There are two distinguishable flagpoles, and there are $ 19$ flags, of which $ 10$ are identical blue flags, and $ 9$ are identical green flags. Let $ N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when $ N$ is divided by $ 1000$.

[hide="Solution. Thought this problem was easier than an average number 12."]
We have two cases: either there are blue flags on one of the poles or both poles. When there are $ k$ blue flags on one pole, green flags can be placed with at most one green flag between each pair of blue poles and thus there are $ k\plus{}1$ spaces to place all the green flags.

In the first case, we have one green flag on one of the poles and everything else on the other pole. There are $ 10$ blue flags and $ 8$ green flags on the other pole, so there are $ 11$ spaces between blue flags and thus $ \binom{11}{8}$ ways to place the green flags after placing the blue flags. The poles are distinguishable, so there are $ 2$ ways to have a green flag on one pole and thus $ 2\binom{11}{8}$ ways to place the flags in this case.

In the second case, we have $ n$ flags on the first pole and $ 10\minus{}n$ on the second pole. As $ n$ ranges from $ 1$ to $ 9$ we have every possible distinguishable arrangement of blue flags, which totals to $ 9$ arrangements. On the first pole there are $ n\plus{}1$ places to place green flags and on the second there are $ 11\minus{}n$ places, so we have $ 12$ places to place $ 9$ green flags. Therefore, there are $ 9\binom{12}{9}$ ways to place the flags in this case.

Between the cases, there are $ 2\binom{11}{8}\plus{}9\binom{12}{9}\equal{}330\plus{}1980\equal{}2310$ and $ 2310\equiv\boxed{310}\mod{1000}$.
[/hide]

I have a pretty cool way to think about this one (I think)
[hide]
imagine that the poles are concatenated with pole 1 on the bottom and pole 2 on top.  Then we can choose a break a point in this "long pole" for every arrangement of flags to determine one distinguishable arrangement.  However, we have to consider 2 cases: 1, when the cut point is between a blue and a green, and 2, the point's between 2 greens.  for case 1, there are a total of $ \binom{11}{9}$ (balls and urns) flag arrangement on the "long pole" and a total of 18 cut points, so for case 1 there's $ 18\binom{11}{9} \equal{} 990$ arrangements.  For case 2, imagine that we combine 2 green flags into one. then there are $ \binom{11}{8}$ flag arrangements; now to make the 2 adjacent flags "pop up" again, we can choose any one of the 8 green flags to turn into the adjacent pair, so there's a total of $ 8\binom{11}{8}\equal{}1320$ for case 2.  Then the total is $ 1320\plus{}990\equal{}2310$ and the remainder then is $ 310$.[/hide]

[hide="I liked this problem, Generalized Solution:"]Consider a (a+1)+b element ordered string of a+1 blues and b greens such that no greens are consecutive, which is commonly known to be $ {(a \plus{} 1) \plus{} b \minus{} (b \minus{} 1)\choose b} \equal{} {a \plus{} 2\choose b}$ such strings.

For each of those strings, pick one of the (a+1) blues, everything before it in order from left to right is the order of flags on the first pole, everything after it is on the second pole - the extra blue is just a place holder so that we can account for the no consecutive green rule. Thus, we now have $ (a \plus{} 1){a \plus{} 2\choose b}$ such strings.

However, we have overcounted the strings where there are no flags on either pole, ie. when the placeholder blue that we pick is at the either end of the string. Thus given the two ways to place the placeholder blue on an end, there are then a more blues and b greens to order such that no greens are consecutive: $ 2{a \plus{} 1\choose b}$ such strings.

Thus the total for generalized a blue, b green is $ (a \plus{} 1){a \plus{} 2\choose b} \minus{} 2{a \plus{} 1\choose b}$, which is $ \boxed{2310}$ for 10 blue and 9 green.[/hide]

Does anyone have a recursive solution to this problem?

I don't see why you'd need one as it's relatively straightforward without recursion.

You could probably come up with a dual recursion based on the number of blue and green flags.

[quote="worthawholebean"]In the second case, we have $ n$ flags on the first pole and $ 10 \minus{} n$ on the second pole. As $ n$ ranges from $ 1$ to $ 9$ we have every possible distinguishable arrangement of blue flags, which totals to $ 9$ arrangements. [b]On the first pole there are $ n \plus{} 1$ places to place green flags and on the second there are $ 11 \minus{} n$ places, so we have $ 12$ places to place $ 9$ green flags.[/b] Therefore, there are $ 9\binom{12}{9}$ ways to place the flags in this case.[/quote]

I don't get this part of the solution, particularly the bolded sentence.  Could someone please explain how it was gotten?  Also, I'm going to assume that the first sentence means $ n$ *blue* flags.  Thanks!

After placing the blue flags, green flags can be placed in each space between blue flags - so if there are $ n$ blue flags, there are $ n\plus{}1$ spaces to place green flags. On the other pole, there are $ 10\minus{}n$ blue flags and therefore $ 11\minus{}n$ places for green flags. This will always total $ n\plus{}1\plus{}11\minus{}n\equal{}12$ spaces for green flags; we have $ 9$ green flags and thus $ \binom{12}{9}$ ways to place them.

[quote="worthawholebean"]After placing the blue flags, green flags can be placed in each space between blue flags - so if there are $ n$ blue flags, there are $ n \plus{} 1$ spaces to place green flags. On the other pole, there are $ 10 \minus{} n$ blue flags and therefore $ 11 \minus{} n$ places for green flags. This will always total $ n \plus{} 1 \plus{} 11 \minus{} n \equal{} 12$ spaces for green flags; we have $ 9$ green flags and thus $ \binom{12}{9}$ ways to place them.[/quote]

Oh, I get it now.  Thanks a lot!  Just wondering, how would I think of this method of representing the flags?

Sorry to revive this (is this considered bad forum-etiquette here?) but isn't this just a ball-and-urn problem?

[hide]
For example, if it's divided 4 green on one pole and 5 on the other:

■G■G■G■G■ | ■G■G■G■G■G■

7 of the blues must go in the inner urns. Then there are 3 more blues, and 11 urns, so it's  13C3 = 286. The divider can occur in 8 places, not including the case in which there are no greens on one flagpole, so 8*268 = 2288. Then it is easy to find that there are 22 ways to divide it with all the greens on one flagpole, for a total of 2310 arrangements. 2310 (mod 10) = 310.
[/hide]

[quote="pianoforte"]Consider a (a+1)+b element ordered string of a+1 blues and b greens such that no greens are consecutive, which is commonly known to be $ {(a \plus{} 1) \plus{} b \minus{} (b \minus{} 1)\choose b} \equal{} {a \plus{} 2\choose b}$ such strings.

[/quote]

I don't understand this. Can someone prove it to me?

The idea to this problem is; if n blue balls are put in a flagpole, then 10-n blue balls are put in the other flagpole, where n ranges from from 0-9

[quote=amuthup]Can someone explain what is wrong with my [hide=solution]We use balls and urns for each distribution of green flags:
$0$ on left pole, $9$ on right: $\binom{1}{0}*\binom{11}{9}+\binom{2}{0}*\binom{10}{9}+\binom{3}{0}*\binom{9}{9}+....+\binom{11}{0}*\binom{1}{9}$

$1$ on left pole, $8$ on right: $\binom{1}{1}*\binom{11}{8}+\binom{2}{1}*\binom{10}{8}+\binom{3}{1}*\binom{9}{8}+....+\binom{11}{1}*\binom{1}{8}$
.
.
.

$9$ on left pole, $0$ on right: $\binom{1}{9}*\binom{11}{0}+\binom{2}{9}*\binom{10}{0}+\binom{3}{9}*\binom{9}{0}+....+\binom{11}{9}*\binom{1}{0}$

Using combinatorial identities, we can simplify this to $11*\binom{12}{9}=2420\equiv 420 (mod 1000).$
[/hide]?[/quote]

Each pole must have at least one flag

I feel like my attitude towards computational combinatorics has generally become "ditch thinking, write a gigantic expression and do algebra." It works surprisingly well.

Suppose we place $k$ green flags and $r$ blue flags on one of the posts. It is well-known that there are ${r+1 \choose k}$ ways to arrange the flags such that no two green flags are consecutive. So the answer is $$\sum_{r=0}^{10} \sum_{k=0}^9 {r+1 \choose k}{11-r \choose 9-k} - 2{11 \choose 9}$$ when we subtract the two cases where one flagpole is empty. By Vandermonde's, this is $$\sum_{r=0}^{10} {12 \choose 9} - 2{11 \choose 9} = 220 \cdot 11 - 110 = 2\boxed{310}$$ arrangements.

Just add an arbitrary flag and put them all on one pole separated by that arbitrary flag.

Pretty similar to some previous solutions..but I liked the problem too much not to post...
Note that if we combine the two poles there is at most $1$ $GG$ block.
Case I
If we combine the two poles there is no $GG$ block.We consider a permutation of $10 \text{B's}$ and $9$ G's . This number is $\binom{11}{9}$.Now we will add a divider between some two flags to divide them into the two flagpoles.Number of gaps = 18 and as every pole must be non-empty hence our required number is $18\times \binom{11}{9} = 990 $
Case(ii)
If we combine two poles there is $1$ $GG$ block.Now note that the whole thing is permutation of $GB$ sequence with only one pair of $G$ adjacent .So , number of ways = $\binom{11}{8}\times \frac{8!}{7!}$( just permutation with repeatations.)= $11\times\binom{10}{7} = 1320$
So, total number of ways$1320+990 = 2310$                                 $\blacksquare$


inequalities
LaTeX
integration
induction
inequalities proposed

Please see attachment.
Good luck ![quote] :yup: [/quote]

Question in Latex:

$ f: [ 0,\infty ) \to R$ such that it is continuous and satisfies the condition: $ 0 \le f(x) \le 2009$ $ \int\limits_{0}^{x}{f(t)}dt,$

Prove that:  $ f(x)\equal{}0$ for all $ x \in [ 0,\infty )$

[b]Solution:[/b]

The condition gives that for all $ 0 \le m \le x$ , $ 0 \le f(m) \le 2009$ $ \int\limits_{0}^{x}{f(t)}dt$ as $ f(x) \ge 0$ for all $ x \in  [ 0,\infty ) (\int\limits_{0}^{x}{f(t)}dt)$ is increasing)

So $ \int\limits_{0}^{x}{f(t)}dt \le x(2009\int\limits_{0}^{x}{f(t)}dt)$ for all $ x$.

But for $ 0 \le x <\frac{1}{2009}$, this can't hold unless $ \int\limits_{0}^{x}{f(t)}dt\equal{}0$,
so $ f(x)\equal{}0$ for $ x \in [0,\frac{1}{2009})$

Now let $ a$ be a very small real.

Then $ \int\limits_{\frac{1}{2009}\minus{}a}^{x}{f(t)}dt \le (x\minus{}(\frac{1}{2009}\minus{}a)) (2009\int\limits_{\frac{1}{2009}\minus{}a}^{x}{f(t)}dt)$

Again this gives $ f(x)\equal{}0$ for $ x \in [\frac{1}{2009}, \frac{2}{2009})$, .......

So by induction we can show that $ f(x)\equal{}0$ for $ x \in  [0, \infty)$


Make simply!!!
$A=\prod\limits_{i=0}^{n}(1+\frac{1}{cos2^{i}a})$
 :D  :rotfl:

$1 + \frac{1}{\cos 2^{k+1}a} = \frac{\cos 2^{k+1}a+ 1}{\cos 2^{k+1}a}$

$= \frac{2\cos^2 \frac{2^{k+1}a}{2} - 1 + 1}{\cos 2^{k+1}a} = 2 \cdot \frac{\cos^2 2^{k}a}{\cos 2^{k+1}a}$

Use that and induction find:

$P_n(a) = \displaystyle\prod_{i=0}^{n} \left(1 + \frac{1}{\cos 2^{i}a}\right) = \frac{2^n}{\cos 2^n a} \cdot \left[\displaystyle\prod_{i=0}^{n-1}\cos 2^i a + \displaystyle\prod_{i=1}^{n-1}\cos 2^i a\right]$

:D Thanks a lot Pork_Chop!!!!How you give your solution??? I don't think you guess that formula.
My solution: Using simplify: $A=\prod \limits_{i=0}^{n} cos(2^i.a)$
$\Leftrightarrow A.sina=\frac{sin(2^{n+1}.a)}{2^{n+1}}$


How many numbers are there such that $10^{18}$ divides into $x^{2}-x$

[hide="Hint"]Unique prime factorization and the fact that $x^{2}-x=x(x-1)\implies\gcd (x,x-1)=1$.[/hide]

Masoud Zargar

[quote="Ilthigore"]Are you asking "find all $x$ such that $x(x-1)|10^{18}$" or "$10^{18}|x(x-1)$"?

If the latter, it's trivially infinitely many, so I'll go for the former.

Obviously, (x,x-1)=1, so we concentrate on $x|10^{18}$, of which by definition there are $\tau(10^{18})=\tau(5^{18})\tau(2^{18})=19^{2}=361$ solutions. We can add 1 to all such solutions to obtain a further 361 solutions in $x-1$, so our total is 722.[/quote]

I must say I feel very doubtful about $x(x-1)|10^{18}\iff x|10^{18}$...

Somehow it seemed to me that the problem comes down to solving $|2^{a}-5^{b}|=1$ where $a,b\leqslant 18$... and it doesn't feel like there are $722$ pairs $(a,b)$ which would satisfy this.

(NOTE: If $x(x-1)|2^{18}5^{18}$, then $x=2^{a}, x-1=5^{b}$ or the other way round.)

The solution i gave in the exam used induction....

The other way involved the chinese remainder theorem...

The answer is 3, by the way. Just in case anyone was interested


function
induction
strong induction
number theory
prime factorization
Functional Equations

Find all surjective functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: \[m \vert n \Longleftrightarrow f(m) \vert f(n).\]

Suggestion: put this in the divisor function section, perhaps. This is not much like a functional equation.

The answer: Let $ P\subseteq\mathbb{N}$ be the set of primes. Let $ g : P\rightarrow P$ be any bijection (i.e. it is a permutation of the primes). Then if $ n \equal{} p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}$, $ f(n) \equal{} g(p_{1})^{e_{1}}g(p_{2})^{e_{2}}\cdots g(p_{k})^{e_{k}}$.

The solution:

Important observation: $ \tau(n) \equal{} f(\tau(n))$, since $ f$ is surjective. First, since $ \tau(n) \equal{} 1$ has only $ n \equal{} 1$ as a solution, $ f(1) \equal{} 1$.

Consider any prime $ p$. Note that $ f(p)$ has only two divisors, so it's also prime. We have $ f(1)\mid f(p)$ and $ f(p)\mid f(p)$, but no other restrictions. Define $ g$ as above (permutation of primes), so that we get $ f(p) \equal{} g(p)$ for all primes $ p$. To show $ g$ must be bijective, it's sufficient to show $ f$ is bijective. We already know it's surjective. If $ f(a) \equal{} f(b)$, then $ f(a)\mid f(b)$ and $ f(b)\mid f(a)$, so $ a \equal{} b$, therefore $ f$ is also injective. Done.

Now we prove by strong induction $ f(p^{k}) \equal{} g(p)^{k}$ for all positive integers $ k$. Base case $ k \equal{} 1$ just done. Suppose it's true for all numbers $ k\minus{}1$. Now $ f(p^{k})$ is divisible by $ 1, g(p),\ldots g(p)^{k\minus{}1}$, and itself, but nothing else. So $ f(p^{k})$ has $ g(p)^{k\minus{}1}$ at least in its prime factorization, and $ \tau(f(p^{k})) \equal{}\tau(p^{k}) \equal{} k\plus{}1$. Now, if $ k > 1$, then having any other primes other than $ g(p)$ in the prime factorization of $ f(p^{k})$ means that $ \tau(f(p^{k}))\geq 2k > k\plus{}1$, contradiction. So $ f(p^{k})$ is a power of $ g(p)$, and since $ f(p^{k})$ has $ k\plus{}1$ divisors, it must be $ g(p)^{k}$.

Suppose $ n$ is any positive integer and $ p$ is a prime that does not divide $ n$. We now prove that $ f(n)f(p^{k}) \equal{} f(np^{k})$ for all positive integers $ k$.

Since $ (n,p^{k}) \equal{} 1$, we have $ \tau(n)\tau(p^{k}) \equal{}\tau(n p^{k})$. We know $ g(p)^{k}\mid f(np^{k})$ and that $ g(p)\nmid f(n)$. So since all divisors of $ f(n)$ divide $ f(np^{k})$, and $ g(p)^{k}$ divides $ f(np^{k})$ also, we have all divisors of $ f(n)$ and $ g(p)^{k}$ in $ f(n p^{k})$. But $ \tau(f(n)g(p)^{k}) \equal{}\tau(np^{k}) \equal{}\tau(f(np^{k}))$. If $ f(np^{k})$ has any other divisors other than those in $ f(n)$ and $ g(p)^{k}$, we get $ \tau(f(np^{k})) >\tau(f(n)g(p)^{k})$ contradicting our previous statement. So $ f(np^{k})$ has its only divisors as exactly the ones in $ f(n)$ and $ g(p)^{k}$, and so by unique factorization $ f(np^{k}) \equal{} f(n)g(p)^{k}\equal{} f(n)f(p^{k})$.

Since we have $ f$ for all prime powers, the claim we just proved is sufficient to define $ f$ for all natural numbers, leading to the answer given above.

[quote="MellowMelon"]Let $ g : P\rightarrow P$ be any bijection (i.e. it is a permutation of the primes).[/quote]Perfect solution. Just a remark though: I think that with permutation, one usually denotes a bijection in which only finitely many elements are changed. So theoretically, it's not necessarily a permutation, we really can have any bijection here. :)

[hide=Waldemar Pompe & Jarosław Wróblewski (Poland)]If $f(m)=f(n)$, then $m|n\wedge n|m$ so $n=m$. So $f$ is a bijection. Hence and from equivalency in first post we obtain that numbers $a$ and $f(a)$ have the same number of different dividers. In particular $f(1)=1$ and $f(p_1),f(p_2),f(p_3),...$ is a permutation of all prime numbers sequence $p_1,p_2,p_3,...$. Denote $f(p_i)=q_i$ for $i=1,2,3,...$.
Assume, that prime $q_j$ divides $f(p_i^k)$ where $k\in Z_+$. Then $$f(p_j)|f(p_i^k)\implies p_j|p_i^k\implies p_j=p_i$$. So the only prime divisor of $f(p_i^k)$ is a number $q_i$. Hence $f(p_i^k)=q_i^l$, and because $p_i^k, f(p_i^k)$ have the same number of different divisors $k=l$. In other words $k\in Z_+\implies f(p_i^k)=q_i^k$.
Let $a=p_1^{k_1}p_2^{k_2}\cdot...\cdot p_n^{k_n}$ be the unique-prime-factorization of $a$. Then $q_i^{k_i}|f(a)$ so $q_1^{k_1}q_2^{k_2}\cdot...\cdot q_n^{k_n}|f(a)$. Because $f(a)$ and $a$ have the same number of different divisors it must be $f(a)=q_1^{k_1}q_2^{k_2}\cdot...\cdot q_n^{k_n}$.
Finally the last functions are such that
$$f(1)=1\wedge f\left(p_1^{k_1}p_2^{k_2}\cdot...\cdot p_n^{k_n}\right)=q_1^{k_1}q_2^{k_2}\cdot...\cdot q_n^{k_n}$$ where $q_1,q_2,q_3,...$ is a predetermined permutation of sequence of prime numbers $p_1,p_2,p_3,...$.
[/hide] gave this task on training camp for IMO 1999.

[quote=ith_power][quote="MellowMelon"]

Important observation: $ \tau(n) \equal{} f(\tau(n))$, since $ f$ is surjective. First, since $ \tau(n) \equal{} 1$ has only $ n \equal{} 1$ as a solution, $ f(1) \equal{} 1$.

[/quote]

Should it not be $ \tau(n) \equal{} \tau(f(n))$? :maybe:[/quote]
It should be.


The function $ 4x^2\minus{}12x\minus{}1$:

$ \textbf{(A)}\ \text{always increases as }x\text{ increases}\\
\textbf{(B)}\ \text{always decreases as }x\text{ decreases to 1} \\
\textbf{(C)}\ \text{cannot equal 0} \\
\textbf{(D)}\ \text{has a maximum value when }x\text{ is negative} \\
\textbf{(E)}\ \text{has a minimum value of \minus{}10}$

[hide="Solution"]The vertex is at $ \minus{}b/2a\equal{}12/8\equal{}3/2$.
$ f(3/2)\equal{}9\minus{}18\minus{}1\equal{}\minus{}10$.

So it's $ \boxed{\textbf{E}}$.[/hide]


If two sides of a triangle are given, then show that the area of this triangle is maximum if the sides are perpendicular to each other!

[hide="Solution"]$ S\equal{}\frac{ab\cdot sin(\alpha)}{2}$ =>...[/hide]


greatest common divisor
number theory
least common multiple

What is the product of the GCF and LCM of 8 and 40?

[hide]320

product of GCF and LCM is equal to product of numbers[/hide]

As a rule, the product of the lcm and gcf is the product of the numbers


[quote="moldovan"]Show that the geometric mean of a set $ S$ of $ n$ positive numbers is equal to the geometric mean of the geometric means of all nonempty subsets of $ S$.
[/quote]

For symetry reason, each member of S has the same power (which depends only on cardinal of $ S$) in the geometric mean of the geometric means of all nonempty subsets of $ S$.
Taking the peculiar case where $ S$ is made wjth exactly $ n$ identical positive numbers "$ a$", we get that the geometric mean of the geometric means of all nonempty subsets of $ S$ is $ a$ and so that the required power is $ \frac 1n$.

Hence the result.


Let 10 be the first term of an arithmetic sequence of 16 terms whose sum us less than 2600. If the common difference of the arithmetic sequence is a positive integer, find the sum of all such distinct arithmetic sequences in which at least one of the first 5 terms is the square of an integer.

I need help treading through the solution.

[b]1[/b]
If $a_{n}=10+(n-1)d$, where $d$ is the common difference, then $\sum_{n=1}^{16}a_{n}=\sum_{n=1}^{16}{[10+(n-1)d]}$
$=160+d\frac{15\cdot16}{2}=160+120d$.

[b]2[/b]
This must be less than $2600$, so $1\leq d\leq 20$.

[b]3[/b]
Use $seq(10+(n-1)d, n, 1, 16) | d=1$, then with $d =2, ..., 20$ to find that the first five terms contain at least one square when $d = 2, 3, 5, 6, 13, 15, 18$.

[b]4[/b]
Sum these successful sequences to get $400, 520, 760, 880, 1720, 1960$ and $2320$, whose sum is $8560$.

How did they come to this:
$=160+d\frac{15\cdot16}{2}=160+120d$.?

If the common difference is $d$, then the sum of the terms of the sequence is
$\sum_{k=0}^{15}10+dk=(\sum_{k=0}^{15}10)+(d\times \sum_{k=0}^{15}k)=$
$160+d\times \left(\frac{15(15+1)}{2}\right)=160+120d<2600\Longrightarrow d\le 20$.

thanks! For my record, the identities you used are:

$\sum_{i=m}^{n}x=(n-m+1)x$
$\sum_{i=m}^{n}i=\frac{(n-m+1)(n+m)}{2}$

[b]3[/b]
???

When $d = 2$ in the equation $10+(n-1)d$ I find the numbers to be:
$0, 12, 14, 18, 20, 22$ none of them is the square of an integer.
What does the whole $seq(...)|d=1$ means?


What is the smallest 3-digit number that can be obtained from 4921508 by crossing out 4 digits? (Note: the order cannot be changed)

[hide] Note that the numbers $ 4$, $ 9$, $ 2$, and $ 5$ can be crossed out to form the number $ \boxed{108}$.[/hide]

Um, you already posted this problem..
so then anyway..[hide]Your answer is 108, because you cross out the 4,9,2, and the 5.[/hide]


inequalities
ratio
inequalities proposed

Hi every one,

In my book (named "SANG TAO BAT DANG THUC") which is published in Viet Nam in the end of July-2006, there are three very nice and hard inequalities which can be solved by SOS method, you can visit it here

http://www.mathlinks.ro/Forum/viewtopic.php?t=80127

Now I will introduce them, and I hope that, after one one month, we can see all complete solutions. It's really worth challenging,

[b]Problem 1(or Problem 84, chapter II, page 176).[/b] Prove that if $a,b,c$ are non-negative real numbers then \[\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}}\ge \frac{a+b+c}{2}.\] [b]Problem 2(or Problem 94, chapter II, page 188).[/b] Prove that for all non-negative real numbers $a,b,c$ then \[\frac{a^{3}}{2a^{2}+b^{2}}+\frac{b^{3}}{2b^{2}+c^{2}}+\frac{c^{3}}{2c^{2}+a^{2}}\ge \frac{a+b+c}{3}.\] [b]Problem 3(or Problem 98, chapter II, page 196).[/b] Prove that for all positive real numbers $a,b,c$ then \[\left(a+\frac{b^{2}}{c}\right)^{2}+\left(b+\frac{c^{2}}{a}\right)^{2}+\left(c+\frac{a^{2}}{b}\right)^{2}\ge \frac{12(a^{3}+b^{3}+c^{3})}{a+b+c}.\] Although many people are belied that, SOS is always mechanical method, solving these problems will indicate that SOS is not only showing the strength but also exacting creativity.

PS. Sometime, for some easy problems (but even very hard one year ago) SOS is "not creative" because the solution is very easy, it's only applying, applying, applying... It's the reason why I don't post SOS-kind problem in Mathlinks and also in Vietnamese forum. But it's not the case of three above problems, the solution is very creative. I think it's three hardese SOS-kind problem up to now.

These problems are not for anyone in Viet Nam who owns the book,

[b]PS. These problems are particularly dedicated to Vasile Cirtoaje, Gabriel Dospinescu, Arqady, Zhao Bin, Albanian Eagle, Sung Joon Kim, Manlio, Perfect_ratio, Ductrung and anyone who interested.[/b]

[b]PS. Flove=Hungkhtn.[/b]

Thanks, Flove. But I do own the book and I like SOS a lot :-)

Hi, Ductrung... so these problems won't be for you :).  :lol:

The first problem was solved by Aqady (if I'm not wrong, but the solution isn't nice as I think). Two other problems were also posted by me in this forum but have't any solution.

Now, I want to see any idea of anyone about these problem? Or a predictable way for solution?  :roll:

... so I should extend the time for them  :blush:  is two months enough? Any idea? After two week, if there is no solution, I will post solution of problem 1, after one month, I will post solution of problem 2 and also two weeks after to post solution of problem 3.

Try more!  :wink:

[quote="silouan"]Vasc have post a beautiful solution to the first one in  a word file in this forum .So it is not unsolved .
Also I remember Arqady solved the second problem also in this forum .So only the final problem remains . :)
And because it is wrong to leave a problem unsolved here is the SOS of the final one .
$(b-c)^{2}s_{a}+(c-a)^{2}S_{b}+(a_{b})^{2}S_{c}\geq 0$
where $S_{a}=\frac{2(b+a)}{c}+b^{2}c^{2}+\frac{4a}{a+b+c}-4$
$S_{b}=\frac{2(b+c)}{a}+a^{2}c^{2}+\frac{4b}{a+b+c}-4$
$S_{c}=\frac{2(c+a)}{b}+a^{2}b^{2}+\frac{4c}{a+b+c}-4$

then it is easy  :wink:[/quote]

Solutions of both the first and the second ones are not beautiful (I can't find the solution to the second one). Moreover, the third is very hard (you solution show... nothing  :lol: ).

I intend to post their solutions some days after, but I must finish training Inequalities in Mathlinks to learn English, so I post them, also SMV theorem. I hope you will complete my idea,

The following is solution of problem 2, (P92)

Why I say it's one of the hardest SOS-problem? Because the solution based on SOS is very elementary and full of aristic value.

[quote="Hungkhtn's book"]Prove that for every positive real numbers $a,b,c$, we have \[\frac{a^{3}}{2a^{2}+b^{2}}+\frac{b^{3}}{2b^{2}+c^{2}}+\frac{c^{3}}{2c^{2}+a^{2}}\ge \frac{a+b+c}{3}.\] Solution. Notice that \[\frac{3a^{3}}{2a^{2}+b^{2}}-a=\frac{a(a+b)(a-b)}{2a^{2}+b^{2}}\]  \[\Rightarrow \frac{3a^{3}}{2a^{2}+b^{2}}-a-\frac{2(a-b)}{3}=(a-b)\left(\frac{(a+b)}{2a^{2}+b^{2}}-\frac{2}{3}\right)=(a-b)^{2}\frac{2b-a}{3(2a^{2}+b^{2})}.\] So the inequality can be rewritten as follow \[\frac{2b-a}{2a^{2}+b^{2}}(a-b)^{2}+\frac{2c-b}{2b^{2}+c^{2}}(b-c)^{2}+\frac{2a-c}{2c^{2}+a^{2}}(c-a)^{2}\ge 0.\] Deperate the problem into two smaller cases

+, The first case. $a \ge b \ge c$, therefore \[\frac{4b}{2a^{2}+b^{2}}-\frac{c}{2c^{2}+a^{2}}\ge 0, \frac{-2a}{2a^{2}+b^{2}}+\frac{2a}{2c^{2}+a^{2}}\ge 0 \Rightarrow \frac{4b-2a}{2a^{2}+b^{2}}+\frac{2a-c}{2c^{2}+a^{2}}\ge 0\]  \[\Rightarrow \frac{4b-2a}{2a^{2}+b^{2}}(a-b)^{2}+\frac{2a-c}{2c^{2}+a^{2}}(c-a)^{2}\ge 0 \quad (1)\]  \[\frac{(4c-2b)b^{2}}{2b^{2}+c^{2}}+\frac{(2a-c)a^{2}}{2c^{2}+a^{2}}\ge 0 \Rightarrow \frac{4c-2b}{2b^{2}+c^{2}}(b-c)^{2}+\frac{2a-c}{2c^{2}+a^{2}}(c-a)^{2}\ge 0\quad (2)\] adding up side by side two results $(1)$ and $(2)$, we have indeed the desired result.

+, The second case. $c \ge b \ge a$. Consider two smaller cases

$(i).$ If $2b \ge a+c$. We will prove two following inequalities \[\frac{2b-a}{2a^{2}+b^{2}}+4\frac{2a-c}{2c^{2}+a^{2}}\ge 0.\] Indeed, the left hand side is an increasing function of $c$, so we only need to prove it for $c=b$, or \[\frac{2b-a}{2a^{2}+b^{2}}+\frac{8a-4b}{2b^{2}+a^{2}}\ge 0.\] This one is equivalent to \[4b^{3}+2ba^{2}-2ab^{2}-a^{3}+16a^{3}-8a^{2}b+8ab^{2}-4b^{3}\ge 0\]  \[\Leftrightarrow 15a^{3}-6a^{2}b+6ab^{2}\ge 0 \Lr 5a^{2}-2ab+2b^{2}\ge 0\] which is obvious. Hence \[\frac{2b-a}{2a^{2}+b^{2}}(a-b)^{2}+\frac{2c-b}{2b^{2}+c^{2}}(b-c)^{2}+\frac{2a-c}{2c^{2}+a^{2}}(c-a)^{2}\]  \[\ge \frac{2b-a}{4(2a^{2}+b^{2})}(c-a)^{2}+\frac{2a-c}{2c^{2}+a^{2}}(c-a)^{2}\ge 0.\] $(ii).$ If $2b \le a+c$, we will prove that \[\frac{2b-a}{2a^{2}+b^{2}}+\frac{6a-3c}{2c^{2}+a^{2}}\ge 0 \quad (3)\] Indeed, because the left hand side is an increasing function of $c$, so we only need to check the case $c=2b-a$. The inequality becomes \[\frac{2b-a}{2a^{2}+b^{2}}+\frac{9a-6b}{8b^{2}+3a^{2}-8ab}\ge 0\]  \[\Leftrightarrow 10b^{3}-15b^{2}a+2a^{2}b+15a^{3}\ge 0,\] which is clearly true by $AM-GM$ Inequality \[10b^{3}+15a^{3}\ge 3\sqrt[3]{5b^{3}.5b^{3}.15a^{3}}\ge 15b^{2}a\] Now, we need to complete the last step \[\frac{2c-b}{2b^{2}+c^{2}}+\frac{3}{2}.\frac{2a-c}{2c^{2}+a^{2}}\ge 0 \quad (4)\] Indeed, because the left hand side is a decreasing function of $a$, so we only need to check it for $a=b$, the inequality becomes \[\frac{4c-2b}{2b^{2}+c^{2}}+\frac{6b-3c}{2c^{2}+b^{2}}\ge 0\]  \[\Leftrightarrow 5c^{3}+2c^{2}b-2b^{2}c+10b^{3}\ge 0 (true!).\] From two above results and notice that $(c-a)^{2}\le 3(b-a)^{2}+\frac{3}{2}(c-b)^{2}$, we conclude \[\frac{2b-a}{2a^{2}+b^{2}}(a-b)^{2}+\frac{2c-b}{2b^{2}+c^{2}}(b-c)^{2}+\frac{2a-c}{2c^{2}+a^{2}}(c-a)^{2}\]  \[\ge \left(\frac{2b-a}{2a^{2}+b^{2}}+3\frac{2a-c}{2c^{2}+a^{2}}\right)(a-b)^{2}+\left(\frac{2c-b}{2b^{2}+c^{2}}+\frac{3}{2}.\frac{2a-c}{2c^{2}+a^{2}}\right)(b-c)^{2}\ge 0.\] The inequality is completely solved and has equality if $a=b=c$. [/quote]

[quote="Flove"]Hi every one,

In my book (named "SANG TAO BAT DANG THUC") which is published in Viet Nam in the end of July-2006, there are three very nice and hard inequalities which can be solved by SOS method, you can visit it here

http://www.mathlinks.ro/Forum/viewtopic.php?t=80127

Now I will introduce them, and I hope that, after one one month, we can see all complete solutions. It's really worth challenging,

[b]Problem 3(or Problem 98, chapter II, page 196).[/b] Prove that for all positive real numbers $a,b,c$ then \[\left(a+\frac{b^{2}}{c}\right)^{2}+\left(b+\frac{c^{2}}{a}\right)^{2}+\left(c+\frac{a^{2}}{b}\right)^{2}\ge \frac{12(a^{3}+b^{3}+c^{3})}{a+b+c}.\] [b]PS. Flove=Hungkhtn.[/b][/quote]

Problem 3 is even more harder, but I'm afraid if I must post its solution. Try more, guy!  :(

[quote="hungkhtn"]
Why I say it's one of the hardest SOS-problem? Because the solution based on SOS is very elementary and full of aristic value.
[/quote]Now I don't want to cause a passionate or polemic discussion about what is [b]artistic [/b]and what isn't... (please, please) But I must say frankly, for me reading & following those long SOS solutions, especially if there are several cases to distinguish, is less pleasant than trying to understand certain pieces of art in other realms... Of course [i]finding [/i]such a solution is very pleasant and rewarding.  :P I would call an "elegant" proof (in whatever sense) more artistic. And I wouldn't be frustrated at all if a so-called "SOS inequality" has a short and ELEGANT non-SOS solution. [hide]and non-(n-1)EV/ABC/SMV/... [b]Cauchy & weighted Jensen forever![/b] :) [/hide]

Yes, it's also what I want to say. But how to prove these problems? I dont think there's any other nicer solution than SOS. Moreover, I want to say that, [i][b]we must prove such nice inequalities[/b][/i], in anyway. If you can prove these problems above by another way different from SOS, which are more beautiful and shorter, I will never use SOS, mention to SOS and anything related to it.

Now, it's the 997th post of mine. I don't want not to keep my word, so I will share the solution for the third one, in PDF file follows


Hi:

Has anyone played the game in the iTunes App store called Strike or Shield? If so, have you been able to solve the difficult level of it? I was able to solve the Medium level of play after a few times, but the diffcult level is driving me crazy? If you have played the Difficult game and won both sides, can you please send me a screenshot?

This should be in Games and Fun Factory.

I have tried playing shield on Itunes... but I gone only to the easy round...
I think that was really hard... that's why I didn't try playing it again..!

______________________
from an avid fan of [url=http://www.dressup.me/][b]dressup games[/b][/url] | [url=http://girls-dressup.dressup.me/][b]dress up girls[/b][/url] society


function
group theory
combinatorics unsolved
combinatorics

let $ X \equal{}${ set of all bijective functions from $ {1,,..,n}$ to $ {1,,..,n}$} let $ f \in x$
we define $ t_{f}(a) \equal{}$ $ 1$ if $ f^{12}(a) \equal{}$  $ a$ else $ \equal{} 0$

find $ \sum_{f \in X}^{} \sum_{i \equal{} 1}^{n}t_{f}(i)$

You can count in two different ways for this.  The sum is equal to the number of pairs $ (f,a)$ with $ f \in S_n$ and $ a \in {1, 2, \ldots, n}$ such that $ f^{12}(a) \equal{} a$.  The sum you wrote counts this by first fixing $ f$ and then enumerating $ a$.  Now fix $ a$ and enumerate $ f$.  This is $ n \cdot ($the number of $ f \in S_n$ such that $ f^{12}(1) \equal{} 1)$

To find the number of $ f$ such that $ f^{12}(1) \equal{} 1$, decompose $ f$ into cycles and note that $ f^{12}(1) \equal{} 1$ iff either $ f \in S_{n\minus{}1}$ (where we think of $ S_{n\minus{}1}$ as those in $ S_n$ fixing $ 1$, or one of the cycles in its decomposition looks like $ (1 \, a_2 \, a_3 \, \ldots \, a_{12})$.  There's none of the latter type unless $ n \geq 12$, in which case there are another $ (n\minus{}1)!$ of them.

I think you should check my counting.

[quote="jinjohn"]You can count in two different ways for this.  The sum is equal to the number of pairs $ (f,a)$ with $ f \in S_n$ and $ a \in {1, 2, \ldots, n}$ such that $ f^{12}(a) \equal{} a$.  The sum you wrote counts this by first fixing $ f$ and then enumerating $ a$.  Now fix $ a$ and enumerate $ f$.  This is $ n \cdot ($the number of $ f \in S_n$ such that $ f^{12}(1) \equal{} 1)$

To find the number of $ f$ such that $ f^{12}(1) \equal{} 1$, decompose $ f$ into cycles and note that $ f^{12}(1) \equal{} 1$ iff either $ f \in S_{n \minus{} 1}$ (where we think of $ S_{n \minus{} 1}$ as those in $ S_n$ fixing $ 1$, or one of the cycles in its decomposition looks like $ (1 \, a_2 \, a_3 \, \ldots \, a_{12})$.  There's none of the latter type unless $ n \geq 12$, in which case there are another $ (n \minus{} 1)!$ of them.

I think you should check my counting.[/quote]
cant understand your idea what are $ S_n$  and  $ S_{n \minus{} 1}$

There is never a good reason to quote the entire post preceding yours.  $ S_n$ is the symmetric group on $ n$ letters, what you called $ X$.  $ S_{n \minus{} 1}$ has the obvious analogous definition.

jinjohn had a decent beginning, but one substantial error.  $ f^{12}(a) \equal{} a$ implies that $ a$ lies in a cycle of order dividing 12, i.e. a cycle of size 1, 2, 3, 4, 6 or 12.  jinjohn only treated the first and last of these.  The general method -- that we want $ n$ times the number of permutations in which $ 1$ lies in such a cycle -- is correct, as were the computations for those two cases.

[quote="JBL"]There is never a good reason to quote the entire post preceding yours.  $ S_n$ is the symmetric group on $ n$ letters, what you called $ X$.  $ S_{n \minus{} 1}$ has the obvious analogous definition.

.[/quote]i dont still get it is jinjohn trying to get a recurrence :?:  :?:  :(  :(

thanks i got the answer on similar lines

[quote="ashwath.rabindranath"]The final answer (and the official one) is $ d_{12}(n)\cdot n!$ where $ d_{12}(n)$ denotes the number of factors of $ 12$ less than or equal to $ n$. You need to switch the summations, and use the fact that $ i$  belongs in a cycle of length $ k$ for $ (n \minus{} 1)!$ permutations. Adding up over all $ i$ and using the fact that $ k|12$ you get it.[/quote]
thanks a lot i got the same answer


summer program
Mathcamp
induction
Pascal\u0027s Triangle

A strings of 10 letters A, B,and C(for example AAABBCCABC). For the next row, if the two consecutive letters are the same, then you write the same letter downIf they are different, then you write the third letter down(for example,A B C C)
                                                 C A C
keep doing that until there is only one letter left. For example              A A A B B C C A B C
                            A A C B A C B C A
                             A B A C B A A B
                              C C B A C A C
                               C A C B B B
                                 B B A B B
                                  B C C B
                                    A C A
                                      B B
                                       B
Prove that no matter what string you start with, the three corners will always be all the same or all different. For what other numbers of string can you start with and will have the same pattern? Have fun!!

Is this question on the mathcamp quiz every year, or is it from only an old year?

This is a very cute problem

[hide]

Let's code A,B and C, 0,1 and 2.

The transformation described by Beta is :

a b -> (a+b)*2 mod 3

It works if a=b or if a!=b.

If I detail for four letters only (mod 3 everywhere)

a, b, c, d -> 2(a+b), 2(b+c), 2(c+d) -> 4(a+2b+c), 4(b+2c+d)-> 8(a+3b+3c+d) = 2(a+d)

Now if a!=d then 2(a+d)mod 3 is != a and d. So I proved it works for 4 letters  :mrgreen:

It's exactly the same for 10 letters, we obtain at the end 512*(a+j)=2*(a+j) different from a and j.  :mrgreen: again

It can be generalized to every 6N+4 number.

[/hide]

I don't think the formula is right. According to the formula, 16 works, but if we start with ABBBBBCBBABBCBBB, it wouldn't work.

I think the formula is [hide]3^n+1 for every positive integer n.[/hide]

And it is connected to the following property of Pascal's triangle  ...

[hide]... the values on row n of Pascal's traingle, apart from the first and last values (whcih are always 1), are all divisble by 3 whenever n is a power of 3[/hide]

[hide]Actually, once you've proven it for 4, you don't need to do the algebra for 10 - just look at numbers 1,4,7,10 on the first row, 1,4,7 on the fourth, 1 and 4 on the seventh, and the bottom. They are also a four-triangle, and the numbers are formed using the given property, QED. You can also use this to find other numbers that work.[/hide]

That's exactly how I did it, Kamior, [hide]except that I couldn't find an elegant approach to prove why four works. I just did when two letters are same and  three letters and four letters are same to prove why four works, then I did what you did to prove 10 works. Then you found out the next number works according to the triangle is 4+(4-1)(10-2)=28=3^3+1. Then I did an induction to prove for every power(n) of 3+1. It obviously works when n=0. For n+1, 3^(n+1)+1+(3^(n+1)+1-1)(3^m+1-2)=3^(mn+n)+1, which is a power of 3+1  [/hide]


Let $ a,b,c$ positive reals such that $ a\plus{}b\plus{}c\equal{}11$. Prove that

$ \sum\frac{b(a\minus{}2)\plus{}c}{a\plus{}b\plus{}2c}\leq2$

[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ a \plus{} b \plus{} c \equal{} 11$. Prove that

$ \sum\frac {b(a \minus{} 2) \plus{} c}{a \plus{} b \plus{} 2c}\leq2$[/quote]

Who can solve this problem?  :oops:


hi im just doing some work at school on the doppler effect for the motion of light and am wondering if anyone could give me some equations to calculate the size, distance, britghness and mass of a star
any help would be great

As compared to the doppler effect in sound, the equations for doppler effect in light are

\[ \nu = \nu_{0} \sqrt{ \frac{c-v}{c+v} } \] , when the relative velocity imples the source and observer are receding etc...


AMC
AMC 12
calculus
Princeton
college
AMC 8
AMC 10

[quote="unl.edu/amc"]The AMC include: the American Mathematics Contest 8 (AMC 8) (formerly the American Junior High School Mathematics Examination) for students in grades 8 and below, begun in 1985; the American Mathematics Contest 10 (AMC 10), for students in grades 10 and below, begun in 2000; the American Mathematics Contest 12 (AMC 12) (formerly the American High School Mathematics Examination) for students in grades 12 and below,[/quote]

The AMC website is located [url=http://www.unl.edu/amc/index.html]here[/url]

[quote="akalra1"]can u do amc 10 in seventh grade?[/quote]
Yes.

And you can do amc12 if you want to also.

[quote="akalra1"]so a seventh grader COULD be at the imo?[/quote]
it's possible

[quote="Xantos C. Guin"]It is possible for a 7th grader to be in the IMO, but highly unlikely.

If you want to increase your chances of going to the IMO, move to a country that has a lack of good math students. Antarctica is trying to find 6 penguins that can write a proof. So far, all attempts have failed.

Or you could simply study really hard. But this is not recomended.[/quote]

Trying to learn math in that country's language would be hard enough.

Also, does anyone know what grade was the youngest IMO participant ever from the US?

[quote="mtms5467"][quote="lotrgreengrapes7926"]Not from the US, but...
[quote="Wikipedia"]Terence Tao from Australia got a gold medal at the age of thirteen in 1988, being the youngest person to have a gold medal. [/quote][/quote]
What grade was he/she in at the time?[/quote]

13? I think that's seventh

[quote="probability1.01"]I think he was in 10th. He must have skipped some grades, or maybe I'm not remembering correctly.[/quote]

he skipped 3 whole grades?

[quote="solafidefarms"]Hm. Were you born then, SplashD? :P[/quote]

Hah, that's good point.  :blush: Though I would expect people to talk about him.

[quote="mtms5467"][quote="lotrgreengrapes7926"]Not from the US, but...
[quote="Wikipedia"]Terence Tao from Australia got a gold medal at the age of thirteen in 1988, being the youngest person to have a gold medal. [/quote][/quote]
What grade was he/she in at the time?[/quote]
To be fair, that was a pretty old IMO. The newer ones are much harder.

Well, at whatever difficullty, he, a 13 year old, won the gold and beat a LOT of older kids. And, since this hasn't been repeated, he had to be REALLY smart. Even if the test was this year's, he probably would have still won.

[quote="saday"]I think that is an understatement-
He was in high-school (full time) at the age of 8!
Also, he took college math courses at 9! :o[/quote]

theres plenty of people who take college courses at that age, my self included, in science.  The guy who got gold was definitely extremly bright in math, but I don't think that this is the explanation.

Smiles,
lyra

Um, great.  No offence, but isn't there something better to do than talk about how "REALLY smart" $x$ is?

I must agree with lyra.  Read "[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]The Calculus Trap[/url]."

By the way, since the proportions of people who get certain types of medals are realtively fixed, the difficulty of an IMO has relatively little to do with how much of an accomplishment it is to get gold in a particular year.

maybe he was just extremely smart

but hey, age doesn't matter all that much

it depends how much time he has spent overall

a high schooler maybe has spent 2 hours on his math
that kid might spend 20 hrs on his math

exactly

So it all goes by grade not age? Also-I think you guys are getting a little crazy just about someone's age... :)

Terence Tao has been years ahead of everyone else his entire life. Tao started taking high school classes at age eight; by 11, he was learning calculus and thriving in international mathematics competitions (he got a bronze medal at 11, a silver medal at 12, and a gold medal at 13).  He was only 21 when he earned his Ph.D. from Princeton University, and joined UCLA's faculty that year. The UCLA College promoted Tao to full professor of mathematics at 24.

He just got the Fields Medal (August 22, 2006) that is considered the Nobel Prize for mathematics.

[quote]If he's that smart, then why is it that I never heard of him before?[/quote]

Hm I like this logic. People can't be smart of SplashD hasn't heard of them :lol:  :P

[quote="anirudh"]can we take both amc8 and amc10 during the smae year[/quote]
Yes, as long as you're in 8th grade or below and you can find a school that lets you take the AMC 10.

Yes.

In some schools, like mine, we have to do well in this previous Math Contest in order to take AMC 12A.  Then we pay money inorder to determine if we can take AMC 10B/12B.

[quote="akalra1"]can u do amc 10 in seventh grade?[/quote]

yes and you can do it in first grade too :P  :P  :P

[quote="akalra1"]well, i wouldnt take my chances there ;)

lol guys wat if a first grader won[/quote]
A 1st grader getting a perfect score?  :P Well, anything is possible...but that's not very likely

[quote="k8reindeer"][quote="akalra1"]so I could take the AMC 8, 10 and 12 next year when I am going into seventh?

cool[/quote]

Only if you qualify and get the oppurtunity. When I took the AMC 8 it said on the packet certain requirements for advancement to the next level (top x%, score of y or better) and I had to take the AMC 10 through the local High School.[/quote]
Really, I took the AMC 10 but got a 20 or 21 on AMC 8.  :blush:

Maybe it's because I took high school classes, they just let me. :maybe:

Wait are you talking about taking the AMC 12?

Have any of you guys gotten into the USAMO?

Anybody young enough can take the AMC 10.  It says that high scorers on AMC 8 are [i]invited[/i] to take the AMC 10, which I suppose means that they specially prompt you to do so, or something :  I don't know; I never took the AMC 8.

But you can find the answers to most of your questions in the FAQ thread at the top of this forum, or on the [url=http://www.unl.edu/amc/]AMC website[/url], I believe, so you should probably check those places before posting a question; it clutters the forum a bit.

buzzer11, many people on this forum have qualified for USAMO and beyond.


When converting from quarts to gallons, Hal multiplied by 4 when he should have divided by 4, and he got answer of 128. How many gallons are in the answer Hal should have gotten to the problem?

[hide]
128/4 = number of quarts = 32
32/4 = number of gallons = 8
[/hide]


function
integration
inequalities
calculus
derivative
real analysis
real analysis solved

Don't you mean (x1-x2)2002 ?

Let F(x)=\int{0->x}f(t)dt. Then F is continuous, so it has Darboux' property, so if there are no x1 and x2 with that property then we may assume that the inequality is always < or always > (let's assume we have |F(x1)-F(x2)| in the inequality).

If it's always < then we get that F is derivable and the derivative in every pt is 0, so it's constant, so F(1)=0, which is false. this means that we always have >, which means that |F(1)|>1, but we have |F(1)|<=1, so we have a contradiction again.

I guess this solves the problem, doesn't it?

If f is continous

G(x,y) =  \int _(x to y) f(t) dt - (y-x)^(2002) is continous functions
in the variable x,y \in [0;1]
G(0,1) = \int _(0 to 1) f(t) dt - 1 < 0 with the hypothesis.

|\int _(0 to 1) f(t) dt |> 0 with mean formulae there exist d \in [0,1]
s.t |f(d)| > 0 , we can suppose f(d) >0 (if not change f to -f)

f is continous for  \varepsilon > 0, there exist k > 0 s.t for any
x \in [d-k;d+k] f(x)  \geq  \varepsilon > 0
with  \varepsilon - 2^(2001)k^(2001) > 0

G(d-k;d+k) =  \int _(d-k to d+k) f(t) dt - 2^(2002)k^(2002)  \geq 2k \varepsilon - 2^(2002)k^(2002) = 2k( \varepsilon - 2^(2001)k^(2001)) > 0

G is continous
G(0,1)<0
G(d-k,d+k)>0

Then there exist a,b such that G(a,b)=0

 \int _(a to b) f(t) dt = (b-a)^(2002)


Let $ ABCD$ be a convex quadrilateral with $ BA$ different from $ BC$. Both diagonals $ AC$ and $ BD$ meet at $ P$. Denote by $ M$ the Miquel point of the complete quadrilateral $ ABCD$. Suppose that there exists a circle $ k$ (center at $ O$) tangent to ray $ BA$ beyond $ A$ and to the ray $ BC$ beyond $ C$, which is also tangent to the lines $ AD$ and $ CD$. Prove that the point $ M$ lies on the line $ OP$.

[quote="Fang-jh"]Denote by $ M$ the Miquel point of complete quadrilateral $ ABCD$.[/quote]
What is a [b]Miquel point[/b]? Can you site it with a diagram please!


geometry
trapezoid
MATHCOUNTS
trigonometry
geometric transformation
rotation
cyclic quadrilateral

Hello.  I have a few questions I was not sure about on the state target competition.  I believe that this is close to the wording, but I may be a bit off.

6. An isoceles trapezoid has legs of 30 centimeters, diagonals of 40 centimeters, and the longest base is 50 centimeters.  What is the area of the trapezoid?

7. In $(2^{1})(2^{2})(2^{3}) . . . (2^{100})$, what is the product of the units and tens digits?

Here my ugly and wrong answers as I try to solve the problems.

[hide="6"]First notice the two right triangles within the trapazoid.  One triangle would have an area of 600.  We can calculate the height of the trapazoid by finding the height of the triangle using 600 as the area and 50 as the base.  So the height is 24.

Then, we can calculate the length of the smaller base by using the pythagorean theorem.

$50-2\sqrt{30^{2}-24^{2}}=14$

So the area is (50+14)*24/2=768 hopefully.[/hide]

[hide="7"]
That long product is $2^{5050}$.  Now I'll look at the powers of 2 in mod 100.
2,4,8,16,32,64,28,56,12,24,48,-4, -8, -16, -32, -64, -28, -56, -12, -24, -48, 4

Using that pattern, we see that $2^{5050}$ is 24 so the answer should be 8. [/hide]

For number 7 you'll find that 2^10=1024.  Only consider the last two digits, since only they will affect the final result.  (2^10)^5 has the same last two digits as 24^5.  The last two digits are 24.  2^50 thus ends in 24.  Take this to the tenth power.  24^10 ends in 76, so 2^500 does, too.  2^5000 by the same method ends in 76.  (2^5000)*(2^50) ends in what 24*76 ends in=24. 2*4=8

You know, a solution is on the mathcounts website. :D

Trig really helped out on #6, but of course, you don't need it to solve the problem.

[quote="acchessor"]Hello.  I have a few questions I was not sure about on the state target competition.  I believe that this is close to the wording, but I may be a bit off.

6. An isoceles trapezoid has legs of 30 centimeters, diagonals of 40 centimeters, and the longest base is 50 centimeters.  What is the area of the trapezoid?

7. In (2^1)(2^2)(2^3) . . . (2^100), what is the product of the units and tens digits?[/quote]
[hide="higher level solution to number 6"]
Note that since it's isosceles it's cyclic.
Use Ptolomey's to find that the shorter leg is 30.
Use Bhramagupta's to find the area of 768.[/hide]

[hide="a good solution to 7 (I think)"]
NOte that this is $2^{5050}$
Also note that the last two digits of $2^{20}=76$
Now also note that $76^{2}=5476$ (omg, last two digits are the same.
So our answer is $76 \cdot 1024=24$ (last two digits) giving us a final answer of 8.[/hide]

[quote="kstan013"]Draw a triangle in the trapezoid with the leg, the longer base, and the diagonal. Notice that the sides are 30-40-50, so it is a right triangle. The area is 600, and the base is 50, so the height of the trapezoid is 24. The triangles are similar, so the one created with the height, the diagonal, and the portion of the base to the right of the height is a 24, 40, 32. The portion cut off by the height that is part of one base but not the other is 50-32=18. Subtract 18 from 32 to get 14 as the upper base. The average of the bases is 50+14/2, which is 64/2, or 32. Multiply this by the height, 24, and we get the answer of 768.[/quote]
Or just use Ptolomey's on the trapezoid.
Call the other leg x.
$50x+900=1600$
$x=14$
Then use Bhramagupta's.

[quote="acchessor"]Hello.  I have a few questions I was not sure about on the state target competition.  I believe that this is close to the wording, but I may be a bit off.

6. An isoceles trapezoid has legs of 30 centimeters, diagonals of 40 centimeters, and the longest base is 50 centimeters.  What is the area of the trapezoid?

7. In (2^1)(2^2)(2^3) . . . (2^100), what is the product of the units and tens digits?[/quote]

For the trapezoid one I quickly realized it would be a 3-4-5 (or in this case 30/40/50 with longest base being hypotenuse..  I just didn't write down 768 for some reason, I know I had the answer  :wallbash: )

For the second one you can see that the one's digit for a power of 2 goes in the pattern 2 4 8 6 and repeats...  The tens digit goes something like 2 4 8 16 32 64 128 256 512 1024 2048 etc, you can get it through a calculator i don't remember what it was.

You can use a calculator on #7:

the last two digits are (for each power starting with 2^1):

02,04,08,16,32,64,28...88,76,52,04 -- keep multiplying until it repeats (except it repeats starting from 04, instead of 02). It's 20 terms long. 5050/20 = 252 R 10. Count down 10 from 04 but you have to remember to go back one because you didn't count 02. This term is 24, and 2*4 = 8 :P

[quote="original question"]7.  When the expression $(2^{1})(2^{2})(2^{3}) . . . (2^{100})$ is written as an integer, what is the product of the tens digit and the ones digit?[/quote][size=75]

[acchessor:  For the future, each question should generally be posted in its own thread.]

[/size]How high do you know your powers of 2?  I know mine to $2^{16}.$  Calculators are fine, but the higher you know your powers, the less work is required for this problem.  Here are two solutions, of increasing sophistication:

[hide="Solution A"]
Notice the product of the terms is $2^{5050}$.  Start doubling:

$\begin{array}{cl}n & 2^{n}\text{ (last two digits)}=2^{n}\mod 100 \\ \hline 1 & 02 \\ 2 & 04 \\ 3 & 08 \\ 4 & 16 \\ 5 & 32 \\ 6 & 64 \\ 7 & 28 \quad{128 \mod 100=28}\\ 8 & 56 \quad{256 \mod 100=56}\\ 9 & 12 \quad{512 \mod 100=12}\\ 10 & 24 \quad{1024 \mod 100 = 24}\\ 11 & 48 \quad{2048 \mod 100 = 48}\\ 12 & 96 \quad{4096 \mod 100 = 96}\\ 13 & 92 \quad{8192 \mod 100 = 92}\\ 14 & 84 \quad{16384 \mod 100 = 84}\\ 15 & 68 \quad{32768 \mod 100 = 68}\\ 16 & 36 \quad{65536 \mod 100 = 36}\\ 17 & 72 \\ 18 & 44 \\ 19 & 88 \\ 20 & 76 \\ 21 & 52 \\ 22 & 04 \end{array}$

This took a while, but the pattern repeats every 20 numbers (from $2^{2}$ to $2^{22}$).  $5050 / 20 = 252 \text{ R }10.$  So $2^{5050}\mod 100 = 2^{10}\mod 100 = 1024 \mod 100 = 24.$  The answer is $2\times 4=8.$
[/hide]

[hide="Solution B"]
Notice the product of the terms is $2^{5050}$.  Start squaring:

$\begin{array}{cl}n & 2^{n}\text{ (last two digits)}=2^{n}\mod 100 \\ \hline 1 & 02 \\ 2 & 04 \quad{(02^{2}= 04)}\\ 4 & 16 \quad{(04^{2}= 16)}\\ 8 & 56 \quad{(16^{2}= 256)}\\ 16 & 36 \quad{(256^{2}= 65536)}\\ 32 & 96 \quad{(36^{2}= 1296)}\\ 64 & 16 \quad{ (96^{2}= 9216)}\end{array}$

This was a little easier.  This shows the pattern repeats every 60 numbers (from $2^{4}$ to $2^{64}$).  $5050 / 60 = 84 \text{ R }10.$  So $2^{5050}\mod 100 = 2^{10}\mod 100 = 1024 \mod 100 = 24.$  The answer is $2\times 4=8.$
[/hide]

after i accidently multiplied the answer by 2...i was so angry

my coach afterwards showed me u could graph it on a ti-89, then go to table

my way was like 1.5 min, his was like 8 seconds

and the trapezoid one...don't ask

jorian

Edit: SORRY GUYS
I'm wrong
and the solution page on MATHCOUNTS is wrong too

I don't know why
but I think all the answers posted to number 7 are wrong
or
I'm just plain blind

[hide]
using rcv's chart
you notice 02 doesn't appear again
and
its a 20 number rotation
the remainder part is right (10)
04
08
16
32
64
28
56
12
24
48

4x8 equals 32
32 is your answer
I either put 8 or 54 for my answer
I MISCOUNTED by ONE (I realized this when I was checking with others)
though i might have put 32... :ninja:
i have bad memory sometimes[/hide]

Answer was 8. I got it. You multiply all of the numbers out by adding (1+2+3+...100). So you have find the tens digit and units digit of 2 to the 2050. This is really easy if you know that the tens digits and units digit repeat after every twenty terrms. So all you need to know are the tens and units digits of 2 to the tenth (which is a multiple of 20 away from 2050). Since they are 2 and 4 then answer is 8. Also check out mathcounts solution
 http://www.mathcounts.org/webarticles/articlefiles/510-07_S_Solutions.pdf

The solution is wrong
Click on the link
even though the answer key is right... :ninja:

[hide]
a little obsesssion with the coach don't you think osama(V 0.5)
[/hide]

What? The answer in the answer key to #7 is not same as the answer you get from the answer explained on the solutions?

[quote="Aznwithabrain"]The solution is wrong
Click on the link[/quote]

The MATHCOUNTS official solution to Target #7 does contain a typographical error:  $\frac{5045}{20}=252\text{ R }10.$  That is not correct.

But their final answer (and most of the solution) is correct.

Using your calculator, $2^{10}=1024$ and $2^{30}=1073741824.$

Wouldn't you expect $2^{50},$ $2^{70},$ and $2^{5050}$ to also end with the digits $24.$  The answer key shows the answer is $8.$

[quote="diophantient"]
Or just use Ptolomey's on the trapezoid.
Call the other leg x.
$50x+900=1600$
$x=14$
Then use Bhramagupta's.[/quote]

What is Ptolomey's formula?

[quote="magixter"]What is Ptolomey's formula?[/quote]
Ptolomey's theorem states that when you have a cyclic quadrilateral, and the sides are a, b, c, and d and the diagonals are e and f, then ac+bd=ef


AMC
USA(J)MO
USAMO
factorial
number theory
relatively prime

Show that if p is prime then ((n^p)*m - (m^p)*n) is divisible by p for any integers m and n.

'cause both terms are mn by Fermat's Little Theorem.

We are trying to prove that (n^p)m - (m^p)n = 0 (mod p) <=> (n^p)m = (m^p)n (mod p). By Fermat's Little Theorem, we have n^p = n (mod p) and m^p = m (mod p). Multiplying, we have our desired result.

But u still need to prove that mn=0(modp)

I know you're losing, but split posting? Fight with honor, young grasshopper :)

Wow, I don't think we're all that shallow.

-interesting_move

[quote="darkquantum"]i wasn't split posting
i realized my mistake right after i hit the "submit" button[/quote]
Oh, OK. Sorry I said you split-posted, but I love to say "young grasshopper" :)

How did you multiply to get the solution?

Anyways, this is how I did it.

[hide](n^p)m-(m^p)n :equiv: 0 mod p

n^(p-1) :equiv: 1 mod p

m^(p-1) :equiv: 1 mod p

Subtract and we obtain our answer...[/hide]

Hmm, that won't quite work because you are assuming that gcd(p, m) and gcd(p, n) are both 1.

When I said that I multiplied, I meant that I multiplied n^p = n (mod p) and m = m^p (mod p) => (n^p)m = (m^p)n (mod p), which implies are desired result.

[quote]m=m^p(mod p)?[/quote]

That's Fermat's Little Theorem. Only when gcd(m, p)=1 is m^(p-1)=1 (mod p) true, since that is the only case when you can divide by m on both sides without needing to change the mod.

Um. I sorta proved this once when I was just fiddling around with a USAMO question. Anyways, consider the integers 1, 2, 3, ... p-1. Call that set S. If we multiply all of them by any number m relatively prime to our prime p, we will get the numbers  m, 2m, 3m, ... , m(p-1). Now, (this is the good part!)  none of those can be congruent to each other mod p, since if am :equiv: bm mod p, we can cancel the m's (remember, it's relatively prime) to get a :equiv: b mod p, which is not true, since both a and b have to be integers from 1 to p-1. So anyways, since none of these numbers are congruent to each other mod p, and none of them are a multiple of p, each one of them must be congruent to a different number in S. So, if we multiply them all up,  they should be equivalent mod p. We get (p-1)! :equiv: m^(p-1) (p-1)! mod p. We can cancel the factorials to get the desired result.

Now, to get the slightly stronger m^p:equiv:m mod p, notice that if m and p are relatively prime, we can use fermat and multiply by m on both sides. If m is a multiple of m, well then, both sides are congruent to 0. Done again. YEAH! :)

I forgot that a=a^p mod p is equal to a^p=a mod p...  :lol:   Whoops.


1. A hexagon is inscribed in a circle. Another circle is inscribed inside the hexagon. What is the ratio of the area of the smaller circle to the area of the larger circle.

2. Find the ratio of the area of a 45 45 90 triangle inscribed in a  circle to the area of the part of the circle remaining when the triangle is removed.

[hide="Problem 1"]
By looking at the hexagon, we know that one of its sides is equal to the outer circle's radius and its apothem is equal to the inner circle's radius. So we find that the outer circle's radius is $r$ and the inner circle's radius is $\frac{\sqrt{3}}{2}r$. So the area of the larger circle is $\pi r^{2}$ and the area of the smaller circle is $\frac{3}{4}\pi r^{2}$. So the ratio is
\[\frac{\frac{3}{4}}{1}=\frac{3}{4}\]
*haha thanks java man
[/hide]

[hide="Problem 2"]
The longest side of the triangle must be the diameter of the circle since it is a $45-45-90$ triangle. By labeling the diameter as $2r$ we find the length of one side to be $\frac{2r}{\sqrt{2}}$. So the area of the triangle is
\[\frac{2r}{\sqrt{2}}^{2}=r^{2}\]
and so the ratio can be found by
\[\frac{r^{2}}{\pi r^{2}-r^{2}}=\frac{1}{\pi-1}\]
[/hide]

[quote="NeverOddOrEven"]
$\frac{\frac{3}{4}}{1}=\frac{4}{3}$
[/quote]

You sure?  :)

hahaha man what has gotten into me?  :rotfl:  :rotfl:


[quote="Beat"]$a,b,c>0$, $\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}=1$. Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\le 3$[/quote]
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\le 3\Leftrightarrow$
$\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1+2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}-1\right)\le 0,$ which true because $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\leq0\Leftrightarrow\frac{ab+ac+bc}{a^{2}+b^{2}+c^{2}}\leq1$
and $\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}-1\leq0\Leftrightarrow\frac{abc}{a^{2}+b^{2}+c^{2}}\cdot\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)-1\leq0\Leftrightarrow$
$\Leftrightarrow a^{2}+b^{2}+c^{2}\geq\sqrt{abc}\left(\sqrt a+\sqrt b+\sqrt c\right),$ which true by Muirhead.

Very Nice. The original solution uses only Causchy-B-Shwartz and AM-GM.


modular arithmetic
algebra unsolved
algebra

Let $ s(x)$=sum of digits of x then:
$ s..s(s(a)*s(b))$=$ s..s(ab)$ we apply s to the number until we get a 1-digit number.
For example $ s(23) \equal{} 5, s(45) \equal{} 9, s(s(23)*s(45)) \equal{} 9 \equal{} s(s(23*45))$

$ s(n)\equiv n\pmod{9}$ for all $ n\in\mathbb{N}$.
For $ a,b\in\mathbb{N}$ take $ k,l\in\mathbb{N}$ large enough with $ s^k(s(a)s(b)),s^l(ab)\in\{1,\ldots ,9\}$.
Then $ s^k(s(a)s(b))\equiv ab\equiv s^l(ab)\pmod{9}$ gives the equality.


function
arithmetic sequence
topology
inequalities proposed
inequalities

Given is a positive real $A>3$ and a function $f :R\rightarrow R$ s.t. if $a<b<c<d$ form an arithmetic progression then $|f(a)-f(d)|\geq A|f(b)-f(c)|$.
Prove that $f$ is constant.

I have proved it thinking that we had "$f$ continuous" in the hypothesis.
Actually I use the fact that $f$ is bounded on a closed set and then I show that it is differentiable with zero as a derivative.
But then I realized that this property (continuous or bounded) isn't in the statement so I'm now trying to solve the case "$f$ discontinuous not bounded."


Assume that  $\frac{5x^2+ax-b}{cx^2-8x+3}$  takes the same value $m$ for three distinct values of $x$,

Prove that the value of this fractional expression is $m$ for any values of $x$ for which the denominator doesn't vanish.

[hide="Solution"]$\frac{5x^2+ax-b}{cx^2-8x+3}=m$

So $5x^2+ax-b=cmx^2-8mx+3m$, and $cmx^2-5x^2+ax-8mx+3m-b=0$. This is a polynomial in x, because a,b,c,m are all constant.

A non-constant second degree polynomial can have at most 2 roots, so the polynomial is constant.

Thus the fraction is equal to m as long as it exists.[/hide]

Yes, that's right, but you made a little mistake in the second line,RC-7th.
Please correct this. :)

kunny


so...
i have a new avatar :P
couldn't get it small enough
but i downloaded the gimp
so now its cool
how do you guys like it?

I like the avatar, and I use the gimp too. (or at least until my laptop's hard drive failed).
What's WoW?  :ninja: I have a feeling it's a video game that I've been completely missing out on.

[quote="junggi"][quote="DiscreetFourierTransform"]yay! My fav. photo editing software.

It's hawt.
It's going to throw me off for a while though.

Note to all. I just got WoW, and was wondering if anyone else from AoPS plays.
If so, we could form a guild, and we could consider ourselves the nerdiest, math-iest guild on there.[/quote]
ADVICE FROM A FELLOW MATHNERD: DONT PLAY GAMES THEY EAT YOUR BRAINS
well so does music and i listen to a lot of music
i guess i am a hyprocrite?
yay

@daermon:what IS that thing[/quote]
i second junggi!  i spent a year of my life on runescape and i'm not feeling any better about it :ninja:

@junggi:  its linux as a dj
or, as my friend from school thinks, a dog. :huh:

[quote="daermon"]
i second junggi![/quote]
YAY i AM a hyprocrite
[/retardedness]

[quote="DiscreetFourierTransform"]I just quit RS. I'm a level 116 with full all barrows, and 28m cash. That's what I did when I didn't do Math.
Three years of life wasted?
No, I met a bunch of great people, and I like to think that I didn't waste my time.[/quote]
Well you ARE a 9th grader...lots of free time
I guess it's ok

[quote="daermon"] its linux as a dj
[/quote]
umm no it's not linux it's tux :wink:


A square of side length $ s$ is inscribed symmetrically inside a sector of a circle with radius of length $ r$ and central angle of measure $ x$, such that two vertices lie on the straight sides of the sector and two vertices lie on the circular arc of the sector. Determine the exact value of $ \frac{s}{r}$

Let OP = OQ = r be radii of the given circle  (O, r), $ \angle POQ \equal{} x.$ Let (O, p) be a circle concentric with (O, r), p < r, intersecting OP, OQ at A, B. Let ABCD be a square with $ C,\ D \in (O, r)$ and s = AB = AD. Using bisector of the $ \angle POQ,$

$ s \equal{} AB \equal{} 2p \sin \frac {x}{2}$

$ s \equal{} AD \equal{} \sqrt {r^2 \minus{} \frac {s^2}{4}} \minus{} p \cos \frac {x}{2}$

Substituting s from the 1st equation to the 2nd one,

$ \sqrt {r^2 \minus{} p^2 \sin^2 \frac {x}{2}} \minus{} p \cos \frac {x}{2} \equal{} 2p \sin \frac {x}{2}$

$ r \equal{} p \sqrt {1 \plus{} 4 \sin \frac {x}{2} \cos \frac {x}{2} \plus{} 4 \sin^2 \frac {x}{2}}$

Dividing the 1st equation by this r,

$ \frac {s}{r} \equal{} \frac {2 \sin \frac {x}{2}}{\sqrt {1 \plus{} 4 \sin \frac {x}{2} \cos \frac {x}{2} \plus{} 4 \sin^2 \frac {x}{2}}}$


Euler
function
Gauss
trigonometry
real analysis
real analysis solved

Prove that

\[\prod^{n-1}_{k=1} \Gamma \left( \frac{k}{n} \right) = \frac{(2 \cdot \pi)^{\frac{n-1}{2}}}{\sqrt{n}}\]

with $n \in \mathbb{Z}^+$ and $\Gamma$ denoting the Euler gamma function.

It follows directly from Gauss' multiplication formula
$\prod^{n-1}_{k=0} \Gamma \left( \frac{z+k}{n} \right) = (2 \cdot \pi)^{\frac{n-1}{2}} \cdot n^{\frac{1}{2}-z} \cdot \Gamma \left( z \right)$
when setting $z=1$.
Do you need a proof for this formula¿

[quote="ZetaX"] Do you need a proof for this formula¿[/quote]

Please write it for the sake of completeness.  :lol:

OK, a try  ;)

Define $f(z):=\frac{\prod^{n-1}_{k=0} \Gamma \left( \frac{z+k}{n} \right)}{n^{\frac{1}{2}-z}}$

[b]i)[/b]
$f(z+1)=\frac{\prod^{n-1}_{k=0} \Gamma \left( \frac{z+k+1}{n} \right)}{n^{\frac{1}{2}-z-1}}=$
$=\frac {\Gamma \left( \frac{z}{n}+1   \right) }{\Gamma \left( \frac{z}{n}   \right)} \cdot   \frac{\prod^{n-1}_{k=0} \Gamma \left( \frac{z+k}{n} \right)}{n^{\frac{1}{2}-z-1}}=   z \cdot \frac{\prod^{n-1}_{k=0} \Gamma \left( \frac{z+k}{n} \right)}{n^{\frac{1}{2}-z}}=z \cdot f(z) $

[b]ii)[/b]
$f(z)$ is bounded in the stripe $1 \leq \Re (z) \leq 2$ because all $\Gamma \left( \frac{z+k}{n} \right)$ are bounded there and $|n^{z-\frac{1}{2}}| = n^{\Re (z)-\frac{1}{2}} $

[b]iii)[/b]
${f(1)=\sqrt{(n)} \cdot \prod^{n}_{k=1} \Gamma \left( \frac{k}{n} \right)}$
Thus ${{f(1)^2=n \cdot \prod^{n-1}_{k=1} \Gamma \left( \frac{k}{n} \right)} \cdot  \Gamma \left( 1- \frac{k}{n} \right)} =  n \cdot \frac{ \pi^{n-1}}{ \prod^{n-1}_{k=1} \sin ( \frac{\pi \cdot k}{n})} = n \cdot \frac{ \pi^{n-1}}{ \frac{n}{2^{n-1} }}=(2 \cdot  \pi )^{n-1} $

Since $f(1)\geq 0$, the formula follows immediatly since every in this stripe bounded function that fulfils $g(1)=1$ and $g(z+1)=z \cdot g(z)$ is the $\Gamma$-function (one of the many characterisations of $\Gamma$) ...


inequalities
limit
inequalities unsolved

[quote="sxgfly"]$ Let\,\, x,y > 0,\,\, such\,\,that \,\,x \plus{} y \equal{} 1.\,\,prove\,\,that \\
\frac {x}{x^2 \plus{} y^3} \plus{} \frac {y}{y^2 \plus{} x^3}\le \frac {8}{3}$[/quote]

Since you know the equality occurs when $ x\equal{}y$ or $ x\equal{} \frac{1}{2}$, the solution will not be difficult. Here is a practical solution.
Let $ y\equal{} 1\minus{}x$
Then
$ \frac {x}{x^2 \plus{} (1\minus{}x)^3} \plus{} \frac {1\minus{}x}{(1\minus{}x)^2 \plus{} x^3} \minus{}\frac{8}{3}$

$ \equal{} \frac{(2x\minus{}1)^2(2x^4\minus{}4x^3\minus{}9x^2\plus{}11x\minus{}5)}{(x^2\plus{}(1\minus{}x)^3)((1\minus{}x)^2\plus{}x^3)} \geq 0$

Applying Ho\''lder inequality, we get

$ 3(4x\plus{}1)(x^2\plus{}y^3) \ge (2x\plus{}y)^3$

and
$ 3(4y\plus{}1)(y^2\plus{}x^3) \ge (2y\plus{}x)^3$

Hence, it is sufficient to demonstrate that

$ \frac{3x(4x\plus{}1)}{(x\plus{}1)^3}\plus{}\frac{3y(4y\plus{}1)}{(y\plus{}1)^3} \le \frac{8}{3}$

However, by homogenization, we obtain the following

\[ \frac{3x(5x\plus{}y)(x\plus{}y)}{(2x\plus{}y)^3}\plus{}\frac{3y(5y\plus{}x)(x\plus{}y)}{(2y\plus{}x)^3} \minus{}8/3 \equal{} \frac{\minus{}(x\minus{}y)^2 (19 x^4\plus{}122 x^3 y\plus{}204 x^2 y^2\plus{}122 x y^3\plus{}19 y^4)}{3 (2 x\plus{}y)^3 (x\plus{}2 y)^3}\]

We complete our proof.

Gibbenergy, I found that your first inequality is true but your second inequality looks like this after full expansion

$ \frac { \minus{} (2 x \minus{} 1)^2 (8 x^6 \minus{} 24 x^5 \minus{} 66 x^4 \plus{} 172 x^3 \minus{} 218 x^2 \plus{} 128 x \minus{} 29)}{3 (x^2 \minus{} x \plus{} 1) (x^3 \minus{} 5 x^2 \plus{} 4 x \minus{} 1) (x^3 \plus{} 2 x^2 \minus{} 3 x \plus{} 1)}$

However
\[ \lim_{x \rightarrow \frac {1}{2}} (8 x^6 \minus{} 24 x^5 \minus{} 66 x^4 \plus{} 172 x^3 \minus{} 218 x^2 \plus{} 128 x \minus{} 29)\equal{} \frac { \minus{} 11}{4}\]
so I think probably this problem has limitation about degrees as well as generalization.

[size=150][color=red] warning:  Tons of tedious computations!!! [/color][/size]

[quote="sxgfly"][quote="Gibbenergy"][quote="sxgfly"]$ Let\,\, x,y > 0,\,\, such\,\,that \,\,x \plus{} y \equal{} 1.\,\,prove\,\,that \\
\frac {x}{x^2 \plus{} y^3} \plus{} \frac {y}{y^2 \plus{} x^3}\le \frac {8}{3}$[/quote]

Actually, there are  interesting facts.
$ Let\,\, x,y > 0,\,\, such\,\,that \,\,x \plus{} y \equal{} 1.\, \\
\frac {x}{x^4 \plus{} y^5} \plus{} \frac {y}{y^4 \plus{} x^5} \leq 2 \left( \frac {x}{x^3 \plus{} y^4} \plus{} \frac {y}{y^3 \plus{} x^4} \right) \\
\frac {x}{x^5 \plus{} y^6} \plus{} \frac {y}{y^5 \plus{} x^6} \leq 2 \left( \frac {x}{x^4 \plus{} y^5} \plus{} \frac {y}{y^4 \plus{} x^5} \right)$[/quote]

[size=150][color=green]Dear Gibbenergy:  How to prove above your inequalities?[/color][/size][/quote]

For the first one,  I did the substitution $ y \equal{} 1\minus{} x$. Then do the factorization.

$ \frac{x}{x^4\plus{}(1\minus{}x)^5}\plus{}\frac{1\minus{}x}{(1\minus{}x)^4\plus{}x^5}\minus{}2 \left( \frac{x}{x^3\plus{}(1\minus{}x)^4}\plus{}\frac{1\minus{}x}{(1\minus{}x)^3\plus{}x^4} \right)$

$ \equal{}\frac{\minus{}(x^8\minus{}4x^7\plus{}14x^5\minus{}31x^4\plus{}34x^3\minus{}21x^2\plus{}7x\minus{}1)(2x^2\minus{}2x\plus{}1)(\minus{}1\plus{}2x)^2}{(1\minus{}3x\plus{}3x^2\minus{}x^3\plus{}x^4)(\minus{}3x^3\plus{}1\minus{}4x\plus{}6x^2\plus{}x^4)(x^2\minus{}x\plus{}1)(x^3\plus{}2x^2\minus{}3x\plus{}1)(x^3\minus{}5x^2\plus{}4x\minus{}1)} \leq 0$


I'm in grade 10 this year, and I want a goal for the Waterloo Seminar. Would a perfect Cayley be enough to get in?

Is there any website I can find out more about the seminar? Also, any people who have got invited in the past, could you please post what scores you got invited for?

Uhh seminar home page?  See I think the thing with seminar is that like they don't really want everyone to know about it..like you get invited because you did well, you don't do well to get invited.  Also, the inviting is really subjective, like my friend who got 126 on Fermat one year (and that was his best score too) got invited.  It's kinda random, and they don't want to make those who they didn't pull out of the hat feel left out.  (well it's not that random, like if you're on IMO you get invited for sure, but you get the point)

Anyways, hope that helps.  Just love math, work hard, and you'll get in :).

Do Grade 9's have a chance?  :ninja:

[off-topic but important?]
For all those amazing mathematicians out there, is math a daily hobby for you? Do you learn more math concepts and/or write contests on a daily basis?

[quote="d.lam.86"]Do Grade 9's have a chance?  :ninja: [/quote]

Course they do!...if you write the gr 10/11/12 contests only.  But it looked like this last year they were phasing out the younger grades....

Cool; personally, I doubt my school will allow me to write Cayley unless I can miracously get exceptional marks. For now, my objective will be for next year, but then I won't meet a majority of ppl in this forum.  :o

What type of marks do you need on the CEMC contests, exactly?

[quote="Singular"]fermat or cayley  perfect or close to perfect, varying depending on location
euclid same thing, comc (? do they take from here?) gotta be past the CMO cut i think[/quote]

They don't take COMC, its too early in the year.


analytic geometry
geometry
geometric transformation
reflection
function
floor function
combinatorics unsolved

A [i]silver[/i] n-path is a lattice pathof n up steps (1,1) and n down steps (1,-1) that starts at the origin and never goes below the x-axis.A two-step sub-path in which the first step is an up-step and the second a own-step is called a [i]peak[/i]. The [i]height[/i] of a peak is the y-coordinate of the right end-point of its up-step. Let  $ {X}_n$  and  $ {Y}_n$ denote the no. of silver n-paths having an even and an odd no. of peaks, respectively, both at [i]even [/i]height. Find  $ {X}_n$  and  $ {Y}_n$ in terms of catalan nos.

[quote="aakansh92"]Let  $ {X}_n$  and  $ {Y}_n$ denote the no. of silver n-paths having an even and an odd no. of peaks, respectively, both at [i]even [/i]height.[/quote]  Does this mean that $ X_n$ is the number of silver $ n$-paths having an even number of peaks at even height and $ Y_n$ is the number of silver $ n$-paths having an odd number of peaks at even height?

After the Indian TST - this was one of the harder problems - our professor told us that for odd $ n$, $ X_n \equal{} Y_n \equal{} \frac {C_n}{2}$. He mentioned a proof using reflections of even peaks to create 'valleys' - can anyone find a solution using that? Or some bijection? The generating functions proof is cool, but I'm struggling to find a bijective proof and would love to see one.

Ashwath

[hide][quote="JBL"]I use everywhere "path" to refer to "Dyck path," which the O.P. termed "silver paths."  It is known that the total number of such paths with $ n$ upsteps and $ n$ downsteps is $ C_n$, the $ n$th Catalan number.

So, $ X_n$ is the number of paths with an even number of peaks at even height and $ Y_n \equal{} C_n \minus{} X_n$.  Let $ U_n$ be the number of paths with an even number of peaks at odd height and $ V_n \equal{} C_n \minus{} U_n$.  We have initial values $ X_1 \equal{} V_1 \equal{} 1, W_1 \equal{} Y_1 \equal{} 0$.  (It seems that starting with $ n \equal{} 0$ gets me into trouble.)

For $ n \geq 2$, consider the number of paths of length $ n \plus{} 1$ with an even number of peaks at even height that first acheive $ y$-coordinate 0 after $ i \plus{} 1$ upstep-downstep pairs, $ 1 \leq i \leq n \minus{} 1$.  The number of such paths is $ U_i \cdot X_{n \minus{} i} \plus{} V_i \cdot Y_{n \minus{} i}$.  For $ i \equal{} 0$ we get $ X_n$ such paths and for $ i \equal{} n$ we get $ U_n$.

It follows that for $ n \geq 1$, $ X_{n \plus{} 1} \equal{} X_n \plus{} U_n \plus{} \sum_{i \equal{} 1}^{n \minus{} 1}U_i \cdot X_{n \minus{} i} \plus{} V_i \cdot Y_{n \minus{} i}$.

Similarly, if we consider the number of paths of length $ n \plus{} 1$ with an even number of peaks at odd height we get $ U_{n \plus{} 1} \equal{} X_n \plus{} V_n \plus{} \sum_{i \equal{} 1}^{n \minus{} 1} X_i U_{n \minus{} i} \plus{} Y_i V_{n \minus{} i}$.

Setting $ U(x) \equal{} \sum_{n \geq 1} U_n x^n$, $ X(x) \equal{} \sum_{n \geq 1} X_n x^n$ and $ C(x) \equal{} \sum_{n \geq 1} C_n x^n \equal{} \frac {1 \minus{} 2x \minus{} \sqrt {1 \minus{} 4x}}{2x}$, this gives

$ X \equal{} x \plus{} x(X \plus{} U \plus{} XU \plus{} (C \minus{} X)(C \minus{} U))$ and $ U \equal{} x(X \plus{} (C \minus{} U) \plus{} XU \plus{} (C \minus{} X)(C \minus{} U))$.  Solving this system (using our known value of $ C$) is ugly; the solutions are
$ X \equal{} \frac {2 \minus{} 2x \minus{} \sqrt {1 \minus{} 4x} \minus{} \sqrt {1 \minus{} 4x^2}}{4x}$ and $ U \equal{} \frac {2 \minus{} 8x^2 \minus{} (1 \plus{} 2x)\sqrt {1 \minus{} 4x} \minus{} \sqrt {1 \minus{} 4x^2}}{4x(1 \plus{} 2x)}$.

This certainly allows you to express $ X_n$ in some form involving Catalan numbers, but I don't want to :)

Of course, all of these sequences can be found in the [url=http://www.research.att.com/~njas/sequences/index.html]OEIS[/url] just by computing a few terms.[/quote][/hide]
i think there is a typo $ X_{n} \plus{} Y_{n} \equal{} C_{n\plus{}1}$

Unless you define Catalan numbers differently, I don't see why that is true.

according to you then $ X_{n} \equal{} \frac {3}{4}C_{n}$
which is obviously false for $ n \equal{} 3,5$ (n is odd in these cases)

gaurav, please go read through the posts again, slowly.  If, after doing that, you have a coherrent comment to make, go ahead.  So far, your comments do not appear to have anything to do with what anyone has written.  (It would also help if you quoted the small section of posts that you are commenting on -- quoting an entire long post and quoting nothing at all are not helpful ways of indicating the statement to which you are objecting.)

ashwath -- are you sure that was for $ n$ odd, not $ n$ even?  If $ n$ is [i]even[/i], every Dyck path has at least one peak or valley at an even height.  So, find the first such peak or valley and reflect it (i.e. replace the DU at that position with UD or vice versa).  This changes by 1 the number of peaks at even height and it's an involution, so it is a bijection between $ X_n$ and $ Y_n$.

This doesn't work if $ n$ is odd because there are many Dyck paths for odd $ n$ with all peaks and valleys at odd height.  (Perhaps one could count these directly, thus giving a complete combinatorial proof?)

I wasn't sure and I've wasted several hours trying to prove something that wasn't true. I never, for a moment, considered the possibility that I might be barking up the wrong tree. I [i]am[/i] an idiot!
 Yes JBL - you're absolutely right and thanks so much. A vague idea Dr.Pranesachar mentioned to me for $ n$ odd (odd I see now, not even!) was some sort of a reduction from $ n$ to $ \frac{n}{2}$. The final expression is something like $ \displaystyle\frac{C_n \plus{} C_{\frac{n}{2}}}{2}$ though now I'm so confused, I don't remember whether that's $ X_n$ or $ Y_n$.  And of course, now, $ \frac{n}{2}$ refers to the floor of $ \frac{n}{2}$. Now that I've not mixed up odd and even, maybe I'll give it a shot!

Can anyone find a solution to this? It seems rather tricky.

So, the pairing argument still works on those paths with either a peak or a valley at even height.  Thus, our problem is solved if we count those Dyck paths with all peaks and valleys at odd height.  We have a bijection between such Dyck paths of length $ n \equal{} 2m \plus{} 1$ and "normal" Dyck paths of length $ m$: since all valleys are at odd height, the path never falls below the line $ y \equal{} 1$ except at its endpoints, so we can delete the initial up step and final down step.  Then, since all peaks and valleys are at even height, we know both up steps and down steps come in pairs, so replace each UU with U and each DD with D.  Thus there are $ C_m$ such paths, where $ m \equal{} \left\lfloor \frac {n}{2} \right\rfloor$.  Each of these paths is counted by $ X_n$ as are half of the other paths, so $ X_n \equal{} \frac {1}{2} (C_n \minus{} C_m) \plus{} C_m \equal{} \frac {C_n \plus{} C_{\frac {n}{2}}}{2}$, as you noted.

It's not too hard to extract this from the expression I derived with generating functions, but this is certainly more elegant.

[quote="JBL"] Each of these paths is counted by $ X_n$ as are half of the other paths,[/quote]
could you please help me in understanding why are half of the other paths are alo counted in $ X_ {n }$ ?

Hi gaurav,

That's the argument from the second paragraph of my preceding post.  The argument is to construct a function from paths to paths, as follows: if the path has all of its peaks and valleys at odd height, it is a fixed point.  Otherwise, it has some peak or valley at even height.  In this case, we "reflect" the first such peak or valley.  For example, if "U" means a step up and to the right and "D" means a step down and to the right and we apply our function to a path that begins UUD..., we get a path that begins UDU... (where the stuff after the "..." hasn't changed).  The effect of this move is to change the number of peaks at even height by exactly 1, in particular changing the parity.  So, this function represents a bijection between (1) paths with at least one peak or valley at even height and an even number of peaks at even height and (2) paths with at least one peak or valley at even height and an odd number of peaks at even height.  In the case that our paths have even length, every path falls into one of these two categories and so we get that half of the paths are in the set counted by  $ X_n$ while half are counted by $ Y_n$.  In the case that our paths are of odd length, there are some with the property that all peaks and valleys are at odd height.  Subtracting these off, we can apply the same argument to the remainder to see that half of the remaining paths are counted by $ X_n$ and the other half by $ Y_n$.

I hope this is more clear.


[quote="AstroPhys"]Are the numbers prime?[/quote]
Then Mr. P would not have said "I don't know what the numbers are..." .

[quote="lingomaniac88"]Well, if the numbers were 2 and 5, both primes, Mr. P could still not know them.  He'd be told "10", so the numbers could be either 2,5 or 1,10.[/quote]

Please read the question:
[quote]none of them are 1. [/quote]

By the way, this should be moved somewhere else, maybe some puzzle section.

We know:
$2\leq n_{1},n_{2}\leq 99$

$2+3\leq n_{1}+n_{2}\leq 98+99$
$5\leq n_{1}+n_{2}\leq 197$

$2 \cdot 3 \leq n_{1}\cdot n_{2}\leq 98 \cdot 99$
$6 \leq n_{1}\cdot n_{2}\leq 9702$

[b]Mr. P : "I don't know what the numbers are..."[/b]
They're obviously not two primes if Mr. P cannot factor them and figure them out.

[b]Mr. S : "I knew you wouldn't know..."[/b]
If the sum were even it could be written as the sum of any two primes and Mr. P could figure it out and Mr. S is definite in the fact that Mr. P would not know so there is no way the sum can be written as the sum of two primes.  We know that the sum is odd and one number must be odd and the other must be even.

Possible sums: $\{11, 17, 23, 27, 29, 35, 37, \ldots \}$

As far as I'm willing to go tonight.  I'll see what I can do tomorrow.

[quote="WatsonLadd"][hide=" Partial Solution"]So Mr. S is able to determine from knowing the sum alone of the numbers that they are not two primes. This means that his sum cannot be even, and cannot be two more then a prime. 17 works. So this is equal to 12+5. Mr. P would have gotten 60 and been unable to tell between 32, 19,17 and 16 as the sums. But all of these are even or two more then a prime other then 17. So when Mr. S tells Mr. P that he knows Mr. P cannot figure out the answer this tells Mr. P that 17 was selected. And so the two numbers are 12 and 5. [/hide] I still need to figure out how Mr. S figures it out.[/quote]

This doesn't work

[hide]Note that $2 \cdot 15 = 5 \cdot 6$.  $2+15=17$ and $5+6=11$ thus they both cannot be feasible solutions because Mr. P would have ruled them out due to $11$ and $17$ are in fact both possible sums.  I don't know.  I hope that made sense.  I think its $4$ and $13$.  I think it must be a prime and an even number that is not $2$ but I don't know how to show this.

Oops I just noticed you said $12$ and $5$ and $12 \cdot 5 = 60 = 3 \cdot 20$ and both $12+5=17$ and $3+20=23$ thus they both cannot be acceptable as I just said above.[/hide]

Skimnc in 6th post;
well,you are just 2 steps away out of 4 steps...
mmm,i don't know if you want me to reveal the solution coz it just need some more thinking and reasoning...however,if you feel just tell me to reveal it...

[hide]as a hint for step 3:
if Mr.P said "I know the numbers" it means that the Product he's dealing with would lead him just to one "possible" case[/hide]

[quote="farbod_mp2002"]Here's a nice puzzle:

3 mathematician were having their time...one of them generated 2 numbers in his mind that both are less than 100 and none of them are 1.
he gave the Sum of the numbers to Mr.S (the second mathematician) and gave the Product of them to Mr.P (the third one). and then,the following discussion started:

Mr.P : "I don't know what the numbers are..."
Mr.S : "I knew you would'nt know..."
Mr.P : "I know the numbers"
Mr.S : "I know the numbers too! "

what are the numbers?  :)[/quote]

[hide="Complete solution"] Let the original numbers be $a, b$.

First statement:  $ab$ is not a prime or a product of two primes.

Second statement:  $a+b$ is not a sum of two primes.  In particular, $a+b$ is not even or two more than a prime, so WLOG $a$ is even and $b$ is odd.

Third statement:  $ab$ is expressible as the product of an even and odd number in exactly one way.  This means that $b$ must be an odd prime and $a$ must be a power of $2$ greater than $2$.

Fourth statement:  $a+b$ is expressible as the sum of an odd prime and a power of $2$ greater than $2$ in exactly one way.

[b]skimnc[/b]'s guess of $\boxed{ (4, 13) }$ works nicely. [/hide]


number theory
greatest common divisor
relatively prime
number theory unsolved

um

rewrite it as $x(x^{2}+x+1)=(y+1)(y-1), gcd(y+1,y-1) \leq 2$ by the alogrithnm
$\iff y-1=x^{2}+x+1 \iff y+1 >x, y=x+1 \iff x=0 \rightarrow y=1$ now to check when the gcd=2 $\iff y=2k+1 \iff x=2m \iff \text{no solution}$ am i right

Actually, we have several solutions.  First, rewrite as

$(x+1)(x^{2}+1) = y^{2}$

$gcd(x+1, x^{2}+1) = 1, 2$; in the first case we can only have $x+1 = 0, \pm 1 \implies (x, y) = (-1, 0), (0, 1)$ (because we cannot have $x+1, x^{2}+1$ simultaneously square except in the case $x = 0$ which has already been covered).  In the second $y$ is odd, so let $y = 2w, x = 2z+1$ and we have

$(z+1)(2z^{2}+2z+1) = w^{2}$

$gcd(z+1, z^{2}+(z+1)^{2}) = 1$; therefore we can only have either $z+1 = \pm 1 \implies (x, y) = (1, 2)$ or else $z+1, z^{2}+(z+1)^{2}$ are relatively prime squares; the latter is a square if and only if $z =-4,-1, 0, 3$ (the first gives no solution, and the second two have already been covered), which gives $(x, y) = (7, 20)$, and so there are no more solutions.

[quote="t0rajir0u"]$z+1, z^{2}+(z+1)^{2}$ are relatively prime squares; the latter is a square if and only if $z =-4,-1, 0, 3$.[/quote]

Sorry, this is incorrect; in fact this is a Pell's equation with infinitely many solutions (not least of which is $z = 20$), but the stipulation that $z+1$ is also a square limits solutions.  I am working on a fix.

[quote="kostas"]solve in intergers the equation

$x^{3}+x^{2}+x+1=y^{2}$[/quote]
This is another problem: Find all  $(x,y,m,n)\in\mathbb{N}^{4}$ such that \[1+x+x^{2}+...+x^{n}=y^{m}.\]

[quote="Rust"]or $x=\frac{(1+\sqrt 2 )^{2k+1}+(1-\sqrt 2)^{2k+1}}{2}$.
Therefore $z^{2}=\frac{(1+\sqrt 2 )^{2k+1}+2-(\sqrt 2-1)^{2k+1}}{4}$.
We must choose k, when z is integer.
I think it is true for odd k.[/quote]

so there are infinite solutions , but these are exactly the conditions  for $x$ ? I mean can we find all $x$ ?

[quote="Rust"][quote="kostas"]so there are infinite solutions , but these are exactly the conditions  for $x$ ? I mean can we find all $x$ ?[/quote]
Algebraic curve (our curve is elliptic) consist only finite integer points, when curve is not special (had not points, were $grad=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})=0$).
Only solutions $(x,y)=(1,2),(7,20)$. There are not another solutions.[/quote]

yes it is Siegel's theorem:the curve $x^{3}+ax+b=y^{2}$ has only finite number of interger points. thanks for the explanation.  :)

I 'll try N.T.TUAN 's equation now .[/hide]

[quote="N.T.TUAN"]
This is another problem: Find all  $(x,y,m,n)\in\mathbb{N}^{4}$ such that
\[1+x+x^{2}+...+x^{n}=y^{m}. \]
[/quote]

This is ridiculously difficult: it is a famous problem, and I don't think that it has been solved completely... Or if it has been solved, the proof most certainly used rather sophisticated techniques such Baker's method. Of course, for any fixed $n \geq 2$ the equation has finitely many solutions $x,m$ with $m \geq 2$.

Here is a related diophantine equation.

This is a classical equation due to Fermat. (Burton)

"Find a cube which when increased by the sum of its proper divisors becomes a square" For example $7^{3}+1+7+7^{2}=20^{2}$.

Fermat considered a simpler case, say, $x$ is an odd prime. Then $x^{3}$ has $1,x,x^{2}$ as its proper divisors and hence, $1+x+x^{2}+x^{3}=y^{2}$.

As torajirou suggested we can factorize $(1+x)(1+x^{2})=y^{2}$. Note that $\gcd(1+x,1+x^{2})=2$ we can write, $uv = \left( \frac{y}{2}\right)^{2}$ with $u = \frac{1+x}{2}$ and $v=\frac{1+x^{2}}{2}$.

[b]Theorem:[/b] If a product of two positive relatively prime integers is a square then each one is a square.

Hence, $u=s^{2}$ and $v=t^{2}$ hence, $1+x=2s^{2}$ and $1+x^{2}= 2t^{2}$. Note the second equation reduces to $x^{2}-2t^{2}=1$ which is Pellian.

If you set $y=x+z$  this equation transforms at $x^{3}-(2z-1)x-(z^{2}-1)=0$ and i prove that for every $z$ there is none or one solution of $x$.As kostas wrote the equation $x^{3}+ax+b=0$ has finite integer solutions....and final the only solutions are $x=-1,0,1,7.$


geometry trigonometry function inequalities

Show that for a triangle with sides $ a$, $ b$, and $ c$:
a) $ a^2\plus{}b^2\plus{}c^2\ge4\sqrt{3}A$, where $ A$ is the area, and
b) Equality exists only in an equilateral triangle.

(Mods, in retrospect, this might belong in Intermediate.)

This was a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14388]question[/url] on an IMO ... (even if it was 1961) For an example of a stronger version, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=229859]this thread[/url].

Prove this problem:
$ a^2\plus{}b^2\plus{}c^2\geq 4\sqrt{3}A\plus{}(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2$ :D

$ a^2= b^2+c^2- 2bc \cos A =( b - c ) ^2 +2bc - 2bc \cos A =( b - c ) ^2+ 2bc (1 - \cos A) = ( b - c ) ^2 +4S\frac{1-cos\ A}{\sin A} , \text{where S is the area of triangle, and } S= \frac{1}{2}bc \sin A$

And
$ \frac{1-\cos A}{\sin A }=\frac{2\sin^2\frac{ A}{2} }{2\sin\frac{A}{2}\cos\frac{A}{2}}= \tan \frac{A}{2}$

therefore,
$ a^2=( b - c) ^2 + 4S \tan \frac{A}{2}$

Similarly,
$ b^2 = ( c- a) ^2 + 4S \tan\frac{B}{2}$
and
$ c^2 = ( a-b ) ^2 + 4S \tan\frac{C}{2}$

$ \text{where A/2 , B/2, C/2 all lie in} [0, \pi/2] ,\text{and  }\tan \text{is convex function in} [0, \pi/2)$
Thus
$ \tan \frac{A}{2} + \tan \frac{B}{2} + \tan \frac{C}{2} \ge 3{\tan \frac{\frac{A}{2}+\frac{B}{2}+\frac{C}{2}}{3}}$

$ = 3 \tan\frac{A+B+C}{6} = 3 \ tan 30^o= \sqrt{3}$

$ \text {hence }$

$ a^2 + b^2 + c^2 \ge ( b - c) ^2 + ( c- a) ^2 + ( a-b ) ^2 + 4S \sqrt{3}$

${ \implies  a^2 + b^2 + c^2 - ( b - c) ^2 - ( c- a) ^2 - ( a-b ) ^2 \ge 4S \sqrt(3}$

$ \text {Where S = A , A in the original inequality . }$

$ \text {So, }$

$ a^2 + b^2 + c^2 - ( b - c) ^2 - ( c- a) ^2 - ( a-b ) ^2 \ge4A \sqrt{3} .....\text{ans.}$

Nice solution,vinskmen!
More problem :D
Prove that:
1/ $ xa^2\plus{}yb^2\plus{}zc^2\geq 4\sqrt{xy\plus{}yz\plus{}zx}S \ \forall x,y,z>0$
2/$ \frac{xa^2}{y\plus{}z}\plus{}\frac{yb^2}{z\plus{}x}\plus{}\frac{zc^2}{x\plus{}y}\geq 2\sqrt{3}S \ \forall x,y,z>0$

Refering to the initial problem we begin with the main idea that we have equality for the equilateral triangle.
I have here a series of proofs:
[b]The first one[/b] :an equilateral triangle with the side $ c$ has the altidude $ \frac {c}{2}\sqrt {3}$.Any triangle with side $ c$ will have an altitude perpendicular on $ c$ with the lenght $ \frac {c}{2}\sqrt {3} \plus{} y$.She will divide $ c$ in two parts $ \frac {c}{2} \minus{} x$ and $ \frac {c}{2} \plus{} x$.
\[ a^2 \plus{} b^2 \plus{} c^2 \minus{} 4\sqrt {3}A \equal{} (\frac {c}{2} \minus{} x)^2 \plus{} (\frac {c}{2} \plus{} x)^2 \plus{} 2(y \plus{} \frac {c}{2}\sqrt {3})^2 \plus{} c^2 \minus{} 2\sqrt {3}c(y \plus{} \frac {c}{2}\sqrt {3}) \equal{} 2x^2 \plus{} 2y^2\geq 0
\]
with equality only if $ x \equal{} y \equal{} 0$
For [b]the second one[/b] let $ a\leq b \leq c$.Build the equilateral triangle $ ABC'$ on $ AB$ and denote $ p \equal{} CC'$ as beeing the deviaton from an equilateral triangle.Relying on the cosines theorem we have
\[ p^2 \equal{} a^2 \plus{} c^2 \minus{} 2ac\cos(\beta \minus{} 60) \equal{} a^2 \plus{} c^2 \minus{} 2ac(\cos \beta \cos 60 \plus{} \sin \beta\sin 60)
\]

\[ p^2 \equal{} a^2 \plus{} c^2 \minus{} ac\cos\beta \minus{} \sqrt {3}ac\sin\beta \equal{} a^2 \plus{} c^2 \minus{} 2\sqrt {3}A \minus{} \frac {1}{2}(2ac\cos\beta)
\]

\[ p^2 \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2}{2} \minus{} 2\sqrt {3}A \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} 4\sqrt {3}A}{2}\geq0
\]
with equality only if $ a \equal{} b \equal{} c$
For[b] the third one[/b] suposes that $ 4A\sqrt {3} > a^2 \plus{} b^2 \plus{} c^2$only if $ 2bc\sin\alpha > \frac {1}{\sqrt {3}}(a^2 \plus{} b^2 \plus{} c^2)$
As a result we have $ 2bc\cos\alpha \equal{} b^2 \plus{} c^2 \minus{} a^2$.We will obtain the next contradiction:
$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 > a^4 \plus{} b^4 \plus{} c^4$ only if $ (a^2 \minus{} b^2)^2 \plus{} (b^2 \minus{} c^2)^2 \plus{} (c^2 \minus{} a^2)^2 < 0$
[b]The forth one [/b] relies on the Heron formula and AM-GM and will result
$ 16A^2 \equal{} (a \plus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)\leq(a \plus{} b \plus{} c)(\frac {a \plus{} b \plus{} c}{3})^3$,
$ 4A\leq\frac {(a \plus{} b \plus{} c)^2}{3\sqrt {3}} \equal{} \sqrt {3}(\frac {a \plus{} b \plus{} c}{3})^2\leq\sqrt {3}\frac {a^2 \plus{} b^2 \plus{} c^2}{3}$ or $ a^2 \plus{} b^2 \plus{} c^2\geq 4A\sqrt {3}$
[b]The fifth[/b] one says that $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca \equal{} 2A(\frac {1}{\sin\alpha} \plus{} \frac {1}{\sin\beta} \plus{} \frac {1}{\sin\gamma})$ and now we will use the fact that $ f(x) \equal{} 1/\sin s$ is convex and with Jensen results imediately.
In [b]the 6th[/b] try to find an inequality that for the equilateral triangle to become equality.This is $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2\geq 0$ and now use the fact that $ ab \equal{} 2A \\
sin\gamma$ etc..

And now to prove The Hadwiger-Finsler inequality that Kyoshiro proposed we will write it in the form
\[ (a \minus{} b \plus{} c)(a \plus{} b \minus{} c) \plus{} (b \minus{} c \plus{} a)(b \plus{} c \minus{} a) \plus{} (c \minus{} a \plus{} b)(c \plus{} a \minus{} b)\geq 4A\sqrt {3}
\]
and then try the substitution $ x \equal{} \minus{} a \plus{} b \plus{} c$, $ y \equal{} a \minus{} b \plus{} c$ and $ z \equal{} a \plus{} b \minus{} c$ and we will obtain
\[ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\geq\sqrt {3(\frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx})}
\]
the same with $ u^2 \plus{} v^2 \plus{} w^2\geq uv \plus{} vW \plus{} wu$

a,b,c are the sides of the triangle and x,y,z are any positive real numbers.

if you want to try a special one in my opinion you can go for this one:
For any triangle with $ a,b,c$ the sides of this triangle we have
\[ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a)\geq 0
\]
[i](Proposed by Klamkin and used for OIM 1983, it was initialy allotted to E.Catalan, Educational Timesc N.S 10,57(1906))[/i]

[quote="vinskman"]

$ a^2 \plus{} b^2 \plus{} c^2 \minus{} ( b \minus{} c) ^2 \minus{} ( c \minus{} a) ^2 \minus{} ( a \minus{} b ) ^2 \ge4A \sqrt {3} .....\text{ans.}$[/quote]

That's Hadwiger- Finsler inequality.

I have forgotten to say that the initial inequality $ \boxed{a^2\plus{}b^2\plus{}c^2\geq 4\sqrt{3}A}$ it is called Weitzenbock inequality, and it's source was in here: $ OF_1^2\equal{}a^2\plus{}b^2\plus{}c^2\minus{}4A\sqrt{3}$ where $ F_1$ is the first point of Fermat.

We need to prove that $ a^2 \plus{} b^2 \plus{} c^2\geq 4\sqrt {3} A$. Now replace $ a^2$ by $ b^2 \plus{} c^2 \minus{} 2bc\cos\alpha$ and $ A$ by $ \frac {bc\sin\alpha}{2}$.

The above is equivalent to $ b^2 \plus{} c^2 \minus{} bc(\sqrt {3}\sin\alpha \plus{} \cos\alpha) \geq 0 \ (*)$.

But $ \sqrt {3}\sin\alpha \plus{} \cos\alpha \leq 2$ because (using $ \sin(\alpha \plus{} \theta) \equal{} ....$)
\[ \frac {\sqrt {3}\sin\alpha \plus{} \cos\alpha}{2} \equal{} \sin(\frac {\pi}{6} \plus{} \alpha) \leq 1.
\]
Now look again at $ (*)$ to see that the inequality is proven.


Does anyone know the relative difficulties of the national olympiads? I've been using a lot of the national olympiad problems on Kalva, but I would like to start from easier problems and work my way up.

Thanks.

[b]Relative[/b]... :roll:

I think CanMO is easier but I'm not sure.

Ehhh... I think the main tough ones are US and China. Russia and Japan are also fairly hard, and I think Korea and India are not that easy either. Oh, Iran is pretty hard. I think all the others aren't that tricky though, and roughly the same level.