1.  Find a six-digit number in which the digits will rotate when the number is multiplied by either two, three, four, five or six.

2.
We have the following information about a one-story apartment:

a) There is maximum one door between any two rooms.
b) There is maximum one door from any room to the outside.
c) There are 12 doors in the apartment.
At least how many rooms are there in the apartment?

3. Find 5 square numbers using every positive digit only once.

[quote="Phelpedo"]#1 reminds me of a problem on the MathPath qualifying quiz...chess64 is that where you got the answer from?[/quote]

 :D yes

wow did you visualize that hexagon apartment in your head?

Well, actually it's a hexagon apartment, and yes, I just visualized it.


calculus
linear algebra
matrix
algebra
polynomial
vector
complex analysis

I've been looking quickly through the existing algebra forum, looking for topics that would be more suitable for linear algebra. I'm sure I missed many, and probably moved a few that shouldn't have been moved. Any other moderator who wants to help sort this out is welcome to try.

I moved a very small number of topics from "calculus-analysis" to "complex analysis" but there aren't nearly as many of them. To be honest, I'm a little skeptical of the demand for complex analysis. There's no such question about linear algebra - the problem traffic will be robust.

Ok, I've made this post into an annoucement on the whole College Playground forums. To all moderators please move all the topics that you see with Matrix - Linear Algebra and Complex Analysis in their respective categories.

The easier (computational only) matrix problems should be moved in the Tutorial forum.

[quote="fedja"][quote="Valentin Vornicu"] To all moderators please move all the topics that you see with Matrix - Linear Algebra and Complex Analysis in their respective categories.
[/quote]
Hmm... What are the exact criteria for the problem to belong to Matrix-Linear Algebra? Does every problem about matrices and linear operators in finite-dimensional spaces belong there now?  :?[/quote]Yeah, anything that deals with Matrix Algebra computations (plain ones, determiants, rank, systems of linear equations), characteristical polynomials, determinants, vector spaces, bases over vector spaces, endomorphisms, etc.

OK, I spent a couple of evenings and moved whatever I found in the "Unsolved and Proposed Problems" sections. I don't know whether somebody wants to do it with the "Solved Problems" sections (certainly, not I  :D ). I left a few posts about which I was in doubt and Kent's post about a moved topic. Also my rule was to move a thread to an "Unsolved Problems" section if I could not understand the solution in 30 seconds, so some of the problems there should, probably, be retired. But otherwise I consider the "great move" complete.

[color=red][size=117][i][b]Please, use the new forums for the problems that belong to them  (they'll be moved there anyway, but it is much more convenient for everyone if you post in the right place from the beginning :) )[/b][/i][/size][/color]


For a poitive integer $ n$, denote $ S(n)$ be the sum of digits of $ n$. Find all positive integers $ n$ satisfying $ n\equal{}(S(n))^2$

Because $ S(n)^2\equal{}n\equiv S(n)\mod 9$ we get $ 9|S(n)$ or $ S(n)\equal{}1\mod 9$. Let $ S(n)\equal{}9k$ or $ S(n)\equal{}9k\plus{}1$, then $ n$ had at least $ k$ or $ k\plus{}1$ digits. It give $ n\ge 10^k\minus{}1$ or $ n\ge 2*10^k\minus{}1$.
But $ 10^k\minus{}1>81k^2\equal{}S(n)^2$ if k>2$ , and$2*10^k-1>(9k+1)^2$ if k>2$.
Therefore we can shek only $ n\equal{}81k^2,(9k\plus{}1)^2,k\le 2$.
0. k=0 give solution $ n\equal{}1$.
1. k=1 give solution $ n\equal{}81$.
2. k=2 $ n\equal{}324,361$ - not solution.


inequalities
function
symmetry
inequalities unsolved

If $ a$, $ b$, $ c$, $ d$ are four positive reals such that $ abcd \equal{} 1$, then prove that

$ \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3} \plus{} \frac {1}{1 \plus{} b \plus{} b^2 \plus{} b^3} \plus{} \frac {1}{1 \plus{} c \plus{} c^2 \plus{} c^3} \plus{} \frac {1}{1 \plus{} d \plus{} d^2 \plus{} d^3} \geq1$

Vasc posted something that could help solving this problem... see his posts! ( I cannot find it now :( )

There was something related to the inequality:

Let $a,b,c,d$ and $r$ be positive numbers so that $r = \sqrt[4]{abcd} \geq 1.$ Prove that: \[\displaystyle \frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} + \frac{1}{(1+c)^{2}} + \frac{1}{(1+d)^{2}}  \geq \frac{4}{(1+r)^{2}}.\]

which is solved and other four inequalities based on this one are posted (this one is between those four) ...

I hope I am not mistaking.. cheers!

I dont know how my Probl1.doc has appeared in this forum. I have deleted it. Anyway , the solution of Probl1.doc is irrelevant for this inequality.

Well you see $ \displaystyle f(x) = \frac 1{1 + e^x + e^{2x} + e^{3x}}$ is convex.
Now suppose $ a = e^s, b = e^t ,c = e^u, d = e^v s + t + u + v = 0$. Now use Jensen inequality.

[quote="darij grinberg"][quote="Vasc"]I have deleted it.[/quote]

Valiowk, but why on the earth did you post that?   :D  :D  :D

  Darij[/quote]

Certainly not because I wanted such a result to happen ;) .

I have an idiot question. Is it for sure that the function is convex? My computations did not reveal that, but they are probably wrong.

I think you are right Harazi. Also my computations say it is not convex.  :?

Excuse me, I had some mistakes in computing.

[quote="Valiowk"]Lagrangia, maybe you couldn't find it because there's a document with the proof in a thread that is totally unrelated to it (for reasons which I myself cannot comprehend :? ).  It's in Probl1.doc in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=19510&highlight=#19510]this[/url] thread![/quote]

I don't know why I thought that there was a solution for the inequality posted by manlio in that document. And I didn't mean that the problem posted by manlio is related to this one (let a,b,c,d,r etc ) .. I meant that there was a problem which is related to "let a,b,c,d,r etc" because I was't sure that this is the exact problem...

 ;)  :D

Well the following inequlity is not correct We see that:
$ \displaystyle \frac 1{1+x} +\frac 1{1+\frac 1x} =1$
So:
$ \displaystyle \frac 1{1+x} +\frac 1{1+\frac 1x}+ \frac 1{1+y} +\frac 1{1+\frac 1y}+ \frac 1{1+z} +\frac 1{1+\frac 1z}+ \frac 1{1+t} +\frac 1{1+\frac 1t}=4$
But we know $xyzt= \frac 1{xyzt}=1$
So we have :
$ \displaystyle \frac 1{1+x} +\frac 1{1+y} +\frac 1{1+z} + \frac 1{1+t} \leq 2$ or:
$ \displaystyle \frac 1{1+\frac 1x} +\frac 1{1+\frac 1y} +\frac 1{1+\frac 1z} + \frac 1{1+\frac 1t} \leq 2$

The inequality
$(1+a+a^2+a^3)(1+b+b^2+b^3)(1+c+c^2+c^3)+(1+b+b^2+b^3)(1+c+c^2+c^3)(1+d+d^2+d^3)+(1+c+c^2+c^3)(1+d+d^2+d^3)(1+a+a^2+a^3)+(1+d+d^2+d^3)(1+a+a^2+a^3)(1+b+b^2+b^3)
\geq(1+a+a^2+a^3)(1+b+b^2+b^3)(1+c+c^2+c^3)(1+d+d^2+d^3)$ can be proved by a number of AM-GM inequalities. I think maybe it can be showed that the inequality can be divided into many AM-GM inequalities for general n, but I fail to show it.

Here is my another solution, which is for general n:

        If $a_1a_2a_3...a_n=1$, prove that
$\frac{1}{1+a_1+...+a_1^{n-1}}+...+\frac{1}{1+a_n+...+a_n^{n-1}}\geq1$

Proof: First notice that the function
        $f(x)=\frac{1}{1+e^x+...+e^{(n-1)x}}$ is convex on $[0, infinity)$.

Thus we may assume that $a_1\leq a_2\leq...\leq a_k<1<a_{k+1}=...=a_n$.
Now we need a lemma:
        $Lemma:$ Let a>1 be an arbitrary real. Then
$\frac{1}{1+a^{-n}+a^{-2n}+...+a^{-n^2}}+\frac{n}{1+a+...+a^n}>1$, for every positive integer n.
        By the Lemma, we have $\frac{1}{1+a_1+...+a_1^{n-1}}+...+\frac{1}{1+a_n+...+a_n^{n-1}}-1\geq \frac{(k-1)n}{n}-\frac{n-1}{1+a_1^{-\frac{1}{n-1}}+...+a_1^{-\frac{n-2}{n-1}}+a_1^{-1}}$
$-...-\frac{n-1}{1+a_k^{-\frac{1}{n-1}}+...+a_k^{-\frac{n-2}{n-1}}+a_k^{-1}}+\frac{1}{1+a_{k+1}+...+a_{k+1}^{n-1}}+...+\frac{1}{1+a_n+...+a_n^{n-1}}\geq 0$, because f(x) is convex on $[0,infinity)$ and $0<a_i^{-\frac{1}{n-1}}<a_{k+1}=...=a_n$ for $1\leq i\leq k$. The equality holds iff k=0, i.e. $a_1=a_2=...=a_n=1$
I shall post the proof of the lemma later...I've gotta go.

Your lemma is very interesting Darkseer but
f(x) as was before was mentioned isn' t convex.

[quote="prowler"]f(x) as was before was mentioned isn' t convex.[/quote]
No doubt that it is not convex. But it is convex on $[0,infinity)$ and I only make use of it. This can be showed by computing f''(x) or just prove that
$\frac{1}{1+a^2+...+a^{2n}}+\frac{1}{1+b^2+...+b^{2n}}
\geq\frac{2}{1+ab+...+a^nb^n}$ when a,b$\geq$1.

Sorry for ignoring your post Darkseer. But I wanted to post my idea.
We can use mixing variables: take a pair (a,b) in a such way
that ab ≥ 1 and (b>a) and we mix (a,b) with (√ab, √ab).
We want to prove:

1/(a+1)(a2+1) +1/(b+1)(b2+1) ≥ 2/(√ab+1)(ab+1) or

(√ab+1)(ab+1)((a+1)(a2+1)+(b+1)(b2+1)) ≥ 2(a+1)(a2+1)(b+1)(b2+1)

(b+1)(b2+1)(√ab+1)(ab+1)-(a+1)(a2+1)) ≥ (a+1)(a2+1)( (b+1)(b2+1)-(√ab+1)(ab+1))

(b+1)(b2+1)(√b-√a)√a (ab+a√ab+a2+a+√ab+1) ≥
(a+1)(a2+1)(√b-√a)√b (ab+b√ab+b2+b+√ab+1)

(b+1)(b2+1)√a (ab+a√ab+a2+a+√ab+1) ≥ (a+1)(a2+1)√b (ab+b√ab+b2+b+√ab+1)

But              (b+1)(b2+1)√a(a2+1) ≥ (a+1)(a2+1)√b (b22+1)
because       (a2+1)(b2+1)(√ab-1)(√b-√a) ≥ 0

                   (b+1)(b2+1)√a √ab(a+1)≥ (a+1)(a2+1)√b √ab (b+1)
because       (a+1)(b+1)(√b-√a)(√ab (b+√ab+a)-1) ≥ 0

                    (b+1)(b2+1)√a (ab+a) ≥ (a+1)(a2+1)√b (ab+b)
because        (ab√ab-1)(√b-√a) ≥  0

May be I have some mistakes?

[quote="prowler"]We can use mixing variables: take a pair (a,b) in a such way
that ab ≥ 1 and (b>a) and we mix (a,b) with (√ab, √ab).
We want to prove:
1/(a+1)(a2+1) +1/(b+1)(b2+1) ≥ 2/(√ab+1)(ab+1)[/quote]
I think you've made no mistake. My computation also show that ab$\geq$1 is enough(instead of a,b$\geq$1) for general n.

Two very nice proofs to kill a very nice ineq.

Prowler proved the Lemma :

[quote="prowler"]We can use mixing variables: take a pair (a,b) in a such way
that ab ≥ 1 and (b>a) and we mix (a,b) with (√ab, √ab).
We want to prove:

1/(a+1)(a2+1) +1/(b+1)(b2+1) ≥ 2/(√ab+1)(ab+1) [/quote]

Using that we can reduce the original ineq to

$(ab-1)(\sqrt{ab}-1) \geq 1 $

which is true as $ab \geq 1$.

But I think it is necessary to prove Prowler lemma also for $cd \leq 1$ as being $abcd =1$ we can put WLOG $ab \geq 1$ but then $cd \leq 1$.

How to prove it for $cd \leq 1$ ?

Thank you very much.

You are right, sorry. But now I cannot how to use Prowler Lemma to prove the inequality.

Thank you very much. :lol:

Thanks manlio for kindly noticing my mistake.
Not mistake but that lemma doesnt give complete solution.
Maybe we can continue with the next idea:

I want to say that using upper thoughts we can say that minimum is attained when
from that four numbers (a,b,c,d) - three of them are bigger or equal then one
and are equal between them.  Let us do WLOG  two cases.

1 case  one number is bigger than 1, a ≥ 1 ≥ b ≥ c ≥ d. then  obviously ab ≥1
             And we can say (√ab,√ab,c,d) better (a,b,c,d)
             √ab ≥ 1 and we go to the second case:

2 case - two numbers bigger than 1,  a ≥ a ≥ 1 ≥ c ≥ d. (we can say a =  √ab )
       Again WLOG we can say ac ≥ 1 or (√ab,√ac,√ac,d) better (√ab,√ab,c,d)
       Then we have  √ab,√ac ≥ 1.
       Applying lemma minimum will be when  a = √ab = √ac

So we have minimum is attained  in (a,a,a,d) or (a,a,a,1/a^3). a ≥ 1
So our inequality reduces to
  3/(1+a)(1+a2) +a^9/(1+a^3)(1+a^6) ≥ 1
or after simple opening parenthesis
 2 - 3a + 2a^3  3a^5+2a^6 ≥ 1
Which is equalent to (a-1)2(2a^4+a^3+a+2) ≥ 0
Please check it, hope to be everything correct.

[quote="Vasc"]Two questions: Why $a_i^{-\frac{1}{n-1}}<a_n$ ? How does it help to prove the above inequality ?[/quote]
I have made a mistake in a inequality. It should be
$1<a_i^{-\frac{1}{n-1}}\leq a_n$ instead of $0<a_i^{-\frac{1}{n-1}}<a_n$

Anyway, $a_i^{-1}\leq a_n^{n-1}$ because $a_i^{-1}\leq a_1^{-1}...a_k^{-1}=a_{k+1}...a_n\leq a_n^{n-1}$
The inequality can then be solved by Jensen's.
More clearly, write $b_1=\frac{-ln(a_1)}{n-1},...,b_k=\frac{-ln(a_k)}{n-1}$ and $b=ln(a_n)$.
The inequality becomes $(nk-n)f(0)+(n-k)f(b)\geq(n-1)f(b_1)+...+(n-1)f(b_k)$. Note that $f(0)=\frac{1}{n}$ . This is true because $\sum(n-1)b_i=(n-k)b$ , $0\leq b_i\leq b$ , and because f is convex on $[0,infinity)$.

Dear Prowler,

I think your proof is very good and very nice, congratulations, but I would like to see a complete proof of Mixing Variables Theorem, so I will post it as a new topic.

Can you post your solution, Vasc, please ?

Yes Darkseer. Your method works in my opinion.  Excellent solution!
But the last inequality you used is not Jensens Inequality, it is the Majorization Inequality (of Karamata).

Prowler's solution is also excellent for n=4.
Unfortunately, it doesnt work for n>4.

Manlio,
My solution for n=4 uses the inequalities
$\frac{1}{1+a+a^2+a^3}+\frac{1}{1+b+b^2+b^3}\geq \frac{1}{1+(ab)^{3/2}}$,
$\frac{1}{1+c+c^2+c^3}+\frac{1}{1+d+d^2+d^3}\geq \frac{1}{1+(cd)^{3/2}}$.

For n>4, my solution is too complicated.

By symmetry of equation,WLOG $0<a\leqq{b}\leqq{c}\leqq{d}$, from $abcd=1$,$abcd\geqq{a}\cdot{a}\cdot{a}\cdot{a}\cdot$,we have $0<a\leqq1$.

Since $(a-1)(a^2+2a+3)\leqq0$, we obtain $(1+a)(1+a^2)\leqq4$,yielding

$\frac{1}{(1+a)(1+a^2)}\geqq\frac{1}{4}$,  i.e. $\frac{1}{1+a+a^2+a^3}\geqq\frac{1}{4}$ ,the equality holds $a=1$.Similarly,changing $a$ to $b,c,d$,

yielding RHS of inequality in question $ \geqq{1}$, the equality holds for $a=b=c=d=1$. Q.E.D.

kunny

You are kidding, right? Anyway, congratulations to Darksheer for the marvelous solution he gave.

What do you mean?This was my first solution.Is my answer wrong?

kunny

I think you made a small mistake.
Whan you use WLOG a ≤ b ≤ c ≤ d then (of course  1 ≤ d)
so you can't use same argument for b,c,d.

Thank you for your comment, prowler.

I could confirm my basically mistakes rather than small mistakes.

kunny

Sorry, Prowler, but I don't understand that step

[quote](b+1)(b2+1)√a (ab+a) ≥ (a+1)(a2+1)√b (ab+b)
because        (ab√ab-1)(√b-√a) ≥  0 [/quote]

You are right manlio I did a small mechanical mistake.
Anyway it doesn't change nothing.

(b+1)(b2+1)√a (ab+a) ≥ (a+1)(a2+1)√b (ab+b)
(b+1)2(b2+1)*a√a  ≥ (a+1)2(a2+1)*b√b   or
(√b+1/√b)2(b√b+1/√b) ≥ (√a+1/√a)2(a√a+1/√a)

But this is right because

√b+1/√b ≥ √a+1/√a  or  (√b- √a)(√ab -1) ≥ 0

b√b+1/√b ≥ a√a+1/√a  or  (√b- √a)(a√ab+ab+b√ab - 1) ≥ 0

Now I hope no mistake.

Thank you very much, Prowler.

Now all is OK and your proof is really very nice.


calculus
integration
function
calculus computations

Calculate the integral below:
$ I \equal{} \int(x^2 \plus{} a^2)^{\frac{1}{3}}dx$

It doesn't have an elementary antiderivative (Mathematica uses hypergeometric function to express it, and it can't be simplified). Is it a random problem, or you're using the integral in a bigger calculation (finding a definite integral of it or something related)?

Thanks for your comment!
Actually, i need calculate this intergral for my papers which use the MQs(Multiquadric functions) $ f(x)\equal{}(x^2\plus{}a^2)^{\alpha}$ with $ \alpha \in R$.
Best, Luan.

If you only need a numerical value, then the explicit form of antiderivative is somewhat overkilling. Or, if you concern about the definite integal over $ [0, \infty)$, then we can have the answer in terms of Gamma function.


You can order chocolates in boxes of 1, 2, or 4.
If you want to order 98 chocolates, how many ways can you do it?

The original problem was 14, I solved it without brute force then I realized how trivial it was with brute force.

I'll post a hint if no one gets it by tommorow.

[quote="kevinchenxi"][hide="my answer"]662[/hide]

Edit:
sorry, a mistake[/quote]
Solution?

hmm... let me gather my thoughts.

[hide="Another miscalculation?"]
You make 98 with 1s, 2s, and 4s.
First, find how many combinations are there with only 1s and 2s. Since there can be from 0 2s to 49 2s, there are 50 ways to do that.

But using only 4s and 1s, there are 24 ways.
Using only 4s and 2s, there are 24 ways.

Using all 3 of the numbers to create 98:
lets say we use 23*4s.
23*4+2*1+2*2.
23*4+4*1+1*2

there are 2 ways to do it.

and there are 4 ways to do it when we have 22*4.

there are 6 ways to do it when we have 21*4.

I'm kinda tired, so I won't elaborate for now, but I saw that, I added 2+4+6+8+....+46.

48*11.5=552.

add all of them together and I got
552+24+24+50=650.

? maybe I had a miscalculation here or before...[/hide]

sorry, I'm real bad at explaining

I didn't use any real method for this problem. I'm sure there's an easier way.

[quote="bpms"]Hm, here was my solution.
[hide]
$x+2y+4z=98$
$x \equiv 0 \mod 2$
$x=2a$
$a+y+2z=49$
$a+y \equiv 1 \mod 2$
There are 2 cases, a is odd, y is even or a is even, y is odd. Since both are equally likely, lets just assume the first and double what we get.
$2b+1+2c+2z=49$
$b+c+z=24$
$\binom{26}{2}\cdot 2=650$[/hide][/quote]
[hide]Bpms' solution is quite nice for this problem. I actually used generating functions to solve this at the contest. Following is how I did it. Though I am listing out the steps, in reality it is much faster. It only took me 30 sec to do it..
\begin{eqnarray*}P_{1}& = & x^{0}+x^{1}+x^{2}+\ldots+x^{15}\\ P_{2}& = & x^{0}+x^{2}+x^{4}+\ldots+x^{14}\\ P_{3}& = & x^{0}+x^{4}+x^{8}+\ldots+x^{12}\\ P_{2}\cdot P_{3}& = & (x^{0}+x^{2}+x^{4}+\ldots+x^{14})+\hspace*{0.25in}\Longrightarrow \text{ 8 terms }\\ & & (x^{4}+x^{6}+\ldots+x^{14})+\hspace*{0.6in}\Longrightarrow \text{ 6 terms }\\ & & (x^{8}+x^{10}+x^{12}+x^{14})+\hspace*{0.5in}\Longrightarrow \text{ 4 terms}\\ & & (x^{12}+x^{14}) \hspace*{1.25in}\Longrightarrow \text{ 2 terms }\end{eqnarray*}
Each of the above 20 terms will multiply with an appropriate term in $P_{1}$ to give $x^{15}$, thus the total possibilities is 20 for the original problem in the contest.[/hide]

[quote="#H34N1"]Where was this from?[/quote]
This years county, only that one said 14 chocolates.

Oh. Yes. I should have knew.

For the real one, I used generatign functions as mythiaga did (I just brushed up on it the day before the competition).

[quote="archimedes1"][quote="bpms"]Hm, here was my solution.
[hide]
$x+2y+4z=98$
$x \equiv 0 \mod 2$
$x=2a$
$a+y+2z=49$
$a+y \equiv 1 \mod 2$
There are 2 cases, a is odd, y is even or a is even, y is odd. Since both are equally likely, lets just assume the first and double what we get.
$2b+1+2c+2z=49$
$b+c+z=24$
$\binom{26}{2}\cdot 2=650$[/hide][/quote]
I understand your solution up to the last step.  How do you get 26C2?  Am I missing something really simple or is there a formula I don't know?[/quote]
For the equation $a+b+c = n$, where $n$ is a positive integer, the number of possible ordered solutions of $(a, b, c)$ in non-negative integers is $\binom{n+2}{2}$.

In general for the equation, $a_{1}+a_{2}+\ldots+a_{i}= n$, the number of ordered solutions in non-negative integers is $\binom{n+i-1}{i-1}$.

[quote="madskillz1"]Extension:

wait, im just curious, what if it was like w+2x+4y+8z=98

and also w+2x+4y=96

How would you do these problems-(the first one has 4 terms and the second one, when you divide by 2, it creates an even number)[/quote]
Let's see.
For the first:
[hide]
w is even.
$a+x+2y+4z=49$
$a+x=1 \mod 2$[/hide]
It looks like it'll get messy without generating functionology, but you can do it.
The second
[hide]
$a+x+2y=48$
Case I:
a and x are odd.
$n+o+y=23$
$\binom{25}{2}$
Case II:
a and x are even.
$n+o+y=24$
$\binom{26}{2}$
$\binom{25}{2}+\binom{26}{2}=.5*25(24+26)$
$625$[/hide]

Thank you very much-I have to really learn mods..


Let $ k \geqslant 1$, $ n \geqslant 2$, and $ a_1, \ldots, a_n$ pairwise distinct integers.
Denote by $ P_k(X) \equal{} (X \minus{} a_1)^{2^k}\cdots(X \minus{} a_n)^{2^k} \plus{} 1$.
Can we find all the $ k\geqslant 1$, s.t. $ P_k$ is irreducible over $ \mathbb{Q}[X]$?

hint:[hide] We can show that $ X^{2^k}\plus{}1$, $ P_1$, and $ P_2$ are irreducible over $ \mathbb{Q}[X]$[/hide]

Yes, we can, but the proof is quite long and uses some nontrivial facts about units in cyclotomic fields. See the paper of I.Seres "Irreducibility of polynomials" in Journal of Algebra 2, pages 283-286, 1965. He proves there a much more general result.

[quote=JC_math]Sounds interesting, I don't have access to this article, do you have a copy?
Thanks[/quote]
You can download it [url=http://www.sciencedirect.com/science/article/pii/0021869365900098/pdf?md5=2ea31940ea876306ea5f928ebd97d765&pid=1-s2.0-0021869365900098-main.pdf]here[/url].


trigonometry
geometry
geometric transformation
rotation
projective geometry
geometry proposed

Let $ C$, $ D$ be two arbitrary points on the semicircle with center $ O$ and diameter $ AB$. Tangent of the semicircle from point $ B$ intersects $ CD$ at point $ P$. Line $ PO$ intersects lines $ CA$, $ AD$ at $ E$, $ F$, respectively. Prove that $ OE = OF$.

I'll post my solution to this nice problem. For point names, look at the picture.

$ SH$ is the polar line of $ G$ wrt $ SA,SB$.
So we have that $ G,T$ divide $ A,B$ harmonically, and now the following relation is immediate:$ \frac{GA}{GO}=\frac{GT}{GB}$
[hide="Method for disbelievers"]Because $ G,T$ and $ A,B$ form a harmonic bundle, we can find that $ GA = \frac{AT.AB}{AB-2AT}$.
Using this relation and the harmonic division mentioned above, we can easiley check that $ \frac{GA}{GO}=\frac{GT}{GB}$.
[/hide]
This tells us that $ AK \parallel PO$.

Because $ K = CD \cap HS$, we know that $ AS,AK,AH,AO$ is a harmonic bundle.
We look at the transversal $ PO$ of this bundle, which happens to be parallel to $ AK$.
The intersection points with the four lines in the bundle have to be harmonic. Since $ PO \cap AK$ is a point on infinity, we must have that $ O$ is the midpoint of $ EF$.

[img]http://img72.imageshack.us/img72/8302/aprilequalsegmentsfj8.jpg[/img]

(sorry jan if this is exactly your solution,because I didnt read yours carefully...  :oops: )
let $ K$ be a point on the semicircle such that $ PK$ is tangent to this circle...
then we have that points $ B,K,D,C$ are a harmonic division i.e $ (BK,DC)=-1$
now transfer these points with $ A$ to line $ PO$ to get that:
$ (OK',FE)=-1$ where $ K'$ is the intersection point of $ AK,PO$ now its sufficient to show that $ K' \to \infty$ i.e. $ AK \parallel PO$
but we have that:
$ \triangle OPD = \triangle OPB \Rightarrow \angle POD=\angle POB \Rightarrow \angle DAP=\angle AFO \Rightarrow AK \parallel PO$
as wanted...

[color=purple][size=134][b]This problem and the its Jan's proof are very nicely ![/b][/size][/color]

[color=darkblue][b]Definition.[/b] [u]A division[/u] is a tidy set of four collinear points $ (A,C,B,D)$ so that $ A\_\_\_\_{C}\_\_{B}\_\_\_\_\_\_{D}$.

The division $ (A,C,B,D)$ is [u]harmonically[/u] ([b]h.d.[/b]) iff $ \frac{CA}{CB}=\frac{DA}{DB}$.

[b]Jan[/b] used the characterization : $ \boxed{\mathrm{(A,C,B,D)\--\ h.d.}\Longleftrightarrow AC\cdot AD=AB\cdot AM}$, where $ M$ is the midpoint of $ [CD]$.

I must recognize that I didn't know this characterization of a harmonical division which was used by [b]Jan[/b].[/color]

[quote="April"][size=117][color=darkred]Let $ C$, $ D$ be two arbitrary points on the semicircle with center $ O$ and diameter $ AB$.

Denote $ P\in BB\cap CD$, $ E\in PO\cap CA$, $ F\in PO\cap AD$. Prove that $ OE = OF$.[/color][/size][/quote]

[quote="Virgil Nicula"][size=117][color=darkred][b]An equivalent enunciation.[/b]  Let $ ABC$ be a triangle and let $ D\in BC$, $ E\in CA$, $ F\in AB$ be the projections of the orthocenter $ H$

to the sidelines of $ \triangle ABC$. Define the points $ \{\begin{array}{c}M\in (BC)\ ,\ MB=MC\\\\ P\in EF\ ,\ PC\perp BC\end{array}$ and $ \{\begin{array}{c}X\in AB\cap PM\\\\ Y\in BE\cap PM\end{array}$. Prove that $ MX=MY$.[/color][/size][/quote]

[color=darkblue][b]Proof[/b] ([u]similarly with [b]Jan's[/b][/u]). Define the point $ \{\begin{array}{c}K\in AD\cap EF\\\\ S\in (AH)\ ,\ SA=SH\end{array}$  and denote $ \{\begin{array}{c}m(\widehat{KBD})=\phi\\\\ m(\widehat{PMC})=\psi\end{array}$. I"ll show $ \phi=\psi$, i.e. $ BK\parallel MP$.

$ 1\blacktriangleright$ Is well-known that $ (A,K,H,D)$- [b]h.d.[/b], i.e. $ DH\cdot DA=DK\cdot DS$ $ \Longleftrightarrow$ $ DK=\frac{DH\cdot DA}{DS}=$ $ \frac{DB\cdot DC}{DS}$ $ \Longrightarrow$

$ \tan\phi =\frac{DK}{DB}=$ $ \frac{2\cdot DC}{2\cdot DS}=$ $ \frac{2\cdot DC}{DH+DA}=$ $ \frac{2}{\frac{DH}{DC}+\frac{DA}{DC}}=$ $ \frac{2}{\cot B+\tan C}$ $ \Longrightarrow$ $ \boxed{\tan\phi =\frac{2\cdot\tan B}{1+\tan B\cdot\tan C}}\ \ (1)$.

$ 2\blacktriangleright$ The Sinus' theorem in $ \triangle CPE$ $ \implies$ $ \frac{PC}{\sin B}=\frac{CE}{\cos (B-C)}=\frac{2\cdot MC\cos C}{\cos (B-C)}$ $ \implies$ $ \tan\psi =\frac{CP}{CM}=\frac{2\sin B\cos C}{\cos (B-C)}$ $ \implies$

$ \boxed{\tan\psi =\frac{2\cdot\tan B}{1+\tan B\cdot\tan C}}\ \ (2)$. From the relations $ (1)$ and $ (2)$ obtain $ \tan \phi =\tan \psi$, i.e $ \phi =\psi$.

In conclusion, $ BK\parallel MP$ and $ \{\begin{array}{c}B(A,K,H,D)\mathrm{\ is\ harmonical\ fascicle}\\\\ M\in BD\ ,\ MP\parallel BK\end{array}\implies MX=MY$.

[b]Remark.[/b] Can use and another characterization of the harmonical division : ${ \boxed{\mathrm(A,C,B,D)\--\ h.d.\Longleftrightarrow\frac{2}{AB}=\frac{1}{AC}+\frac{!}{AD}}}$.[/color]

Let $ K$ be the midpoint of $ CD$.
We have $ OK$ and $ CD$ are perpendicular so $ O,B,P,K$ are cyclic
Thus $ \angle BKD=\angle BOP=\angle AOE$
otherwise $ \angle EAO=\angle KDB$ so triangles $ EAO$ and $ BDK$ are similar
thus $ EA/BD=AO/DK=AB/CD$
So triangles $ EAB$ and $ BDC$ are similar
then $ \angle DAB=\angle ABE$.
Thus $ AD//BE$.
so $ OE=OF$ $ q.e.d$

Let $ C'$ be the point diametrically opposite $ C$, and let $ F' = BC' \cap DA$.  Then $ P = BB \cap DC$, $ F' = BC' \cap DA$, and $ O = CC' \cap BA$, so by Pascal's theorem, $ P$, $ F'$, and $ O$ are collinear.  In other words, $ F'$ is the intersection of $ OP$ and $ AD$.  But so is $ F$, so points $ F$ and $ F'$ coincide.

Since $ A$ is diametrically opposite $ B$, and $ C$ is diametrically opposite $ C'$, $ BC'$ is the image of $ AC$ under a rotation around $ O$ by $ 180^\circ$.  Therefore, points $ E = OP \cap AC$ and $ F = OP \cap BC'$ are symmetric around $ O$.

I've seen this before on a competition, but I don't remember where.


solve this equaltion:

$tg^4xsin^2x+cotg^4xcos^2x+\frac{1}{tg^2x+cotg^2x}=\frac{1}{2}(tg^2x+cotg^2x+1)$

 :D  :D  :lol:  :blush:

this is my hint
[hide]cauchy_schwart[/hide]

Let $y=\tan ^2 x$, we can rewrite your equation into $2y^7-2y^6-y^3-y+2=0\Longleftrightarrow (y-1)(y+1)(2y^5-2y^4+2y^3-2y^2+y-2)=0$

Therefore some of roots is $x=\pm \frac{\pi}{4}+n\pi\ (n=0,\pm 1,\pm2,\pm3,\cdots)$

By the way, How many roots do your equation have at $0\leqq x\leqq \pi$ ?

i used cauchy_schwart ineq

and have a nice proof :D


How long have you guys played your instrument?

What kind of instrument do you play?

As for me, I play violin, and been playing for nearly 3 years..(3 and 1/6 years to be precise.. :lol: )

I started out learning to play the trumpet and now I'm a trombonist. I have been playing since 3rd grade so I guess about 6 years now.

I play the Piano! Must be around 9 or 10 years now I think.

Piano, 6 years. I know this person who's played four years and is better than me (he practices five hours a day), a person who's played 8 years and is worse than me (he practices thirty minutes a day) and a person who's played 7 years and is worse than me (she practices one hour a day).

I play violin just for about 5 months. But I think I can learn it fast and I will play it well. :P

I started out on trumpet. But I saw the light, and switched to the French Horn. I 've been playing for close to 7 years now.

Piano.  3.5~4 years.  Recently got a new teacher from Saint Petersburg Russia.

Piano, 12 years now (since I was 4). I've done all the exams 'n stuff. I now play for leisure. It's great to be able to sight-read almost anything instantly. Although, thinking back now, I would rather have used all that money on math lessons instead, so I can be geniuses like you people.

[quote="Identity"] Although, thinking back now, I would rather have used all that money on math lessons instead, so I can be geniuses like you people.[/quote]

Being a genius is a natural thing. Although I would say most people on this forum are hard working (very few are actually genius).

drums for about 2 or 3 years

Go percussion!

well actually how do u play since birth??

anyway... i've played violin for like 4 3/4 years and erhu for 2years......... yeah :icecream:  :juggle:  :jump:

[quote="BanishedTraitor"]very few are actually genius[/quote]
OK, then what is actually genius?

I play piano for 2.8333333.... years

[quote="serialk11r"]Genius is like, having extremely good sense of what the music is supposed to be like or something...I think its safe to say that I have not met any musical genii yet.[/quote]

Yes, but I think BanishedTraitor is talking about math genius in this site, for which he think there is only few.

They do. But sometimes it can be ten times as hard as piano competitions.

Is it because the saxophone is a jazz-oriented instrument and requires lots of improvisation. Plus I bet there aren't as many people as the piano ones.

I live in a "jazz-oriented" area, with a bunch of sax players, but not a lot of pianists. :(

I've played the piano for around eight or nine years, the violin for two, and the harp for one. My harp teacher died, so I dropped out.

I play the violin and I have been playing for 3 years! :)

I've played piano for maybe 6 years. I practice 2 1/2 hours a day
and trombone :lol: for about two years six months

i've been playing the piano for around 9-10 years now, since i was 4. but i think i might have to quit soon to focus on mathand tennis and ap stuff more ):

Played the violin for 1/2 year, took tabla lessons for 1/2 month. My cousin says I will do guitar 1/2 week. I haven't even done guitar before and don't have any plans to do it. Taking indian classical vocal for over a yaer now.

I've been playing for 4 years. Can do George Winston's arrangement on Canon now. I only practice like an hour a day...

I play piano and have benn playing since about 3 1/2. I'm 11. Do the math.
[hide="If you're too lazy to do the math, click this underlined blue text."]7 1/2 years![/hide]

[quote="Readingrocks88"]I play alto sax and piano.[/quote]

Wow did you seriously have to revive this...please don't.

Same thing happens with kwausouq, both of whom I captained this year (State Champs!)

But anyways, Piano for 6 years (quit), and Percussion for 3, at which I consider myself to be extremely great at.

6-8 months. But also, that's a bad idea for a poll because it changes over time


Q: Why are some of the 1 point questions I recieve quite difficult while some 2 point questions are incredibly easy? For me, the 3 and 4 point ones seem reasonable in terms of this.

Maybe the 1 poiners are my opinion, but they are certainly much more difficult than 2 pointers like this-

2 points

Compute 10C4.

A problem's point value is based not on its difficulty, but on the probability you'll get it right, which is based on how well you've done previously on the topic.  :lol:

At least, I think so...  :huh:

http://www.mathlinks.ro/viewtopic.php?t=255930

Getting 1pt questions is a GOOD thing. It means that you are mastering the topic quite well.

james4l is right. It is good, but not if you want a lot of points.

Also, just a clarification, there are no 3-pt questions.

[quote="connaissance"]james4l is right. It is good, but not if you want a lot of points.[/quote]

Of course, doing Alcumus for the points is both a waste of such a great program and a waste of whoever's doing that's time. Thus, hopefully, there aren't that many users doing Alcumus with that objective in mind.

[quote="weird888"]ummm, if you get a problem wrong, why does it sometimes give you 2 points, than sometimes it only gives you 1? :huh:[/quote]

Huh? If you get a problem wrong, you shouldn't get any points.

It's probably a bug in the system.
And why do topics about 1-point problems keep popping up? After reading other 1 point topics you should know a lot.


hihihihi:D

okieee there is this one problem that im having trouble with
i asked my mathcounts friend && she got 11

but for some reason tat doesn't seem right

 problem:
during a special football game, a team scores 7 points for a touchdown & 3 points for a feild goal. There is no limit on the number of points a team can score. Using any combinations of touchdowns&&feild goals what is the most number of points a team CANNOT score. Explain why you think your answer is correct solution to the problem.

can u guys please help me! :]
thankss:]]

Ooh the chicken-nugget theorem!!!

The Chicken Nugget Theorem states that the highest number you CAN'T get with two numbers m and n is $ mn\minus{}(m\plus{}n)\equal{}7*3\minus{}(7\plus{}3)\equal{}21\minus{}10\equal{}\boxed{11}$.  So your mathcounts friend was right  :D

ARE YOU KIDDING ME?  I always have to use so much number theory and manipulation to get these problems (like the one I did recently on this forum), and you can do it with a simple formula? =/

For example, for this problem, you know that if the sum has at least 2 7's, you can replace them with 5 3's to get the next number.  Meanwhile, if you have all 3's, then if you have 2, you can replace these with a 7 to get the next highest number, and then depending on what remains, you can do use these methods to continue.  So, the biggest number you could not obtain would be the one that does not contain 2 7's and 2 3's, namely: 7+3.  The next number up could hence not be gotten, so the answer is 7+4=11 (The next number 12, can then be replaced by 4 3's, and so on).  This wasn't that hard, but when the numbers are bigger, the process is much more tedious.

of course you could list possible scores until u get 3 in a row because all other scores can be made by adding 3('s) to any one of these scores

that would work here, and i think it might be faster. what if you had "a touchdown is 17 points and a feild goal was 11"

Hmm, this problem seems familiar...
I think I've seen it before...
I wonder where...

[quote="infinity4ever"]Ooh the chicken-nugget theorem!!!

The Chicken Nugget Theorem states that the highest number you CAN'T get with two numbers m and n is $ mn \minus{} (m \plus{} n) \equal{} 7*3 \minus{} (7 \plus{} 3) \equal{} 21 \minus{} 10 \equal{} \boxed{11}$.  So your mathcounts friend was right  :D[/quote]

remember, however, this only applies if $ m$ and $ n$ are relitively prime.


function
limit
induction
geometry
3D geometry
real analysis
real analysis unsolved

Prove that the sequence $ x_{1},x_{2},x_{3}...$ of real numbers given by $ x_{1}\equal{} 1$ and $ x_{n\plus{}1}\equal{} x_{n}\plus{}\frac{1}{x_{n}^{2}}$ for each $ n \equal{} 1,2,3...$ is unbounded.

I'm trying to do proof by contradiction. Saying there is a bound M and than either showing there is an $ x_{n}$ that is bigger than it, or showing that the existence of a bound makes the recursive definition not true. I haven't been able to get any where. Please help! Thanks!!

If a sequence if real numbers is bounded above, it has a least upper bound. Use that.

A more general version, which isn't any harder:
Suppose $ f$ is a continuous function from $ \mathbb{R}$ to itself, and $ f(x)>x$ for all $ x$. If $ x_{0}$ is arbitrary and $ x_{n\plus{}1}\equal{}f(x_{n})$ for $ n\ge 0$, $ \lim x_{n}\equal{}\infty$.

I'm still kind of confused as what to do. Do I assume that the LUB is a and then show at some value of n, the sequence is bigger than it? Even if that is what you are supposed to do, I'm still kind of confused as to how do go about doing that.

$ x_{n\plus{}1}>x_{n}$
If $ x_{n}$ is bound then $ \{x_{n}\}$ converge.
Call limit is l then
$ l\equal{}l\plus{}\frac{1}{l^{2}}$ tradition.
So $ x_{n}$ unbound.

Hmm, I used to do that stuff for like Calc BC and stuff, but is that really rigorous? I just feel that argument wouldn't fly with my teacher. Is that the only way to do it?

other method you can prove that
$ x_{n}\geq\sqrt [3]{n}$

You can use induction
$ f(x)\equal{}x\plus{}\frac{1}{x^{2}}$
$ f(x)\equal{}1\minus{}\frac{2}{x^{3}}$
So $ f'(x)>0$ for $ x>\sqrt[3]{2}$
check n=1,2 then induction
note :I like first method .It will prove for general problem.

If there's a least upper bound, you can get close to it. If you're close enough, the next step takes you over.

But why does it take you over. I see in my mind why it does, but how would you write it down mathematically. Like what does 'close enough' mean mathematically.

TTsphn, thanks for your help too!

What "close enough" means depends on the function. Using the $ f(x)$ version, suppose the least upper bound is $ A$. There is a unique $ B<A$ with $ f(B)\equal{}A$; if $ x_{n}>B$, $ x_{n\plus{}1}>A$.

[quote="jmerry"]What "close enough" means depends on the function. Using the $ f(x)$ version, suppose the least upper bound is $ A$. There is a unique $ B < A$ with $ f(B) \equal{} A$; if $ x_{n}> B$, $ x_{n\plus{}1}> A$.[/quote]

So:

There is a unique $ B < A$ with $ f(B) \equal{} A$

is true if $ f(x)$ has a least upper bound?

Also, how do you know that $ x_{n}> B$ is ever true?

$ B$ has nothing to do with least upper bounds; it's about $ f$. I should have qualified that, actually. There are two possibilities: either $ f(B) \equal{} A$ for some $ B < A$ or $ f(x) > A$ for all $ x$. This comes from $ f$ being an increasing continuous function with $ f(x) > x$ everywhere, and the proof comes from the Intermediate Value Theorem.

If $ x_{n}$ were never greater than $ B$, $ B$ would be an upper bound- and then $ A$ wouldn't be the least.

Also, there's no need to quote entire posts, especially the post immediately preceding yours. Unless stated otherwise, it's assumed you're responding to that post anyway.

Ok, so for this function, the Intermediate Value Theorem says that either either $ f(B) \equal{} A$ for some $ B < A$ or $ f(x) > A$ for all $ x$?

If so, we haven't done that yet so I don't think I would be able to use it : )

Also, how does the least upper bound theorem play into this?

You need something like the IVT for the general case. You don't need it for this special case: if $ f(x)\equal{}x\plus{}\frac{1}{x^{2}}$, $ f(A\minus{}\frac{1}{A^{2}})>A$, which is good enough.

Least upper bounds and completeness are essential in the general case; I could construct an increasing continuous function $ f$ from $ \mathbb{Q}$ to $ \mathbb{Q}$ with $ f(x)>x$ that generates a bounded sequence.

An example: $ f(x)\equal{}\begin{cases}x\plus{}\frac{1}3& x\le 1\\ \frac{4x}{2\plus{}x^{2}}& 1<x<\sqrt{2}\\ \frac{2x^{3}}{2\plus{}x^{2}}&x>\sqrt{2}\end{cases}$
If $ x_{0}\equal{}1$, $ x_{n}<\sqrt{2}$ for all $ n$. The natural extension of $ f$ to $ \mathbb{R}$ has a fixed point at $ \sqrt{2}$, but that doesn't exist in $ \mathbb{Q}$. Anything greater than $ \sqrt{2}$ is an upper bound, but there is no least upper bound.

In any specific example, we could avoid using least upper bounds by constructing explicit estimates as TTsphn did. That isn't usually worth the effort.

With what TTsphn did, how can you just say $ a_{n}\equal{} l$. If l is the least upper bound, isn't it true that $ a_{n}$ doesn't necessarily ever equal $ l$ (just gets really really close to it?) Do you have to do some type of epsilon thing to do what he did 'rigorously'?

I never said that $ a_{n}$ is equal to the least upper bound $ A$- just that it's greater than some strictly smaller $ B$. That comes straight from the definition of [i]least[/i] upper bound.
TTsphn's method is completely different; for $ f(x)\equal{}x\plus{}\frac{1}{x^{2}}$, we have $ f(t^{1/3})\equal{}t^{1/3}\plus{}t^{\minus{}2/3}>(t\plus{}3)^{1/3}$, so $ x_{n\plus{}1}\equal{}f(x_{n})\ge f(\sqrt[3]{3n\minus{}2})\ge\sqrt[3]{3n\plus{}1}$, and $ x_{n}\ge\sqrt[3]{3n\minus{}2}$ by induction.
(Details improved- $ \sqrt[3]{n}$ is a pretty weak estimate)

[quote="vertigo74"][color=darkred]Prove that for a continue function $ f:\mathcal R\rightarrow (0,\infty )$, the recurrent sequence $ (x_{n})_{n\in N^{*}}$
given by $ x_{1}> 0$ and $ x_{n\plus{}1}\equal{} x_{n}\plus{}f(x_{n})$, $ n\in N^{*}$ is unbounded.[/color][/quote]
[color=darkblue][b]Proof.[/b] This exercise is very easily ! The first [b]TTsphn's reply[/b] "killed" it. Observe that
$ x_{n}\nearrow$ (increasing) $ \implies$ $ x_{n}\nearrow l\in (0,\infty )$ or $ x_{n}\nearrow\infty$. If $ x_{n}\nearrow l\in (0,\infty )$, then
from the recurrent relation (for $ n\rightarrow\infty$) obtain $ l \equal{} l\plus{}f(l)$, falsely because $ f(l) > 0$.
In conclusion, $ x_{n}\nearrow\infty$, i.e. this sequence is increasing and unbounded.[/color]

jmerry: I was talking about when he said l = l + 1/l^2 to show that the series did not have a limit. I understood the being bigger than cube root n stuff. So would comparing it to a another unbounded sequence be the best way to go?

... I'll just link to what I've said [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=162858]before[/url].


[quote="DenPoNaEi"]Parathrhsa oti olloi xrhsmopoioitai ena programa pou grafei me ma8hmatika sumbola.mhpos sas kanei kopo na mou dosete ena link me auto to pragrama?
euxaristo prokatavolika  :P[/quote]

 Τα παιδιά ξέρουν να σου απαντήσουν καλύτερα , αλλά σίγουρα , αν ψάξεις εδώ στο ελληνικό forum λίγο προς τα πίσω τα μηνύματα θα βρεις πολύ καλές συμβουλές. Νομίζω όμως ότι στο artofproblemsolving έχει πολλές πληροφορίες για το Latex.
 Πάτα και ψηλά στο LATEX Help.
 Sorry που αυτή τη στιγμή δε μπορώ να ψάξω περισσότερο.
 Μπάμπης

DenPoNaEi εχω βρει ενα πολυ καλο προγραμμα για LATEX:texaide.Ψαξε το στο google νομιζω οτι θα εχεις αποτελεσμα. :wink:


Done this a ton of times
If you're willing to give up points (or your opponents are weak or your test is hard) then just take your time.

or read the article the resources section...

before: practice.
during: check.
simplest way to drastically reduce stupid mistakes.

concentrate, stay focused, eat a good breakfast, and try to understand the question the best you can :)

Make sure your answer actually makes sense in the context of the problem--for example, if it's about the U.S. states, don't answer 140. (That's what I did at chapter countdown :blush:)

Cast out nines for arithmetic.

After you solve a problem, try quickly solving it in a different way.

and also.... [wait.. wat am i saying?! i'm the one that makes mistakes at practice]:

don't put down an answer that has never appeared in the history of mathcounts. If you find an answer like that, you should know that you did something wrong.

For example, infiniti.... (one of our teammates did it on the countdown... *cries and laughs at the same time*

But otherwise, just take everyone else's advice on doublechecking everything.

I used to zip through the problems once and then go back and redo them to compare answers, but that isn't a very good way. I suggest you take your time and don't feel like you have to rush. Write out your solution and calculations neatly on your scratch paper so you can just go back and check your work for logic and accuracy instead of having to redo the problem.

Also, practice. Do practice problems, check your answers, go back and redo the ones you get wrong, and make sure you understand why you got them wrong.

[quote="mysmartmouth"]At the 3 minute call on Sprint, I always panic and can't do (concentrate on) any more problems.  So I find this an excellent time to go back and check, and to ensure inches are expressed as inches, radicals as radicals, etc.[/quote]

That's what happened to me last year: I didn't pace myself and I really panicked at the 3-minute call, because I only finished about 22 problems. And I ended up getting only 23 points total (out of 46) :( Remember to bring a watch.

Yeah.  I always panic at 3 minutes.  Any suggestions on how to stay concentrated.

I know this seems kind of hard to do, but just let yourself tune out.  Make it so that all you see is the math, and let nothing else distract you.  You can even practice.  Turn on the t.v. and see if you can hear the words without understanding them.  Even if the 3-minute warning startles you, don't let it get you down, and continue as necessary.  I found it best not to check your answers unless you're completely done, because then you could be passing up some easy points.  If you feel like you need to check the problem, just do it right after you write in the answer.  Same goes for target and team rounds as well.  Good luck to everyone at states!

There is another topic on careless mistakes here:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76592]www.artofproblemsolving.com/Forum/viewtopic.php?t=76592[/url]


Show that if y = mx+k is a tangent to the ellipse at $\frac{x^{2}}{a^{2}}$+$\frac{y^{2}}{b^{2}}$ = 1, then $m^{2}a^{2}+b^{2}=k^{2}$

soln.
The ellipse has parametric eqn $x=acos\theta$ and $y=bsin\theta$.

Hence $dy/dx$ = $\frac{-bcos\theta}{asin\theta}$. If y =mx+k at $P(acos\delta,bsin\delta)$, then

m=$\frac{dy}{dx}$ at P $=\Rightarrow $ ma$sin\delta$ + $bcos\delta$ = 0
.
.
.
how did they obtain the last line?

O.K. You can use Implicit differentiation.

[quote="kunny"]Differentiate with respect to $x$ both sides of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.[/quote]

$\frac{2}{a^{2}}x+\frac{2}{b^{2}}y\frac{dy}{dx}=0$. Substitute $x=a\cos \delta ,\ y=b\sin \delta$.

If you're going to post it in intermediate, don't use calculus. You can solve this problem pretty quickly with transformations and the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1020599223&t=140526]"Stretch Method"[/url]


inequalities
combinatorics unsolved
combinatorics

Let$p$be a prime integer and$S=\{1,2,..p-1\}$Prove that for any patitions S into three nonempty sets$A,B,C$,there exist elements$a\in A,b\in B,c\in C$such that$a\equiv b+c$or$b\equiv a+c$or$c\equiv a+b(mod p)$

[quote="mumath"]Let$p$be a prime integer and$S=\{1,2,..p\}$Prove that for any patitions S into three nonempty sets$A,B,C$,there exist elements$a\in A,b\in B,c\in C$such that$a\equiv b+c$or$b\equiv a+c$or$c\equiv a+b(mod p)$[/quote]
this problem is wrong.please check it.(let A={p} & B={1,p-1} & C={2,..,p-2}.) :?:

[quote="jensen"][quote="mumath"]sorry,i edited,can you solved this?[/quote]
you're welcome.this problem is right and nice and i will send the solution later. :D[/quote]

Please give me your solution, [b]jensen[/b].

there was a basic problem with my solution....i have taken it off. it only occured to me today afternoon.


conics
parabola
symmetry
algebra
linear equation

Find the vertex, focus and directrix of the given parabola. Graph the equation.

y^2 - 4y + 4x + 4 = 0

MY WORK:

1-I subtracted -4x -4 from both sides and got:

y^2 -4y = -4x - 4

2-I then completed the square on the left side and factored on the right side and got:

(y^2 -4y + 4) = -4(x + 1), which became (y - 2)^2 = -4(x + 1)

Is this okay so far?

3-I expressed (y - 2)^2 = -4(x + 1) in the form
(y - k)^2 = -4a(x - h)....From this information, I found that the vertex is
(h, k) = (2, -1)....Is this the right vertex?

4-I found that a = 1 by solving for a in the linear equation: -4a = -4, then a = 1.

I also know this:

focus = (h - a, k)

Directrix is x = a + h

Axis of symmetry is the line y = k

I found the focus using (h - a, k) = (-1-1, 2) = (-2, 2)

I found the directrix using x = a + h, which is x = 1 + (-1) = 0, which is the x-axis itself, but this does make sense.

The axis of symmetry is found by using y = k, which to me is y = 2

This parabola opens to the left.

However, the math book tells me that I am way wrong.

Can someone check my work and tell me what I did wrong?

Thank you
symmetry

In your second step, you did not add 4 to both sides.

You should have:
\[ y^2 \minus{} 4y \equal{} \minus{} 4x \minus{} 4 \\
y^2 \minus{} 4y \plus{} 4 \equal{} \minus{} 4x \minus{} 4 \plus{} 4 \\
(y\minus{}2)^2\equal{}\minus{}4x
\]
Note that you need to add four to both sides, not just one to keep equality.  Whenever you do something to one side of an equation, you must do it to the other as well.  Try reworking it now.

[i]2-I then completed the square on the left side and factored on the right side and got:

(y^2 -4y + 4) = -4(x + 1), which became (y - 2)^2 = -4(x + 1) [/i]

When you complete the square, you forgot to add the $ 4$ to the right side when adding it to the left. So the RHS should be $ \minus{}4(x)$ instead of $ \minus{}4(x\plus{}1)$

After that, it's $ (y\minus{}2)^2\equal{}\minus{}4x$
The vertex should be at $ (0,4)$, and the focus should be in the direction that it's opening differing by the coefficient of $ x$ divided by $ 4$, so it should be $ (\minus{}1,4)$
The directrix is on the other side of that, $ x\equal{}1$


How do you define Free Trade?

I have read articles that define it as something other than simply letting people trade goods and services across borders freely.

Anyway, what's your opinion?

I wrote an article in my college paper, actually it should be out this Friday regarding free trade. I present two opposite arguments. I think that since a lot of economists agree with it, quite likely it is right. But at the same time, you will probably say that you can't believe everything you here. Based on the information, there are benefits like comparative advantage, specilization making an increase in economic growth. Yet, one example that I use in my paper which I can post on Friday is that NAFTA had negative effects according to the paper I read in the Economic Policy Institute. It brought a lot of people to lose their jobs, etc etc.

Free international trade, in the long run, typically provides lower prices on consumer goods since the country with the highest comparative advantage will provide these goods at the lowest oppurtunity cost.  However, unemployment, as you mentioned with NAFTA, is also a consequence of free international trade worth considering.

Recently in my hometown, Charleston, SC, tariffs on shrimp imports have been lifted, allowing Asian seafood companies to begin to edge out local shrimpers because of legislation concerning local fishing limits and such.
Though this imported shrimp is much cheaper, several local restaurants and providers have resisted this market domination, running "No import" ad campaigns and such.  However, in most situations, the local shrimpers would be forced to find more profitable ventures, either by moving or by becoming employed in another trade (which would probably be difficult, considering that most of them lack the skills to join another industry profitably).  So it seems that both free trade and restriction have their negative consequences, consumer benefits on one side, structural unemployment on the other.  Therefore, it is probably best to implement whichever would fit the situation.

It seems to me that artificial barriers are short term solutions to local jobs.

Besides that, it doesn't matter to me so much that a job is taken by an American or by someone else.  Let the world catch up economically.  That makes for a better community.

People learn new job skills -- if they didn't, there would never be need to fear new competition.

A small pool of unemployed workers is healthy for an economy and for the workers in the long run.  Innovation requires retooled employees for working new jobs.  The more advanced our economy becomes, the more often workers must learn new skills -- and have incentive to learn new skills.

Perhaps what we really need is a better system for providing occupational services to workers looking for jobs they are capable of educating themselves to work.  Hopefully the internet will make it easier to provide such services.

It's obviously terrible and tragic when a local economy breaks down, but that may simply be an unavoidable part of the growing pains of finding equilibrium with natural resource supplies.

I definetly agree with Mr Crawford that it is important that people in other countries get jobs and become employed, it essential for the future growth of the world economy.

You might consider posting your papers in this forum.  They might make for good discussions and you might get good feedback about them.

Ok, here is my article, I hope it is not very horrible.

http://www.thetriangle.org/media/paper689/news/2005/06/03/EdOp/Free-Trade.Confusing.But.Not.Beyond.Understanding-955462.shtml

You are free to discuss it in this forum.

[quote="hello"]I was wondering what all of you think about free trade? I have done a little research and see that almost all economists support it but there are some activists against it and other people too. :arrow:  :surf:   :rotfl:[/quote]

Michael,

I am a physicist by training, but I like all kinds of math applications so when it comes to things like Economics I dabble. I say that to say this, please take my opinion with a grain of salt. (Be advised that I might not have the slightest clue what I'm talking about.)

I'd like to offer an analogy. The 2nd Law of Thermodynamics says that it is impossible to extract energy from any thermodynamic system with 100% efficiency. The corollary to this is that the entropy (a measure of energy wasted) in a closed thermodynamic system can never decrease. In an open system this is not necessarily so. The difference between an open system and a closed system is that an open system allows material flux across its boundaries, but a closed system does not.  I think you can see where this is leading...

If one views the nation in the role of a thermodynamic process then to guarantee increasing economic inefficiency, however one wants to measure that, close the borders to material flux (trade). However, if one wishes to allow for the possibility of decreasing inefficiency the borders should not be closed.

Just because the borders are open doesn't necessarily imply prosperity though. If the borders are open and you have more trade leaving than coming in, then entropy (wasted value) still increases.

PS I read your paper, it was well written.

[quote="blahblahblah"]I think that one of the big problems with free trade is the human cost.  In North America, this is not such a big deal - increased competition due to free trade can displace people from their jobs, but, since North America is quite prosperous, there are other jobs, which may not pay as much nor be as attractive, but which can still easily provide a living.

In the Third World, the situation is considerable more dire.  It is routine for farmers to spoil their crops because crops coming from other countries where agriculture is government-subsidized can be sold for a much lower price.  Free Trade essentially puts all sellers on an equal footing in the marketplace, regardless of whether one group has received government subsidies or whether they have access to superior technology or whatever.  We may reap the benefits of Free Trade - that is, the benefits consumers receive from competition in the marketplace - but there are people whose livelihoods, and, by extension, their lives - are threatened by Free Trade.  You should probably look up some information on Fair Trade if you are interested in what I have said.[/quote]

That's always been the problem with Econ, hasn't it? balancing everyone's interests simultaneously. On the whole however if a nation can acquire more goods at a lower price by opening its borders than by shutting them then it ought to open them.

[quote="blahblahblah"]I think that one of the big problems with free trade is the human cost.[/quote]

The short term human cost may seem large.  Perhaps liberal trade laws between nations displaces industries and worker while making the whole process more efficient.  Perhaps then the problem is finding a way to employ the displaced workers during the transition process.

I think many of the effects of globalization center around this kind of issue.  In the process of generating more efficient economic processes, short term costs are large.  When those problems become political, even more enmity is generated regarding the process.

It also doesn't help that there are issues relating to free trade that large nations like the U.S. often don't address.  It's one thing to say we freely trade food with most nations.  It's another to point out that we subsidize our food industries to the tune of billions of dollars.  These subsidies can crush farming economies of our neighbors and free trade only accelerates the process.

If we're going to play the game of opening economic borders (and I think we should), it seems to me we need to reevaluate many related policies.

Wow! I can't believe almost all of you had positive things to say about free trade. I hate free trade with a passion. I got very angry just reading your posts. Have you never heard of Oxfam or Fair Trade? Their Make Trade Fair campaign is huge.

I got most of my information from Oxfam's propaganda they gave me when I went to the local office to sign up to volunteer for them. Yes they are biased, but which bias whould you trust more? Oxfam, a non-profit organization or economists who have money invested in the corporations who benefit most from free trade.

Obviously the rich countries benefit more from the unfair trade rules of the free global market. I can't even take the time to write down all the negative aspects of free trade because it makes me too angry.

When it comes to food, I try to buy as much organic and fair trade products as my small income will allow. Which is usually well over half my grocery bill spent on these products. I don't care if I pay two or three dollars more for things. At least I know that my dollar is going to make someone else's life better and the food I'm buying is of better quality.

I think the global economy is a bad idea and I can't wait until it come crashing down.

The Cons
-transportation costs of transporting all our cheap labour goods over the world hurt the environment and are wasteful.
-when you buy locally, you are supporting a stronger community and helping the farmers in your neighbourhood.
-less globalization means more local culture and more diversity. Nature prefers diversity, not conformity.
-there's more but this has made me angry at how stupid and ignorant are when it comes to: "oh this imported, factoy farmed good costs less than something at the farmer's market, I should buy it." and "if it makes the fortune 500 coroporations richer it's good for the economy." Seriously, people need to open their eyes.

Yes, I am a hippie, a vegetarian (soon to be vegan), anti-consumer, an anarchist and a hypocrite. I'm insanely left-wing and impractical and that's the way I want to live my life.

[quote="annabelle's temptation"]
I think the global economy is a bad idea and I can't wait until it come crashing down.[/quote]

That might be the worst catastrophe in the history of the world.  Shut down global trade and more people will die than in all of the wars in the history of the word combined.

[quote="annabelle's temptation"]
The Cons
-transportation costs of transporting all our cheap labour goods over the world hurt the environment and are wasteful.
[/quote]

If you're worried about environmental effects of transportation, there is a much easier way to make a dent in the problem -- people should stop using fossile fuels as much and accept the clean and cheap if irrationally scary alternative of building more nuclear plants.

The global economy will gradually balance out and resource transportation will become less common and necessary as it will also become cheaper and cleaner.

What's wasteful is not letting somebody who can make clocks in country A trade for food in country B.  Trade allows people to rise out of poverty.  Just look at the gains made in Asia over the past few years alone.  The middle classes are growing rapidly and the proportion of impoverished is sinking.

[quote="annabelle's temptation"]
-when you buy locally, you are supporting a stronger community and helping the farmers in your neighbourhood.
[/quote]

So it's better to pay Bob 50 cents for an apple than to pay Jose 25 cents for one because Bob lives in the neighborhood?

Astoundingly, Bob is often subsidized heavily to grow and even not grow food in the first place resulting in crushing of neighboring farm economies.

It's wasteful not to take advantage of the fact that different regions have different resources and laborers with different skill sets.  Trade only leads to optimization of production making everyone wealthier along the way.  If it didn't, people wouldn't trade.

[quote="annabelle's temptation"]
-less globalization means more local culture and more diversity. Nature prefers diversity, not conformity.
[/quote]

Nature doesn't prefer diversity through artificial injunction.  Nature thrives by competition.  Local culture isn't lost so much as it is held up to a candle.

[quote="annabelle's temptation"]
-there's more but this has made me angry at how stupid and ignorant are when it comes to: "oh this imported, factoy farmed good costs less than something at the farmer's market, I should buy it." and "if it makes the fortune 500 coroporations richer it's good for the economy." Seriously, people need to open their eyes.
[/quote]

It's not about support large corporations.  We should definitely take a serious look at the nature of corporate structure, but in some sense, corporations are mini-governments and it's about time we simply view them this way and create realistic approaches to dealing with the downsides of corporate interests.  As of just a few months ago, the majority of the world's largest economies are corporations (there are at least 51 corporations larger than the 50th wealthiest nation).

[quote="annabelle's temptation"]
Yes, I am a hippie, a vegetarian (soon to be vegan), anti-consumer, an anarchist and a hypocrite. I'm insanely left-wing and impractical and that's the way I want to live my life.[/quote]

I'm not at all sure what to say to this.  If you're happy as a hypocrite then I suppose there's no point in pointing at that nearly every computer in this global internet was corporate built and this forum is operated by an incorporated entity.

All I would suggest is to take at economic structures without labels like "government" or "corporation."  Anarchists like Caplan or Tolkien seem to take this approach and found themselves with philosophies that are probably more practical for their day and age (and without just being mad at everybody).

[quote="hello"]Do you think that economists favor free trade because they invest in these companies? I don't think economists are all that rich.

This has little to do with the topic of free trade, but somehow, I could never understand why saving is so good and why it equals investment in a closed economy, can someone explain with very simple language? Thanks[/quote]

I doubt that most economists are subject to a serious amount of moral hazard when they take positions.  If they were more interested in money they'd be running hedge funds.

Savings provides a liquid pool of capital for fresh investment.  Most people would rather put their money to work than stuff it in a mattress.

However, saving too much money might sometimes mean a lack of money being spent on new products and services, so the investment would be for naught.

I think there is some equilibrium and I don't know how to define it.

Of course, take that with a grain of salt as I'm no economist.

You completely ignored my comment on fair trade.

Also, you said, you think people should stop using fossil fuels so much but buying internationally only supports our dependance on them. Do you really think the oil companies are just going to give up and accept that a renewable energy source would completely put them out of business.

I don't like how you seem to accept the free market equilibrium as a law. I can't wait to take second year microeconomics next year at university so I can argue with the prof about how economists do not take into account all the important factors in life when deriving their equilibrium.

The one thing that keeps me sane is the knowledge that I can fight back with my dollar, which I am currently accumulating by selling my labour to a local restaurant (non-chain thankfully) washing dishes. I'll admit the computer I'm using isn't very fair trade and there were probably human rights violations that went into my ability to type this message to you. I bought this computer before I cared and questioned the world so I consider it a sunk cost but if I have to buy anything new technologically I will definitely research the company thoroughly before I deem whether or not the company is worth my dollar.

What is economic stability if you've sacrificed your health and your environment so you can no longer enjoy these financial benefits the market equilibium has created. Seriously, money along with slavery and poverty are all human creations and we can live without them.

I know I don't have good arguments against what you said to me and I need to learn to stop arguing on the internet. I'm not going to change anyone's opinion and I could have just spent this time doing something meaningful and productive.

Disregard this post and do something you enjoy while you still have a world you can enjoy it in.

[quote="annabelle's temptation"]

What is economic stability if you've sacrificed your health and your environment so you can no longer enjoy these financial benefits the market equilibium has created. Seriously, money along with slavery and poverty are all human creations and we can live without them.

[/quote]

Economics is not solely about money. It is concerned with the production and distribution of goods and services for the benefit of society. By definition the goal of economics is to produce the best possible outcome for the greatest number of people. Excuse me for saying so, but to decry economic stability as bad for society is manifest self-contradiction.

[quote="annabelle's temptation"]You completely ignored my comment on fair trade.[/quote]

No, I just didn't consider it substantive enough to provide a response.

I'm perfectly happy with more equitable trade rules such as getting rid of tariffs and subsidies, but that's askew from the rest of the conversation.

However, it should be noted that the only way to halt globalization would be unfair trade in which some vendors are silenced in some markets to provide an artificial disconnection, thus preventing trade.

The biggest thing to remember in all this is that every trade [freely entered into by all parties] is at the very least perceived to be mutually beneficial and that's probably the case far more often than not.

But overall I'm bored with theoretic conversations that seem to be nothing more than the venting of unqualified and unquantified philosophy.

If you have any proposed solutions to the world's economic problems or can explain your yearning for a fundamental paradigm shift, I'll be happy to discuss it.

[quote="annabelle's temptation"]I don't like how you seem to accept the free market equilibrium as a law. I can't wait to take second year microeconomics next year at university so I can argue with the prof about how economists do not take into account all the important factors in life when deriving their equilibrium.[/quote]

I think you have a fundamental misunderstanding of the study of economics.

Economics is the study of scarce resources and the value of these resources is almost universally viewed in terms of all forms of utility, not just money.  In that sense, economics takes all important factors into account that we can identify and even often finds ways to take into account those we cannot identify.  In fact, economists go out of their way to include any ordering of value preferences in their models that anyone can suggest.

If you feel very emotional about these issues, it's probably best to continue with your own studies before suggesting that so many intelligent people who study these problems are completely ignorant of our impending doom.

I'm also tired of arguing politics and ideas on the internet. My goal is never to prove my point to others but sort of play the devil's advocate to see what the other side has to say and then just dismiss it internally. Besides I have no discovered that what one thinks and believes does not matter at all. Only what we do. I'm constantly improving my lifestyle and I can't expect to change anyone else's so why should I argue ideals. I might come back to this board in the fall when Ihave more math related questions. Anyway, thanks for helping me out in the other topic.

In general, free trade is a good thing. If a country lost from trade, why would it. Trade represents an expansion in the consumption possibilities of a country. Gains from trade are realised from the exploitation of comparative advantages which countries have over each other in the production of certain good.

There are also gains from trade through the exploitation of (internal) economies of scale.

There is a good argument for the short-term subsidisation  or protection of an industry to help it compete with imports. This is when average cost is decreasing in scale (external/industry economies of scale). When the industry has grown sufficiently, these can be relaxed.

 [quote]I hate free trade with a passion. I got very angry just reading your posts. Have you never heard of Oxfam or Fair Trade? Their Make Trade Fair campaign is huge.
[/quote]
Free trade = bad, but fair trade = good?
A problem arises when large trading blocks eg U.S. impose tarriffs on imports. These are imposed to increase a country's 'terms of trade' - export prices/import prices.
Small nations cannot do much to counter this, but their best response is to impose tarrfs on imports from eg U.S.
This is a zero-sum game, the pareto-optimal outcome is of no tarrifs.
I agree that trade needs to be made fair, by allowing poorer nations fair access to U.S. etc markets


Does someone know where can the e-mails of the guides be obtained? Or can someone post them?

I only have the Email ID of our Guide and the guide of Singapore as she recieved us from the airport.
whose email id do u want, can u specify ?

I am interested in the mails of the guides of latin america.


How do you do it?

8.	 	To	transmit	a	positive	integer	less	than	1000,	the	Networked
Number	Node	offers 	two	options.
	 Option	1:		Pay	$d	to	send	each	digit	d.	Therefore,	987	would
		 	 cost	$9	+	$8	+	$7	=	$24	to	transmit.
	 Option	2:	 Encode	integer	into	binary	(base	2)	first,	and	then
	 	 pay	$d	to	send	each	digit	d.	Therefore,	987	becomes
	 	 		 1111011011	and	would	cost	$1	+	$1	+	$1	+	$1	+
	 	 		 $0	+	$1	+	$1	+	$0	+	$1	+	$1	=	$8.
	 What	is	the	largest	integer	less	than	1000	that	costs	the	same
whether	using	Option	1	or	Option	2?

Heh, a friend of mine did that... you have to realize that increasing an even number by one increases the sum of the digits by one as well as the binary representation, so the number has to be [u]odd[/u].

[quote="illusiat"]a bases program would make this really easy to brute force checking, starting from 512 down. grrr if i got this right i would've made nats[/quote]

I checked from 500 up.  :wallbash: . I didn't even think about the fact that the answer should have been odd.

I did this after the competition and realized I should have done it at the competition.
Base 2: 111111111=sum of 9()=511 in base 10.
So we have to find a number under 511 subtracting a power of 2 and the digits have to equal 8.

so brute force obviously:
511-1=510 sum of 6
511-2=509 sum of 14
511-4=507 sum of 11
511-8=503 sum of 8. yay!!!

We can obviously see that the greatest sum of the digits possible is $ 9$, and the numbers that will satisfy this are $ 1111011111_2\equal{}991$.
The others are $ 111111111_2,1011111111_2,1101111111_2,1110111111_2$.
Continue for the sum is 8 to get 503.  Then you check that 511 doesn't work, which is obvious, and that 510 doesn't work, since those are the only two numbers above 503 that have a sum below 8.

Another way:
It is easy to see that any number greater than 512 won't actually work.
Thus attempting again for a sum of 9.  The largest is 504.  Doesn't work.
Then attempt sum of 8. 503 works.
Sum of 7, 511 does not work.

Sum of 6, 510 does not work.

Thus largest is 503

Wait, I'm confused.  Why isn't it 502?  5+0+2=1+1+1+1+1+0+1+1+0=7, and 502 is less than 503.  Or is there some part of the directions that I overlooked?

[quote="mathcrazed"]You want the greatest possible value.[/quote]

Oooooohhh....  Reading the directions helps, doesn't it? :wink:

lol. i started with 511, went down to 504, found it didnt work, ran outa time and put 503 down

[quote="mathcrazed"]We can obviously see that the greatest sum of the digits possible is $ 9$, and the numbers that will satisfy this are $ 1111011111_2 \equal{} 991$.
The others are $ 111111111_2,1011111111_2,1101111111_2,1110111111_2$.
Continue for the sum is 8 to get 503.  Then you check that 511 doesn't work, which is obvious, and that 510 doesn't work, since those are the only two numbers above 503 that have a sum below 8.

Another way:
It is easy to see that any number greater than 512 won't actually work.
Thus attempting again for a sum of 9.  The largest is 504.  Doesn't work.
Then attempt sum of 8. 503 works.
Sum of 7, 511 does not work.

Sum of 6, 510 does not work.

Thus largest is 503[/quote]

Exact way I did it. *hi fives*

[quote="ZhangPeijin"][quote="dynamo729"][quote="distracted523"]I did it with calculator, brute force-ish

I put 502. Guess what the answer was... :o[/quote]

i did the same thing!!!!!!! I was so mad![/quote]
ME TOO!
Join the club.
I just got.
502.
And I'm like
Ok that's the highest I'm gonna go.
So then I stopped testing other numbers.[/quote]

You're kidding me, right.  :|

I put 502... and I would've made nats if I had've got it right.

[quote="Ryuk"][/quote][quote="ZhangPeijin"][quote="dynamo729"][quote="distracted523"]I did it with calculator, brute force-ish

I put 502. Guess what the answer was... :o[/quote]

i did the same thing!!!!!!! I was so mad![/quote]
ME TOO!
Join the club.
I just got.
502.
And I'm like
Ok that's the highest I'm gonna go.
So then I stopped testing other numbers.[/quote][quote="Ryuk"]

You're kidding me, right.  :|

I put 502... and I would've made nats if I had've got it right.[/quote]

GAH! But you will do good in high school. Most people who get so close work harder and improve more than those who dont.

heres something:
you know it has to be odd, because if n is even, and n works, then n+1 must work. (1 is added in base 2 at the end, and 1 is added to the base 10 numberal)

[quote="dnkywin"][quote="distracted523"]I did it with calculator, brute force-ish

I put 502. Guess what the answer was... :o[/quote]

I put $ the$ $ same$ $ thing!!!$  :cursing: I would have moved one place up if:
I checked one more time [b]and/or[/b] The guy above me didn't cheat with his TI-89!!! :starwars:  :spider:[/quote]

No u wouldn't

I would've won the tiebreaker  :P

on the TI-89 titanium, you can set it on binary... so i just went

511 nope
510 nope
jumps down to 505
504 nope
503 YAYAYAYAY

lol

Lucky. So lucky that TI-89s don't have qwerty keyboard... -_-

lols

Calvin (dnkywin) also has a TI-89.....

[quote="PI-Dimension"]lols

dnkywin also has a TI-89.....

blah blah blah I am such a filthy cheater :D [/quote]

My 89 ran out of batteries... :maybe:

i dont get to have an 89 until next year. :( (or maybe even the year after, if i dont pass for pre calc.)

Could you guys please get back on topic? :) thanks.


Parabola $y=x^2+ax+a$ passes through a fixed point for any value of $a$. Find the fixed point.

[hide]Since the parabola passes through the fixed point for all $a$, set $a=0$ and $a=1$. The fixed point is the intersection of these parabolas.
$x^2=x^2+x+1$
$x=-1\rightarrow y=1$

$\fbox{(-1,1)}$

(Check: $1=1-a+a$)[/hide]

[hide="Another solution"]$y^2 = x^2 + ax + a$
$(x + 1)a + x^2 - y^2 = 0$ for every $a$, then...

$\left\{\begin{array}{c}{a} x + 1 = 0 \ x^2 - y^2 = 0\end{array} \quad \Rightarrow \quad \left\{\begin{array}{b} x = -1 \ y=1 \end{array}$[/hide]


analytic geometry
vector
linear algebra
combinatorics proposed
combinatorics

We call a set $ A$ to be [i]even[/i] if it has an even number of elements. Suppose that $ n$ is an even positive integer. Suppose that $ S_{1}$, $ S_{2}$, ... $ S_{n}$ are $ n$ even subsets of the set $ \left\{1;2;...;n\right\}$. Prove that there are two different numbers $ i$ and $ j$ from the set $ \left\{1;2;...;n\right\}$ such that the union of the sets $ S_{i}$ and $ S_{j}$ is also even.

This is a very cute problem for an application of linear algebra onto combinatorics.
With a friend, we published a small paper (in french, and not available on internet :( ) which describes some kind of links between the two areas. In it we prove some connected results :

Let a subset $S$ of $\{1,2,...,n \}$ be even (resp. odd) if $|S|$ is odd (resp. even).
Thus :
a) The maximal number of odd subsets with pairwise even intersection is $n$

b) The maximal number of even subsets with pairwise odd intersection is the greatest odd integer less than or equal to $n$.

c) The maximal number of even subsets with pairwise even intersection is $2^{[ \frac {n} {2} ]}$

d) The maximal number of odd subsets with pairwise odd intersection is $2^{ [ \frac {n-1} {2}] }$.

Reference :
P Bornsztein, X. Caruso, 'Des formes bilinaires en combinatoire', R.M.S., May 2004, p.35-44.

Pierre.

In fact the paper is available here (in french) :
http://www.rms-math.com/article.php3?id_article=118&Num=0&Type=enonce

Pierre.

Pierre, could you please post the solution (for this problem) in English? Sorry, i cant read French ... i dont know much about linear algebra and would like to know how it is really used. Thanks.

I know the sol to this nice prob .without linear algebra the solution becomes really long .if u want I'll post it.

If you have a linear algebra solution it is sometimes not too difficult to translate it into 'purely' combinatorics words, and it sounds quite magical then. We have done it at the end of the paper for one of the problems. But, if you want to understand 'why' it works and 'how' to find such a solution, the conceptual view with linear algebra gives the keys....

I will translate the solution of the initial problem in english later (sorry, always busy...).

Pierre.

Well i am no sure that if we want to know why the solution works is better to look at the linear algebra.....  Well anyway...... the linear algebra aproach is really nice.....directly from the book!
 Here is what i did:
 Number the element of the original set with the numbers from 1 to n. With each subset asociate avector in Z2^n with a one in i-th coordinate if the number "i" is in the subset and 0 otherwise.
 Let V1, V2, V3,...Vn be the vectros for the n subsets. It is easy to show that wqe are looking for two vectors such that Vi * Vj = 0 . (where * is the usual scalar product).  Suppose the contrary....that is Vi*Vj=1 for each i and j (if i different from j). And Vi*Vj =0 if i=j since Vi*Vj is the number of elemtns of the subset which is even for every i.
 Know it is easy to show that the n vectors cant be linearly independt since in that case they would span all Z2^n but for instance 100..0 cant be reached summing some of the Vi (since if N(V)=the sum of the coordinates then N(Vi)=0 mod 2 but   N(10...0)=1 mod 2 absurd!!
 Now...since they arent indepndt then we sholud have some relation
 a1*V1+a2*V2+....+an*Vn=(0....0)...... where ai =0 or 1 not all zero. Put a1=1 now usin the linearlity of the scalar product. we reach the following: Suppose ai=0 then 0= a1*V1*Vi+a2*V2*Vi+...+an*Vn*Vi=a1+a2+..+a(i-1)+a(i+1)+...an (since Vi*Vj=1 if i<>j and 0 if i=J.
 BUt 0= a1*V1*V1+a2*V2*V1+...+an*Vn*V1=a2+...an
From where 1=a1=ai=0 mod 2...... absurd!! We win....

 Ok it isnt very hard to take away the linar algebra..... ie we need it only for showing that there was some relation.....but the scalar product is just the number of element of the intersection......

 You posted us 102 combinatorics problems Is there any list of problems? Can you send it to me (gausseuler2001@yahoo.com ) thanks

[quote="Charlydif"]
 Here is what i did:
 Number the element of the original set with the numbers from 1 to n. With each subset asociate avector in Z2^n with a one in i-th coordinate if the number "i" is in the subset and 0 otherwise.
 Let V1, V2, V3,...Vn be the vectros for the n subsets. It is easy to show that wqe are looking for two vectors such that Vi * Vj = 0 . (where * is the usual scalar product).  Suppose the contrary....that is Vi*Vj=1 for each i and j (if i different from j). And Vi*Vj =0 if i=j since Vi*Vj is the number of elemtns of the subset which is even for every i.
 Know it is easy to show that the n vectors cant be linearly independt since in that case they would span all Z2^n but for instance 100..0 cant be reached summing some of the Vi (since if N(V)=the sum of the coordinates then N(Vi)=0 mod 2 but   N(10...0)=1 mod 2 absurd!!
 Now...since they arent indepndt then we sholud have some relation
 a1*V1+a2*V2+....+an*Vn=(0....0)...... where ai =0 or 1 not all zero. Put a1=1 now usin the linearlity of the scalar product. we reach the following: Suppose ai=0 then 0= a1*V1*Vi+a2*V2*Vi+...+an*Vn*Vi=a1+a2+..+a(i-1)+a(i+1)+...an (since Vi*Vj=1 if i<>j and 0 if i=J.
 BUt 0= a1*V1*V1+a2*V2*V1+...+an*Vn*V1=a2+...an
From where 1=a1=ai=0 mod 2...... absurd!! We win....

 Ok it isnt very hard to take away the linar algebra..... ie we need it only for showing that there was some relation.....but the scalar product is just the number of element of the intersection......
[/quote]

Thanks, it seems very nice. But I dont understand some things ...sorry for my idiocy :
(1) What do you mean that the vectors are 'linearly dependant'?
(2) How do you get from: a1*V1+a2*V2+....+an*Vn=(0....0) to
0= a1*V1*Vi+a2*V2*Vi+...+an*Vn*Vi
(3) How do you get this: 0= a1*V1*V1+a2*V2*V1+...+an*Vn*V1?

Ok......i am not very good for definitions an all that, and i dont know very much about linear algebra but here is something:
  The n vectors V1,V2,...,Vn are linearly independt if the following holds:
a1V1+a2V2+....+anVn=0 implies a1=a2=a3=.....=an=0.
 Where ai are reals and Vi are vectors in Z^m. If they arent independent we say they are dependent.... that is that  any there is a vector which is a linear combination of the others, that is Vn=a1*V1+...+a(n-1)V(n-1)....
 Now take a set of vectors A with the following properties is called a subspace of Z^m:
If V is in A then V is in A for any real number a,
If V1 and V2 is in A then V1+V2 is in A.
 now take the maximal number of mutually  independent vectors in A. This is the dimension of A (prove that is unique and finite). And if V1, V2,....Vn are this mutually independt vectors then every vector in A can be represnted as a1V1+...+anVn.....
  Now... the space spanned by n vectors V1,...Vn is the vectorsd of the form a1V1+...+anVn (check that this is a subspace)
  The scalar product of two vectors is the following: V1*V2 = sum ai*bi where V1=a1,a2,..am and V2=b1,b2,....bn .

 When i obtain that relation i multiplied by Vi the first time and by V1 the second
note that (v1+v2)v3=v1v3+v2v3
 Also all about the subspace and dimension is do it in the same way....(we just look everything mod 2)
 I think it should be better to read something on any book ........there a lots of books with some introduction.....

[quote="pbornsztein"]In fact the paper is available here (in french) :
http://www.rms-math.com/article.php3?id_article=118&Num=0&Type=enonce[/quote]

Pierre if I go there I get the stupid error message:
[quote]Cette partie du site est rserve aux abonns.
Vous devez entrer un identifiant et un mot de passe valide[/quote]

Is the article expired or do we need to sign up anywhere?

You are right. This is not a free journal, so they have locked it....

Pierre.

Well, perhaps you can copy-paste it in here? :)

I will try. Let' see...does this work?

Pierre.


How did Georgians do on the AMC 12A?

I got a 132.

144 here.

I am from 1200 miles from your place at least.

135 :| I wish I had done a bit better, but to be honest, I'm not that invested in my individual score.

I'm pretty invested in Lassiter's team score, though.  I'm pretty sure we broke 400  :D  :D

[quote="Xantos C. Guin"](22,1,2)=133.5

#14: Didn't read the word distinct.
#24: Didn't realize that when n=4j+1, there are only n solutions.
#25: Didn't have time to do.[/quote]

It was interesting, I took the 10 (taking the 12 in two weeks) but after the test I "took" the twelve (at least the problems I hadn't seen before) and it seemed like 25 was too easy for a 25, yet 24 was too hard for even a 25...
but that's just me.

I thought I got a perfect on the 10 until my friend spotted my stupid mistake... man that was the only reason to take it this year...

Oh, well 24 was actually kind of stupid...

[hide]Essentially, you think about the graph (or intermediate value theorem if you must be that way) and the fact that $\sin (nx) = \sin x$ has $n+1$ intersections hits you like a bucket of bricks, except then at the last minute you realize that when $n$ is 1 mod 4, you only have $n$ intersections.  So you want $\sum_{k=3}^{2008}k-2004/4$, which gives you some horrifically long answer which just happens to be choice (D) 2016532.[/hide]

Problem 25 was kind of intimidating, until you spotted the recursive relation (or the neat method that seemed to involve generating functions, which JSteinhardt posted in the AMC forum).

I was pretty sure of my score right after the competition, as Phillip, Nathan, and I confirmed each other's answers.

And then Mr. Slater had the solution booklet yesterday afternoon.

And yes, answers are posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132366]here[/url].

Wheeler gave out results.
We have 4 AIME qualifiers. If the cuttofs for the AMC 10/12 A were raised 0.5 points, we'd only have 1.


[quote="D.P.L"]If i have only enough money to get one of these books, which do you think is the most easiest/elementary of them all:
Mathematical Olympiad Challenges
Mathematical Olympiad treasure
Problem solving through problems
the mathematical olympiad handbook, british olympiad
Mathematical Miniatures
Any help would be greatly appreaciated :)[/quote]
I have no experience with any of these but the first & and third sound good. :)

[quote="D.P.L"]whatabout mathematical olympiad handbook, Mathematical Olympiad treasure Mathematical Miniatures ?[/quote]
Try http://www.amazon.com. You can read what they have to say. I was just judging the books by their names. :D

olympiad challenges and olympiad treasures by andreescu (and some other guy?) are excellent. i just got treasures from amazon a few days ago and it is fairly easier than challenges. im not sure but i think miniatures is slightly more advanced.

I would say Mathematical Olympiad Treasures.

Just another question:

What is the so called chinese book in this link:

http://www.kalva.demon.co.uk/books/engel1.html


Find the following functions correct to five decimal places:

(1) sin 22 degrees 43'

(2) cos 15 degrees 12' 51'

What can I do?

What does the apostrophe mean after the numbers 43, 12 and 51 in trigonometry?

43' means 43 minutes, or $\frac{43}{60}$ of a degree. To put it into a calculator, just do $\sin(22+43/60)$.

Just figure out what $\frac{43}{60}$ is, append it to the end of $22$, and type it into your calculator.


algebra
polynomial
quadratics
number theory
number theory theorems

Mods, if this is the wrong place, please move this thread.

So far, there are a lot of methods to solve Diophauntine equations. The most powerful is the use of the mod. However, if there -are- actually small solutions, it would be fruitless to use the mod isn't it? But I don't really know any other methods... bounding between squares/cubes (whichever it may be?) what else is there?

[quote]
ZetaX
PostPosted: Wed Mar 09, 2005 1:40 am    Post subject:
I would say 'infinite descent' is one possibility (but as much as I know you can only apply it when there is only one solution...)
[/quote]

What actually is infinite descent?

infinite descent:

If you have an equation in integers, assume that it has a 'postive' solution and proof out of this fact that it has a 'smaller' 'positive' solution.
Here 'positive' and 'smaller' can be of various meaning (for example when you have an equation in $x,y,z$, you can define 'positive' as 'not all numbers are zero' or as 'none of the numbers is zero' and 'smaller' as 'take the biggest/smallest/one special (eg $x$) (when necessary modulus) of these numbers and compare them')

One easy problem you can do this way is proving that $x^4+y^4=z^2$ has only solutions with $x$ or $y$ being zero.

I also have a question:

Can someone show me how to use p-adic numbers in attacking Diophantine equations ? Not "forced" solution just to show the solution with p-adic numbers but some more "natural" (I hope you understand what I actually mean  :D ). What I've found on Internet is vastly general equations which are beyond me for now. If one can solve olympiad-level equation with p-adics I'll be blissed  :lol:

The use of $p$-adic methods to attack diophantine equations in the integers is called the local-to-global principle. For example, if you have a polynomial $P(x_1,\ldots, x_n)$ with integer coefficients, you could try to see if it has solutions over $\mathbb R$ and over $\mathbb Q_p$ for each prime $p$. Then one could ask if it has solutions over $\mathbb Z$.

If $P$ is a quadratic form (a homogeneous polynomial of degree 2), then it is true that $P$ has roots over each local field if and only if it has solutions over the global field (Hasse-Minkowski). Otherwise, the failure to pass from the local case to the global case is measured by cohomology.

If $P$ is the equation of an elliptic case, say $P(x,y)=y^2-x^3-x-1$, then the failure of the local-to-global principle is measured by the Shafarevich-Tate group, $III=\textrm{Ker}(H^1(\mathbb Q,E)\to \oplus H^1(\mathbb Q_v,E))$.

The best reference here is Mazur, Barry. On the passage from local to global in number theory, Bull. Am. Math. Soc. vol. 29, No. 1, 1993, pp. 14-50.

Andrei

Oh, I reread the previous post and realized that I answered only the first part of the problem.

If you want to solve olympiad equations with $p$-adic methods, chances are few that you'll succede. The reason is that the general style of proof in the local cases is by reducing to equations in $\mathbb Z/p\mathbb Z$ and work there (using quadratic reciprocity, perhaps) and then use Hensel's lemma to lift to the $p$-adic case. So you gain nothing by reverting the process.

Andrei

Andrei, thank you very much for enlarging upon this.  Can I find somewhere this article which you give reference to? On AMS site it seems that they only have articles from 1996 on. :(

Here is a copy of the article. The article goes comprehensibly in the begining, but then goes off to talk about Kolyvagin systems and a lot of goodies that went into the proof of Fermat's last theorem, so it's quite hard to read.

Andrei

Wow, thanks a lot, Andrei, for the great article!!!  :D

Thanks Andrei as well :) , though as for now I'm only capable to understand just a small part  :(

Surely, infinte descent is more applicable than to just one solution.
Its usable to prove no solutions, 1 solution or infinetely many solutions.

Bomb

Its my fave way of proof. :)  :D


algebra
polynomial
quadratics
pigeonhole principle

[b]There exists a quadratic polynomial with integer coefficients such that (a) both of its roots are positive integers, (b) the sum of its coefficients is a prime, and (c) for some integer k, P(k)=-55. Show that one of its roots is 2, and find the other root.[/b]

[hide="Solution"]Let $ P(x) \equal{} a(x\minus{}m)(x\minus{}n) \equal{} ax^2 \minus{} a(m\plus{}n)x \plus{} amn$. Then the sum of the coefficients is a prime, or $ p \equal{} a[1 \minus{} m \minus{} n \plus{} mn] \equal{} a(m\minus{}1)(n\minus{}1)$. Since all three terms are nonnnegative integers, it follows that two of the three terms must be $ 1$, with the third being $ p$. By pigeonhole at least one of the two roots is $ 2$.

Without loss of generality let $ m\equal{}2$; then the third condition gives us that $ a(k\minus{}2)(k\minus{}n) \equal{} \minus{}1 \cdot 5 \cdot 11$.

If $ n\equal{}2$, then $ a(k\minus{}2)^2 \equal{} \minus{}55$, and since no two prime factors of $ 55$ are repeated, $ k\minus{}2$ must be $ \pm1$. It follows that $ a \equal{} \minus{}55$, which is not prime (contradiction).

If $ a\equal{}1$, then $ (k\minus{}2)(k\minus{}p\minus{}1) \equal{} \minus{}55$, and the difference between the two factors is $ p\minus{}1$. Pairing up factors, we see that of the possible pairs $ (\minus{}1,55)(\minus{}55,1)(\minus{}5,11)(\minus{}11,5)$, only the latter two might work ($ p\equal{}17$). This yields the polynomial $ P(x) \equal{} (x\minus{}2)(x\minus{}\boxed{18}) \equal{} x^2 \minus{} 20x \plus{} 36$, and indeed $ P(7) \equal{} \minus{}55$.
[/hide]


geometry
geometric transformation
reflection
circumcircle
projective geometry
perpendicular bisector
geometry proposed

[color=darkblue]Let $H$ be the orthocenter of the tringle $ABC$ and two points $U\in AB$, $V\in AC$ so that $H\in UV\ .$
Denote the reflection $W$ of the point $A$ w.r.t. the circumcenter of the given triangle and the intersections $X\in BC\cap WU$, $Y\in BC\cap WV$. Prove that $UV\parallel BC$ $\Longleftrightarrow$ $\widehat{AWU}\equiv\widehat{AWV}$ and in this case $XB=XU$ , $YC=YV\ .$[/color]

Let the B-, C-altitudes meet the circumcircle again at E, F. Let FU, EV meet at W. Since the intersections $U \equiv AB \cap FW,\ V \equiv AC \cap EW,\ H \equiv BE \cap CF$ are collinear,  by Pascal's theorem, W lies on the triangle circumcircle. Since $\angle ABE = \angle ACF = |90^\circ-\angle A|,$ WA bisects the angle $\angle UWV \equiv \angle FWE.$ WLOG, let $\angle B < \angle C,$ so that $\angle B < 90^\circ$ is acute.

Assume $UV \parallel BC.$ Then $\angle UHF = \angle BCF = 90^\circ-\angle B.$ Since F is a reflection of H in AB, the triangle $\triangle UHF$ is isosceles, so that $\angle UFH = 90^\circ-\angle B$ as well. But $\angle AFH \equiv \angle AFC = \angle ABC \equiv \angle B,$ so that

$\angle AFW \equiv \angle AFU = \angle AFH+\angle HFU = \angle B+90^\circ-\angle B = 90^\circ.$

Consequently, AW ia a diameter of the triangle circumcircle. Conversely, if AW is a diameter of the triangle circumcircle, then the angle $\angle AFW = 90^\circ$ is right, the angle $\angle HFU = 90^\circ-\angle B$ and from the isosceles trianhgle $\triangle UFH,$ $\angle UFH = 90^\circ-\angle B,$ which means that $UV \equiv UH \parallel BC.$

From the isosceles triangle $\triangle UFH$ with the base angles equal to $90^\circ-\angle B$ and with the perpendicular bisector AB of its base HF, we get $\angle XUB = \angle FUA = \angle B = \angle XBU,$ so that the triangle $\triangle XBU$ is isosceles with XB = XU. Similarly, YC = YV.


calculus
integration
real analysis
real analysis unsolved

I think I might have found a typo in my calculus text, "Forgotten Calculus". It would not be the first.  :mad:  Anyway, if it is I and not the author who is wrong, then I'm missing something fundamental, and would appreciate guidance.

The problem asks to find the integral of

(10x - 3) [(5x^2 - 3x + 17)^(1/7)] dx

via substitution.

My solution is

(7/8) [(5x^2 - 3x + 17)^(8/7)] + C.

The author's is

(7/8) [(5x^2 - 3x - 17)^(8/7)] + C.

Note that we differ only over whether one should add or subtract 17.

Who's right?

If my computations are good, then you're right. :)

Anyway, I don't think the author has to be blamed for that :
1/ it is just a typo
2/ he is not the one who typed the book.

Thanks for taking the time to check my work.

A note on typos: whoever is to blame, the author or the publishers, they a) are  unprofessional and b) drive students like myself crazy, needlessly. This book is designed to be a self-study guide. Typos in such books are especially bad, because I think the first reaction of solo students is to blame themsleves for the mistake, and to spend some time trying to figure out why.

I've found 10 so far, I am sure, either because they were very self-evident (one of her solutions was based on the idea that when 24 = 6x, then x = 3) or because I've asked in numerous math forums, and the responses were always unanimous that the book was wrong.

Here is an example of the kind of error which is simply inexcusable. In the first pages of the first chapter on integration, the first rule she provides is

Integral x^n  dx = { [ n ^ (n+1)] / [n+1] }  + C,  when n does not equal -1.

When in fact, the base of the numerator should be x, not not n.  It took me a few very confused minutes to figure this out via her examples.  Some students might have taken longer, believing themselves to be idiots. Not good for morale.

Yes, I understand that.

But, you know, most of the time, the people who type the books don't know anything about the subject, they type a math book as if it was a cooking book or w/e.
So, even if it's re-read, some typos still remain.

In a way, the fact that you notice these typos prove that you got the notion. :)


inequalities
linear algebra
matrix
inequalities proposed

Let a,b,c are side-lengths of triangle such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
$ \frac {1}{(a\plus{}1)^2}\plus{}\frac {1}{(b\plus{}1)^2}\plus{}\frac {1}{(c\plus{}1)^2}\ge \frac {9}{4(ab\plus{}bc\plus{}ca)}$
Oh, i think it is simirlaly with Iran inquality 1996.:)

[quote="thegod277"]Let a,b,c are side-lengths of triangle such that $ a + b + c = 3$. Prove that
$ \frac {1}{(a + 1)^2} + \frac {1}{(b + 1)^2} + \frac {1}{(c + 1)^2}\ge \frac {9}{4(ab + bc + ca)}$
Oh, i think it is simirlaly with Iran inquality 1996.:)[/quote]
We have the stronger result
\[ \frac{1}{(a+1)^2} +\frac{1}{(b+1)^2}+\frac{1}{(c+1)^2} \ge \frac{9}{4(ab+bc+ca)}\]
For all $ a,b,c \in [0,2]$ such that $ a+b+c=3$.
Indeed, setting
\[ \left\{ \begin{matrix} a+1 &=y+z \hfill\\ b+1&=z+x \hfill\\ c+1& =x+y \hfill\\ \end{matrix} \right.\]
Then
\[ \left\{ \begin{matrix} x &=2-a \ge 0 \hfill\\ y&=2-b \ge 0 \hfill\\ z& =2-c \ge 0 \hfill\\ \end{matrix} \right.\]
And
\[ x+y+z=3\]
We have
\[ ab+bc+ca= \sum (2-x)(2-y)=\sum (2-x)(2-y) =12-4%Error. "sumx" is a bad command.
+\sum xy =\sum xy\]
Hence, the inequality becomes
\[ \frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}+\frac{1}{(x+y)^2} \ge \frac{9}{4(xy+yz+zx)}\]
Which is Iran 1996. :)