
infinitely many positive integers $k$.

I was just kind of screwing around with this problem and I guessed

[hide]$r(2^{k})=r(2^{k}-1)$[/hide]

which I don't know is right or not, and if it is I don't know how to prove it... :maybe:

This problem has been posted at least five times on this forum:
http://www.mathlinks.ro/viewtopic.php?t=125625
http://www.mathlinks.ro/viewtopic.php?t=108343
http://www.mathlinks.ro/viewtopic.php?t=37133
http://www.mathlinks.ro/viewtopic.php?t=5866

A follow up question: Prove that $ \frac{n^2}{10}<r(n)<\frac{n^2}{4}$ for all integers $ n\ge7$.


Let $ a,$ $ b,$ $ c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca>0.$ If $ k \ge \frac{64}{105},$ then
$ a\plus{}b\plus{}c \ge \sqrt{\frac{a^2\plus{}k}{1\plus{}k}}\plus{}\sqrt{\frac{b^2\plus{}k}{1\plus{}k}}\plus{}\sqrt{\frac{c^2\plus{}k}{1\plus{}k}}.$

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca > 0.$ If $ k \ge \frac {64}{105},$ then
$ a \plus{} b \plus{} c \ge \sqrt {\frac {a^2 \plus{} k}{1 \plus{} k}} \plus{} \sqrt {\frac {b^2 \plus{} k}{1 \plus{} k}} \plus{} \sqrt {\frac {c^2 \plus{} k}{1 \plus{} k}}.$[/quote]

$ k\geq 2$

$ (a \plus{} b \plus{} c)^2 \ge 3\sum {\frac {a^2 \plus{} k\frac {(ab \plus{} bc \plus{} ca)^2}{(a \plus{} b \plus{} c)^2}}{1 \plus{} k}}$.


after my failed last tourny... :(  i've decided to make another one
hopefully one that'll do better
anyway, here are the rules

this will be a 16-player tourny, open to all players
when you sign up, plz give me your highest rating, as seeding will be done by this
It will be a normal game, 10 questions, 12 seconds, time based
whether it's rated or non-rated, will be up to the players

plz sign up here and i wish all participants good luck!!!

pinkpiglet signs up!!!! highest: 1323 :blush:

[quote="mewto55555"]WHEN WILL THE MADNESS END???? JUST SO YOU KNOW I, MEWTO55555, WILL BOYCOTT [b][u]ALL FUTURE TOURNAMENTS[/b][/u] UNTIL THE EXCESS AMOUNT OF TOURNAMENTS ENDS.[/quote]

....didnt u start the first tourney...

[quote="ernie"]Please ignore this post.[/quote]

u just [i]had[/i] to spam in my topic didn't u?
well, i'll do that 2 urs to :D
MHAHAHAHAHA (i can't find an evil scientist's face...)

OK
this is the bracket

ROUND 1
game 1: person 1 vs. person 16
game 2: person 2 vs. person 15
game 3: person 3 vs. person 14
game 4: person 4 vs. person 13
game 5: person 5 vs. person 12
game 6: person 6 vs. person 11
game 7: person 7 vs. person 10
game 8: person 8 vs. person 9

WINNER'S BRACKET
ROUND 2
game 9: winner of game 1 vs. winner of game 8
game 10: winner of game 2 vs. winner of game 7
game 11: winner of game 3 vs. winner of game 6
game 12: winner of game 4 vs. winner of game 5

ROUND 3
game 13: winner of game 9 vs. winner of game 12
game 14: winner of game 10 vs. winner of game 11

ROUND 4
game 15: winner of game 13 vs. winner of game 14

LOSER'S BRACKET
ROUND 2
game 16: loser of game 1 vs. loser of game 8
game 17: loser of game 2 vs. loser of game 7
game 18: loser of game 3 vs. loser of game 6
game 19: loser of game 4 vs. loser of game 5

ROUND 2b
game 20: winner of game 16 vs. loser of game 9
game 21: winner of game 17 vs. loser of game 10
game 22: winner of game 18 vs. loser of game 11
game 23: winner of game 19 vs. loser of game 12

ROUND 3
game 24: winner of game 20 vs. winner of game 23
game 25: winner of game 21 vs. winner of game 22

ROUND 3b
game 26: winner of game 24 vs. loser of game 13
game 27: winner of game 25 vs. loser of game 14

ROUND 4
game 28: winner of game 26 vs. winner of game 27

ROUND 4b
game 29: winner of game 28 vs. loser of game 15

CHAMPIONSHIP ROUND
game 30: winner of game 15 vs. winner of game 29

ooh i wanna join
TOURNEY FEVER :O

highest 1404

ok,
soooo, pplz participating so far

#1. Xoangieexo…….1404
#2. Johnnybeard……1367
#3. Mathblitz……….1357
#4. Pinkpiglet……….1323

oops, right
srry sicarius

#1. Xoangieexo…….1404
#2. Sicarius1029……1369
#3. Johnnybeard……1367
#4. Mathblitz……….1357
#5. Pinkpiglet……….1323

we just need 11 more pplz to get started!!
tell all ur friends!!!!

changes in bracket
winners will be decided by the coin toss

winner's bracket round 3 and loser's bracket round 2 must be completed july 30th
next deadline will be after the weekend, during that time, i will not be here because i'm going to acadia(not sure if that's how u spell it)
i will give out instructions and pms on July 31st, before i leave

ROUND 1.
#5. AIME_is_hard……...1375 vs #9. Mathblitz(winner due to coin toss)………..….1357 (game 1)
#6. AIME15 (winner)……....………1370 vs#10. Pinkpiglet…………1323 (game 2)
#7. Sicarius1029(winner due to ducky's FTW failing)……..…1369 vs#11. Myyellowducky82...1200 (game 3)
#8. Johnnybeard(winner due to coin toss)……..…1367 vs#12. Pacman2812……….1200 (game 4)

WINNER'S BRACKET
ROUND 2.
Kl2836 vs. Johnnybeard(winner due to kl2836 not responding) (game 5)
Budi713 vs. Sicarius1029(winner due to coin toss) (game 6)
Xoangieexo vs. AIME15(winner) (game 7)
Bisbmard9(winner due to coin toss) vs. Mathblitz (game 8)

ROUND 3
[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b](game 9)
[b]Sicarius1029 [/b]vs. [b]AIME15[/b](game 10)

ROUND 4
winner of game 9 vs. winner of game 10 (game 11)

LOSER'S BRACKET
ROUND 2
[b]AIME_is_hard [/b]vs. [b]Mathblitz [/b](game 12)
Pinkpiglet(winner) vs. xoangieexo (game 13)
[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b](game 14)
[b]pacman2812 [/b]vs. [b]Kl2836 [/b](game 15)

ROUND 2a
winner of game 15 vs. winner of game 14 (game 16)
winner of game 12 vs. [b]Pinkpiglet [/b](game 17)

ROUND 3.
winner of game 16 vs. loser of game 9 (game 18)
winner of game 17 vs. loser of game 10 (game 19)

ROUND 3a
winner of game 18 vs. winner of game 19 (game 20)

ROUND 4.
winner of game 20 vs. loser of game 11 (game 21)

CHAMPIONSHIP ROUND

WINNER OF GAME 21 vs. WINNER OF GAME 11

IF THE WINNER OF THE LOSER'S BRACKET WINS THE CHAMPIONSHIP ROUND, THEN THE ROUND MUST BE PLAYED AGAIN

[b][i][u]games that can be played right now[/u][/i][/b]
[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b] (bisbmard9 has responded, if the game does not take place for any reason, and johnnybeard does not respond, then bisbmard wins)
[b]Sicarius1029 [/b]vs. [b]AIME15[/b] (sicarius1029 has responded, if the game does not take place for any reason, and AIME15 does not respond, then sicarius1029 wins)
[b]AIME_is_hard [/b]vs. [b]Mathblitz [/b] (BTW, i haven't seen AIME_is_hard on in a long time, so if this game doesn't get completed by the deadline, mathblitz wins)
[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b]
[b]pacman2812 [/b]vs. [b]Kl2836 [/b]

i will send out reminders tomorrow as well, when i get on AoPS on july 30, any games not played will be decided by the above criteria and the coin toss

Vallon- if you do a coin toss, wouldn't that just defeat the whole purpose of an FTW tourney?  Unless you're doing a coin-tossing tourney...  :rotfl:

[quote="vallon22"]ya, to a certain extent, but i'm just trying to get the tourney to keep moving, that way, the people actually participating can play new pplz
anyway, i have told pplz that if they respond, but their opponent doesn't, they get the win, so....
it's not completely coin toss
the people that play move on, the people that don't play lose[/quote]

But that [b]completely[/b] would defeat the purpose of any math tourney.  This is (no offense) like a probablity/social reply tourney.

well, i just wanna get this tourney moving
hopefullly, some pplz will actually play

on that note, i have postponed the deadline for the games in winner's bracket round 3 and loser's bracket round 2 until August 1

good luck with your games

Bah, I need to prepare to a selection test and make homeworks. No time to FTW

kl(something) wins.

ok
pacman forfeits so....

ROUND 1.
#5. AIME_is_hard……...1375 vs #9. Mathblitz(winner due to coin toss)………..….1357 (game 1)
#6. AIME15 (winner)……....………1370 vs#10. Pinkpiglet…………1323 (game 2)
#7. Sicarius1029(winner due to ducky's FTW failing)……..…1369 vs#11. Myyellowducky82...1200 (game 3)
#8. Johnnybeard(winner due to coin toss)……..…1367 vs#12. Pacman2812……….1200 (game 4)

WINNER'S BRACKET
ROUND 2.
Kl2836 vs. Johnnybeard(winner due to kl2836 not responding) (game 5)
Budi713 vs. Sicarius1029(winner due to coin toss) (game 6)
Xoangieexo vs. AIME15(winner) (game 7)
Bisbmard9(winner due to coin toss) vs. Mathblitz (game 8)

ROUND 3
[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b](game 9)
[b]Sicarius1029 [/b]vs. [b]AIME15[/b](game 10)

ROUND 4
winner of game 9 vs. winner of game 10 (game 11)

LOSER'S BRACKET
ROUND 2
[b]AIME_is_hard [/b]vs. [b]Mathblitz [/b](game 12)
Pinkpiglet(winner) vs. xoangieexo (game 13)
[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b](game 14)
pacman2812 vs. Kl2836(winner) (game 15)

ROUND 2a
[b]Kl2836 [/b]vs. winner of game 14 (game 16)
winner of game 12 vs. [b]Pinkpiglet [/b](game 17)

ROUND 3.
winner of game 16 vs. loser of game 9 (game 18)
winner of game 17 vs. loser of game 10 (game 19)

ROUND 3a
winner of game 18 vs. winner of game 19 (game 20)

ROUND 4.
winner of game 20 vs. loser of game 11 (game 21)

CHAMPIONSHIP ROUND

WINNER OF GAME 21 vs. WINNER OF GAME 11

IF THE WINNER OF THE LOSER'S BRACKET WINS THE CHAMPIONSHIP ROUND, THEN THE ROUND MUST BE PLAYED AGAIN

[u][i][b]games that can be played right now [/b][/i][/u]
[b]Johnnybeard [/b]vs. [b]Bisbmard9 [/b](bisbmard9 has responded, if the game does not take place for any reason, and johnnybeard does not respond, then bisbmard wins)
[b]Sicarius1029 [/b]vs. [b]AIME15 [/b](sicarius1029 has responded, if the game does not take place for any reason, and AIME15 does not respond, then sicarius1029 wins)
[b]AIME_is_hard[/b] vs. [b]Mathblitz [/b](BTW, i haven't seen AIME_is_hard on in a long time, so if this game doesn't get completed by the deadline, mathblitz wins)
[b]Myyellowducky82 [/b]vs. [b]Budi713 [/b]

[quote="myyellowducky82"]I forfeit. (spelling?)[/quote]

Spelling is correct. :wink:

YAY!  My 800th post!

And vallon, no offense, but this tourney is in serious need of cancelling.

ya, ur prob right
but right now, hardly any tournies are up and going
hmm...
i'll think of something l8r

THIS TOURNEY IS OFFICIALLY CANCELLED  :(  :(  :(

since its cancelled, admins (no mods for this forum) please lock this thread?

admins hardly come in here
the easiest way is for pplz to just stop posting
dang, [i]i'm[/i] posting....

I'LL JOIN.THOUGH I'LL LOSE IN THE FIRST ROUND. HIGHEST RATING 1321

It's canceled, Andrew. At least read the post right before you.


trigonometry
geometry
trig identities
Law of Sines
geometry unsolved

I need help, this is the only topic I am really struggling at.
This may seem rude as its my first post, but if anyone is kind enough! :)

--------

[img]http://i14.tinypic.com/2heatyd.jpg[/img]

Show that, for triangle RST, t = r sin (alpha + beta) [b]/[/b] sin alpha

iv tried using the sine rule,
t [b]/[/b] sin(180 - alpha - beta)   [b]=[/b]   r [b]/[/b] sin alpha
t = r / sin alpha [b]x[/b] sin(180 - alpha - beta)

After that, i get confused because usually i would expand the sin bracket. ie:
[b]sin(45+45) = sin 45 cos 45 + cos 45 sin 45[/b]

but in this case, i have 180, alpha, and beta in one bracket.......

Am i even taking the right steps?
Please help.

The angle measure of $T=180-(a+b)$

By law of sines,:
$\frac{r}{\sin a}=\frac{t}{\sin (180-(a+b))}$

For any number $x$, $\sin x = \sin (180-x)$

So you get:
$\frac{r}{\sin a}=\frac{t}{\sin (a+b)}$
$\implies \boxed{t=\frac{r\sin (a+b)}{\sin a}}$


Please help me. I'm struggling over math.

Kindly show the complete solutions.. THANKS ;]

ok we can rewrite n(n) as $ n^2$. it is a well-known theorem that the minimum of a parabolic function (or maximum of a negative one) comes when $ x\equal{}\minus{}\frac{b}{2a}$ which in this case is $ n\equal{}\boxed{600}$ plugging that in, we get $ c\equal{}\boxed{600}$ as well. coooooollll....

We see this by completing the square (never liked formulas anyway :) ):
\begin{align*}
c &= n^2 - 120n + 4200 \\
&= n^2 - 120n + 3600 + 600 \\
&= (n - 60)^2 + 600.
\end{align*}
Since something squared is always nonnegative, we easily see that the minimum value of c is 600, since the smallest $ (n-60)^2$ can be is 0.


search
combinatorics unsolved
combinatorics

A palindrome is a number or word that is the same when read forward and backward, for example, "176671" and "civic." Can the number obtained by writing the number from 1 to $ n$ in order ($ n>1$) be a palindrome?

This is Russia 1996.  It has appeared many times on the forum, and it's easy enough to find with a search for the word "palindrome."
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128508
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19872


geometry
circumcircle
induction
extremal principle
combinatorics unsolved
combinatorics

[hide="Hint"]Take a circle that covers the polygon and try to make it get smaller until it becomes a circumcircle of some triangle.[/hide]

I don't think that hint will work for me. I mean.... how will you show that that circle will always become some triangle's circumcircle?

i would say try to find a triangle (general) whose circumcircle covers the n gon and then try to create more triangles

[quote="aakansh92"]Let $ A_1A_2...A_n$ be a convex polygon. Show that there exists an index $ j$ such that the circum-circle of the triangle $ A_j A_{j \plus{} 1} A_{j \plus{} 2}$ covers the polygon (here indices are read modulo n).[/quote]

[hide="Hint to the solution"]
Prove that the biggest circle possible through any 3 vertices is
i) :arrow:  covering the polygon
ii) :arrow: passing through three consequtive points.

both can be proved by contradiction method. :) [/hide]

Please be more clear and give full solution.

Gaurav, could you do that problem during the postals? If yes, I am eager to look at the solution.

Here's my (not well-written  :oops: ) solution. Hope you understand the idea.

[hide]I don't know if the following requires proof. :|
[b]Lemma 1:[/b]
Suppose $ S$ is a circle which contains $ Y,Z$ but doesn't contain $ X$.
Then the portion of the circumference of $ S$ that lies opposite $ X$ wrt $ YZ$ lies outside the circumcircle of $ XYZ$
 [geogebra]8311777ddf6748db7b929f40c8136d68ab2b9775[/geogebra]

Suppose the problem statement is true for all convex $ n$-gon for some $ n\ge 4$
(Base case $ 4$ is easy to prove)
Assume there exists a convex $ n \plus{} 1$-gon $ A_1,A_2,\cdots A_{n \plus{} 1}$ such that the statement isn't true.

By induction hypothesis on the polygon $ A_1A_2\cdots A_n$, there exists $ A_j,A_{j \plus{} 1},A_{j \plus{} 2}$ such that exactly one vertex of the polygon, $ A_{n \plus{} 1}$ lies outside their circumcircle.

By induction hypothesis on the polygon $ A_1A_2\cdots A_jA_{j \plus{} 2}\cdots A_{n \plus{} 1}$, exactly one vertex of the polygon, $ A_{j \plus{} 1}$ lies outside a circle $ S$ that has on it's circumference 3 consecutive points from the polygon.
 [geogebra]a1b4053b21d62e5195188b71d634b48bdaf91f91[/geogebra]
But none of these three points lie on the same side as  $ A_{j \plus{} 1}$ wrt $ A_jA_{j \plus{} 2}$. But since $ A_{j \plus{} 1}$ lies outside $ S$, by lemma 1 such three points can't lie inside the circumcircle of $ A_jA_{j \plus{} 1}A_{j \plus{} 2}$. So these 3 points must be $ A_j,A_{j \plus{} 2},A_{n \plus{} 1}$. But these are consecutive only if $ n \equal{} 3$. Which isn't true because $ n\ge 4$[/hide]
Do reply if you understand or not. :oops:

Ah... Sorry... When I replied, I was thinking in another problem.

Akashnil, I think you should replace three of your $ A_{j \plus{} 1}$'s by $ A_{j \plus{} 2}$'s. Anyway, what is wrt?

wrt = with relation to, or something similar.

wrt=with respect to.
$ A$ is opposite $ B$ wrt $ CD$ means the points $ A$ and $ B$ lie on oposite sides of the line $ CD$.
Yes I messed up some of $ A_{j\plus{}1}$ and $ A_{j\plus{}2}$s . Sorry. I've edited now. Thanks.

And once again do reply if you understand it now or not.

[quote="aakansh92"]Please be more clear and give full solution.

[/quote]
Sorry for incomplete post, actually i dont like Latexing much. anyway [hide="hare goes the proof"]
Among the finitely many circles through any 3 vertices, there is[b] [u]a maximal one, denote it by $ \tau$[/u][/b].

as said earlier we prove:
1.  :arrow: $ \tau$ covers n-gon.
2.  :arrow: $ \tau$ passes through 3 consequtive vertices.

$ 1.$ Suppose point A' lies outside the maximal circle about $ \triangle ABC$ where $ A,B,C,A'$
are vertices of polygon so that they form a convex quadrilateral. Then circumcircle of $ \triangle A'BC$ has larger radius than $ \triangle ABC$. :!: CONTRADICTION. :)
$ 2.$ Let $ A_i,A_j,A_k$ be vertices on the MAXIMAL circle. Let $ A_l$ lie between $ A_j \& A_k$, not on $ \tau$. Because of (1), it lies inside $ \tau$, but then circle about $ \triangle A_jA_lA_k$ is larger than maximal circumcircle. :!:  CONTRADICTION. :)  [/hide].
please verify the solution.

[quote="Akashnil"]I don't know if the following requires proof. :|
[b]Lemma 1:[/b]
Suppose $ S$ is a circle which contains $ Y,Z$ but doesn't contain $ X$.
Then the portion of the circumference of $ S$ that lies opposite $ X$ wrt $ YZ$ lies outside the circumcircle of $ XYZ$
 [/quote]
P.S.-
above lemma can be easily proved since two circles(not identical) can intersect at only two points.
P.P.S.- AKashnil, your proof is really very good. :coolspeak:

those two solutions are god damn BEAUTIFUL
hats off to you both ith_power and Akashnil
my solution amounted to atleast 5 whole written pages in Ideas but yours(both) is just beauty

It is very unfortunate that this problem was taken from the very well known Arthur Engel book's Extremal principle chapter. Infact it is an example problem. Too bad of the problem setters in India.
Anyway, akashnil's solution is superb as it is a different one .
ith_power ur solution is superb as u have used the extremal principle well. I also solved it by that way and even in the book they solved it in the same way.

Moderators if this is an unnecessary post sorry but just wanted to say that it was copied thats all.


calculus
integration
floor function
calculus computations

Here's another one that was interesting:

Determine whether the improper integral

\[ \int_0^\infty (\minus{}1)^{\lfloor x^2 \rfloor}dx
\]

converges or diverges.

[hide]The question is equivalent to the convergence of the series $ \sum_{k \equal{} 0}^{\infty} ( \minus{} 1)^k(\sqrt {k \plus{} 1} \minus{} \sqrt {k})$ (here $ k$ replaces $ \lfloor x^2\rfloor$. The series converges by the alternating series test.[/hide]


I want to make my reference goes like this:

Amacc, D and JXIK, L. 2009, paper compilation, journal of xx, 202-241

Bxxx, L and  wxx, D. 2008, xxxx, .

Cxxx, ..

Basically, I want the references ordered by the alphabetical order of authors' last name.(.Amacc/0 is last name..and then first name.

which bibliography style should I use with \usepackage{natbib} ?

I download the APA from CTAN..But when I tried to install it to my TexnicCenter, it fails..No ideas what I should do now..pls help me with the procedure of downloading new bibliography style.

[quote="weng"]I download the APA from CTAN..But when I tried to install it to my TexnicCenter, it fails..No ideas what I should do now..pls help me with the procedure of downloading new bibliography style.[/quote]This makes no sense. TexnicCenter is merely an editor so you can't install anything to it. What is "the APA"? What does "it fails" mean? MiKTeX already comes with APA like styles in it such as apacite and apalike so there's nothing to install. See MiKTeX, Browse Packages for what is installed.

Also look at the sort-by-letters styles already included in MiKTeX. These are described [url=http://tug.ctan.org/tex-archive/biblio/bibtex/contrib/sort-by-letters/README]here[/url].

when i used \bibliographystyle{apa}

the page of Bibliography just disappear..

I tried to install apa into MikTek....

[quote="weng"]when i used \bibliographystyle{apa}

the page of Bibliography just disappear..[/quote]You should know by now that you must give a minimal example to have any chance of analyzing the problem.
[quote="weng"]I tried to install apa into MikTek....[/quote]Why? It's already installed assuming you have the full version. Again you haven't given any information to work with. What did you do to install it? Didy you use the MiKTeX install procedure in Browse Packages? Did you install it manually? If so, did you refresh the database?

Thank you for clarifying those to me!!!!!
I get the problem fixed..


I pm-ed Paul about ARML but he hasn't replied. Does anyone know who I should contact and what the requirements are?

For Iowa ARML, you're looking for James Kirpes. Email is: kirpes.james@iccsd.k12.ia.us.  Hm...I'm not entirely sure of requirements.  Usually you have to have scored well (placed) in GPML meets.  AMC scores are also factored in.  I've never had too much trouble getting on a team.  If you're a regular on this site, I doubt you will either.

Oh, looks like Darren took care of it. I probably should have just given you the email from the beginning, Penguin. <_<;;

Oh well.

Sorry to call up an old post, but who in here is going to the Iowa City ARML this June?

For those of you that have done it before, how are the three teams divided up?

Well, I'm going.  :lol:

There's a Central Academy team, I'm pretty sure. And then the other two teams are divided according to how Kirpes views their mathematical ability. (based on experience with the team and AMC scores and such)

I think. Darren or Nick, do you know anything?

woohoo! im going. hope to c u guys there   :)

Yeah, you're pretty much right Paul.  CA's got their team.  And then Kirpes divides people based on who's good and who's not.  More or less.

Generally speaking, what is the cutoff for the two teams (based on AMC)? This could be skewed, as many people (like me) who take a lot of time suck at speed-type contests (over 1 hour).

I know from central 3 8th graders are going (me being one of them :P ), 1 or 2 freshmen, and 6ish seniors and juniors.

I don't think there is a specific cutoff. It's mostly qualitative, rather than quantitative. Don't worry! If you guys aren't on the first team this year, then chances are that a decent performance at ARML will put you on the top team. At least, that's what I think is true.

Darren, Allendorf, some help here?


Each page number of a 599-page book is printed one time in the book. The first page is page 1 and the last page is page 599. When printing all of the page numbers, how many more 5's are printed than 8's?

Hide your solutions

Billy

[color=indigo]
[hide]There are 100 more 5's than 8's. First i counted the number of 5's: Between 1-99 there are 20. Between 100-199 there are 20. Same for the 200's, the 300's and the 400's. Then in the 500's there are 120 5's. THat mkes a total of 220 fives. Then the 8's are pretty much the same except they don't get the extra 100 in the 500's so there are 120 8's. :D [/color][/hide]

[hide]If you realize, the 5's have 500-599. 8 dont.
Otherwise, they're the same
There are 100 pages in 500s
>>100<<[/hide]

[hide]From 1-599, 8 shows up 9 * 6 times in the ones spot, 9 * 6 times in the 10s spot and 0 times in the hundreds spot. 5 shows up 9 * 6 times in the ones spot, 9 * 6 times in the 10s spot and 100 times in the hundreds spot, so its 100 more. [/hide]

so why don't you edit your post, or reenter another to fix it?


function
search
complex analysis
complex analysis unsolved

Suppose that $f,g$ are  an analytic functions on $\mathbb C$. Prove that if $f,g$ have no common roots, then there exist analytic functions $A,B$ whcih $Af+Bg=1$.

Leva proved it (on the forum) a while back. I'm sure it's easy to locate using the search function.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46388]The topic grobber was referring to.[/url]


trigonometry
calculus
calculus computations

I need to solve a problem from physics, but first i need to solve these differential equations to find the frequence of two moving rods.
$\begin{array}{l}\left\{ \begin{array}{l}\varphi_{1}= \frac{{\varphi_{0}}}{2}[\cos \omega_{1}t+\cos \omega_{2}t] \\ \varphi_{2}= \frac{{\varphi_{0}}}{2}[\cos \omega_{1}t-\cos \omega_{2}t] \\ \end{array}\right.\\ \end{array}$
where
$\omega_{1}^{2}= \omega_{0}^{2}+\omega^{'2}\\ \omega_{2}^{2}= \omega_{0}^{2}-\omega^{'2}$

Shouldn't this be done using the addition of fazors, in a fazorial graph?

[quote="Nea"]What's fazorial graph? I need to solve the differential equations.[/quote]
Your original post does not contain any differential equations. It is unclear what you are asking.

:blush:
I've erroneously written the previous equations. You arrive to them after some substitutions to these:
$\begin{array}{l}ml^{2}\varphi_{1}^{''}+mgl\varphi_{1}+ca^{2}(\varphi_{1}+\varphi_{2}) = 0 \\ ml^{2}\varphi_{2}^{''}+mgl\varphi_{2}+ca^{2}(\varphi_{1}+\varphi_{2}) = 0 \\ \end{array}$
where:
$\begin{array}{l}\varphi_{1}+\varphi_{2}= \varphi \\ \varphi_{1}-\varphi_{2}= \varphi \\ \end{array}$
Anyway only these equations, that I wrote now,  count. I've arrived at the previous equations because of the initial conditions I had in the problem.

[quote="akech"]I was wondering if you needed $mgl\varphi_{1}'$ as opposed to $mgl\varphi_{1}$.
You wanted to let people know what $m,l,c,a$ are (constants I guess).
I was wondering if something is wrong with the following array because after solving it for $\varphi$ something is not true about $\varphi_{2}$
$\begin{array}{l}\varphi_{1}+\varphi_{2}= \varphi \\ \varphi_{1}-\varphi_{2}= \varphi \\ \end{array}$[/quote]

1. $mgl\varphi_{1}$ is correct not $msgl\varphi_{1}'$
2. Yes $m,l,c,a$ are constants
3. About the "array problem" you mentioned, the two equations are true after approximations.


It seems police practices have gotten out of control in the US with  constant racial profiling always looking to pulling over black people to try and catch them ridin' dirty as well as tasering college students for not producing an ID or getting up on time when they want to use the library:

http://helpychalk.blogspot.com/2006/11/ucla-student-repeatedly-tasered-for-not.html

More taser fun here

http://portland.indymedia.org/en/2006/01/332829.shtml

Tasers have been common now for police to use probably because they are claimed not to be dangerous at all.

How brutal are the police in other countries? Would actions such as these be frowned down upon or are they common?

Our police are known for their alternating acts of heroism and callousness. Oh, and they have never heard of tasers.

Tasers are small weapons that disperse electricity to stun somebody, they are considered a "safe" alternative to standard practices and a good way to disarm people.

In Florida, it's even legal to taser minors.

They're either illegal in some 40 states (USA) or legal in some 40, can't remember which. I know that they are used commonly in prisons to break up fights and stuff. I think as long as they don't get so advanced that they start actually doing damage to people, they are fine, but should by no means be available to the public, even with a liscense (unlike handguns)

The media is out to sell stories, not report the truth.

The ones that have gotten confused about this are no longer in business, as the public won't support a press or something that isn't what is demanded.

Most of the police in the USA, as elsewhere, are people just trying to do the best job they can.  Some are drawn to the job for poor reasons, such as: they like to have power over others, they are part of a gang that use the police power to further gang goals, etc.

Racial profiling is used for one reason; it is effective.  For most purposes it isn't allowed to be used to directly or indirectly lead to a conviction of a criminal.

But let's think about this, what if your loved one was kidnapped by someone who was observed.  Wouldn't you want the police to stop those of the same race/height/weight/gender as the witness identified?  I would, but this is profiling.

If the police are stopping all black men drivers just because they are black, they are not being very productive.  This will also clearly show up in their arrest records, so I think the actual occurrances of this, while real, are perhaps overstated.

Let's be rational, how easy would it be for a defense counsel, looking for any way out for his client, to find out that a traffic officer issued 2 or 3 times as many tickets to purple people, or any other class than exist in the general population or than other officers did.  The feds and ACLU would both love to have such a case.  I don't believe that such things are still occurring.  I will believe that those with the same race as the officer might be treated a bit different, as would an attractive young driver, by an officer who happened to be of the other gender.  Also, what paper wouldn't want to run such a story?

I won't argue that it doesn't occur, but I think the prevelence is far overstated by the press.

A Taser is a less-lethal method of stopping someone.  Wouldn't you want of officer to Tase a kid before he shoots one?  I would.  It should be legal, but just like shooting -- only in appropriate circumstances.

(I understand that a Taser fires two darts that penerate the skin and a battery that powers an electric shock that is very painful, ususally causes the equivalent of a short seizure in a person hit by one and is usually not lethal.)

Our local police force is obtaining them.  Several officers asked to be shot themselves so they would know what they were subjecting people to if they decided to use it on a member of the public.  Admirable.  One was quoting saying that they wouldn't want to do anything they didn't experience themselves.  (I don't think he extended this to shooting.)

I hope this helps.

[quote="wizard2378"]
 Several officers asked to be shot themselves so they would know what they were subjecting people to if they decided to use it on a member of the public.  Admirable.  One was quoting saying that they wouldn't want to do anything they didn't experience themselves.  (I don't think he extended this to shooting.)[/quote]

In my state, i'm pretty sure it's required for police officers to go through the effects of thier various weapons, so that in court they can justify any action they may have taken with the weapon because they've felt it too. This includes tasers and pepper-spray, but of course not guns.

a) What you have provided is very small evidence for making a general statement about police practices in the U.S.

b) Did you even read this part? [quote]Predictably, the coroner says the taser had nothing to do with his death.[/quote]

c) Edit: after looking at it again, I change my mind - I think they did step over reach their boundaries and shouldnt' have gone that far.


[quote="xp"]Καλο Φθινόπωρο σ΄ όλους

Σκαρωσα μια (Για ευκολη την πηγα)

Να αποδειχθεί η ανισότητα

$ (\frac {\sqrt {a^4 \plus{} b^4} \minus{} \sqrt {b^4 \plus{} c^4} \minus{} \sqrt {c^4 \plus{} a^4}}{2})^2 \plus{} \frac {(|ab| \minus{} |bc| \minus{} |ca|)^2}{2}\geq{c^4}$
για καθε $ a,b,c$ στo $ R^{*}$[/quote]

Counterexample: $ (a,b,c)\equal{}(1,0.5,0.45)$

παρε ως προυπόθεση τα α,b,c τετοια ωστε οι βασεις των τετραγωνων να είναι θετικες.

[quote="xp"]Καλο Φθινόπωρο σ΄ όλους

Σκαρωσα μια (Για ευκολη την πηγα)

Να αποδειχθεί η ανισότητα

$ \left[\frac {\sqrt {a^4 \plus{} b^4} \minus{} \sqrt {b^4 \plus{} c^4} \minus{} \sqrt {c^4 \plus{} a^4}}{2}\right]^2 \plus{} \frac {(|ab| \minus{} |bc| \minus{} |ca|)^2}{2}\geq{c^4}$
για καθε $ a,b,c$ στo $ \mathbb{R^{*}}$[/quote]

Θα χρησιμοποιήσουμε την ακόλουθη ανισότητα:

$ \left(a^{p}_{1} \minus{} a^{p}_{2} \minus{} a^{p}_{3}\right)^{\frac {1}{p}} \plus{} \left(b^{p}_{1} \minus{} b^{p}_{2} \minus{} b^{p}_{3}\right)^{\frac {1}{p}}\geq \left[\left(a_1 \plus{} b_1\right)^{p} \minus{} \left(a_2 \plus{} b_2\right)^{p} \minus{} \left(a_3 \plus{} b_3\right)^{p}\right]^{\frac {1}{p}}$,
με
$ 0 < p < 1$

Οπότε για:

$ a_1 \equal{} a^4 \plus{} b^4,a_2 \equal{} b^4 \plus{} c^4,a_3 \equal{} c^4 \plus{} a^4\wedge b_1 \equal{} 2a^2b^2,b_2 \equal{} 2b^2c^2,b_3 \equal{} 2c^2a^2\wedge p \equal{} \frac {1}{2}$

έχουμε ότι

$ \left(\sqrt {a^4 \plus{} b^4} \minus{} \sqrt {b^4 \plus{} c^4} \minus{} \sqrt {c^4 \plus{} a^4}\right)^{2} \plus{} \left[\sqrt {2}\left(\left|ab\right| \minus{} \left|bc\right| \minus{} \left|ca\right|\right)\right]^{2}\geq \left(\sqrt {\left(a^2 \plus{} b^2\right)^{2}} \minus{} \sqrt {\left(b^2 \plus{} c^2\right)^{2}} \minus{} \sqrt {\left(c^2 \plus{} a^2\right)^{2}}\right)^{2}$$ (1)$

Διαιρώντας την $ (1)$ με το $ 4$ παίρνουμε:

$ \left[\frac {\sqrt {a^4 \plus{} b^4} \minus{} \sqrt {b^4 \plus{} c^4} \minus{} \sqrt {c^4 \plus{} a^4}}{2}\right]^{2} \plus{} \frac {\left(\left|ab\right| \minus{} \left|bc\right| \minus{} \left|ca\right|\right)^{2}}{2}\geq \frac {\left(a^2 \plus{} b^2 \minus{} b^2 \minus{} c^2 \minus{} c^2 \minus{} a^2\right)^{2}}{4} \equal{} c^{4}$

Ωραίος
Η ανισότητα λεγεται Βelman και ισχύει για θετικές βασεις .Βλεπεις το βιβλιο που είχε την Βelman το είχε ξεχάσει αυτο, για αυτό και εμφανίστηκε το προβληματακι......

Ας δώσουμε λοιπόν και τη γενικευμένη ανισότητα $ Bellman$.

Έστω $ 0<p<1$ και $ a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}\in \mathbb{R}$ τέτοια ώστε $ a^{p}_{1}\geq a^{p}_{2}\plus{}a^{p}_{3}\plus{}...\plus{}a^{p}_{n}$ και $ b^{p}_{1}\geq b^{p}_{2}\plus{}b^{p}_{3}\plus{}...\plus{}b^{p}_{n}$.

Τότε ισχύει η ακόλουθη ανισότητα:

$ \left(a^{p}_{1}\minus{}a^{p}_{2}\minus{}...\minus{}a^{p}_{n}\right)^{\frac{1}{p}}\plus{}\left(b^{p}_{1}\minus{}b^{p}_{2}\minus{}...\minus{}b^{p}_{n}\right)^{\frac{1}{p}}\geq \left[\left(a_1\plus{}b_1\right)^{p}\minus{}\left(a_2\plus{}b_2\right)^{p}\minus{}...\minus{}\left(a_n\plus{}b_n\right)^{p}\right]^{\frac{1}{p}}$


AMC
AIME
trigonometry
floor function
probability
geometry
trig identities

In triangle ABC, it is given that $\angle B = \angle C$. Points P and Q lie on AC and AB, respectively, so that $AP = PQ = QB = BC$. Angle ACB is r times as large as angle APQ, where r is a positive real number. Find the greatest integer that doesn't exceed 1000r.

Huh? This doesn't seem right... I obviously made a mistake somewhere, but where?

$\angle APQ$ is between two same-length segments. $\angle QBC$ is also between two same-length segments, and these two segments are also the same length as the other two segments, so $\angle QBC=\angle APQ$. We already know that $\angle QBC=\angle ACB$, so $\angle APQ=\angle ACB$. Therefore, $r=1$, and the greatest integer that does not exceed that is $1000$...

The other problem with this is that, if $\angle APQ=\angle ACB$, then $PQ\parallel BC$, but then they cannot be equal, or we wouldn't have a triangle.

" \angle APQ is between two same-length segments. \angle QBC is also between two same-length segments, and these two segments are also the same length as the other two segments "

So let's take some two arbitrary isosceles triangle ABC and XYZ.  You are only given the fact that AB=AC=XY=XZ.

this doesn't imply any relationship between angle BAC and angle YXZ.

a nice diagram will show this.

note: check out my exceptional paint and grammar skills

Angle chase.  Let $\angle PAQ = x$.  Then $\angle PQA = x \implies \angle QPC = 2x \implies \angle QCP = 2x \implies \angle CQB = 3x \implies \angle CBQ = 3x \implies \angle ACB = 3x \implies \angle QCB = x$.  Also note that from triangle QCB, $7x = 180$.  Now we have $\frac{\angle BCA}{\angle APQ} = \frac{3x}{180-2x} = \frac{3x}{5x} = \frac{3}{5}$ so the answer is 600.

I believe the answer is actually 571.

let $\angle QPB=k$
let $AP=l$
$\angle QPB=k \Rightarrow \angle PQA=2k$
$AQ=2l\cos2k \Rightarrow PC=2l\cos2k$ due to ABC being isosceles.
note that $\angle PCB=90-k$ , $\angle PBC=90-2k$, and $\angle BPC=3k$

$BP=2l\cos k$
$BC=l$ is given
$BC=BP\cos(90-2k)+PC\cos(90-k)=2l\cos k\sin(2k)+2l\cos(2k)\sin k$
so
$l=2l\cos k\sin(2k)+2l\cos(2k)\sin k$
$\frac{1}{2}=\cos k\sin(2k)+\cos(2k)\sin k$
$\frac{1}{2}=\sin(k+2k)=\sin(3k)$
so $3k=30$, $k=10$

by angle chasing, we arrive at $\angle APQ=180-4k$, and $\angle PCB=90-k$
since k=10, we have $\angle APQ=140$, and $\angle PCB=80$

$\lfloor 1000\times \frac{80}{140} \rfloor=571$

I don't understand those relationships. probability, how did you get any of these steps: $\angle QPC=2x \Rightarrow \angle QCP=2x \Rightarrow \angle CQB=3x \Rightarrow \angle ACB=3x \Rightarrow \angle QCB=x$

Nosepuppy, how did you get: $\angle QPB=k \Rightarrow \angle PQA=2k$

[quote="Nosepuppy"]I believe the answer is actually 571.

let $\angle QPB=k$
let $AP=l$
$\angle QPB=k \Rightarrow \angle PQA=2k$
$AQ=2l\cos2k \Rightarrow PC=2l\cos2k$ due to ABC being isosceles.
note that $\angle PCB=90-k$ , $\angle PBC=90-2k$, and $\angle BPC=3k$

$BP=2l\cos k$
$BC=l$ is given
$BC=BP\cos(90-2k)+PC\cos(90-k)=2l\cos k\sin(2k)+2l\cos(2x)\sin x$
so
$l=2l\cos k\sin(2k)+2l\cos(2k)\sin k$
$\frac{1}{2}=\cos k\sin(2k)+\cos(2k)\sin k$
$\frac{1}{2}=\sin(k+2k)=sin(3k)$
so $3k=30$, $k=10$

by angle chasing, we arrive at $\angle APQ=180-4k$, and $\angle PCB=90-k$
since k=10, we have $\angle APQ=140$, and $\angle PCB=80$

$\lfloor 1000\times \frac{80}{140} \rfloor=571$[/quote]

In my opinion, it is correct, although a mysterious $x$ slipped in there for $k$ at one point. Alternatively: Take $AP = PQ = QB = BC = 1$ and $\angle{A} = x$ so that $\angle{C} = 90-\frac{x}{2}$. Now, by the law of sines, $\frac{1}{\sin(x)} = \frac{1+2\cos(x)}{\sin\left(90-\frac{x}{2}\right)}$. Use that $\sin(x) = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$, $\cos(x) = 1 - 2\sin^2\left(\frac{x}{2}\right)$ and $\sin\left(\frac{3x}{2}\right) = 3\sin\left(\frac{x}{2}\right) - 4\sin^3\left(\frac{x}{2}\right)$ to obtain $\frac{1}{2} = \sin\left(\frac{3x}{2}\right)$; whence, $x= 20$, $\angle{C} = 80$, and $\angle{APQ} = 140$, for an answer of $\left\lfloor \frac{4000}{7}\right\rfloor = 571$.

To answer your question Jesusfreak, he used the external angle theorem on the isosceles $\triangle{PQB}$.

What's the external angle theorem? Checked Mathworld and can't find it.

Perhaps that isn't its primary name. Anyway, it simply says that in a triangle $ABC$, the external angle at $A$ is equal in measure to $\angle{B} + \angle{C}$.

[quote="Mildorf"]

In my opinion, it is correct, although a mysterious $x$ slipped in there for $k$ at one point.

[/quote]

:O

man, i need to start proofreading my solutions. fixed


you could create a macro that logs in as you and then loops through a number of people...but that'd take like 50 years to make

[quote="vallon22"]is it possible to send a pm to more than one person?
maybe a team for a FTW tourny or reminding a group of pplz to do something?

edit: actually, now that i look at the site and gerneral information tab closer...
i think this should go under questions, suggestions, and annoucements place
could someone move this?[/quote]
I think there should be a feature that allows this.


geometry
geometric transformation
rotation
trigonometry
AMC

I recently took a very hard math tournament and I encountered the following problem:

 In an equilateral triangle ABC, there exists a point P inside the triangle so that AP=3, BP=4, and CP=5. Find the side length of the triangle.

If anyone could please help me solve this I would be very grateful. Thank you very much

Ahhh this is a "classic."

[hide="Hint 1: bananas"]$3^2 + 4^2 = 5^2$[/hide]
[hide="Hint 2: plantains"]Use a rotation[/hide]

I've worked over and over on this problem and still can't get it.

Can someone please post a solution?

Thanks a lot.

this problem and its variations have been posted millions of times...search

Consider a rotation about B taking C to A.  It takes A to A' and P to P'.  Then PP' = 4, because PBP' = 60 deg by our rotation, and BP = BP' = 4.

So PAP' is a 3-4-5 triangle, and thus P'PA is 90 degrees.  So in APB, we have side lengths 3 and 4 contained by a 150 deg angle.  We can find answer w cosine law.


In any convex $ 2n$-gon, there is a diagonal not parallel to any side.

[hide="A bit confused"] I am a bit confused, why should each side can have at most $ n\minus{}2$ diagonals parallel to it. :maybe: [/hide]

Moonmathpi496 wrote
[quote]I am a bit confused, why should each side can have at most n-2 diagonals parallel to it.[/quote]

Draw a picture may help you understand it .
Let the polygon be $ CDA_1 A_2 \dots A_{2n\minus{}2}$, where $ CD$ is the side, then for any vertice $ A_x$you can find at most 1 vertice in $ \{A_n\}$ say $ A_k$ such that $ A_x A_k // CD$ so there are at most $ n\minus{}1$ diagonals $ // CD$ when all the vertices are utilized.Now let $ A_p A_q$ be the parallel diagonals having the longest distances from $ CD$ , if it's one of the sides of the polygon, then done.If it isn't , there must be a point beyond the parallel line , not all of the vertices are utilized, done also.


Okay guys, finally it's here. To mark the quarter century I am bringing in a brand new set of roles. Hope it turns out great, eh? :)
_____________________________________________________

[u][b]Story:[/b][/u]

So there is this settlement with all its cheerful people, and they go about doing things that cheerful people in settlements usually do. Let's not bother about the details - as they are irrelevant, and, more importantly, thoroughly uninteresting.

However, in the province in which our small settlement is situated, there are two rival gangs of buddies ... er, i mean, baddies (sorry, I have just watched the awesome movie Air Buddies, it ended around ten minutes ago). These gangs go by the fanciful names "The Cobras" and "The Scorpions" (cliched, no doubt, but very strong nevertheless, and anyway you can't reasonably expect more innovation from these baddies can you?). As always happens in mafia game stories, they decide to send their representatives to our settlement, three each.

Now what do these newcomers do? Elementary, my dear Watson. They look at the list of inhabitants and each independently chooses to bump one off and then impersonate him. Now, fortunately for them, (and to keep the game straight,) no two choose the same person. However none of the six have any idea at all as to which of the other people are baddies in disguise.

In the days to come, as the two gangs try to take control of our settlement, it is going to turn into one huge bloody mess ... ooh, I can't even imagine it. Thankfully I am not in there, so for all those of you who are, best of luck!
_____________________________________________________

[b][u]Roles:[/u] [/b]

3 cobras
3 scorpions
2 inspectors
2 doctors
2 vigilante (should I put an 's' here to indicate the plural? :maybe: )
10 civilians
_____________________________________________________

[b][u]Player list:[/u] [/b]

b-flat
PeterVDD
Fanatic
mustafa
Lena
probability1.01
Jashper
jackdillon
Ars
junggi
pianoforte
egghead91
perfect628
Vihang
life=tennis+math
kheg_k
Sly Si
baseballfreak2
Captain Sugar
weiquan
laughinghead505
Melissa

_____________________________________________________

[u][b]Rules:[/b][/u]

1) Each night, each of the special roles sends in an independent choice.
2) If two people of the same group visit the same person on a particular night, they will be connected to each other.
3) Two rival baddies visiting each other on the same night will both fail, but will get to know each other.
4) Each day will be exactly 96 hours long, irrespective of what goes on during the day. Each night will be 48 hours long. (You can think of time having been slowed down by a factor of six.) The ONLY exception to this rule will be if the site's server is down or the forum is locked, in either of which case I'll make appropriate adjustments.
5) Anybody who has been accused and seconded TWICE (all by different persons) is declared to be on the block.
6) Any number of people can be put on the block on a single day.
7) Once you are on the block, you don't get off it till the day is over.
8) At the end of the day, a guy on the block is executed if and only if more than half the current living population have voted on him, and there are more ayes than nays.
9) The accusers and the seconders are automatically assumed to vote unretractable ayes, irrespective of their statements later on. (Note that very late in the game this may imply that if you get on the blocked, you are automatically lynched. This is a bug that has made me unhappy, so I may change some of the rules later.)
10) No one is allowed to accuse oneself, secodn oneself's accusation, or vote on oneself.
11) Everyone else may change their votes or even retract them any number of times before the days is over.
12) Since time is limited, you need not wait for someone's defense to start voting on him.
13) Any number of people can be executed at the end of the day.
14) Since there is no way I can stop you from interacting through PMs during the day, I make it legal.
_____________________________________________________

I am sure I have not covered everything, and you will have a lot of questions. So go ahead, the game doesn't start immediately (I have not even decided upon the final list of players), have your appetite full of questioning. However, I am not changing the above rules, with the possible exception of 4, where I am ready to make each day to 144 hours, if may of you wish to have it so.

Looking forward to a grand game! :)

w00t first post!

I'm kidna confused about the powers of the baddies? help?

[quote="bubka"]
14) Since there is no way I can stop you from interacting through PMs during the day, I make it legal.[/quote]
er what do you mean by this? we can discuss the game through PMs? :huh:

If two baddies are conencted, they can discuss the game.
Each baddy gets one kill per night, he'll pray it's not one of his own gang.

so each night there's going to be 6 kills, one from each baddie?
erm... :maybe: and do they know who the fellow baddies are?
and what's that part about impersonating?

Vihang: what's that part about impersonating?
so each night there's going to be 6 kills, one from each baddie?

Ritam: plus two from the vigilante

Vihang: eeeks
8 !

Ritam: add to it around four lynches a day

Vihang: is the game 2 -3 days long ?

Ritam: do you want it to be longer?
and die eventually?

Vihang: nope
this is good fast and furious

Ritam: no mafia game lasts beyond three days and remains full of activity
_____________________________________

That answers a lot. As for knowing each other, read through the story carefully again.
_____________________________________

Ritam: as for impersonating, the baddies start the game impersonating the people who play them

Vihang: ok

Ritam: the idea is that these people were inhabitants of the settlement
so now nobody notices the difference

Vihang: yeah got it
_____________________________________

That answers the rest.

erm...
if they haven't confirmed by now, might as well kick them out...it just means that they're going to be inactive in the game

Firstly, this game is a bit robust to inactives.
Secondly, the situation's a bit different.
I can't explain it.
Don't worry, I'll try and hurry!

I would request that days be made at least 144 hours, if not longer, given that there will be very few of them.

96 hours? What the hell? How about one week?

Also, when do we find out the roles of people?

This game seems way too swingy.

Why don't we just start now? The people who haven't confirmed yet can be moved up to 26.

[quote="bubka"]Role PMs??
Dream on boy!
I don't even have the final list of players yet!

Why, they maybe even a week off![/quote]I really suggest making the current list final... this looks so good that we want to start right away! :)

And I agree 144 hours is much better.

I know I haven't signed up, but THESE NEW ROLES SOUND GREAT!!! :)

P.S. This doesn't mean I'm signing up... :|

Okay 144 hours it be.
And I'll try and hurry up the role PMs.
But lazy as I am, they are a torture. :(

By the way life=tennis+math is away on a ten-day easter break and he has asked me not to replace him. What do I do with him?

Okay now for the night events:

The purple cobra made short work of perfect628, a mere civilian.
The red scorpion got the eluding green cobra Sly Si finaly in the small of the back, and squared things up.
The second vigilante went back seething after finding perfect628 dead already.
As a parting shot, Sly Si disposed of the second fairy laughinghead505 who has long since bid this forum goodbye.
Finally, the impatient purple cobra killed me.  :(

It shows the lengths to which I go for you people, that even after being ungratefully killed by one of you, I am continuing the game for you.

'It's a new day' - [i]Bryan Adams, Here I am, Spirit: Stallion of the Cimarron soundtrack[/i]

Next night you should agree on who to kill, this was a waste of 2 kills. =)

Omg Peter VDD has dropped VDD and become simply Peter.

hmm the purple cobra....
sounds familiar

Sorry but it is my game. It's still open.

who are the people who are still playing?

i am the purple cobra

hissssssssssssssss

this is really confusing

game title said game abandoned


[quote="Allnames"]I had "Contests Around the World " volumes: 1998-1999; 1999-2000; and 2000-2001. I searched (by google ; and some sites in Vietnam ) but I couldn't found another volumes ( since 2001-2002 till now). Who have one of them (e-book versions  :) ) so please help me!
Thank you![/quote]

I want to purchase the original copy of :

Mathematical Olympiads: Problems and Solutions from around the World, 1995-1996

Can any one tell me where i can find. I already tried Amazon and all other online book stores no one has this book in their database.

I don't claim to know whether these books are still copyrighted or not (however I believe they are), but do heed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126338]this[/url].


ratio
vector
algebra
polynomial
limit
induction
absolute value

In this infinite tree, degree of each vertex is equal to 3. A real number $ \lambda$ is given. We want to assign a real number to each node in such a way that for each node sum of numbers assigned to its neighbors is equal to $ \lambda$ times of the number assigned to this node.
Find all $ \lambda$ for which this is possible.

What have I misunderstood? $ \lambda/3$ to every node means any $ \lambda$ seems to work. EDIT: Oh, ratios of sums.

That gives a ratio of $ 3$, not $ \lambda$.  In fact, this is an eigenvalue problem:  if $ \mathbf{A}$ denotes the adjacency operator of the graph (acting on the vector space consisting of linear combinations of its vertices), we want to find the set of real eigenvalues corresponding to real eigenvectors.

Now, I'm also getting that every real $ \lambda$ works:  assign every vertex a weight $ w_n$ depending only on its depth $ n$ (where the root has depth $ 0$, each neighbor has depth $ 1$, and so forth) such that

$ w_0 \equal{} 1$
$ w_1 \equal{} \frac {\lambda}{3}$
$ w_n \plus{} 2w_{n \plus{} 2} \equal{} \lambda w_{n \plus{} 1} \Leftrightarrow$
$ w_{n \plus{} 2} \equal{} \frac {\lambda w_{n \plus{} 1} \minus{} w_n}{2}.$

[b]Edit:[/b]  Did you mean to add the condition that the weights must be bounded?  The growth rate of $ w_n$ is controlled by the roots of the characteristic polynomial $ 2x^2 \minus{} \lambda x \plus{} 1$, which are

$ x \equal{} \frac {\lambda \pm \sqrt {\lambda^2 \minus{} 8}}{4}.$

We want both roots to have absolute value less than or equal to $ 1$.  The larger root is less than or equal to $ 1$ if and only if $ \lambda \le 3$, and the smaller root is greater than $ \minus{} 1$ if and only if $ \lambda \ge \minus{} 3$, so the spectrum by this construction is $ [ \minus{} 3, 3]$.  (Note that $ x$ doesn't have to be real.)  We claim that this is best possible.

By the spectral radius formula, the spectrum of $ \mathbf{A}$ is contained in the ball of radius $ \lim_{n \to \infty} \parallel{} \mathbf{A}^n \parallel{}^{\frac {1}{n} }$, where $ \parallel{} \mathbf{A} \parallel{}$ is the operator norm on $ \ell^{\infty}(V)$.  But since the degree of each vertex is $ 3$, by analogy with the finite-dimensional case it follows that $ \parallel{} \mathbf{A}^n \parallel{} \equal{} 3^n$, so the conclusion follows.  (I can give a more fleshed-out argument, but I hope it's not really necessary.)

[b]Edit #2:[/b]  Okay, so a more elementary argument is possible.  Namely, if the weights are distributed such that they do not depend only on depth, show by induction that for every $ n$ there exists a vertex of depth $ n$ whose weight has absolute value greater than or equal to the absolute value of $ w_n$ from the above construction (essentially, from each vertex pick the child with the larger weight in absolute value).  If $ \lambda \not \in [\minus{}3, 3], w_n$ is unbounded and the conclusion follows.


function limit calculus derivative integration trigonometry algebra

What is l'hospital? I tried googling it, but I still don't understand what it is. Thanks in advance!

(Also, could someone please provide a few examples? Thank you.)

St. Andrews history site link for the [url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/De_L%27Hopital.html]Marquis de L'Hôpital.[/url]

Variations in both the spelling of French over the centuries and for fashion in transliteration into English mean you may see it as L'Hospital, L'Hôpital, L'Hopital, and possibly some other variants.

Here's the [url=http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule]Wikipedia link.[/url]

OK, neither link is posting correctly. I'll put them up as is while I explore what's wrong.

I think it's a matter of underscores not interacting well with the url tags.  i recall that darij knows a way around it but i don't remember what it is offhand, so you may want to pm him.

[quote="Kent Merryfield"]St. Andrews history site link for the [url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/De%5FL%27Hopital.html]Marquis de L'Hôpital.[/url]

Variations in both the spelling of French over the centuries and for fashion in transliteration into English mean you may see it as L'Hospital, L'Hôpital, L'Hopital, and possibly some other variants.

Here's the [url=http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s%5Frule]Wikipedia link.[/url]

OK, neither link is posting correctly. I'll put them up as is while I explore what's wrong.[/quote]

OK, hex ASCII rules. Just substitute apostrophe with %27 and underscore with %5F.

Any quote ' should be escaped as %27. You cannot have quotes in an URL (legally.)

I'm going to assume you want to know what the rule is?  Suppose you have two functions, $ f(x)$ and $ g(x)$, and
$ \lim_{x\rightarrow a}f(x)\equal{}0$ and $ \lim_{x\rightarrow a}g(x)\equal{}0$.  Then what is
$ \lim_{x\rightarrow a}\frac{f(x)}{g(x)}$?
If you just stick in yourt two limits, you're going to get $ 0/0$, which is called an [i]indeterminant[/i].  So to evaluate this limit, simply take the derivative of $ f$ and $ g$, and then take the limit.  That is,
$ \lim_{x\rightarrow a}\frac{f(x)}{g(x)}\equal{}\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$
Note, however, that you may still have an indeterminant integral so you might have to do it again.  This technique also holds for when the quotient is $ \infty/\infty$.  But if you don't have such an indeterminant expression, the rule will not work.

Here's an example:
Find $ \lim_{x\rightarrow 1}\frac{x^{3}\plus{}x\minus{}2}{x^{2}\minus{}3x\plus{}2}$.

[hide="Solution"]
$ \lim_{x\rightarrow 1}\frac{x^{3}\plus{}x\minus{}2}{x^{2}\minus{}3x\plus{}2}\equal{}\lim_{x\rightarrow 1}\frac{3x^{2}\plus{}1}{2x\minus{}3}\equal{}\frac{4}{\minus{}1}\equal{}\minus{}4$
If you try to use it again, you will get the wrong answer since what we have now is not an indeterminant.[/hide]

I generally argue for the minimization of the use of L'Hôpital's rule. There are a few places in which it is the most convenient thing to use, and I'll use it there, but in many other cases, I will use other arguments. These arguments include:

[b]1. Knowing familiar limits related to derivatives:[/b]

Examples here would include $ \lim_{x\to 0}\frac {\sin x}{x},$ which you would need to know before you compute the derivative of the sine in the first place. It's circular reasoning to use anything involving the derivative to obtain it.

[b]2. Algebra:[/b]

I would not have used L'Hôpital for JRav's example. Once I have verified that this polynomial divided by a polynomial is a $ \frac00$ case as $ x\to 1,$ then I know that $ (x \minus{} 1)$ is a factor of both the numerator and denominator. Since I know that is a factor, I should be able to factor both quite easily (by synthetic division if I have to) and eliminate the common factor.

[b]3. The heirarchy of size:[/b]

I do not want to refer to L'Hôpital every time I face $ \lim_{x\to\infty}x^3e^{ \minus{} x/2}$ or $ \lim_{x\to0^ \plus{} } \sqrt {x}\,\ln x.$ I know that both are zero because I know that exponentials grow faster than powers of $ x$ and logarithms grow slower than powers of $ x.$

Similarly, we can decide that $ \lim_{n\to\infty}\frac {3n^4 \plus{} 3n \minus{} 5}{n^4 \plus{} n^2 \plus{} 1} \equal{} 3$ and $ \lim_{n\to\infty}\frac {n^4 \minus{} 3^n}{n^4 \plus{} 3^n} \equal{} \minus{} 1$ by simply paying attention to what parts of each numerator and denominator are important.

[b]4. Taylor polynomials and power series:[/b]

A famous geometry limit problem leads to this: $ \lim_{\theta\to0}\frac {\theta \minus{} \sin\theta}{(1 \minus{} \cos\theta)\sin\theta}.$ Is this a good place to use L'Hôpital? Not really: L'Hôpital will be quite messy, and will need to be done more than once. Given my familiarity with the power series for sine and cosine, I can do this [i]in my head[/i]. Try it! See if you can do it in your head. [hide="Informal solution."]$ \theta \minus{} \sin\theta$ looks like $ \frac {\theta^3}6$ and $ 1 \minus{} \cos\theta$ looks like $ \frac {\theta^2}2.$ Throw in $ \sin\theta\approx\theta$ and we have $ \frac {\theta^3}{\theta^3}$ - the exponents match. (And I know why L'Hôpital would have to be done several times.) Paying attention to the coefficients I see $ \frac16$ divided by $ \frac12.$ Do that, and the limit is $ \frac13.$[/hide]

Some posters around here are fond of posting things like "Find $ \lim_{x\to 0}\frac {e^x \minus{} 1 \minus{} x}{x^2}$ and don't use Taylor series or L'Hôpital." And I immediately go away - why tell me not to use Taylor series when I can see [i]at a glance[/i] that that limit is $ \frac12?$

Here's another one: $ y \equal{} \lim_{m\to\infty}\left(1 \plus{} \frac rm\right)^{mt},$ from a compound interest problem. Since this is a $ 1^{\infty}$ indeterminate case, we take the logarithm:
\[ \ln y \equal{} \lim_{m\to\infty}mt\ln\left(1 \plus{} \frac rm\right).
\]
This is now an $ \infty\cdot 0$ indeterminate case. Do we find a way to turn it into a quotient so we can use L'Hôpital? I'd rather not - it's fairly easy to make mistakes that way in handling the extraneous variables $ r$ and $ t.$ Instead, let's use the Taylor series for $ \ln(1 \plus{} x) \equal{} x \minus{} \frac {x^2}2 \plus{} \cdots.$

$ \ln y \equal{} \lim_{m\to\infty}mt\left(\frac rm \minus{} \frac {r^2}{2m^2} \plus{} \cdots\right)$

$ \equal{} \lim_{m\to\infty}\left(rt \minus{} \frac {r^2t}{2m} \plus{} \cdots\right) \equal{} rt,$ from which $ y \equal{} e^{rt}.$

===================

There are cases in which L'Hôpital is the most efficient method. Here's one, which makes use of the fact that L'Hopital works for complex analytic functions:

If $ f(z) \equal{} \frac {e^{iz}}{z^4 \plus{} 4},$ find the residue of $ f$ at $ z \equal{} 1 \plus{} i.$ This amounts to finding $ \lim_{z\to 1 \plus{} i}\frac {e^{iz}(z \minus{} 1 \minus{} i)}{z^4 \plus{} 4} \equal{} e^{i(1 \plus{} i)}\lim_{z\to 1 \plus{} i}\frac {z \minus{} 1 \minus{} i}{z^4 \plus{} 4}.$

Now, I could say about this what I said in #2: it's a rational function, I know the factors, and I can do this by algebra. But there are several factors with some complex mulitplications. Let's just use L'Hôpital:

$ \equal{} e^{ \minus{} 1 \plus{} i}\lim_{x\to1 \plus{} i}\frac {1}{4z^3} \equal{} \frac {e^{ \minus{} 1 \plus{} i}}{4(1 \plus{} i)^3} \equal{} \frac {e^{ \minus{} 1 \plus{} i}}{8 \minus{} 8i},$ which can then be simplified further.

(Further simplification will be dictated by the larger problem this is part of. That larger problem might be the computation of $ \int_{\minus{}\infty}^{\infty}\frac{\cos x}{x^4\plus{}4}\,dx.$ )

[quote="Kent Merryfield"]
I would not have used L'Hôpital for JRav's example.
[/quote]
I probably would not use it either but that is a problem I took from the Spivak book, that emphasizes the fact the the rule doesn't work when we don't have an indeterminant.


[quote="nkht-tk14"]Let a,b,c are positive real numbers such that abc=1
prove that $a^{2}+b^{2}+10(ab+bc+ca) \geq 11(a+b+c)$[/quote]
It's wrong.
Maybe $a^{2}+b^{2}+c^{2}+10(ab+ac+bc)\geq11(a+b+c)$ $?$
If so, see here
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80197


geometry
trigonometry
circumcircle
AoPS Books

I am currently in a math class and I get all of the  material, but the tests are SOOOO HARD!They are almost always curved and I got a 74, 68, and 88 for the quiz grades. I got 79 and 70 for tests grades so my grade is like a C+. I just took another insane quiz today and I will tell you about it Monday.

So if the tests are really hard, and the homework and classwork are [i]easy[/i] in comparison to the test problems, what should I do? I bought a AOPS textbook and it seems to help but I think I should do more

And is it possible to still get an A-?The semester ends January 21st.

What math class are you in?  Does your school average the quarter grades or is it cumulative?

And to prepare for your tests, think about this:

It's highly unlikely that your teacher would give you something on the test that you had not even learned. (Right???  :huh: )
So the most likely reason that your tests are harder than the homework is that they apply the same things, but in different ways. So make sure you understand how and when to apply the concepts, in [i]all[/i] situations.  :wink:

In most cases, solving the "C" level exercises (the last 3-4 problems in each exercise test) helps you prepare for the harder problems on most tests.

I had one of these insanely hard teachers before for geometry. She would give out the most insanely hard tests for us. What i did to get prepared was that i did all of the C level problems and i also got my mom to help me on some of the proofs i didnt get  :P

Volume 1 would really help, because it has a lot of information on Trig, and it has lots of problems that they don't teach you the solutions to. You have to kind of think about what you know and think out of the box. The solutions to the problems would probably help you a lot with some of the problems on your tests.

I suggest he get Volume 1, and the solution manual, if that's what you were wondering AIME.

No, I was saying that all the problems have solutions in the solutions manual...

Some stuff you learn in AoPS  books won't help you much in school, but will hlep you in contests.

for example, some circumcircle properties and basic cyclic quadrilateral's are almost never covered in some geometry courses.

[quote="AIME15"]Which AoPS book did you buy?  Geometry?[/quote]
me? i just used the one our teacher gave us


Look:  http://mathworld.wolfram.com/GreatestCommonDivisor.html
Here is a formula for $gcd(ab,cd)$, but I found counterexample. Could you explain how should I read what is written there.

Talking about the 14th formula?

They might have forgotten a "GCD"..

yes, I guessed, but try according to this formula calculate
$gcd(2\cdot5, 4\cdot6)$


geometry
geometric transformation
reflection
AMC
AMC 12
MATHCOUNTS

Please do not discuss difficulty, your scores, problems, or anything else of the sort [EXCEPT ONLY your and your team's  placement - NOT the placements of, say, Jonathan Horton's or Wenyu Cao's unless you are one of those people] until after Feb. 25. Remember that some schools do determine their MC chapter participants by the School Round, and therefore if you inform a contestant who has not taken the test that the problems were easy/medium/hard and that you got a xx they have an unfair advantage.

(Feb. 25 is the date of the last chapter competition; to be safe we may assume that schools may choose their competitors by the school round administered anytime up to the day before their chapter competition. It's only 20 days from now!)

Thanks for cooperating and making AoPS a great place to be!

Billy

[size=75][Edited to update the date to the present year.][Edited to update the title to reflect chapter too.][Edited to clarify that your rank is all right][/size]

If they took those tests before you did and told you those things about that test, you would know that the tests were not super-super-impossibly hard. And if someone else said so, and they were a somewhat active poster, one could judge how good they were and thus how hard the test is.

If I told you that the school round was super-impossibly-hard, would it really give them an unfair advantage to know that information?

They would know to study really hard and to work really hard, whereas if you told them it was really easy they'd know that they didn't need to study. Let's say that your friend takes it and tells you that you don't need to study, it's really easy, and you know that he is truthful and also pretty stupid. You know that it's a really easy test. On the other hand, if your really super smart friend, let's say Andrew Ardito, tells you that the school round is really hard and you really ought to study for it. You know that you really have to study hard for it and you need to make sure the ones you do are right, not answer all of them. But in both cases you have an unfair advantage.

Whether you agree with solafidefarms is irrelevant.  The rule of this forum is that there shall be NO discussion of any MATHCOUNTS Competition until the window for taking the competition is closed.  End of discussion.

Do not say anything about the 2006 School Round until after February 26.

I also agree with rcv and solafidefarms.
If you see a scores discussion, PM the moderators quick, as it gets along very fast.
Also, aren't we allowed to discuss our place in the school and whether we are a team member or individual?

[quote="BFG"][quote="solafidefarms"]if you inform a contestant who has not taken the test that the problems were easy/medium/hard and that you got a xx they have an unfair advantage.

Billy[/quote]

[size=150]I do not think so. [/size]If Joe Jia or Alberta Ni informed you that AMC 12 problems are easy and they all got perfect scores two or three times, OR  if  Patricia Li  or Gregory Gauthier informed you that the school rounds are very easy and they all got 46, do you really have an unfair advantage when you take AMC12 or Mathcounts school rounds?[/quote]

There's no point in making an argument.  The discussion to prevent discussion is one made by AoPS and requested by the MATHCOUNTS organization on a yearly basis.

There are plenty of past problems that can be discussed for math's sake.

Whoa, feb 26? That's weeks after my Chapters.

Some schools give the school test immediately before the chapter test. The last chapter test is Feb. 25. Ergo some schools may give the school test Feb. 25. Thus you should not discuss the school round until Feb. 26.

We should have a chapters topic up as well, I know our region had their chapter today.

Can we say whether we're making the next level, even though our regionals are like 2 weeks from now?

I pm'ed nebula and he said that we can say our place, just not our score

[quote="bpms"]Can we discuss our ranking?
Also, is it normal for someone who made nationals to not make states the year after :oops:[/quote]

I guess, as long as you do not discuss scores, it is perfectly fine to discuss the ranking.

[quote="jhredsox"]is it after the 25th in case there is "inclement" weather and it will have to be held the 25th, not the 24th in some areas?

-jorian[/quote]
No, it is after mathcounts posts county on their site.

This thread has been superseded by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132111]my official announcement[/url] of the discussion rules.


real analysis
advanced fields
advanced fields unsolved

Let A and B be two disjoint sets in the interval (0, 1) . Denoting by μ the Lebesgue measure on the real line, let μ(A) > 0 and μ(B) > 0 . Let further n be a positive integer and λ=$ \frac{1}{n}$. Show that there exists a subinterval (c, d) of (0, 1) for which μ(A$ \cap$ (c, d)) = λμ(A) and
μ(B $ \cap$(c, d)) = λμ(B) . Show further that this is not true if λ is not of the form $ \frac{1}{n}$.

As a general preliminary remark we should additionally assume $ A$ and $ B$ be measurable sets. [quote="Mimmzy"]Show further that this is not true if λ is not of the form $ \frac {1}{n}$.[/quote]I don't understand, since there is an easy counterexample for every real $ \lambda\in(0,1)$:

Then let $ A\equal{}(a_1,a_2)$ and $ B\equal{}(b_1,b_2)$ be two intervals with $ 0<a_1<a_2<b_1<b_2<1$. For every real $ \lambda\in(0,1)$ the claim can be fulfilled with  $ c_\lambda\equal{}\lambda a_1\plus{}(1\minus{}\lambda)a_2\in A$ and $ d_\lambda\equal{}(1\minus{}\lambda)b_1\plus{}\lambda b_2\in B$.

Or am I missing something here?

Where is the counterexample? :maybe:
The statement claims that there $ exists$ an interval but not for $ all$ intervals or I'm missing something.

I interpret the last sentence as that if there are $ c$ and $ d$ with the property given above given $ A$, $ B$ and $ \lambda$  then necessarily $ \lambda\in\frac {1}{\mathbb N}$, in particular $ \lambda\in\mathbb Q$. For this I provided a counterexample that $ \lambda$ need not necessarily be rational.
... or is the last statement to be read in another way?  :(

I guess it should be readed that if for every pair $ A,B$ the statement is true then $ \lambda$ is of the form $ \frac {1}{n}$ (I did not think about the problem so maybe I'm wrong).

I had the same puzzle as Third Edition, that's why I post it here.

OK, if I understand it correctly then it is to be shown that, given a real number $ \lambda\in(0,1)$, the following assertions are equivalent (the case $ \lambda\equal{}1$ is trivial, hence excluded):

1 - $ \lambda\in\frac{1}{\mathbb N}$;

2 - For every pair $ A,B$ of disjoint measurable subsets of $ (0,1)$ with positive Lebesgue measure, there is an interval $ C\equal{}(c,d)\subset(0,1)$ such that $ \mu(A\cap C)\equal{}\lambda\mu(A)$ and $ \mu(B\cap C)\equal{}\lambda\mu(B)$.


calculus
integration
trigonometry
absolute value
calculus computations

Integrate the following \[\int{\sqrt{1-\sin x}}\,dx\]

$\sqrt{1-\sin x}=\frac{|\cos x|}{\sqrt{1+\sin x}}.$

Try making the substitution $u=\frac{\pi}2-x,$ which should turn the problem into

$\int\sqrt{1-\cos u}\,du$

Then use a half-angle identity.

All of these things work. Also note something that is explicit in kunny's post (with the absolute value) and hidden but still implicitly there in the other two posts: you will have to be careful about signs in different quadrants. In a definite integral, you will probably need to break the integral into separate pieces.

many thanks for all ,

[quote="Kent Merryfield"]will probably need to break the integral into separate pieces.[/quote]

[b]How?[/b]

Make de substitution

$\sin x = u$,

thus

$\int \sqrt{1 \pm \sin x}\, dx \, = \, \int \sqrt{1 \pm u}\, \frac{du}{\sqrt{1-u^{2}}}\, = \, \int \frac{du}{\sqrt{1 \mp u}}\, =$

$= \, \mp 2 \sqrt{1 \mp u}+C \, = \, \mp 2 \sqrt{1 \mp \sin x}+C$


hello to everybody

there is a circle S with radius r and centre O.
it has a chord(not diameter) with points A,B on S.
The altitude from O to AB cuts AB and S respectively
at the points P and Q.
Take an another chord(not diameter and AB) which
goes through P and cuts S at M and N.( PM<PN )
if PM=a and PQ=b
prove that a=b.cos(ANB)

thanks...

Are you sure? I'm pretty sure PM>PQ, so PM can't be PQ*cos(something)... Maybe I'm wrong.

What this wrong problem is still doing here???? :?

So you agree with me, Fedor? Did ne1 else check this out?

Fatih....I guess Grobber is correct

If we start-off with the assumption that a < b......

We have a.PN=b.(2r-b) as a < b

We have PN> 2r-b...results in stating that diameter is not the longest chord in a circle.

hence....a=b*cos(ANB) is not possible.....

Please check the problem.

It's easy to see that if A and B are 2 pts on a circle and M is their midpt, then max{MX|X on the circle}=MS and min{MX|X on the circle}=MT, where S and T are the mdidpts of the larger/smaller arc AB, respectively.