function algebra domain parameterization USAMTS

Can someone tell me why I only got a 3 on question 5? I don't really want to post this on Grading Issues because there is probably some obvious error I don't really see. I checked the grading rubric but I'm not really sure...

The grader's comment was about why that equation at the very bottom of the first page is true... Do I really have to explain that?

Any explanation would be GREATLY appreciated. Thanks.

I'm not sure if this is why, but I think you made a rather big jump from your big radical equation to the next one. I don't know that much about infinite radicals though :maybe:

You got a 3 for that? I did something similar (with powers, instead of radicals --- I did it the opposite way of you), and I got a 2...should I have gotten a 3?

It looks like this only works when all $ a_i$ are equal, but you don't prove that this must be the case.

I don't know much about the rigorous definition of limit and infinity, so it seems true.

What I did is very like yours, but takes longer time. I first define an inverse of the original function with the restricted domain, then prove for any $ x$ within the domain, the inverse raised to any power is defined. After that, I try to find the interval of $ x$ with the parameters of some input. Then I input the given and got the similar big radical equation, but my equation doesn't have a indefinite variable inside the infinite radical operations, and can be simply solved by squaring.

[quote="Don Quixote"]The grader's comment was about why that equation at the very bottom of the first page is true... Do I really have to explain that?[/quote]

Absolutely.  You haven't proven that your expression converges.  This is a necessary step.

Edit:  Here is an example of what I mean.  Try to solve the equation

$ x^{x^{x^{...}}} \equal{} 3$

You might naively say that $ x^3 \equal{} 3$, hence $ x \equal{} \sqrt[3]{3}$.  In fact, if you actually plug in $ x \equal{} \sqrt[3]{3}$, the LHS evaluates to $ 2.478...$ and [b]not[/b] to $ 3$.

The reason is that $ x^{x^{x^{...}}}$ attains a maximum value of $ e < 3$.  The above equation actually has no solution.

[quote="Don Quixote"]wait why do they all have to be equal? all i'm doing is substituting in the expression for a sub i for a sub (i-1).

 :huh: I'm confused...[/quote]

You can show that all $ a_i$ are equal because $ a_n$ tends to drift farther and farther away from $ \minus{}\frac{1}{2}$ unless $ c\equal{}\minus{}\frac{1}{2}$.  That's what I did, and I got a 5.

You can't tell by glancing at the problem that all $ a_n$ are equal.  It could be possible (even though it wasn't so) that you have a loop, as in, there exists some $ k>1$ such that $ a_n\equal{}a_{n\plus{}k}$.  In other words, you only showed that $ c\equal{}\minus{}\frac{1}{2}$ works, but not that the others don't.

Actually, no offense intended or anything, but I don't think this solution gets anywhere. USAMTS graders are generally very nice with these sorts of grades, though on real contests proofs along this line would probably reach a 0/7. Again, I really don't mean to put you down or anything, but just a note to look forward to the future.

Let $ f(x) \equal{} 2x^2 \minus{} 1$. You need to find all $ c$ such that $ f_n(c) < 0$ for all $ n$, where $ f_k(x)$ means to do $ f(x)$ $ k$ times.

In your proof, you say that (at the line in question) $ c \equal{} f^{ \minus{} 1}(f^{ \minus{} 1}(\ldots f^{ \minus{} 1}(a_b)\ldots ) )$, for some term $ a_b$, your proof seems to put $ b \rightarrow \infty$. So up to this point, you basically just restated the definition of $ a_b \equal{} f_{b \minus{} 1}(c)$, just written as inverses instead.

Then you jump to the conclusion that $ a_b \equal{} c$ for all $ b$ (in your proof, it literally says $ \boxed{f^{ \minus{} 1}_{b \plus{} 1}(a_b) \equal{} c \Rightarrow f^{ \minus{} 1}(c) \equal{} c}$: 1st part is the definition, 2nd part is the thing to prove. Everything in between skipped!). You might as well have just put this in one line and avoided typing any radicals, since though it looks like it does something it really just creates a harder-to-see logical fallacy. In fact, since you radicals just keep on going, I'm not sure where the assumption that $ c \equal{} \minus{} \sqrt{\frac {1 \plus{} c}{2}}$ came from. You never did approach any limit with $ a_i$s so you can't assume that at some point the sequence will return to $ c$ underneath the radicals. The condition that all of the terms had to be negative was barely even touched (do you see why now?)

You can take a look at the solutions thread, where similar radical limit arguments exist where it is proven that the sequence will diverge given any $ c \neq \minus{} \frac {1}{2}$.

Final note, it appears that this might be your first round of the USAMTS. Don't be discouraged, this was a tough problem to rigorize, and there's still plenty of rounds to go. The goal after all with contests like these is to learn from these mistakes.  :) [and if my explanation doesn't make sense, just let me know. I'm terrible at explaining stuff]

[quote="123456789"]lingomaniac, could I see your solution please?[/quote]

Here's a brief description of it:

I first show that $ 4x^4\minus{}3x^2\minus{}x\geq 0$ when $ x<0$.  Then I use this to show that if $ a_n<0$ and $ d(n)\equal{}\left|n\plus{}\frac{1}{2}\right|$, the distance between $ n$ and $ \minus{}\frac{1}{2}$, then $ d(a_{n\plus{}1})\geq d(a_n)$ with equality iff $ a_n\equal{}\minus{}\frac{1}{2}$.  Therefore if $ c\neq \minus{}\frac{1}{2}$, it tends to drift further away from $ \minus{}\frac{1}{2}$, so it must become positive at some point.

Yes, my proof was the same as yours as well, but I proved that it converged. In doing so, I used some approximations to prove that certain quantities were larger than others which apparently did not "constitute a rigorous proof." Decimal approximations apparently did, however, "constitute a rigorous proof" in number three when comparing $ \arctan(\frac{1}{4})$ and $ \frac{\pi}{12}$ :maybe:
Ironically, judging by your score, lingomaniac88, I could have left out everything about converging and gotten a 5. Oh well, a 4 is good enough for me I guess. :)

[quote="lingomaniac88"][quote="diophantient"]That was my solution too, but it isn't rigorous because there's no guarantee that the rate of change is fast enough, it could still theoretically converge.[/quote]

I guess you're right.  But I got a 5 on it, and I'm happy with that.[/quote]
Gah, I got a four for the same solution.

[quote="diophantient"][quote="lingomaniac88"][quote="123456789"]lingomaniac, could I see your solution please?[/quote]

Here's a brief description of it:

I first show that $ 4x^4 \minus{} 3x^2 \minus{} x\geq 0$ when $ x < 0$.  Then I use this to show that if $ a_n < 0$ and $ d(n) \equal{} \left|n \plus{} \frac {1}{2}\right|$, the distance between $ n$ and $ \minus{} \frac {1}{2}$, then $ d(a_{n \plus{} 1})\geq d(a_n)$ with equality iff $ a_n \equal{} \minus{} \frac {1}{2}$.  Therefore if $ c\neq \minus{} \frac {1}{2}$, it tends to drift further away from $ \minus{} \frac {1}{2}$, so it must become positive at some point.[/quote]
That was my solution too, but it isn't rigorous because there's no guarantee that the rate of change is fast enough, it could still theoretically converge.[/quote]

Right, but you can make that solution work by proving a lower bound of $ d_{n}$.  I think I proved in my solution that if all $ a_{i}$ are negative, then $ d_{n}\ge 2d_{n\minus{}2}$.  You can also prove that the sequence $ \{a_{n}\plus{}1/2\}$ has alternating signs.  Then we can find a large value of $ n$ such that $ a_{n}\plus{}1/2$ is positive and $ d_{n}>1/2$, therefore there are positive terms in the sequence unless $ c\equal{}\minus{}1/2$.


geometry
perimeter
inequalities
triangle inequality

Here is an interesting problem:

Suppose you have a pentagon and you connect each vertex to each other non adjacent vertex to get an inscribed star (or pentagram). The inscribed star will have an inscribed pentagon in it. Prove the sum of the perimeters of the two pentagons is less then the perimeter of the star.

[hide]Take away the sides of the smaller pentagon from the pentagram. This leaves you with a 10-sided start shape, which forms 5 small non-overlapping triangles with the larger pentagon. You can apply the triangle inequality to each of these 5 triangles to get the result.[/hide]

This shows that the star has greater perimeter then the larger pentagon, but what about the sum of the perimeters of the larger pentagon and the smaller pentagon?

He included 5 internal segments of the pentagram (star) as part of the pentagram. By perimeter, do you want to include those internal segments, or just the outer edges?

Interesting, indeed. It is easily proven that the perimeter of the pentagram is greater than the perimeter of the smaller and the larger pentagon, but can we prove that it is greater than the sum of both? :P

[b]Zuton[/b] — the sides of the inscribed pentagon is not considered when measuring the perimeter of the pentagram, as [b]MellowMelon[/b] suggested (so, yes, only the 10 outer edges).


geometry parallelogram rhombus AMC AIME USA(J)MO USAMO

6. Let $ABCD$ be a quadrilateral with an incircle that touches $AB$, $BC$, $CD$, and $DA$ at $P$, $Q$, $R$, and $S$, respectively. Provided that $AC$, $DB$, and $PQ$ bisect the area of $ABCD$, prove that $ABCD$ is a rhombus.

EDIT: $PQ$ should be $PR$...$PR$ divides the area in half, sorry

Because of the diagonals $AC,$ $BD,$ bisect the area of $ABCD,$ we conclude that it is a parallelogram $($ From $(ABD) = (CBD)$ $\Longrightarrow$ $AA' = CC',$ where $A',$ $C',$ the orthogonal projections of $A,$ $C$ respectively, on $BD$ and so, from the congruent right angled triangles $\bigtriangleup AA'P,$ $\bigtriangleup CC'P,$ where $P\equiv AC\cap BD,$ $\Longrightarrow$ $PA = PC$ and similarly $PB = PD$ $).$

Because of now ( as a parallelogram ), $ABCD$ is a circumscribed quadrilateral, we conclude that it is a rhombus ( easy to prove ) and the proof is completed.

Kostas Vittas.

sorry this problem was way too easy, i was thinking of using the AC, BD, PR concurrency [the proof for that was new to me, so i thought it would be difficult, but it seems to be way more well-known than i thought], i will still be trying to pose "olympiad-level" geometry problems...i enjoy making problems and i am working on moving from posing aime level problems to posing usamo level problems

--Alex


Giving your answer in simplest form, find the value of $ (64)(64)\minus{}(67)(61)$.


lol, I played this before, but I prefer Stepmania.

if jesus played computer games he'd play stepmania as well as doing 5-man naxx runs with noah and a couple of his other wow buddies

ffr is fun though, i play every now and then, more used to stepmania though so my scores tend to blow

http://www.flashflashrevolution.com/profile/Potatoes/


find all natural numbers (a,m,n) that
a^m +1|(a+1)^n

look at this http://www.mathlinks.ro/viewtopic.php?p=1243223#1243223


Hi.
On the application it says that the first recommendation should be from a teacher; the second should "be from another adult who knows you personally, [...] preferably someone outside of school and not a relative".  If there are no adults outside of school and family that do not know me well enough to write a recommendation, should I get another teacher or a parent to do it?  (I do not attend church, so 'pastor' is not an option; all sports I play are coached by school faculty; I have no job (yet ;) ) )

how about your guidance counselor? I know it says "outside of school" but it may be better than a relative or another teacher :)

[quote="Geehoon"]What do they mean by "adult"

can a friend write one also?[/quote]

Hmm...I am not so sure that your friend could write a reccommendation ;)

[quote="Danbert"]If there are no adults outside of school and family that do not know me well enough to write a recommendation, should I get another teacher or a parent to do it?  (I do not attend church, so 'pastor' is not an option; all sports I play are coached by school faculty; I have no job (yet ;) ) )[/quote]

Another teacher would be fine, as long as you give them the instructions for the second rec letter and not the teacher letter (i.e. as long as you ask them to comment on your personal qualities, rather than on your academic achievements).  A family friend would also be fine, for example.

[quote="xxreddevilzxx"]Could it be somebody that i got to know well..., for example, while i was working??[/quote]

Sure.

[quote="ztbb"]A family friend would also be fine, for example.[/quote]
How about a parent?  I've been turned down a couple of times, and I'm getting worried and looking for a last resort.

[quote="ztbb"][quote="Danbert"]How about a parent?  I've been turned down a couple of times, and I'm getting worried and looking for a last resort.[/quote]

It's very strongly preferable to us that it not be a relative.[/quote]

It's all right now, parents decided that I can't go anyway :(


[color=darkblue]Let $ABC$ be a triangle. Prove that for any point $M\in [BC]$ there is the inequality

$\boxed{\ 1+\frac{MA^{2}}{MB\cdot MC}\ge \left(\frac{b+c}{a}\right)^{2}\ }\ .$ When we have the equality ? [/color]


calculus
integration
derivative
parameterization
calculus computations

I was wondering if there is a good way to evaluate this integral, I was able to show that it's equal to $\sum_{n=1}^\infty \frac{1}{n^2}$, but can it be evaluated any other way?

\[ -\int_0^1\frac{\ln(1-x)}{x} \ dx  \]

I couldn't figure out how to do it except by using a series expansion

Mathematica's Integrator calls it "Li2(x)" and refers me to http://mathworld.wolfram.com/Polylogarithm.html
which is the same series expansion

So I'm betting my money that the only way to do it is with a series

Well, there probably isn't an elementary anti-derivative, but I was wondering if anyone knows of any technique such as introducing a parameter that can be used to evaluate the integral.  Thanks.

See post 2: Taylor Series seem to do the trick.

[quote="Wumbate"]How did you show it is equal to that sum?[/quote]

I started with the sum $\sum_{n=1}^\infty \frac{1}{n^2}$, then I used a double integral $\int_0^1\int_0^1 (xy)^{n-1} \ dxdy = \frac{1}{n^2}$ and the monotone convergence theorem to show:
\[ \sum_{n=1}^\infty \left(\int_0^1\int_0^1 (xy)^{n-1} \ dxdy \right) = \int_0^1\int_0^1\sum_{n=1}^\infty (xy)^{n-1}\ dxdy = \int_0^1\int_0^1\frac{dxdy}{1-xy} \]

I've seen a way to evaluate this integral with a Jacobian transformation, but I just wanted to see if there was another way, and the above integral reduces to what I have in the first post.

Also, to show that that integral (the one in the first post) is equal to the sum, just use a taylor series.

[quote="puuhikki"]The Fourier series of $f(x)=x^2$ in the range $[-\pi,\pi]$ and Dirichlet theorem yields the identity
\[ \frac{\pi^2}{3}+4\sum_{k=1}^\infty \frac{(-1)^k}{k^2}\text{ cos }kx=x^2  \]
for all $x\in [-\pi,\pi]$. Now substitute $x=\pi$ and you can find that $\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$.[/quote]

Thanks, but I knew how to evaluate the sum ( by the way this is my favorite method), I just wanted to know how (if) there was another way to evaluate that integral, but I guess there isn't.


induction
number theory proposed
number theory

Find the positive integers $a$ with the following property: any positive integer has at least as many divisors of the form $ak+1$ as of the form $ak-1$.

Only those with $\gcd(x,a) \implies x^{2}\equiv 1 \mod a$.

Let $a= \prod_{i}p_{i}^{v_{i}}$.
If $\gcd(x,a)$ does not imply $x^{2}\equiv 1 \mod a$, we know that we can already find one of those prime powers $p_{i}^{v_{i}}$ such that this does already not hold $\mod p_{i}^{v_{i}}$. Let $k$ be one of these $i$ with that property.
Now we take, by Dirichlets theorem, a prime $r \equiv-1 \mod p_{i}^{v_{i}}$ for all $i \neq k$ and $\equiv 1 \mod p_{k}^{v_{k}}$.
We also choose two primes $q_{1}\equiv q_{2}\mod a$ such that they are $\equiv 1 \mod p_{i}^{v_{i}}$ for all $i \neq k$ and such that $q_{1}^{b}\equiv q_{2}^{b}\equiv-1 \mod p_{k}^{v_{k}}$, but $q_{1}^{s}\equiv q_{2}^{s}\not\equiv-1 \mod p_{k}^{v_{k}}$for some $b>1$ and all $0 < s < b$ (residues of the last kind exist by assumption, e.g. we could choose $q_{1},q_{2}$ as primitive roots $\mod p_{k}^{v_{k}}$).
Finally, set $n=q_{1}^{b}\cdot q_{2}\cdot r$.

Let $d$ be a divisor $\equiv 1 \mod a$ of $n$. If $p_{i}| d$, then $d \equiv-1 \mod p_{i}^{v_{i}}$, which contradicts $d \equiv 1 \mod a$ except if $p_{i}=2, v_{i}=1$ and there are no other prime divisors except $p_{k}$ (in the last case, we simply consider $n=q_{1}^{b}\cdot q_{2}$ instead of our previous $n$).
Similar, we exclude that $q_{1}$ or $q_{2}$ divides $d$ (we need that $q_{1}^{s}\not\equiv 1 \mod p_{k}^{v_{k}}$ for $0<s \leq b+1$, which is true by assumption again). Thus in total there is only one divisor $\equiv 1 \mod a$: $1$ itself.
But there are at least two that are $\equiv-1 \mod a$, namely $n=q_{1}^{b}\cdot r$ and $n=q_{1}^{b-1}\cdot q_{2}\cdot r$.

So we are left to take a look at the remaining cases, which are the divisors of $24$. I think this could be done by induction over the number of prime divisors (here some different cases occure) if one extends it to:
There are at most as many divisors $\equiv m \mod a$ of $n$ than such ones $\equiv 1 \mod a$ for all $m$ coprime to $a$.
Well, it would be better to handle all this as the group $(\mathbb{Z}/2\mathbb{Z})^{k}$.
But I'm not passionated enough to go through this casework now and hope there is a better way.

Yep, that's it. It is true that proving the assertion for all divisors of $24$ is particularly disgusting, but I think it is much more interesting to prove that only the divisors of $24$ work.


geometry
conics
ellipse
trapezoid
geometry unsolved

1. What is the area of the largest ellipse that can be inscribed in a 13-14-15 triangle? (with proof)
(also generalize to any triangle with side length $ a,b,c$)

Another random problem (geometric combo):
At most, how many regions do 10 lines and 10 circles divide the plane?
(also generalize to $ m$ lines and $ n$ circles)

(I posted this in Pre-Olympiad but no one responded :( )

One more: In trapezoid $ ABCD$ with $ AB\parallel CD$ and $ AB < CD$, let $ P$ be the intersection of diagonals $ \overline{AC}$ and $ \overline{BD}$. It is given that the sum of the areas of $ \triangle APD$ and $ \triangle BPC$ is 1996. If the lengths of both bases are integers and the distance between them is an integer, compute the minimum area of $ ABCD$.

[quote="mathwizarddude"]1. What is the area of the largest ellipse that can be inscribed in a 13-14-15 triangle? (with proof)
(also generalize to any triangle with side length $ a,b,c$)[/quote]

Read http://mathworld.wolfram.com/SteinerInellipse.html . It says that the Steiner inellipse is the inscribed ellipse with maximum area. There is also a formula to compute the area of this ellipse by knownig the sides of the reference triangle.
$ \text{Area ellipse} \equal{} \frac {\pi \Delta}{3\sqrt {3}}$ . where $ \Delta$ is the area of the refernce triangle.


Dear Vasc,

At first, nice inequality! I won't write a boring solution this time. I have some kind of generalization instead. The calculations got terrible (but straightforward nevertheless) so maybe there is some mistake. But I think the idea is clear:

For $ a_1,\ldots,a_n>0$ with $ a_1\ldots a_n\equal{}1$ and $ p,q>0$ with $ (p\plus{}1)(q\plus{}1)\ge (p\plus{}q\plus{}1)n$ we have \[ \sum_{i\equal{}1}^n\frac1{(a_i\plus{}p)(a_i\plus{}q)}\le\frac{n}{(p\plus{}1)(q\plus{}1)}.\]

I think your ineq was this with $ p\equal{}2n\minus{}2$, $ q\equal{}2n\minus{}1$.  :lol:

Thanks for all your beautiful inequalities!  :)
Juan Dodyk

You are right, Bimbo. However, the general inequality is as follows:

Let $ a_1,a_2,...,a_n$ be positive real numbers such that $ a_1a_2...a_n \equal{} 1$. If $ p$ and $ q$ are nonnegative real numbers such that $ p \plus{} q \le \frac 1{n \minus{} 1}$, then

$ \frac 1{1 \plus{} pa_1 \plus{} qa_1^2} \plus{} \frac 1{1 \plus{} pa_2 \plus{} qa_2^2} \plus{} ... \plus{} \frac 1{1 \plus{} pa_n \plus{} qa_n^2} \le \frac n{1 \plus{} p \plus{} q}$.

This is also nice.

If $ a_1,a_2,...,a_n$ are positive real numbers such that $ a_1a_2...a_n \equal{} 1$, then

$ \frac 1{2n \minus{} 2 \plus{} a_1 \plus{} a_1^2} \plus{} \frac 1{2n \minus{} 2 \plus{} a_2 \plus{} a_2^2} \plus{} ... \plus{} \frac 1{2n \minus{} 2 \plus{} a_n \plus{} a_n^2} \le \frac 1{2}$.

Vasc said:
[quote]If $ a_1,\ldots,a_n$ are positive real numbers such that $ a_1\ldots a_n\equal{}1$, then \[ \frac 1{2n \minus{} 2 \plus{} a_1 \plus{} a_1^2} \plus{} \frac 1{2n \minus{} 2 \plus{} a_2 \plus{} a_2^2} \plus{} ... \plus{} \frac 1{2n \minus{} 2 \plus{} a_n \plus{} a_n^2} \le \frac 1{2}.\][/quote]

I proved this immediate generalization:

If $ a_1,\ldots,a_n$ are positive real numbers such that $ a_1\ldots a_n\equal{}1$ and $ p,q$ are nonnegative numbers such that $ p\plus{}q\equal{}2$, then \[ \frac 1{2n \minus{} 2 \plus{} pa_1 \plus{} qa_1^2} \plus{} \frac 1{2n \minus{} 2 \plus{} pa_2 \plus{} qa_2^2} \plus{} ... \plus{} \frac 1{2n \minus{} 2 \plus{} pa_n \plus{} qa_n^2} \le \frac 1{2}.\]
Thanks for your inequalties!


analytic geometry
graphing lines
slope
trigonometry
geometry
geometric transformation
reflection

[quote="Omid Hatami"]Let $P$ be a point in a square whose side are mirror. A ray of light comes from $P$ and with slope $\alpha$. We know that this ray of light never arrives to a vertex. We make an infinite sequence of $0,1$. After each contact of light ray with a horizontal side, we put $0$, and after each contact with a vertical side, we put $1$. For each $n\geq 1$, let $B_{n}$ be set of all blocks of length $n$, in this sequence.
a) Prove that $B_{n}$ does not depend on location of $P$.
b) Prove that if $\frac{\alpha}{\pi}$ is irrational, then $|B_{n}|=n+1$.[/quote]

What's a block?!  :maybe:

Every block of length $n$ is a $n$-tuple such as $(a_{k},a_{k+1},\dots,a_{k+m-1})$.

[quote="Omid Hatami"]Let $P$ be a point in a square whose side are mirror. A ray of light comes from $P$ and with slope $\alpha$. We know that this ray of light never arrives to a vertex. We make an infinite sequence of $0,1$. After each contact of light ray with a horizontal side, we put $0$, and after each contact with a vertical side, we put $1$. For each $n\geq 1$, let $B_{n}$ be set of all blocks of length $n$, in this sequence.
a) Prove that $B_{n}$ does not depend on location of $P$.
b) Prove that if $\frac{\alpha}{\pi}$ is irrational, then $|B_{n}|=n+1$.[/quote]
I have proved that : if $\tan\frac{\alpha}{\pi}$ is irrational, then $|B_{n}|=n+1$,So I believe there is a typo.
I am sorry that I am not able to post my solution,becuase my english is too bad ,and this is a combinatorics problem.......... :blush:
@Omid Hatami,Could do please check if there is a typo?And do you have an official solution.Thanks :)

Yes you're right. It is $\tan \frac{\pi}{\alpha}$

I guess that $ \alpha$ (the slope) should be irrational, otherwise just take a line parallel to the diagonal and we have $ |B_{n}|=2$.

As in most reflection problems, instead of reflecting the path we reflect the square. Thus we construct a lattice grid and for each ray with slope $ \alpha$ we associate the sequence of $ 0$ and $ 1$. Clearly any translation with integer coordinates preserves the sequence so we may consider points which are lattice translates of each other equivalent.

a) Assume that $ \alpha$ is irrational. Then the ray $ r$ will pass as close as wanted to a point equivalent to $ Q$ whatever may $ Q$ be. Indeed, if $ P$ has coordinates $ (x,y)$ but $ Q$ has coordinates $ (x',y')$ the ray $ r$ will pass through points of coordinates $ (x+k,y+k\alpha)$. If we take $ k=x'-x+n$ where $ n\in N$ then $ r$ will pass through $ (x'+n, y+(x'-x)\alpha+n\alpha$. As $ \{n\alpha\}$ is dense in $ [0;1]$ this point can be made as close to a point equivalent to $ Q$ as we wish. Now if we have a block of $ n$ digits for $ Q$, if the ray $ r$ passes through a point $ Q'$ close enough to a translate of $ Q$, this block will also appear on ray $ r$, so in the block for $ P$. So the set of blocks is independent of $ P$.

Assume that $ \alpha$ is rational. Then if a ray with slope $ \alpha$ meets a lattice point it meets infinitely many lattice points. Draw through each lattice point a ray with slope $ \alpha$. It is easy to see that $ R^{2}$ is then partitioned into infinitely many discrete strips (the region between two lines parallel to $ \alpha$) that do not contain lattice points in the interior. If $ P,Q$ are in the same strip we can move $ P$ to $ Q$ and in the process the sequence of $ 0$ and $ 1$ will not change (it will change if and only if the ray with slope $ \alpha$ passes through a lattice point, but there are no such points inside a strip). So $ |B_{n}|$ is the same for $ P$ and $ Q$. Finally if $ P$ and $ Q$ are in different strips then there exists a lattice translation that maps $ P$ to a point in the same strip with $ Q$ and lattice translations do not significantly affect the (periodic) sequence of $ 0$ and $ 1$.

b) As we have proven in a), every sequence of $ 0$ and $ 1$ that may EVER appear for some point $ Q$ appears also for every other point $ P$. So we have to prove there are exactly $ n+1$ differenct sequences of length $ n$. Let $ OACB$ be a unit square. Assume that $ O$ is the origin, $ A$ has coordinates $ (0,1)$ and $ B$ has coordinates $ (1,0)$. We may assume we start counting from a point inside it and that the ray $ r$ with slope $ \alpha$ intersects $ [AC]$ or $ [BC]$ for every such point. It is clear we may move the point on the line $ r$ as long as it remains inside the square so we may assume it is on $ [OA]$ or $ [OB]$. Now let's draw the staircase with alternating horizontal and vertical unit steps connecting the points with coordinates $ (x,y)$ where $ x,y \in \mathbb{N_{0}}$ and $ x+y=n+1$. A ray through $ P$ intersects this staircase at a point $ Q$ and the block of $ n$ digits is uniquely defined by the alternation of horizontal and vertical segments on the segment $ (PQ]$. Now let's move $ P$ continuously from $ B$ to $ O$ then to $ A$. The point $ Q$ will also move continuously from a point $ K$ to a point $ L$ and $ KLAB$ is a parallelogram. Next the sequence changes if and only if $ (PQ]$ passes through a lattice point inside $ KLAB$ (or on the boundary, but not in a vertex). When it does, the change that occurs if of the following type: take a $ 0$ followed by a $ 1$ and exchange their places. Only one counter example: when it passes (once) through the lattice point between $ K$ and $ L$ on the staircase the operation will change the last $ 0$ into $ 1$, and this point is outside $ KLAB$. Thus if we have $ l$ lattice points inside $ KLAB$ we will have $ l+1$ changes so $ l+2$ blocks. These blocks will be different because each change increases by $ 1$ the monovariant which is the sum, for all unities, of $ n+1-p$ where $ p$ is the position of that unity in the block. So we have $ |B_{n}|=l+2$. It only remains to count $ l$. $ KLAB$ is sandwiched between the strip defined by the conditions $ 1\leq x+y\leq n+\delta$ where $ 0<\delta<1$. However, of each set of lattice points whose sum of coordinates is $ k$ ($ 1<k\leq n$) exactly one point belongs to $ KLAB$: this is because thes point lie on a line parralel to $ AC$ and $ KL$ at distances $ \sqrt 2$ apart, whereas $ KLAB$ intersects this line in a nontrivial segment of length $ \sqrt 2$. As $ k$ may take $ n-1$, values, $ l=n-1$ so $ |B_{n}|=n+1$.


do there exist pairwise relatively prime integers $a,b,c>1$ such that$a|2^{b}+1\ ,\ b|2^{c}+1\ ,\ c|2^{a}+1$

This is from problem 25 of Russian 2000.


For how many two-digit positive integers is the product of the digits
12?


search
inequalities
probability
inequalities proposed

Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$ prove that :
$3(a^4+b^4+c^4+d^4)+4abcd \geq (a^3+b^3+c^3+d^3)$

First assume a<=b<=c<=d("<=" means no greater than) :D
we all know that (a^4+b^4+c^4+d^4)/4  >=  abcd
which the question turns out to be :
prove that 16abcd >=  a^3 + b^3 + c^3 + d^3
since a+b+c+d = 1 && a is the smallest number in set M {a,b,c,d} so that a <= 1/4
then we just need to prove a^3 + b^3 + c^3 + d^3 <= 4 (a^4 + b^4 + c^4 +d^4)
then that's it

[quote]First assume a<=b<=c<=d("<=" means no greater than)
we all know that (a^4+b^4+c^4+d^4)/4 >= abcd
which the question turns out to be :
prove that 16abcd >= a^3 + b^3 + c^3 + d^3
since a+b+c+d = 1 && a is the smallest number in set M {a,b,c,d} so that a <= 1/4
then we just need to prove a^3 + b^3 + c^3 + d^3 <= 4 (a^4 + b^4 + c^4 +d^4)
then that's it
_________________
-T_I-G_E-R
[/quote]
I think it is not a good idea. Because $16abcd \geq a^3 + b^3 + c^3 + d^3$is not correct!!!

http://www.mathlinks.ro/Forum/viewtopic.php?t=20851

Using theorem from here the problem becomes very easy (I mean not general one).

[quote="Ramtin Madani"][quote="hungkhtn"]The general has been posted before:

$(n-1)(a_1^n+a_2^n+...+a_n^n)+na{}_1a{}_2...a_n \ge (a_1+a_2+...+a_n)(a_1^{n-1}+a_2^{n-1}_...+a_n^{n-1})$

Harazi posted his exellent solution by intruction.[/quote]
Wonderful :lol:
Now please tell me where can I find the solution of this?[/quote]

Search for Suranyi's inequality in this forum and find harazi's proof.

I'll give this a try, even though I will not be solving this using inequality theoroms.  Since we have $a+b+c+d=1$, and since the right hand side is $a^3+b^3+c^3+d^3$, we can say,
$a^3+b^3+c^3+d^3=(a+b+c+d)^2=1$ On the left hand side, we have $3(a^4+b^4+c^4+d^4)+4abcd \geq (a^3+b^3+c^3+d^3)$.  Since a,b,c, and d have a greater power, and are multiplied by 3, then added $4abcd$, the probability is very high that $3(a^4+b^4+c^4+d^4)+4abcd \geq (a^3+b^3+c^3+d^3)$.  Sorry for the partially incorrect proof, however, I'll try to solve this using inequalities rather than pure verbal logic. :)

Masoud Zargar

Oh  :D
If I am not wrong then there are 124 number $a$. ($b,c,d$ is the same amount)

 :rotfl  :rotfl:


Solve the equation :

$ \sin \theta \plus{}\sin 2\theta \equal{}\sin 4\theta \left(0<\theta <\frac{\pi}{2}\right).$

that is:
$ \Leftrightarrow sin(x)=sin(4x)-sin(2x)\\
\Leftrightarrow sin(x)=2cos(3x).sin(x)\\
\Leftrightarrow sin(x)(2cos(3x)-1)=0\\
\Leftrightarrow \[\begin{array}{lll}
sin(x)&=&0\\
cos(3x)&=&\frac{1}{2}.
\end{array}\\
\Leftrightarrow \[ \begin{array}{lll}
x&=&k\pi\\
x&=&\pi/9+k2\pi/3\\
x&=&-\pi/9+k2\pi/3.
\end{array}$
So solution is $ x= \frac{\pi}{9}$
he he. :rotfl:
it is normal.

[quote="minhphuc.v & my EDIT"]
For $ k$ an integer:

$ \Leftrightarrow \[ \begin{array}{lll} x & = & k\pi \\
x & = & \pi/9  + >> k2\pi/3 << \\
x & = & - \pi/9  + >> k2\pi/3. << \end{array}$
[/quote]

The k variable is not being shown multiplied here by the 2 in the lowest
two lines (of the quote box), despite your intentions.

My suggestion is that for each respective portion of each of the lowest
two lines, you show it as one of these to be correct:

$ k(2\pi/3)$ or as $ 2\pi {k/3}$.

[quote="minhphuc.v"]that is:
$ \Leftrightarrow sin(x) = sin(4x) - sin(2x) \\
\Leftrightarrow sin(x) = 2cos(3x).sin(x) \\
\Leftrightarrow sin(x)(2cos(3x) - 1) = 0 \\
\Leftrightarrow \[\begin{array}{lll} sin(x) & = & 0 \\
cos(3x) & = & \frac {1}{2}. \end{array} \\
\Leftrightarrow \[ \begin{array}{lll} x & = & k\pi \\
x & = & \pi/9 + k2\pi/3 \\
x & = & - \pi/9 + k2\pi/3. \end{array}$
So solution is $ x = \frac {\pi}{9}$
he he. :rotfl:
it is normal.[/quote]

What's $ k$? You get 4ponit/10. :wink:

Because the paper isn't logic in mathematicaly.

yes. I sorry.
he he.
this is $ k \;\in\mathbb{Z}$


If $ a,b,x,y \in \mathbb{R}$ and
$ ax \plus{} by \equal{} 3$
$ ax^2 \plus{} by^2 \equal{} 7$
$ ax^3 \plus{} by^3 \equal{} 16$
$ ax^4 \plus{} by^4 \equal{} 42$
Find the value of $ ax^5 \plus{} by^5$.

This problem is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=459558]AIME 1990 #15[/url].

It's a well-known property of sequences of this form.  See [url=http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums]Newton sums[/url] and the general discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490#comments]here[/url].


The results are in.

First place goes to [url=http://www.consolmath.com/news.php]thoobik's webpage[/url].

Second place goes to [url=http://www.mathmatix.com/]joml88's webpage[/url].

Third place goes to Cicatriz's webpage, though I need an updated link to display it here.  Where'd it go?  Please provide us with an updated link.

These webpages looked good and will hopefully serve the math programs at the schools that run them.

Thanks again to Rep123max for judging the webpages and running the contest!

Please send me your name and mailing addresses at crawford@artofproblemsolving.com so that we can send you your gift certificates as soon as possible.


OK, so I have state in less than 2 weeks, and i have no clue how to study!!!!
Can someone give me any suggestions on how to study.
Should I do past contests?
Should I read the aops book volume 1?
Should I play FTW?
Or should I do al of the above?
If I should do all of the above....which one should I do the most?
How would you study?!?!?!?!?1
Oh yeah, and btw I live in Michigan...tough competition....hope I make it to nats.... :winner_first:

Since you have so little time left, I would recommend doing old mc state/national tests. That way you will have a good strategy for the real test. If your state has an official countdown, play on AoPS FTW. You should also start AoPS Vol. 1 and do as many of the problems as you can. Even if you don't get through the whole thing, you could probably do about half of it. (or more, depending on whether you know some of the topics yet or not.)

well, the guy b4 me already said most of it. But also, start posting in forums. Like the Mathcounts Marathon! You learn a whole bunch from your peers. (Ahem! Not me!!) But don't worry about FTW. If it is official, worry about it until the LAST moment because Countdown is not nearly as important as written.

LOL AIME guys

guess who AIME's competition is

ME, NIKE, DAFJBOMB, and a whole bunch of other ppl lol

Nike is gonna own us all  :(

Dont worry too much on CD until you are sure you can actually make it. I was stupid and practiced a lot of CD and didnt even make it. Just do a lot of old problems and you will be fine.


For $ a>0,\ b>0,\ c>0$ with $ a\plus{}b\plus{}c\equal{}1$, prove : $ \frac{bc}{a\plus{}bc}\plus{}\frac{ca}{b\plus{}ca}\plus{}\frac{ab}{c\plus{}ab}\geq \frac{3}{4}$.

[quote="kunny"]For $ a > 0,\ b > 0,\ c > 0$ with $ a \plus{} b \plus{} c \equal{} 1$, prove : $ \frac {bc}{a \plus{} bc} \plus{} \frac {ca}{b \plus{} ca} \plus{} \frac {ab}{c \plus{} ab}\geq \frac {3}{4}$.[/quote]
We have
\[ LHS\equal{}\sum\frac{bc}{a(a\plus{}b\plus{}c)\plus{}bc}\equal{}\sum\frac{bc}{(a\plus{}b)(a\plus{}c)}\equal{}\frac{\sum bc(b\plus{}c)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\]
Hence, it suffices to show that
\[ 4\sum bc(b\plus{}c) \ge 3(a\plus{}b)(b\plus{}c)(c\plus{}a)\]
\[ \Leftrightarrow \sum bc(b\plus{}c) \ge 6abc\]
which is true by the AM-GM inequality. :)

[quote="kunny"]For $ a > 0,\ b > 0,\ c > 0$ with $ a \plus{} b \plus{} c \equal{} 1$, prove : $ \frac {bc}{a \plus{} bc} \plus{} \frac {ca}{b \plus{} ca} \plus{} \frac {ab}{c \plus{} ab}\geq \frac {3}{4}$.[/quote]
By Cauchy-Schwarz inequality, we have:
$ \frac {bc}{a \plus{} bc} \plus{} \frac {ca}{b \plus{} ca} \plus{} \frac {ab}{c \plus{} ab}\ge \frac{(ab\plus{}bc\plus{}ca)^2}{3abc\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2}$
We need to prove that
$ \frac{(ab\plus{}bc\plus{}ca)^2}{3abc\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2} \ge \frac{3}{4}$

$ \Leftrightarrow 4(a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}2abc(a\plus{}b\plus{}c)) \ge 3(3abc\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2)$

$ \Leftrightarrow a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge abc$

Which is clearly true because $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2 \ge abc(a\plus{}b\plus{}c)\equal{}abc$

[quote="kunny"]For $ a > 0,\ b > 0,\ c > 0$ with $ a \plus{} b \plus{} c \equal{} 1$, prove : $ \frac {bc}{a \plus{} bc} \plus{} \frac {ca}{b \plus{} ca} \plus{} \frac {ab}{c \plus{} ab}\geq \frac {3}{4}$.[/quote]

setting : $ a\equal{}\frac{x}{x\plus{}y\plus{}z}$ etc..
the inequality becomes:

\[ \sum{\frac{yz}{x^2\plus{}xy\plus{}yz\plus{}xz}} \geq \frac{3}{4}\]

by cauchy shwarz : \[ LHS((xy\plus{}yz\plus{}xz)^2\plus{}xyz(x\plus{}y\plus{}z)) \geq (xy\plus{}yz\plus{}xz)^2\]

hence it sufficies to prove that:
\[ (xy\plus{}yz\plus{}xz)^2\plus{}xyz(x\plus{}y\plus{}z) \leq \frac{4}{3}(xy\plus{}yz\plus{}xz)^2\]

wich is equivalent to : \[ (xy\plus{}yz\plus{}xz)^2 \geq 3xyz(x\plus{}y\plus{}z)\]
we put $ a\equal{}xy,b\equal{}yz,c\equal{}xz$
so the inequality becomes: \[ (a\plus{}b\plus{}c)^2\geq 3(ab\plus{}ac\plus{}bc)\]
or : \[ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(a\minus{}c)^2 \geq 0\] wich is true!  :)


probability
real analysis
real analysis theorems

Suppose $A_1,A_2, \ldots, B_1,B_2,\ldots$ are random events of a same sample space such that $P(A_n) \rightarrow 1, P(B_n)\rightarrow p$. Show that $P(A_n\cap B_n)\rightarrow p$ for large $n$.

obs.: $P$ is a probability.
---------------------------------------------------------
My idea was trying to show that $A_n$ increases to the sample space, but I'm actually having some difficulties with the mathematical's rigor.

Try focusing on the complement of $A_n.$ After all, it has small measure, hence $B_n\cap (A_n')$ has small measure.

[I probably should have said "probability" instead of "measure." Not that it makes any difference.]


let f be a function of intergers and assumes it satisfies the following properties-1) f(0) is not equal to 0.f(1)=3. f(x)f(y)=f(x+y)+f(x-y).find f(7) andf(8).

Start by making x=1 and y= 0 to get f(0) = 2. Then make:

1) x = y = 4
2) x = y = 2
3) x = y = 1

[hide="And so"]$ f(8) \equal{} [([f(1)]^2\minus{}2)^2 \minus{} 2]^2 \minus{} 2$[/hide]

With the above you cal also get f(2), f(3), and f(4). To find f(7) then make x = 4 and y = 3.

[quote="Carcul"]Start by making x=1 and y= 0 to get f(0) = 2. Then make:

1) x = y = 4
2) x = y = 2
3) x = y = 1

[hide="And so"]$ f(8) \equal{} [([f(1)]^2 \minus{} 2)^2 \minus{} 2]^2 \minus{} 2$[/hide]

With the above you cal also get f(2), f(3), and f(4). To find f(7) then make x = 4 and y = 3.[/quote] butmakingx=1and y=0f(0)=f(1)+f(1)=6

[quote="21 7 15"]butmakingx=1and y=0f(0)=f(1)+f(1)=6[/quote]

Making $ x\equal{}1,y\equal{}0$ gives $ f(0)f(1)\equal{}f(1)\plus{}f(1) \Rightarrow f(0)\equal{}2$


quadratics
inequalities
algebra proposed
algebra

How can you solve for x in:

$ \minus{}3 \leq \frac{|x^2\plus{}1|\minus{}1}{|x^2\plus{}1|\minus{}2} \leq 3$?

[quote="exoeragonius"]How can you solve for x in:

$ \minus{} 3 \leq \frac {|x^2 \plus{} 1| \minus{} 1}{|x^2 \plus{} 1| \minus{} 2} \leq 3$?[/quote]
It is too easy for olympic forum,Notice $ x^2\plus{}1>0$

hello, because $ x^2 \plus{} 1\geq1 \forall x \in \mathbb{R}$ you have to consider
$ \minus{} 3\le\frac {x^2}{x^2 \minus{} 1}\le 3$.
Sonnhard.


ARML MATHCOUNTS rotation search geometry AMC AIME

I thought it might be interesting for those of us with access to this forum to put up what we know of local competitions in our areas. I'm curious about how the opportunities and organizations vary around the country. I'll lead off with Southern California, and try to set a format:

Name: CSULB Math Day at the Beach
Since: 2000
Type: Individual & team event for high schools.
When: A Saturday in March (will be March 19, 2005)
Where: Campus of California State University, Long Beach
Eligibility: 6 students per high school, teams chosen by high school teachers. Limited to ~25 schools. Some other students attending ARML practices may be invited as "at large" competitors.
Structure: All short-answer questions. Starts with individual round, 20-30 questions, some multiple-choice, some free-response. Followed by team round: ~8 free-response short-answer questions for the 6 team members to work together. After lunch and speeches, an oral round (modeled on Mathcounts Countdown) for the top few individuals. Ends with awards ceremony.
Information and contact: [url]http://bosna.natsci.csulb.edu/mathday/[/url], Kent Merryfield
Results are used to help pick the ARML team.

Name: Orange Coast College Math Meet
Since: 1957
Type: Individual & team event for high schools.
When: A Friday afternoon, often in May
Where: Campus of Orange Coast College
Eligibility: 4 students per high school, plus alternates, chosen by high school teachers. For high schools in Orange County.
Structure: Individual multiple choice exam. Ends with awards ceremony.
Information and contact:
[url]http://www.occ.cccd.edu/departments/instruction/mathsci/math/mathmeet.html[/url]

Name: Bay Math League
Since: ?
Type: Individual & team series of event for high schools, large teams.
When: Four Wednesday late afternoons per year.
Where: Rotates among participating high schools.
Eligibility: Each team has 10 students, and must include a mixture of grades. Some schools bring multiple teams. Participating high schools have been from Los Angeles County.
Structure: Rounds tied to curriculum and grade level; topics specified in advance. Each student competes in his own grade level round and in some of the other grade level rounds. Each round is 3 short-answer questions in 10 minutes. Team round at end has questions from each grade level round. Questions selected by participating teachers.
Information and contact: Kent Merryfield could relay messages to the organizing teachers.

Pomona College Math Talent Search
Since: ? (probably 1980's) Related to/inspired by Wisconsin Math Talent Search
Type: Mail-in proof contest, about four times a year.
Where: Mail-in. An awards ceremony is held at Pomona College on a Saturday in the spring.
Eligibility: High school (or younger) students in Southern California
Structure: Mail-in, 4 or 5 proofs per round. Students may compete either as individuals or cooperating groups.
Information and contact: [url]http://www.math.pomona.edu/talentsearch.html[/url]

There are, or were, others. I don't have current information about events at either Cerritos College or Los Angeles Community College but believe those events are ongoing.

The was the statewide contest sponsored by the UCSB College of Creative Studies. I just checked their website, and the only link was to a list of awards which includes the names of students who were HS seniors in 2002-2003. That is, that information is more than a year old.

Upstate NY: (Apologies for not following format here, partly because there would be a lot of blanks).  As far as I know,  there are a handful of regional math leagues -- Southern Tier, DUSO (Dutchess-Ulster-Sullivan-Orange) are the ones I know of, although I know there are others.  I don't know how they work, however.  In my area, there is a community college (I think it's Schenectady County community college) that sponsors a math competition, but I don't know very much details.  I believe each school gets to send two students in each of two divisions (divisions determined by how much math you've taken, instead of grade level) and that it's multiple-choice.

The major New York State math competition is of course NYSML (http://www.nysml.org): New York State Math League.  I believe it's older than ARML, and it's run in a similar format (there are slight differences with scoring and time limits, but the rounds are the same format).  It's held at a different high school in the state every year, and like ARML teams are 15 students, generally from a single high school, math league, or area math club.

Sorry for not following the format, but this is the current list in the Alabama community:

[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 3 -- January 11, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] Log1 Topic Tests -- January 13, 2005
Gainsville -- January 22, 2005
[url=http://www.mandelbrot.org]Mandelbrot[/url] Round Three -- January 24-28, 2005
[url=http://rocketcitymath.org/]Rocket City Math League[/url] Round Two -- January 24-28, 2005
[url=http://www.unl.edu/amc/]AMC[/url]-10/12 Contest A --  February 1, 2005
[url=http://www.natassessment.com/]National Assessment & Testing[/url]: Four-by-four -- February 3, 2005
Auburn High School -- February 5, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 4 -- February 8, 2005
[url=http://www.unl.edu/amc/]AMC[/url]-10/12 Contest B --  February 16, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] Log1 Ciphering -- March 3, 2005
Hoover High School -- March 5, 2005
[url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] I -- March 8, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 5 -- March 8, 2005
[url=http://rocketcitymath.org/]Rocket City Math League[/url] Round Three --March 7-11, 2005
[url=http://www.mandelbrot.org]Mandelbrot[/url] Round Four -- March 14-18, 2005
Alabama State Written - March 19, 2005
[url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]AIME[/url] II -- March 22, 2005
[url=http://www.mathleague.com/]Ala-La-Miss Math League[/url]  Contest 6 -- April 5, 2005
[url=http://www.natassessment.com/]National Assessment & Testing[/url]: Collaborative -- April 13, 2005
Grissom High School -- April 16, 2005
Alabama State Ciphering -- April 23, 2005
[url=http://www.arml.com]ARML[/url] -- June 3-4, 2005
[url=http://mualphatheta.org/]Mu Alpha Theta[/url] National Convention in Hawaii -- August 1-6, 2005

The format for most local competitions in Alabama is 25 multiple choice problems in an hour plus a ciphering event in which students get points for solving problems in faster time intervals.  It's pretty good training for the AMC10 and AMC12 (except that not all AMC12 topics are covered), though it's not great training for the AIME or ARML -- problems that challenge a student to think for more than 2 or 3 minutes at a time and don't have all the clues that multiple choice presents.

Name: Minnesota Junior High Mathematics League
Since: 1986
Type: tournaments with individual and team events for junior high schools.
When: five meets each school year, from September to January
Where: in geographical divisions throughout Minnesota
Eligibility: 6 students per team, teams chosen by teachers. Administrative limit of 15 schools and 185 students per division, but each school may bring as many teams as it likes.
Structure: two individual rounds with five short-answer questions each. Followed by team round: ten short-answer questions for the 6 team members to work together.
Information and contact: http://members.aol.com/mathleague/
Many Minnesota schools use this competition to help select their MATHCOUNTS teams. The founder of this competition actively promotes participation in AMC 8 and MATHCOUNTS.

Not nearly complete...

Name: Montgomery County Middle School Math League
Since: ?
Type: Individual & team event for middle schools. Used to select ARML.
When: On a wednesday, 4 meets throughout the year
Where: Random middle schools
Eligibility: 1 team of 6 middle schoolers, plus 2 secondary teams. Chosen by coach.
Structure: 6 Individual short answers in 24 minutes. 6 team questions in 20 minutes. 2 relay questions. (questions where one person's result feeds into the next person)
Information and contact: ?

Name: Montgomery County Interscholastic Mathematics League.
Type: Used to select ARML
Elegibililty: High schoolers chosen by coach
(No Details)

Name: Howard County Math League
Since: 1978
Type: Individual & team event for high schools. Used to select ARML.
When: On a wednesday, 8 meets throughout the year
Where: Random high schools
Eligibility: 2 teams of high schoolers. Chosen by coach
Structure: 6 Individual short answers in 30 minutes. 5 team questions in 15 minutes.
Information and contact: http://www.howard.k12.md.us/math/hcml/hcml.html

Name: Univ. of MD math competition: Part I
Since: 1979
Type: Individual event for high schoolers.
When: October 10 this year
Where: Local high schools
Eligibility: Anyone not graduated from high school
Structure: 25 AMC10 style questions, 75 minutes
Information and contact: http://www.math.umd.edu/highschool/mathcomp/

Name: Univ. of MD math competition: Part II
Since: 1979
Type: Individual event for high schoolers.
When: December 3 this year
Where: Local high schools OR Univ. MD campus
Eligibility: Top scorers (usually around 230) in Part I
Structure: 5 proof style problems, 2 hours
Information and contact: http://www.math.umd.edu/highschool/mathcomp/

Not-quite complete list for Michigan. Excludes AMC/Mathleague/ARML contests.

Name: Math Field Day
Since: 1968 (38th Annual Competition)
Type: Individual and Team Event for High Schoolers
When: Tuesday, March 1, 2005
Where: University of Michigan-Flint
Eligibility: 3-6 students per high school act as a team. Over 50 teams are generally invited to the tournament. Teams are picked by high school coaches/advisors
Structure: Competition runs around team events. There are individual events and team events. Individual events include the Mad Hatter, a speed-based individual contest, a two-person problem solving contest with more difficult questions, and a presentation event where a small speech is given on a pre-selected topic by a student. Team events include relays, ARML-like power rounds, and a Guess the Function-style game involving the whole team. The competition is won by the best overall team result.
Information and contact: http://www.umflint.edu/departments/math/Field_Day/

Name: Saginaw Valley State University Math Olympics Competition
Since: 2000 (?)
Type: Individual and Team Event for High/Middle Schoolers
When: March 18th, 2005
Where: Saginaw Valley State University
Eligibility: Any high school students.
Structure: Two multiple-choice exams similar to the AMC-10 and AMC-12. The more difficult exam contains questions with higher prerequisites.
Information and contact: Tom Zerger, http://www.svsu.edu/%7Ezerger/Math%20Olympics/Math%20Olympics.htm#Schedule

Name: LTU Annual Mathematics Competition
Since: 2000 (?)
Type: Individual Competition for High Schoolers
When: Sunday, April 25, 2004
Where: Lawrence Tech. University
Eligibility: Any High School students.
Structure: A multiple-choice exam, similar to the AMC-12.
Information and contact: Michael Merscher, http://www3.ltu.edu/ltucomp/

Name: Kettering Math Olympiad
Since: 2001
Type: Individual Competition for High Schoolers
When: Unknown so far  Around early May, 2005
Where: Kettering College
Eligibility: Any High School students.
Structure: A 3-hour, 6-question exam resembling more difficult math Olympiads.
Information and contact: Dr. Ruben Hayrapetyan, http://www.kettering.edu/~acheng/Olympiad/olympiad.html

Name: Michigan Mathematics Prize Competition (MMPC)
Since: 1959
Type: Individual Competition for High/Middle Schoolers
Where: Taken at local high schools.
When: Early October and Early December 2005
Eligibility: Any High School-age or younger students are eligible for Part I. High scorers on Part I are invited to take Part II.
Structure: Part I is an 100 minute, 40 question multiple choice exam. About 1000 of the top scorers on Part I are invited to take Part II. Part II is a proof/written exam with 5 questions and 100 minutes working time. The top 100 are considered winners and are invited to a banquet in March/April to receive prizes.
Information and contact: David Redman, http://www.delta.edu/math/mmpc/

Potential Schedule:

Michigan Math League: Contest 3 -- January 11, 2005
Mandelbrot Round Three -- January 24-28, 2005
AMC-10/12 Contest A -- February 1, 2005
Michigan Math League: Contest 4 -- February 8, 2005
AMC-10/12 Contest B -- February 16, 2005
Math Field Day  March 1, 2005
AIME I -- March 8, 2005
Michigan Math League: Contest 5 -- March 8, 2005
Mandelbrot Round Four -- March 14-18, 2005
Math Olympics Competition  March 18, 2005
AIME II -- March 22, 2005
Michigan Math League: Contest 6 -- April 5, 2005
USAMO  April 19/20, 2005
LTU Annual Competition  April 25, 2005
Kettering Math Olympiad  Early May, 2005
ARML -- June 3-4, 2005

Known Colorado competitions:

[b]Name:  CSU Math Day[/b]
Since:  1977
Size:  About 600 to 700 students
Type:  Individual (written) and Team (oral); with large-school and small-school divisions
When:  a Friday in early November
Where:  Campus of Colorado State University, Fort Collins
Eligibility:  Max 15 students per high school.  32 teams per division.
Structure:  Individual:  Problem-solving and proofs.
Structure:  Team:  College-bowl format.  3 persons per team per round.  True double-elimination.  Early in the elimination rounds, there are many, many simultaneous contests occurring in different rooms.  Most questions are in the style of MATHCOUNTS Countdown questions, but may require higher math.  History of mathematics is also fair game.

[b]Name:  UNC Mathematics Contest[/b]
Since:  About 1992
Size:  About 2000 students
Type:  Individual (written)
When:  First round:  A 3-day window in November
When:  Final round:  A Saturday in late January / early February
Where:  First round:  Your school
Where:  Final round:  Campus of University of Northern Colorado, Greeley
Eligibility:  Colorado students in grades 7-12
Structure:  First round:  Problem solving.  Approximately 350 students are invited to the final round.  Cutoffs vary by grade level.
Structure:  Final round:  Problem solving, with some proofs.

[b]Name:  Colorado Mathematical Olympiad[/b]
Since:  1984
Size:  About 600-1000 students
Type:  Individual written essay
When:  A Friday in April
Where:  Campus of University of Colorado at Colorado Springs
Eligibility:  High school or below
Structure:  Olympiad.  5 problems, 4 hours.  All solutions must be justified.  The "problems require ... a minimal amount of knowledge and a great deal of creativity, originality, and analytical thinking."

[b]Name:  Aims Math and Science[/b]
Since:  About 1986
Size:  About 400 students
Type:  Individual written contests in Algebra I, Algebra II, Calculus, Geometry, and "Senior Math", as well as several science contests.
When:  Late April.  May vary.
Where:  Greeley campus of Aims Community College
Eligibility:  A student is eligible to enter one math contest, corresponding to the course he/she is taking.
Structure:  Multiple choice tests, lasting about 1 hour, followed by a science "open house", with several demonstrations.

[b]Schedule of known competitions:[/b]

Nov 04, 2004 -- [url=http://www.math.colostate.edu/~franklin/mathday/]CSU Math Day[/url]
Nov 05-07, 2004 -- [url=http://www.cs.unco.edu/Other/Mathcontest.html]UNC Mathematics Contest[/url] First Round
Nov 16, 2004 -- [url=http://www.unl.edu/amc/]AMC 8 Contest[/url]

Jan 29, 2005 -- [url=http://www.cs.unco.edu/Other/Mathcontest.html]UNC Mathematics Contest[/url] Final Round (by invitation)

Feb 01, 2005 -- [url=http://www.unl.edu/amc/]AMC 10/12 Contest A[/url]
Feb 16, 2005 -- [url=http://www.unl.edu/amc/]AMC 10/12 Contest B[/url]

Mar 08, 2005 -- [url=http://www.unl.edu/amc/]AIME I Contest[/url] (by invitation)
Mar 19, 2005 -- State MATHCOUNTS Competition
Mar 22, 2005 -- [url=http://www.unl.edu/amc/]AIME II Contest[/url] (by invitation)

Apr 19, 2005 -- [url=http://www.unl.edu/amc/]USAMO Day 1[/url] (by invitation)
Apr 20, 2005 -- [url=http://www.unl.edu/amc/]USAMO Day 2[/url] (by invitation)
Apr 22, 2005 -- [url=http://www.uccs.edu/~asoifer/olympiad.html]Colorado Mathematical Olympiad[/url] competition
Apr 29, 2005 -- [url=http://www.uccs.edu/~asoifer/olympiad.html]Colorado Mathematical Olympiad[/url] awards ceremony

May 05-08, 2005 -- [url=http://www.mathcounts.org]National MATHCOUNTS competition[/url] (by invitation)


geometry
circumcircle
search
geometric transformation
homothety
ratio
incenter

The incircle (I) of triangle ABC touches BC,CA,AB at A',B'C'. Prove that the circumcenters of triangles AIA',BIB',CIC' are collinear.

nice problem   :)
let $ C''$ the intersection of $ AB$ and the circumcircle of $ CIC'$ , his center is the midle of $ [IC'']$ ,and we define with the same way the points $ B''$ and $ A''$ .
to conclude it suffice to prove that $ A'', B''$ and $ C''$ are collinear .
from the fact that $ CC''$ is an exterior bissector of $ <BCA$ we have $ \frac{C''A}{C''B}\equal{}\frac{CA}{CB}.$
 and with the same way for $ \frac{B''A}{B''C}$ and $ \frac{A''B}{A''C}$ and by menalus we get the desired result.

If I am not mistaken,this problem was posted at least twice before.
I remember that [b]pohoatza[/b] proposed outstanding solution with using inversion(exactly my solution,but I was the one who posted the problem) and several other solutions were posted as well,one of them contains using of [b]Menelaus[/b] theorem.
Hmm...It looks like I was wrong :) ,but nevertheless,the problem mentioned above was generalization of this problem,if I am not mistaken,the goal was to prove that these three circles intersect at most in two points,what is obviously stronger result than the one posted above by [b]mr.danh[/b].
[b]Proof:[/b]
Invert with center in $ I$.The rest follows from the fact that three medians in a triangle are concurrent.
Edited(after 15 minutes)
That's a miracle: I've just casually found this problem,here is the link:
http://www.mathlinks.ro/viewtopic.php?search_id=1663197197&t=181827

This problem is a problem of Singaporean TST 1998 and the author's solution is the proof with [b]using inversion[/b].  :)

Dear Mathlinkers,
the circumcircles of triangles AIA',BIB',CIC' are concurrent at the Schröder's point of ABC (See a nice paper of Darij Grinberg on his website) ; it result that the circumcenters are collinear.
Note that this point of concurrence is on the OI line of ABC.
Sincerely
Jean-Louis

The external angle bisectors of triangle ABC cut the opposite sides at 3 collinear points. A homothety center I and ratio ½ brings A, B, and C to M, N and P, the midpoints of AI, BI and CI respectively.
The center of excircle of triangle AIA’ is the intersection of the perpendicular bisector of AI with MN and the centers of the other 2 triangles are obtained similarly. Due to above mentioned homothety, these three points are collinear.

Best regards,
sunken rock

Inversion WRT incircle $(I)$ takes $A'$ into itself and $A$ into the midpoint $ M$ of $\overline{ B'C'}.$ Hence, circles $ \odot (AIA'),$ $\odot (BIB')$ and $\odot(CIC')$ are taken into the medians of  the intouch triangle $ \triangle A'B'C'.$ Thereby, the said circles pass through the incenter $I$ and  the inverse of the centroid $ G'$ of $ \triangle A'B'C'$ under inversion WRT $(I),$ i.e.  they are coaxal.

[quote][color=darkred][b][u]An easy extension[/u].[/b] Let $ P$ be a point from the plane of the triangle $ ABC$ . Denote

the projections $ X$ , $ Y$ , $ Z$ of the point $ P$ to the sidelines $ BC$ , $ CA$ , $ AB$ respectively.

Prove that the circumcenters of the triangles $ APX$ , $ BPY$ , $ CPZ$ are collinearly.[/color][/quote][hide="An indication."][u][b]BOBILLER[/b] theorem[/u] (without inversion).[/hide]


geometry
3D geometry
LaTeX
limit
number theory proposed
number theory

Let a be a positive integer. Prove that if 4*((a^n) + 1) is a cube for all natural number n then a=1.

$ a>1$
$ a\plus{}1\equal{}2x^3$
$ a^4\plus{}1\equal{}2y^3$
$ 8x^{12}\minus{}16x^9\plus{}12x^6\minus{}4x^3\plus{}1\equal{}y^3$
$ x>1$
now we manually found counter examples for $ x\leq6$ and
for x>6 we
have
$ (2x^4\minus{}4\frac{x}{3})^3<y^3<(2x^4\minus{}4\frac{x}{3}\plus{}1)^3$
now we only check if it is possible that $ y\equal{}2x^4\minus{}4\frac{x}{3}$ its supposed that this fracture represents the least integer greater or equal to it (I don't know the latex command)
and that is pretty easy so we are done.
(If I have made some mistake in calculations please correct me but this idea surely solves the task)

[u]Hint:[/u] denote $ x_k \equal{} 2(a^k \plus{} 1).$
$ \mathop {\lim }\limits_{k \to \plus{} \infty } (x_{k \plus{} 3} \minus{} ax_k ) \equal{} 0$


Let $P_0(x) = x^3 + 313x^2 - 77x - 8$.  For integers $n \ge 1$, define $P_n(x) = P_{n - 1}(x - n)$.  What is the coefficient of $x$ in $P_{20}(x)$?

Notice that $P_n(x)=P_0(x-\triangle{n})$ when $\triangle{n}$ denotes the $n$th triangular number.


The extremities of a rope with length $L$ and mass $m$ are fixed at two points with the same height, and the distance between the points is $l\,(l<L)$. What and how are the the forces acting on each point on the rope? Besides, what is the equation of the curve formed by the rope?

Moreover, how about the condition that the two points have different heights? (Suppose the difference between their heights is $d$)

Um... I posted this [b]very[/b] recently...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131636&sid=061b2988f0678bf09ae6bb22d11b46b6[/url]

[quote="Djole"]Um... I posted this [b]very[/b] recently...[/quote]
I don't know because I do not often browse the physics forum... :blush:

Just look three or four places below your post :[b] shape of  hanging rope[/b]


symmetry
geometry
geometric transformation
reflection

Five people are sitting at a round table. Let $ f \geq 0$ be the number of people sitting next to at least one female and $ m \geq 0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $ (f,m)$ is

$ \textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11$

Casework. We use symmetry and just double our final answer. WLOG more males than females.

5 of one, 0 females: Only possibility is (0,5).

4 male, 1 female: All 5 are next to male, 2 are also next to female so (2,5).

3 male, 2 female: Let a female be on top. Then, we sort the others 4 ways (going clockwise): FMMM, MFMM, MMFM, MMMF. Note that the last 2 are just reflections across the y-axis of the table so we only need to consider the first 2.

FMMM: 4 people are next to females, 5 to males. (4,5)

MFMM: 3 next to female, 4 next to male. (3,4)

So there are 4 when there are more males, and then 4 when there are more females, so 8 total, i.e. B


group theory
abstract algebra
superior algebra
superior algebra unsolved

Is there a group $ G$ with normal subgroup $ N$ such that there is a $ g \in G$ with $ gNg^{\minus{}1} \subsetneq N$?

No. Suppose there is such a $ g$, then replacing $ g$ by $ g^{ \minus{} 1}$ we see that the conjugate is equal to the subgroup.


this inequality is very easy???????????why you don't post?/////////


probability
combinatorics unsolved
combinatorics

its for the informatic olympiad but its pure combinatorics

let $\ A_i $ and $\ B_i $ be sets where  $ 1 \leq x \leq n $ with the following conditions:

1_Each set containds exactly r elements

2_$ A_i \cap B_j =\oslash $

if and only if $ i=j $

Prove that $n \leq \left(\begin{array}{cc}2r\\r\end{array}\right) $

i think this problem has two part !!
this is the second part ,and i think this part will be solve simply by using double counting and using first part ;)

this can be solved by the so-called probabilistic method: let $X=\bigcup (A_i\cup B_i)$. take a random linear arrangement of the elements of $X$. let $N$ denote the number of indices $i$ for which all the elements of $A_i$ precede the elements of $B_i$.clearly $N=\sum N_i$ where $N_i=1$ if all the elements of $A_i$ preced all elements of $B_i$ and $0$ otherwise.then $E(N)=\sum E(N_i)$; also $E(N_i)=\frac{1}{\binom{2r}{r}}$; but also by the given condition if $N_i=1,N_j=1,i<j$ then we have that all elements of $A_i$ precede all elements of $B_i$ and also all elements of $A_j$ precede all elements of $B_j$. but this cannot simultaneously happen since both $A_i\cap B_j\neq\emptyset, A_j\cap B_i\neq \emptyset$. hence with probability one $N\leq 1$.
so,$\sum_{i} \frac{1}{\binom{2r}{r}} \leq 1$ and that gives what you want.

actually this is well known as Bollobas's inequality.


Here is a strange problem for me. It says exactly:
"A man visit a city every 5 months. Is it posible for him to visit the city on March?"
So now, "A man visit a city every 5 months" means for example on January and on May, or on January and on June?

January and June. And he definately can if he starts on March.
And if he starts on January, he still can.
Let January be month 1, february be month 2, etc.

We are looking for an integer n such that 1 + 5n = 3 (mod 12)
5n = 2 (mod 12)
5n - 2 = 0 (mod 12)

So we are looking for a multiple of 12 that is 2 less than a multiple of 5; that is, it must end in 3 (impossible) or 8. 48 is one such possiblity. So he can in fact come in March.

While you're at it, notice that every number can be written as 5n, 5n-1, 5n-2, 5n-3, or 5n-4.

5n = 0 (mod 12) : possible solution: 5n = 0, n = 0
5n - 1 = 0 (mod 12) : 5n - 1 = 24, n = 5
5n - 2 = 0 (mod 12) : 5n - 2 = 48, n = 10
5n - 3 = 0 (mod 12) : 5n - 3 = 12, n = 3
5n - 4 = 0 (mod 12) : 5n - 4 = 36, n = 8

Thus, starting from January, you can get to any month, which means that starting from any month, you can also get to any other month.


Find the integers a and b, given that
 $a^{4}+(a+b)^{4}+b^{4}$
 is a square number.

We must have $a=b=0$.  To prove this, suppose that w.l.o.g. $|a|>0$ and consider the equation above modulo $4$.  Since all squares are only $0$ or $1$ mod 4, after checking all cases we can deduce that $2|a$ and $2|b$.

But if we let $a=2a_{1}$ and $b=2b_{1}$, then we can substitute to obtain $4^{2}(a_{1}^{4}+(a_{1}+b_{1})^{4}+b_{1}^{4})$ must be a perfect square, meaning $a_{1}^{4}+(a_{1}+b_{1})^{4}+b_{1}^{4}$ is a perfect square.  But from this we obtain an infinite progression of positive integers such that

$|a|>|a_{1}|>|a_{2}|>\cdots$

which is obviously a contradiction.  Hence $|a|=|b|=0$, meaning $a=b=0$ as desired.

The same proof works for the stronger result $a^{2}+(a+b)^{2}+b^{2}$ I believe.


Can anyone tell me how to print greek letters like
\varepsilon,\eta etc in non-italic fonts. I know that default for printing these is italic. But is there a way to get non-italic characters.

viswanath

The problem is that the upright version of those symbols aren't available in the standard LaTeX fonts so using \mathrm won't help. However, give the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=upgreek]upgreek package[/url] a try and see if that will give you what you want. If you are using MiKTeX then use the Package Manager to install the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=was]was[/url] package, which includes upgreek.


[quote="SAPOSTO"]If $ x \plus{} y \plus{} z\le 1$ and $ x,y,z\ge 0$ than prove that
$ x^{2}y \plus{} y^{2}z \plus{} z^{2}x \plus{} xyz\le\frac {4}{27}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960

[quote="SAPOSTO"]If $ x \plus{} y \plus{} z\le 1$ and $ x,y,z\ge 0$ than prove that
$ x^{2}y \plus{} y^{2}z \plus{} z^{2}x \plus{} xyz\le\frac {4}{27}$.[/quote]

We have:
$ x^{2}y \plus{} y^{2}z \plus{} z^{2}x \plus{} xyz\le\frac {4}{27}(x\plus{}y\plus{}z)^3$


arithmetic sequence
number theory
algebra unsolved
algebra

[quote="ayush_2008"]three numbers are in ARITHMETIC PROGRESSION such that sum of the squares of the number is 1994. how many such triplets are formed.[/quote]

The three numbers are $ x-y,x,x+y$ and so the equation is $ 3x^2+2y^2=1994$

So we obviously have infinitely many such triples if numbers $ \in\mathbb R$ (default value without any precision, since we are in algebra forum and not number theory one)

If numbers are magically supposed to be integers, then $ x=2z$ and $ 6z^2+y^2=997$ and $ |z|\le\sqrt{\frac{997}6}$ and so $ |z|\le 12$, so $ 13$ tests to do and no solution.

Please, try to copy the full problem statement, without forgetting any word, even if you think the missing word is useless .... :(


[b][size=150]Selected Problems from Open Contests[/size][/b]

[b][size=150]Juniors problems[/size][/b]

[b]Problem 1.[/b] http://www.mathlinks.ro/viewtopic.php?t=217731
[b]Problem 2.[/b] http://www.mathlinks.ro/viewtopic.php?t=217733
[b]Problem 3.[/b] http://www.mathlinks.ro/viewtopic.php?t=217734
[b]Problem 4.[/b] http://www.mathlinks.ro/viewtopic.php?t=217735
[b]Problem 5.[/b] http://www.mathlinks.ro/viewtopic.php?t=217736
[b]Problem 6.[/b] http://www.mathlinks.ro/viewtopic.php?t=217737
[b]Problem 7.[/b] http://www.mathlinks.ro/viewtopic.php?t=217738
[b]Problem 8.[/b] http://www.mathlinks.ro/viewtopic.php?t=217739
[b]Problem 9.[/b] http://www.mathlinks.ro/viewtopic.php?t=217740

[b][size=150]Seniors problems[/size][/b]

[b]Problem 1.[/b] http://www.mathlinks.ro/viewtopic.php?t=217741
[b]Problem 2.[/b] http://www.mathlinks.ro/viewtopic.php?t=217742
[b]Problem 3.[/b] http://www.mathlinks.ro/viewtopic.php?t=217744
[b]Problem 4.[/b] http://www.mathlinks.ro/viewtopic.php?t=217745
[b]Problem 5.[/b] http://www.mathlinks.ro/viewtopic.php?t=217746
[b]Problem 6.[/b] http://www.mathlinks.ro/viewtopic.php?t=217748
[b]Problem 7.[/b] http://www.mathlinks.ro/viewtopic.php?t=217750
[b]Problem 8.[/b] http://www.mathlinks.ro/viewtopic.php?t=217751
[b]Problem 9.[/b] http://www.mathlinks.ro/viewtopic.php?t=217753
[b]Problem 10.[/b]  http://www.mathlinks.ro/viewtopic.php?t=217754

[size=150][b]Selected Problems from the Final Round of National
Olympiad[/b][/size]

Grade 9.

Pro 1. http://www.mathlinks.ro/viewtopic.php?t=217755
Pro 2. http://www.mathlinks.ro/viewtopic.php?t=217756
Pro 3. http://www.mathlinks.ro/viewtopic.php?t=217758
Pro 4. http://www.mathlinks.ro/viewtopic.php?t=217759

Grade 10

Pro 1. http://www.mathlinks.ro/viewtopic.php?t=217760

Grade 11.

Pro 1. http://www.mathlinks.ro/viewtopic.php?t=217761
Pro 2. http://www.mathlinks.ro/viewtopic.php?t=217762
Pro 3. http://www.mathlinks.ro/viewtopic.php?t=217763
Pro 4. http://www.mathlinks.ro/viewtopic.php?t=217764
Pro 5. http://www.mathlinks.ro/viewtopic.php?t=217766

Grade 12.

Pro 1. http://www.mathlinks.ro/viewtopic.php?t=217767
Pro 2. http://www.mathlinks.ro/viewtopic.php?t=217768
Pro 3. http://www.mathlinks.ro/viewtopic.php?t=217769
Pro 4. http://www.mathlinks.ro/viewtopic.php?t=217770
Pro 5. http://www.mathlinks.ro/viewtopic.php?t=217772

the above posts collections have been created, here they are
[url=https://artofproblemsolving.com/community/c1090795_2006_estonia_math_open_junior_contests]2006 Estonia Math Open Junior Contests[/url],
  [url=https://artofproblemsolving.com/community/c1090800_2006_estonia_math_open_senior_contests]2006 Estonia Math Open Junior Contests[/url],
 [url=https://artofproblemsolving.com/community/c1090801_2006_estonia_national_olympiad]2006 Estonia National Olympiad[/url]

PS. I have asked to create the corresponding folder for these problems in Aops Contests Collections [url=https://artofproblemsolving.com/community/c40244h1064480p14240613]here[/url]


calculus
integration
function
calculus computations

Hello
If the function f:[0,8]-> R is differentiable then for 0<a.b<2,
find $ \int_0^{8}f(t)dt$
Thank u.

Do you mean differentiable? And what are a and b?

hello
Ya it is differentiable.Sorry for the mistake.
Well in the question it was written nothing about a and b.
Let c what comes if we consider a and b to Real nos.
Thank u.

I think what he meant was, what do $ a$ and $ b$ contribute to the problem?

We have to find the definite integral of some differentiable function. All we know about it is that it's differentiable. Without more information we can't really go much further.

Hello
Well I was clueless when I saw this question as I don't know how to proceed for this question from this much of information only.It was given that the product of a and b lies b/w 0 and 2.
I got the options for this particular question,may be this will help u::
1.$ 3[a^3f(a^2)\plus{}b^2f(b^2)]$

2.$ 3[a^3f(a)\plus{}b^3f(b)]$

3.$ 3[a^2f(a^3)\plus{} b^2f(b^3)]$

4.$ 3[a^2f(a^2)\plus{}b^2f(b^2)]$

Thank u.


modular arithmetic
number theory proposed
number theory

Prove that if $p\equiv 1\pmod{6}$ and it is a prime then product of the $\frac{p-1}{3}$ cubic residues modulo $p$ is congruent to $-1$ modulo $p.$

It's quite obvious and you don't need that $p \equiv 1 \mod{3}$ (since in other case it is the Wilson Theorem). Just note that if $a$ is cubic residue then $\frac{1}{a}$ is a cubic residue as well. Since $a \equiv \frac{1}{a}$ only for $a \equiv 1$ or $a \equiv-1$ we are done.


algebra
polynomial
induction
calculus
derivative
Galois Theory
number theory

Suppose $\displaystyle P(x)= \prod _{i,gcd(i,n)=1} (x-\omega ^i)$
Prove $P \in  \mathbb{Z}[x]$ and $P$ is irreducible in $\mathbb{Z}[x]$

What is $\omega$?

And to indicate the product, it's \prod: $\prod$, not \pi: $\pi$.

I think he wants people to show that the cyclotomic polynomials have integer coefficients and are irreducible.

its not hard to show that they have integer coefficients (use induction + long division)

but this problem might be moved to superior algebra, because i believe the proofs of cyclotomic polynomials being irreducible use Galois Theory (or at least some ideas of ring theory).

are you sure this is advanced rather than college playground? :?

Well I still want to know why this polyniomial is irreducible.

My teacher gave me a subject from ENS exam where, among other things such as a proof of Dirichlet theorem on arithmetic progressions there was also a proof for this one, an indirect one. The ideas are the folloswing.
  1) Consider the minimal unitary polynomial with rational coefficients of the primitive root of unity,call it epsilon and this polynomial P.
  2) Prove that this polynomial has in fact integer coefficients
  3) Using Dirichlet theorem and the fact that $ Q^p(X)-Q(X^p) $ has all coefficients divisible by p for any polynomial with integer coefficients and any ^prime p, deduce that $ epsilon^k$ is a root of P for all k relatively prime with n.
 4) Deduce an estimation for the degree of P.
 5) Conclude.
  I just adore the way french prove difficult results.

a prove of irreducibility:

Let be $\Phi _n (x)$ be the $n$-th cyclotomic polynomial (the polynomial that has exactly all primitive roots of unity as roots).
As said by zscool you can show that it has integer coefficients by induction.

Lemma: If $\Phi_n(x) = f(x)g(x)$ is a factorisation of the $n$-th cyclotomic polynomial (over $\mathbb{Q}$) and $p$ a prime that doesn't divide $n$, then $f(\zeta)=0$ implies $f( \zeta ^p)=0$.

Proof:
Assume that $\zeta ^p$ is not a root of $f$, so it's a root of $g$.
So consider the polynomials modulo $p$ (this is possible since all coefficients are integers) to get (by writing out the polynomial in term's of binomial coefficients using the binomial theorem) that $f(\zeta ^p) \equiv f(\zeta)^p \equiv 0 \mod p$ (think of $\zeta \mod p$ as some number 'added' to the 'normal' residue classes $\mod p$). But since also $g(\zeta^p)\equiv 0 \mod p$, there is some root occuring twice $\mod p$. But this is impossible, since the derivative of $x^n-1 \mod p$ is $nx^{n-1}$ and so different from zero for $z\not\equiv 0 \mod p$ (it's the same like in the real numbers: you only can have a double root if the polynomial and it's derivative are both zero, the proof for that isn't that hard). This contradiction implies that $f(\zeta ^p)=0$.

To prove the theorem now, simple take some representing integer $k$ of a prime residue class $\mod n$ and factor it into primes (not dividing $n$). Now use the lemma inductively on the factors of $k$ to prove that with $\zeta$ being a root of $f$ also $\zeta^k$ is necessary a root and by this [u]all[/u] primitive $n$-th roots of unity are roots of $f$. But this shows that $f=\Phi_n$, so the cyclotomic polynomial is irreducible.

Something similar and very nice:

Find all pairs $(m,n) \in \mathbb{Q}^2$ so that $\Phi_n$ is irreducible in $\mathbb{Q}[\zeta_m]$ ($=\mathbb{Q}[x]/\Phi_m$)
 :)

(here $\Phi_n$ is the $n$-th cyclotomic polynomial and $\zeta_m$ is a $m$-th primitive root of unity)

The degree of an irreducible factor of $\Phi_{n}$ over $\mathbb{Q}(\zeta_{m})$ is

$[\mathbb{Q}(\zeta_{m})(\zeta_{n}) : \mathbb{Q}(\zeta_{m})] = [\mathbb{Q}(\zeta_{kgV(m,n)}) : \mathbb{Q}(\zeta_{m})] = \frac{[\mathbb{Q}(\zeta_{kgV(m,n)}) : \mathbb{Q}]}{[\mathbb{Q}(\zeta_{m}) : \mathbb{Q}]}= \frac{\phi(kgV(m,n))}{\phi(m)}= \frac{\phi(n)}{gcd(m,n)}$

We conclude, that $\Phi_{n}$ is irreducible over $\mathbb{Q}(\zeta_{m})$ if and only if $\phi(gcd(m,n)) = 1$, i.e. $gcd(m,n) \in \{1,2\}$.


If $ x\equal{}6\plus{}4\plus{}3\plus{}1\plus{}\frac x3$, what is the value of $ x$?

How do you get from $ x\equal{}6\plus{}4\plus{}3\plus{}1\plus{}3$ to $ \frac{2}{3}x\equal{}14$?

Ernie wrote it wrong, the first two steps should have been:

$ x\equal{}6\plus{}4\plus{}3\plus{}1\plus{}\frac{x}{3}$.

$ x\equal{}14\plus{}\frac{x}{3}$.

first,

you would add and then subtract then divide again,

x = 14+(x/3)

(2x/3) = 14

x=21

how do you guys do the bold-looking thing?


Consider two triangles $ACB$ and $ADE$ such that $\angle DEA$ and $\angle ACB$ are right angles and $\angle DAE = \angle CAB$ (where the angles are directed), and such that $E$ and $C$ are not the same point. The line $l$ passes through the point $D$ and is perpendicular to the line $EC$. Let $L$ be the point of intersection of the lines $l$ and $AC$. Prove that the points $L$, $E$, $C$, $B$ are concyclic.

It is sufficient to show that $\measuredangle{BEL} = 90^0$.

 Let $U, V, W$ be the midpoints of $BA, AD, DB$. Then triangles $CUW$ and $WVE$ are congruent. Hence $CW = WE$. Thus $CB^2 + CD^2 = EB^2+ED^2$. According to Carnot's theorem, we obtain that the line $CA, EL$ and the line through $D$ perpedicular to $CE$ are concurrent. This implies that $\measuredangle{BEL} = 90^0$ and we are done.

Carnot's Theorem? http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Carnot&t=45859

How is it used here, I cannot see.

Consider two triangles $CBE$ and $EDC$ that has the relation $CD^2 - CC^2 + BC^2 - BE^2 + EE^2 - ED^2 = 0$. Hence they are orthologic.


Χαίρετε Χριστός Ανέστη .Παρακαλώ τους διαγωνιζομένους να βάλλουν όλα τα παραπάνω που αναφέρονται στον τίτλο και να κάνουμε μια ευχάριστη ορθή και μαθηματική συζήτηση .Περιμένουμε όλοι     :D  :D  :lol:

Silouan, do you have subjects given at Balkan Olympiad? :)

ΑΛΗΘΩΣ ΑΝΕΣΤΗ ΑΓΑΠΗΤΕ ΜΟΛΙΣ ΤΑ ΧΟΥΜΕ ΤΑ ΒΑΖΟΥΜΕ ΤΑ ΘΕΜΑΤΑ ΤΩΡΑ ΤΙΣ ΛΥΣΕΙΣ ΑΠΟ 2 ΜΕΡΕΣ ΚΑΙ ΜΕΤΑ ΓΙΑ ΕΠΙΣΗΜΕΣ ΑΝ ΕΙΝΑΙ ΓΙΑ ΑΝΕΠΙΣΗΜΕΣ  :rotfl:  :rotfl:
ΧΑΙΡΕΤΩ ΜΕ ΑΓΩΝΙΣΤΙΚΟΥΣ ΧΑΙΡΕΤΙΣΜΟΥΣ ΚΑΙ ΜΕ ΠΟΛΥ ΘΕΡΜΕΣ ΕΥΧΕΣ ΓΙΑ ΤΗΝ ΕΛΛΑΔΑ ΚΑΙ ΤΗΝ ΚΥΠΡΟ ΜΑΣ

Μπορεί κάποιο από τα παιδιά να γράψει την άσκηση γεωμετρίας της ΒΜΟ 2006 ??

Γιατί στο αγγλικό φόρουμ πολύ αργούν..... :D

Άκουσα ότι η ομάδα πήγε καλά(για Ελλάδα μιλάμε, που ότι κάνουν οι μαθητές , οφείλεται μόνο στη δική τους μελέτη και στο ταλέντο τους). Βέβαια , η γενική κατάταξη όπως έγραψε ο start ,δεν είναι καλή ή τουλάχιστον όπως θα έπρεπε να είναι(το ξαναλέω : δε φταίνε οι μαθητές για αυτό ,  αλλά η μαθηματική κοινότητα - όχι μόνο η ΕΜΕ -  που κοιμάται κανονικά και δε νοιάζεται για τίποτα άλλο παρά μόνο για τα θέματα εξετάσεων στη Γ΄Λυκείου και για τα ιδιαίτερα μαθήματα). Ωστόσο , όλοι πρέπει να το ξέρετε πια, ο αγώνας θα συνεχιστεί σε ατομικό πια  επίπεδο  και όσοι μπορούν να βοηθήσουν - ο καθένας από το πόστο του - ας το κάνει.
  Πρέπει σιγά σιγά να φτιάξουμε γερή ολυμπιακή ομάδα και να ανεβάσουμε τη μαθηματική κουλτούρα στην Ελλάδα , όχι μόνο για τα μετάλλια , αλλά κυρίως γιατί χωρίς μαθηματικά δεν έχουμε κανένα μέλλον στην οικονομία και την ανάπτυξη.

   Συγχαίρω όλα τα παιδιά της ολυμπιακής ομάδας. Καλή συνέχεια !

   Μπάμπης

  [i]( Με το φίλο μου Νίκο Σκουμπρή χαρήκαμε για τα μετάλλια και είπαμε - για να δυναμώσουμε την ομάδα -  να γράψουμε ένα βιβλίο στη θεωρία αριθμών που να είναι το Ευαγγέλιο κάθε Έλληνα διαγωνιζόμενου σε ολυμπιάδα.Θα βάλουμε θεωρία , πολύ καλά παραδείγματα  και  θέματα διαγωνισμών , από Junior  έως ΙΜΟ και ακόμα παραπέρα. Ας ελπίσουμε ότι θα βρούμε τη δύναμη να ολοκληρώσουμε αυτό το έργο. Η βοήθειά σας - με όποιαδήποτε μορφή- θα είναι πολύτιμή).[/i]

ΧΑΙΡΕΤΩ ΟΛΟΥΣ ΤΟΥΣ ΕΛΛΗΝΕΣ ΚΑΙ ΞΕΝΟΥΣ ΜΑΘΛΙΝΚΕΡΣ ΚΑΙ
ΓΙΑ ΤΟΝ ΣΙΛΟΥΑΝΟ
ΠΡΟΒΛΗΜΑ 2 (ΓΕΩΜΕΤΡΙΑΣ)
ΕΣΤΩ ΑΒΓ ΤΡΙΓΩΝΟ ΚΑΙ Μ ΜΙΑ ΕΥΘΕΙΑ ΠΟΥ ΤΕΜΝΕΙ ΕΣΩΤΕΡΙΚΑ ΤΙΣ ΑΒ ΚΑΙ ΑΓ ΣΤΑ Δ ΚΑΙ Ζ ΑΝΤΙΣΤΟΙΧΑ ΚΑΙ ΤΗΝ ΠΡΟΕΚΤΑΣΗ ΤΗΣ ΒΓ ΣΤΟ Ε ΩΣΤΕ ΤΟ Γ ΑΝΑΜΕΣΑ ΣΤΟ Β ΚΑΙ ΣΤΟ Ε ΟΙ ΠΑΡΑΛΛΗΛΕΣ ΑΠΟ ΤΗΥΝ Μ ΑΠΟ ΤΑ Α,Β,Γ ΤΕΜΝΟΥΝ ΤΟΝ ΠΕΡΙΓΓΕΓΡΑΜΜΕΝΟ ΚΥΚΛΟ ΤΟΥ ΑΒΓ ΓΑΙ ΔΕΥΤΕΡΗ ΦΟΡΑ ΣΤΑ ΣΗΜΕΙΑ Α1,Β1,Γ1 ΑΝΤΟΙΣΤΙΧΑ ΝΑ ΑΠΟΔΕΙΞΕΤΕ ΟΤΙ ΟΙ ΕΥΘΕΙΕΣ Α1Ε Β1Ζ ΚΑΙ Γ1Δ ΣΥΝΤΡΕΧΟΥΝ
ΓΙΑ ΤΟΥΣ ΥΠΟΛΟΙΠΟΥΣ ΓΙΑ ΝΑ ΕΛΕΓΞΕΤΕ ΤΗΝ ΟΡΘΟΤΗΤΑ ΤΩΝ ΛΥΣΕΩΝ ΜΟΥ ΔΙΟΤΙ ΒΙΑΖΟΜΑΙ ΚΑΙ ΕΠΕΙΔΗ ΘΑ ΗΘΕΛΑ ΤΙΣ ΛΥΣΕΙΣ ΚΑΙ ΤΩΝ ΤΕΣΣΑΡΩΝ ΕΝΩ ΔΗΜΟΣΙΕΥΩ ΤΙΣ ΛΥΣΕΙΣ ΜΟΝΟ ΤΩΝ ΤΡΙΩΝ
ΠΡΟΒΛΗΜΑ 1
ΑΝ Α,Β,Γ>0 ΝΔΟ
1/Α(1+Β)+1/Β(1+Γ)+1/Γ(1+Α)>=3/1+ΑΒΓ
ΛΥΣΗ
ΛΟΙΠΟΝ ΑΠΟ ANDREESCU  ΣΗΜΕΙΩΣΗ ΓΙΑ ΓΡΗΓΟΡΗ ΚΑΤΑΝΟΗΣΗ  Α+Β+Γ+ΑΒ+ΒΓ+ΓΑ=Α(1+Β)+Β(1+Γ)+Γ(1+Α)
Α ΜΕΛΟΣ>= 9/Α+Β+Γ+ΑΒ+ΒΓ+ΓΑ (ΑΡΚΕΙ ΝΔΟ)>=3/1+ΑΒΓ (1)
ΓΙΑ ΝΑ ΙΣΧΥΕΙ Η (2) ΠΡΕΠΕΙ ΚΑΙ ΑΡΚΕΙ 3(ΑΒΓ+1)>=Α+Β+Γ+ΑΒ+ΒΓ+ΓΑ (3)
ΤΩΡΑ ΑΝ ΔΕΝ ΙΣΧΥΕΙ Η (3) ΘΑ ΙΣΧΥΕΙ  3(ΑΒΓ+1)<Α+Β+Γ+ΑΒ+ΒΓ+ΓΑ
ΘΑ ΙΣΧΥΕΙ Η ΖΗΤΟΥΜΕΝΗ ΠΡΟΣ ΑΠΟΔΕΙΞΗ ΔΙΟΤΙ  ΑΝ ΠΡΑΓΜΑΤΙΚΑ 1/Α(1+Β)+1/Β(1+Γ)+1/Γ(1+Α)>=3/1+ΑΒΓ
(Α ΜΕΛΟΣ)(Α(1+Β)+Β(1+Γ)+Γ(1+Α))>=(3/1+ΑΒΓ)3(ΑΒΓ+1)=9
ΠΟΥ ΙΣΧΥΕΙ ΑΠΟ CAUCHY-SCHWARZ
ΠΡΟΒΛΗΜΑ 2(ΒΛΕΠΕ ΠΙΟ ΠΑΝΩ ΚΑΙ ΤΟ ΣΧΗΜΑ)
ΛΥΣΗ
ΕΣΤΩ Ζ ΗΤΟΜΗ ΑΑ1 ΚΑΙ Γ1Δ ΚΑΙ Θ Η ΤΟΜΗ ΓΓ1 ΚΑΙ ΑΒ ΕΣΤΩ ΟΤΙ ΟΙ Β1F ΚΑΙ Η ΕΑΙ ΤΕΜΝΟΝΤΑΙ ΣΤΟ R ΘΑ ΔΕΙΞΟΥΜΕ ΟΤΙ ΤΑ R Γ1 ΚΑΙ Δ ΕΙΝΑΙ ΣΥΝΕΥΘΕΙΑΚΑ
ΕΧΟΥΜΕ ΟΤΙ ΑΒΒ1=ΑΔΕ ΕΝΤΟΣ ΕΚΤΟΣ ΚΑΙ ΕΠΙ ΤΑ ΑΥΤΑ
ΕΧΟΥΜΕ ΟΤΙ ΑΒΒ1=ΑΔΖ+ΔΖΕ
ΕΧΟΥΜΕ ΟΤΙ ΑΔΖ+ΔΖΕ+ΕΔΒ=180 ΛΟΓΩ ΤΩΝ ΠΑΡΑΛΛΗΛΙΩΝ
ΘΕΛΟΥΜΕ ΝΔΟ
Γ1ΔΒ+ ΔΖΕ+ΕΔΒ=180 ΑΡΑ ΑΡΚΕΙ ΝΔΟ
Γ1ΔΒ=ΖΔΑ
ΣΤΑ ΤΡΙΓΩΝΑ ΑΔΖ ΚΑΙ ΓΙΔΒ
Η ΔΓ1Η=ΑΖΔ ΕΝΤΟΣ ΕΝΑΛΛΑΞ ΚΑΙ
ΖΑΔ=ΔΗΓ1    ΕΝΤΟΣ ΕΝΑΛΛΑΞ ΑΡΑ
Γ1ΔΒ=ΖΔΑ ΑΡΑ ΙΣΧΥΕΙ ΤΟ ΖΗΤΟΥΜΕΝΟ
ΠΡΟΒΛΗΜΑ 3ΝΑ ΠΡΟΣΔΙΟΡΙΣΕΤΕ ΟΛΟΥΣ ΤΟΥΣ ΘΕΤΙΚΟΥΣ ΡΗΤΟΥΣ (Μ,Ν,Ρ) ΕΤΣΙ ΩΣΤΕ ΟΙ
Μ+(1/ΝΡ) Ν+(1/ΜΡ)  Ρ+(1/ΜΝ) ΝΑ ΕΙΝΑΙ ΑΚΕΡΑΙΟΙ
ΛΥΣΗ
ΑΠΟ ΤΟΥΣ ΑΡΙΘΜΟΥΣ ΛΑΜΒΑΝΟΥΜΕ ΟΤΙ ΜΝΡ/ΜΝΡ+1 ΑΡΑ ΕΠΕΙΔΗ (ΜΝΡ,ΜΝΡ+1)=1 ΜΝΡ=1  ΚΑΙ ΑΝ ΘΕΣΟΥΜΕ
Μ=Α/Β (Α,Β)=1
Ν=Γ/Δ  (Γ,Δ)=1
Ρ=Ε/Ζ  (Ε,Ζ)=1
ΚΙ ΟΙ ΑΡΙΘΜΟΙ ΓΙΝΟΝΤΑΙ
2Μ  2Ν  2Ρ
ΑΡΑ Β,Δ,Ζ ΑΝΗΚΟΥΝ ΣΤΟ {1,2} ΚΑΙ ΜΕ ΔΟΚΙΜΕΣ ΔΙΑΠΙΣΤΩΝΟΥΜΕ ΟΤΙ ΟΙ ΖΗΤΟΥΜΕΝΕΣ ΤΡΙΠΛΕΤΕΣ ΕΙΝΑΙ ΟΙ
(1,1,1) (2,1,1/2) (4,1/2,1/2) ΚΑΙ ΟΛΟΙ ΤΟΥΣ ΟΙ ΣΥΝΔΥΑΣΜΟΙ
ΠΡΟΒΛΗΜΑ 4 ( ΠΑΡΑΚΑΛΕΙΤΑΙ ΚΑΠΟΙΟΣ ΝΑ ΤΟ ΕΠΙΛΥΣΕΙ)
ΕΣΤΩ Μ ΘΕΤΙΚΟΣ ΑΚΕΡΑΙΟΣ ΝΑ ΠΡΟΣΔΙΟΡΙΣΕΤΕ ΟΛΟΥΣ ΤΟΥΣ ΘΕΤΙΚΟΥΣ ΑΚΕΡΑΙΑΟΥΣ Α ΩΣΤΕ Η ΑΚΟΛΟΥΘΙΑ Η ΟΠΟΙΑ ΟΡΙΖΕΤΑΙ ΩΣ Α0=Α ΚΑΙ
ΑΝ+1=ΑΝ/2 ΑΝ ΑΝ ΑΡΤΙΟΣ
ΑΝ+1=ΑΝ+Μ ΑΝ ΑΝ ΠΕΡΙΤΤΟΣ
ΓΙΑ Ν=0,1,2,,, ΝΑ ΕΙΝΑΙ ΠΕΡΙΟΔΙΚΗ ΑΚΟΛΟΥΘΙΑ ΔΗΛΑΔΗ ΟΙ ΟΡΟΙ ΤΗΣ ΝΑ ΕΠΑΝΑΛΑΜΒΑΝΟΝΤΑΙ ΑΡΧΙΖΟΝΤΑΣ ΑΠΟ ΤΗΝ ΜΟΡΦΗ Α0,.....ΑΚ ΓΙΑ ΚΑΠΟΙΟ Κ
ΤΕΛΟΣ ΘΑ ΗΘΕΛΑ ΝΑ ΜΕ ΕΝΗΜΕΡΩΣΕΤΕ ΓΙΑΤΙΣ ΒΑΘΜΟΛΟΓΙΕΣ ΚΑΙ ΤΑ ΜΕΤΑΛΛΙΑ ΤΗΣ ΟΜΑΔΑΣ ΜΑΣ
ΧΑΙΡΕΤΩ

Να τα μετάλλια [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=74042[/url]

Γιατί συνεχίζεις να γράφεις με κεφαλαία ?? Είναι κουραστικά στο διάβασμα  :)

Συγχαρητήρια στήν ελληνική ομάδα (επιτυχόντες και μη ) και εις ανώτερα  :)

Ξέρει κανείς πως τα πήγαν οι αναπληρωματικοί ?

 Ποιοί θα καλύψουν τις 3 θέσεις , του Κουφογιάννη (πάει στο ΜΙΤ με υποτροφία) και των Δεσποτάκη και
 Παπαγιαννάκη που δεν τα πήγαν καθόλου καλά ?

Αμβρόσιε σόρρυ αλλά η λύση στην ανισότητα είναι λάθος . Δεν αποδεικνύονται έτσι οι ανισότητες .... :blush:
Τώρα στην θεωρία αριθμών θα ήθελα  να γράψεις αναλυτικά τη λύση σου για να τη δω . Οκ ?(Εκεί μάλλον είσαι πιο σωστός)!!!
Της γεωμετρίας τη λύση θα την κοιτάξω μόλις την προσπαθήσω και εγώ....
 :D

Αν σας ενδιαφέρει μια λύση χωρίς απαλοιφή και πράξεις (γιατί από ότι έμαθα οι περισσότεροι έτσι την έλυσαν ) δείτε εδώ . http://www.mathlinks.ro/Forum/viewtopic.php?p=499588#p499588

Το λινκ περιέχει και μια ισχυροποίηση της ανισότητας

[quote="ambros"]ΣΤΑ ΤΡΙΓΩΝΑ ΑΔΖ ΚΑΙ ΓΙΔΒ
Η ΔΓ1Η=ΑΖΔ ΕΝΤΟΣ ΕΝΑΛΛΑΞ ΚΑΙ
ΖΑΔ=ΔΗΓ1    [/quote]

Σορρύ αλλά και εδώ είσαι λάθος .Δεν μπορείς να πεις ότι ΔΓ1Η=ΑΖΔ ΕΝΤΟΣ ΕΝΑΛΛΑΞ

αφού έχεις μεν παράλληλες ευθείες αλλά δεν έχεις ευθεία που να τις τέμνει .!!!!!!!
καθώς η Γ1Ζ  δεν είναι ευθεία ......(θέλουμε να δείξουμε ότι είναι ευθεία)
Συνεπώς χρησιμοποίησες αυτό που θες να αποδείξεις άρα λάθος .

ΣΙΛΟΥΑΝΕ ΣΤΟ ΠΡΩΤΟ ΕΧΕΙΣ ΔΙΚΙΟ ΟΜΩΣ
ΑΝΤ'ΑΥΤΟΥ ΘΕΩΡΗΣΕ ΤΙΣ ΑΔΖ=ΑΔΗ ΩΣ ΚΑΤΑΚΟΡΥΦΗΝ ΑΦΟΥ ΑΠΟ ΥΠΟΘΕΣΗ Ζ,Δ,Γ1 ΣΥΝΕΥΘΕΙΑΚΑ   :D  :D
ΤΩΡΑ ΣΤΗΝ ΑΝΙΣΟΤΗΤΑ ΕΧΩ ΔΙΚΙΟ ΚΑΙ ΤΟ ΞΕΡΕΙΣ ΜΗΝ ΚΟΙΤΑΣ ΤΗΝ ΜΕΘΟΔΟ ΤΟΥ ΜΠΑΚΑΛΗ ΜΕ ΤΗΝ ΟΠΟΙΑ ΤΟ ΑΠΕΔΕΙΞΑ ΔΕΝ ΕΙΝΑΙ ΜΑΘΗΜΑΤΙΚΑ ΛΑΘΟΣ ΔΙΟΤΙ ΑΠΕΞΕΙΞΑ ΟΤΙ Η ΑΝΙΣΟΤΗΤΑ ΙΣΧΥΕΙ ΑΦΟΥ ΥΠΑΡΧΟΥΝ 2 ΠΙΘΑΝΕΣ ΠΕΡΙΠΤΩΣΕΙΣ 'Η
3(ΑΒΓ+1)>=Α+Β+Γ+ΑΒ+ΑΓ+ΒΓ ΄'Η
3(ΑΒΓ+1)<Α+Β+Γ+ΑΒ+ΑΓ+ΒΓ
ΑΡΑ ΑΝ ΔΕΙΞΕΙΣ ΟΤΙ ΙΣΧΥΕΙ ΚΑΙ ΓΙΑ ΤΗΝ ΜΙΑ ΠΕΡΙΠΤΩΣΗ ΚΑΙ ΓΙΑ ΤΗΝ ΑΛΛΗ ΚΑΘΑΡΙΣΕΣ
ΟΚ?
ΧΑΙΡΕΤΑΙ  :D  :D

Καταρχήν πάλι στη γεωμετρία μου έβαλες λάθος γράμματα αλλά αν κατάλαβα καλά εννοείς ότι  ΑΔΖ=Γ1ΔΗ . Πώς όμως μπορείς να λες ότι τα σημεία Γ1 ,Δ, Ζ είναι συνευθειακά αφού αυτό θες να αποδείξεις . :? Κοίτα το καλύτερα

Όσο αναφορά την ανισότητα τώρα  . Όταν λες ότι υπάρχουν δύο περιπτώσεις δεν είσαι σωστός . Μπορεί η μία η σχέση να ισχύει για κάποιες τριάδες (α,β,γ) και η άλλη να ισχύει για κάποιες άλλες τριάδες  (α,β,γ)  δεν είναι υποχρεωτικό να έχεις μία από τις δύο .
Λοιπόν για την ανισότητα που μου ζήτησες έχω βάλλει μια λύση και την ισχυροποίησή της . Διάβασέ την προσεκτικά.

Και κάτι άλλο . Αν η γεωμετρία έβγαινε τόσο εύκολα δε θα έμπαινε στην ΒΜΟ αλλά στο Θαλή της Γ' Γυμνασίου . :rotfl:
Και αν θεωρείς όλες αυτές τις παρατηρήσεις που σου κάνω μούφαρα τότε μπες και γράψε τις λύσεις σου στο αγγλικό φόρουμ να δούμε και εκεί τι θα σου πούνε οι πιο ειδικοί από εμένα . Μπορεί να έχεις τη δυνατότητα να λύνεις πολύ εύκολα τα πάντα . :D  .Σκέψου όλα τα παραπάνω.
Φιλικά
Σιλουανος

Αγαπητέ σιλουνέ το ζητόύμενο είναι ΝΔΟ οτι Γ1,Δ,R είναι συνευθειακά ΟΚ? και δεν επικαλούμαι καμία εξωανθρώπινη μαθηματική ικανότητα διότι δεν την έχω αλλά ισως αν θές την τεμπελιά μου που με οδηγεί στο να προσπαθώ να τα λύσω όλα απλά και  σύντομα αύτο είναι το έγκλημά μου :rotfl:  :rotfl:  :rotfl:  και τέλος  στην ΒΜΟ δεν μπαίνουν έυκολα θέματα ΠΟΤΕ? αφού τα περισσότερα είναι τετριμμένα ?
Χαιρετώ φιλικότατα

Οκ θες να δείξεις ότι τα Γ1, Δ, R είναι συνευθειακά . Για να το κάνεις αυτό δεν θέλεις  να αποδείξεις(αυτό δείχνει ο τρόπος σου) ότι Γ1ΔΗ+ΒΔΕ+RΔΕ=180  ??

Αν είναι έτσι τότε θεωρείς ότι οι Γ1Δ , ΔR είναι ευθείες και όχι η Γ1Ζ.......
Ελπίζω να κατάλαβες.......

ΦΙΛΕ ΑΜΒΡΟ,
 εχω την εντυπωση οτι η λυση που δινεις λιγα μνμ πιο πανω για το 2ο θεμα δεν ειναι ολοκληρωμενη.
 Εκει που λες οτι (mnp,mnp+1)=1 νομιζω οτι υπαρχει προβλημα.Ο λογος ειναι οτι δεν γνωριζεις αν αν ο mnp ειναι ακεραιος ;) Αν οντως ειναι, τοτε προφανως mnp=1
Αν ομως δεν ειναι, υπαρχει προβλημα γιατι πολυ απλα δεν εχει νοημα τοτε ο συμβολισμος (mnp,mnp+1)=1 αφου mnp εQ.
Βεβαια τελικα αποδεικνυεται οτι οι λυσεις που βρισκεις ειναι και οι μοναδικες αλλα αυτο βγαινει συμπτωματικα....
Ουσιαστικα δηλαδη,λυνεις το προβλημα m,n,p ε Q και mnp=1 που ειναι διαφορετικο απο το προβλημα οπως δινεται....
Αρκει τωρα να αποδειξεις οτι δεν υπαρχουν m,n,p εQ με mnp διαφορετικο του 1 ωστε οι 3 ζητουμενοι αριθμοι να ειναι ακεραιοι,και η λυση τοτε θα ειναι πληρης.Βεβαια φαινεται λιγο δυσκολο.....

 ΦΙΛΙΚΑ,

 ΡΑΛΛΗΣ :D

φίλε ράλλη αν κοιτάξεις και στις λύσεισ του βαλεντίν θα δείς οτι έχω δίκο αλλά όπωσ και να'χει για να διαιρεί ο α τον α+1 πρεπει και αρκεί α=1 για α ανήκει στο  Ρ πραγματικοι
χαιρεται

Ερώτηση: Γιατί δεν υπήρχε πρόβλημα Συνδυαστικής στη ΒΜΟ2006;
Απάντηση: Διότι το μόνο πρόβλημα Συνδυαστικής που προτάθηκε ήταν ελληνικό και οι διωργανωτές πανικοβλήθηκαν (3 προβλήματα από την ίδια χώρα!).
Ερώτηση: Πώς έφθασε στη διατύπωση του πρώτου προβλήματος ο προτείνων το πρώτο πρόβλημα;
Απάντηση: Ξεκίνησε από μια άλλη γνωστή ανισότητα (βρέστε ποιά) και έκανε το μετασχηματισμό α στο 1/α κλπ. Χωρίς να το καταλάβει βρέθηκε σε άλλη γνωστή.
Ερώτηση: Το πρόβλημα 2 ήταν όπως προτάθηκε αρχικά;
Απάντηση: Όχι. Αρχικά ήταν κάτι σαν αντίστροφο. ΟΙ Ρουμάνοι επέμεναν στην αλλαγή.
Ερώτηση: Πόσο γνωστό ήταν το 4ο Πρόβλημα που προτάθηκε από τη Βουλγαρία;
Απάντηση: Εμφανίστηκε σε Γαλλικό Διαγωνισμό.
Ερώτηση: Ποιός ήταν ο Πρόεδρος της Επιτροπής Επιλογής Προβλημάτων;
Απάντηση: Γνωστός Καθηγητής του Παν/μίου Κρήτης που θέλει να δωθούν οι Μαθηματικοί Διαγωνισμοί στη νεότευκτη "Ένωση Ελλήνων Μαθηματικών" του Β. Σπανδάγου και έχει χρισθεί Αρχισυντάκτης Μαθητικού Μαθηματικού Περιοδικού που πρόκειται προσεχώς να εκδώσει η "ΕΕΜ".

Stay tuned...more gossip is coming...

Πολύ ενδιαφέρουσες οι ερωτο-απαντήσεις .

 Να προσθέσω και εγώ δύο ερωτήσεις :

 1. Στην επιτροπή διαγωνισμών της ΕΜΕ δεν αναφέρεται το όνομα του κ. Ζώτου που ήταν υπαρχηγός  της αποστολής
    στην Κύπρο .Γιατί ?

 2.Γιατί παραιτήθηκε ο κ.Μπόλης απο πρόεδρος της επιτροπής  διαγωνισμών ?

Φιλοι μου η συνδυαστική επειδη δεν διδάσκεται καθόλου στο λύκειο ενω στις αλλες χωρες διδάσκεται είναι άδικο για μας αρα δεν μας συμφέρει
Χαρετώ

Προς: Start

   Σύμφωνα με τις άκρως μυστικές πηγές μου ο κ. Ζώτος περνούσε από το διάδρομο της ΕΜΕ όταν ψάχνανε για υπαρχηγό, όπως ακριβώς ο Καθηγητής κ. Γιάκοβλεφ έτυχε να περνάει από το Γραφείο του Καθηγητή κ.Μήστσενκο στο Ινστιτούτο Στεκλώφ όταν ψάχνανε για Πρόεδρο της ΙΜΟ Site Committee, ο οποίος έμεινε γνωστός στην Ιστορία της ΙΜΟ ως ο "Μπουένος Άϋρες" (έτσι αποχαιρέτησε την ξεναγό του στην Κούβα!). Για τη Μαθηματική Ικανότητα του κ. Ζώτου σάς παραπέμπω στο άρθρο του στο τελευταίο τεύχος του Ευκλείδη Β', ιδιαίτερα στην εισαγωγή...
   Όσο για την παραίτηση του κ. Μπόλη από τη θέση του Προέδρου της Επιτροπής Διαγωνισμών της ΕΜΕ, η μοναδική πηγή μου μέ πληροφορεί ότι οφείλεται σε "αυστηρώς προσωπικούς λόγους". Ο αντιπρόεδρος της Επιτροπής Επίκουρος Καθηγητής του ΕΜΠ κ. Ανάργυρος Φελλούρης δεν εμφανίστηκε ποτέ στις συνεδριάσεις από τον τελευταίο Μάϊο λόγω του ότι αποκλείστηκε από Αρχηγός της Ομάδας στην ΙΜΟ2005. Έτσι η Επιτροπή τώρα έχει ως κεφαλή τον Α' Αντιπρόεδρο Δρ. Δ. Κοντογιάννη με Γραμματέα τον Δρ. Π. Βλάμο.

Lancelot

Πες μας για σένα Dulac, πως τα έμαθες όλα αυτά?

Η δουλειά μου είναι να μαθαίνω...

Μια ακόμα παρατήρηση. Το θέμα 3, το πρότεινε ο φίλος Valentin Vornicu. Είναι αυτός που αν ήθελε η ΕΜΕ θα τον είχαμε εδώ ως προπονητή της ομάδας μας. Aλλα....έφαγε πόρτα. Τα δικά μας τα θέματα ποιός τα πρότεινε; To πρώτο θέμα βέβαια είχε εμφανιστεί ως 2362 προβλημα στο crux.[/code]

Και τα δύο θέματα ήταν του k.Κοντογιάννη .

Λοιπόν όπως έχει προαναφερθεί τα δύο θέματα που προτείναμε δυστυχώς ήταν γνωστά . Το πρώτο ειδικά..... :rotfl: Είναι πασύγνωστο .Αλλά και το αντίστροφο να προτεινόταν (στη Γεωμετρία) πάλι το έχουν οι Ιησουίτες ως γνωστό θεώρημα .Είναι το θεώρημα Aubert τόμος δεύτερος σελ 589 ασκ 1251 γ Τελος πάντων.....

Όσο για τον Κύριο Μπόλη .Από πηγές μου γνωρίζω ότι παραιτήθηκε για δύο λόγους .
Πρώτον : Ζήτησε από τον κύριο...... τάδε να βάλλει τα θέματα στον Αρχιμήδη και του επέστησε την προσοχή ώστε αυτά να είναι προτότυπα .Αυτός του το υποσχέθηκε και το βράδυ πριν τον Αρχιμήδη τον πήρε τηλέφωνο και του είπε να είναι ύσυχος ότι τα θέματα είναι δικά του , είναι προτότυπα κτλ. Προς έκπληξη όλων την άλλη μέρα το πρωί διαπιστώθηκε ότι ήταν αντιγραφές από γνωστό Αμερικάνικο βιβλίο......

Δεύτερον : Το λάθος θέμα που δώθηκε στον Μεσογειακό...
Για το δεύτερο δεν είμαι σίγουρος.......

The official results of the BMO2006 are posted at http://www.cms.org.cy. Don't be surprised with the plethora of medals! The BMO rules are different than the IMO rules. If IMO rules were to apply, the results would have been: COUNTRY:(Gold,Silver,Bronze):
ALB: (1,0,3), BUL:(0,2,2), CYP:(0,0,1), HEL:(0,2,2), MCD:(0,0,1), MLD:(2,0,1), ROM:(1,4,0), SCG:(0,1,1)
GUESTS: KAZ:(1,2,2), UNK:(0,0,0)
Compare this with the official BMO results:
ALB: (1,2,2), BUL:(0,2,3), CYP:(0,1,0), HEL:(0,3,1), MCD:(0,0,1), MLD:(2,1,2), ROM:(2,3,1), SCG:(1,1,3)
GUESTS: KAZ:(2,3,1), UNK:(0,0,3)

SUCCESS RATE (OFFICIAL PARTICIPANTS ONLY): PROBLEM 1: 68,33%, PROBLEM 2: 37,08%, PROBLEM 3: 43,96%, PROBLEM 4: 23,75%

Ε,ΚΑΙ ΚΑΤΙ ΑΛΛΟ!!!!
ΟΠΟΙΟΣ ΕΧΕΙ ΤΟ ΒΙΒΛΙΟ ΤΟΥ ΜΠΑΜΠΗ ΣΤΕΡΓΙΟΥ<ΑΛΓΕΒΡΙΚΕΣ ΑΝΙΣΟΤΗΤΕΣ>
ΑΣ ΤΟ ΑΝΟΙΞΕΙ ΣΤΗΝ ΣΕΛΙΔΑ 304 ΚΑΤΩ ΚΑΤΩ ΧΑΜΗΛΑ ΝΟΜΙΖΩ ΑΣΚΗΣΗ 5.120!!!!
[b]ΝΑΙ ΚΑΛΑ ΒΛΕΠΕΙ!!!![/b]
[b]Ειναι το 1ο θεμα!!!![/b]

ΡΑΛΛΗΣ :D