algebra
polynomial
irrational number
Rational Root Theorem

If $ p$ is an integer and $ \sqrt {p}$ is a rational number, then $ \sqrt {p}$ must be an integer (cannot be a terminating decimal number like 5.62)?

Is this true or false?

So basically the square root of an integer will either be an integer or an irrational number that never ends?

[hide="Yes"] This is a consequence of the [url=http://en.wikipedia.org/wiki/Rational_root_theorem]Rational root theorem[/url] applied to the polynomial $ x^2 \minus{} p \equal{} 0$.  The generalization to $ x^n \minus{} p \equal{} 0$ is obvious. [/hide]


A long, straight wire containing a semicircular region of radius 0.95m  is placed in a uniform magnetic field of magnitude 2.20 T. What is the net magnetic force acting on the wire when it carries a current of 3.40 A?

depends on dirn of mag field

i told the force on semicircular portion only :P


make y the "bottom" face. fold the other sides in and leave y the bottom face, such that G is "left" of y and O is "right of Y and w is below Y and b is above y

if u picture it right, you;; see the w is across from b , so the ans is b

there are many other ways to picture it

The folded net would yield the cube shown below where the center square on each side of the cube identifies the color of the corresponding side. We see that white is opposite to the color blue thus the answer is $ \boxed{B}$


1)prove that $ {2222}{}^5{}^5{}^5{}^5$+$ {5555}{}^2{}^2{}^2{}^2$ is divisible by 7.
2)Factorise ${ (x-y)}^3$+$ {(y-z)}^3$+$ {(z-x)}^3$.
3)find all (x,y) such that $ 2{x}^3$+$ {y}^2$=375
this are sample questions for the tournament of pure mind PM me to participate[quote]Einstein said"a question that sometimes drives me hazy is am i or the others crazy"

[hide]
2.
3(xy(y-x)+zy(z-y)+zx(x-z)). I do not know if I am done.[/hide]

No, I think the desired answer is [hide]$ \minus{}3(x\minus{}y)(x\minus{}z)(y\minus{}z)$[/hide].

hello, to 2) the right answer is
$ (x\minus{}y)^3\plus{}(y\minus{}z)^3\plus{}(z\minus{}x)^3\equal{}\minus{}3(x\minus{}y)(x\minus{}z)(y\minus{}z)$.
Sonnhard.


Show that the equation $a^{2}b^{2}+b^{2}c^{2}+3b^{2}-c^{2}-a^{2}=2005$ has no integer solutions.

[i]Arnoldo Aguilar, El Salvador[/i]

Oh well, we indeed get \[2002= (b^2 - 1)(a^2 + c^2 + 3)\]

One of the factors on the right hand side must be even because the left hand side is, but it can't be the first one since then $b$ must be odd and the right hand side will be a multiple of 8 while the left hand side obviously isn't.
So the second factor must be even and the first one must be odd. But it is impossible for the second factor to be equal to 2 (mod 4) because the square residues (mod 4) are 0 and 1. So there are no solutions.

Yes, you're right! Indeed, the official solution is esentially the same: factorize the expression, and then compare the form of the factors in order to obtain a contradiction.

The worst thing about this pretty equation is that it succumbs to a straightforward modulo 4 approach... You even don't need to factorize it :(

[quote="shyong"][quote="siuhochung"]$\frac{2002}{a^2+c^2+3}$ is a perfect square.
We only need to check the 8k+3, 4, 7, 0 factors of 2002.
They are, 7, 11, 91, 143.
Then it remains simple checking.[/quote]

How come $\frac{2002}{a^2+c^2+3}=b^2-1$ would be a perfect square ? :?[/quote]
oops sorry, I got a mistake in calculations.

We can rewrite the equation as

$(a^{2}+c^{2}+3)(b^{2}-1)=2002$

Now, the divisors of $2002$ are $ 1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,2002 $, and the only one that can be written as $b^{2}-1$ with $b$ integer is $143$ what leaves us with $a^{2}+c^{2}=11$ and this last equality has no integer solutions for both $a,c$.

Note: the post just above mine is probably the best-[b]written[/b] solution to this problem. Just in case you needed help with this problem, like I did.  :)

Note: the post just two above mine is probably the best-written solution to this problem. Just in case you needed help with this problem, like I did.  :)


x,y,z different positive integers ;IF $ y^{2}\minus{}zx\equal{}\minus{}103 , z^{2}\minus{}xy\equal{}22$
$ x^{2}\minus{}yz\equal{}?$

Trying $ y\equal{} 1, 2, 3$ in $ y^{2}\minus{}zx\equal{}\minus{}103$,

$ y\equal{}3 \implies zx\equal{}112 \equal{} 8\times 14$

$ 8^{2}\minus{}3\times 14 \equal{} 22$.

Therefore $ x\equal{}14, y\equal{}3, z\equal{}8$

and so

$ x^{2}\minus{}yz\equal{}196\minus{}24\equal{}172$

Subtracting and factoring yields that either
z-y=5 and x+y+z=25
or
z-y=1 and x+y+z=125

Plugging z=y+5, x=25-y-(y+5)=20-2y yields desired result, the other branch has irrational roots.


inequalities
geometry
perimeter
inequalities proposed

In Mathematics and Youth Magagine, Vol 346 in 2006, I proposed the following inequality

[u][b]Problem 1.[/b][/u] Prove that for all non-negative real numbers $a,b,c$ such that $a+b+c=1$ then
\[ \frac{ab+bc+ca}{a^2b^2+b^2c^2+c^2a^2} \ge 8(a^2+b^2+c^2).  \]

It's not hard problem, I think. But there are some other nice and hard one. First, try to prove problem $1$ and a strange inequality as follow

[u][b]Problem 2.[/b][/u] Prove that for all non-negative real numbers $a,b,c$ such that $a+b+c=3$ then
\[ \frac{ab+bc+ca}{a^3b^3+b^3c^3+c^3a^3} \ge \frac{a^3+b^3+c^3}{36}.  \]

After that, I will post some nice other. Hope you will like it.

 :lol:

first one :
Let $ab+bc+ca=x$ , $abc=y$ then we have

$\sum (ab)^2=x^2-2y$ , $\sum a^2=1-2x$

so question equivalent to

$\frac{x}{x^2-2y}\geq 8(1-2x)$ $\Longleftrightarrow 16x^3-8x^2-32xy+x+16y\geq 0$
with $x\leq \frac{1}{3}$

But LHS = $16y(1-3x)+16xy+x(4x-1)^2$ .
So we are done .

Here is the second extension

[b]Problem 3.[/b] Prove that for all non-negative real numbers $a,b,c$ which add up to $1$ then
\[ \frac{ab+bc+ca}{a^2+b^2+c^2+16abc} \ge 8(a^2b^2+b^2c^2+c^2a^2).  \]
And prove that $16$ is the best constant (if we replace $16$ by a bigger one, it's no true).

I solved also the second problem!
 (the solution is similar to shyong's above for the first one, if you want I can post it)
the equality case is $(0,1,2)$ :)

Very nice problems hungkhtn! :)

(I'm working on the third problem now)

The sign for the third one is reseverd. Sorry, now I had corrected it.

[quote]I solved also the second problem!
(the solution is similar to shyong's above for the first one, if you want I can post it)
the equality case is $(a=2,b=1,c=0)$.

Very nice problems hungkhtn!

(I'm working on the third problem now)[/quote]

Yes, but expanding way isn't the good solution?

for the second one, we only need to use mixing variables to prove it

Oh, Lovarz.! Mixing variable is work here and $(2,1,0)$ is also the unique case of equality,  :lol:

[quote="Lovasz"]Actually, I'm bad at mixing variables.  :oops:
Can you post solution? Thanks.[/quote]
Just let
$f(a,b,c)=\frac{ab+bc+ca}{(a^3+b^3+c^3)(a^3b^3+b^3c^3+c^3a^3)}$,$(a\ge b \ge c)$
It is easy to prove:$f(a,b,c) \ge f(a,b+c,0)$

Yes, Zhaobin. It's also my solution. But what's your idea for the third ineq? It's more difficult.

[quote="hungkhtn"]Here is the second extension

[b]Problem 3.[/b] Prove that for all non-negative real numbers $a,b,c$ which add up to $1$ then \[\frac{ab+bc+ca}{a^{2}+b^{2}+c^{2}+16abc}\ge 8(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}).\] And prove that $16$ is the best constant (if we replace $16$ by a bigger one, it's no true).[/quote]
Hi,hungkhtn:
I use the method of shyong's to prove the inequality:
Let $x=3(ab+bc+ca),y=27abc$
we will get the inequality is equal to:
$256y^{2}-8(48x^{2}+36x-54)y+27x(4x-3)^{2}\ge 0$
(we have  $0 \le y \le x \le 1$ and also $y \ge 4x-3$(by schur's))
Then discuss for $0\le x \le \frac{3}{4}$ and $\frac{3}{ 4}\le x \le 1$
I hope it is ok.
And for $16$ is the biggest one,
Let $a=b=\frac{1}{ 2}-k$,$c=2k$
let $k \to 0^{+}$
Is it OK?

Yes, very nice, Zhaobin. It's also my solution which shared in M&Y 7.2006. There are also two nice improvements for $n$ numbers

Problem 5. \[\frac{\sum_{i<j}a_{i}a_{j}}{\sum_{i<j}a_{i}^{2}a_{j}^{2}}\ge 8\sum_{i=1}^{n}a_{i}^{2}\] (if also $a_{1}+a_{2}+...+a_{n}=1$ and $a_{1},a_{2},...,a_{n}\ge 0$.

Problem 6. Prove that \[\frac{ab+bc+ca}{a^{2}b^{2}+b^{2}{c}^{2}+c^{2}a^{2}+\frac{19}{8}abc}\le 8(a^{2}+b^{2}+c^{2})\] if $a,b,c\ge 0$ and $a,b,c$ are three lengths of a triangle with perimeter $1$.

Could you prove it, Zhaobin?

[quote="zhaobin"][quote="Flove"]
Problem 5. \[\frac{\sum_{i<j}a_{i}a_{j}}{\sum_{i<j}a_{i}^{2}a_{j}^{2}}\ge 8\sum_{i=1}^{n}a_{i}^{2}\] (if also $a_{1}+a_{2}+...+a_{n}=1$ and $a_{1},a_{2},...,a_{n}\ge 0$.

Could you prove it, Zhaobin?[/quote]
I think it is also mixing variables :)

For Problem 6:First there is a typo(I think you know)
And I guess It also can be proved by let $x=3(ab+bc+ca),y=27abc$
and find some contralation on $x,y$ by $a,b,c$ are sides of triangle.Anyway I don't have my pan now,I'm sorry if wrong :)[/quote]

Solution for problem 5 is very simple. I don't use mixing variables.


Any Photoshop users know of a good beginners guide to teach ppl to use the program?...(I mean commercial...not stuff Adobe provides with the software)...

And then, how about ImageReady?

...And Illustrator, GoLive, and InDesign?

best way to learn these stuff is have a friend teach you or just simply play around with it..
reading computer books is soo boring.. but if u could get past tat, its good info in there..

This is one site I rely on...

http://www.good-tutorials.com


Each interior angle of a regular n-gon measures x degrees. This n-gon has y diagonals. If x+y=295, find n.

could someone show me step-by-step how to do this problem?

[hide="solution"]
The number of diagonals in a regular n-gon is
$ \frac{n(n\minus{}3)}{2}$
The measure of an interior angle of a regular n-gon is
$ \frac{180(n\minus{}2)}{n}$

If you need me to, I can explain why these formulas are true.
Now, we can just plug in and solve.
$ \frac{n(n\minus{}3)}{2}\plus{}\frac{180(n\minus{}2)}{n}\equal{}295$
Multiply both sides by 2n
$ n^3\minus{}3n^2\plus{}360n\minus{}720\equal{}590n$
$ n^3\minus{}3n^2\minus{}230n\minus{}720\equal{}0$
We find $ \boxed{n\equal{}18}$. I did this by guess and check, but there might be a better way.
[/hide]

how would u factor that though.... that is if u dont do guess and check...

$ n^3\minus{}3n^2\minus{}230n\minus{}720\equal{}0\implies(n\minus{}18)(n^15n\plus{}40)$ The whole number solution is 18.

you could for instance use the rational roots theorem and see that the number (if integer root) has to be a factor of 720  :)

Meh I factored it wrong it should be:
$ n^3\minus{}3n^2\minus{}230n\minus{}720\equal{}0\implies(n\minus{}18)(n^2\plus{}15n\plus{}40)$


How many pemutations $ \pi$ of set $ \{1,2,...,n\}$ exist such that $ \pi(1)<\pi(2)>\pi(3)<\pi(4)>...$  or $ \pi(1)>\pi(2)<\pi(3)>\pi(4)<...$  (alternative less and bigger, you understand   :) ) ?

"Alternating" (or "up-down," "down-up," "zig-zag," etc.) is the word you want.  This is a very famous problem, and it's sequence [url=http://www.research.att.com/~njas/sequences/A000111]A000111[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  (Well, that counts either the up-down or the down-up permutations -- double it to get [url=http://www.research.att.com/~njas/sequences/A001250]A001250[/url], counting both types together.)


abstract algebra
group theory
Galois Theory
superior algebra
superior algebra unsolved

Show that every sub group of order < 60 is solvable

You probably will need to work on a case by case basis.

If you remmeber that abelian -> nilpotent -> soluble, you can knock out quite a few cases immediately.

Also remember that every p-group is nilpotent.

You will probably need to use a few more theorems and results - perhaps Feit and Thompsons result which says that every group of odd order is soluble.

Look on Google books for : A Course on Group Theory by John S. Rose. I'm sure this will help you through the question.

There are the following theorems:
Let $ p,q,r$ be distinct primes and $ a,b\in\mathbb{Z}_ \plus{}$. Then every group of order $ p^aq^b$ is solvable, and every group of order $ pqr$ is solvable. Thus the smallest non-solvable group has order at least $ 2^2\cdot 3\cdot 5\equal{}60$. In my opinion, Feit-Thompson theorem is too strong theorem for this problem.


A simple magic square of order $ n$ is an arrangement of numbers $ 1,2,...,n^2$ in a square such that the $ n$ numbers in all rows, all columns, and both diagonals sum to the same constant. Find this constant.

[hide]Consider the sum of all n rows. It is equal to n times this constant. However it is also equal to the sum of all the numbers, so the constant is:
$ \frac{n(n^2\plus{}1)}{2}$[/hide]


search
number theory
algebra proposed
algebra

Find $x,y,z,n \in Z^{+}$ such that:
$n^{x}+n^{y}=n^{z}$

It was solved a little time ago by pco, in the number theory forum I think, you could search for it.
edit: http://www.mathlinks.ro/Forum/viewtopic.php?t=150284

and also two similar problems:
$i)$find all the natural solutions of $n^{x}+n^{y}+n^{z}=n^{t}$
$ii)$find all the natural solutions of $4^{x}+4^{y}+4^{z}=4^{t}$

i, natural solutions of this equation are $n=2, x=y, z=x+1, t=x+2$ and $n=3, x=y, x=z, t=x+1$  ($x\leq y\leq z<t$)

yeah and for $ii)$ you can easily prove it using $i)$...

I don't see any reason for you to be posting point a), but heh, you've got your reasons. Just scroll up a little.
For point b) since they are all natural numbers it follows quickly that $x,y,z\leq 9$ and a "brute-force" computation will five the solutions quite easily.(for example: $512+32+8=552$)


function
floor function
algebra unsolved
algebra

The function $f: N \rightarrow N$ is defined recursively with $f(1) = 1$ and

$f(n+1) = f(n) + 2$ if $n = f(f(n)-n+1)$
$f(n+1) = f(n) + 1$ otherwise

for all $n \geq 1$.

a) Prove that $n \leq f(f(n) - n + 1) \leq n+1$ for all $n$.
b) Find an explicit formula for $f$.

first notice that $f$ is well defined.

Now put $g(n)=\left\lfloor \vartheta n\right\rfloor$ where $\vartheta=\frac{1+\sqrt{5}}{2}$, and verify $g(n)$ satisfy the given recurrence and initial values of $f(n)$ and hence $f(n)=g(n)$


inequalities
geometry
trigonometry
inequalities proposed

$x_i,y_i$ are reals and $i=1,2....,n$ and $n\in N*$.
If $\sum x_i^2=\sum y_i^2=1$ prove that
$(x_1y_2-x_2y_1)^2\leq 2|1-\sum x_iy_i|$

From Korea.
By Lagrange's Identity,\[(x_1y_2-x_2y_1)^2\le\sum(x_iy_j-x_jy_i)^2=\sum x_i^2\sum y_i^2-(\sum x_iy_i)^2=(1+\sum x_iy_i)(1-\sum x_iy_i)\le2|1-\sum x_iy_i|\]

The right-hand side equals \[2\left|1 - \sum_{i = 1}^nx_iy_i\right| = \left|\sum_{i = 1}^nx_i^2 - 2\sum_{i = 1}^nx_iy_i + \sum_{i = 1}^ny_i^2\right| = \left|\sum_{i = 1}^n\left(x_i - y_i\right)^2\right| = \sum_{i = 1}^n\left(x_i - y_i\right)^2.\]

Now, it is sufficient to prove that if $x_1^2 + x_2^2 \leq 1$ and $y_1^2 + y_2^2 \leq 1$ then the inequality \[\left(x_1y_2 - x_2y_1\right)^2 \leq \left(x_1 - y_1\right)^2 + \left(x_2 - y_2\right)^2\] holds. We use a geometric idea to do this.

Consider the points $A\left(x_1,x_2\right)$ and $B\left(y_1,y_2\right)$ in the plane.
These points lie inside the unit circle centered at the origin $O$.

The area of triangle $ABO$ equals $\frac{1}{2} \cdot \left|x_1y_2 - x_2y_1\right|$, and $AB = \sqrt{\left(x_1 - y_1\right)^2 + \left(x_2 - y_2\right)^2}$.

Hence, we need to prove that \[2\left[AOB\right] \leq AB.\] That is easy though: \[2\left[AOB\right] = AB \cdot OA \cdot \sin{\angle{OAB}} \leq AB \cdot OA \leq AB\] since $A$ lies inside the unit circle, and we are done.

Edit: you beat me, mecrazywong :D

Not really. This is only because we started from different sides of the inequality. Also, the geometric interpretation in your solution is indeed surprising.

See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=327849&sid=45bbb8d3c80c1a891bf4227455290b38#p327849]http://www.mathlinks.ro/Forum/viewtopic.php?p=327849&sid=45bbb8d3c80c1a891bf4227455290b38#p327849[/url](only for Arne).


Prove that :

$ \{x\in \mathbb{R}\ |\ [x]\cdot \{x\}\equal{}x\}\equal{}\{\frac{p^2}{p\minus{}1}\ |\ p\in \mathbb{Z},p\le 0\}$

[hide]$ \lfloor x \rfloor \{x\} \equal{} \lfloor x \rfloor \plus{} \{x\} \Longrightarrow \{x\} \equal{} \frac{\lfloor x \rfloor}{\lfloor x \rfloor \minus{} 1}$. As $ 0 \le \{x\} < 1$, it follows that $ \lfloor x \rfloor \le 0$. Then $ x \equal{} \lfloor x \rfloor \plus{} \frac{\lfloor x \rfloor}{\lfloor x \rfloor \minus{} 1} \equal{} \frac{p^2}{p\minus{}1}$ for $ 0 \ge p \in \mathbb{Z}$. This is sufficient, as $ \frac{p^2}{p\minus{}1} \equal{} p \plus{} \frac{p}{p\minus{}1}$, where $ 0 \le \frac{p}{p\minus{}1} < 1$ for $ p \le \minus{}2$, and $ \lfloor x \rfloor \{x\} \equal{} p \times \frac{p}{p\minus{}1}$; and we easily verify that the statement is true for $ p \equal{} \minus{}1,0$.[/hide]


Normally I can decifer this kind of internet speaking, but xD has really got me stumped.  Can anyone help?


function
analytic geometry
graphing lines
slope
real analysis
real analysis unsolved

Prove that every real $f \in C([0,1])$ can be uniformly approximated by piecewise linear function $g$ whose linear pieaces, finite in number, have a slope $\pm a$ for a prescribed value $a$.

Weierstass-Stone theorem can be aplied. Or am I wrong?

[quote="mlok"]Slope $\pm a$ would be more correct. And $a\ne 0$, of course. W-S does not help, since such functions don't form an algebra.

But I may be reading this wrong: if piecewise linear functions are not required to be continuous, then approximation is possible with fixed slope $a$, including $a=0$.[/quote]

Yes, these piecewise functions are not required to be continuous.
And could you please explain you construction, how can we do this.

So there are two problems here:

#1. Approximate $f\in C[0,1]$ by a function $g$ such that $g'(x)=a$ for all but finitely many values of $x$.

Solution: For an integer $n$ define $g_{n}(x)=f(k/n)+a(x-k/n)$ for $\frac{k-1}{n}<x\le \frac{k}{n}$, $k=1,\dots,n$. Extend $g_{n}$ to $0$ by continuity. The difference $|f(x)-g(x)|$ does not exceed $|a|/n+\omega_{f}(1/n)$, where $\omega_{f}$ is the modulus of continuity of $f$. Thus $g_{n}\to f$ uniformly.

#2 Approximate $f\in C[0,1]$ by a function $g\in C[0,1]$ such that $|g'(x)|=a$ for all but finitely many values of $x$. Here $a\ne 0$.

Solution: :?:

take $\epsilon > 0$ and $\eta$ a $\epsilon$-modulus of continuity such that $|a| \eta < \epsilon$ : $|x-y|<\eta$ implies $|f(x)-f(y)|<\epsilon$.
 Then, for $x \in [k \eta; (k+1) \eta[$, define $g(x) = f(k \eta)+(x-k \eta)a$.
Hence, if $k \eta \leq x < (k+1) \eta$:
$|g(x)-f(x)| \leq |g(x)-g(k \eta)|+|g(k \eta)-f(k \eta)|+|f(k \eta)-f(x)| \leq |a| \eta+\epsilon \leq 2 \epsilon$

edit: same solution mlok  :lol:  and #2 is clearly impossible if one imposes continuity

My bad: in #2 $f$ must be assumed Lipschitz with constant at most $a$: $|f(x)-f(y)|\le a|x-y|$ for all $x,y\in [0,1]$.


algebra
polynomial
limit
function
pigeonhole principle
inequalities
geometry

The 2009 CMO will be written tomorrow, March 25, 2009! Good luck to all the participants! :D  :D

Also, let's withhold any discussion until [b]Friday evening (7 pm EST)[/b]. If anyone feels it should be earlier/later, please say so.

yes good to you all...eventhough im not writing it... :(

i think we can start discussing now, since there has been a thread on compsci.ca since yesterday
so, how did everyone do?

Yeah, I think we can start discussing now...

Well... I did better than last year, that's for sure... but I'm kind of mad at myself for not leaving enough time to finish 1 (I left the last 20 mins... i.e. I didn't finish it). 2 was very easy, and 3 I think the upper bound was 9/8, and the lower bound was 1, although I stupidly wrote 0 on the contest, and 4 was not bad... though I had a really long solution, and 5, I tried (and sort of failed) in using the Extremal principal, and afterwards I reviewed my notes, and realized that it was trivial with Helly's theorem... oh well...

How did everyone else do?  :maybe:

I think I got #1; I'm not sure, and #2 was pretty easy. For #3 I got a lower bound of one and an upper bound of 2. I got a bit of #4, hoping for a 2 or 3 on it, and I didn't get #5 at all.

Solutions to 1,2,3

1:
[hide]
We find a recurrence (big surprise). Let $ f(n)$ be the number of good $ 2$ by $ n$ boards.

Consider the furthest most right column of the $ 2$ by $ n\plus{}1$ board.

If the top right and bottom right squares are white, we have $ f(n)$ boards.
If the top right and bottom right squares are black, we have $ f(n)$ boards.
If the top right is black, bottom right is white, we have $ f(n)$ boards.

If the top right is white, bottom right is black, then we have all other squares on the bottom being black, and the remaining top squares are therefore arbitrary, giving us $ 2^n$ boards.

Hence, $ f(n\plus{}1)\equal{}3f(n)\plus{}2^n$

So $ f(n)\equal{}3f(n\minus{}1)\plus{}2^{n\minus{}1}$ and $ 2f(n\minus{}1)\equal{}6f(n\minus{}2)\plus{}2^{n\minus{}1}$ so subtracting yeilds

$ f(n)\minus{}2f(n\minus{}1)\equal{}3f(n\minus{}1)\minus{}6f(n\minus{}2)$

or $ f(n)\equal{}5f(n\minus{}1)\minus{}6f(n\minus{}2)$ which is a recurrence relation with characteristic polynomial $ x^2\minus{}5x\plus{}6\equal{}(x\minus{}2)(x\minus{}3)$, so has the form $ A\cdot 2^n\plus{}B\cdot 3^n$

Now, we also know that $ f(1)\equal{}4$ and $ f(2)\equal{}14$ so

$ 2A\plus{}3B\equal{}4$
$ 4A\plus{}9B\equal{}14$

yielding $ A\equal{}\minus{}1$ and $ B\equal{}2$, so $ f(n)\equal{}2\cdot 3^n\minus{}2^n$
[/hide]
2)
[hide]
Double counting trivializes the problem. Consider the sum over all possible arrangements of the number of sectors which line up with the same color.

Counting one way, we have this number to be $ 100*100\plus{}100*100\equal{}20000$ since the number is simply the number of pairs of sectors of the same color, one from the large circle, and one from the small.

On the other hand, if every sum is $ <100$ then the sum will be less than $ 200*100\equal{}20000$, contradiction!
[/hide]
3)
[hide]
$ 1\plus{}\frac{xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\le 1\plus{}\frac{xyz}{2\sqrt{xy}2\sqrt{xz}2\sqrt{yz}}\equal{}\frac{9}{8}$

The lower bound of 1 can never be attained, but can be reached arbitrarily close from above, by considering $ \lim_{x\minus{}>0} f(x,1,1)\equal{}\lim_{x\minus{}>0} 1\plus{}\frac{x}{2(x\plus{}1)^2}\equal{}1$, but can never be attained since $ \frac{xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}>0$

Since $ f(x,y,z)$ is a continuous function on the positive reals, it must therefore attain every value in the interval $ (1, \frac{9}{8}]$.[/hide]

what do you think cutoffs will be for group 1 and 2?

I think that this year was relatively easier than most years, because the problems were really standard... like pigeonhole = #2, inequality = #3, number theory = #4, so maybe the cutoff for group 1 will be like 25-28, and group 2 might be 20-21? I might be way off with these, but that would be my guess... What does everyone else think?

Also, for people who didn't write the contest, I typed up the problems below (but note that I don't have a question paper -- they didn't give them this year, so these are from memory). To others who wrote the contest, please tell me if I made an error.

1. Colour a mxn grid in black and white. Call a black square stranded if there is a white square above it and to the left of it. Determine a closed formula for the number of 2xn grids that have no stranded squares.

2. Given two disks, one smaller than the other. Each disk is divided into 200 congruent sectors. In the larger disk 100 sectors are chosen arbitrarily and painted red; the other 100 sectors are painted blue. In the smaller disk each sector is painted either red or blue with no stipulation on the number of red and blue sectors. The smaller disk is placed on the larger disk so that the centers and sectors coincide. Show that it is possible to align the two disks so that the number of sectors of the smaller disk whose color matches the corresponding sector of the larger disk is at least 100. (copied from cut-the-knot website)

3. Determine all real numbers r for which it is possible to find a triplet $ (x,y,z)$  of positive reals such that \[ r \equal{} f(x,y,z) \equal{} \dfrac{(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\].

4. Find all integers $ a$ and $ b$ such that $ 3^a \plus{} 7^b$ is a perfect square.

5. Given a set of $ n \ge 3$ points in the plane such that any three can be covered by a disk of radius 1, prove that all n points can be covered by a disk of radius 1.

[hide="Hidden"]
For #1, is it anywhere to the left, or immediately to the left? Just now I assumed the latter and got $ f(n) \equal{} \frac{(2\plus{}\sqrt{2})^n(1\plus{}\sqrt{2})\plus{}(2\minus{}\sqrt{2})^n(1\minus{}\sqrt{2})}{2}$, which seems an unlikely #1.

#4. I miss these ad-hoc number theory problems. Suppose $ 3^a \plus{} 7^b \equal{} k^2$. First note $ a,b \ge 0$. My first reaction is to try difference of squares. $ k^2 \equiv 0,1$ mod $ 4$, so $ a$ and $ b$ have opposite parity.

If $ a$ is even, say $ a\equal{}2c$, then $ 7^b \equal{} (k\minus{}3^c)(k\plus{}3^c).$ Both terms on the RHS are powers of 7. But if 7 divides both terms, then we get that 7 divides their difference $ 2\cdot3^c$, which is impossible. Hence $ k\minus{}3^c\equal{}1$, giving $ 7^b\minus{}1\equal{}2\cdot 3^c.$ If $ c \ge 2$, then considering both sides mod 9 gives that 3 divides $ b$; noting that 19 divides $ 7^3 \minus{} 1$, we then see that $ 7^b \minus{} 1$ is always divisible by 19. Thus $ c < 2$, giving the solution $ (a,b) \equal{} (2,1)$.

If $ b$ is even, say $ b \equal{} 2d$, then as above we get $ 3^a\minus{}1\equal{}2\cdot7^d$. As above (this time considering mod 7 and then mod 13), we get that $ d \equal{} 0$. This gives the solution $ (a,b) \equal{} (1,0)$.[/hide]

@Karpo
for number 1 the definition for stranded in exact words is:
... is stranded if there is some square to its left in the same row that is white and there is some square above it in the same column that is white
so i think it's the the first one u said but now that i read it i think i misread it on the test  :(

oh crud

i misinterpred #1 and got exactly the same answer as carpo.
i hope i get a partial mark for solving a completely different question =-=... which is slightly harder than the original question :S

i did that too for #1, and i know a few people who made that mistake. since the question was stated unclearly, do you think they would accept either solutions, or would they just give everyone who did that 2 marks =_= ?

I used the same interpretation as Carpo, so I guess its good that I didn't spend so much time on #1...  :maybe:  :maybe:

Aww, I should have gotten 2 and 4.

Well, I did get 1, so that makes me feel better.

What do you guys think your total scores are? I think mine is around 23ish.

Is there a way to solve number 5 without hellys theorem? Could one use extremal principle?

How does one use Helly's Theorem?

Given blootzboy's wording of #1, there is no ambiguity. However the CMO markers are usually generous, so if you solved the different (and more difficult) version you will probably get 4-5 marks.

For #5, I did not know Helly's theorem, although if I did, I could have pwned it so much easier, and alot less sketchy...

Construct a circle of radius 1 around every point. The intersection of every three of these circles is nonempty (or else there is no circle of radius 1 covering the three points). Hence Helly's theorem guarantees an intersection to all the circles. Take a circle of radius 1 w/ center inside this intersection. Done.

:rotfl: thats pretty sad...
However, I don't think that this was much help to some winter camp ppl who weren't able to solve #5

I bet you will be able to find this problem posted some time ago on aops :P

[quote="Nawahd Werdna"]Wow, this is sort of cheap... I just found the winter camp 2009 website, and on the combinatorial geometry handout that was presumably given to everyone, page 8, problem 2 under coverings, is EXACTLY the same as the CMO #5... that's such an advantage... :mad:

Oh well.

/rant

Link:
[url]http://www.stanford.edu/~darthur/wc09/combinatorics.pdf[/url][/quote]

Well, I didn't have an advantage.  He never really showed any of us how to do it per se.  I also didn't do this handout, so I didn't really feel an advantage.

But yeah it is sad.

"Unofficial" solutions are up at the CMS website...
[url]http://www.cms.math.ca/Competitions/CMO/2009preliminary.pdf[/url]

For the interested individuals, I would like to mention two theorems, one of which is a generalization of problem 5, and another is the generalized version of Helly's theorem with compact convex sets.

[u]Jung's Theorem:[/u] [i]Every subset of $ \mathbb{R}^d$ with diameter at most $ 1$ lies in a closed ball of radius $ \sqrt{\frac{d}{2(d\plus{}1)}}$. [/i]

[u]Remark 1[/u]. In the case of problem 5, you may scale the set by a factor of $ \frac{1}{\sqrt{3}}$ via a homethety.  Then we have the result after multiplication by $ \sqrt{3}$.

[u]Generalized Helly's Theorem[/u]. [i]If $ A\equal{}\{A_{\alpha}\}$ is a collection of compact convex sets in $ \mathbb{R}^d$ such that any $ d\plus{}1$ of them have a common point, then all of them have a common point.[/i]

[u]Remark 2[/u].  Now you may generalize problem 5 using the generalized version of Helly's theorem.

[u]Remark 3[/u]. This problem is not suitable to be an Olympiad problem; you need not study combinatorial geometry very much to encounter stronger versions of this problem.


In the shoe store there are $ n$ identic pairs of white shoes and $ n$ identic pairs of black shoes.  If $ n$ consecutive customers buy 2 pairs of shoes, each, randomly, find the probability of the event A: Every costumer buys 2 different colored pairs of shoes.

I don't know whether my intuition is right.

Let $ B_i$ be the event that the $ i$ th customer buys 2 different colored pairs of shoes.
Then we are interested in
\begin{eqnarray*} P(B_1\cap B_2\cap \dots \cap B_n)&=&P(B_n|B_2 \cap \dots \cap B_{n-1})\dots P(B_3|B_1\cap B_2)\cdot  P(B_2|B_1)\cdot P(B_1)\\&=&\left(\frac{1}{3}\right)^{n-1} \end{eqnarray*}

I think the answer should be $ 1\cdot \frac{2}{3} \cdots \frac{n\minus{}1}{2n\minus{}3}\cdot \frac{n}{2n\minus{}1}\equal{}\frac{2.(n!)^2}{(2n)!}$


A bag contains balls of 2 differrent colours. What is the minimum number of balls must be drawn out from the bag, without looking, so that among these balls there are 4 balls of same colour :?:

[hide=" click to reveal hidden content"]

I think the answer is $ 7$, because in the worst-case scenario, we draw alternate colours.

[/hide]

[hide]It's 7, as AndrewTom has stated. The worst-case scenario is where you draw 3 of each color. Then, the 7th ball as to make 4 of one of the colors.[/hide]


1)Determine $ x>0$ such that $ y\equal{}x^2\plus{}2\lfloor x\rfloor$ is the square of an integer number .

2)Solve this equation :

$ \left\lfloor \frac{6x\minus{}5}{4}\right\rfloor \plus{}\left\lfloor \frac{3x\plus{}1}{2}\right\rfloor \plus{}\left
\lfloor \frac{6x\plus{}15}{4}\right\rfloor \plus{}\left\lfloor \frac{3x\minus{}6}{2}\right \rfloor\equal{}2x\plus{}3$

3)Solve this equation :

$ x^3\minus{}\lfloor x\rfloor \equal{}3$

4)Solve the following equation :

$ \left\lfloor \frac{2x}{2\minus{}x}\right\rfloor\equal{}\left\lfloor \frac{2x}{x^2\minus{}x\plus{}1}\right\rfloor$

1. Since $ 2\lfloor x \rfloor$ is an integer and all squares are integers, $ x^2$ must be an integer, or $ x \in \mathbb{N}$. Then we have $ x^2<x^2\plus{}2\lfloor x \rfloor<x^2\plus{}2x\plus{}1\equal{}(x\plus{}1)^2$, so there are no solutions.

[quote="Brut3Forc3"]1. Since $ 2\lfloor x \rfloor$ is an integer and all squares are integers, $ x^2$ must be an integer, or $ x \in \mathbb{N}$. Then we have $ x^2 < x^2 \plus{} 2\lfloor x \rfloor < x^2 \plus{} 2x \plus{} 1 \equal{} (x \plus{} 1)^2$, so there are no solutions.[/quote]
I thought this at first.  But note x can be written in the form $ \sqrt{k}$, and we have $ \sqrt{2}$ a solution.  But I do not know how to proceed to finding the other solutions.

From $ x>0$ and $ x<\lfloor x\rfloor\plus{}1\Rightarrow \lfloor x\rfloor^2<y<(\lfloor x\rfloor\plus{}2)^2\Rightarrow y\equal{}(\lfloor x\rfloor\plus{}1)^2$ and $ x^2\equal{}\lfloor x\rfloor^2\plus{}1$ .

Denoting $ \lfloor x\rfloor\equal{}k\in \mathbb{Z}\Rightarrow x\equal{}\sqrt{k^2\plus{}1}$

Problem 3 was posted and solved [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=75659]here[/url].

for $ 4$ I think $ x\equal{}1$ is the only solution.

[hide="Problem 2"]Putting $ u\equal{}{3x\over 2}$ and extracting the integers out of the floor signs (e.g. $ \left\lfloor u\plus{}{15\over 4}\right\rfloor\equal{}\left\lfloor u\minus{}{1\over 4}\right\rfloor\plus{}4$ etc.), we get

$ 2\left\lfloor u\minus{}{1\over 4}\right\rfloor\plus{}\left\lfloor u\plus{}{1\over 2}\right\rfloor\plus{}\lfloor u\rfloor\equal{}{4u\over 3}\plus{}3$

Therefore $ {4u\over 3}$ must be integer. Write $ u\equal{}{3\over 4}(4k\plus{}r)\equal{}3k\plus{}{3r\over 4}$, where $ k,r\in\mathbb{Z},0\leqslant r\leqslant 3$. Then the equation becomes

$ 6k\plus{}2\left\lfloor{3r\minus{}1\over 4}\right\rfloor\plus{}3k\plus{}\left\lfloor {3r\plus{}2\over 4}\right\rfloor\plus{}3k\plus{}\left\lfloor {3r\over 4}\right\rfloor\equal{}4k\plus{}r\plus{}3$

or

$ 2\left\lfloor{3r\minus{}1\over 4}\right\rfloor\plus{}\left\lfloor {3r\plus{}2\over 4}\right\rfloor\plus{}\left\lfloor {3r\over 4}\right\rfloor\equal{}\minus{}8k\plus{}r\plus{}3$

Quick checking for $ r\equal{}0,1,2,3$ shows that only $ r\equal{}2$ gives an integer value for $ k$, which is $ k\equal{}0$, hence the only solution for $ u$ is $ u\equal{}3k\plus{}{3r\over 4}\equal{}{3\over 2}\iff x\equal{}1$[/hide]

[hide="Problem 4"]Let's analyze $ \minus{} 1 < {2x\over x^2 \minus{} x \plus{} 1} < 2$

Since $ x^2 \minus{} x \plus{} 1$ is always positive for real $ x$, the LHS yields $ \minus{} x^2 \plus{} x \minus{} 1 < 2x\iff x^2 \plus{} x \plus{} 1 > 0$, which holds for all real $ x$, and the RHS yields $ 2x < 2x^2 \minus{} 2x \plus{} 2\iff 2(x \minus{} 1)^2 > 0$, which holds for all $ x\neq 1$. We check that $ x \equal{} 1$ satisfies the given equation, and from now on we must examine only the cases $ \left\lfloor{2x\over x^2 \minus{} x \plus{} 1}\right\rfloor \equal{} 0$ and $ \left\lfloor{2x\over x^2 \minus{} x \plus{} 1}\right\rfloor \equal{} 1$

By solving the systems of inequations

$ 0\leqslant{2x\over 2 \minus{} x} < 1\land 0\leqslant{2x\over x^2 \minus{} x \plus{} 1} < 1$

and

$ 1\leqslant{2x\over 2 \minus{} x} < 2\land 1\leqslant{2x\over x^2 \minus{} x \plus{} 1} < 2$

(which is easy, but tedious, so I'll skip an exercise in LaTeX-ing here)

and taking $ x \equal{} 1$ into account, we find that the solution set is $ x\in\left[0,{3\minus{}\sqrt{5}\over 2}\right)\cup\left[{2\over 3},1\right]$.[/hide]


Show that:

 $ sup \bigcup_{n \equal{} 1}^{\infty}X_{n} \equal{} sup\lbrace supX_{n}\rbrace$.

Where $ X_{n}$ is a nonempty set with upper bound

--------

According to the given problem I have 1 simple question to You: what does exactly mean $ sup\lbrace supX_{n}\rbrace$  ?? I don't get it. Is it equal just to $ supX_{n}$ ?? Please help me. Thanks

$ \{ \sup X_n \}$ is a set of supremums, one for every $ X_n$.  One then takes the supremum of this set.


I have basic math problems that I need help on. I can't remember them nor I think I learn them.

This is something about tessellation.

The explanatio of it can be found on:
http://mathworld.wolfram.com/Tessellation.html

But I can't answer these questions:

1. Name two regular polygons that tessellate the plane.
[hide="my_solution"]Hexagon and equilateral triangle[/hide]
2. All quadrilaterals will tessellate. Explain why.
[hide="hmmmmm"]$\text {I really don't have any clue for this one.}$
[/hide]
3. Will a scalene triangle tessellate? Explain why or why not.
[hide="um"]I don't think they are because each side doesn't have any same length. But again, I'm not so sure. [/hide]
4. Cut out a non-regular quadrilateral and make a tessellation withi it.
[hide]NO CLUE. Maybe I should try to like cut out a pentagon and divide it into a equilateral triangle and square?[/hide]
5. Why doesn't the regular pentagon tessellate? Explain.
[hide]Well, I'm not $\text {SO SURE}$ about this one but here's my guess:

To be able to tessellate, you have to have a repeating shape, like if you connect two shapes, four shapes, six shapes, etc, you'll get congruent shapes. I think that's how it works for triangle and quadrilateral, isn't it?[/hide]
6. Name another regular polygon that will not tessellate the plane.
[hide="MAYBE_THIS_ONE?"]Heptagon? (Or septagon, they basically mean same thing) $\bigstar \clubsuit \spadesuit$[/hide]

For these tesselation problems try to figure out why a certain polygon can tesselate the plane.  Focus in on a single vertex and find out what must be true about the angles for that certain polygon to tesselate the plane.

For the quad question you should start with parallelograms and quads that have special properties and than try to think about it generally.

For the scalene triangle think about it this way: You can make a scalene triangle by cutting a parallelogram in half along the diagonal. So since you can tile parallelograms you can tile scalene triangles because they fit in parallelograms that tessellate.

[quote]3. Will a scalene triangle tessellate? Explain why or why not. [/quote][hide]
Yes, join two identical scalene triangles like this:[code]
       * - - - - - - - *
     /   \            /
   /        \       /
 /             \  /
* - - - - - - - *[/code]
And tesselate the plane with these parallelograms.[/hide]

[quote]5. Why doesn't the regular pentagon tessellate? Explain.[/quote] [hide]
At every 'vertex' of the tesselation the angles must sum to 360 degrees.
Three regular pentagons take up 324 degrees with no room for a fourth.[/hide]


modular arithmetic
floor function
abstract algebra
number theory
relatively prime
group theory
Congruences

Suppose that $m>2$, and let $P$ be the product of the positive integers less than $m$ that are relatively prime to $m$. Show that $P \equiv -1 \pmod{m}$ if $m=4$, $p^n$, or $2p^{n}$, where $p$ is an odd prime, and  $P \equiv 1 \pmod{m}$ otherwise.

We first observe that this is true for $m=1,2$. Let now $m\geq 3$.

If some $x$ with $(x,m)=1$ has an multiplicative inverse $x^{-1}$ with $x\not\equiv x^{-1}\pmod m$, they drop out of $P$, so we have
\[P=\prod_{\substack{1\leq x \leq m\\ (x,m)=1}}x\equiv \prod_{\substack{1\leq x\leq m\\ x^{2}\equiv 1 \bmod m}}x \equiv \prod_{\substack{1\leq x \leq \lfloor \frac{m}{2}\rfloor\\ x^{2}\equiv 1 \bmod m}}-x^{2}\equiv \prod_{\substack{1\leq x \leq \lfloor \frac{m}2 \rfloor\\ x^{2}\equiv 1 \bmod m}}-1 \equiv (-1)^{\frac{a(m)}{2}}\pmod m\]

where $a(n)$ denotes the number of $1\leq x \leq m$ so that $x^{2}\equiv 1 \pmod m$.

[color=brown]Lemma 1:[/color] $a(1)=1, a(2)=1, a(4)=2$.
Proof: Inspection. $\Box$

[color=brown]Lemma 2:[/color] $a(2^{k})=4$ for $k\geq 3$.
Proof: We have
\[x^{2}\equiv 1 \pmod{2^{k}}\\ \Leftrightarrow (x-1)(x+1)\equiv 0 \pmod{2^{k}}\\ \Leftrightarrow x\equiv-1,+1 \pmod{2^{k-1}}\\ \Leftrightarrow x\equiv 1, 2^{k-1}-1,2^{k-1}+1, 2^{k}-1 \pmod{2^{k}}\]
$\Box$

[color=brown]Lemma 3:[/color] $a(p^{k})=2$ for all odd primes $p$.
Proof: We have
\[x^{2}\equiv 1\pmod{p^{k}}\\ \Leftrightarrow (x-1)(x+1)\equiv 0\pmod{p^{k}}\\ \Leftrightarrow x\equiv-1,+1 \pmod{p^{k}}\]
$\Box$

[color=brown]Lemma 4:[/color] $a$ is multiplicative, ie $a(mn)=a(m)a(n)$ for all coprime $m,n$.
Proof: Since $m$ and $n$ are coprime, we have $x^{2}\equiv 1 \pmod{mn}$ if and only if $x^{2}\equiv 1\pmod m$ and $x^{2}\equiv 1 \pmod n$.
Let $y_{1},\ldots,y_{a(m)}$ be those residues modulo $m$ so that $y_{i}^{2}\equiv 1\pmod m$ and let $z_{1},\ldots,z_{a(n)}$ be those residues modulo $n$ so that $z_{i}^{2}\equiv 1 \pmod n$.
Then, by the chinese remainder theorem, we have $x^{2}\equiv 1\pmod{mn}$ if and only if

$x\equiv y_{i}\pmod m$ and $x\equiv z_{j}\pmod n$

for some $1\leq i\leq a(m)$ and $1\leq j \leq a(n)$. Clearly, we have $a(m)a(n)$ possibilities to choose such $i,j$ and since we get another $x$ modulo $mn$ for each such possibilities, we obtain $a(mn)=a(m)a(n)$.
$\Box$

[color=brown]Corollary 1:[/color] $a(m)$ is not divisible by $4$ if and only if $m=1,2,4,p^{k}, 2p^{k}$ where $p$ is an odd prime and $k$ is a positive integer.

This solves the problem.

Any link between this and primitive roots?

Yes and no: the correct generalisation is that the product (or if you prefer additive notation: sum) of all elements in an abelian group is the neutral element $e$ if and only if $x^2 = e$ has exactly two solutions. If it would have more than two solutions, then it cannot be cyclic, and if the group has an even number of elements, then it has at least two solutions by Cauchy's theorem. But not every group satisfying this property is cyclic as the additive group $\mathbb Z/2 \times (\mathbb Z/3)^2$ shows.


[b]Euler's Formula:[/b]  $ e^{ix} \equal{} \cos x \plus{} i \sin x$

Hence $ e^{i \frac {\pi}{2} } \equal{} \cos \frac {\pi}{2} \plus{} i \sin \frac {\pi}{2} \equal{} i$.

For all intents and purposes at this point, the above can be considered a definition of the exponential function for complex values.  The real motivations behind that definition are complicated.


The series $ \sum_{n \equal{} 1}^{\infty}$ converges. Must $ \sum_{n \equal{} 1}^{\infty} \frac {x_n}{n}$ also converge?

[color=green][Moderator says: while this is obviously incorrectly stated, please don't fix it. The same problem is posted in another recent topic. Look for that.][/color]

Two proofs were posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315256]here[/url].


function
algebra
polynomial
calculus
calculus computations

Find, with explanation, the maximum value of the function f(x) = x^3 − 3x on the set of
all real numbers x satisfying the condition x^4 + 36<=13x^2.

$ x^{4}-13x^{2}+36\leq 0 \Leftrightarrow (x-2)(x+2)(x-3)(x+3)\leq 0 \Leftrightarrow x\in [-3,-2]\cup [2,3]$ , and then find the maximum.
The obvious candidate is $ x=3$

More generally, assume that the condition is $ p(x)\leq0$ where $ p$ is a polynomial of even degree. Then $ B\equal{}\{x\in\mathbb{R}|p(x)\leq0\}$ is a compact subset and $ \exists{x_{0}}\in{B}$ s.t. $ f(x_{0})\equal{}sup_{\{x\in{B}\}}f(x)$. If $ p(x_{0})<0$ then it's a free maximum, $ f'(x_{0})\equal{}0$ and $ x_{0}\equal{}\minus{}1$. If $ p(x_{0})\equal{}0$ then $ x_{0}\in{U}$ where $ U$ is the set of the $ p$-roots.
Conclusion: if $ \minus{}1\in{B}$ then $ f(x_{0})\equal{}sup\{f(\minus{}1), sup_{\{u\in{U}\}}f(u)\}$, else $ f(x_{0})\equal{} sup_{\{u\in{U}\}}f(u)$,


Let $ a,b$ be positive real numbers such that $ a^2 \plus{} b^2 \equal{} a^9 \plus{} b^9$. Prove that $ a^{9002} \plus{} b^{9002}\geq a^{2009} \plus{} b^{2009}$.

yes, that's one way  :P  when i created it, i thought using holder like this,

$ (a^9 \plus{} b^9)^7 \equal{} (a^2 \plus{} b^2)(a^9 \plus{} b^9)^6\geq(a^8 \plus{} b^8)^7$, so $ a^9 \plus{} b^9\geq a^8 \plus{} b^8$...

from cauchy-schwarz we have that $ \dfrac{a^{10} \plus{} b^{10}}{a^9 \plus{} b^9}\geq \dfrac{a^9 \plus{} b^9}{a^8 \plus{} b^8}\geq 1$, and so we can continue analogously to prove that $ a^{9002} \plus{} b^{9002}\geq a^{9001} \plus{} b^{9001}\geq ... \geq a^{2009} \plus{} b^{2009}$...  :D


It was a lot of fun!  See pictures at our website:

[url]http://www.mathzoom.org[/url]

I recognize [url=http://picasaweb.google.com/ren.kelly/MathZoomSummer2008/photo?authkey=H2bJ0YbqMuY#5225720833111391170]that guy on the right[/url].


Let $G=\{k\in\mathbb{Q}: k=a^2+b^2, a,b\in\mathbb{Q}^*\}$

a) Prove that if $k\in G$, then $\frac{1}{k}\in G$.
b) Let $k$ be a prime of the form $M(4)+3$. Prove that $k\notin G$
c) $6,8,11\in G$?

I just proved part b) in a particular case: when $a,b\in\mathbb{Z}$.

[hide="Part a"]
$k$ can be expressed in the form $\frac{a^2 + b^2}{d^2}, a, b, d \in \mathbb{Z}$.  $\frac{1}{k} = \frac{d^2}{a^2 + b^2} = \frac{(da)^2 + (db)^2}{ (a^2 + b^2)^2 }$, the same form.  Hence $\frac{1}{k} \in G$.
[/hide]

[hide="Part b"]
$d^2 \equiv 0, 1 \bmod 4$.  So if $k \equiv 3 \bmod 4$ then $a^2 + b^2 \equiv 3 \bmod 4$.  But $a^2 + b^2 \equiv 0, 1, 2 \bmod 4$.
[/hide]

[quote="hpoincare"]thanks, I did exactly that for part b) but it is not complete because I must show that does not exist $a,b\in\mathbb{Q}^*$

For part a) I found sth really easy now:

if x=a^2+b^2, then 1/x = x/x^2 = (a/x)^2+(b/x)^2[/quote]

My definitions are different; I have the two fractions $\frac{a}{d}, \frac{b}{d}$ with $a, b \in \mathbb{N}$ rather than $\mathbb{Q}$.  My solution to the second part is complete.  :)

[hide="Part c"]
The integer members of $G$, expressible in the form $\frac{a^2 + b^2}{d^2}$, must themselves be expressible as a sum of integer squares $u^2 + v^2$ (this follows from the expansion $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad + bc)^2$) so $6, 8, 11 \not \in G$.
[/hide]

Thanx for your hints

$8\in G$ cause $8=(2/5)^2+(14/5)^2$ (in fact, $2=1^2+1^2$ and $4=(6/5)^2+(8/5)^2$

For part b) you solve equation $a^2+b^2=kd^2$

k=3 mod 4, d^2=0,1 mod 4. If d^2=1 mod 4, kd^2=3 mod 4: no problem.
But what if d^2=0 mod 4? Then a^2+b^2 = 0 mod 4

The general version is: $r \in \mathbb{Q}^+$ can be written as $r=a^2+b^2$ with $a,b \in \mathbb{Q}$ (or $\mathbb{Q}^*$) iff every prime factor $\equiv 3 \mod 4$ occurs an even number of times in the decomposition of $r$. There are $\infty$ many solutions then.

1) Zeta: you wrote $r=a^2+b^2$ is a rational number, so what you mean about the descomposition of a rational number?

2) We are trying to solve the equation $a^2+b^2=kd^2$
$k\equiv 3\mod 4$

The problem is the following: suppose $d^2=0\mod 4$, then $a^2+b^2=0\mod 4$ so $a=2A$, $b=2B$ thus eq. becomes $A^2+B^2=kD^2$, $A<a, B<b, D<d$. Continuing this way we arrive to a contradiction. Is this right?

As every integer can be represented by a product of primes in an unique way, so this holds for the rationals:

For any rational $r\neq 0$, you can find integers $v_i$ (when $r$ is an integer itself all nonnegative) such that $r= \pm \prod_{i \in \mathbb{N}} p_i^{v_i}$ with $p_i$ being the $i$-th prime (by some random order).
E.g.:
$0.2 = 5^{-1}$
$\frac{123}{456}=2^{-3} \cdot 19^{-1} \cdot 41^1$

You can get the representation of $\frac{a }{b}$ with integers $a,b$ out of the factorisation of $a,b$ itself:
Let $a= \pm \prod_{i \in \mathbb{N}} p_i^{a_i}$ and $b= \pm \prod_{i \in \mathbb{N}} p_i^{b_i}$, then $\frac{a}{b} = \pm \prod_{i \in \mathbb{N}} p_i^{a_i-b_i}$, thus $v_i =a_i-b_i$ works and is the only way you can do it.

Thanks Zeta for your reply. It never crossed my mind that simple idea that even a rationa number can be represented by a product of primes in an unique way.

I'd like to know if this proof is correct.
It goes:

$a^2+b^2=kc^2$ WLOG we suppose a,b,c coprime (otherwise simplify). $x^2=0,1\mod 4$, then $a^2+b^2=0,1,2\mod 4$; also $kc^2=3c^2=0,3\mod 4$. So the only way is that $a^2+b^2=c^2=0\mod 4$ This is imposible since it'd imply a,b,c not coprime
Right?

I think we need that statement in order to prove your general statement.
First of all, I see that every square is in S. In fact,

$(\frac{3k}{5})^2+(\frac{4k}{5})^2=k^2$

In your representation of a rational number we easily see that we can divide in three parts r =k·l·m, where k is a square, l are primes of =2,1 (mod 4),  and the rest primes =3 (mod 4) with exponent 1.
I think it is proved that primes of the form =1 mod4 are sums of two squares (in a unique way) so $k,l\in S$, then in order to prove that $r\in S$, we must show that $m\in S$

Now I'm lost again...

$S=G$ ¿

Yes, your proof for the frisrt part is ok.

Well my claim can be written in your way in the following way:
Let's write $r=k^2 \cdot q_1 \cdot q_{-1}$ where $q_1$ (your $l$) contains the prime factors $\equiv 1,2 \mod 4$ with exponent $0$ or $1$ and $q_{-1}$ (your $m$) contains the factors $\equiv -1 \mod 4$ (also with exponents $0$ or $1$ only).
Then $r \in S \iff q_{-1} =1$.

The equivalence comes from the fact that the factor $k^2$ doesn't change the parity of the $v_i$. Since my claim was that every $v_i$ belonging to a prime $\equiv -1 \mod 4$ has to be even, we get that this means here that all primes $\equiv -1 \mod 4$ occure an even number of times in $q_{-1}$, giving (since they only occure $0$ or $1$ times, thus $1$ excluded because it's odd) that $q_{-1}=1$.

So when using that every prime $\equiv 1 \mod 4$ can be written as sum of two squares, we know that $r$ is sum of two squares when $q_{-1}=1$ (since all other factors are in $S$).
The converse can shown by looking $\mod$ the primes $p \equiv -1 \mod 4$ (using that there is no $x$ with $x^2 \equiv -1 \mod p$: such $x$ would give $-1 \equiv (-1)^{\frac{p-1}{2}} \equiv x^{p-1} \equiv 1 \mod p$, a contradiction).


απδείξτε ότι: $ (x \minus{} 1)^{n} \plus{} 2(x \minus{} 2)^{n}(x \minus{} 1)^{n} \plus{} (x \minus{} 2)^{n} \minus{} x^{n} < 0$ για κάθε $ x,n > 2$ και $ x,n$ ακέραιοι

Ummm, είσαι σίγουρος για τη διατύπωση? Πχ, για $ x\equal{}4, n\equal{}3$ (βασικά όποτε το $ x$ είναι αρκετά μεγάλο) έχεις ένα αντιπαράδειγμα...

Και ούτε αν αντιστρέψεις την ανισότητα ισχύει, οπότε μάλλον δεν είναι εκεί το typo...

Cheerio,

Durandal 1707

Όντως ο Durandal έχει δίκιο μήπως έχεις κάνει κανένα λάθος στη διατύπωση

Έχετε όντως δίκιο και οι δυό σας. Δεν είναι δική μου η άσκηση< κατασκευή ενός φιλού>. Sorry από εμένα και απο αυτόν εαν σας παιδέψαμε παιδιά. Γιαυτό προσπαθήστε να λύσετε αυτή: Έστω $ n$ θετικός ακέραιος. Βρείτε το άθροισμα όλων των θετικών ακεραίων μικρότερων του $ 2n$ και σχετικά πρώτων με το $ n$. :)


analytic geometry
parameterization
number theory
least common multiple
ratio
calculus
integration

Consider all points with integer coordinates in the carthesian plane. If one draws a circle with M(0,0) and a well-chose radius r, the circles goes through some of those points. (like circle with $r=2\sqrt2$ goes through 4 points)

Prove that $\forall n\in \mathbb{N}, \exists r$ so that the circle with midpoint 0,0 and radius $r$ goes through at least $n$ points.

Let $x= \frac {2t} {1+t^2}$ and $y = \frac {1-t^2} {1+t^2}$ be a parametrization of the unit circle.
Clearly, if $t$ is rational then $M(x(t),y(t))$ is a rational point.
Choose $n$ rational values for $t$ such that to obtain $n$ pairwise distinct rational points $M(x_k,y_k)$.
Now let $m$ be the lcm of the denominators of the $x_k,y_k$'s for $k=1, \cdots ,n$.
Now use an homothetie with centre $O$ and ratio $m$. Thus the images of all the $M_k$ becomes pairwise distinct integral points, and we are done (note that it may be  some other integral points than these ones).

Pierre.

Yes. My solution was a little the same:

(There are infinitely many prime - pythagorean triplets)

consider: $a^2+b^2=c^2$ and $d^2+e^2=f^2$
there are points on the circle with radius c, points on the circle with radius b, so since we take them relatively prime, there are more points on the ricle with radius c.f

continuing like this we can create circles through arbitrarily many points.


function
limit
real analysis
real analysis unsolved

Let $ f$ be a non-decreasing function, defined on $ [0,1]$, such that:

[b]1.[/b] Show that for each $ a\in(0,1)$, $ \lim_{x\rightarrow a^\plus{}}f(x)$ and $ \lim_{x\rightarrow a^\minus{}}f(x)$ both exist.

[b]2.[/b] For any $ \epsilon>0$, there are only finitely many numbers $ a\in(0,1)$ with
\[ \lim_{x\rightarrow a^\plus{}}f(x) \minus{} \lim_{x\rightarrow a^\minus{}}f(x) >\epsilon\]
[b]3.[/b] Prove that the set of points at which $ f$ is discontinuous is countable.

What will be erken's next post? He posted Banach's fixed point theorem, and now Froda's theorem, stating that for any real function, the set of points of discontinuity of first species is at most countable! (in a particular setting for monotonous functions).


Στο οξυγώνιο τρίγωνο $ABC$ φέρνουμε τα ύψη $AA_{1}, \, BB_{1}, \, CC_{1}$ .Στα τμήματα $A_{1}C_{1}, \, B_{1}C_{1}$ παίρνουμε σημεία $K$ και $M$

τέτοια ώστε η γωνία $\measuredangle MAK$ να είναι ίση με τη γωνία $\measuredangle CAA_{1}$ .Να δείξετε ότι η $AK$ είναι διχοτόμος της γωνίας $\measuredangle C_{1}KM$ .

 :)

Πολύ ωραίο και κλασσικό θέμα. Δείτε εδώ
http://www.mathlinks.ro/Forum/viewtopic.php?p=3673#p3673

3 χρόνια  πρίν :blush:


algebra
binomial theorem
Pascal\u0027s Triangle
binomial coefficients

Is there a way to do find the thousandths' digit of 11 ^ 86 without using a calculator? I can't seem to find any pattern with the lower powers of 11.

\begin{eqnarray*} 11^1 & = & 11 \\
11^2 & = & 121 \\
11^3 & = & 1331\\ 11^4 & = & 14641 \end{eqnarray*}

Look Familiar?

[hide="Solution"]
This is the numbers of the Pascals triangle so:

$ \binom{1000 - 1}{2} = 498501\implies\boxed{5}$
[/hide]

[hide="other way"]11=10+1, so you can use the Binomial Theorem.[/hide]

The first method is incorrect.  And you mean the "thousands" digit.  The "thousandths" digit is $ 0$.

Binomial theorem is probably the best way, as if you try using pascal's triangle for $ 11^5$, you get $ 15101051$, which clearly isn't $ 11^5$.

The "Pascal's triangle" approach is the same approach as binomial theorem, except that once the binomial coefficients have more than one digit everything carries.

That's the hundreds digit.

Let's be very specific here:  what we want to do is use the Binomial Theorem to write

$ (10 \plus{} 1)^{86} \equal{} 1 \plus{} 10 \cdot {86 \choose 1} \plus{} 10^2 \cdot {86 \choose 2} \plus{} 10^3 \cdot {86 \choose 3} \plus{} 10^4 \left( \text{Stuff} \right)$

so that we can find the thousands (that is, the fourth-to-last) digit, which is the first digit of the remainder upon division by $ 10000 \equal{} 10^4$.  Three of the first four terms (not the $ 1$ term) need to be considered because of carrying.

[quote="Fraenkel"]I get $ 3$.

$ \binom{86}{85} \equal{} 86$
$ \binom{86}{84} \equal{} \frac {86*85}{2} \equal{} 3655$

Thousands digit is $ 8 \plus{} 5\mod{10} \equal{} 13\mod{10} \equal{} 3$[/quote]

The answer is [hide]$ 6$[/hide].


inequalities
calculus
derivative
algebra
polynomial
Galois Theory
inequalities open

Let $ a,b,c$ be nonnegative real numbers which sum 3. Find the minimum and maximum value of the folloeing expression

$ P\equal{}\frac1{a^2\plus{}b\plus{}1}\plus{}\frac1{b^2\plus{}c\plus{}1}\plus{}\frac1{c^2\plus{}a\plus{}1}$

I post this problem in Vietnam Inequality Forum but it's still Open Questions  :mad:

http://batdangthuc.net/showthread.php?p=2742#post2742

I believe that P is convex over positive a,b,c since second derivatives in all variables are positive. Thus it attains its maximum at its endpoints, namely (3,0,0), (0,3,0) or (0,0,3). Plugging in gives a maximum of $ \frac {27}{20}$.

Finding the minimum seems more difficult and I'm not entirely sure how to do it... I would conjecture that it's simply $ 1$ though since it's symmetric in a,b,c.

Actually, I believe convexity implies that the minimum is obtained at $ x\equal{}y\equal{}z$ giving $ 1$ for the minimum, right?

[quote="kops723"]Actually, I believe convexity implies that the minimum is obtained at $ x \equal{} y \equal{} z$ giving $ 1$ for the minimum, right?[/quote]

Sorry,prediction is not true. Try $ a\equal{}0,b\equal{}1,c\equal{}2$  :wink:

[quote="dduclam"]Let $ a,b,c$ be nonnegative real numbers which sum 3. Find the minimum and maximum value of the folloeing expression

$ P \equal{} \frac1{a^2 \plus{} b \plus{} 1} \plus{} \frac1{b^2 \plus{} c \plus{} 1} \plus{} \frac1{c^2 \plus{} a \plus{} 1}$
[/quote]

Altheman's answer seems correct. Have a look at the polynomial \[ 1753490160\,{k}^{8}\minus{}7242829056\,{k}^{7}\plus{}11338524468\,{k}^{6}\minus{}8267726008\,{k}^{5}\plus{}2802538373\,{k}^{4}\minus{}513184720\,{k}^{3}\plus{}132643611\,{k}^{2}\minus{}8858430\,k\plus{}2068173\] It has 2 real roots - one is about 0.9167 and other is about 1.3532. First is the minimum and second is the maximum of your expression. Galois group of this polynomial is $ S(8)$, so it's roots cannot be expressed using radicals. Are you sure the answer for your problem can be expressed in better terms?

Sergey Markelov
markelov@mccme.ru


$f: R\rightarrow R$

$f(x+y+f(xy))=f(f(x+y))+xy$

Find all such $f$.  :lol:

Yes,It is not hard.I complete the solution( I hope it is correct)
y=0 gives that $ff(x)=f(x+f(0))$
then $f(x+y+f(0))+xy=f(x+y+f(xy))$
$x=-y$ gives that $ff(-x^{2}+f(0))=f(f0))-x^{2}$
$u=f(0)-x^{2}$
then for all $u\leq f(0)$ $f(u)=ff(0)-f(0)+u=u+l$
$\forall t$Let $x$ satisfies $x(t-x-f(0))\leq f(0)$,$t-f(0)+x(t-x-f(0))+ff(0)-f(0)\leq f(0)$(we can let $x\to-\infty$)
then let $x=x,y=t-x-f(0)$
then we have $f(t)=t+2ff(0)-3f(0)=x+a$
where a is a constant.

[quote="Hawk Tiger"]
$\forall t$Let $x$ satisfies $x(t-x-f(0))\leq f(0)$,$t-f(0)+x(t-x-f(0))+ff(0)-f(0)\leq f(0)$(we can let $x\to-\infty$)
then let $x=x,y=t-x-f(0)$
then we have $f(t)=t+2ff(0)-3f(0)=x+a$
where a is a constant.[/quote]

What do you mean by this line? Please write clearer.

oh,If you ignore this line and go to the next line directly and have some calculation(I didn't post the detail)then you can find the use of the line you don't understand . :)
Maybe there is another better way to prove it.
To rust ,Do you have another solution?

Yeah I also want to see Rust's solution.

By the way, Hawk Tiger, I still don't get why when letting $x=x, y=t-x-f(0)$ you get that equation...


symmetry
arithmetic sequence
combinatorics unsolved
combinatorics

Each of the numbers 1q,2,3,....,N is colored  black or white .We are allowed  to simultaneously change the color of any three numbers in arithmetic progression .
For which numbers N can we always make all the number white

An observation:

The question is reduced to this: given $N$, for $i\in\overline{1,N}$, we will call $e_i$ the configuration where $i$ is black and the rest of the numbers are white, and $0$ the all-white configuration. The problem is reduced to finding those $N$ for which $e_i$ can be reduced to $0$ for all $i$. Thanks to this, we can also notice that if $N$ is good, then $N+1$ is good, so all we need to do is look for the smallest $N$ which works.

I think $N=8$ works, and I'm not sure about $N=7$.

:huh:
I remember we have already discussed it. I wrote my solution, and there were other solutions. I think grobber wrote his solution too.

I can't remember it :?.

Anyway, here's a proof that $8$ does indeed work. When I say that $e_i\equiv e_j$ I mean that the configurations $e_i,e_j$ can be turned into one another by means of allowed moves (and the same holds for $e_1\equiv 0$).

When we have $N=7$, $e_1,e_7,e_4\equiv 0$. Since $e_1\equiv e_4$ by performing the moves $123,234$ (in fact, for $k\le N-3$, we have $e_k\equiv e_{k+3}$ by the same type of move), and $e_1\equiv 0\iff e_7\equiv 0$ by symmetry, it means that all we need to show is $e_1\equiv 0$. This is achieved through $123,135,147,234,357$.

Now, when we go to $N=8$, $e_1,e_4,e_7\equiv 0$ because they're part of the first $7$ numbers, then, by symmetry, $e_8,e_5,e_2\equiv 0$, and we are left with $e_3\equiv e_6$, so we need to show that $e_3\equiv 0$. This is done by performing the move $357$, which shows that $e_3\equiv e_5+e_7$, and both $e_5,e_7\equiv 0$.

For $N=7$, it's easy to see that $e_2\equiv 0$ iff all $e_2,e_3,e_5,e_6\equiv 0$, so we only need to look at $e_2\equiv 0$. I think it doesn't hold (so the minimal $N$ should be $8$ in this case), but there's some case consideration involved.

An easy problem from ARO fourth round (it is not a final)
http://www.mathlinks.ro/Forum/viewtopic.php?t=28309


Suppose that $ I$ is incenter of triangle $ ABC$ and $ l'$ is a line tangent to the incircle. Let $ l$ be another line such that intersects $ AB,AC,BC$ respectively at $ C',B',A'$. We draw a tangent from $ A'$ to the incircle other than $ BC$, and this line intersects with $ l'$ at $ A_1$. $ B_1,C_1$ are similarly defined. Prove that $ AA_1,BB_1,CC_1$ are concurrent.


Find the real numbers $ a,b,c$ such that :

$ |ax^2\plus{}bx\plus{}c|\le |cx^2\plus{}bx\plus{}a|\ ,\ (\forall)x\in \mathbb{R}$

Let $ f(x)\equal{}|ax^2\plus{}bx\plus{}c|$ and $ g(x)\equal{}|cx^2\plus{}bx\plus{}a|$

$ f(0)\leq g(0)$ $ \implies$ $ |c|\leq|a|$
$ f(x)\leq g(x)$ when $ x\to\plus{}\infty$ $ \implies$ $ |a|\leq |c|$

So $ |a|\equal{}|c|$

1) $ a\equal{}c$ and $ f(x)\equal{}g(x)$ so OK

2) $ a\equal{}\minus{}c\neq 0$. (case 0 is the previous one)Then $ f(x)\equal{}|ax^2\plus{}bx\minus{}a|$ and $ g(x)\equal{}|ax^2\minus{}bx\minus{}a|$
$ g(x)$ has two real roots and if we want $ 0\leq f(x)\leq g(x)$ when $ g(x)\equal{}0$, we need that $ f(x)$ has the same roots. So $ b\equal{}0$

And so the only solutions :
$ a\equal{}c$
or
$ a\equal{}\minus{}c\neq 0$ and $ b\equal{}0$


superior algebra
superior algebra unsolved

The same question as everytime: what have you done so far¿
MathLinks/AoPS is not a "we-do-your-homework" machine...

sry. I have

1) Units +/- 1 and +/- i
2) +/-1 and +/- zeta squared and +/- zeta
 3) i'm not really sure

sry, but could you give me a hint about what the algebraic integers are in the fields?

You got 1 and 2 correct (I assume that $ \zeta \equal{} e^\frac{\pi i}{3}$).
For general squarefree $ d$, the ring of integers of $ \mathbb{Q} [\sqrt d]$ is (note that I use $ d$ where you used $ \minus{} d$ for covenience):

If $ d \equiv 1 \mod 4$, the ring of integers is $ \mathbb Z\left[ \frac {1 \plus{} \sqrt d}{2} \right] \equal{} \left\{a \plus{} b \cdot \frac {1 \plus{} \sqrt d}{2} | a,b \in \mathbb Z \right\}$.
If $ d \not \equiv 1 \mod 4$, the ring of integers is $ \mathbb Z\left[\sqrt d \right] \equal{} \left\{a \plus{} b \cdot \sqrt d | a,b \in \mathbb Z \right\}$.

heh...the third one has no more units than +/-1
but y?...

 : )

No, it has only the two canonical units $ \pm 1$.
But for $ d>0$, the ring of integers of $ \mathbb{Q} [\sqrt d]$ has $ \infty$ units. See either Pell equations or Dirichlet's theorem on units.


Some of you may know Coach Monk from [url=http://math.scranton.edu/monks/courses/ProblemSolving/mathcountsplaybook.pdf]this[/url] MathCounts playbook.  Or his High School Playbook.  But [url=http://math.scranton.edu/dept/careers.html]here[/url] is a listing of mathematical careers from his website.

There is also a nice list here:

http://www.ams.org/careers/archived.html

Further resource [url=http://dmoz.org/Science/Math/Employment/]in the ODP[/url].


[quote="bryantacademy"]the sum of tom and bill are is 138 ponds and one boy is 34 pounds heavier than the other. how much does the heavier boy weigh?

I need help.
 :blush:  :blush:  :blush:[/quote]

lets say tom is x and bill is x+34 ( we are assuming bill is heavier- it doesn't really matter who is heavier in this case) SO we know that x+x+34 is 138 ( becuz the sum of the 2 boys wight is 134) so 2x=104 and dividing both sides by two we know x= 52.
THe lighter boy is 52 pounds so the heavier one is 52+34= 86

the heavier boy is 86 lbs.

Can you explain your solution using NO ALGEBRA, yet making it clear why  your method has to work?

[quote="bryantacademy"]the sum of tom and bill are is 138 ponds and one boy is 34 pounds heavier than the other. how much does the heavier boy weigh?

I need help.
 :blush:  :blush:  :blush:[/quote]

[hide="Using Algebra"]
t+b=138
t=b+38

3b=100
b=50
t=88

The heavier boy is 88 pounds.[/hide]

[hide="no algebra"]if the two boys + 38 pounds have to weigh 138 pds, then 138-38 is 100. divide that by 2(2 boys) to get 50. Add 38(because you subtracted that in the beginning) to get 88.[/hide]

138-34=104
104/2=52
52+34=86pounds
wats pounds anyway? i only learn kg

[hide]
You get the equation $x+34+x=138$.
$2x=104$
$x=52$
So, the heavier boy wieghs $52+34=86$.[/hide]

[quote="risinglegend"]
wats pounds anyway? i only learn kg[/quote]

pounds are are unit of weight equal to 16 ounces that are used in the US---I dont know where else they are used.

If you're wondering, approximately 2.2 pounds is equal to a kilo.

Using algebra is the easiest method to solve the problem. You  could use the guess and check method though. It just kinda makes it harder!!!!!!!!!!  :)


[quote="Ji Chen"]If $ x,y,z$ be positive numbers such that $ x \plus{} y \plus{} z \equal{} 1$, then

$ \left(\frac {x}{y \plus{} z}\right)^z\left(\frac {y}{z \plus{} x}\right)^x\left(\frac {z}{x \plus{} y}\right)^y\leq \frac {3}{2}(yz \plus{} zx \plus{} xy),$

with equality if and only if $ x \equal{} y \equal{} z \equal{} \frac {1}{3}.$[/quote]

I think it's wrong,try $ x\equal{}y\rightarrow 0.5;z \rightarrow 0$  :wink:

[quote="akai"]try $ x \equal{} y\to \frac{1}{2},z \to 0$[/quote]:arrow: $ \left(\frac {x}{y \plus{} z}\right)^z\left(\frac {y}{z \plus{} x}\right)^x\left(\frac {z}{x \plus{} y}\right)^y\rightarrow 0 < \frac{3}{8}\leftarrow \frac {3}{2}(yz \plus{} zx \plus{} xy).$


Hello everyone, I just registered and I had a few question about some math stuff I'm having a hard time understanding.

My teacher gave this assignment to me and I can't seem to figure a couple out:

[quote]Write and equation in slope-intercept form for the line that contains the points (1,2).

          - Parallel to 3x - 6y=12
          - Perpendicular to 3x -6y = 13
[/quote]

[quote]Write in standard form.

          - Contians the points (2,-3) and has a slope of 5.
[/quote]

and the last problem...

[quote]Crosses the x-axis at x=-3 and the y-axis at y=1.

          - Write an equation for the line that contains each pair of points.
                          - (4,3) (-2,-9)
                          - (-7,2) (5,-1)

 [/quote]

Alright, thats all. I would love it if someone could tell me how to do these. :)

Parallel lines have the same slope. Product of slopes of two perpendicular lines is equal to $-1$
Slope Intercept Form : $y=mx+a$ $m - slope$
If we have two distinct points: $A=(x_1;y_1) \quad B=(x_2;y_2)$ Then a line through this two points have the equation:

$(x_2-x_1)(y-y_1)=(y_2-y_1)(x-x_1)$

This should be enough to solve this problems.
If you still don't know how to do it. Give a sign, then I'll post a solution.

No idea what you're talking about man, sorry.

I was told to do this:

[quote]You have a line equation in standard form. Solve it for y; Since the lines are parallel, the slopes are the same(3). Use the slope-intercept form and enter in the values you were given(1,2), solve for b. You now have your equation.

For the perpendicular line, use the negative reciprocal of the slope of the line given. [/quote]

But I don't know how to "solve for y".

Add six on both sides?

[b]First problem[/b]:
$a)$
$3x-6y=12 \iff y=\frac{1}{2}x-2$

The line parallel line through point $A=(1,2)$ has the form:

$y=\frac{1}{2}x+b$
$2=\frac{1}{2}+b \iff b=\frac{3}{2}$ hence our line is :$y=\frac{1}{2}x+\frac{3}{2}$

$b)$
$3x-6y=13 \iff y=\frac{1}{2}x-\frac{13}{6}$
A perpendicular line through point $A$ has the form:
$y=-2x+b$
$2=-2+b \iff b=4$. Hence the line is :$y=-2x+4$

[b]Second problem (generalization)[/b]
Given the point $A=(x_0;y_0)$. Find the equation of a line through point $A$ which has slope $m$

Answer:
$y-y_0=m(x-x_0)$

[b]Third problem:[/b]
Find the equation of a line through two points.
Answer is in my previous post


Did mathcounts changed Target problem time from 2 minute per problem to 3 minutes this year?

it has been 6 minutes for a set of 2 problems for a while now... :huh:

The Target Round time changed in [b][i]1990[/i][/b] from four minutes a pair to six minutes a pair.


Assume A and B are n×n E is unit (detE=1)matrix. Show that    det(AB-E)=det(BA-E)
Itis easy problem

[quote="Orif"]Assume A and B are n×n E is unit (detE=1)matrix. Show that    det(AB-E)=det(BA-E)[/quote]

Do you mean $A,B\in \mathcal l M_n(\Re)$ and $E=I_n$ ......? Because if this is what you mean then proving that $\det(AB-I_n)=\det(BA-I_n)$ is equivalent to  $\det(I_n +AB)=\det(I_n+BA)$ which is clearly true.


Let $ p$ be prime number,  $ x, y \in Z$ ,  $ a, b \in N^*$
Prove that:
If
$ (x^a \minus{} y^a) \vdots p$
$ (x^b \minus{} y^b) \vdots p$
then
$ (x^d \minus{} y^d) \vdots p$
where $ d \equal{} gcd(a,b)$

we know that there exist $ r,s\in\mathbb{N}$ such that $ ra\equal{}sb\plus{}d$
if $ p|y$ then $ p|x$ and $ p|x^d\minus{}y^d$
therefore suppose that $ (p,y)\equal{}1$
$ x^a\equiv y^a \mod p \Rightarrow x^{ra}\equiv y^{ra} \mod p \Rightarrow (x^b)^s.x^d\equiv y^{sb\plus{}d} \mod p \Rightarrow (y^b)^s.x^d\equiv y^{sb\plus{}d} \mod p
\Rightarrow x^d \equiv y^d \mod p$

it's a famous problem:

[b]for natural numbers $ a,b,m,n$ if $ \gcd (a,b)\equal{}1$ and $ \gcd (m,n)\equal{}d$ then $ \gcd \left(a^m\minus{}b^m,a^n\minus{}b^n\right)\equal{}a^d\minus{}b^d$[/b]


Solve in natural numbers: $(x+y)(1+xy)=2^{z}$


Prove that $ \log_{2}{5}$ is a irrational number.


Show that every subset of {1,2,...,99,100} with 55 elements contains at least 2 numbers with a difference of 9.

We divide the set in 9 arithmetical progressions $9k+i, i=\overline {0,8}$. If we choose more than 6 elements from every progression, we will choose 2 consecutive terms of the progression, and this means their difference is 9. So we could choose maximum $6\cdot 9=54<55$, numbers such that we don't have a $9-$ difference.

Here's another solution - we can organize 92 out of the 100 numbers in pairs in the following way: $ (18k \plus{} i, 18k \plus{} 9 \plus{} i), 0 \le k \le 4, 1 \le i \le 9,$ and $ (91,100)$. The remaining numbers are 92,...,99. All the rest occur exactly once in these pairs. Number of pairs: 9*5+1=46. If we take 55 numbers, a of them are in the set { 92,...,99}, (55-a) of them are in the pairs. $ a \le 8 \implies 55 \minus{} a \ge 47 > 46$, therefore there are 2 numbers that are in the same pair, which means that there are 2 numbers with difference 9.

Another question - how many subsets of {1,2,....,100} with i elements, $ 1 \le i \le 54$, don't contain 2 numbers with a difference of 9? (it's easy to compute for i=1,2,54, but what about other values?)

Here is another solution.
For every 18 consecutive integers, we can only choose at most 9 integers so that there are no pairs with difference of 9.
Thus, for the first 90 integers from 1 to 90 we can choose at most $ 9*5 \equal{} 45$ numbers.
In addition, we can choose at most 9 out of 10 numbers from 91 to 100 so that there are no pairs of difference of 9. Therefore we can choose at most $ 45 \plus{} 9 \equal{} 54 < 55$


Find:

$\sum_{i,j=1}^{n}x^{i}y^{j}$
where $x$ and $y$ are real numbers.

I thought the same thing,but  $x,y\in %Error. "mathdsl" is a bad command.
{R}$.So,take $x=1$ (or $y=1$) ,and the whole thing doesnt work.Maybe there is a another way around.

\[\sum_{i,j=1}^{n}x^{i}y^{j}=\sum_{i=1}^{n}x^{i}\cdot\sum_{j=1}^{n}y^{j}\]
Both factors are simple to evaluate


[hide]could be 8, 9, or 10.

so 3/10.[/hide]

[quote="colts18"]Moh's hardness scale ranges from 1 to 10 and describes different minerals in terms of their hardness. Gypsum is the softest mineral on the scale (hardness of 1) and diamond is the hardest mineral on the scale (hardness of 10). In a collection of 10 minerals, each with an integer hardness value different from any other mineral in the collection, what is the probability of randomly choosing a mineral from the collection that has a hardness greater than 7 ?[/quote]

There are only 3 hardnesses greater than 7, so the probability would be $\frac{7}{10}$

[hide] $ 8, 9, 10$, 3 choices out of a possible 10, $\frac{3}{10}$[/hide]

[hide]
3 favorable over 10 total choices.
$\frac{3}{10}$

[/hide]


What is the probability of getting x numbers of wins on a 3X3 Scratch Card in under y turns? (There are only one "win" in each card and you can only scratch once per card.)

Scratch Card: Used in campaigns like "Instant Scratch and Win!" by McDonalds. You scratch the silver coating over a card with a coin and voila! The silver stuff comes off and it reveals a picture or a message which usually says something along the lines of "Sorry, try again"
It is usually 3X3, which means three squares horizontally and vertically, forming a 3 by 3 grid with 9 squares. There are also 4X4 scratch cards which is 4 by 4 and have 16 squares. [url=http://www.ace-cgi.jp/cgi/asobi/scratch/sample/sample.html]Example[/url] (click on the link to the left, then enter a name and a made-up e-mail address; this is just the test version of the original cgi script used in conpetitions, so the name and the e-mail address doesn't really matter)

If you still don't understand the problem, feel free to ask.

Summary of the above: What is the chance of you winning x number of times under y number of tries if the odds of winning is 1/3*3?

[color=cyan]You're going to have to explain more about what exactly that means for me to understand it -- I don't get the real-life reference.[/color]

It would help if you shorten your questions... or like just make it clear... so we don't have to read through it... thanks

[color=cyan]If you would restate the actual process of what you are doing -- that is, it isn't clear from your question at all what hte process I am doing with these scratch cards is.  There is only one win per card?  I can only scratch once per card?  How many cards am I scratching from?[/color]

:?: Now I'm confused...I've already stated the answer to all three of the questions in the problem...
Answer 1 and 2: [quote="I"](There are only one "win" in each card and you can only scratch once per card.)[/quote]
Answer 3: [quote="I"]under y turns?[/quote] meaning "under y cards" or "under y number of tries"

I tried to make a table with y going from 1,2,3..m and x going from 0,..m.

eg. with y=1, it's 8/9 for x=0 and 1/9 for x=1
after writing this for y=2 and 3, it appears each term resembles the terms in the binomial expansion of (8/9 +1/9) raised to y

my guess for x hits is the valueof the term with (1/9)raised to x.
ie., ((8/9) raised to (y-x)) * ((1/9) raised to x ) * yCx

am i on the right track?

Yes, you're getting the general idea. Keep condensing it...

[quote="astatine"]Sorry, couldn't progress beyond one of the denominator factors being 9^y.

Out of curiosity, how long does one get to solve a problem such as this in the actual Mathcounts contest?[/quote]N
No clue for this is my first year (and my last :( )
Besides, these won't be on MATHCOUNTS since they're either
a.) confusing
or
b.) too difficult/easy

You're very close HariSeldon88, now try to write that without combinatrics.

Astatine - this is harder than a MATHCOUNTS problem.  A similar type problem for MATHCOUNTS might be:

Tom flips a fair coin repeatedly until he gets a total of 3 heads.  What is the probability that he only flips the coin a total of at most 5 times?

This would be a somewhat tough MATHCOUNTS problem, though the experienced, top students would knock it off in less than two minutes, I think.

I am horrible at this "appropriate level" thing...My school's not that interested in MATHCOUNTS so they haven't gave us anything yet...*sigh*

Hmm, is the answer to your question, Mr. Rusczyk [hide]11/16[/hide]?

I thought it was [hide]1/2 since the probability of not getting three heads in six turns is a half...oh well, I'm probably wrong again, so ignore me.[/hide]


$p$ is prime number find the number of elements of the set

$F=\{M\in M_n(Z/pZ) \;such \;that \; M^2=M \}$

--------------------------------------------------------------
Here it was the number of symetry
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19337

Probably I am misunderstanding something or I am really idiot, but this is not just a simple consequence of the problem regarding symmetries? I mean, can't we use the close connection between symmetries and orthogonal projections? PS: all these things work for odd p. For p=2, it is again a good exercise!  ;)


[quote="t0rajir0u"][hide="1"] The set of Pythagorean triplets is given by $ (2kmn, km^2 \minus{} kn^2, km^2 \plus{} kn^2)$, where $ k$ is not divisible by primes congruent to $ 1 \bmod 4$, $ (m, n) \equal{} 1$, and $ m, n$ are of opposite parity.  Whether $ a$ is less than $ b$ depends on the values of $ m, n$.  [/hide]
As to your second question, is $ c$ fixed?  I've already directed you to the [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]sum of squares[/url] function.[/quote]
I think you misunderstand the problem, t0rajir0u! The problem asks for what possible values of a, the equation a^2+b^2=c^2 has at least 1 solution(b,c) satisfying a<b! IT MEANS WHEN YOU FIX a under some restrictions, the equation always has at least 1 solution(b,c) satisfying a<b. The problem is finding the restriction of a. For example of contradiction, you can replace a=4, which corresponds to 2kmn=4 in your solution!
PS: happyme: Are you doing some contests?If you are, I think you should stop posting the problem now! :wink:

Edit: opps, I think your logic is right, t0rajir0u! :blush: It's still right because it covers all the conditions of sufficient and necessary.

Its from a Number Theory book (I have it in djvu format if you want, but I'm not sure if its completely legal to send it to you :lol:)

EDIT:

Just to get some cold hard numbers, I tried plugging in values of $ (m,n)$ to see if maybe there was a pattern way to determine the "breaking point" at which $ 2kmn < k(m^2 \minus{} n^2)$...no pattern so far.

Also, the parameterization of Pythagorean quadruples (according to [url=http://mathworld.wolfram.com/PythagoreanQuadruple.html]Mathworld[/url]) does not lead to certain values that work, is there a method to find those values? (Just kidding, realized that only occurs when you take $ q\equal{}0$)

EDIT 2:

Is this problem on a contest? If it is, I'll more than gladly remove it and wait till the contest is over.


linear algebra
matrix
algebra
polynomial
induction
geometry
geometric transformation

[quote="xxl"]...(A^m+1) -k*(A^m) - k*A + In=0 ...[/quote]
Hi xxl,

Just a question (A^m+1) ... do you mean A^(m+1) or A^m + In

Moubinool

he means A^(m+1).. i think.. :)
cheers! :D :D

the problem is rather easy.
consider the polynomial f(x) = xm+1 - kxm -kx +1. We know that f(A)=0 thus all the proper values of A are roots of f. It is easy to check that if \lambda is a root of f of multiplicity k then 1/\lambda is also a root of f of multiplicity k. (f is a reciprocial polynomial).

So all it remains to prove is that g | f then g is also a reciprocal polynomial, and as the characteristic polynomial of A divides f, we are done.

The proof of the above fact follows easy by induction on the degree of g, or simple computation.

[quote="xxl"]Let A be a n*n matrix in R.If exists m>n>=3(m in N) and k in R, |k|<=1 so (A^m+1) -k*(A^m) - k*A + In=0 , prove that detA=1.(In is the unit matrix). :D[/quote]

hi xxl,

Something strange in this problem,

k could be 0, n could be 3, m odd integer say 2p+1>n=3,

A^(2p+2) + I_3 = 0
A^(2p+2) = -I_3
det (A^(2p+2)) = det(-I_3)
(det A)^(2p+2) = (-1)^3

now  LHS  \geq 0, RHS is -1  :?

the fact is that if n is odd then the char. polynomial of A has at least a real root, which does not happen when k=0.

if think that a typing error is the cause of this. for |k|\geq 1 we are assured upon the existance of the real root of f, as f(-1) <0 and the continuity of f.

I think k must be <>0 and it's no need to add a condition on m(to be even ).

the main problem  is solved. if you want another problem, try finding out for which values of k, |k|\leq 1 the polinomial f (as I defined it above) has at least one real root.

I have a question. Can smb please help me? I always thought that if f(A)=0n with f a polynomial, then the MINIMAL polynomial of A divides f, not the CHARACTERISTIC polynomial. I didn't like linear algebra very much in the 11'th grade, so I might have some distorted notions. Can you make this a bit clearer for me?

you are right grobber ... I hurried a little bit last night :)
the minimal polynomial is reciprocical.

this means that my proof stands only for k=1 :)
for all other values I have to re-think it.

Here's another reason why I think there is sth wrong here:

What if m and n are even? Then -1 is a root of the polynomial and A could very well be -In, so detA=-1 (instead of 1)

The problem is like so:

Let A be a n*n matrix in R.If exists m>n>=3(m in N) and k in R, |k|<=1 so A^(m+1) -k*(A^m) - k*A + In=0 , prove that [b]|detA|[/b]=1.

cheers! :D :D

Let x1,x2,.., x_(m+1) be the solutions of the equation! We know that |x_k|=1 and the apply Frobenius!

We know that det(-A+XI_n)=(X-x_1)^(a1)*..*(X-x_(m+1))^(a_(m+1)) with a1,.., a_(m+1) natural numbers and their sum add up to n so we get that:
(-1)^n*det(A)=(-x_1)^(a1)*..*(-x_(m+1))^(a_(m+1))
so we get that |det(A)|=1.

problem solved!!!
cheers! :D :D

Frobenius:

Let K be one of {N,Z,Q,R,C}. If A is in M_n(C), the the characteristic and the minimal polinomial of A have the same ireductible divizors over K.

Hey, Lagrangia, maybe you skipped some steps and that's what got me confused:

How do you know that the roots of the equation have modulus 1?

Yes grobber I skipped the big part of the problem, I shall post it later! The solution for that is huge! :D at least my solution! :D

I mean we here in Romania do that problem in the 10th grade(at least I have :) )  and I considered that everyone has done it before! :D
It was given at a contest don't know the year and place, but I know the solution! :)
Don't worry I shall post it for everyone, but a bit later! :D :D

So, if i understand correctly, the problem is this?

The polynomial x^(m+1)-k*x^m-k*x+1 has all roots with modulus equal to 1? I didn't know that...

also grobber keep in mind that |k|<=1, and yes you did understand it corect! :D

cheers!

I think N shouldn't be in that list Lagrangia. It's not even a ring.
What do you think? Does x 2+x+1|x 4+x 2 +1 in N[X]?

You are 100% right amfulger, the theorem is only good for a ring (K is a ring! ), sorry about that!

uu...
i didn't make many problmes with polynoms ... so i do not know what means:
the [b]characteristic[/b] and the [b]minimal[/b] polinomial of A have the same ireductible divizors over K.
and ... what means "the proper value for a matrix"??
where can i find problems with polynoms...(matrices)
thx

I will send you some problems with matrices (they are written in Romanian). Proper value is in fact a mot-a-mot (wrong!) translation from Romanian which is supposed to mean eigenvalue. An eigenvalue of a matrix A is a number x for which det (A-xI) = 0. Obivously there aren't many numbers that fulfill that. In fact they are at most n, if A is a square matrix with n lines  and n columns.

The polynomial whoose roots are the eigenvalues of A is called the minimal polynomial of A. It is known that mA(A)=0n.

Also pA(x) = det (A-xI) is a polynomial, called the characteristical polynomial of A. It's roots (maybe some at high multiplicity) are the eigenvalues of A, and thus mA (x) | pA(x) => (this is actually Hamilton - Cayley theorem) pA(A)=0n.

Using Viete relationships one can notice that the sum of the eigenvalues (added with the corresponding multiplicity) is exactly TrA (that is the sum of the elements found on the main diagonal of A).

The product of the eigenvalues is obivously det A (it's pA(0)!).

[quote="grobber"]The polynomial x^(m+1)-k*x^m-k*x+1 has all roots with modulus equal to 1[/quote]
Also add |k| \leq 1 and k is real.
--------------------------------------------------------------------------------
Proof. Let P(X)=X m+1-k*X m-k*X+1.
Let f(x)=(k*x-1)/(x-k).
Denote by x' the complex conjugate of x.

If k=1, then P(X)=(X-1)(X m-1), so P has all roots of modulus 1.
If k=-1, then P(X)=(X+1)(X m+1), so P hass all roots of modulus 1.

Case |k|<1.
Notice that z is a root of P iff f(z)=z m.
Lets investigate f.

If |x|=1 then |f(x)|=|k*x-1|/|x-k|=|k*x'-1|/|x-k|. Since |x|=1, we get x'=1/x, so |f(x)|=|k-x|/(|x|*|x-k|)=1.
So if |x|=1, then |f(x)|=1.

If |x|<1 then I'll prove |f(x)|>1.
|f(x)|>1 <=> |k*x-1|>|x-k| <=> (kx-1)(kx'-1)>(x-k)(x'-k) <=> k 2|x| 2 -k(x+x')+1 > |x| 2 -k(x+x')+k 2 <=> (k 2 -1)(|x| 2-1)>0. The last is true because k 2<1 and |x| 2<1.

Similarly show that if |x|>1 then |f(x)|<1.

Now suppose P has a root z with modulus <>1. Then f(z)=z m.
If |z|>1 then |f(z)|<1, so |z| m<1 and |z|>1. This is not possible since m is a positive integer.
If |z|<1 proceed analogously or notice that 1/z is a root.

-----------------------------------------------------------------------------------
I had this problem (polynomial problem not matrix problem) 4 years ago in a contest, but I couldn't solve it then. I solved it today while "attending" an analysis class. Thanks for bringing this problem back to my attention.
That is because unsolved problems in the past just keep reaching back to you. This does not apply only for maths.  :D

it is rather strange as I had this problem at the same contest and remember that I've solved it (well in fact I don't remember it, but I solved all the problems then - with some minor glitches - and this problem seems to have been at that contest :D - strange)


Fie a,b,c numere reale pozitive si $abc=1$. Sa se demonstreze ca: \[(1+\frac{1}{a})(1+\frac{1}{b})(\frac{1}{a}+\frac{1}{b})^{2}\geq 32\frac{c^{2}}{1+c}.\]

Dupa ce gasim inegalitatile
$\ 1+\frac{1}{a}\geq\ 2\sqrt{\frac{1}{a}}$
$\ 1+\frac{1}{b}\geq\ 2\sqrt{\frac{1}{b}}$
$\frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b}$,
putem studia semnul trinomului de gradul 2.

hai ca transcriu eu :

Inlocuim abc cu 1 si obtinem:
$(1+bc)(1+ac)(bc+ac)^{2}\geq32\cdot\frac{c^{2}}{c+abc}$
$(1+bc)(1+ac)(a+b)^{2}\geq\frac{32}{c+abc}$
$(1+ab)(1+ac)(1+bc)c(a+b)^{2}\geq32$
$1+ab\geq2\sqrt{ab}$
$1+bc\geq2\sqrt{bc}$
$1+ac\geq2\sqrt{ca}$
$(a+b)^{2}\geq4ab$
Rezulta:
$(1+ab)(1+ac)(1+bc)c(a+b)^{2}\geq8abc\cdot 4abc=32$ q.e.d.

Sorin!!!De ce nu inmultesti relatiile obtinute din inegalitatea mediilor?????Iti zic eu ca dupa aceea iese usor!!!!!

Din $abc=1 \Rightarrow  c=\frac 1{ab}\Rightarrow \frac{c^{2}}{1+c}=\frac{\frac1{a^{2}b^{2}}}{1+\frac1{ab}}=\frac1{(ab)(ab+1)}$ deci trebuie sa demonstram o inegalitate in a si b:
$\frac{(a+1)(b+1)(a+b)^{2}}{a^{3}b^{3}}\geq \frac{32}{(ab)(ab+1)}\iff \frac{(a+1)(b+1)(a+b)^{2}}{a^{2}b^{2}}\geq \frac{32}{ab+1}$.Iar cum $a+b\geq 2\sqrt{ab}\Rightarrow \frac{(a+1)(b+1)(a+b)^{2}}{a^{2}b^{2}}\geq \frac{(a+1)(b+1)4ab}{a^{2}b^{2}}=\frac{(a+1)(b+1)\cdot 4}{ab}$.Acum vom demonstra ca $\frac{(a+1)(b+1)\cdot 4}{ab}\geq \frac{32}{ab+1}\iff$

$\frac{(a+1)(b+1)}{ab}\geq \frac 8{ab+1}\iff (a+1)(b+1)(ab+1)\geq 8ab$.Aceasta din urma este evidenta,caci rezulta din inmultirea inegalitatzilor:
$a+1\geq 2\sqrt a$
$b+1\geq 2\sqrt b$
$ab+1\geq 2\sqrt{ab}$
End of story

si eu am gandit-o cam la fel! Am scris inegalitatea dintre Ma si Mg pentru cele trei paranteze.Le-am inmultit,iar dupa aceea am mers pe echivalente
am obtinut ca (1+1/a)(1+1/b)(1/a+1/b)>=16/abradical din ab. deci trebuie sa demonstram ca 16/ab radical ab>=32c^2/1+c <=>
1/ab radical ab >=2/a^2b^2 inmultit cu ab/ab+1 <=>
ab+1>=2 radical ab (evident adevarata) => inegalitatea a fost rezolvata.

Imi cer scuze pentru felul in care am redactat rezolvarea deoarece nu stiu latexul :!:

hint :
vedeti butonu de Latex Help de sus ? intrati pe acolo si cititi cu rabdare. pe bune ca nu e deloc greu. eu de acolo am invatat.

Mersi fain, Maky, pt. ajutor   :) .La momentul in care scriu mesajul de fata am reusit, in sfarsit, sa invat $\ Latex$. :showoff:

Eu in general nu prea am rabdare.Nu-mi place sa citesc instructiunile la un program ,ci sa le descopar singur :)  :!:


For any number $ x$, we are told that $ x\&\equal{}7\minus{}x$ and $ \&x \equal{} x \minus{}7$. What is the value of $ \&(12\&)$?

[quote="GameBot"]For any number $ x$, we are told that $ x\& \equal{} 7 \minus{} x$ and $ \&x \equal{} x \minus{} 7$. What is the value of $ \&(12\&)$?[/quote]

12& = 7 - 12 = -5

&(-5) = 7 - (-5) = $ \boxed{12}$

I think $ \boxed{ \minus{} 12}$.

[b]Remark[/b]
$ \&(x\&) \equal{} (7 \minus{} x) \minus{} 7 \equal{} \minus{} x$

It's &(&x) not &(x&).  :)

EDIT: Fixed a typo - that was supposed to be &. All better now.

EDIT 2: Oh no wait huh...


Does there exist a positvie integer $ a$ such that all the numbers $ a,2a,3a,...,1997a$ are perfect powers (i.e. numbers of the form $ m^k$, where $ m,k \in \mathbb{N}$, $ k \ge 2$)?

Yes.  Let $ k_1, ... k_{1997}$ be distinct primes.  We claim that for each $ i, 1 \le i \le 1997$ we can have $ ia \equal{} m_i^{k_i}$ for some $ m_i$.  Writing

$ a \equal{} p_1^{e_1} p_2^{e_2} ...$

we see that the set of conditions $ ia \equal{} m_i^{k_i}$ gives, for each $ j$, a system of congruences for $ e_j$.  For example, for $ i \equal{} 1$ and $ p_i \equal{} 2$ the system is

$ e_1 \equiv 0 \bmod k_1$
$ e_1 \equiv \minus{} 1 \bmod k_2$
$ e_1 \equiv 0 \bmod k_3$
$ e_1 \equiv \minus{} 2 \bmod k_4$

and so forth.  More generally we have

$ e_j \equiv \minus{} v_{p_j}(i) \bmod k_i$

where $ v_p(i)$ is the greatest power of $ p$ dividing $ i$.  By CRT each of these systems has a positive integer solution.


Hi
Either something is still missing on this wonderful forum or I simply don't know where to look

Is there a place where you can go and ask questions about countries or regions in the world?  I know there is round table for non math stuff but they are not really geography related.  The national forums exist already, but they are mostly for math purposes, and although they seem to be open to foreigners, there is the language issue

So, if I may be very explicit, suppose I was wondering about the issue in Korea of unification, and the nuclear threat there, and I would just like to know what some people thought of that (with this many nationalities like Chinese, Japanese, South Korean, American here... that should be interesting) where should I go?

You may say that's more politics than mathematics, but having a mathematical mind gives you a certain different view on things, at least different enough to be interested in fellow mathematicians opinions on such matters as well (of course nonmath people can have just as interesting opinions)

So I repeat my statement : is there such a place here, or is it deliberately avoided to avoid discussions that would needlessly create conflicts or to avoid distraction of the actual purpose of mathlinks?

With most respect and thanks,
Fred

Hello,

I would not vote for such an idea: there are plenty of other places where such subjects can be fully discussed. Personally, I even believe that the Chemy/Physics forums alienate the nature of this website by attracting mostly people who want to have their homework done (other people going more likely to specialized websites). I have nothing against the lower math levels though, for it is good for the younger to be in contact with various levels.

Good point, I don't want such a forum that much that I want the essence of Mathlinks to go lost (Mathlinks is a safe haven that must be protected at all cost...)
But isn't there less risk in alienation, knowing that nobody would go there just to get homework done?

It would just be nice, if you see all these people from all these countries you could once in a while ask them something else (don't ge me wrong the math is what I came here for and what I will stay for)

But hey, I'm not the boss here

There are hundreds of different forums on this website :) so everyone has it's own little corner.

For the main question about this topic, there is the [url=http://www.mathlinks.ro/Forum/index.php?f=219]Other Topics[/url] forums. You can discuss all those issues (some of them have been discussed before even) in the [url=http://www.mathlinks.ro/Forum/index.php?f=138]Round Table[/url].

Julien: don't worry about it. There are plenty of forums in which the young ones solve easier problems (the Middle School forums, the High School forums, the Tutorial forums, etc. etc. etc.)

That shouldn't affect your wanting to discuss hard problems ;)

PS In the National and Local communities you can discuss of course other topics than mathematics, provided it is somewhat connected to that community. There are cases like this in many national communities ;)


Does countdown round ever count? Because it doesn't in my region.

In some chapter and state meets, it does.  Usually it does at both state and chapter (correct me if wrong?) but in many it doesn't (like mine!!!! California fails.)

Countdown round can be official or not. It was official for me, for example.

I refer you to [url]https://mathcounts.org/Page.aspx?pid=289[/url] which has information.

Basically, if a CD is official, it must be done in a certain way, and must be official for all chapters in the state. If not, it can be done in basically any way.

What do you mean by the way "Does Countdown Round count?" Because I thought all the states and regions participating in Mathcounts always use countdown.

Yes, all states use countdown, but whether it is [i]official[/i] or not is the state's choice. For example, the first and second place written round winners could compete against each other in an unofficial countdown with the written round's first place losing. However, the final first place award still goes to the first person in the written round. In other states, the first place trophy would go to the person that won the countdown, thus, in that state, the countdown is called [i]official[/i].[/i]

for example, my chapter uses countdown to determine official results, while AIME15's uses written scores.

Also, Nats uses Countdown for placing winners.

[quote="jonathanchou711"]This is new to me. OK always use countdown to determine our official ranking. So does that mean in AIME15's state they just take the written score and just play out the countdown for no reason at all? Dang.. That stinks. Countdown is my favorite round.[/quote]

Yup, many of the states do this.

[quote="nikeballa96"]for example, my chapter uses countdown to determine official results[/quote]

Unfortunately...


Let $ \{\alpha_1,\alpha_2,...,\alpha_n,\beta_1,\beta_2,...,\beta_n \} \in \mathbb{R}^{2n}$ such that $ \tan(\alpha_i \plus{} \beta_j) \neq 0$ ,$ \tan(\alpha_i) \neq 0,\tan(\beta_j) \neq 0,\tan(\alpha_i) \plus{} tan(\beta_j) \neq 0$ , for all $ 1 \leq i,j \leq n$
 and $ A \equal{} [a_{ij}]_{1 \leq i ,j \leq n} \in M_n(\mathbb{R})$ such that : $ a_{ij} \equal{} cotg(\alpha_i \plus{} \beta_j)$ , for all $ 1 \leq i,j \leq n$
Calculate Determinant : $ \det(A)$
[hide=" Times"]A result very nice  :) [/hide]

Answer of Me :
 $ \det(A) \equal{} |cotan(\alpha_p \plus{} \beta_q)|_{1\leq p,q \leq n} \equal{} \frac{1}{2}|\frac {1}{tan(\alpha_p) \plus{} tan(\beta_q)}|_{1\leq p, q \leq n}\{\prod_{k \equal{} 1}^{n}{(tan(\alpha_k) \plus{} i)(tan(\beta_k) \plus{} i)} \plus{} \prod_{k \equal{} 1}^{n}{(tan(\alpha_k) \minus{} i)(tan(\beta_k) \minus{} i)}\}$ ($ i^2 \plus{} 1 \equal{} 0$)
 From here easy calculate $ \det(B)$ , $ b_{pq} \equal{} tan(\alpha_p \plus{} \beta_q)$  .....
  Read Hint in Books: Determinants and their Applications in Mathematical physics (Robert Vein and Paul Dale) !

Determinants and their Applications in Mathematical physics (Robert Vein and Paul Dale)
Springer (Applied mathematical sciences vol 134)

Maybe Quybac tought

§4.1.5

page 58 exercice 2


Given scalene triangle ABC, how do you construct (with compass and straightedge) an equilateral triangle with the same area as ABC?


Your first inequality is equivalent to
$\dfrac{a}{b}+\dfrac{b}{a}\geq\dfrac{1-a}{1-b}+\dfrac{1-b}{1-a}$
$\Leftrightarrow (\dfrac{a}{b}+\dfrac{a-1}{1-b})+(\dfrac{b}{a}+\dfrac{b-1}{1-a})\geq 0$
$\Leftrightarrow \dfrac{a-b}{b(1-b)}+\dfrac{b-a}{a(1-a)}\geq 0$
$\Leftrightarrow \dfrac{a-b}{b(1-b)}\geq\dfrac{a-b}{a(1-a)}$
$\Leftrightarrow a(1-a)(a-b)-b(1-b)(a-b)\geq 0$
$\Leftrightarrow (a-b)(a-a^2-b+b^2)\geq 0$
$\Leftrightarrow (a-b)^2(1-a-b)\geq 0$
Since $a+b\leq \dfrac{1}{2}+\dfrac{1}{2}=1$, we get the desired inequality.

good! Now try the second one which is bit more difficult than this.

We know that  1> a + b

 1-(a+b) + ab > ab

 (1-a)(1-b) > ab

  Since 0< a < 1/2, 1/2 <1-a<1

 also 1/2 <1-b<1

 From  (1-a)(1-b) > ab

 1/ab > 1/(1-a)(1-b)

since 1-a,1-b are positive

   We have (a-b)^2 = (b-a)^2

 Multiply this into the last inequality

 (a-b)^2 / ab >= (b-a)^2/(1-a)(1-b)

 Add  2 to both sides and simplify the left side

 (a^2 + b^2)/ab  >=[[(1-a)-(1-b)]^2 +2(1-a)(1-b)]/(1-a)(1-b)

 Simplfy the right side

  (a^2 + b^2)/ab  >= [(1-a)^2 + (1-b)^2]/(1-a)(1-b)


For which real numbers c is 1/2 (e^x+e^(-x))<=e^(cx^2) for all real x?

Is-it a challenge (and I will let people search - just to make sure : [hide]c >= 1/2 ?[/hide]) or are you looking for the answer ?

Let's make it a challenge I guess. That's correct, by the way.

Makes no difference, being in this forum..

[quote]Is-it a challenge...or are you looking for the answer?[/quote]

Hmm...has Simon ever posted a problem for which he was actually looking for the answer?  :roll: [/quote]

I don't think there is a problem out there where Simon [i]doesn't[/i]know the answer to....

[quote]I don't think there is a problem out there where Simon doesn'tknow the answer to....[/quote]

Hmm...maybe we should all combine our powers and come up with a problem that will stump him.  :twisted:

I have posted problems I have been unable to solve before. I posted a Mandelbrot problem I thought was tough once, and I posted a problem I made up once that I couldn't solve (it turned out it was really hard though). More often, when I post stuff, I post problems that come from books or past contests, and those usually have solutions, so even if I can't figure them out, you don't know that since I can read the solutions. Seriously though, there are tons of problems I can't solve. You can look through the list of IMO problems, and you can be sure that I can't solve half them or maybe more.

Ohh, I know some of those :) but they're lying around the house somewhere.. don't know where.

About two years ago, I had some urge to find all the olympiad problems I could and put them into a big binder. So now I have a big binder full of problems, many of which I cannot solve.

OK guys, Simon has just revealed his weakness! LEt's get him!

BUt please, no Irish or Canadian MO's. Even i can do them
Give him RUssian, Romanian, Bulgarian, or CHinese ones. Let him have some funn.. :twisted:

[quote]BUt please, no Irish or Canadian MO's. Even i can do them
Give him RUssian, Romanian, Bulgarian, or CHinese ones. Let him have some funn..[/quote]

Yeah, but solutions to those can be found on kalva, so we wouldn't know whether he knew how to do a problem or whether he simply looked up the answer. We gotta find something harder...

Oh come on. You didn't really think I could solve all olympiad problems, did you?

I don't know, it might be possible.

Hmm, perhaps maybe we should wait until Joel comes back....

Who cares how many problems he can solve? Dang. i/3, what method did you use? I used series expansion (first term is cosh), and didn't bother finding any other way of solving the problem after that (which I usually do).

yeah thats what i was thinking, but dont know where to go from there.

[quote="Mindspa"]what method did you use?[/quote]

I used series near 0 to guess that c >= 1/2 was the answer. Then  x - x^2/2 <= log(1+x) <= x to finish

Yes, that's the solution in the book too.