Notwithstanding the fact it must be well known prove the following: every even perfect number is of the form $2^{p-1}(2^p-1)$ where $2^p-1$ is a Mersenne prime. (A perfect number is equal to the sum of its divisors excluding itself). I've tried searching the forum but found only the applications of this fact.

This theorem is classical, here is the proof I found from the Hardy's and Wright's book

Theorem 1: If $2^{n+1}-1$ is prime, then $2^n(2^{n+1}-1)$ is perfect.

Proof: Write $2^{n+1}-1=p,N=2^np$. Then
\[ \sigma(N)=(2^{n+1}-1)(p+1)=2^{n+1}(2^{n+1}-1)=2N,  \]
so that $N$ is perfect.

Theorem 2. Any even perfect number is of the form $2^n(2^{n+1}-1)$, where $2^{n+1}-1$ is prime.

Proof: We can write any such number in the form $N=2^nb$, where $n>0$ and $b$ odd. Now $\sigma$ is multiplicative, and therefore
\[ \sigma (N)=\sigma (2^n)\sigma (b)=(2^{n+1}-1)\sigma (b).  \]
Since $N$ is perfect, $\sigma (N)=2N=2^{n+1}b$ and so
\[ \frac{b}{\sigma (b)}=\frac{2^{n+1}-1}{2^{n+1}}.  \]
The fraction on the right-hand side is in its lowest terms, and therefore $b=(2^{n+1}-1)c$ and $\sigma (b)=2^{n+1}c$, where $c$ is an integer.

If $c>1$, $b$ has at least the divisors $1,b,c$, so that
\[ \sigma (b)\geq b+c+1=2^{n+1}c+1>2^{n+1}c=\sigma (b),  \]
a contradiction. Hence $c=1,N=2^n(2^{n+1}-1)$, and
\[ \sigma (2^{n+1}-1)=2^{n+1}.  \]
But, if $2^{n+1}-1$ is not prime, it has divisors other than itself and 1, and
\[ \sigma (2^{n+1}-1)>2^{n+1}.  \]
Hence $2^{n+1}-1$ is prime, and the theorem is proved.


I'm planning on sending my solutions by mail for this round. Can I send them on exactly Monday?

[quote="gauss1181"]Yeah, why not? That's the [b]postmark[/b] date. :wink:[/quote]

I'm just wondering just in case they don't take the mail then.

Why would they not take the mail?  Its the postmark date...

Besides, for the last 3 rounds, I mailed them on the very last day...this round I mailed like Thrusday this week since I was done early.

im basically done but im gonna wait till monday anyways

To once again clarify: Monday March 9 is the [b]postmark[/b] deadline.  If you put it in the mailbox on Monday, it's fine.

Also, another reminder: if you are planning to upload your solutions on the website, the deadline is [b]3 PM Eastern / Noon Pacific[/b] on Monday.


How to write this Function in LATEX:
[img]http://www.part.lt/img/277678728adcdd3bbdc854dd35f16c6e65.JPG[/img] :maybe:  :?:

hello, see here:
http://www.math.uiuc.edu/~hildebr/tex/displays.html
Sonnhard.

Here is the standard way.
\[ \left\{\begin{aligned}
x\plus{}3&\equal{}y\plus{}2
\\
x^2\plus{}y^2&\equal{}0
\end{aligned}\right.\]
Note that you need both \left\{ and \right. though only \left\{ produces a symbol that is printed and that you can put & wherever you want resulting in different alignments (usually it makes sense to put it either in the beginning of each equation or at the beginning of the equality sign if you do a system).

If you are defining a function, the cases environment is better:
\[ f(x)\equal{}\begin{cases}
x\plus{}3,& x>0;
\\
x^2,&x<0;
\end{cases}\]

fedja your first problem has no real solution  :P


analytic geometry
superior algebra
superior algebra unsolved

Let $ k$ be an algebraically closed field and $ X$ and affine variety in $ k^n$ such that $ A$, its coordinate ring, is not null. Prove that there exists $ L$ a (linear) subspace of $ k^n$ and a $ k$-linear map from $ k^n$ onto $ L$ which maps $ X$ onto $ L$.

Can you give a more detailed proof? I want to see where is the algebraically closed condition used.

The proof of the Noether normalization lemma given in Atiyah-Macdonald assures that the algebraic independent variables can be chosen as linear combinations of the original ones when the defining field k is infinite. This is certainly the case if k is algebraically closed.

Can you show that $ X\rightarrow L$ is surjective on the k-rational points without any additional assumptions on k given $ L$? I am asking this since here we are at the border line between the world of classic algebraic varieties over k and schemes of finite type over k.


$ \tan^3x \minus{} 9\tan^2x \minus{} 3\tan x \plus{} 3 \equal{} 0$

So let $ u \equal{} \tan x$.
The original equation becomes $ u^3 \minus{} 9u^2 \minus{} 3u \plus{} 3$.
Look up the formula for cubic equations and find the roots, and take the arctangents, and add $ n\pi$ for arbitrary $ n$. That, or I'm missing an obvious rational root because I'm too sleepy, or I made a typo.

(response to post conflict: That reminds me of HTML, or XML, or XHTML. Um. Oh, MathML. I really don't know what to say.)
EDIT: Something strange just happened. Never mind.

[quote="math_explorer"]
(response to post conflict: That reminds me of HTML, or XML, or XHTML. Um. Oh, MathML. I really don't know what to say.)
EDIT: Something strange just happened. Never mind.[/quote]

Yes, I typed a bit MathML in the previous post. But, then it didnt work here. So, I edited it.

PS: No latex yet :(

Oh, they don't have the "Last edited by xxx, xxx times total" in the post preview, and edits to the last post don't trigger the conflict thing.

Copied from my TI89:
Zeroes of the cubic equation:
-.73041580968956
.44220005778938
9.2882157519002
Arctangents of them:
-.63084896039718
.41634859079942
1.463546141996
and, of course, add arbitrary multiples of $ \pi$.

[quote="akma327"]tan^3X - 9tan^2X - 3tanX + 3 = 0

solve for all values of X[/quote]
[img]http://www.mathmontada.net/vb/uploaded/24_1260955629.jpg[/img]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318846
Please don't double post  :wink:


geometry
geometric transformation
reflection
rotation
puzzles

How many different designs can be formed by painting 3 squares(1x1) of a 5x5 grid with black?

If a design can be formed by rotating another one by several times,then these are considered to be the same designs, but reflection is not allowed.

For instance,

BBWWW          WWWWB        WWWWW         WWWWW
WWWBW         WWWWB         WWWWW        WBWWW
WWWWW        WWWWW        WWWWW         WWWWW
WWWWW        WWWBW         WBWWW         BWWWW
WWWWW        WWWWW        WWWBB          BWWWW

are all the same designs  but

BBWWW                     WWWBB
WWWBW                    WBWWW
WWWWW       and       WWWWW
WWWWW                   WWWWW
WWWWW                   WWWWW

are considered to be different.

cooley, you have made double posting!

Japanese communities moderator

kunny

I've seen this question before, I think -- what's the source?

Should easily die with Burnside lemma. (no?)

[quote="xsquaredster"][hide]If rotations were counted as distinct designs, then the total number would be $\left(\begin{array}{c}25 \\ 3 \end{array}\right) = \frac{25!}{22!3!}= \frac{25 \cdot24 \cdot 23}{6}= 2300$ designs. However, rotations are not counted as distinct designs, so each true distinct design appears 4 times, one for each orientation. Therefore, the number of distinct designs is [b]575[/b].[/hide][/quote]

Consider
[code]
WWWWB           BWWWW
WWWWW           WWWWW
WWBWW    and    WWBWW
WWWWW           WWWWW
BWWWW           WWWWB
[/code]
There are two of them, and then the number is divided with 4. So, this solution is wrong. To get around this problem, you should use burnside lemma. \[\frac{_{25}C_{3}+12}{4}=578\]

The two designs you gave are nonetheless rotations of each other. Does Burnside's lemma account for that?

Perhaps you could explain Burnside's lemma briefly?

And i think that if i did forget to account for cases like that, the answer should be lower than the answer i got...

Yes,it can be solved by Burnside's Lemma but i found the answer as 581.
Only 12 of the designs(ones in which there is one black square in the top three rows, one in the bottom three rows and one in the middle row) turn out to be the same after rotating 180 degrees and this makes 24 of all the designs. Each of the rest of the designs(2276) are formed by rotating themselves, and this makes 569 designs. So, the answer is
12 + 569 = 581.

578 is the correct answer.
581 is wrong ,sorry  :(


IMO 2007-Hanoi, Vietnam was ending. If you have IMO shortlist 2007, please upload to the forum  :) Thanks

There is a rule saying that the shortlist will NOT be uploaded until the next IMO is over.

Maybe IMO short list 2007 can't be obtained until next IMO.

Something less impudent: Will we ever see the proposed solutions of the IMO? I know that getting them up is somewhat more difficult this time than it has been most years (don't ask for explanation), but up to now it seems that MathLinks is the only site to have solutions at all.

  Darij

I got it, if anyone want them, may be I could scan it.

I think you shouldn't be proposing this... the IMO shotrlist is confidential until next IMO...please don't post it

I have some algegras problem in My Blog:
http://www.mathlinks.ro/weblog.php?w=1139
Problem A1 -> A7.


From a set of 3 capital consonants, 5 small consonants and 4 small vowels, how many words can be made each starting with a capital consonant and containing 3 small consonants and 2 small vowels?
a.	3600
b.	7200
c.	21,600
d.	28,800

The answer depends on whether we can use a letter from the a set more than once. For example, could we use a particular small vowel twice?

[hide="If No"] To start with, we have 3 choices for the first letter. Now look at the other 5 letters. There are ${5 \choose 3}$ different sets of small consonants we can choose, ${4 \choose 2}$ sets of small vowels we can choose, and then $5!$ ways to order these small letters. So the total number of "words" we can make is:
$3*{5 \choose 3}{4 \choose 2}* 5! = 21,600$
So $c$ is the answer.
[/hide]

That probably is the way since you found an answer choice that was given in the problem.

can we put a capital in the middle of a word?  :maybe:

I don't think so. They way I interpreted the problem, you have to have to have the exact amount with no extra stuff in the middle.


number theory
relatively prime
Additive Number Theory

Let $a$ and $b$ be positive integers with $\gcd(a, b)=1$. Show that every integer greater than $ab-a-b$ can be expressed in the form $ax+by$, where $x, y \in \mathbb{N}_{0}$.

This is a basic result about [url=http://mathworld.wolfram.com/FrobeniusNumber.html]Frobenius number[/url] of two integers.

Can we extend this result for three pairwise relatively prime positive integers, $ a,b,$ and $ c$ such that $ ax\plus{}by\plus{}cz\equal{}n$ has non negative integer solutions $ (x,y,z)$ for integers $ n>abc\minus{}ab\minus{}bc\minus{}ca\plus{}a\plus{}b\plus{}c$?

Can this be done for $ k$ pairwise relatively prime positive integers?


Find all triples $(a,b,c)$ of positive integers such that $2^{c}-1$ divides $2^{a}+2^{b}+1$.

Let $ n \equal{} 2^c \minus{} 1$, and $ a \equal{} kc \plus{} r,\ r < c$, then $ 2^a\equiv (2^c)^k\cdot 2^r\equiv 2^r\pmod n$.
If $ r_a \equal{} r_b \equal{} c \minus{} 1$, then $ 2^a \plus{} 2^b \plus{} 1\equiv 2\mod 2^c \minus{} 1$.
If $ c > 3$ and $ r_a < c \minus{} 1$ or $ r_b < c \minus{} 1$, then $ 2^{r_a} \plus{} 2^{r_b} \plus{} 1 < 2^c \minus{} 1$. Therefore for all solutions $ c\le 3$.
1. $ c \equal{} 1$ a,b - any.
2. $ c \equal{} 2$ a,b are even.
3. $ c \equal{} 3$ $ a \equal{} 1\mod 3,b \equal{} 2\mod 3$ or $ a \equal{} 2\mod 3,b \equal{} 1\mod 3$.

ModEdit: correct but sloppy, needs rewrite.


\[ \text{Each letter represents a non-zero digit}.  \text{  What is the value of } t ?
\]
\begin{align*}
c + o &= u\\
u + n &= t\\
t + c &= s\\
o + n + s &= 12
\end{align*}

Plugging the first equation into the second gives $ c\plus{}o\plus{}n\equal{}t\implies o\plus{}n\equal{}t\minus{}c$. Substituting this into the last equation, we get $ t\minus{}c\plus{}s\equal{}12$, and since $ t\plus{}c\minus{}s\equal{}0$ by equation $ 3$, adding results in $ 2t\equal{}12\implies t\equal{}\boxed6$.


inequalities
calculus
modular arithmetic
ratio
algebra
polynomial
induction

Am I allowed to post such topic?

If yes, then here are the rules:

1) Solutions, not only answers should be posted
2) The solver should post a new problem
3) You can not post new problems until the old are unsolved
4) Keep the problems pre-olympiad problems level.

And the 1st problem:

Prove an inequality: $ (a\plus{}b)(a\plus{}c)\ge2\sqrt{abc(a\plus{}b\plus{}c)}$ , a,b,c - positive real numbers.

Expanding the LHS, we are set to prove that
$ a(a\plus{}b\plus{}c)\plus{}bc \ge 2\sqrt{abc(a\plus{}b\plus{}c)}$ which is a direct consequence of the AM GM inequality.

Next Problem: Given $ a \ge b\ge 0$, prove that
$ a\plus{} \frac{1}{b(a\minus{}b)} \ge 3$

$ \frac {a}{2}\ge\sqrt {b(a \minus{} b)}$   (AM-GM)

$ b(a \minus{} b)\le\frac {a^2}{4}$

$ \frac {1}{b(a \minus{} b)}\ge\frac {4}{a^2}$

$ a \plus{} \frac {1}{b(a \minus{} b)}\ge{a \plus{} \frac {4}{a^2}}$

$ (a\plus{}\frac{4}{a^2})'\equal{}1\minus{}\frac{8a}{a^4}\equal{}0$

a=2;

$ 2 \plus{} \frac {4}{2^2} \equal{} 3$

New problem:

abc=1;

Prove that $ \frac {1}{1 \plus{} a \plus{} ab} \plus{} \frac {1}{1 \plus{} b \plus{} bc} \plus{} \frac {1}{1 \plus{} c \plus{} ca} \equal{} 1$

We substitute $ a\equal{}\frac{y}{x}, b\equal{}\frac{z}{y}, c\equal{}\frac{x}{z}$ as $ abc\equal{}1$.

Now,
$ \frac {1}{1 \plus{} a \plus{} ab} \equal{} \frac{1}{1\plus{}\frac{y}{x}\plus{}\frac{y}{x}.\frac{z}{y}} \equal{} \frac{x}{x\plus{}y\plus{}z}$

Similarly,

$ \frac{1}{1\plus{}b\plus{}bc}\equal{}\frac{y}{x\plus{}y\plus{}z}$ and
$ \frac{1}{1\plus{}c\plus{}ca}\equal{}\frac{z}{x\plus{}y\plus{}z}$.

Adding the three, we have the required result.

New Problem: Prove that $ 71|61!\plus{}1$ and $ 71|63!\plus{}1$.
Here $ a|b$ means $ a$ divides $ b$.

[hide="Regarding previous problem"]I was hoping for a solution without any calculus but Bachukas' solution is pretty neat as well. The same solution can be ended in AM-GM as
$ \frac{a}{2}\plus{}\frac{a}{2}\plus{}\frac{4}{a^2} \ge 3$

My solution was pure AM-GM as follows
$ (a\minus{}b)\plus{}b\plus{}\frac{1}{b(a\minus{}b)} \ge 3$

[/hide]

[hide="Solution"]$ 70!\equiv \minus{}1\pmod{71}$ by Wilson's theorem.

$ 70\cdot 69\cdots 62\equiv \minus{}1\cdot \minus{}2\cdots \minus{}9\equal{} \minus{}362880\equiv \minus{}70\equiv 1\pmod{71}$.

Therefore, $ \frac{70!}{61!}\cdot 61!\equiv 61!\equiv \minus{}1\pmod{71}\implies 61!\plus{}1\equiv 0\pmod{71}$.

$ 70\cdot 69\cdots 64\equiv \minus{}1\cdot \minus{}2\cdots\minus{}7\equal{}\minus{}5040\equiv 1\pmod{71}$.
$ 70!\equiv 63!\cdot 1\equiv \minus{}1\implies 63!\plus{}1\equiv 0\pmod{71}$.[/hide]

New problem.

Let $ S_n$ be the minimum value of $ \sum_{k\equal{}1}^{n}\sqrt{(2k\minus{}1)^2\plus{}a_k^2}$, where $ a_1,a_2,\cdots,a_n$ are positive, real numbers, and $ \sum_{k\equal{}1}^{n}a_k\equal{}17$. Find a value $ n$ such that $ S_n$ is an integer.

[hide="solution"]
Suppose we let the ratio $ a_1: a_2: a_3...: a_n \equal{} 2(1)\minus{}1: 2(2)\minus{}1: ...2n\minus{}1$ then the expression:
$ \sum_{k\equal{}1}^{n}\sqrt{(2k\minus{}1)^2\plus{}a_k^2}$ can be represented by many similar right-angle triangles.
If we stack them on top of eachother in a "staircase pattern", then we can use the fact that the shortest distance between two points is a straight line.
Thus the expression is minimised at

$ \sqrt{\left(\sum_{k\equal{}1}^n 2k\minus{}1 \right)^2\plus{} \left(\sum_{k\equal{}1}^n a_k\right)^2} \equal{}\sqrt{\left(\sum_{k\equal{}1}^n 2k\minus{}1\right)^2 \plus{} 17^2}$

$ \Rightarrow \sum_{k\equal{}1}^n k \equal{} \frac{n(n\plus{}1)}{2} \Rightarrow \sum_{k\equal{}1}^n 2k\minus{}1 \equal{} n(n\plus{}1)\minus{}n \equal{} n^2$

So we need $ n$ such that $ n^4 \plus{} 17^2$ is an perfect square. So $ a^2 \minus{} n^4 \equal{} 17^2$ for some $ a$

Therefore $ (a\minus{}n^2)(a\plus{}n^2) \equal{} 17^2 \Longrightarrow a\minus{}n^2\equal{}1, a\plus{}n^2\equal{}17^2 \Rightarrow 2n^2 \equal{} 17^2\minus{}1$

$ \Rightarrow n^2 \equal{} 144 \Rightarrow n\equal{}\boxed{12}$

[/hide]

New Problem
Suppose a Polynomial $ P(x)$ takes rational values for all rational $ x$ and irrational values for all irrational $ x$.
(1)  Show that the coeficients of $ P(x)$ must be rational.
(2)  Show that all such Polynomials must be linear.

[quote="Bachukas"]Am I allowed to post such topic?

If yes, then here are the rules:

1) Solutions, not only answers should be posted
2) The solver should post a new problem
3) You can not post new problems until the old are unsolved
4) Keep the problems pre-olympiad problems level.

And the 1st problem:

Prove an inequality: $ (a \plus{} b)(a \plus{} c)\ge2\sqrt {abc(a \plus{} b \plus{} c)}$ , a,b,c - positive real numbers.[/quote]

I think for you idea is needed  a forum , not just a topic.

[quote="ocha"]
New Problem
Suppose a Polynomial $ P(x)$ takes rational values for all rational $ x$ and irrational values for all irrational $ x$.
(1)  Show that the coeficients of $ P(x)$ must be rational.
(2)  Show that all such Polynomials must be linear.[/quote]

It's not hard. I can't solve most Pre-oly problems, but I can solve this.  :)

[hide="(1)"]
Suppose there is a coefficient of $ P(x)$ that is irrational. If there is only one, then obviously for nonzero rational $ x$ the polynomial is irrational, since a rational added to or multiplied by an irrational results in an irrational. Thus there must be at least two; let these be $ a_i$ - note that the constant coefficient cannot be irrational, or else $ P(0)$ would be irrational. We need $ \sum a_i$ to be rational in the case when $ x\equal{}1$, so suppose this is rational. Now, write the terms of the polynomial with the irrational coefficients to be $ \left[\left(\sum a_ix^{b_i}\right)\plus{}a_n\right]/x^{b_n}$ for nonzero $ x$, where $ a_n$ is the term in the series corresponding to the lowest power of $ x$, the sum goes to $ n\minus{}1$, and $ b_i$ are the powers of $ x$ corresponding to $ a_i$. For this to be rational, we need the summation irrational for nonzero $ x$. But with $ x\equal{}1$, we have that the summation is just the sum of all the coefficients minus the last one of lowest degree - since a rational minus an irrational is irrational, we have a contradiction.
[/hide]

I'll take a look at (2) later.

[hide="or even easier..."]
$ P(1),P(2),...,P(n)\in \mathbb{Z}$ where $ n\equal{}degP(x)$
We have a system of equation and basing on Cramer's formula's we have... :winik:
[/hide]

[quote="pythag011"][hide="easier solution to 1"]
Lagrange Interpolation gives that the coefficients are rational pretty easily.[/hide][/quote]

Anyone mind explaining more (i.e. more complete proof with Lagrange interpolation)?
I guess you would treat the rational number values as "points" through which the polynomial passes...?

[hide="(2)"]Proceed by sort-of-strong induction to prove that there's a irrational number so that $ P(x)$ is rational for $ \deg (P(x)) \equal{} n\ge 2$. This number is easily derived for $ n \equal{} 2$ - do it yourself. Now assume it's true for $ 1,2,\ldots,n \minus{} 1$. Take a polynomial of degree $ n$ (just assume its first coefficient is $ 1$; I'm too lazy to type $ a$ before the polynomial each time and it doesn't change anything) $ (x \minus{} r_1)(x \minus{} r_2)\ldots(x \minus{} r_{n \minus{} 1})(x \minus{} r_n)$. By our inductive hypothesis, the first $ n \minus{} 2$ terms of this product have at least one irrational number so that the product is rational - call this $ i$. Then $ ai/b$ for any rationals $ a,b$ also gives a rational result from the first $ n \minus{} 1$ terms. We need to show that we can choose $ a,b$ so that $ (ai/b \minus{} r_{n \minus{} 1})(ai/b \minus{} r_n)$ is rational. Expanding and getting rid of the irrelevant terms which are always rational, we have that $ a^2i^2 \minus{} abr_ni \minus{} abr_{n \minus{} 1}i$ needs to be rational. If we let $ b \equal{} \frac {1}{r_n \plus{} r_{n \minus{} 1}}$, we see this becomes rational; hence by our inductive hypothesis and base case (2) is true and $ P(x)$ is linear, as desired.

Note that this proves that there exists an *algebraic* irrational number so that the polynomial of higher degree is rational. Bonus: does there exist a transcendental for all such polynomials satisfying the conditions? Any of them?
[/hide]

On solutions to (1); pythag's might be shorter, but is less elementary and probably harder to see - polskimisiek's is neither shorter, more elementary, nor easier. I believe my solution is easier than both of the "easier" solutions.

Anyways, new problem:
$ a,b,c\in \mathbb{R}^ \plus{}$ satisfy $ ab \plus{} bc \plus{} ca \equal{} abc$. Show that

$ \frac {a^4 \plus{} b^4}{ab(a^3 \plus{} b^3)} \plus{} \frac {b^4 \plus{} c^4}{bc(b^3 \plus{} c^3)} \plus{} \frac {c^4 \plus{} a^4}{ca(c^3 \plus{} a^3)} \geq 1$

From an old Polish olympiad.

EDIT: Just realized that because I use $ n \minus{} 2$ in my proof of (2), I need to prove it for $ n \equal{} 3$ as well. Oh well; that's not difficult - figure it out yourself.

[quote="Bachukas"]Isn't this problem to hard for pre-olympiad level? I guess this is for olympiad section.[/quote]
I guess it depends on your definition of "pre-olympiad level."  It is actually an extension of BAMO (Bay Area Mathematical Olympiad) 2002, number 5.

[quote="Temperal"]$ (x \minus{} r_1)(x \minus{} r_2)\ldots(x \minus{} r_{n \minus{} 1})(x \minus{} r_n)$. By our inductive hypothesis, the first $ n \minus{} 2$ terms of this product have at least one irrational number so that the product is rational - call this $ i$. [/quote]
Doesn't this assume that your polynomial has all rational roots?  It might be fixable in the sense that the polynomial can be decomposed into other polynomials with rational coefficients, but you still assume that the polynomial can be factored.  I think you have to patch up your proof.

Anyways, a solution to the second part of the problem:
[hide="Solution"]
Assume that a polynomial exists with degree greater than or equal to $ 2$.  First, take $ P(x)$ and shift it down by it's constant term, $ c$.  This guarantees the constant term of $ P(x)$ is now $ 0$.  Multiply $ P(x)$ by an integer so that all of the coefficients are integers, but do not share a common factor.  Let the leading coefficient of the polynomial be $ a$ (where $ a > 0$).  We know that $ P(x)$ is continuous, so it reaches every value from $ K$ to $ \infty$ for some constant $ K$.  Let us consider all primes $ p_i$ so that $ p_i\ge a$.  There exists an $ x_i$ for each $ i$ so that $ P(x_i) \equal{} p_i$.  By the conditions, we may not have an irrational $ x$ take $ P(x)$ to a rational value.  Thus, all $ x_i$ are rational.  We now employ the Rational Roots Theorem.  Notice that a root of $ P(x) \minus{} p_i$ is $ x_i$.  Since $ P(x_i) \minus{} p_i \equal{} 0$, $ x_i \equal{} \frac {\pm p_i}{d}$ or $ \pm \frac {1}{d}$, where $ d$ is a divisor of $ a$.  Assume that the second case happens for infinitely large $ p_i$.  Then, we notice that $ P(x)$ can only reach a finite level when $ x\le |1|$.  However, $ p_i$ can become infinitely large, which is a contradiction.  It follows that, when $ i > t$, where $ t$ is a constant, $ x_i \equal{} \frac {\pm p_i}{d}$.  Notice that the absolute value of the leading term of $ P(\frac {\pm p_i}{d})$ will eventually become larger than that of any of the other terms of the polynomial.  Thus, when the degree of $ P(x)$ is greater than or equal to $ 2$, the leading term of $ P(\frac {\pm p_i}{d})$ will overtake the linear term, so for sufficiently large $ i$; we will have that $ |P(\frac {\pm p_i}{d})| > |p_i|$, which is a contradiction.  It follows that $ P(x)$ is linear.  [/hide]

[quote="Temperal"]

Anyways, new problem:
$ a,b,c\in \mathbb{R}^ \plus{}$ satisfy $ ab \plus{} bc \plus{} ca \equal{} abc$. Show that

$ \frac {a^4 \plus{} b^4}{ab(a^3 \plus{} b^3)} \plus{} \frac {b^4 \plus{} c^4}{bc(b^3 \plus{} c^3)} \plus{} \frac {c^4 \plus{} a^4}{ca(c^3 \plus{} a^3)} \geq 1$

From an old Polish olympiad.[/quote]

[hide="Solution"]
All sums from now on are symmetric.  Dividing everything by $ abc$, we have that the condition is equivalent to $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 1$.  Substituting $ x \equal{} \frac {1}{a}$, $ \frac {1}{b} \equal{} y$, and $ \frac {1}{c} \equal{} z$, we have that $ x \plus{} y \plus{} z \equal{} 1$ and we want for $ \sum \frac {x^4 \plus{} y^4}{x^3 \plus{} y^3} \ge 1$.  Now, notice that\[ \frac {x^4 \plus{} y^4}{x^3 \plus{} y^3} \ge \frac {x \plus{} y}{2}\iff\]\[ 2x^4 \plus{} 2y^4 \ge x^4 \plus{} y^4 \plus{} x^3y \plus{} xy^3\iff\]\[ x^3(x \minus{} y) \minus{} y^3(x \minus{} y)\ge 0\iff\]\[ (x \minus{} y)^2(x^2 \plus{} yx \plus{} y^2)\ge 0\]which is true.  Thus,\[ \sum \frac {x^4 \plus{} y^4}{x^3 \plus{} y^3} \ge \sum \frac {x \plus{} y}{2} \equal{} \sum x \equal{} 1\][/hide]

Yes; I assumed it has all rational roots because if it didn't, then of course then the irrational root would result in a rational answer.  :)

New problem, anyone?

[b]New problem [/b]

Let be $ a\equal{}1\plus{}\left\lfloor \left(\sqrt{13}\plus{}3\right)^{2004} \right\rfloor$ . Prove that $ 2^{2004}\ |\ a$ .

[quote="alex2008"][b]New problem [/b]

Let be $ a \equal{} 1 \plus{} \left\lfloor \left(\sqrt {13} \plus{} 3\right)^{2004} \right\rfloor$ . Prove that $ 2^{2004}\ |\ a$ .[/quote]
Let us define $ p \equal{} 3 \plus{} \sqrt {13}$; $ q \equal{} 3 \minus{} \sqrt {13}$.
$ p \minus{} q \equal{} 2\sqrt {13}$
$ p \plus{} q \equal{} 6$
$ pq \equal{} 4$.
Now, define $ x_n \equal{} p^n \plus{} q^n$. We have; $ x_1 \equal{} 6, x_2 \equal{} 28$.
Also $ x_3 \equal{} (p \plus{} q)(p^2 \plus{} q^2 \minus{} pq) \equal{} 6(x_2 \minus{} 4) \equal{} 6x_2 \minus{} 4x_1$.
Proceeding, we get the recurrence relation : $ x_{n \plus{} 2} \equal{} 6x_{n \plus{} 1} \minus{} 4x_{n}$ (which can be proved using induction on $ n$).

We have already seen that $ x_1 \equal{} 6; x_2 \equal{} 28$. We guess that $ 2^n|x_n$. To prove this we can again use induction.
[hide="Proof"]
$ x_n \equal{} 2^n k$, it ias reqd to prove that $ x_{n \plus{} 1} \equal{} 2^{n \plus{} 1}l$.
$ x_{n \plus{} 1} \equal{} 6x_n \minus{} 4x_{n \minus{} 1} \equal{} 6\cdot2^{n} \minus{} 4\cdot2^{n \minus{} 1}$ which is divisible by $ 2^{n \plus{} 1}$; QED.

[/hide]
$ x_{n \plus{} 2} \equal{} 2^{n \plus{} 2}(3l \minus{} k)$
$ x_n\in\mathbb{N}$.Also as $ n$ is an even integer; so $ 0 < (3 \minus{} \sqrt {13})^n \equal{} (p \plus{} q)^n < 1$ (putting $ n \equal{} 2004$.

Hence $ 2^{2004}|x_{2004} \equal{} 1 \plus{} \lfloor (3 \plus{} \sqrt {13})^{2004}\rfloor$.

[b]NP[/b]
Two acute angles $ \alpha, \beta$ satisfy $ \sin^2\alpha \plus{} \sin^2\beta \equal{} sin (\alpha \plus{} \beta)$. Prove that $ \alpha \plus{} \beta \equal{} \frac {\pi}{2}$.

[quote="Potla"]

[b]NP[/b]
Two acute angles $ \alpha, \beta$ satisfy $ \sin^2\alpha \plus{} \sin^2\beta \equal{} sin (\alpha \plus{} \beta)$. Prove that $ \alpha \plus{} \beta \equal{} \frac {\pi}{2}$.[/quote]

Potla you posted this easy problem just to keep pre-olympiad marathon alive. :)

[hide]We have the equation $ \sin^2 \alpha \plus{} \sin^2 \beta \equal{} sin(\alpha \plus{} \beta)$

Adding to both sides $ \minus{} 2\sin\alpha \cos \beta \plus{} \cos^2 \beta \minus{} 2\sin\beta \cos \alpha \plus{} cos^2 \alpha$

Then $ LHS \equal{} (\sin\alpha \minus{} \cos \beta)^2 \plus{} (\sin \beta \minus{} \cos \alpha)^2$ and $ RHS \equal{} \cos\alpha(\cos \alpha \minus{} \sin \beta ) \plus{} \cos\beta(\cos \beta \minus{} \sin\alpha)$

Factoring it we have :$ (\sin\alpha \minus{} \cos \beta)(\sin \alpha \minus{} \cos\beta \plus{} 1) \plus{} (\sin \beta \minus{} \cos \alpha)(\sin \beta \minus{} \cos \alpha \plus{} 1) \equal{} 0$

Obviously $ \sin \alpha \minus{} \cos\beta \plus{} 1 \neq 0$ and $ \sin\beta \minus{} \cos \alpha \plus{} 1 \neq 0 \implies$$ \sin \alpha \minus{} \cos \beta \equal{} \sin \beta \minus{} \cos \alpha \equal{} 0 \implies \alpha \plus{} \beta \equal{} \frac {\pi}{2}$[/hide]

New Problem :

Find all functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ continous in $ \mathbb{R}$ such that $ f(x\plus{}f(y))\equal{}f(x)\plus{}y$ for all $ x,y\in \mathbb{R}$

[hide="Solution"]$ x \equal{} y \equal{} 0\Rightarrow f(f(0)) \equal{} f(0)$; additionally; suppose $ x \equal{} y$; then $ f(x \plus{} f(x)) \equal{} x \plus{} f(x)$. Finally, let $ f(y) \equal{} x$; then $ f(2x) \equal{} 2f(x)$ (1).

Now we claim that $ f$ is one-to-one; let $ f(x) \equal{} f(y)$. Then $ f(x \plus{} f(y)) \plus{} y \equal{} f(y \plus{} f(x)) \plus{} x\Rightarrow f(x \plus{} 2f(y)) \equal{} f(y \plus{} 2f(x)) \equal{} f(y \plus{} 2f(y))$, so by (1), $ x \equal{} y$. This means that $ f(0) \equal{} 0$. Then letting $ x \equal{} 0$, we see that $ f(f(y)) \equal{} y$, so $ f$ is an involution.

To wrap up, substituting $ 2y' \equal{} y$, using (1), and then using the same method to prove (1) repeatedly, we see that $ f(nx) \equal{} nf(x)$; letting $ f(1) \equal{} k$, and $ x \equal{} 1$, we see that $ f(n) \equal{} nk$ for all integers $ n$, and by substituting $ y \equal{} x/n$ we get that it's true for rationals as well. Since $ f(x)$ is continuous, it's true for all reals as well (I'm not going to delve into this in a high school forum, but this can be proven using intermediate value theorem). Now, since $ f$ is an involution, $ n^2x \equal{} x\Rightarrow n \equal{} \pm1$. Our solution set is therefore $ f(x) \equal{} \pm x$.[/hide]

Prove that it is impossible to dissect a rectangle into acute triangles using only lines (i.e. no segments!).

OK, I give up
What is the solution to this problem, Temperal?
It's interesting.
[hide]SAVE THIS MARATHON; READ MY POST BEFORE & KNOW HOW YOU CAN SAVE IT!  :(  [/hide]

we have  $ \frac {1}{1 \plus{} b \plus{} bc} \equal{}$\frac {a}{1+a+ab} and  $ \frac { 1}{1 \plus{} c \plus{} ca}  \equal{}$\frac {ab}{1+a+ab}
then we have

 $ \frac {1}{1 \plus{} a \plus{} ab} \plus{}$\frac {1}{1 + b + bc} +  $ \frac {1}{1 \plus{} c \plus{} ca} \equal{}$\frac {1}{1 + a + ab} +  $ \frac {a}{1 \plus{} a \plus{} ab} \plus{}$\frac {ab}{1+a+ab} =  $ \frac {1 \plus{} a \plus{} ab}{1 \plus{} a \plus{} ab} \equal{} 1$

[quote="ouboujlal"]we have  $ \frac {1}{1 + b + bc} =\frac {a}{1+a+ab}$ and  $ \frac { 1}{1 + c + ca} =\frac {ab}{1+a+ab}$
then we have

 $ \frac {1}{1 + a + ab} + \frac {1}{1 + b + bc} +  \frac {1}{1 + c + ca} = \frac {1}{1 + a + ab} +  \frac {a}{1 + a + ab} +\frac {ab}{1+a+ab} =  \frac {1 + a + ab}{1 + a + ab} = 1$[/quote]

Fixed your $ \text{\LaTeX}$ :wink:
You forgot to post a new problem...
[hide]This stuff is kinda hard for me..so I'll just go to another forum :) Just browsing :wink: [/hide]

[quote="KING1"][quote="ouboujlal"]we have  $ \frac {1}{1 \plus{} b \plus{} bc} \equal{} \frac {a}{1 \plus{} a \plus{} ab}$ and  $ \frac { 1}{1 \plus{} c \plus{} ca} \equal{} \frac {ab}{1 \plus{} a \plus{} ab}$
then we have

 $ \frac {1}{1 \plus{} a \plus{} ab} \plus{} \frac {1}{1 \plus{} b \plus{} bc} \plus{} \frac {1}{1 \plus{} c \plus{} ca} \equal{} \frac {1}{1 \plus{} a \plus{} ab} \plus{} \frac {a}{1 \plus{} a \plus{} ab} \plus{} \frac {ab}{1 \plus{} a \plus{} ab} \equal{} \frac {1 \plus{} a \plus{} ab}{1 \plus{} a \plus{} ab} \equal{} 1$[/quote]

][/quote]
Actually it is straightforward using $ a\equal{}\frac{x}{y},b\equal{}\frac{y}{z},c\equal{}\frac{z}{x}$

new problem

If $ a,b,c>1$ and $ n\in \mathbb{N}$ ,then prove that :

$ \frac{1}{(\log_{bc}a)^n} \plus{}\frac{1}{(\log_{ca}b)^n} \plus{}\frac{1}{(\log_{ab}c)^n} \geq 3\cdot 2^n$

[hide="Solution"]
The given expression can be written as $ \frac {{(\log_{a}bc)^{n}} \plus{} {(\log_{b}ca)^{n}} \plus{} {(\log_{c}ab)^{n}}}{3}\geq 2^{n}$
We know
$ \frac {(\log_{a}bc)^{n} \plus{} (\log_{b}ac)^{n} \plus{} (\log_{c}ba)^{n}}{3}\ge[\frac {(\log_{a}bc) \plus{} (\log_{b}ac) \plus{} (\log_{c}ba)}{3}]^{n}$
now$ {(\log_{a}bc) \plus{} (\log_{b}ac) \plus{} (\log_{c}ba)}\ge6$
 using $ A.M\ge G.M$

Hence Proved[/hide]

Prove for each positive integer $ n$
$ (2n^{2}\plus{}3n\plus{}1)^{n}\ge6^{n}(n!)^{2}$

[hide="solution"]

The inequality is equivalent with :

$ \left(\frac {2n^3 \plus{} 3n^2 \plus{} n}{6}\right) \geq n \sqrt [n]{(n!)^2}$

but $ \left(\frac {2n^3 \plus{} 3n^2 \plus{} n}{6} \right) \equal{} \sum n^2$

Now by Am-Gm $ LHS \equal{} 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \geq n \sqrt [n]{1^2\cdot 2^2 \cdots n^2}$ as desired.[/hide]

[hide="new problem"]

If all roots of the equation

$ x^6 \minus{} 6x^5 \plus{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} 1 \equal{} 0$ are positive real numbers,then find $ a,b,c,d$  [/hide]

[hide="Solution"]Let the roots be $ r_1 \ldots r_6$. We have $ \sum r_i \equal{} 6 \implies \text{AM} (r_1 \ldots r_6) \equal{} 1$
We also have $ \prod r_i \equal{} 1 \implies \text{GM} (r_1 \ldots r_6) \equal{} 1$.

Since $ r_i$ are positive reals, AM-GM applies, and we know $ r_1 \equal{} \ldots \equal{} r_6 \equal{} 1$, and hence, the polynomial is $ (x\minus{}1)^6$. Expand to find the coefficients.[/hide]

I'll re-post a problem from earlier in the thread that was never solved:
[quote="Temperal"]Prove that it is impossible to dissect a rectangle into acute triangles using only lines (i.e. no segments!).[/quote]

[hide]Four of the lines must go through each of the rectangle's right angles, or else right (not actue) triangles will exist in our dissection.  We have two cases: when two of these lines meet at one point on one side, and the other two on the opposite side, or when two of them do not meet on one of the sides.  For the first case, there will be 2 intersections in the interior of the rectangle; these intersections will create either a right or obtuse angle with each; drawing a line through both intersections will solve this, but then where they hit the sides of the rectangle will create new obtuse angles, and the process will continue ad infinitum.  For the latter case, where these two lines don't meet on one of the sides, they will create obtuse angles, and no matter how many lines we draw to break them up there will still always be another right or obtuse angle, continuing ad infinitum once again.  So, there can't be a finite number of drawn lines because one of them will always create an obtuse angle with one of the sides, and an obtuse triangle will be created.  If you draw another line to break up this angle, it too will create an obtuse angle, and so on. Therefore, since we have covered both cases of how these lines must dissect the right angles, we are finished.

[/hide]
Problem: 500 basketball players are divided into 250 two-person teams for a tournament.  On each day of the tournament, the teams are rearranged such that no two people ever play together twice.  What is the longest possible such tournament (and prove your answer!)?


on average, how many hours of math per day do you guys do (i mean competition math, not school work)?

well, on school days, i usually dont do any except for during math period (w00t intermediate algebra :D )
thats 40 minutes
and i have some math activity a lot so maybe that adds an average of 20 minutes
for 1 hour during weekdays

weekends...
well depending on my homework/activities load
1-6 hours a day perhaps

What about none? :P

Two weeks before a contest, I average an hour a day, and on the final seven-day stretch, I do about two or three hours a day.

But then again, I kinda suck either way. :roll:

uhh average 1-2 hours a day. on a weekend i mite go crazy and sepnd 4 hours reading a calc book but im too lazy most of the time  :D

most of the time exactly one hour

coincidentally, math class is exactly an hour long

I usually do ~3 hours of math. Nothing better to do (not that math is boring).

I practiced a lot before USAMO and the AIME roughly 1-2 hours a day depending on hw and normally other than that like 2 hours a week or so depending on projects etc


{ 3/4x + 2/3y = 7
{ 3/5x - 1/2y = 18

I'm suppose to find out what x and y are.

Again the answer is:

[hide="click to reveal"]
(20,-12)
[/hide]

I think my problem here is trying to take the fractions and turn them into regular numbers. Maybe that's not what I need to do. I did this problem a few times and I got these whacked out, far fetched answers for x and y so I gave up.

I probably am missing something and solving this is simpler than I think.
I'm guessing the substitution "injection" method would be used here.

-note to self-
page. 168
number 27.

If you're good with fractions, solve it like it is. If not, change them to whole numbers.

Multiply the top equation by 12 to get $9x+8y=84$.
Multiply the bottom equation by 10 to get $6x-5y=180$

Now we can solve.

Multiply the top equation by 2 and the bottom by 3 to get
$18x+16y=168$
$18x-15y=540$

Subtract.

$31y=-372$
$y=-12$

Substitute into the first equation.

$9x+8(-12)=84$
$9x-96=84$
$9x=180$
$x=20$

Answer: (20, -12)

[quote="236factorial"]
$18x+16y=168$
$18x-15y=540$

Subtract.

$31y=-372$
$y=-12$

Answer: (20, -12)[/quote]

Why do you get rid of the $18x$s but add the $16y$ and $15y$.
From the looks of it, you are suppose do this:

$16y-15y=1y$

The book didn't cover this, this book sucks.
Know of any weblinks that discuss the subtraction method?

Actually, since we subtracted, $16-(-15) = 16+15 = 31$


modular arithmetic
Euler
number theory
relatively prime

this is a question from the mock amc 2 and the problem is as follows
find the units digit of $1^1 + 2^2 + 3^3 + ... + 2004^{2004}$
i don't reall have any experience with number theory, but it seems that there would be a way to simplify $x^x$

Since we're dealing with the unit digits, don't we need to consider them in mod 10?

For instance, $11^{11} \equiv 1^{11} \equiv 1 \mod 10$ and do the same thing to others.

I'm not sure if there is any simple ways to do them. There is a short way to write them as:

\[\sum^{2004}_{k=1} k^k\]

But that's far as I can go.  :P It seems that this is very hard sum since the base and power changes as it goes.

right...but this was on an amc, so there should be something that blows the problem apart (like a theorem)
is there an application of fermat's little theorem here?
$(x-1)^{x-1} \equiv 1 \mod{x}$ the only thing is that x is not always prime

This was actually a problem right off of a Mandelbrot from previous years (though I believe it was up to $2003$). I remember my solution was just quickly finding $n^n\pmod{10}$ for $n={1,2,\ldots,9}$, and then finding $n^{10}\pmod{10}$. Once you look it in terms of $n^n, n^{n+10}, n^{n+20},\ldots \pmod{10}$, it really quickly cancels itself out $\pmod{10}$. A bit brute force of a solution, but it can quickly be done in 4-5 minutes.

If you notice that $\phi (10) =4 $ so that you know that  last digit of  $n^{a+4k} $ is same as $n^a$

(Btw  this is same as  what many people know :  fifth power of any number has the same unit digit as the number)

and notice the HCF or 10 and 4 is 20 and thus notice $(n+20)^{n+20}$ has the same last digit as $n^n$
so if last digit of $(1^1+2^2 \dots + 20^{20}) $  is  $x$   (One need not care/calculate  what x is )
Required answer is:
 last digit of ( $100 x + 1^1+2^2+3^3+4^4$  )  or $\equiv 1+4+7+6 \equiv 8 $

good answer...i don't really know much of this number theory stuff, but this is helping, mildorf's site is really helpful
why did you consider cycles of 20, here is a simpler solution, if it is the same, my bad
as you said, $n^{a+4k} \equiv n^a \mod{10}$ since $\phi (10) = 4$ (using eulers generalization of fermat's little theorem)
so there are cycles of 4 that repeat, and $\frac{2004}{4} = 501$ cycles of the same units digit
so multiplying by 501 will give the sum off all the cycles, all cycles being
$501(1^1 + 2^2 + 3^3 + 4^4) \equiv \mod{10}$
$(1^1 + 2^2 + 3^3 + 4^4) \equiv \mod{10}$
which is easily solved, the answer being $\boxed{8}$

here is another problem that i thought of, which is a general result that could be shown after understanding the previous problem
find $\left( \sum_{k=1}^{n}k^k \right) \mod{10}$ in terms of $n$

[quote="Altheman"]g
so there are cycles of 4 that repeat, and $\frac{2004}{4} = 501$ cycles of the same units digit
so multiplying by 501 will give the sum off all the cycles, all cycles being
$501(1^1 + 2^2 + 3^3 + 4^4) \equiv \mod{10}$
$(1^1 + 2^2 + 3^3 + 4^4) \equiv \mod{10}$
which is easily solved, the answer being $\boxed{8}$

here is another problem that i thought of, which is a general result that could be shown after understanding the previous problem
find $\left( \sum_{k=1}^{n}k^k \right) \mod{10}$ in terms of $n$[/quote]

You have to be a little careful, the cycle does not repeat in 4 (but rather 20),  for example if you had 8 terms the answer is NOT  8*2 but only 8....you were just lucky that the number 500 is divisible by 5 too.

BTW - if  you take the first 20 numbers, the answer is 4,(  which of course is not 5*8)  ... this is the value of "x" in my previous post.

[b]Here is a little harder (but not much harder) question:

what are  the last 2 digits  of the above sum: (how about last 3 digits) [/b]

i kind of see this, but not really, is the idea that
$(x +n)^{x+n} \equiv x^x \mod{10}$ for some $n$
and n has to be divisible by 10 for the base, and 4 for the exponent

and one other question that i had is what about the restriction on the use of this formula
$a^{\phi(n)} \equiv 1 \mod{n}$ if $gcf(a,n) =1$
but if we have like $30^{30} \mod{10}$, 30 and 10 are not relatively prime, so it is a misuse of the formula

[quote]
and n has to be divisible by 10 for the base, and 4 for the exponent  [/quote]

Exactly, this is the idea I used, that is  $(x+20)^{x+20} \equiv x^x   \bmod 10 $  as long as $x$ is a positive integer. ($x>0$ )

[quote]
and one other question that i had is what about the restriction on the use of this formula
$a^{\phi(n)} \equiv 1 \mod{n}$ if $gcf(a,n) =1$
but if we have like $30^{30} \mod{10}$, 30 and 10 are not relatively prime, so it is a misuse of the formula[/quote]

That's why  instead of saying $a^4 \equiv 1 \bmod 10$     I said       $a^5 \equiv  a \bmod 10 $  The former equation  is true only when $gcf(a,10)=1$ while the second is true  for other numbers too. This is somewhat clever  apolication of the same rule. Another ""trick"  one employs is to consider the value mod $2$ and mod $5$ (instead of $10$)  and in that case, for a number   like $2$ , $2^{(4k) } $ would be $0$  mod 2 but $1$  mod 5  (which is$6$)  etc.

Similarly if one is using mod 100,    $a^{20} \equiv 1 $ if    $a$   does not have a factor of 2 or 5 .
You would still get to say for all integers $m >1 $
$a^{20k+m} \equiv a^m  \bmod 100$
Hope this helps.
.

how would you prove that $a^5 \equiv a \mod{10}$
you could break it down to
$a* a^{\phi(10)} \mod{10}$ but you could only say that is true if $gcf(a,10) = 1$
or would you prove it for 1 through 9, and all the others are the same because only the units digit matters when dealing with mod 10

[quote="Altheman"]how would you prove that $a^5 \equiv a \mod{10}$
[/quote]
Yes you can, as you said,  go and check from $0$  to $9$, or  use  Fermat's $a^p  \equiv a \bmod p $ for prime p  (which is true for any a) and doing   this for  two cases .. for p=2 and  p=5 and applying CRT.


When are we informed whether we have qualified for the USAMO or not :?:

April 10th, according to the USAMO Teachers Manual, which is available on the front page of the [url=http://www.unl.edu/amc]AMC website[/url].


let $ a,b,c$is no-negative discriminate,$ k$is constant.prove that:
$ \frac{a^2b^2\plus{}k^2}{(a\minus{}b)^2}\plus{}\frac{b^2c^2\plus{}k^2}{(b\minus{}c)^2}\plus{}\frac{c^2a^2\plus{}k^2}{(c\minus{}a)^2} \ge \frac{3k}{2}$

[quote="tuantam1lan"]let $ a,b,c$is no-negative discriminate,$ k$is constant.prove that:
$ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {b^2c^2 \plus{} k^2}{(b \minus{} c)^2} \plus{} \frac {c^2a^2 \plus{} k^2}{(c \minus{} a)^2} \ge \frac {3k}{2}$[/quote]

Assume : $ c\equal{}min\{ a;b;c\}$

We have : $ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {b^2c^2 \plus{} k^2}{(b \minus{} c)^2} \plus{} \frac {c^2a^2 \plus{} k^2}{(c \minus{} a)^2} \ge \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {k^2}{b^2} \plus{} \frac {k^2}{a^2}$

We have to prove that : $ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {k^2}{b^2} \plus{} \frac {k^2}{a^2} \geq \frac{3k}{2}$

$ \Leftrightarrow k^2(\frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{a^2}\plus{}\frac{1}{b^2})\minus{}\frac{3k}{2}\plus{}\frac{a^2b^2}{(a\minus{}b)^2} \geq 0$

We have : $ 4(a\minus{}b)^4 \delta \equal{}\minus{}(a\minus{}b)^2[7(a^2\plus{}b^2)\plus{}18ab]\minus{}16a^2b^2 < 0$

We have done ! :)

@ : $ \delta$ = delta :D

[quote="Materazzi"][quote="tuantam1lan"]let $ a,b,c$is no-negative discriminate,$ k$is constant.prove that:
$ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {b^2c^2 \plus{} k^2}{(b \minus{} c)^2} \plus{} \frac {c^2a^2 \plus{} k^2}{(c \minus{} a)^2} \ge \frac {3k}{2}$[/quote]

Assume : $ c \equal{} min\{ a;b;c\}$

We have : $ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {b^2c^2 \plus{} k^2}{(b \minus{} c)^2} \plus{} \frac {c^2a^2 \plus{} k^2}{(c \minus{} a)^2} \ge \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {k^2}{b^2} \plus{} \frac {k^2}{a^2}$

We have to prove that : $ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {k^2}{b^2} \plus{} \frac {k^2}{a^2} \geq \frac {3k}{2}$

$ \Leftrightarrow k^2(\frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{a^2} \plus{} \frac {1}{b^2}) \minus{} \frac {3k}{2} \plus{} \frac {a^2b^2}{(a \minus{} b)^2} \geq 0$

We have : $ 4(a \minus{} b)^4 \delta \equal{} \minus{} (a \minus{} b)^2[7(a^2 \plus{} b^2) \plus{} 18ab] \minus{} 16a^2b^2 < 0$

We have done ! :)

@ : $ \delta$ = delta :D[/quote]
 nice solution!
my proof uses equality is:
 $ \sum_{cyc}{\frac {ab \plus{} k}{a \minus{} b}\frac {bc \plus{} k}{b \minus{} c}} \equal{} k$
but your proof better than my proof

[quote="tuantam1lan"]let $ a,b,c$is no-negative discriminate,$ k$is constant.prove that:
$ \frac {a^2b^2 \plus{} k^2}{(a \minus{} b)^2} \plus{} \frac {b^2c^2 \plus{} k^2}{(b \minus{} c)^2} \plus{} \frac {c^2a^2 \plus{} k^2}{(c \minus{} a)^2} \ge \frac {3k}{2}$[/quote]
Why it does not hold for all distinct $ a,$ $ b$ and $ c$ $ ?$


Find the mean of all solutions for $ x$ when $ x^3 \plus{} 3x^2 \minus{} 10x \equal{} 0$.

[size=134][b]Method 1 (Find the roots):[/b][/size]

$ x^3 \plus{} 3x^2 \minus{} 10x \equal{} 0$

$ x(x^2 \plus{} 2x \minus{} 10) \equal{} 0$

$ x(x \plus{} 5)(x \minus{} 2) \equal{} 0$

$ x \equal{} 0, \minus{} 5, 2$

The average being the sum (0+-5+2) divided by the number of the roots (3), we can find that quickly as $ \boxed{ \minus{} 1}.$

[size=134][b]Method 2 (Use a clever Vieta's Formula):[/b][/size]

We have a cubic equation of the form-
$ ax^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} 0, d \equal{} 0$

By Vieta's formula, the sum of the roots would be in this case:

$ \dfrac{ \minus{} b}{a} \equal{} \dfrac{ \minus{} 3}{1} \equal{} \minus{} 3$

A cubic can have at most 3 roots if there is no multiplicity. Here, by looking at the equation, I could tell by the coefficients that no root would be the same (formally called multiplicity.) Thus it would be the sum (found above) divided by 3 roots, for an answer of $ \boxed{ \minus{} 1}$

Here, Vieta's formulas proves the more powerful.

do you have a proof for vieta's formula? just curious.

Not really a proof, but expand $ (x\minus{}a)(x\minus{}b)(x\minus{}c)$ for a cubic and the coefficient for the x^2 term is $ \minus{}(a\plus{}b\plus{}c)$.


let $ v$ be positive root of the equation $ t^2\minus{}1998t\minus{}1 \equal{}0$. Define the sequence $ x_0,x_1,...$ by setting

$ x_0, x_{n\plus{}1}\equal{} [v.x_n] . (n \ge 0)$

Find the remainder when $ x_1998$ is divided by $ 1998$

http://www.mathlinks.ro/Forum/viewtopic.php?p=484476#484476
See there

And please be careful, you have posted lots of equal posts, that's not necessary

Don't EVER again post the same eight times :!:  :mad:


Ok, I am going to the MathCounts Nationals as a guest.

I need another person to come with me to lower the costs (or 3 more if you guys want to share double beds)

Here are the costs for the hotel:
1 person: $\$$171.35 each night, $\$$514.05 for the whole stay
2 people: $\$$85.68 each night, $\$$257.03 for the whole stay
4 people: $\$$42.84 each night, $\$$128.52 for the whole stay

get what I mean by I need someone with me now?

[hide="room description"][url]http://marriott.com/reservation/viewRoomPool.mi?marshaCode=DFWDT&roomPoolCode=GENR&roomPoolDescription=Quality%20Room&displayRoomPoolList=Y&roomPoolListSrc=mat[/url][/hide]
[hide]
there's more, but i only think i'm doing the following

Saturday banquet dinner - 50 dollars

Stockyards Station event (includes Friday dinner) - 60 dollars

The registration fee for the National Competition is 35 dollars per guest. (CountDown)

Round trip (50 dollars) (shuttle)

Guests may pre-order the National Competition T-shirt for 15 dollars per shirt.

Thursday buffet dinner - 45 dollars[/hide]

Anybody interested?
Jorian Hoover

If so, please post here and then we can talk about it privately.

I might be - lets see if I make nats first though  :lol:

I don't know if you want a stupid person with you, lol, but I might, depending on whether or not I make it to nats this saturday

Yeah, as more states go on, more people would probably be willing

well, i would, but

1. i might make nats
2. my states are on the 30th

ok, i've been too bored over the last two days so it's seemed like about 30!

this will change soon i hope.

thanks everyone
still apply!

-jorian

[quote="jhredsox"]ok thank you

when is your state competition?

jorian[/quote]

This saturday (The 10th)

what state do you live in?

because I might make it to states...

haha if you get to Nationals by competition don't you have an all expense payed stay?

ya, i would've but i did some silly things

-jorian

[quote="13375P34K43V312"]I'm pretty sure that you cannot room with a national competitor if you are not a relative of the national competitor or are a national competitor yourself. I'm not sure how many other people going who are not national competitors you will find.[/quote]
[quote="frost13"]In fact National competitors must room with another national competitor.  A couple of years ago I had a protective parent that wanted her 6th grader rooming with the parents and Nationals said then he could not compete.[/quote]

Umm.....when did this topic start?

Ok, I think mathmatatics just bailed on me.

Taking new roomies :P !

-jorian

Ok, this post is to move this up in the forum, but mostly to emphasize that I DO need someone to room with. Please volunteer or say you're going but need a roommate.

-jorian

true, i see it as a better experience for those who have a chance next year...but maybe if they don't show it on tv...

Speaking from personal experience, you're wrong. It cost me about 500 dollars last year. If you're sharing a room with others, it'll cost maybe 50 dollars a night, plane tickets you can get for about 189 dollars round trip, and who says you're going touring?

[quote="13375P34K43V312"]I'm not buying the 650 dollars thing...

Hotel room-3 nights: 450 at the least
Round trip Flight: 800 at the least
Food: 150 if you're getting some of your meals from Mathcounts
Touring: 100

1500?[/quote]

flight=97.50 (frequent flier miles, minor fee, processing fee)
hotel=250 (if i get someone to share with)
shuttle=50 (rdtrip)
meals (total)=250

so 650

jorian

I don't think CT takes non-highschoolers for their ARML team. It's hard enough as a 9th or 10th grader underclassman as they take the top 3 upperclasman scores in each local math league even though our local math league this year had zero eligible people in the top three. So instead of taking the ninth grader that finished 3rd, they take 4th, 5th, and 7th places. Underclassman have to take the ARML Runoff and I don't even think you can get into Team A from that, only Teams B and C.  :mad:  We'll have to see though.

Why don't they just take the 45 highest scores, like normal people?

oh I see, frequent flyer miles.

Yeah, sorry, the confusion was my fault.

Sorry :blush:

Jorian

Even w/o frequent flyer, my math teacher got round trip to fort worth (from Indiana) for 189 dollars this year...

Ok, I'm pretty positive I am NOT going (please do not close this topic though). My dad has really crappy reasons for me not to go, but he's hard to convince. :(  :(  :(

So, I can probably now go to MathPath (Yay!)

And I'm sad that I will probably not meet you guys now.
Can you guys tell me if you're going, or where else I could meet you guys?

-Jorian

arml is fun.  it has, depending on where you go, better entertainment.  i went last year as a 7th grader and got 6/8.  i almost guessed a 7th.

ARML is at least as much fun as MC nats, at least if you make it fun.  You can be antisocial and just hole up in your room the whole time and not have any fun, but if you actually make the best of your time, it's a blast.  It's shorter than nats for most teams, though.

how good do you have to be to go to nats with arml in 9th grade (i'm in ct)?

jorian

You should get in touch with the coaches from your team about that.  Some teams allow anyone at all to go to ARML if they want (like SFBA); others require a tryout of some kind.  There may be a specific tryout contest (Chicago does this), or it may be based on local leagues or other contests (Minnesota does this).  Among those that require you to get invited to the team, the selectivity can vary quite a bit.  In any case, you should get in touch with coaches now, and find out if you can go this year, and if not, what you need to do to go next year.

You might also ask on the "How to join an ARML team" thread in the ARML forum.

i'm in 8th grade this year, so i guess i'll contact them

thanks sly si

jorian


Does anyone know of any good internet- or mail-based math contests (i.e. ones that don't require my school to sign up)?  So far, I know about the USAMTS, the Wisconsin Math, Science, and Engineering Talent Search, and the Ole Miss Problem of the Week Contest, but that's about it.

mathlinks has one...first round due this sunday i think...

Can you program? You can do USACO entirely over the internet without your school signing up. It's pretty mathematical.

there are a lot of "Problem of the Week" type contests on the web.  also you can get the them from magazines.

http://matholymp.com/index.html has a monthly math olympiad contest.

I'm a graduate student at the University of Texas and an alum of Mathcounts, the American Mathematics Competitions, and ARML.  I run an online math competition primarily for high school students in Texas, but it is open to any high school student or college undergraduate anywhere who wants to compete.  Check us out at http://www.codybowl.com.

Hi Cody!

I should have asked you more about Codybowl when I saw you at Rice. I just went to your website. It looks cool, but I didn't see any sample problems or a description of exactly how this works. What other contest is Codybowl comparable to? And would team-members have to be physically in the same place to work on it? I'd like to pass along info to the AAH team.

I was trying to figure out how to get involved in your contest, but there didn't seem to be any obvious way to join or an operating FAQ.  Help, anyone?

Last night (after I sent in some answers to the individual round) Cody sent me a document with some information/rules.  Since I can't find the FAQ anywhere on Cody's website, I've posted it (temporarily) [url=http://mathcontests.0catch.com/CodyBowlIIILeagueRules.pdf]here[/url].  I hope Cody doesn't mind - it's only a list of rules.  I'll take it down once he provides us with the official location.

Mathfanatic - the link didn't work. Could you cut and past some info?

It worked for me after a long time.  Try pasting the url directly into Acrobat.  (Go to open file, and paste the url there.)

Oh, I see - you're not allowed to link directly to a PDF; you have to go through a 0catch page.  Okay, try going [url=http://mathcontests.0catch.com/codybowl_rules_link.htm]here[/url] and following the link.

Or, you could just save the file to your hard drive.  I think that is allowed, although I'm not sure.

Hi everyone!  Sorry, I forgot to give you a contact so that you can get in the game.  If you want to play, e-mail me at cpatters@math.utexas.edu.

Thanks.  (And thanks for the PDF, mathfanatic.)


geometry
symmetry
ratio
inequalities
trigonometry
number theory

I thought #5 was a lot of fun, and also quite challenging - it made me think for quite a while before I made it to a solution.

Yeah problem 3! Any geometry problem I can solve is my favorite.

I really liked number 3.  If I hadn't brute forced #5 and rather found a nice solution for it, I would probably like it more.  Otherwise, 1 was okay, 2 was pretty cool, and 4 involved a little bit too much brute force (for my solution at least).

Yeah, my solutions this round were worse than usual and definitely longer than they've been the last two rounds.

#3 was interesting except once you connect the points, it became obvious that the triangles are identical and it's easy to be proven too. #1 seemed a little redundant in all the case analysis. the game in #5 was great, but the solution isn't that ingenious. for #4, we can all use some computer programs and find the solution.

#4: I wrote a calculator program to find the answer.  Then, I decided to work out a solution with mods and casework and got the same solution.

#1: Not much to say.  Basic rules, and easy problem

#2: Straightforward, but too easy to really be interesting

#5: Part a was easy.  I was trying to do a general solution for part b using parity and symmetry, but then I got lazy, so I wrote out all 18 possibilities (well technically, 72 possibilities, but I compacted them) and assigned a "Disorder"  number to each: that is, from one with disorder n, you can get to a configuration of disorder n-1 in 1 move.  Then I showed that the highest possible disorder there was 8.

#3: I liked this one.  My solution was a bit complicated, but meh...just a little bit more than the area ratios one.

[quote="SamE"]1 and 4 were all case analysis. 2 was brute force. 5 was annoying and the solution wasn't all that exciting.[/quote]
I agree 4 boiled down to casework, but I thought there was enough interesting analysis there that the casework didn't bother me.  I thought that the solution for #5 was quite interesting - the math jam didn't do it justice because there wasn't time to extend the given solution very much, and much of a math jam is exploring the problem so that you can motivate a solution; however, if you check out my problems in one of the solution threads (the one I started), you'll see that without delving into any specific cases except the base n = 2 case, you can logically figure out how to find a solution.

I mostly share zanttrang's opinion on 4 and 5.  The casework on #4 was varied enough that it actually became my favorite problem, and #5 was a very good problem... now that I have seen the solution.  Unfortunately for me, I ruined #5 by turning it into a lot of casework... I wasn't creative enough, I guess =p.  numbers 2 and 3 were both fun, intermediate difficulty problems, which I certainly enjoyed without any worries (as opposed to 4 and 5, which I didn't think that I'd manage to solve at first... and in a way, I didn't solve #5 at all... brute force sucks).

And yeah, #1 was very excessive, in my opinion.  It took less than 5 minutes to construct/disprove circles with n = 3 to 6, and n = 7, probably the hardest case on there from a number-theoretical standpoint, was fairly easy to stumble upon by just trying out a few cases.  In any case, #1 could have been "construct disprove n = 7, 8, 9," and stayed roughly equivalent in difficulty.  On the other hand, it's always good to have a few easier portions of problems/problems so as to ensure that people who are unable to solve the more difficult ones walk away motivated to keep improving, instead of simply disappointed with their score.  All in all, I think that the easiest problems in this round were easier than in the other rounds, but the hardest ones were much much harder (Round 2 #5, the 3-D geo problem, probably takes third behind numbers 4 and 5 in this round)

Darn ... someone beat me to it ...

I usually put up this topic =)

JB

well...I can't really say anything about number 5...(brute force case work blah blah blah for 4 pages...)

I like number 4 cause it's interesting and involve number theory and inequality, one thing they didn't mention is that using parity we can deduce that 3 of the roots are even, which eliminates choices.

But I cant believe I miss the AM-GM.....

[quote="pkerichang"]I like number 4 cause it's interesting and involve number theory and inequality, one thing they didn't mention is that using parity we can deduce that 3 of the roots are even, which eliminates choices.[/quote] I mentioned it in the Math Jam, I think. ;)

I loved #2. It may well be the hardest problem I have ever managed to solve. #1 was also cool, #3 was hopeless since my badness at geometry is so extreme, #4 looked simple and doable on first sight but I couldn't get anywhere at all, and I hate, hate, hate #5 because I got so close to solving it completely and then hit a wall. If I solve 5a and get a point due to my blithering in 5b, will that be three points?

#3! First time I've solved a geometry problem at this level all by myself! (in case you're wondering I got a 1 on both the geometry problems of round 1 and 2)

I did use some trig but even so the trig wasn't very messy.

[quote="dts"][quote="K81o7"]#5: Part a was easy.  I was trying to do a general solution for part b using parity and symmetry, but then I got lazy, so I wrote out all 18 possibilities (well technically, 72 possibilities, but I compacted them) and assigned a "Disorder"  number to each: that is, from one with disorder n, you can get to a configuration of disorder n-1 in 1 move.  Then I showed that the highest possible disorder there was 8.[/quote]

I think your mistake was trying to catch all even $n$ for a general case; I'm not sure if it's possible for Lisa to win in, say, the $n=6$ case.  You can, however, induct $n$ over the powers of two (see if you can see how)....

Anyway, I agree with most of you on #1, #2, and #4.  I bet I'm not the only one who used a WLOG argument for #4.

For #3, I could have had a solution about half a page long if only I was brave enough to invoke $\cot A \cot B+\cot A\cot C+\cot B\cot C=1$ (A, B, C angles of a triangle) without proof.[/quote]

Hmm...it seems we looked at #3 the same way.  Mine is a trig solution.
But yeah, for #5, Lisa uses a series of moves to get symmetry or a pattern, and then works from there.


function
quadratics
induction
algebra
algebra proposed

Consider the sequence $a_n$ defined by $a_0=1$ and
\[a_{n+1}=\sum_{k=0}^na_k a_{n-k}.\]

Does there exist $R$ such that
\[a_n\leq R^n\]
for every $n$?

kick me if I'm wrong cause I'm not sure it's correct:

Cauchy's Theorem states that if sequences $(a_n)$ and $(b_n)$ are arbitrarily convergent then
$(\sum_{i=0}^{\infty}a_i)(\sum_{i=0}^{\infty}b_i)=\sum_{i=0}^{\infty}c_i$, where
$c_i=\sum_{j=0}^{i}a_jb_{i-j}$.
Suppose that there exists such $R$ that $a_n < R^n$ for all $n$ ($R>0$).
Let's define sequence $(q_i)$ such that $q_i=\frac{a_i}{(2R)^i}$, so $q_i<\frac{1}{2^i}$ where $(a_i)$ is the sequence from the problem's
text. Thus for all $n$ is $q_n=\frac{1}{2R}\sum _{i=0}^{n}q_iq_{n-i}$.

So from Cauchy's Theorem:
$(\sum_{i=0}^{\infty}q_i)(\sum_{i=0}^{\infty}q_i)=\sum_{i=0}^{\infty}c_i$ where
$c_i=\sum_{j=0}^{i}q_jq_{i-j}$, so $\frac{c_i}{2R}=q_i$, so $c_i=2Rq_i$, so
$\sum_{i=0}^{\infty}c_i=2R\sum_{i=0}^{\infty}q_i$, so if $\sum_{i=0}^{\infty}q_i=T$ then we must have
$T^2=2RT$, so $T=2R$, but $T=\sum_{i=0}^{\infty}q_i<\sum_{i=0}^{\infty}\frac{1}{2^i}=2$,
so having $R>1$ we have a contradiction.

It is the sequence of Catalan numbers, and you are asking whether their generating function has a positive radius of convergence.   The recurrence implies that this generating function is the solution of a quadratic equation, which does not have a double root at $z=0$, so that the square root in the solution is analytic near 0.   So yes, $R$ exists, and in fact $R=2$.

beautifull my mistaken proof is :),

actually the $n$-th $(n \geq 1)$ Catalan number is form $\frac{{2n \choose n}}{n+1}$, simply by inducton
so $R=4$ holds for sure.

[quote="fleeting_guest"] and in fact $R=2$.[/quote]

$R=4$, if I'm not wrong. In any case, your approach by generating functions is wonderful. My proof was by induction, but I had to guess (what now I know to be) the asymptotics for Catalan numbers!

Right, $R=4$.  I confused $n$ and $2n$ in the binomial coefficient, where ${{2n} \choose {n} } \sim 2^{2n}/\sqrt{\pi n}$.
Alternatively, the square root appearing in the generating function is $\sqrt{1-4z}$, so the radius of convergence is $1/R = 1/4$.

if u set $f(X)=\sum_{k=0}^{\infty}a_k X^k$

we will have that $f^2(X)=\frac{f(X)-1}{X}$

let's say that $f(X)=\frac{1+\sqrt(1-4X)}{2}$

if you develop th $\sqrt$ on 0 you ll find your sequence


The equilateral triangle $ABC$ has sides of integer length $N$. The triangle is completely divided (by drawing lines parallel to the sides of the triangle) into equilateral triangular cells of side length $1$. A continuous route is chosen, starting inside the cell with vertex A and always crossing from one cell to another through an edge shared by the two cells. No cell is visited more than once. Find, with proof, the greatest number of cells which can be visited.

From inspection , it is $N^2-N+1$ . If this relationship only holds until the length $k$ , then at length $k+1$ , since the last cell visited should be at the right/left most of the $k$ length , the next cell must be the one below it ( triangle has 3 side and one the common side with the big triangle and another one already been used for the second last cell to enter ) . At the $k+1$ length , its left/right only has one cell while its other side has $2(k+1)-3$ cells (since each row has $2n-1$ cells) . To maximise the number of cell , we have to give up the one cell and choose the $2(k+1)-3$ cells which make the total number of cells visited is now

$k^2-k+1+[2(k+1)-3]+1=k^2+k+1=(k+1)^2-(k+1)+1$ which we see that it the relationship still holds at length $k+1$ . Hence contradict with our assumption and the relationship holds for all $N$ .  :)

Color the unit triangles white and black alternately, then the number of the triangles with these two colors differ N. Since the route can only pass through different color triangles one by one, so the largest route will be no longer than $N^{2}-N+1$.


Let n be a natural number.To proof that,if $ n^5\plus{}n^4\plus{}1$ has six diferent natural divisors,$ n^3\minus{}n\plus{}1$ is a perfect square.

One has $ n^5\plus{}n^4\plus{}1 \equal{} (n^3\minus{}n\plus{}1)(n^2\plus{}n\plus{}1)$. The only numbers that have exactly 6 natural divisors are of the form $ p^5$ or $ p^2q$. As $ \gcd(n^3\minus{}n\plus{}1,n^2\plus{}n\plus{}1) \mid 7$, the only possibility for the first case is for $ p\equal{}7$, with no solutions. For the second case, we cannot have $ n^2\plus{}n\plus{}1 \equal{} p^2$, since then $ n^2 < p^2 < (n\plus{}1)^2$, so we must have $ n^2\plus{}n\plus{}1 \equal{} q$, whence $ n^3\minus{}n\plus{}1 \equal{} p^2$.
Also, we cannot have the two factors sharing $ p$, since that would again imply $ p\equal{}7$, with no solutions.
An example is for $ n\equal{}3$.

It says that $ n^5\plus{}n^4\plus{}1$ has exactly six [b]diferent[/b] natural numbers.It seems that  the solution is wrong.

No?

For example, 32=2^5 has 6 different factors (1,2,4,8,16,32)

Also 2^2x3=12 has 6 factors (1,2,3,4,6,12)

i'm sorry mavropnevma,the solution is right,but i have a question:why $ \gcd(n^{3}\minus{}n\plus{}1,n^{2}\plus{}n\plus{}1)\mid 7$?

You can use the Euclidean algorithm.

The GCD of $ n^3 \minus{} n \plus{} 1$ and $ n^2 \plus{} n \plus{} 1$ is the same as the GCD of $ n^2 \plus{} n \plus{} 1$ and the quotient of $ n^3 \minus{} n \plus{} 1$ and $ n^2 \plus{} n \plus{} 1$, or $ \minus{}n\plus{}2$. The GCD of those two is then $ \minus{}n\plus{}2$ and $ 7$. Now, the only factors of $ 7$ are $ 1$ and $ 7$, so the GCD must divide $ 7$.


4) find all polynom with coeffs a permutation of $[1,...,n]$ and all roots rational

Because all coefficients are positive, so all roots are negative. Suppose that $\sum_{i=0}^{n}a_{i}x^{i}=a_{n}(x+\beta_{1})\dots(x+\beta_{n})$. Beacuse there exists $i$ that $a_{i}=0$, so one of the roots is zero and $a_{0}$ is zero. So $p(x)$ has exactly one zero root. Suppose that $\beta_{n}=0$. Now $(\sum_{i=1}^{n-1}\beta_{i})^{n-1}\geq(n-1)^{n-1}\beta_{1}\dots\beta_{n-1}$. So $a_{n-1}^{n-1}\geq (n-1)^{n-1}a_{1}a_{n}^{n-2}$. Because one of $a_{1},a_{n}$ is at least $2$, then $a_{n-1}\geq n$. So $a_{n-1}=n$. Also $3\geq\left(\frac{n+1}n\right)^{n}\geq a_{1}a_{n}^{n-2}$.
If $n\geq4$ we must have $a_{n}=1$ and all roots are integers so $a_{n-2},a_{n-3}\geq\binom{n-1}{2}\geq n-1$ and this is not possible because $a_{i}$ are different. So for $n\geq4$ there does not exist such polynomial.
If $n=2$ then $x^{2}+2x$ and $2x^{2}+x$ are polynomials satisfying the conditions.
If $n=3$ then $x^{3}+3x^{2}+2x$ and $2x^{3}+3x^{2}+x$ are polynomials satisfying the conditions.

The degree must be $n-1$ (there are $n$ coefficients). Notice also that there cannot be a nonnegative root, so all the roots are negative. The assumption that the polynomial splits over $\mathbb{Q}$ implies it also splits over the integers (Gauss's lemma), so we can write
\[p(x)=\prod_{i=1}^{n-1}(a_ix+b_i)\]
where $a_i, b_i$ are positive integers (the sign comes from the fact that all roots are negative). This implies
\[\frac{n(n+1)}{2}=p(1)=\prod_{i}(a_i+b_i)\geq 2^{n-1}\]
which already implies $n\leq 4$. In addition $n=4$ is impossible: we have $10=p(1)\geq \prod_{i=1}^3 (a_i+b_i)$ which implies all $a_i=b_i=1$. Thus $p(x)=(x+1)^3$ which doesn't satisfy the condition.
It remains to check $n=1,2,3$ which I leave to the reader.


Find all polynomials $P$ with real number coefficients such that if $x+y$ is rational, then $P(x)+P(y)$ is also rational.

set $y=0$ from now on to get:
setting also $x=0$ gives you that $P(0)$ is rational, so everytime when $x$ is rational, also $P(x)$ is rational.
Now some interpolation formula gives you that all coefficients are rational, so this are all such polynomials.

คนไทยรึป่าวเนี่ย อุตส่าถามเป็นภาษไทย ก้อไม่ตอบ


function
real analysis
real analysis unsolved

Define B(s) = 1-(1/3)^s+(1/5)^s-....
B(1) = (pi)/4, B(3) = (PI)^3/32.
Does this trend continue for all odd n? I'd be much obliged if I get a simple yes or no for an answer, for I'd like to try it myself.


$ ABCD$ is a rectangle (see the accompanying diagram) with $ P$ any point on $ \overline{AB}$. $ \overline{PS} \perp \overline{BD}$ and $ \overline{PR} \perp \overline{AC}$. $ \overline{AF} \perp \overline{BD}$ and $ \overline{PQ} \perp \overline{AF}$. Then $ PR \plus{} PS$ is equal to:
[asy]defaultpen(linewidth(.8pt));
unitsize(3cm);

pair D = origin;
pair C = (2,0);
pair B = (2,1);
pair A = (0,1);
pair P = waypoint(B--A,0.2);
pair S = foot(P,D,B);
pair R = foot(P,A,C);
pair F = foot(A,D,B);
pair Q = foot(P,A,F);
pair T = intersectionpoint(P--Q,A--C);
pair X = intersectionpoint(A--C,B--D);

draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(P--S);
draw(A--F);
draw(P--R);
draw(P--Q);

label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$P$",P,N);
label("$S$",S,SE);
label("$T$",T,N);
label("$E$",X,SW+SE);
label("$R$",R,SW);
label("$F$",F,SE);
label("$Q$",Q,SW);[/asy]
$ \textbf{(A)}\ PQ\qquad \textbf{(B)}\ AE\qquad \textbf{(C)}\ PT \plus{} AT\qquad \textbf{(D)}\ AF\qquad \textbf{(E)}\ EF$

$ \begin{cases} \triangle AQT \sim \triangle PRT \\ AT\equal{}PT (\angle TAP\equal{} \angle TPA)  \end{cases} \Rightarrow \triangle AQT \cong \triangle PRT$

So $ PR\equal{}AQ$. Also $ PS\equal{}QF$.

$ \Rightarrow PR\plus{}PS\equal{}AQ\plus{}QF\equal{}AF$. The answer is $ \text{(D)}$

[hide="Cheap way"]

Assume $ P$ is the same point as $ B$, it becomes clear that $ AF\equal{}PR$ in that case.

[/hide]


Can someone tell me some useful things about the isodynamic points of a triangle? :)

Hmm... try searching the web... Well, Clark Kimberling has a [url=http://faculty.evansville.edu/ck6/tcenters/class/isodyn.html]definition with a figure[/url]... then there is a nice [url=http://woopy.woo.org/math/isodyn1.htm]page about them by Peter Woo[/url]... and somewhere else you might find these nice properties: If ABC is a triangle, then the A-Apollonian circle is the locus of all points P such that BP : CP = c : b. And hence, the only points P satisfying the property AP : BP : CP = (1/a) : (1/b) : (1/c) are the points of intersection of these three Apollonian circles, i. e. the two isodynamic points of triangle ABC. Or you could find out that the two isodynamic points are mutually inverse with respect to the circumcircle of triangle ABC; hence, the line joining them passes through the circumcenter O of triangle ABC. This line also passes through the symmedian point K of triangle ABC. The two isodynamic points are the isogonal conjugates of the two Fermat points. Finally, the pedal triangle of a point P with respect to the triangle ABC is equilateral if and only if P is one of the two isodynamic points of triangle ABC. This is already quite a number of results, isn't it?

BTW, I was rather unprecise when I wrote that they're so popular in IMOs. Actually the two most popular geometric themes are isodynamic points and pedal triangles. But this all is interconnected, since the pedal triangles of the isodynamic points are equilateral  :D

  Darij

Do you know where the name "isodynamic" comes from?
The word "Isodynamic" suggests that some forces are equal. Which ones? This must be related to the property of distances to the vertices.
Does someone know a reference?

Thanks!


As some users in this forum (stergiu, Ahiles and me) are trying to gather JBMO Shortlists, maybe they could be added in the Resources section?

Sure, have them PM me so I can give them the details about how to do that (I think Sergiu already knows.) Basically I need you to post the problems on the Forums first (each in its own topic.)

Sure :)
As for now, we have 2 shortlists, gathering more.


A circle is inscribed in a triangle ABC. It touches AB at D. AD=5cm, DB=3cm and angle A = 60 degree. Then side BC = _____cm?

Let the circle touches AC at E, BC at F. Then AE=5,BF=3, Let CF=CE=x.
Then from the cosine theorem, (3+x)^2=8^2+(5+x)^2-2*8*(5+x)*\cos 60. x=10, BC=13.


analytic geometry
projective geometry
geometry
trigonometry
function
geometry unsolved

Dear Luis and Mathlinkers,
it seem that the Vecten's figure is inexhaustible..
Let ABC be a triangle,
CB'B"A, AC'C"B, BA'A"C three squares erected externally on the sides of ABC,
and X, Y respectively the points of intersection of AA' and CC', AC and BB".
Prove that XY is parallel to BB'.
Sincerely
Jean-Louis

We use barycentric coordinates WRT $\triangle ABC,$ keeping in mind Conway's notation

$ B'' \ (S_C \plus{} S: \minus{} b^2: S_A) \ , \ C' \ (S_B \plus{} S: S_A: \minus{} c^2)$

$ A' \ ( \minus{} a^2: S_C \plus{} S: S_B) \ , \ B' \ (S_C: \minus{} b^2: S_A \plus{} S)$

$ AA' \equiv S_By \minus{} (S_C \plus{} S)z \equal{} 0 \ , \ CC' \equiv (S_B \plus{} S)y \minus{} S_Ax \equal{} 0$

$ BB'' \equiv S_Ax \minus{} (S_C \plus{} S)z \equal{} 0 \ , \ BB' \equiv (S_A \plus{} S)z \minus{} S_Cz \equal{} 0$

Infinite point of $BB''$ is then $ T_{\infty}(S_C: S \plus{} S_A \plus{} S_C: S_A \plus{} S)$

$X \equiv CC' \cap AA' \equiv ((S_C \plus{} S)(S_B \plus{} S): S_A(S_C \plus{} S): S_BS_A)$

$Y \equiv AC \cap BB'' \equiv (S_C \plus{} S: 0: S_A)$

$\Longrightarrow \ XY \equiv \minus{} S_Ax \plus{} Sy \plus{} (S_C \plus{} S)z \equal{} 0$

Infinite point of $XY$ is $ P_{\infty}(S_C: S \plus{} S_A \plus{} S_C: S_A \plus{} S) \Longrightarrow \  T_{\infty} \equiv{} P_{\infty}.$

Dear Luis and Mathlinkers,
I found a synthetical proof based on Desargues and Pappus that I will write next.
Sincerely
Jean-Louis

Dear jayme，this problem is very easy .
Actually as long as we draw a complete map，use the methods we already have，maybe we can Infer many conclusions.
We can solved it with several methods and deal with different ways ，you know.
Next，I use area and sine function what is I often use to solved：

$ AR/A'R \equal{} AC/A'C*sin$∠$ ACC'/sin$∠$ DCA'$.......$ (1)$.

$ AQ'/A''Q' \equal{} AB/A''B*sin$∠$ ABB''/sin$∠$ DBA''$.......$ (2)$.

$ (1)/(2) \equal{} >$

$ AR/A'R*A''Q'/AQ' \equal{} AC/AB*sin$∠$ ACC'/sin$∠$ ABB''$

$ \equal{} AC/AB*sin$∠$ ACC'/sin$∠$ AC'C \equal{} AC/AB*AB/AC \equal{} 1$.

So，$ AR/A'R \equal{} AQ'/A''Q'$.

We also can think aboat Macro approach，such as “$ Desargues theorem$”.

Dear Lym and Mathlinkers,
Luis is right, but now if you applied the Pappus's theorem, you will find my result. Nice you have found the beginning.
Sincerely
Jean-Louis

Sorry，Dear jayme，I don't know the Pappus's theorem. Can you introduce to me？
I will use different methods to solve it.

I didn't found a good way to prove it，
I just use Menelaus theorem to solve it(easy)，and I have a Desargues's method，but is not very simple.
If anybody need，I can put it.

So，I think useing Pappus's theorem is very ingenious，I look forward to jayme's method，

I will tell you Pappus'Theorem.
If you have 2 lines r and s, and choose 3 points in each: A, B and C at r and D, E and F at s in that order, then the intersections of AE and BD, AF and CD and BF and  CE are collinears.

Regards!

The [b][size=100]Pappos theorem[/size][/b] is also true, for any order of the triads of arbitrary points, considered on two given lines.

So, if $ A,\ B,\ C,$ are three arbitray points on the line $ (e)$ in any order and $ A',\ B',\ C',$ are three arbitrary also points on the line $ (e')$ in any order , the [b][size=100]Pappos theorem[/size][/b] states that the points $ X\equiv AB'\cap A'B$ and $ Y\equiv AC'\cap A'C$ and $ Z\equiv BC'\cap B'C,$ are collinear.

Kostas Vittas.

Dear Mathlinkers,
an article concerning the Vecten's figure and its developpement can be seeing on my site
http://perso.orange.fr/jl.ayme , la figure de Vecten  vol. 5 p. 42
Sincerely
Jean-Louis


Let $p, q$ be two distinct primes and $n,m\in \mathbb{N}$ be the smallest natural numbers such that $r=mp+1$ and $s=nq+1$ are primes. Is it true that we can never have $r=s$?

Let $f(p)$ be the smallest positive integer $n$ such that $np+1$ is prime.
It seems plausible that $f(p)\leq p$ for all prime $p$ and that gives affirmative answer to the original problem.
Indeed, if $r=s$ then $mp=nq$ implying $p\leq n=f(q)$ and $q\leq m=f(p)$. Without loss of generality we assume $p<q$. Then $f(p)\geq q>p$, a contradiction.

I do not have a proof for ineqality $f(p)\leq p$ but it is true for all prime $p<10^5$.


$ 2(3^x) \equal{} 162 \Rightarrow 3^x\equal{}81 \Rightarrow x\equal{}4$


geometry
quadratics
trapezoid
analytic geometry
rectangle
algebra
quadratic formula

[hide]First, we can find the area of this pentagon. We can use the shoelace method to find the area and it's equal to $\displaystyle \frac{80}{2}=40$ So, since the area of pentagon $ABCDE$ is $40$ we have to divide it into two equal polygons of an area of $20$. First, we can divide this pentagon into two quadrilaterals by letting point $(6,0)$ be $F$. We see that the polygon $FBCD$ has an area of $10$, so the line $x=k$ must lie in the interval $(0,6)$.

Next, we can find the equation of the line from $D$ to $E$. After calculation, we see that this line is $y=-x+8$. Now, we can observe that when the line $x=k$ is drawn, it will cut the polygon at two points, namely $(x,0)$ and $(x,-x+8)$. Now, we can also use points $A$ and $E$ to calculate the value of $x$. Using the shoelace method and setting the area equal to $20$, we see that $|-x^2+16x|=40$. So, we see that the possible values of $x$ are found when $-x^2+16x=40$ and when $-x^2+16x=-40$. But, we can see that the former equation yields imaginary roots and we can throw that out. Using the quadratic formula on the second equation we see that the roots are $x=8\pm 2\sqrt{6}$. We know that $x$ has to be in the interval $(0,6)$ so we can throw out $x=8+2\sqrt{6}$. Thus, $k=8-2\sqrt{6}$ and $(a,b)$ is $\boxed{(8,-2)}$[/hide]

[hide]The area of trapezoid $AFDE$, where $F$ has coordinates $(6,0)$ has area 30, while the rectangle $FBCD$ has area $10$.  Thus, we need $k$ to be such that it cuts the trapezoid into two trapezoids of areas 10 and 20.  The area of the trapezoid which should have area 20 can be expressed as \[ \frac{(8+[8-k])k}{2} \] because $k$ is the "height" while the two bases have length 8 and 8-k.  Solving for k, \[ k(16-k)=40 \implies k^2-16k+40=0 \] \[ \implies k=\frac{16 \pm \sqrt{256-160}}{2}=8\pm2\sqrt{6} \]  However, $k$ canot be greater than 6, or it would not cut the trapezoid, but rather, the rectangle.  Thus, $k=8-2\sqrt{6}$, so our pair is $\boxed{(8, -2)}$[/hide]

the shoelace method is a way to find the area of a polygon when you're given the vertices in the form $(a,b)$. Now, to find the area of that pentagon for example, we have points $A(0,0), B(11,0), C(11,2), D(6,2)$ and $E(0,8)$. We form a column thing, like the following:

0   0
11 0
11 2
6   2
0   8
0   0

Note that you have the same point at the top and the bottom of the "shoe''

now, we can "tie" the shoe by cross multiplication like the following:

-    0   0    -
0   11 0    0
0   11 2    22
12 6   2    22
0   0   8    48
0   0   0    0

See that we're multiplying the terms diagonally. Now, we add the terms in the leftmost column, which is equal to $12$. and then we add the terms in the rightmost column, which is $92$. We then take the absolute value of their difference, $80$, then divide that value by $2$ and we get $40$, which is the area of the pentagon  :D . hope that's clear enough

thanks a lot man. really appreciate it but does it work for any polygon?

yeah, as long as you put the coordinates in order  :D

in order of.....wat if ur just given some random polygon with coordinates on a graph

you'll have to plot the points so you can see which points are next to each other. for example, if you were given the coordinates $(0,0), (1,1),(0,1), and (1,0)$ and you were asked to find the area of the square, you cant just put those coordinates in the way they're listed. You have to go "around" the polygon when you're plugging in in the coordinates. So, you'd have to list them like $(0,0),(0,1),(1,1), and (1,0)$ if you want to get the right area. Just plot the points and pick a point, list that one, and work around the polygon, that's what i do. Ask again if it's still not clear

and u use (0,0) all the time for beginning and end depsite the coordinates which are given? For example (2,3), (4,3), (4,6), and (2,6)

no, you use the one you started with. like, for you example, if you choose the point (2,3) to begin with, the column thing you come up with could be

2  3
4  3
4  6
2  6
2  3

Okay, so just to clarify (i've never seen this method before, looks extremely useful):
If you had, say, a hexagon with coordinates (0,1), (5,1), (6,10), (4,13), (2,11), (1,2),
Then you'd do
0  1
5  1
6  10
4  13
2  11
1  2
0  1

Then,
5    0
6    50
40  78
26  44
11  4
0    1

Then the two numbers you get are 177 and 88, so you get the area is half of 89, or 44.5?


quadratics
ratio
algebra
quadratic formula

From destination A a bicycle left to point B. At the same time a motorcycle left from point B to point A. When the motorcycle had traversed 1/3 of the way, the bicycle still had 26km to go. When the bicycle had traversed 1/3 of the way, the motorcycle had 5km to go.Ffind the distance between the 2 points.

[hide="Hm, I got..."] 30 [/hide] but I did it by guessing and checking. I did set up a whole bunch of equations, but after working on them for like an hour I just gave up. Then I tried guessing and got the answer in 2 minutes :)

[hide="Solution"]
Call $d$ the distance, $B$ the speed of the bicycle and $M$ the speed of the motorcycle.  Then

$\frac{d}{3M} = \frac{d-26}{B}$
$\frac{d}{3B} = \frac{d-5}{M}$

For simplicity, we set $B = 1$.  This gives us, after cross-multiplying,

$d = 3M(d-26)$
$3(d-5) = dM$
$M = 3\frac{d-5}{d}$
$d = 9\frac{(d-5)(d-26)}{d}$
$d^2 = 9d^2 - 9(31)d + 9(130)$
$0 = 8d^2 - 9(31)d + 9(130)$

I'm too lazy to guess-and-check solutions, so quadratic formula ( :P ) gives the answer $d = \boxed{30}$.

Melissa, I don't see how you got 39 by guess-and-check.  This gives the ratio of the speeds as 1 for the first case, but $\frac{34}{13}$ for the second case.  ...?[/hide]


inequalities
calculus
derivative
integration
parameterization
geometry
real analysis

maybe an edit is necessary since the image is not uploaded and we can't prove a red cross image!!!

right. you can upload files now pideltaphi, that's why I put that option, for not proving red-crosses :D:D

You cant prove red cross problems?You disappoint me  :?

Maybe it is not suitable for the beginners corner?

indeed. I must make some kind of notice. Anything that implies use of derivatives, integrals, groups, rings, field, parameter surfaces, etc. is going to the college playground. including this :D

f=(x^4-1)^(1/4) and g=(x^4+1)^(1/4) are inverse of each other so we can think of some Young inequality like problem (http://mathworld.wolfram.com/YoungsInequality.html).

So I1=int(f,0..3), I2=int(g,1..3)

It would be easier with a drawing but : take the symetric of f ./. y=x, the area above f^1=g is the same as I1. So I1+I2 contains entirely the square 3x3 : I1+I2 > 9

There is a little bit outside the square which is easy to majorize : it's smaller than the square (g(3)-3)x(g(3)-3) = .0000849...

So 9 < I1+I2 < 9.00009  :D


Find the coefficient of  in the expansion of

           \[ \left(x\plus{}\frac{1}{x}\right)^{6} \]

hello, we have $ \left(x\plus{}\frac{1}{x}\right)^6\equal{}20\plus{}\frac{1}{x^6}\plus{}\frac{6}{x^4}\plus{}\frac{15}{x^2}\plus{}15x^2\plus{}6x^4\plus{}x^6$.
Sonnhard.

Or, instead of multiplying out, we see that, using the binomial theorem, the combination of the two terms that produce a constant is:

$ \dbinom{6}{3}x^3\times \left(\dfrac{1}{x} \right)^3\equal{}20$

To learn more about the Binomial theorem, click [url=http://en.wikipedia.org/wiki/Binomial_theorem]here[/url].


Let a,b,c are possitive numbers in [0;1]
Prove: \[ \frac{a}{{b^3  \plus{} c^3  \plus{} 7}} \plus{} \frac{b}{{c^3  \plus{} a^3  \plus{} 7}} \plus{} \frac{c}{{a^3  \plus{} b^3  \plus{} 7}} \le \frac{1}{3}
\]

Should this be moved to the preolympiad forum?

i am going to guess AM-GM. but thats just me.


Well, Walton is [i]not[/i] going because Furman is on that day. State Mathcounts is also on that day... and ARML Local was also planned for that day. GAH TOO MUCH MATH TO DO NOT ENOUGH TIME

W00T alabama FTW!!!
the alabama pizitz team is goign to PWN!!!!
and cullman will be going ;]

Lassiter.  I take it you are from Pizitz in Alabama?

By the way, do you know whether Vestavia is coming to GACS or going to Furman?

Vestavia's not going to GACS but I'm not to sure about them going to Furman... pizitz will pwn :D

[quote="chestnut"]... pizitz will pwn :D[/quote]

To bad Dickerson isn't going! And yes, Piztiz probably will pwn. All the good Georgia teams will be doing MATHCOUNTS on Saturday.

and/or Furman.

EDIT: Actually, it sounds like you're talking about middle schools. Are you talking about middle schools?

Yes Allen, I'm talking about middle schools. Thus the MATHCOUNTS reference!  :)

I'm going, but I don't know who else from Northview's going...>.<

Gluck to all who are State MC-ing! :D

Just to add two more questions (Sorry, I don't have the answer to yours, Boru):

Are the individual scores calculated only by the written score? This seems to be the case of last year's competition.

Is there a website for the competition? I could only find an old website with past results and another one for 2006's competition.

[quote="llolloll"]Hi all,

My name is Boru Wang and I go to Parkview High School. Does anyone know if individuals are allowed to register onsite and compete tomorrow? And if so, what are the costs?

Thanks  :)
Boru[/quote]

I emailed the tournament director Angelique Allen: aallen@greateratlantachristian.org earlier this week.

I was told that individual registration was 5 dollars each.

Hi, Oliver and Boru, Website for the GACS

[url]http://tiger.greateratlantachristian.org/sites/dir/teams/vmathteam/gacstourn/default.aspx[/url]

Huh.. very long to type... Good Luck to all :lol:

Can anyone post the results for those who didn't go? Thanks.

I don't remember the JV results :blush:
But for Varsity I think:
Santhosh Karnik  :first:
Miles                 Second
Sitan       Third
Oliver       Fourth
Eric Morphis        Fifth

I think I probably switched the fourth and fifth places. :lol:

[quote="azn6021023"]I am highly disappointed that I was not considered for memory in the top 10, especially since I was 6th (one after your gap).

Of course, that's because Eric beat me by like 20 points.[/quote]

Sorry.  :blush: But usually everyone posts the top five ranks for a tournament. Anyway Congrats for your sixth place.

Hi folks. Congrats to all that beat Eric (and others also of course, but beating eric is notable! :-)

From Furman:
Claus was 6th over all losing a tiebreak for 5th.
Henry, Allen, Maja, and I think Billy were all Honorable Mention.

Walton was 3rd plane team behind AC Flora (1st) and NC School of Science and Math (2nd). Alpharetta was 4th. Augusta Area Homeschoolers (guess who was on that team) was 2nd in Division II.

Take care,
Tom

Good job at Furman!

And good job to everybody at Walton.

Yes, I was also an HM. I ... somehow won the most butchered name award: Bijjy Oormeni. Seemingly I bubbled 'L's as 'J's, but my last name got butchered without my help.

(Also, yes, Maja [i]was[/i] there, but her name was pronounced decently properly - just -ska instead of -ski.)

At Gacs, Chestnut won first place individual in algebra and i won third place individual XD


Why in the world is the notation $ \left(\frac{\partial T}{\partial V}\right)_S$ so ubiqutous in thermodynamics when it means exactly the same thing as $ \left(\frac{\partial T}{\partial V}\right)$, that is the partial of T with respect to V. The definition of a partial is that all the other variables are held constant! Is the problem just that physicists need a constant reminder of what a partial derivative is or what!?!

In the first notation we have the information that the state function T is being considered a function of S and V, while in the second notation we don't know what are the other independent variables of T (besides the volume), unless the context makes it obvious. This is important since, in general,

$ \left(\frac{\partial Y}{\partial Z}\right)_X \neq \left(\frac{\partial Y}{\partial Z}\right)_W$,

where W, X, Y, Z are state functions.

That kind of makes sense. I wish my thermodynamics textbook (Kittel) would give more examples of that and not just gloss over that notation because it is extremely important.


Let $ x,y\geq 0$, s.t. $ x^2 \plus{} y^3\geq x^3 \plus{} y^4$. Prove that:
$ x^3 \plus{} y^3\leq 2$

[hide="Hint"]Look at the title[/hide]

We have \[ x^2\minus{}x^3\ge y^4\minus{}y^3\ge y^3\minus{}y^2\] so\[ x^2\plus{}y^2\ge x^3\plus{}y^3\]But follwing CBS and Cebasev inequalities \[ \frac{1}{2}(x\plus{}y)(x^2\plus{}y^2)\le x^3\plus{}y^3\le x^2\plus{}y^2\le \sqrt{(x\plus{}y)(x^3\plus{}y^3)}\]we have\[ x^3\plus{}y^3\le x\plus{}y\le 2\]q.e.d.

Nice proof, Marius Mainea.
Here is another solution
[hide="Solution"]From CS we get: $ (x\plus{}y^2)(x^3\plus{}y^4)\geq (x^2\plus{}y^3)^2$ so $ x\plus{}y^2\geq x^2\plus{}y^3$
From AM-GM we get $ 1\plus{}x^2\plus{}1\plus{}y^4\geq 2(x\plus{}y^2)\geq (x^2\plus{}y^3)\plus{}(x^3\plus{}y^4)\iff 2\geq x^3\plus{}y^3$ [/hide]

[quote="Inequalities Master"]Let $ x,y\geq 0$, s.t. $ x^2 \plus{} y^3\geq x^3 \plus{} y^4$. Prove that:
$ x^3 \plus{} y^3\leq 2$
[/quote]
The following inequality is true too.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x^2 \plus{} y^3\geq x^3 \plus{} y^4$. Prove that:
\[ x^5\plus{}y^5\leq2\]

[quote="Inequalities Master"]OK, arqady.Please post the solution.[/quote]
Maybe somebody wants to solve it. :wink:

[quote="arqady"][quote="Inequalities Master"]Let $ x,y\geq 0$, s.t. $ x^2 \plus{} y^3\geq x^3 \plus{} y^4$. Prove that:
$ x^3 \plus{} y^3\leq 2$
[/quote]
The following inequality is true too.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x^2 \plus{} y^3\geq x^3 \plus{} y^4$. Prove that:
\[ x^5 \plus{} y^5\leq2
\]
[/quote]
Here is the solution I found for these types. Denote $ x \equal{} ky$, where $ k > 0$. Then from the given condition, we get
\[ k^2 \plus{} y \ge k^3 y \plus{} y^2
\]

\[ \Leftrightarrow y^2 \plus{} (k^3 \minus{} 1)y \minus{} k^2 \le 0
\]

\[ \Leftrightarrow y \le \frac { 1 \minus{} k^3 \plus{} \sqrt {k^6 \minus{} 2k^3 \plus{} 4k^2 \plus{} 1}}{2}
\]
Hence
\[ x^5 \plus{} y^5 \equal{} (k^5 \plus{} 1)y^5 \le (k^5 \plus{} 1) \left( \frac { 1 \minus{} k^3 \plus{} \sqrt {k^6 \minus{} 2k^3 \plus{} 4k^2 \plus{} 1}}{2}\right)^5
\]
It suffices to prove that
\[ (k^5 \plus{} 1) \left( \frac { 1 \minus{} k^3 \plus{} \sqrt {k^6 \minus{} 2k^3 \plus{} 4k^2 \plus{} 1}}{2}\right)^5 \le 2
\]
The rest is not hard and you can do it by yourself. :)


a)Find out the general term of the sequence defined like so:

X is in M_k(R), k natural<>0. We define X_0=I_k, X_1=I_k, X_(n+2)=X_(n+1)+X_n, for any n natural.

b)Find out the general term of the sequence defined like so:

X,A,B is in M_k(R), k natural<>0. We define X_0=O_k, X_1=I_k, X_(n+2)=AX_(n+1)+BX_n, for any n natural.
cheers!!

For (a) (I just wanted to make an extra post  :D ). X_n is, of course, F_n*I_k.


trigonometry
geometry
Triangle
Law of Sines
IMO Shortlist
IMO Longlist

In a triangle $ ABC,$ let $ D$ and $ E$ be the intersections of the bisectors of $ \angle ABC$ and $ \angle ACB$ with the sides $ AC,AB,$ respectively. Determine the angles $ \angle A,\angle B, \angle C$ if $ \angle BDE \equal{} 24 ^{\circ},$ $ \angle CED \equal{} 18 ^{\circ}.$

Let the intersection of $ CE$ and $ BD$ be $ I$, the incenter.  We have that the altitudes from $ I$ to $ AB$ and $ AC$ are equal.  That is,

$ IE\sin BEC \equal{} ID\sin CDB$

From $ BEC \equal{} 180 \minus{} C \minus{} \frac {1}{2}B$, $ CDB \equal{} 180 \minus{} B \minus{} \frac {1}{2}C$, and $ \frac {IE}{ID} \equal{} \frac {\sin 24}{\sin 18}$ (Law of Sines), we get

$ \frac {\sin 24}{\sin 18} \equal{} \frac {\sin(180 \minus{} B \minus{} \frac {1}{2}C)}{\sin(180 \minus{} C \minus{} \frac {1}{2}B)} \equal{} \frac {\sin(B \plus{} \frac {1}{2}C)}{\sin(C \plus{} \frac {1}{2}B)}$

Furthermore, by analyzing triangle $ BCI$, we have $ \frac {1}{2}B \plus{} \frac {1}{2}C \equal{} 180 \minus{} BIC \equal{} 18 \plus{} 24 \equal{} 42$.  It follows that both $ C \plus{} \frac {1}{2}B$ and $ B \plus{} \frac {1}{2}C$ are acute.  One solution to the equations

$ \frac {\sin 24}{\sin 18} \equal{} \frac {\sin(B \plus{} \frac {1}{2}C)}{\sin(C \plus{} \frac {1}{2}B)}$

$ \frac {1}{2}B \plus{} \frac {1}{2}C \equal{} 42$

is (and can be checked to be) $ B \equal{} 72,C \equal{} 12$, using $ \sin 18 \equal{} \frac {\sqrt {5} \minus{} 1}{4}$.  Suppose there was another solution, and without loss of generality suppose $ B \plus{} \frac {1}{2}C > 78$ then $ C \plus{} \frac {1}{2}B$ would have to be greater than $ 12$.  It follows that $ \frac {1}{2}B \plus{} \frac {1}{2}C > 42$, a contradiction.

Does anybody have a solution more geometric than this?  I would love to see one.

hi

my ans is
angleA = 96
angleB = 12
angleC = 72

(in degrees) :D

Is it correct? :huh:

Sorry for reviving a ten year old topic..
Let $T$ be the intersection of $BD$ and $CE$.
$\angle TCB + \angle TBC = 42^{\circ}$
 so $\angle BAC = 96^{\circ}$
Let $M$ be the reflection of $D$ across the line $CE$ and let $N$ be the reflection of $E$ across $BD$. Since $BD$ and $CE$ are bisectors of $\angle C$ and $\angle B$, it follows that the points $M$ and $N$ are located on the line $BC$. Let the line $BD$ meet $EM$ at $R$.
$\angle ERB = \angle RED + \angle EDR = 60^{\circ}$
$\angle BRN = 60^{\circ}$,$\angle MRN = 60^{\circ}$
and $\angle NDM = 24^{\circ}= \angle RDN.$
Now $N$ is the excenter of the triangle $ RDM$ and
$\angle DMC= \angle RMN = \frac{\angle DMC+\angle RMN}{2}= 54^{\circ}$
So $\angle A=96^{\circ}$,$\angle B=12^{\circ}$, and $\angle C=72^{\circ}$


trigonometry
complex numbers
calculus
calculus computations

show :D

$ \sum_{k\equal{}0}^{\infty}\;(\minus{}1)^{k}\cdot\dbinom{s}{2k}\;\;\equal{}\;\;\boxed{2^{s/2}\cdot\cos\left(\frac{\pi s}{4}\right)}$

and

$ \sum_{k\equal{}0}^{\infty}\;(\minus{}1)^{k}\cdot\dbinom{s}{2k\plus{}1}\;\;\equal{}\;\;\boxed{2^{s/2}\cdot\sin\left(\frac{\pi s}{4}\right)}$

In the binomial series $ \sum_{k\equal{}0}^{\infty}{s\choose k}z^{k}\equal{}(1\plus{}z)^{s}$ set $ z\equal{}i$ then

$ \sum_{k\equal{}0}^{\infty}{s\choose 2k}(\minus{}1)^{k}\plus{}i\sum_{k\equal{}0}^{\infty}{s\choose 2k\plus{}1}(\minus{}1)^{k}\equal{}(1\plus{}i)^{s}\equal{}\left(\sqrt{2}\, e^{\frac{i\pi}{4}}\right)^{s}$

After comparing real and imaginary part

$ 2^{s/2}\cos\frac{s\pi}{4}\equal{}\sum_{k\equal{}0}^{\infty}(\minus{}1)^{k}\frac{(s)_{2k}}{(2k)!}$ and $ 2^{s/2}\sin\frac{s\pi}{4}\equal{}\sum_{k\equal{}0}^{\infty}(\minus{}1)^{k}\frac{(s)_{2k\plus{}1}}{(2k\plus{}1)!}$