AMC
AMC 12
AIME
USA(J)MO
USAMO
MIT
college

What is the policy on having food/water for the AMC12, AIME, and USAMO? I am pretty sure they all allow water, but I am not sure about the food.

i've seen people eating during the AIME...i dunno if they're supposed to, but they say it helps

I'm pretty sure it's OK since so many do, but
a.) It's not OK if there's any math-related stuff on it, and I'm not sure if Nutritional Facts pass that requirement since it has percentage*
b.) It will probably get checked to make sure it's not a cheat-sheet...er, food :lol:
c.) What's the point of having food anyways? I mean candies help but a full-course dinner...? :lol: (an insider joke to all who visited/went to MIT)

1) You may eat and drink on AMC/AIME/USAMO. If you go to a school that cares, it might even be provided. Food and water were provided at my high school on the AIME and USAMO.

2) The content was only what happened in 2001. The AMC did not make any guarantee that a similar breakdown will occur at any time in the future.

Ok, thanks for the feedback. I know there was a thread a long while back about drinking water during competitions, but do any of you eat food as well?

ive heard that eating andes or mint stuff helps you think

During AIME, only candies and water were provided. I had a candy or two and my own sparkling water. On the USAMO, we had full lunches since there were only two of us rather than 21 or something for the AIME.

But do you think the candy actually helped?

what kind of candy are we talking about here?

We take the test in the nicest place in the school, so they dont allow food / drink there.

Mmm, both statistically and biochemically, yes. For me, I've never tried it.

Given my score on the AIME, I don't think the candy helped.

im not exactly sure why it would... for me it'd probablly be more of a distraction.  I could see it being helpful if it's something you always do when you do math, but otherwise...

I suggest eating the food before the test, preferrably around 5 minutes beforehand. If you eat after that, then it can be a distraction and/or it won't get digested in time, and if after you might have the urge to go to the bathroom :)
The reason why candy is better than other food is because it contains lots of sugar in forms of glucose or fructose, which is very easy to digest. They are also small in volume so it won't cause any urge to go to the bathroom.
As I said before, the temporal blood-sugar boost will make you excited and sugar-high, which will in turn quicken your thinking, kind of like amphetamine except its effects are miniscule compared to the real thing. It is a double-edged sword though, for the excitement can lead to a lot of calculation error/careless miss.

maybe i'm mixing somethings up, but I think that chewing gum helps.

yea
i think its scientifically shown that gum helps us think

As much as I hate to edit things, we don't want to spread misinformation.  So I appended a little note to the end of Tare's "Question-Analysis" post.  If anyone feels the note should be omitted, I will do so.

Er, if it concerns you that much I'll just delete it.

i think sugar in great amounts helps a lot. dont eat a donut like i did though. i spent half an hour during the AIME trying to clean up the mess i caused by dropping it onto my desk (and my test, paper, etc...).

I'm quite sure it helps. I always eat a lot of candies, chocolate, ... during an exam. And some coke may help also. At the IMO you really need to have much sugar since the exam lasts for 4.5 hours, which is long ; sugar helps you to keep your concentration. It's also no bad idea to go to the toilet during the exam, even if you don't really have to. At the IMO, Kaloyan Slavov, seated very close to me in the exam room, went more than 5 times to the toilet each exam. I think it helped since he got 41/42 :)

...
.....
...er, right...

[quote="tetrahedr0n"]dont eat a donut like i did though. i spent half an hour during the AIME trying to clean up the mess i caused by dropping it onto my desk (and my test, paper, etc...).[/quote]
"That donut's haunted! It turned on me, on[b] me[/b]!!!" :lol:

Hey, I know Kaloyan -- he's the roommate of a good friend of mine.  Did you meet Dobrimir (not sure I spelled this correctly), also from the Bulgarian team?  He's a pretty cool guy -- we're in the same math class.

I know Dobromir Rahnev was in the Bulgarian team but I don't think I talked to him (maybe a few words).


limit
inequalities proposed
inequalities

Let $a;b$ are real numbers satisfying $a \geq \frac{-1}{2}$ and $\frac{a}{b}>1$.Find min:$A=\frac{2a^{2}+b}{b(a-b)}$


MATHCOUNTS
geometry
3D geometry
probability
AMC
AIME
number theory

It was really easy. I couldn't doubt anyone getting a good score on it.

It was waaaay too easy for even a chapter level test. In my opinion, the school target was even harder.

guess wat i voted. :rotfl:  :rotfl:  :rotfl:

Who didn't think that it was the easiest target round that they had ever seen? If you did say that, then which one did you take that was easier?

Yeah, I agree. Most of the questions were simple calculations. The hardest problem was probably the ammonia one, but that one was still pretty easy.
3 of my 4 teammates got a perfect score, and the 4th one messed up on the ammonia problem and the handshake problem.

lol i thought u were talking about nats for a sec..... but yes i got a lower score on school target than chapter. the ammonia one (the only one i missed)i had narrowed it down to the right answer and a wrong one. wrote in the right, erased it, wrote in the wrong, and wrote the right at the bottom with a question mark.

lol I got the exact same score on both tests (26/16) with like so many careless mistakes on both (well, 4...)

And then I phailed states with--a really bad score. :blush: (oh well, 5th written 1st cd)

It was really easy, and if it weren't for me and misreading problems, then, well...

[quote="Yongyi781"]Argh. I thought this was about the Nationals Target and voted Extremely Difficult, only to see that it was about the Chapter Target, for which I should've voted "Easy." :mad:[/quote]

the nats target wasn't hard  :)  I just made tons of stupid mistakes. The only hard one there was 5 (if you didn't know how to do it) and 8 was semi-hard.

Are you kidding me? 5 was super easy (if you mean the triangles one), but you know I stink at counting/probability problems. And there were 4 of them. #8 was really hard at the moment, but after the test I beat myself realizing how easy it was. The dominoes one was pure brute force, the cube one was evil, and #1 was evil as well. Not to mention the ultimate evilness of #2, when I thought "sales" meant profit. I thought #3 and #4 were the easiest, though. Thus for the reason I failed miserably at target (only 3 out of 8).

I stink at counting/probability problems too! thats why i missed the last problem and some others on the state sprint :mad:  :mad:  :mad:

[quote="levans"]Hint: (I'm giving a lot of hints today) - If you finished number 62 among all middle school students in the U.S., you aren't stupid but rather the exact opposite of stupid.[/quote]

Hint: The same could be said of the person who got 228th :D

But really, 62nd!!! That's better than me, at least, and all the AoPSers think I'm good at math for some reason (:D)

Anyways, Nationals this year really focused on counting and probability (19 out of 38 questions were Counting and Probability) and if you're like me, you're bad at those questions and you bomb it. But that was just one test, and if you didn't do well because... well... yeah.

I suck at c and p problems too :mad:

I would phail with you :wink:

I'm pretty good at counting and probability, I just make a bunch of careless mistakes.

Wow...you're opposite of me...Then which subject is hardest for you?

The topic that is hardest for me is probably geometry. But usually I do pretty good at that too. I just miss the hard AIME ones or the really hard ones at Nats.

My weakest topic is geometry as well. I'm okay at counting/probability, but I'm not so great at Number Theory. My best topic is algebra. :)

[quote="(^_^)"]hmph....i didnt think the chapter target was that easy...[/quote]
[sarcasm]Really?! I didn't know that![/sarcasm]

The target basically decided if you were in countdown or not. The target hated me.

My worst topic is Number Theory, but I don't see why I got the ammonia problems wrong on target if I do welll in algebra.

But I PWN IN GEOMETRY :D

I think I'll start another poll for this topic.

I was in the sixth grade this past year.

[quote="isabella2296"][quote="Poincare"]well now i am stupid, i was far away from even making nats. I didn't even reach cd in chapter. But then i got Aops books and got 18th at state  :)  but still didn't make nats  :wallbash_red:  :wallbash_red:[/quote]

How did you do MathCounts for two years if you were in 5th grade this past school year?[/quote]
I think he was just talking about if he really puts his shoes in the feelings of an actual State competitor. :)

Oh I thought she was asking me that question. I didn't see the quote from Poincare.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Suppose we have a finite group $ G$ with exactly one element $ f$ of order 2, then $ \prod_{g\in G}g\equal{}f$.

This seems like a really easy problem but I just can't figure it out.

[quote="coquitao"]$ \prod_{g \in G} g \equal{} \left(\prod_{g \in G\setminus \{e,f\}} g \right)( f)$.

In the first factor we can pair every element with its inverse. Since $ e$ and $ f$ are the only elements of $ G$ that are equal to their inverses, we can assure that this pairing lists every element of $ G \setminus \{e,f\}$ only once. Hence $ \left(\prod_{g \in G\setminus \{e,f\}} g \right) \equal{} e$ and we're done.[/quote]

This was exactly my argument but I wasn't convinced. We are not assuming the group is abelian and the product is in any arbitrary order you can think of so why does pairing each element with its inverse work?

Never mind. I thought about it some more and you're right. $ gh\equal{}hx$ where $ x\equal{}(h^{\minus{}1}gh)$, so we can keep moving around $ h$ in the product $ \prod_{g\in G}g$ until it is next to its inverse at which point we can cancel it and move on to the next element. Since every $ h\in G$ other than $ f$ has an inverse not equal to itself we can cancel all of them except $ f$.

[quote="davidk01"]Never mind. I thought about it some more and you're right. $ gh \equal{} hx$ where $ x \equal{} (h^{ \minus{} 1}gh)$, so we can keep moving around $ h$ in the product $ \prod_{g\in G}g$ until it is next to its inverse at which point we can cancel it and move on to the next element. Since every $ h\in G$ other than $ f$ has an inverse not equal to itself we can cancel all of them except $ f$.[/quote]

Well, try that on abAB where capitals are inverses.  You get aAb^AB = b^A B, but there is no reason to think that b^A and B match up as inverses in general (b^A's inverse could have been a!).  Of course {a,b,A,B} is not the entire group (minus the identity and the order 2 guy), but I think if you check the group SL(2,3) of order 24, you'll find the product depends on the order. The nonabelian group of order 12 with a single element of order 2 is also an example, <a,b:a^4=b^3=1,ba=abb>.

Wonder why it was your teacher decided to drop the condition $ G \in \mathbf{Ab}$...

[quote="JackSchmidt"]

Well, try that on abAB where capitals are inverses.  You get aAb^AB = b^A B, but there is no reason to think that b^A and B match up as inverses in general (b^A's inverse could have been a!).  Of course {a,b,A,B} is not the entire group (minus the identity and the order 2 guy), but I think if you check the group SL(2,3) of order 24, you'll find the product depends on the order. The nonabelian group of order 12 with a single element of order 2 is also an example, <a,b:a^4=b^3=1,ba=abb>.[/quote]

I'm not saying that they will match up but $ gh \equal{} hx$ for some $ x\in G$ and the product doesn't change when we perform this shuffle. Since all the elements of $ G$ appear in the product $ \prod_{g\in G}g$ at some point $ h$ must meet its inverse and be annihilated at which point we are left with fewer elements and can perform the same procedure. In the end only $ f$ will be left since it is its own inverse and every other element has an inverse that does not coincide with itself. I don't see anything wrong with this argument.

[quote="ZetaX"]You change a lot of factors on the way, which may cause some to be there twice whereas others are missing.
By the way, it's simpler to just show that the product _must_ change if you swap two neighbouring noncommuting elements.[/quote]

Ok, I see that commuting elements changes what happens among the other factors but it does not change the product. If $ s \equal{} ght$ and $ gh \equal{} hx$ then swapping $ gh$ with $ hx$ will not change the product since $ hxt \equal{} (hx)t \equal{} (gh)t \equal{} s$. I still think this line of reasoning will work, unless someone has an actual counterexample. All the groups I could think of worked.

[quote="davidk01"]unless someone has an actual counterexample. All the groups I could think of worked.[/quote]

I did give two explicit counterexamples, SL(2,3) of order 24 and 4|x3 of order 12.

(edited to add the following)

Here is the too explicit (parental caution advised) version: Let G=<a,b:aaaa=bbb=1,ba=abb>.  Then G is the set { 1, b, a, bb, ab, abb, aa, aab, aabb, aaa, aaab, aaabb }.  If I multiply this set in that particular order, I get aabb as the final reduced answer.  This is an element of order 6.  However, G has a unique element of order 2, aa = 1*b*bb*a*ab*abb*aa*aab*aabb*aaa*aaab*aaabb.  In one order you get the unique elment of order 2, but in another you get an element of order 6.

In some sense ZetaX is exactly right, if I change the order ( b,a was in the first list, but a,b in the second list), then I introduce a commutator.  If the product of all the elements of the group does not depend on the order, then every commutator must be the identity, and so the group must be abelian.  In other words, every nonabelian group with a unique element of order 2 is a counterexample.

For instance, the quaternion group of order 8 is { 1, -1, i, -i, j, -j, k, -k } and 1*-1*(i*-i)*(j*-j)*(k*-k) = 1*-1*1*1*1 = -1 is the unique element of order 2.  However, 1*-1*(i*-i)*(j*k)*(-j*-k) = 1*-1*1*i*i = -1*-1 = 1 is not of order 2.

I just happened to try SL(2,5) first and then simplified the example down SL(2,3) and then to 4:3.

[quote="JackSchmidt"]I did give two explicit counterexamples, SL(2,3) of order 24 and 4|x3 of order 12.

(edited to add the following)

Here is the too explicit (parental caution advised) version: Let G=<a,b:aaaa=bbb=1,ba=abb>.  Then G is the set { 1, b, a, bb, ab, abb, aa, aab, aabb, aaa, aaab, aaabb }.  If I multiply this set in that particular order, I get aabb as the final reduced answer.  This is an element of order 6.  However, G has a unique element of order 2, aa = 1*b*bb*a*ab*abb*aa*aab*aabb*aaa*aaab*aaabb.  In one order you get the unique elment of order 2, but in another you get an element of order 6.

In some sense ZetaX is exactly right, if I change the order ( b,a was in the first list, but a,b in the second list), then I introduce a commutator.  If the product of all the elements of the group does not depend on the order, then every commutator must be the identity, and so the group must be abelian.  In other words, every nonabelian group with a unique element of order 2 is a counterexample.

For instance, the quaternion group of order 8 is { 1, -1, i, -i, j, -j, k, -k } and 1*-1*(i*-i)*(j*-j)*(k*-k) = 1*-1*1*1*1 = -1 is the unique element of order 2.  However, 1*-1*(i*-i)*(j*k)*(-j*-k) = 1*-1*1*i*i = -1*-1 = 1 is not of order 2.

I just happened to try SL(2,5) first and then simplified the example down SL(2,3) and then to 4:3.[/quote]
Ok, I'm convinced. Thanks.

[quote="davidk01"][quote="ZetaX"]I gave you a counterexample: any nonabelian group. And just the product not changing doesn't give any reason why this should actually work.[/quote]

Saying any non-abelian group is not a counter-example. The conditions are: $ G$ is a finite group, $ G$ has exactly one element of order 2. So you need to produce a non-abelian $ G$ with exactly one element of order 2 and you need to show that $ \prod_{g\in G}g\neq f$. You did no such thing, hence you didn't produce a counter-example.[/quote]
I just didn't want to chew it for you. Finding such a group is really not hard and showing that this product is not constant (especially not always $ f$) was already explained by me.


one more for ya.....can't use calculus and have to use geometry.

A rectangular sheet of paper is to contain 72 square inches of printed matter with 2-inch margins at the top and bottom and 1-in margis on each side. What dimensions for the sheet will use the least paper?


Let x,y,z >0 such that x+y+z=1.Prove that
$ \sum\frac{x}{x\plus{}yz}\leq\frac{9}{4}$

[quote="DuongChinh_K41SPHN"]Let x,y,z >0 such that x+y+z=1.Prove that
$ \sum\frac {x}{x \plus{} yz}\leq\frac {9}{4}$[/quote]
$ x\plus{}yz\equal{}(x\plus{}y)(x\plus{}z).$ :wink:

[quote="arqady"][quote="DuongChinh_K41SPHN"]Let x,y,z >0 such that x+y+z=1.Prove that
$ \sum\frac {x}{x \plus{} yz}\leq\frac {9}{4}$[/quote]
$ x \plus{} yz \equal{} (x \plus{} y)(x \plus{} z).$ :wink:[/quote]
 and Hint:
$ (x \plus{} y)(y \plus{} z)(z \plus{} x) \geq \frac {8(xy \plus{} yz \plus{} zx)}{9}$
$ xyz \leq \frac {xy \plus{} yz \plus{} zx}{9}$
i'm sorry i cut my post.
but i don't understand why everyone don't post solution, and only ...vie to  denote small  mistake.

[quote="Hong Quy"]
$ xyz \geq \frac {xy \plus{} yz \plus{} zx}{9}$[/quote]
I think you want to say $ xyz \le \frac {xy \plus{} yz \plus{} zx}{9}$   :)

[quote="kunny"]you are right.

$ (x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)\geq 9xyz$ by $ AM \minus{} GM$. Isn't it?[/quote]
Yes  :)


I want to attend a math camp over summer and I thought SUMaC would be a good choice because I would like to go to Stanford in the future...but it seems like the curriculum is too specific.  My intention is to get a solid understanding of all of the precalculus topics, such as algebra, combinatorics, geometry, etc.  Is SUMaC a good idea, or do you recommend something else?

Thanks.

hm.. so I don't think you should pick a camp simply because you'd like to attend the university later on, but thats just me. obviously, stanford is great place and sumac is a great program probably because its in such a great place, but the college shouldnt determine it.

you say you want to get a good feel for precalculus topics like combinatorics and such... but SUMaC program I is pretty much Abstract Algebra (groups, rings, fields, and such) and program II is Algebraic Topology... both topics are pretty difficult to completely grasp. i wouldn't say there's much of a pre-requisite to sumac, but there are some precalculus topics you're expected to know. What i'm trying to get at is that summer camps like SUMaC are to learn stuff youll never encounter in high school, and most people dont encounter in college. If you're trying to get ahead for next year by doing precalculus over the summer consider EPGY (which is also stanford) or self-studying. If you want math that's above and beyond, go to a camp.

[quote="Chion"]What i'm trying to get at is that summer camps like SUMaC are to learn stuff youll never encounter in high school, and most people dont encounter in college. If you're trying to get ahead for next year by doing precalculus over the summer consider EPGY (which is also stanford) or self-studying. If you want math that's above and beyond, go to a camp.[/quote]

I agree.  The national-level summer mathematics programs are all about enrichment rather than acceleration.  That is, you won't learn high-school math; you'll learn parts of the college and graduate mathematics curriculum along with mathematics rarely taught in any college/graduate courses.

Also, all of the national-level programs require doing well on an exam of some sort to get in.  I don't know that any of the programs require precalculus before entrance, but they all require that level of mathematical maturity; that is, you'd have to be at least ready for precalculus-level ideas.


Assume [tex]x>1[/tex], [tex]y>1[/tex],[tex]x^xy^y=z^z[/tex], p is prime, p|x, show that p|y.
goodluck

[quote="mozilla"][quote="nguyenquockhanh"]Assume [tex]x^xy^y=z^z[/tex], p is prime, p|x, show that p|y.
goodluck[/quote]

Give an example of such x,y and z. Do they exists? If they do not exist then the problem is wrong.[/quote]
yes,
[b]PRO[/b]:[b]there exist infinite triples satisfying that condition[/b]
HINT:for pro "exist" we use form $2^a\cdot{(2^n-1)^b}$

[quote="Lopes"]x < z, p|z => p^z|z^z => p^z|(x^x).(y^y) => p|y^y => p|y

Sorry, but i don´t speak english very well and i don´t know TeX  :([/quote]

[b]why $p|y^y$[/b][/b]

sorry..
my conclusion is not correct  :(

okie
in my solution, i have some mistake and i've corrected it
and now, check its


I was unsure of four problems that was on my test.  The first three is finding the derivatives. The last one is finding the limit. I want to make sure that I had the right answer.

1.

(3x+3)   ^5
_______
x-2

2.

 xy+y=2

3

1
_____
(5x-3)^5

4. sin x
_______
    3x

limit is 0.


inequalities
AMC
AIME
pigeonhole principle
modular arithmetic
number theory
relatively prime

I have observed the lessons on inequalities and think that it would be nice to extend the concept to another less than mainstream topic: Number Theory. Problems and theorems useful for >=  AIME level math would be appreciated.

One of my favorites is Wilson's Theorem which states that if we have a prime $p$ then $(p-1)!+1$ is divisible by $p$ while if $p$ is composite then $(p-1)!+1$ is not divisible by $p$.

[quote="chess64"]When you state the theorem or whatever, could you also give a problem using it, if possible? Thanks. That would help to understand it.

For example, what are the applications of Wilson's Thoerem?

Can someone please explain the Chinese Remainder Theorem? I looked it up but I couldn't understand it really.[/quote]
Apperently some usage of that relies on tables, so I don't think you will find many problems using it.

Prove Wilson's Theorem!

Let $p$ be a prime number which yields the remainder $1$ upoin division by $4$. Prove that there exists an integer $x$ such that $x^2+1$ is divisible by $p$.

This is not quite going how I expected. I guess I'll start it off. Perhaps you can provide a proof DPopov?
Problem:
find all n such that
13|(n^5+5^n)

Proof of Wilson's theorem: (tell me if this works, I'm a little sleepy)
Lemma: if (x,p) =1, then there exists some integer y such that
xy = 1 mod p, and y is one of the smallest p-1 remainders mod p
Proof:
Contradiction. Suppose there does not exist such a y. Then consider the first p-1 multiples of x mod p
x, 2x, ...., (p-1) x. In some order these are congruent to 2, 3, ... but not 1. Thus by pigeonhole principle, two of these must be congruent mod p. so ax = bx mod p. Factoring, we find x(a-b)=0 mod p, and thus
p|(a-b). But since (a-b)x is one of the multiples of x under consideration, p does not divide a-b. ><
Since we have proven the first part of the lemma, we can find an integer <= p-1 congruent to y, so that it is one of the first p-1 remainders mod p.
Now we apply our lemma to the product
1 * 2 * ... (p-1)  mod p
Each of 2, 3, must be paired with another integer in the product. Also, p-1 = -1 mod p
Combining these two facts, we find that the middle p-3 integers 'pair' with each other (they can't pair with 1 or -1) to produce a product of 1, leaving a total product of -1.

Heh, I am also sleepy but I am 99.9% sure that you have given a correct proof.

[quote="DPopov"]Let $p$ be a prime number which yields the remainder $1$ upoin division by $4$. Prove that there exists an integer $x$ such that $x^2+1$ is divisible by $p$.[/quote]

Lemma: If $GCD(a,p) = 1$, $a^{\frac {p-1}2} \equiv 1 \Leftrightarrow a$ is a square mod $p$ (quadradic residue)

If $a$ is a square mod $p$ then there exists some $u$ with $GCD(u,p)=1$ such that $u^2 \equiv a \pmod p$. By Fermat's Little, $u^{p-1} \equiv 1 \pmod p \Rightarrow a^{\frac {p-1}2} \equiv 1 \pmod p$.

If $a^{\frac {p-1}2} \equiv 1$ then there exists some primitive root $g$ such that $g^k \equiv a \pmod p$ (A primitive root is a number relatively prime to the mod, $p$ in this case, so that when the number is raised to all the powers 1 to $\phi (p-1)$, all of the possible remainders relatively prime to the mod are obtained. There is a proof that there always exists a primitive root mod $p$ where $p$ is a prime, but I will not explain it right now.)

Thus $g^{\frac {k(p-1)}2} \equiv 1 \pmod p$. Because $g$ is a primitive root, $(p-1)|\frac {k(p-1)}2$ (I will not say why right now). Therefore $\frac {k(p-1)}2 = (p-1)q$ for some integer $q$ and we find that $k = 2q \Rightarrow a$ is a square since $g^{2k} \equiv a \pmod p$.

Now if $-1$ is a square mod $p$, $(-1)^{\frac {p-1}2} \equiv 1 \pmod p$ and the only way this can happen is if $\frac {p-1}2$ is even, $\frac {p-1}2 = 2r \Rightarrow p= 4r+1 \Rightarrow p \equiv 1 \pmod 4$.

Edit: Oops I went the wrong way. The acutal proof starts at the last "paragraph". Reverse the steps.

[quote="chess64"]When you state the theorem or whatever, could you also give a problem using it, if possible? Thanks. That would help to understand it.

For example, what are the applications of Wilson's Thoerem?

Can someone please explain the Chinese Remainder Theorem? I looked it up but I couldn't understand it really.[/quote]

The Chinese Remainder Theorem is a method of finding $x$ when

$x \equiv r_1 \pmod {m_1}$

$x \equiv r_2 \pmod {m_2}$

...

$x \equiv r_n \pmod {m_n}$

where $m_1, m_2, ..., m_n$ are pairwise relatively prime.

[quote="DPopov"]The problem I gave has a proof using an application of Wilson's Theorem. However, I don't have time now but if no one has posted a similar proof I will post one later tonight.[/quote]

$\left(\frac {p-1}2\right)! \equiv (-1)^{\frac {p-1}2} \cdot \frac {(p-1)!}{\left(\frac {p-1}2\right)!} \pmod p$

$\left(\frac {p-1}2\right)!^2 \equiv (p-1)! \cdot (-1)^{\frac {p-1}2} \equiv (-1)^{\frac {p+1}2} \pmod p$

Thus if $p \equiv 1 \pmod 4$, $\frac {p+1}2$ is odd and $\left(\frac {p-1}2 \right)!^2 \equiv -1 \pmod p$ and$\left(\frac {p-1}2 \right)!^2 + 1 \equiv 0 \pmod p$

Some important numerical exercises.
Solve:
1235x+45=9090(mod24)
1235x+45=9090(mod11)

(separately)

solve the system of congruences:
x=1(mod3)
x=4(mod5)
x=2(mod7)
x=9(mod11)
x=3(mod13)

also, the problem
find all n such that
13|(5^n+n^5)
remains.


Prove that:
$\frac{a}{\frac{1}{b}+\frac{1}{c}+1}+\frac{b}{\frac{1}{c}+\frac{1}{a}+1}+\frac{c}{\frac{1}{b}+\frac{1}{a}+1}\leq 1$
where $a;b;c$ are non-negative

[quote]Prove that:
$\frac{a}{\frac{1}{b}+\frac{1}{c}+1}+\frac{b}{\frac{1}{c}+\frac{1}{a}+1}+\frac{c}{\frac{1}{b}+\frac{1}{a}+1}\leq 1$
where a;b;c are non-negative[/quote]
um, let $a, b, c \to \infty$ then $LHS \to \infty$


function
calculus
integration
trigonometry
algebra unsolved
algebra

Find the necessary and sufficient conditions on values$a$ and $b$ if the real function($R$ to $R$)$f(x)$ is periodic with an integral period(non-zero) and
$f(x+1)$ $=$ $af(x)$ $+$ $bf(x-1)$ for all real $x$

For any a,b, exist periodic f, satisfyed your condition, but f(x) can be is not continiosly and had irrational period. If f(x) continiosly or had rational period, then b=-1 $a=2$ or $a=0$ or $a=\pm 1$.

Not quite.
$f(x)=\sin{2\pi\over n}x$ is continuous with period $n$, satisfying $f(x+1)=2cos{2\pi\over n}f(x)-f(x-1)$.

And $f\equiv 0$ satisfies the conditions and is periodic for any $a,b\in\mathbb{R}$.
And $f\equiv 1$ satisfies the conditions and is periodic for any $a,b\in\mathbb{R}$ with $a+b=1$.
And $f(x)=\{{1\atop-1}{\text{for }[x] \text{ even}\atop \text{for }[x] \text{ odd}}$ satisfies the conditions and is periodic for any $a,b\in\mathbb{R}$ with $b-a=1$.
And $f(x)=\sin\pi x$ also satisfies the conditions and is periodic for any $a,b\in\mathbb{R}$ with $b-a=1$.

Yes.
1. If f(x) had period 1, then work a+b=1.
2. If f(x) had period 2, and f(x+1)=-f(x), then work b-a=1.
3. If b=-1 work $|a|\le 1$, for example f(x)=sin(cx), a=2cos c, b=-1.


1 Unable to move uploaded file. Please try again.

Is what my page tells me.

The regular upload link isn't showing up, and the alternate page loads, then turns completely blank. Is this supposed to happen? I uploaded my round 1 solutions without any problems.

Apparently the website crashed last night shortly before the deadline.

Anyone who tried to upload last night and failed can email their solutions to usamts@usamts.org before 6 PM today and we will still accept them.  [b]You are on the honor system not to continue to work on your solutions but to email the same file that you tried to upload last night.[/b]

[quote="andersonw"]I uploaded my solutions last night, but when I look at the PDF file I uploaded (on the My USAMTS page), it seems that the file only includes the top-left corner of each page.
Is there something wrong?

Edit: Oh, and I haven't received a checkmark either.[/quote]Hard to tell, but if you uploaded and didn't get a checkmark, it probably didn't go through properly.  Email us the file so that we can be sure.


inequalities
floor function
number theory proposed
number theory


geometry
function
3D geometry
prism
trigonometry
probability
algebra

This is the official mock amc B solutions topic.
Please keep the solutions in order.
Thanks.

[i]Moderator's note: title edited - I don't like capital letters.[/i]

1.  Mode is the number that occurs most.  Thus x+10=20 so x=10.  The median is the middle number(or average of two middle numbers).  They are 19 and 19 which when averages equals 19.  (B)

2.  Moving the decimal point to the right four is equivalent to multiplying by 10 000. So we have \[10 000x=4(1/x)\Rightarrow x^2=\frac{4}{10000}\Rightarrow x=\pm \frac{2}{100}\].  It is positive so the answer is .02  (C)

3.  The circles can intersect in two places among themselves.  The lines can intersect each of the circles twice making 3*2*2=12.  Finally the lines make C(3,2) or 3 intersections among themselves for a grand total of 17 (B)

4. A five digit looks like $a_1a_2a_3a_2a_1$.  There are 9 choices for $a_1$ and 10 for $a_2$ and $a_3$ each.  So there are 9*10*10=900 5 digit palindromes.  A 3 digit looks like $a_1a_2a_1$.  Ther are 9 for $a_1$ and 10 for $a_2$ making 90 3 digit palindromes.  900-90=810 (D)

1. The mode of x+1, x+1, x+7, x+9, x+9, x+10, x+10, x+10 is 20. What is the median?
(a) 10 (b) 19 (c) 20 (d) 17.125 (e) 19.5

The mode is x+10, and it equals 20.  So x=10.  Our numbers are 11, 11, 17, 19, 19, 20, 20, 20.  The median is thus the average of 19 which is [b]19 B[/b].

2. When the decimal point of a certain positive decimal number is moved four places to the right, the resulting number is four times the reciprocal of the original number. What is the original number?
(a) 2 (b) 0.2 (c) 0.02 (d) 0.002 (e) 0.0002

x=original number.  Thus, 10,000x=4(1/x). Solving gives [b]x=0.02 or C[/b].

3. Two circles and three lines lie in the same plane. If neither the circles nor the lines are coincident, what is the maximum possible number of points of intersection?
(a) 16 (b) 17 (c) 18 (d) 19 (e) 20

Each line can hit each circle in 2 places.  Thus 3 lines can hit 2 circles in 3*4=12 ways.  Two circles can intersect with each other in 2 places.  Three lines can intersect in 3 places.  12+2+3=[b]17 or B[/b]

4. What is the positive difference in the number of five digit palindromes and the number of three digit palindromes?
(a) 9 (b) 81 (c) 900 (d) 810 (e) Not Here

There are 9*10*10*1*1=900 5 digit palindromes and 9*10*1=90 3 digit palindromes.  900-90=[b]810 or D[/b]

Here are some more

5.  We have that xyz=6783=3*7*17*19
Letting one of x,y,z be 21, another be 17 and the last be 19 we have 57 as the smallest (B)

6.  6000=2^4*3*5^3

    6000^2=2^8*3^2*5^6

Let's represent a factor of 6000^2 by 2^a*3^b*5^c.  a can be 4,5,6,7 or 8.  b can be 1 or 2 and c can be 3,4,5, or 6.  Therefore there are 5*2*4=40 numbers that fit our requirements (A)

[edit]7.  $x^{yz}=2^6$ is basically what it tells us.  Let's start by having x=2
(y,z) can be (1,6) (2,3) (6,1) or (3,2).  Letting x=4 (y,z) can be (3,1) or (1,3).  If x=8 (y,z) can be (2,1) or (1,2). Finally if x=64 (y,z) can only be (1,1) this gives us 9 (E)

joml88--you forgot (64, 1, 1), so there are actually 9 ordered pairs.

I also got D for Question 7

But the answer listed officially was E

Can someone who got "E" explain?

EDIT: Nevermind

8.) You have x^2 + y^2 = 25^2, so find the pythagorean triples that work. x = 7, y = 24, and x = 15, y = 20, are two possible pairs. You reverse that, and you have 4 pairs in the first quadrant. There are four quadrants, so 4*4 = 16. THen, there is (25,0),(-25,0),(0,25), and (0, -25). 16 + 4 = 20. (B)

9.) f(-3) = 81a - 9b + 2 = 2, so you have 81a - 9b = 0.
You want f(3) = 81a - 9b + 8, which is 0 + 8 = 8. (E)

10.) xy = 40, zy = 80, xz = 200. By a little substitution, x = 10, y = 4, and z = 20. 10+20+4 = 34. (B)

11.) In the triangle ABC, we see by comparison of areas that all three medians divide the triangle into 6 equal areas, and the quadralateral is 1/3 of the entire area, which by Heron's theorem, is 84. 1/3 * 84 = 28. (C).

white_horse_king88, for number 8, how did you know that the solutions to x and y in x^2 + y^2 = 25^2 would be on the circle?  Also, how did you find the solutions?

And for number 9, functions don't always have to equal each other right?  This was just a special case?

14.) Think of this problem as a rectangular box. You have one side as 8, and the other as 1. Then, you have the diagonal of the rectangular prism as 9. So, [tex]\sqrt{8^2 + 1^2 + x^2} = 9^2[/tex]. So, x = 4, or [tex]\sqrt{16}[/tex], which is (D)

15.) The number of diagonals in any polygon, is determined by the formula n(n-3)/2, and n(n-3)/2 + n = n(n-1)/2. So, [tex]D_n = \displaystyle\frac{n(n-1)}{2}[/tex]. [tex]\frac{1}{D_n} = \frac{2}{n(n-1)}[/tex]. You may realize, that all the terms can by expressed this way - so the sum of 1/(D_n) + 1/(D_(n+1)) .... + 1/(D_(n+2000)) = -1(2/n - 2/(n-1) + 2/(n+1) .......... +2/(n+1999) - 2/(1998) + 2/(n+2000) - 2/(n+1999)), which since n = 4 is -1( - 2/3 + 2/ 2004) = -1((-668 + 1) / 1002) = 667 / 1002. So, 667 - 1002 = -335, which is not to be found so (E). I think that's right, but I'm not sure.

For #15 you can just realize that the total number of diagonals plus number of sides would be how many ways you can connect 2 points out of n points, or ${n \choose 2}$.

To add to nightflarer's question, how do you find those pythagorean triplets? To answer nightflarer's other question-the points on a circle with a center at the origin is in the form of x 2 +y 2 = radius 2 . So in this case, it would be x 2 +y 2 = 25 2 .

To FermatPrime:

Number 19) You have 12 quarters and you give them amongst A, B, C, D, and E. That is 5 different people. Set A = 2 quarters to start. So, there are still 5 people but 10 quarters left. There is a clever formula that I memorized for item distribution. Suppose you have n items to be distributed amongst m people. It is [tex]\binom{n + m - 1}{n}[/tex]. So, you have 10 items, 5 people, so [tex]\binom{14}{10} = 1001[/tex] using the calculator. (D). :lol:

To add on - and answer the question. Visit [url]http://mathworld.wolfram.com/PythagoreanTriple.html[/url]

[quote="Fermatprime"]How'd you do 19 and 21?[/quote]

21)Let RS=x, 17+x=21+16-x, yields x=10, BS=6. Now you can use Stewart
s Theorem or Law of Cosines. Since I forgot Stewart's Theorem, I have to do it with Law of Cosines. Let $RSH= \theta$, $HSB=180-\theta$
$10^2+HS^2-20HS\cos\theta=17^2$
$6^2+HS^2+12HS\cos\theta=21^2$

5 times the second equation+3 times the first yields
$480+8HS^2=3072$
HS^2=324, HS=18

Could someone explain(or provide a link) for why white_horse_king's formula works?

12. Suppose that we choose 12 integers at random, without replacement, from the set {1, 3, 5, ..., 2005}. The probability that 1979 is chosen can be written as , where (m, n)=1. Find mn.
(a) 1015 (b) 13039 (c) 12024 (d)10030 (e) Not Here

Since 2005=2(1003)-1 there are 1003 numbers in that set.  The total number of ways to choose 12 from it is C(1003,12).  The total number of ways to choose 12 with 1979 is C(1002,11) because we already know that 12 is one of them so we have to choose the other 11 from the remaining 1002.  Thus the probability is \[\frac{{1002\choose 11}}{{1003\choose 12}}\]  Using a calulator this comes out to $\frac{12}{1003}$ and since it is looking for mn and not m+n :-x the answer is 12*1003=12036 (E)

I got (A) too because I thought Rep was looking for m+n.  :D

Could someone inform me on how to do number 16? I tried that one, but I resorted to lots of trig - and somewhere along the way I got disorganized.

[quote="beta"][quote="joml88"]Could someone explain(or provide a link) for why white_horse_king's formula works?[/quote]
Look at this problem, 10 quarters to distribute to 5 people.
Look at the sequence
QQQQQQQQQQ and you have 4 dividers to divide the sequence into 5 part, the number of Q's in each part correponds to the number of quarters each person has.
For example Q_Q_QQQ_QQQ_QQ. So our problem corresponds to how many ways can we divide the sequence into 5 parts(it's possible that someone can have 0 quarters). Thus the total number of ways correponds to ${14 \choose 4}$.[/quote]

Yeah, that's right. I should've put 1 as an answer choice (because someone has to get at least 2 quarters.  :P )

Heh, everyone got the same solutions I got when I wrote the questions. Yeah, those two were mine. I was sort of hoping for something completely unexpected.

(BTW, Stewart's Theorem is c2 n + b 2 m = d 2 a + mna)

Suppose we have a triangle with vertices at points A, B, and C.  Let a, b, and c be the lengths of sides BC, AC, and AB.  Then the angle bisectors of the triangle meet at the point (aA + bB + cC) / (a + b + c).

Was that your question?

24.

I think this way works.
There's a theorem that states that if a is a root of f'(x) and a is a root of f(x)
then a is a multiple root of the polynomial f(x).
Now if we find roots of the derivative of the original polynomial, we find that 1.5 is a root. Then we can check to see if it works for the original polynomial.

I'm not completely sure it works though

[quote="Rep123max"]Anybody have a solution to #22 without using $\sqrt{R(R-2r)}$? I guess you might be able to do it analytically.?[/quote]

what formula is this?  i guess it's pretty well known, but i've never encountered it >_<

i did this the tedious coordinates way- expressing the perpendicular bisectors of 2 sides as linear equations, seeing where the two equations intersected, then pythagorean's formula to find the distance between that and the incenter.

Can someone please show how to do questions 23 and 24? Thanks!

Also, can someone tell how to do number 16?

It said:

Triangle ABC has <A = 60, <B = 45, and the bisector of <A meets BC at T, where AT = 24. The area of ABC = a :sqrt: b + c where b is not divisible by the squares of any prime. What is a/b+c

To get better help, here is what I did.

Since AT is a bisector of angle A, <BAT :cong: <CAT = 30 degrees. And since <B = 45, then <BTA = 105. With that, <ATC = 75 (supplementary) and therefore, <ACT = 75. Now, if I draw an altitude from <BTA to such that BA, let's call point Q, that BA :perp: QT, then I can find the following lengths.

BQ = 12
AQ = 12 :sqrt: 3
BT = 12 :sqrt: 2
AT = 24
QT = 12
AC = 24
and if the altitude from angle A meet at BC at Z, then:

BZ = 12 + 12 :sqrt: 3

So, I can find the area by Heron's Formula but this doesn't help me any. Can anyone post solution to this?

Thnx. :P  :lol:

Did anybody have a nice solution to #17?  Mine was really ugly and involved some weird formulas.

I started by getting the semiperimeter

s=(1/2)(a+b+c+d)
s=(1/2)(25+39+52+60)
s=88

Then I got the circumradius with

R=(1/4)sqrt(((ab+cd)(ac+bd)(ad+bc))/((s-a)(s-b)(s-c)(s-d)))
R=65/2

So the diamater was 65.

But how else could you solve this?  I had to go to mathworld to get the formulas for this.

By the way, I did this Mock AMC in school in 68 minutes and got a 123.  Then I came back and without a time limit, solved a few extra problems to get a score of 137.  After that I checked my answers.  Think that's good enough to get >=100 on the AMC12?  I'm taking it tomorrow... :P

Can anyone show me how to do #24?? The only way I can think of is graphing it or plugging in the answers.  :? Oh, and how do you do #17?

thanks!


recurrence relation
combinatorics
system of equations
IMO
IMO 1967

In a sports meeting a total of $m$ medals were awarded over $n$ days. On the first day one medal and $\frac{1}{7}$ of the remaining medals were awarded. On the second day two medals and $\frac{1}{7}$ of the remaining medals were awarded, and so on. On the last day, the remaining $n$ medals were awarded. How many medals did the meeting last, and what was the total number of medals ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[b]Solution.[/b]  Let $ m_{k}$ be the number of medals awarded on the $ k^{\text{th}}$ day.  Notice that after the $ \ell^{\text{th}}$ day, $ 6(m_{\ell}-\ell)$ medals are left for day $ \ell+1$.  By the given information in the problem,
\[ \begin{eqnarray*}m_{x+1}&=&x+1+\frac{6(m_{x}-x)-x-1}{7}\\ &=&\frac{6}{7}(m_{x}+1)\\ &=&\left(\frac{6}{7}\right)^{x}(m_{1}-6)+6\\ &=&\frac{6^{x}}{7^{x+1}}(m-36)+6,\]
using $ m_{1}=1+\frac{m-1}{7}$.

Since $ m_{n}=n$, $ 7^{n}(n-6)=6^{n-1}(m-36)\implies 6^{n-1}|n-6\implies n=6\wedge m=36$.

As a result, $ (m,n)=(36,6)$. $ \Box$

[hide="small correction"]
Assume that $(m, n) = (36, 6)$. Then there will be $\frac{36-1}{7}=5$ medals after the first day, and then $\frac{5-2}{7} = \frac{3}{7}$ medals after the second day, impossible as there must be an integer number of medals.

Thus, we conclude the only answer to be $(m, n) = (1, 1)$[/hide]

EDIT: not sure what I was thinking, the above is utter nonsense. Apologies.

[quote="johnmichaelwu"]
Assume that $(m, n) = (36, 6)$. Then there will be $\frac{36-1}{7}=5$ medals after the first day, and then $\frac{5-2}{7} = \frac{3}{7}$ medals after the second day, impossible as there must be an integer number of medals.

Thus, we conclude the only answer to be $(m, n) = (1, 1)$[/quote]
You read it wrong. With $(m,n)=(36,6)$, there will be $1 + \frac{36-1}{7}=6$ medals awarded the first day, thus $36-6 = 30$ left. Then there will be $2 + \frac{30-2}{7}=6$ medals awarded the second day, thus $30-6 = 24$ left. And so on, everything working fine.


I heard of this game in a magazine.  You define a word in a funny, although somewhat true way.  For example, for "water" you could say "a colorless odorless chemical that can cause death if inhaled" or you could just describe someone.  WHen someone else gives the "daffynition" of a word, they have to provide a new word as well.  Let's start with(and this may get interesting):
[b]politicians[/b]

a synonym of actors

[b]statisticians[/b]

someone who fumbles with the government and SafeAuto

[b]athletes[/b]

Please, don't let this game die so early.........

Fine:  people who get paid to exercise

[b]CNN[/b]

great story tellers....

[b]Valentin Vornicu[/b]

I think this is going down the wrong path...........

I don't get it and it is pretty confuisng...

A word that may be used in spelling contests

[b]maybach[/b]

a car rich people buy and don't drive(they get chaueffered)

[b]Ted Stevens[/b]


If X  dollars will buy  Y  pounds of chopped meat, how many dollars are need to buy Z  pounds of the same meat?

a   XZ/Y   b  XYZ  c) Z/XY  d) (X+Y)/Z    e) XY/Z''.

Price is $ \frac{X}{Y}$ dollars per pound. There are $ z$ pounds of meat, so the price for $ Z$ pounds of meat is $ \frac{XZ}{Y}$.


:D I have a money distribution calculation problem.  I know someone is up to the challenge. I have $44,564.00 inheritance to split between 4 people. Person G borrowed (in the past) $11,800.00   Person Z borrowed (in the past) $ 4,300.00  Person T and person B have never borrowed any. I have tried to add dedections to the inheritance amount, then deduct the amount loaned. I have tried to split up each way possible.  I cannot come up with a fair distribution that pleases my brothers.  Can anybody help in the most fair way to split up the remaining money 4 ways, taking into consideration past loans?  Thank you

[quote="rewritten to avoid TeX display problems from dollar sign..."]
I have USD 44,564.00 inheritance to split between 4 people. Person G borrowed (in the past) USD 11,800.00   Person Z borrowed (in the past) USD 4,300.00  Person T and person B have never borrowed any. I have tried to add dedections to the inheritance amount, then deduct the amount loaned. I have tried to split up each way possible.  I cannot come up with a fair distribution that pleases my brothers.  Can anybody help in the most fair way to split up the remaining money 4 ways, taking into consideration past loans?  Thank you[/quote]

Your calculation sounds right.
There were originally 44564 + 11800 + 4300 dollars = 70664.
Everyone is (I assume) entitled to a 1/4 share of that = 17666.
The ones who took advances on their shares through borrowing, receive the remainder (i.e. their 1/4 share, minus amount borrowed), as you calculated.

What are the brothers' objections?  A similar computation could be done with interest on the borrowings, if that is an issue.  The parties that didn't borrow could of course decide to cancel some or all of the loans due to them, but (legally speaking) that is not a decision that the borrowers can make unilaterally.  The loan doesn't die with the lessor, generally it is an asset that passes to the heirs.  There may be questions of taxation that are affected by the difference between past loans and present dollars, which may make it more favorable to the inheritors to carry out some other arrangement than the above.


This problem you have to post in box Algebra, i think  :wink:

we need to proove inequality. $inequality \in algebra$ anyway


inequalities
Cauchy Inequality
inequalities proposed

Let $ a$, $ b$, $ c$ be positive real numbers satisfying $ a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \left(bc\plus{}ca\plus{}ab\right)\left(\left(\frac{2a\plus{}b}{b}\right)^{2}\plus{}\left(\frac{2b\plus{}c}{c}\right)^{2}\plus{}\left(\frac{2c\plus{}a}{a}\right)^{2}\right)\geq 9$.

I have a solution by brute force but it is quite long.Anyone has a nice solution for this.Thank you

Oh, sorry I proved something else. I will try to solve later your inequality again.

[quote="bui_haiha2000"]I think ab+bc+ca<=1/3 is true because a+b+c=1.Can you have the full solution,Beat and Romano[/quote]
 Yes, It is true that $ab+bc+ca \leq \frac13$ but from the sols of Beat, he can't prove  your inequality because the reverse sign at the end :D .
  Probably, you should give us more time for the nice inequality! :lol:

Yes,of couse because i need the second solution for this very nice ine.My solution is quite long.I hope every body will help me :D

[quote="bui_haiha2000"]No one help me?????????? :o[/quote]

  I'm always here to help you...especially if u are a girl  :D ...
OK i solve this problem in three steps:
1) $( \sum ab ) ( \sum ( \frac{2a+b}{b} )^2 ) \geq 9$ is equivalent with $( \sum ab )( 3+ 4 \sum (( \frac{a}{b})^2+ \frac{a}{b} ) \geq 9$
Due to Cauchy inequality we have
$\sum (( \frac{a}{b})^2+ \frac{a}{b}) \geq \sum ( \frac{a}{b} + \frac{b}{a} )$ ( self prove )
But we also have:
$\frac{a}{b} + \frac{b}{a} = \frac{ a^2+b^2 }{ab}$
So the RHS is equal to : $\sum ( \frac{a^2+b^2}{ab})$
2)From Bunhiakowski inequality: $RHS( \sum ab ) \geq ( \sum( \sqrt{a^2+b^2}))^2$ expanding  the LHS
Continue by using Minkowski inequality to prove that:
$\sum ( \sqrt{a^2+b^2} \sqrt{a^2+c^2} ) \geq \sum (a^2) + \sum(ab)$
3)It's very simple now,u pair $\sum (ab)$ and $\sum(a^2)$ to have $( \sum(a))^2$ and remember that $\sum a =1$ and  $\sum(a^2) \geq \frac{1}{3}$....

For such long time that i've not touched inequality, hope my solution not too bad....


videos
linear algebra
matrix
geometry
geometric transformation
reflection
Gauss

POst why you think the person above you chose their avatar (it doesn't have to make sense!)

because mewto sucks, just like mewto55555.

[quote="mewto55555"]HAAHAHAHA ITS A DECAPITATED PACMAN SKULL! IT HAS NO EYEHOLES![/quote]

1st of all: Pacman doesn't have body at the original game
2nd of all: Pacman have small eyes

1=2 chose a fake avatar because his username is fake

pacman got eaten by the evil pacman eating things in the game pacman. then they took a picture of him and made him use it as his avatar :D

guineapiggies has a guinea piggy because (no duh) SHES GUINEAPIGGIES

o and dont try mine trust me there was no reason for my avatar just that i wanted one and i got a fractal

edit: just changed my avatar

<-- guess who this is (hes from a video game not main character but ally)

Err you think he resembles your personality.

Umm... you made that avatar because you were bored and you still use MS Paint, the worst image editor ever.

herefishyfishy1 chose their pic 'cause...um...they...couldnt come up with an idea for the pic.

HINT for my avatar: i like people to realize that im a girl! :mad:

Because she needed a flower to make herself feel better because she hadn't yet signed up for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=220270]BB2[/url] in the games forum, duh.

o...THAT. what r the rules??? MAYBE i'll join.

And you just have a simple avatar. too simple. you are a simple person.

You have an undefined avatar. Nobody knows what it is except you. You can only understand yourself.

it is a colorful blob of blobbing blob blobbiness

Lstar please stop reviving old threads, the last post was in 2009
PhirekaLk6781 lock this now

Lstar I have counted 16 revivals now.
STOP IT!
this has gotten really annoying for PhirekaLk6781 now so stop it.
IF YOU LIKE A GAME MAKE A NEW TOPIC SAYING "game name(revival)"
DON't REVIVE OLD THREADS!
ALSo if a mod locks one of your old posts, then make sure that you don't try it again, they have reasons to lock threads.
So do Phirekalk6781 a favor and stop making SPAM!
Just stop reviving old threads.
A majority of AOPS WOULD BENEFIT IF YOU DID!
sorry for yelling :wink:

Yes. Lstar, you cannot post in a thread if the last post was more than a month ago. Please stop reviving, or you will be banned. Mods, please lock.

Q.E.D.

Firebubble revived it by about 4 hours
YOu revived it by 10 months

I REVIVED IT FIRST!!!

this is making me angry

Where did this pop up from?
From what i've seen,  this should be locked.
Also, um, Lstar, are you trying to take credit for reviving something when you ought not to?

[quote="PhireKaLk6781"]Nice to know. Just make sure you return to the place from which you came and detonate yourself where you won't be hurting others, please.

Also, reporting another user for "being mean to me," especially when they're just stating the rules is the worst abuse report I've seen so far, even among the majority of Fun Factory abuse reports being improper.[/quote]
What the heck, my post got reported for me stating the rules?

Don't worry, there won't be any discrediting to you for that.

Yeah, time to lock this thread before it starts dancing on its belly.


geometry
3D geometry
sphere
calculus
integration
function
conics

I know that the equation of the cone is Z = R, the bounds of my integral relative to (theta) are 0 and 2 pi, but I do not have the clue as to what is the function to integrate or the bounds relative to R.

However, I do know that an element of a polar surface is R DR D(theta). And the sphere is radius 1.

The familiar sno-cone problem: Let the sphere have radius $ R$ and the cone be the usual $ z^2 \equal{} x^2 \plus{} y^2$ so that we have the sphere being $ r^2 \plus{} z^2 \equal{} R^2$ and the cone is $ z \equal{} r$. The upper bound on $ r$ is the projection of the curve of intersection of the sphere and the cone in the $ xy$-plane, namely $ r^2 \plus{} r^2 \equal{} R^2\Rightarrow r \equal{} \frac {1}{\sqrt {2}}R$. The lower bound of $ r$ is $ 0$. Also, $ z$ ranges from the cone to the sphere, which is $ r\leq z\leq\sqrt {R^2 \minus{} r^2}$
\[ V \equal{} \int_{\theta \equal{} 0}^{2\pi}\int_{r \equal{} 0}^{\frac {1}{\sqrt {2}}R}\left(\sqrt {R^2 \minus{} r^2} \minus{} r\right)r\, dr\, d\theta
\]
[hide="Click for calculation of integral"]\[ V \equal{} \int_{\theta \equal{} 0}^{2\pi}\int_{r \equal{} 0}^{\frac {1}{\sqrt {2}}R}\left(\sqrt {R^2 \minus{} r^2} \minus{} r\right)r\, dr\, d\theta \equal{} 2\pi\cdot \int_{u \equal{} \frac {1}{2}R^2}^{R^2}\sqrt {u}\, \frac {du}{2} \minus{} 2\pi\cdot\int_{r \equal{} 0}^{\frac {1}{\sqrt {2}}R}r^2\, dr
\]
\[ \equal{} \frac {2\pi}{3}\left[ u^{\frac {3}{2}}\right|_{u \equal{} \frac {1}{2}R^2}^{R^2} \minus{} \frac {2\pi}{3}\left[ r^3\right|_{r \equal{} 0}^{\frac {1}{\sqrt {2}}R} \equal{} \frac {2\pi}{3}R^3\left( 1 \minus{} \frac {1}{2\sqrt {2}}\right) \minus{} \frac {2\pi}{3}\cdot\frac {1}{2\sqrt {2}}R^3
\]

\[ \equal{} \boxed{\frac {\pi}{3}R^3\left( 2 \minus{} \sqrt {2}\right) }
\]
I did, BTW, verify this calculation by adding the volume formulas for a cone and a spherical cap appropriate to our situation.[/hide]

(i) Does  Z = R means the generatrix makes 45° with the axis?

(ii) The sphere has unitary radius?

(iii) The sphere is tangent (i.e. inscribed ) to the conic surface?

(iv) What is sno-cone?

(i) Yes.

(ii) Yes.

(iii) No, the sphere is not tangent to the cone. Its center is the vertex of the cone, so it's actually orthogonal rather than tangent.

(iv) A sno-cone is an old-fashioned American snack food, associated with fairs, carnivals, and outdoor gatherings in general. A ball of shaved ice is placed into the top of a paper cone, and flavored (and colored) syrup is poured onto the ball of shaved ice. The angular opening of the cone is probably only about half of the 45 degrees of this problem, and the ball in the sno-cone is pretty close to tangent to the cone, so it isn't all that accurate an image for this problem.

[quote="Kent Merryfield"](iv) A sno-cone is an old-fashioned American snack food, associated with fairs, carnivals, and outdoor gatherings in general. A ball of shaved ice is placed into the top of a paper cone, and flavored (and colored) syrup is poured onto the ball of shaved ice. The angular opening of the cone is probably only about half of the 45 degrees of this problem, and the ball in the sno-cone is pretty close to tangent to the cone, so it isn't all that accurate an image for this problem.[/quote]  This put me in a really, really good mood.

Oh - did you also go for the cotton candy? (Since I've had a beard for my entire adult life, that would make cotton candy a really bad idea for me - but I didn't particularly like the stuff when I was a kid, anyway.)

Thank you... now I can eat a sno-cone knowing what its volume will be.


Kazakhstan IMO team:

Daliyev Asset(10th grade)
Bayev Alen(11th)
Stassiy Igor(11th)
Ismailov Adilzhan(10th)
Diyarov Arman(11th)
and me-Mussagaliyev Asset(11th)
Mojno(ili daje nujno) zdes pisat po-russki

Otkuda pishesh ACET? Vi ushe vo Vietname? Udachi na olympiade. :)Ne zabudte zavtra napisat kak proshel 1 tur.

Pochemu ne vse reshili 2 zadachu :| ?Ona ved ne slojnaja.Udachi vo 2 ture :)

[quote="ACET"]Segodnia napisali 1-iy tur:
Predvaritelnie rezultati(sootvetstvenno #1,#2,#3):
Aset Junior:+/-,-,-;
Alen:+,+,-:
Igor:+/-,+,-;
Adiljan:+/-,-,-;
Arman:+/-,-,-;
Ya:+,-,-.[/quote]

Could you please continue to post [url=http://www.mathlinks.ro/viewtopic.php?p=893786#893786]your rumours in this thread.[/url]


Diophantine equation
number theory solved
number theory

equation $x^2+y^2+1=2z^2$. how many integer solutions?

[u][b]The author of this posting is : fgomezr[/b][/u]
____________________________________________________________________

[quote="harazi"]Take y=0 and an infinite number of solutions of the Pell equation x^2+1=2z^2. You will find that your equation has an infinite number of solutions.[/quote]

What with solutions in Z+. I havent try it but maybe then, it is harder.

[u][b]The author of this posting is : Anh Cuong[/b][/u]
____________________________________________________________________

Um,I think if y=1 and we still have a Pell equation x^2+2=2z^2 with the smallest solution is (4,3).So it has an infinite number of solutions too.  :)


[quote="shinwoo"]I think it would be cool if we had a graphing function in AoPS so we could solve graphing problems and graph things .....dahdahdahdah.[/quote]

It would be and hopefully we will one day.  But it's not near the top of our priority list right now.

[quote="math92"]what is on the top of the priority list? :)
just wondering[/quote]That is a classified information  ;)  :lol:


[hide]Call the rate of the slower cyclist $x$. We can set up the equation $9(x+x+3)=171$. Solving we find the slower cyclist travels at $\boxed{\text{8 mph}}$.[/hide]

[hide]the two cyclists together are going $\frac{171}{9}$mph.

So $x+y=19$
$x+3=y$
$2x+3=19$
$x=8$

$\boxed{\text{8 mph}}$[/hide]

[hide="Here is the answer"]Let x = speed of slow cyclist, y=speed of fast cyclist.
y=x+3

Every hour, their distance increases by 171/9 = 19 miles. Therefore,
x+y=19

Substituting y=x+3,

2x+3 = 19

2x = 16
[b]x = 8 miles per hour[/b][/hide]

Good job. You are all correct.  :winner_first: :)

[quote="nonie"]Two cyclists start at the same time and travel in opposite directions. One travels at 3 miles per hours faster than the other. In 9 hours, they are 171 miles apart. What is the speed of the slowest cyclist?[/quote]

[hide]
y=x+3
9x+9y=173
solve from heere[/hide]


function
integration
probability
calculus
calculus computations

How would you solve this?
Let $X$ be a random variable that is uniformly distributed on $(a,b)$ with $a<b$. Let $a<c<b$ find the expression for
$P(X|X>c)$
[hide="what i have"]
$P(X>c)=\frac{b-c}{b-a}$ and $P(X\cap X>c)=\frac{1}{b-c}$
What did I do wrong since it is clearly wrong.
[/hide]

The density function is not a constant- it's a constant times the characteristic function of a set. Include that, and you'll be right.

According to what you've written $P(X > c)$ tends to infinity as $c$ tends to negative infinity.

I think the answer is right.  The density function for a Uniformly distributed random variable is constant and is given by $\frac{1}{b-a}$.  A univariate conditional distribution is given by:
$f(X|A) = \frac{f(x)}{P(A)}$.  In your case $P(A) = P(X>c) = \int_{c}^{b}\frac{1}{b-a}dx = \frac{b-c}{b-a}$. The probability becomes $\frac{\frac{1}{b-a}}{\frac{b-c}{b-a}}= \frac{1}{b-c}$.
This is the conditional density of $X$, which is also Uniformly distributed.

ok but how do you get that $P(X|A)=\frac{P(x)}{P(A)}$?

It's a definition. See this [url=http://www.math.uah.edu/stat/dist/Conditional.xhtml]link.[/url]


I've been accepted to the University of Rhode Island (with a full tuition scholarship).  I've also been accepted to these local Massachusetts schools:

UMass Lowell
UMass Dartmouth
Stonehill College
Worcester Polytechnic Institute (WPI)

I will be hearing from Wheaton College in Massachusetts.

I've received good scholarships to all the college I've been accepted to so far.  My number one choice is WPI because I really like it.  My brother majored in chemical engineering there; he's now going for a Ph.D in the same field at the University of California at Santa Barbera.  WPI has offered me their Chemistry and Biochemistry Scholarship; it provides me with cool benefits like research experience over the summer.

Does anyone have any opinion on any of these colleges?  Do you have a story to share about them?  Are any of you going or have any of you gone to any of them for any time?  Did you consider them at some point?  I'd love to hear any information or advice that any of may have.

I really like chemistry, and I really like math (particularly calculus so far)... so I'm interested in majoring in either chemistry or chemical engineering.  A double major is out of the question because I'm perfectionistic, unfortunately (that's also why I didn't apply to colleges like MIT).  Regardless of my college choices or regarding my college choices, do any of you have any information or advice about these majors?  Are majoring in or have you majored in any of them?

Thanks!!!

w00t!!!  I got accepted to Wheaton College with a good scholarship!  I've been accepted to all 6 colleges I applied to!

Does ANYONE have ANY comments on ANY of these schools?  Please share any insights you may have!

Thanks!

Does anybody have anything to say about any of these colleges and/or majors?

I thought other participants would reply by now. I think WPI is the one best known outside its region (where I am) for technical disciplines.

Well, I decided to go to WPI!  w00t!

Please let me know if you have been to WPI, are going to WPI, or will be going to WPI!  Maybe we can discuss all things WPI...

Nice :)  That would have been my choice...even though I don't know much about any of those schools  :oops:

Have fun in college.  It'll be awesome, I'm pretty sure.


I have a couple of topics for my language arts class that I have to write on.  I wanted to hear your opinions.

These are summaries of the prompts:

1) Is it necessary to have guidelines on student participation in after school sports (or other extracurricular activities), concerning the students GPA?  To what extent should those guidelines go?

2) Does a office referral deter future troublemaking?  What type of punishment should an administrator employ?  What is the most effective means a school has to encourage good behavior amongst students?

3) There are numerous shows on television and radio designed to embarass a person for the purpose of entertaining an audience that is in on the joke.  Is this a bonafide form of entertainment?

Thanks in advance for your input!

ugh.  these are tricky.  the office referral one really depends on the ways of the people, like the environment.

personally, i think that there should be a GPA requirement, and i dont think that a 2.0 is very hard.  and you need skills like writing to write applications, and speeches. you should be able to understand the fine print on a contract, understand what 'compounded interest' means, things like that.

think about if a student can't pass all their classes, hows that going to affect them later on?

i think that a kid getting bad enough grades not to make the team, and then caring about it should work harder.

hope this helps.

thanks!

if anyone has experience relating to one of these things, it would be appreciated.

1. There does need to be GPA requrements to protect the students.  Absent these types of protections, students will be pushed into spending the time required for sports at the expense of their studies.  (See lawsuit where 5 students that played basketball for USC and graduated saying that they couldn't even read and write their name.)  The students tried to take advantage of the system, but they can't make these types of decisions.

2. Hard to tell what will encourage proper behavior, but usually it is only imposed after classroom methods have failed or take too much class time.

3. I don't think so.  I think this is cruel, similar but more moral than the killing in the arena in Rome.

thanks, wizard2378, but i cant find a link for the USC basketball players.  sounds a little far-fetched to me, not being able to read their own name.

i don't watch that much TV, so can anyone give me some examples of the entertainment that is mentioned in #3?

Thanks so much, star99 and wizard2378!  You are a big help!

1) No. These guidelines shouldn't go anywhere.

2) What is an office referral? Punishment should be given depending on the severity of the offense committed. The most effective (but not most practical) means a school has to encourage good behavior is to turn private and only accept goody two shoes.

3) BORAT!!!!!! What does bonafide mean?

[quote="webspiral"]1) Is it necessary to have guidelines on student participation in after school sports (or other extracurricular activities), concerning the students GPA?  To what extent should those guidelines go?[/quote]

School is there to give you an education - not to provide sports teams, debate clubs, etc.  These can be viewed as 'perks' - that is why they are called [i]extra[/i]curricular activities.  I see it that you should not be allowed the reward - sports, clubs, and the like unless you are proficient at what you are really there to do - learn.

How many people would misuse the system if there were no guidelines?  I think my school has a system where you cannot be failing any classes.  This makes sure that you actually do the work in the classes, even if you are struggling, just so long as you try you should be able to maintian a passing grade.

Fundamentally, I would argue that schools are in place to make sure that you get an education, and limiting extras unless you are working hard makes sense - ever had the no TV/computer/candy or whatever reward until you're homework is done pulled on you when you were younger?  I view it as essentially the same concept.  Maybe schools should go further to make sure that you're catching up - provide mandatory tutoring for a class you're failing, weekly parent/councelour/student conferences, and so on.  But these cost money - something that many schools are lacking.


group theory
superior algebra
superior algebra unsolved

Let $G$ be a finite group of order $p+1$ with $p$ prime. Show that $p$ divides the order of $Aut(G)$ if and only if $p$ is Mersenne prime, that is, of the form $2^{n}-1$, and $G$ is isomorphic to $(Z/2)^{n}$.

Probably something for grobber, harazi or zetax ;)

I once tried this, but I couldn't find it. Since it was part of a competition, which is now over, I couldn't ask for hints! So... surprise me :)

The problem is wrong- it misses a case.

The automorphism group of $G$ naturally injects into the permutation group on the non-identity elements of $G$. The only elements of order $p$ in that group are $p$-cycles which move every element; we therefore have that the automorphisms of $G$ act transitively on the non-identity elements if $p$ divides the order of the automorphism group.
Automorphisms preserve order, so every non-identity element of $G$ has order $q$. This $q$ must be prime, and $|G|$ is a power of $q$. There are two cases; either $p=2$ and $G$ is cyclic with three elements, or $p$ is odd and $q=2$. In the second case, all elements of $G$ other than the identity have order 2. By [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=68000]standard arguments[/url], $G$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^{n}$.


is there any possible way to solve for x albegriacally for the following problem?: log base 2x 216 = x ??  thanks!! =]

Note that if $x=y$ then $b^{x}=b^{y}$. You're given an equation, so try to find b.

This probably isn't the best solution, but it's been a while since I've worked with logs.

[hide="Well..."] We have $216 = (2x)^{x}$.  You should know that $216 = 6^{3}$, so $x = 3$.

I do not think there is a way to generally solve $a = x^{x}$. [/hide]

$a = x^{x}$ has no general elementary solution, but it can be given a closed solution using lambert's omega function. [url]http://en.wikipedia.org/wiki/Lambert%27s_W_function[/url]  :maybe:
but one's probably better off using numerical methods to solve this general form.

thanks everyone!! =]...i'm new to this forum...don't really know how to use the fancy math symbols, but i'll start looking into it....and i think this problem should've gone into the beginners section...=P.


Find the number of ordered triples $(x,y,z)$ ($x$, $y$, $z$ are real) such that $x=yz$, $y=xz$ and $z=xy$.

multiply them to get

$xyz=(xyz)^2\implies xyz(xyz-1)=0$

then $x=0 \implies y=z=0$

$xyz=1\implies yz=\frac{1}{x}$, substitute

$x=\frac{1}{x} =\implies x=\pm 1$, then we get

$(-1,-1,1)$ (with permutations),  $(1,1,1)$ and $(0,0,0)$

[quote="b-flat"]Altheman, it asks for the number of solutions, which is
[hide]5[/hide][/quote]

Who cares.


I'm just wondering how much you guys spend on math each day for many of you to get as good as you are.  Not sure if this is the right forum for this to be in, so move it if necessary.
For the most part, I spend between 1/2-1 hour each day doing old problems and going through some textbooks (excluding homework for math at school)

I've been doing 8-10 hours for the past two weeks :lol: Don't know about my chances of getting into IMO, though.

8-10 hours when it is break...
about 3-4hours in regular school days

I have calculus for two periods but I usually don't spend the entire time working.  Usually about an hour and a half for calc on math.  Then about 30 more minutes interspersed throughout the school day on math.  When I get home after track/XC I would say I average about 3 hrs more on math.  So about 4.5 hrs a day.  [Much] More on weekends and during break though.

Depends on the day, on average 2 hours, because sometimes I'll come home and just sleep,  other times I'll spend 6-8 hours on math stuff for math team. So it's pretty sparatic, but I usually get my couple of hours in.

Depends for what, if its for school usually only the time I spend at class. If it's something else like a competition or something for fun, a couple of hours i guess.

wow you guys do alot of math... it's very inspirational to me, on average i only get in 1 hour... :(

I try to do as much as I can.  Calculus class is hard to pretend to pay attention in, unfortunately.  Calculus homework then takes a while, since we have to write out a lot of stuff.  We had to do a full set of AP free response per day for a while, which is like 90 minutes.  Of course I only spent around 45 :P.  Remaining time minus maybe 1 hour a day of other homework is probably half math and half whatever else.

about $5$ for me, since i'm pretty busy in school now...

Too much :D At least if browsing the Fun & Games forum counts as math... after all, it is on a math forum ;)

about 1 to 2 hours... i have heavy school work and i got physic olympiad....

About 2 hours at home and there's always time at school to do math.

The trick is to have a few maths books lying on your table, when your bored in your room you just pick one up, flick through it, see something interesting and read that chapter! Tis what i do, and im going to the IMO this year with ireland, so im doing something right!

Well, Practice makes perfect, I used to be okay in Math but then I started doing about 3-4 hours of math every day for like the last year, going through the AOPS books and got a whole lot better.  :)

When I have school, I do around 2 hours.

When I have break, I do a lot more.  (I've been on AoPS all day toady except for eating :D )

When in school I've been doing a lot of math cause of AP Calc... though i'm really lame when it comes to math contests (can't even get into AIME). Probably an hour a day :D


Find the value of $ x$ for which the function $ f(x)\equal{}2^x\minus{}4^x$ reaches of a maximum.

[quote="PhireKaLk6781"]Find the value of $ x$ for which the function $ f(x) \equal{} 2^x \minus{} 4^x$ reaches of a maximum.[/quote]

[hide]Note that $ (2^x)^2 \equal{} 2^{2x} \equal{} (2^2)^x \equal{} 4^x$

Let $ u \equal{} 2^x$.
Then $ u^2 \equal{} 4^x$.

$ f(u) \equal{} u \minus{} u^2$

$ f(u) \equal{} \minus{} u^2 \plus{} u$

Coefficients:
---------------
$ a \equal{} \minus{} 1$
$ b \equal{} 1$

$ u \equal{} \frac { \minus{} b}{2a}$

$ u \equal{} \frac12$

$ u \equal{} 2^x$

$ \frac12 \equal{} 2^x$

$ 2^{ \minus{} 1} \equal{} 2^x$

$ \minus{} 1 \equal{} x$
$ x \equal{} \minus{} 1$

$ f(x) \equal{} 2^x \minus{} 4^x$ reaches a maximum at $ x \equal{} \minus{} 1$.

[/hide]

[hide=" Alternatively"]

[Wrong forum, but we can check by]

differentiating, f(x) = $ 2^{x} \ln 2 \minus{} 4^{x} \ln 4 \equal{} 0$

$ \implies \frac {4^{x}}{2^{x}} \equal{} \frac {\ln 2}{\ln 4}$

$ \implies 2^{x} \equal{} \frac {1}{2}$

$ \implies x \equal{} \minus{} 1$ and so $ f(x) \equal{} \frac {1}{4}$

[/hide]


modular arithmetic
combinatorics unsolved
combinatorics

Let n and k are naturals numbers such that $ 1\le k\le n$.Write a n-digits natural number by digits 0 or 1. How many that numbers which such that exist k digits 1 consecutive in its?
hope a nice solution!

[hide="click"]lets denote k consecutive 1's by letter x, so the question reduces to find all numbers of length n-k+1 , having one letter x and other places filled by 0 or 1.

Hence required number =  $ \dbinom{n\minus{}k\plus{}1}{1} * 2^{n\minus{}k}$[/hide]

No, that massively over-counts: any word with several different runs of $ k$ 1s is counted many times in that expression.

It's easy to write a recursion for the complement of what you're counting (i.e., the number of sequences of 0s and 1s of length $ n$ with no $ k$ consecutive 1s): break every such word up according to its leftmost occurrence of 0 to see that $ a_n \equal{} a_{n \minus{} 1} \plus{} a_{n \minus{} 2} \plus{} \ldots \plus{} a_{n \minus{} k}$ with $ a_j \equal{} 2^j$ for $ 0 \leq j \leq k \minus{} 1$.  This assumes that an $ n$-digit natural number may have leading 0s, but it's not hard to adjust for the other situation.

To the original poster: please give your threads more descriptive names.

[quote="JBL"]To the original poster: please give your threads more descriptive names.[/quote]
sorry, I will attention more.
can you post a solution for it?thanks

if there are no runs of $ k$ ones in the number then it has form
\[ 1^{s_1} 0 1^{s_2} 0 1^{s_3} 0 \dots 0 1^{s_t}
\]
where $ t$ is the number of runs of ones, $ 0\leq s_i < k$ is the length of i-th runs, and
\[ t \minus{} 1 \plus{} s_1 \plus{} s_2 \plus{} \dots \plus{} s_t \equal{} n.
\]
For a fixed $ t$, the number of such numbers is the coefficient of $ x^{n \minus{} t \plus{} 1}$ in
\[ (1 \plus{} x \plus{} \dots \plus{} x^{k \minus{} 1})(1 \plus{} x \plus{} \dots \plus{} x^{k \minus{} 1})\dots(1 \plus{} x \plus{} \dots \plus{} x^{k \minus{} 1}) \equal{} \left(\frac {x^k \minus{} 1}{x \minus{} 1}\right)^t.
\]
or equivalently the coefficient of $ x^{n \plus{} 1}$ in $ \left(\frac {(x^k \minus{} 1)x}{x \minus{} 1}\right)^t.$

The total number of such numbers is the coefficient of $ x^{n \plus{} 1}$ in
\[ \sum_{t \equal{} 0}^{\infty} \left(\frac {(x^k \minus{} 1)x}{x \minus{} 1}\right)^t \equal{} \frac {1}{1 \minus{} \frac {(x^k \minus{} 1)x}{x \minus{} 1}} \equal{} \frac {1 \minus{} x}{1 \minus{} x \plus{} x^{k \plus{} 1}}
\]
which equals to $ c_{n \plus{} 1} \minus{} c_n$ where
\[ c_{m,k} \equal{} \sum_{t\equiv m\pmod{k}} \binom{t}{\frac {m \minus{} t}{k}} ( \minus{} 1)^{\frac {m \minus{} t}{k}}\qquad\text{is the coefficient of}\,x^m\,\text{in}\qquad\frac {1}{1 \minus{} x \plus{} x^{k \plus{} 1}}.
\]
Therefore, the answer to the original question is
\[ 2^n \minus{} c_{n \plus{} 1,k} \plus{} c_{n,k}.
\]


induction
combinatorics proposed
combinatorics

Let $ a_1,a_2,\dots,a_n$ be a sequence of $ n$ integers so that their sum equals to $ 1$. Among the $ n$ cyclic sequences,
\[ \{a_1,\dots,a_n\},\{a_2,\dots,a_1\},\dots,\{a_n,\dots,a_{n\minus{}1}\}\] there is exactly one sequence whose every partial sum is positive.
(A $ k$- partial sum of a sequence $ \{b_1,b_2,\dots,b_n\}$ is $ \sum_{i\equal{}1}^k b_i$)

I think you mean at least one. Cause, for example, {1/n, 1/n... 1/n} has all n cyclic sequences being postive.

Now suppose there exist a sequence of size n such that none of its cyclic sequences has sum over n. Then we prove we can find a sequence of size n-1 with the same property. There must be one element that is less than 0 in the sequene, call it a. Now add a to the element in front of it (i.e. b). There are now n-1 elements in the sequence with the above property (easily verified by checking cases) Now apply induction on n, we get that we can find a sequence of length 1 with the same property, but a = 1. Contradiction. So there can be no sequences with this property.

Instead of = 1, we can make it = any postive integer.

[quote="srikanth"]Let $ a_1,a_2,\dots,a_n$ be a sequence of $ n$ [b]integers[/b]...[/quote]
So there is a little more work to be done.

Consider the following visualization.  Extend the sequence $ a_k$ to be a periodic sequence with period $ n$, let $ s_n \equal{} \sum_{k \equal{} 1}^{n} a_k$, and consider the plot of the points $ (n, s_n)$ (with broken lines between points for effect).  This plot is "almost" periodic in the sense that $ (k, s_k) \equal{} (k \plus{} n, s_k \plus{} 1)$.  We call a period an interval of the form $ (pn, (p \plus{} 1)n]$.  This plot clearly attains a minimum value some number of times on $ (0, n]$; we will refer to the points such that this occurs as minima.

Now the claim that a given permutation contains only positive partial sums is precisely the claim that if we draw a horizontal line from a given point on this graph of length $ n$, it will not intersect the graph, and we claim that this is true only of the [b]rightmost minimum[/b] on $ (0, n]$.

If we draw the line from any point to the left of the rightmost minimum, it will intersect the graph around the rightmost minimum, and if we draw the line from any point to the right of the rightmost minimum, it will intersect the graph around the rightmost minimum of the next period (since that minimum is exactly $ 1$ unit higher than the previous, and any points to the right of the rightmost minimum are at least $ 1$ unit higher than the rightmost minimum).

On the other hand, if we draw the line from the rightmost minimum, in one period the rest of the points will be strictly above the line (by construction; this minimum is rightmost) and in the next period the rest of the points are higher by $ 1$ than they were in the previous period, so the minima (the lowest points) will also be strictly above the line.

[hide]
Since the sum of the sequence is positive, there must exist at least two adjacent terms $ a_k, a_{k + 1}$ such that $ a_k > 0$ and $ a_k + a_{k + 1} > 0$ (where we define $ a_{n + 1} = a_1$). We then replace the the two terms $ a_k, a_{k + 1}$ with the positive term $ a_k + a_{k + 1}$.

We repeat this process with the $ n - 1$ numbers and so forth until there are only two numbers, one of which is positive, and the other which is non-positive (since the sum is always $ 1$). Obviously the non-positive term was also a term $ a_j$ in the initial sequence, since it was never the result of a replacement. Then we claim that the cyclic sequence starting with term $ a_{j + 1}$ is the unique desired sequence with positive partial sums.

Since in the process we eventually replaced the term $ a_{j + 1}$ with the term $ a_{j + 1} + a_{j + 2}$ (here $ a_{j + 2}$ does not necessarily denote the original term there), it follows that $ a_{j + 1} > 0$ and $ a_{j + 1} + a_{j + 2} > 0$, and so forth. It now follows that the partials sums are positive by induction; indeed, if at some point $ a_{j + 2}$ had been replaced by terms $ a_{j + 2} + a_{j + 3}$, then by the definition of our construction, $ a_{j + 2} > 0$, and so forth. Uniqueness follows also as in our construction, there eventually exists only one $ a_k$ satisfying $ a_k>0, a_k + a_{k+1} > 0$, and the term $ a_j$ was never collapsed.
[/hide]

Woops sorry didn't see the integers.

If it is integers then it's exactly one. Suppose we have a contradiction and there's 2 or more sequences. Let us say that the first element of the second sequence start in the ith position of the first sequence. Then the first element of sequence 1 up to i-1 is positive. But i to the last element must also be positive (due to the property of second sequence). Both sums are integer, and add up to 1. Thus one of them must be not positive. Contradiction. So there's at most one.


ARML geometry perimeter inradius circumcircle

Define angle A to be "nice" if both SinA and CosA are rational.
Define a convex polygon "nice" if all of its interior angles are "nice"

A convex quadrilateral has the following properties:
1. Its sides are integers whose product is a square
2. It can be inscribed in a circle and can be circumscribed about another circle.

Prove this quadrilateral is nice.

wierd, i did this set of power questions but i dont remember this one.  i suppose from the difficulty it was part C?

The sum of two nice angles is again nice (easy to prove).
However I don't know whether that may help for the solution.

[hide]i missed the word quadrilateral. now everything is easier

it has inscribed circle, so divide its primeter into a,a,b,b,c,c,d,d

so its sides are a+b, a+d, d+c, b+c

its area is sqrt((a+b)(c+d)(b+c)(a+d)) (cyclic quad)

hence its area is an integer

the area can be expressed as:

0.5((a+d)(a+b)sin(k)+(b+c)(d+c)sin(180-k))

and similarly another way

sin(180-k) is obviously sin (k)

so sin of the angles is rational

for the cos, for simplicity let the sides be w,x,y,z

then w^2+x^2-2wxcos(k)=y^2+z^2-2cos(180-k)

cos(180-k)=-cosk

solving for cos(k)

hence cos(k) is alo rational[/hide]

[color=cyan]There may have been an earlier part of this question that made that easier.  One earlier part, for example, was that a triangle with rational sides and one nice angle has all nice angles.  You can also show that the area, inradius and circumradius of such a triangle are rational.  I don't know if any of these things is very helpful, but on the actual question they may have given you tools to work with.[/color]

I also tried to prove it for any polygon ...
That's probably why I didn't manage to prove it :?
The word quadrilateral makes it easier !


factorial
algebra
system of equations
number theory proposed
number theory

Let $ a,b,c,x,y,z$ satisfy followings :

$ a x \plus{} b y \plus{} c z \equal{} 1 ,$
$ a x^2 \plus{} b y^2 \plus{} c z^2 \equal{} 2 ,$
$ a x^3 \plus{} b y^3 \plus{} c z^3 \equal{} 6 ,$
$ a x^4 \plus{} b y^4 \plus{} c z^4 \equal{} 24 ,$
$ a x^5 \plus{} b y^5 \plus{} c z^5 \equal{} 120 ,$ and
$ a x^6 \plus{} b y^6 \plus{} c z^6 \equal{} 720.$

Then find the value of $ a x^7 \plus{} b y^7 \plus{} c z^7.$
 :idea:

the relation is a cyclic relationship

have you noticed that the eqns follow the following sequence1!,2!,3!,4!,5!,6!
therefore the answer is 7! or 1540
{here ! denotes factorial;n!=n(n-1)(n-2)....321}

[quote="Rust"]Let $ u_n = ax^n + by^n + cz^n$, then $ u_{n + 3} = \sigma_1 u_{n + 2} - \sigma_2 u_{n + 1} + \sigma_3u_n$,
were $ \sigma_1 = x + y + z,\sigma_2 = xy + yz + zx,\sigma_3 = xyz$.
It give system of equations (n=1,2,3) for $ \sigma s_i$:
$ (n + 3)! = (n + 2)!\sigma_1 - (n + 1)!\sigma_2 + n!\sigma_3$ or
$ 6\sigma_1 - 2\sigma_2 + \sigma_3 = 24,$
$ 24\sigma_1 - 6\sigma_2 + 2\sigma_3 = 120,$
$ 120\sigma_1 - 24\sigma_2 + 6\sigma_3 = 720$.
It give $ \sigma_3 = 12,\sigma_2 = 36,\sigma_3 = 24$ (x,y,z are roots $ t^3 - 12t + 36t - 24 = 0$).
Therefore $ u_7 = 12u_6 - 36u_5 + 24u_4 = - 2880$.[/quote]

Oh! Good idea.

My idea is the same  as yours.

But you have a little mistake , I think. So I would like you to compute again. :wink:


These have been in ML for a while, but just for the sake of completness I will post them here so that they can be added to the resources sections, as well as the other Argentina's TST
Day one

1- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=35067
2- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=35064
3- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=35065

Day two

4-
5- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1089870445&t=35066
6- http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129882059&t=34564

I would like to, but I can't find it, I will ask someone today if they have it

Here is problem $ 4$
(I can't edit the original post)

http://www.mathlinks.ro/Forum/viewtopic.php?p=1633744#1633744


A plane parallel to the base of a pyramid seperates the pyramid into two pieces with equal volumes.  If the height of the pyramid is 12, find the height of the top piece.(please show your work!!!)

I posted an image of what the diagram looks like, but i think you might be able to do it without the diagram.

[hide]
The total height is $12$. The height of the smaller pyramid is $12-x$

Call the base of the big one $y$

The ratio of the heights is $\frac{12-x}{12}$, so the ratio of bases is $\frac{(12-x)^2}{144}$.

so $(12-x)^2$ is the base of the smaller pyramid and $144$ is the base of the bigger one.

Applying the formula for volume
So $(12-x)^2\cdot \frac 13 (12-x)=(12-x)^3\cdot\frac 13$ is half of $144\cdot12\cdot\frac 13=576$

Set up the equation:
$\frac 13 (12-x)^3=\frac 12 576$
we are looking for $12-x$
$(12-x)^3=864$
$12-x=6\sqrt[3]{4}$
[/hide]

[quote="biffanddoc"]
So $(12-x)^2\cdot \frac 13 12-x=(12-x)^3\cdot\frac 13$ is half of $144\cdot12\cdot\frac 13=576$
  [/quote]
i don't get what you did here...

The method I use for these type of problems is really weird... but it's effective and fast   :D (in my opinion)

Call the height x. Since all dimensions of the small pyramid is similar to the big pyramid, you can compare x with 12 directly (without any other variables). Since this is 3-dimensional, call the area of the small pyramid $x^3$ and the area of the big pyramid $12^3$. Since you want the area of the small pyramid to be half of the big pyramid, set $x^3$ equal to half of 1728 (864) and simplify, which gives x to be $6\sqrt[3]{4}$

[quote="frost13"][hide]Nice work Biffanddoc, maybe this will seem a little easier though.

If the volumes are equal, then the volume of the smaller pyramid is $\frac 12$ of the whole thing.  If the ratio of the Volumes is $\frac 12$ then the ratio of the lengths, heights, or any linear comparison will be $\sqrt[3]{\frac 12}$.  That simplifies to $\frac{\sqrt[3]{4}}{2}$.  If that is the ratio, multiply it by 12, the height of the whole thing, and you get $6\sqrt[3]{4}$ the height of the smaller pyramid.

Hope that helps,

Frost

[/hide][/quote]
Nice solution  :D .


trigonometry
calculus
integration
algebra
function
domain
modular arithmetic

find all triple positive integers such that:
$ x
y\plus{}xz\plus{}yz\minus{}xyz\equal{}2$

[quote="asintota"]find all triple positive integers such that:
$ x y \plus{} xz \plus{} yz \minus{} xyz \equal{} 2$[/quote]

[hide="approach"]
call xy+xz+yz=a
this is like finding the roots of $ x^3\plus{}ax\plus{}2\minus{}a\equal{}0$

1, 1, 1 obviously works  :D
[/hide]

[hide="Big Hint on Simplification"]
This problem begs for some sort of factoring solution. Since we have xyz terms and -xy aso. terms, our best bet is to use:
[b]
(x-1)(y-1)(x-1)[/b]
= xyz-xy-yz-zx+x+y+z-1
= -(xy+yz+zx-xyz) + x+y+z-1
= x+y+z-3
[b]= (x-1)+(y-1)+(z-1)[/b]

Define r=x-1, s=y-1, t=z-1, each a non-negative integer. Then:
[b]
rst=r+s+t[/b]
[/hide]

[quote="asintota"]find all triple positive integers such that:
$ x y \plus{} xz \plus{} yz \minus{} xyz \equal{} 2$[/quote]
Diophantient's gonna try out this Diophantine.
[hide]
$ xyz\minus{}xy\minus{}yz\minus{}xz\equal{}\minus{}2$
$ (x\minus{}1)(y\minus{}1)(z\minus{}1)\equal{}x\plus{}y\plus{}z\minus{}3$
Increment them all.
$ abc\equal{}a\plus{}b\plus{}c$
Except now they're nonnegatives.
If anyone is 0, the only soln is (0,0,0) so now they're positive again.
Let $ a\equal{}\tan A$, $ b\equal{}\tan B$, and $ c\equal{}\tan C$.
This gives us that $ A\plus{}B\plus{}C\equal{}180$.
So, the only positive integral values of $ \tan \theta$ in the domain given for rational $ \theta$, which contains all numbers in its range, are when $ \theta\equal{}0,45,180$.
The only one which works out is 0/0/180, which corresponds with 1,1,1 in the original.[/hide]

Um like I'm pretty sure permutations of $ (2,3,4)$ is also a solution.

[quote="diophantient"][quote="asintota"]find all triple positive integers such that:
$ x y \plus{} xz \plus{} yz \minus{} xyz \equal{} 2$[/quote]
Diophantient's gonna try out this Diophantine.
[hide]
$ xyz \minus{} xy \minus{} yz \minus{} xz \equal{} \minus{} 2$
$ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \equal{} x \plus{} y \plus{} z \minus{} 3$
Increment them all.
$ abc \equal{} a \plus{} b \plus{} c$
Except now they're nonnegatives.
If anyone is 0, the only soln is (0,0,0) so now they're positive again.
Let $ a \equal{} \tan A$, $ b \equal{} \tan B$, and $ c \equal{} \tan C$.
[b]This gives us that $ A \plus{} B \plus{} C \equal{} 180$.[/b]
So, the only positive integral values of $ \tan \theta$ in the domain given for rational $ \theta$, which contains all numbers in its range, are when $ \theta \equal{} 0,45,180$.
The only one which works out is 0/0/180, which corresponds with 1,1,1 in the original.[/hide][/quote]

explain the bolded part?

Mmmhmm... looks like the trig argument isn't complete  :P

Continued from my previous post:
[hide="Done Algebraically"]
So we have $ rst\equal{}r\plus{}s\plus{}t$ for $ r,s,t>0$.

If any one equals zero, the rest must also, so $ (r,s,t)\equal{}(0,0,0)$ is a solution.

Otherwise, they are all positive. wlog let $ r \leq s \leq t$. Now write:

$ r(st\minus{}1) \equal{} s\plus{}t \implies st\minus{}1 \equal{} \frac{s\plus{}t}{r} \leq s\plus{}t$
$ \iff st\minus{}s\minus{}t\minus{}1 \leq 0 \iff st\minus{}s\minus{}t\plus{}1 \leq 2 \iff (s\minus{}1)(t\minus{}1) \leq 2$

If neither $ s$ nor $ t$ is $ 1$, then we have that $ t\minus{}1 \leq \frac{2}{s\minus{}1} \leq 2 \implies t \leq 3$.

If $ s\equal{}1$, then $ r\equal{}1$, and we have $ t\equal{}2\plus{}t$, a contradiction. $ t\equal{}1 \implies r,s\equal{}1$, which leads to the same contradiction.

This gives three possible considerations: $ (s,t)\equal{}(2,2) \qquad (s,t)\equal{}(2,3) \qquad (s,t)\equal{}3,3$

The only one that gives an integer solution in $ r$ is $ (s,t)\equal{}(2,3) \implies r\equal{}1$, so $ (r,s,t)\equal{}(1,2,3)$ is the only other solution.

Removing ordering and translating to $ x,y,z$, we have that the solutions are $ (x,y,z)\equal{}(1,1,1)$ and all permutations of $ (x,y,z)\equal{}(2,3,4)$
[/hide]

[quote="t0rajir0u"][b]diophantient[/b], there's no reason to assume that the degree measure of the angles is rational.  The trig substitution can be made to work, but it's less effort just to bound.[/quote]
Whoops, I totally missed that :(

How would you complete the argument?

Another way to finish from $ abc\equal{}a\plus{}b\plus{}c$:

[hide]Assume WLOG that $ a\le b\le c$.  Taking the equation $ \pmod{c}$, we get $ a\plus{}b\equiv 0\pmod{c}$.  Therefore , $ a\plus{}b\equal{}0,c,2c$.  Now we're just a little easy casework away from the answer.  :) [/hide]


Okay, the point of the problem is to solve who likes who

Linda and Louise work at the candy store
Mike's sister, Jane, is dating the check-out clerk at McCaffrey's
Bruce likes tall girls
Bob is going to art school out-of-town, so he only gets to see his girlfriend on weekends
Louise broke up with Richard last year because he didn't want to quit his job to go to college
Linda just got accepted to law school
Mike never seems to run out of Candy bars...
Because of her physique, Jeri may be the first woman basketball player at her school
On weekdays, sometimes Sue goes out with guys other than her boyfriend!! (Shhhhhhh, he doesn't know about it !!)
Only one boy has a steady job (he works for a living)
Chuck has a sweet tooth, but his girlfriend works in a natural foods store and only brings him tofu!
Henry never like girls who were smart
Mary's main hobby is yoga

I know that I'm supposed to make one of those charts...
Here are the things that I know

Mike likes either Linda or Louise
Mike can't like Jane
Bob can't like Sue
Chuck can't like Linda or Louise
Henry can't like Linda

That's as far as I know... So who likes who? (Please show you logic, I'm confused with this problem  :blush: And I usually solve them real fast...)

The problem is certainly ill-posed for many reasons, but, stretching the logic a bit, we get

Mike-Linda
Bob-Sue
Bruce-Jeri
Henry-Louise
Chuck-Mary
Richard-Jane

This requires interpreting "Bruce likes tall girls" as "Bruce likes Jeri" (which is a probable, but formally incorrect conclusion because we are told absolutely nothing about the height of any other girl) and "Sue goes with other guys on weekdays" as "Bob likes Sue" (which is also a probable but formally incorrect conclusion). Adding the correct "Richard likes Jane" to this list (Jane likes a working boy and Richard has a job and only one boy has a job), we reduce the problem to 3 boys and 3 girls about which we have enough information to fill the table in a unique way.

Shouldn't this be in "Puzzles and Brainteasers"?

Ah... Heh, I didn't know where to put this so... Well anyways, I solved now...  :D


Let $S$ be a subset of $N$ with the following properties:

a)Among each $2003$ consecutive natural numbers there exists at least one contained in $S$;

b) If $n\in S$ and $n>1$ ,then $[\frac n2]\in S$. ( where $[x]$ denote the greatest integer less than or equal to $x$).

prove that $S=N$.

supose that $ n\in N $ but $n \not\in S$.
we will work  in binary base  and $x_2$ will mean the representation of $x$ in binary base
so  $[ \frac{x}{2}]_2$ is obtained by deleting the last digit of $x_2$
let $p_2$ be a number formed by adding 11 of 0 at the end of $n_2$and
$q_2$ formed by adding 11 of 1 to the end of $n_2$ between $p_2$ and $q_2$ we have 2048 numbers ($2^{11}$) so at least one of them will be in $S$(because 2048>2003 :D). let it be $a_1$ let $a_{n+1}=[\frac{a_n}2]$ so $a_{12}$ will be n cause you practicaly delete the last digit from the representation in binary base of $a_n$ every time you apply that.
so $n \in S$.


inequalities
calculus
integration
algebra
polynomial
complex numbers
inequalities proposed

Let $a_1,a_2,\ldots,a_n$ be complex numbers of unit modulus. Prove that there is a complex number $z_0$ with $|z_0|=1$ such that
\[
|z_0-a_1|\cdots|z_0-a_n|\geq 1.
\]

Hyper-classical problem. Here is a nice short solution: use the modulus inequality in the integral $\int_{0}^{2\pi}{p(e^{it})e^{-int}dt}=2\pi$. Here p is the polynomial having roots $a_i$. We don"t need the assumption that $a_i$ have modulus 1.


trigonometry
geometry
cyclic quadrilateral
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

Inscribed in a circle is a quadrilateral having sides of lengths 25, 36, 52, and 60 taken consecutively. Find the diameter of the circle.

[hide]I believe that the answer is 28.5, but I did it in such a weird way I don't remember all the steps.  All I know is that I used trig and ratios.[/hide]

Since the quadrilateral is inscribed in a circle, it is cyclic.

Hence, we're allowed to use Brahmagupta's Formula which states that for cyclic quadrilateral, its area $K$ is defined as:

\[\sqrt{(s-a)(s-b)(s-c)(s-d)}\]

For more information, go
[url=http://mathworld.wolfram.com/BrahmaguptasFormula.html]Here[/url]

Actually, the question asks for the diameter of the circle, not the area. ;)

My bad.. Should've read the question more carefully.

You can find the length of a diagonal by using law of cosines twice.
Then use the formula $R=\frac{abc}{4K}$.
$K$ is the area of triangle, which can be calculated by Heron's formula etc.


[quote="qiuxuezhe"]Prove that$a,b,c>0 and a+b+c=1$

$a/(b^{2}+c)+b/(c^{2}+a)+c/(a^{2}+b)\geq 3/(4(ab+bc+ca))$[/quote]
It's wrong! Try $a=\frac{1}{3}$, $b=\frac{2}{3}$ and $c\rightarrow0^+$.
[quote="Flove"]A long time ago, I also conjecture a nice inequality

[i]Prove or disprove that that if $a,b,c$ are non-negative real numbers and[/i] $a+b+c=3$ then \[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\] I hope to see a nice solution to it.[/quote]
Today it's very easy! :lol:

[quote="Flove"]A long time ago, I also conjecture a nice inequality

[i]Prove or disprove that that if $a,b,c$ are non-negative real numbers and[/i] $a+b+c=3$ then \[\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge \frac{3}{2}.\] I hope to see a nice solution to it.[/quote]

by Cauchy-Bunjakowski-Schwarz we get
$\frac{a^2}{ab^2+ac}+\frac{b^2}{bc^2+ba}+\frac{c^2}{ca^2+cb}\geq \frac{(a+b+c)^2}{ab^2+bc^2+ca^2+ab+bc+ca}=\frac{9}{ab^2+bc^2+ca^2+ab+bc+ca}\geq \frac{3}{2}$
this is equivalent to
$3(ab^2+bc^2+ca^2)+3(ab+bc+ca)\le 18$
$3(ab^2+bc^2+ca^2)=(a+b+c)(ab^2+bc^2+ca^2)=ab^3+bc^3+ca^3+(ab+bc+ca)^2-abc(a+b+c) $
$ab+bc+ca=p $ and $abc=r $
By Vasc's ineq $ 3(ab^3+bc^3+ca^3) \le  (a^2+b^2+c^2)^2  $
we need to prove  $ 27p+9r-7p^2-27\ge 0 $
if  $ 18/7\le p\le 3 $
By Schur's ineq $ 9r \ge 12p-27 $
need to prove  $ 27p-7p^2-27+12p-27=39p-7p^2-54\ge 0 $
that's  $ (7p-18)(p-3)\le 0 $ ,it's sure
if  $ p\le 18/7$, How to solve $ 27p+9r-7p^2-27\ge 0 $???

[quote="arqady"]No, yanzongz. Your way is wrong after first step. :wink:[/quote]

We need  to prove \[ \frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b}\ge\frac{3}{2}. \]

by Cauchy-Bunjakowski-Schwarz
$ \frac{a^2}{ab^2+ac}+\frac{b^2}{bc^2+ba}+\frac{c^2}{ca^2+cb}\geq\frac{(a+b+c)^2}{ab^2+bc^2+ca^2+ab+bc+ca}=\frac{9}{ab^2+bc^2+ca^2+ab+bc+ca} $

So we can prove  $ \frac{9}{ab^2+bc^2+ca^2+ab+bc+ca}\geq\frac{3}{2} $
What i write is easy , it's right  :(

[quote="yanzongz"]
So we can prove  $ \frac{9}{ab^2+bc^2+ca^2+ab+bc+ca}\geq\frac{3}{2} $
What i write is easy , it's right  :([/quote]
This inequality is wrong. Try $c\rightarrow0^+$. :wink: