Hi, I'd like to know 2 things:

1) most important: If I cannot find a professor to accompagny us, can I still participate in the contest? Do I just have to pay something extra or can I not participate at all? Me and a friend would like to go but our university doesn't send a professor so we'll have to go on our own...

2) also a good question: how do you best prepare for this? I see a lot of things (like well-known summations the solutions use) that I have never heard of in my life, what sources do I use best for prep? Any good book regarding IMC-preparation (not some russian or alike please)?

[ not just for training, this is the sort of things our university lacks most, and that limits our mind-span ]

Peter,

Sure you can come on your own. That's how we went last year (by we I mean claude, amfulger, me and a few other colleagues). Of course it's best to have a teacher in the committee to represent you in case your paper gets marked badly :) :?

For the preparation I must admit that most of the Romanians usually use the knowledge we gathered in high-school regarding the subjects (as IMC is very  closely related to the Romanian Olympiad for 11-12th grades). So I suggest you start looking up in the College Playground for past IMC questions and past Romanian 11-12th grade olympiads, and also Putnam problems :)

We can go freely? That's terrific news! :D

On the training, it's not only about finding questions (I have over 100 problems in that style already, more I can't make between exams and contest)

I mainly need:
* [asap] A good source for theorems, since many problem solutions start with "it is well know that -1 = sum ( 1 / log (...) )" , and I have never heard of an alike formula. So a place where I can find the "well-knowns" in calculus/analysis/combinatorics

* [much less urgent] Generally some good book on the matter, with a lot of IMC-like problems, but also the theorems. Just to fill the inventivity-gap in my mind, caused by our university. (man I hate "state and prove" questions)

On another note, what does IMC participation cost? (next to plane flight and hotel?) Is it free? And how expensive is a stay there? (since many imc's have been there yet, I guess you guys may know :))

There is no such book, you must take your classes at college and solve lots of problems. In time some things will become "well-known". I suggest you study previous IMC competitions ...

You have to find out the costs your self. I would look at around 600 euros (plus minus 100 euros) per total (including flight and an extra-meal a day in case you're not satisfied with cafeteria food in the Balkans :) ). But you might find yourself cheeper flights though. The organizers will tell you how much there is to pay for food&lodging.

2 more questions :)
1)To go to IMC shall I be studing math?
2)What ages are accepted?

[quote="thatin"]1)To go to IMC shall I be studing math?[/quote]Handy, but not necessary. I did not study mathematics last year (studied engineering) and I even won a medal, so it's not necessary.
[quote="thatin"]2)What ages are accepted?[/quote]You must not have studied mathematics for more than 4 years, but there's no limit on age.

so... if I want to take part next year what should I do? (I am not studing maths)

Just do the normal thing, register on the website and go. :)

so simble!?! :)  And when it is usually held, and how long it lasts?


number theory
relatively prime
number theory proposed

[color=red][size=150][b]IMO 2003[/b][/size][/color]
[b][color=red]B3[/color][/b]
Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^p-p$ is not divisible by $q$.

[ I am trying to prove it myself and i am confused in one of my cases please help me if you can?]

[u][b]Proof :[/b][/u]

When the prime $p=2$ then we can take $q=3$, which leads to contradiction.
Now we are going to analyse the cases $p>2$, which means p is odd, since the smallest even prime is $2$.

[color=blue][u]Case 1:[/u][/color] let n be any even number, thus we get $n^p-p$ is always odd. Taking $q=2$ we are done. Since even number never divides an odd number.

[color=blue][u]Case 2:[/u][/color] let n be an odd number and relatively prime to p, $(n,p)=1$
In this case we will take $q=p$, since $n^p-p=n^p-n+n-p=n(n{}^p{}^-{}^1-1)+n-p$ $(*)$ we know that $(p,n)=1$,so p never divides that equation.

[color=blue][u]Case 3:[/u][/color] let $(n,p)=d$, for $d$ is different form $1$. Which means $d=p$. Thus $n=pk$ for $k$ any natural number.

[i]note:[/i] third case is missed  :(

Davron

You can't pick a different $q$ for each case...


trigonometry
geometry
3D geometry
sphere
function
percent

I set up a matematically challange for myself, but can't figure it out. I hope someone can help me with it - maybe you find it easy and maybe i just haven't looked enough at the problem, but at least i've looked at the problem more than once.

In Dubai it is now being considered to built a tower rising 1.2km above the ground. The reason you can't see all the way to America from Europe when looking west is of course because of the curvature of Earth. But my question is how far away could you be standing and (theoretically) see the tower? I know hot air can bend light, but let's keep this physical aspect out of the question. Starting from the tower you would of course not be able to see more and more of the lower part of the tower and finally you can't see nothing but the tip of the antenna. So how can i compute this? I know i need to use something like $\cos$ and $\sin$, but still i can't do it myself. I tried making a drawing of the problem but didn't help.

Some info:
Height of the tower (Al Burj): 1.2km
Radius of Earth: 6348km

Please show how you calculated it. Thx in advance.

well, earth isn't a perfect sphere, so we need too know the curvature of this

[hide]K so draw a circle, this is the earth (you need to imagine cutting the earth so the plane you look at is a circle). First off, let the point where the tower touches the ground be B and the tip of the tower be C. Let the center of the circle (earth) be O. Draw a ray from the tip of the tower that is tangent to the circle (earth). Let this point of tangency be A. Draw radii to A as well as to B. Now we have right triangle OCA with legs 6348 and x and hypotenuse 6349.2. Let <AOB be a. cos(a) = 6348/6349.2 so do the inverse function of cos to find a, and a = 1.11397373. Therefore, you can see the tower from (1.11397373/360)*6348*2*pi = 123.4212315 km away (in all directions). I think.
[/hide]
Maybe this is right? o-0

[img]9160[/img]
[hide="Solution"]
So we have a right triangle formed by the tangent line segment C (which is what we want to know) and the radius B and hypotenuse B + A (the height of the tower plus radius of the earth)

So we have $b^{2}+c^{2}= (a+b)^{2}$ and $c^{2}= a^{2}+2ab$
So $c = \sqrt{a^{2}+2ab}$

Plug in the numbers given:
$a = 1.2, b = 6348$
$c = \sqrt{1.44+15235.2}$
$c = 123.46 km$
[/hide]

OK, thx a lot for your help and your drawing. As i expected it wasn't that difficult, i just couldn't find the triangle to work with, but i can easily follow you here.

And yes, Earth isn't spherical, that's also true. So looking at it from one direction you would be able to see it farther away, while another direction you wouldn't be able to see it that far away. So if we, on a day where the light ray isn't bended because of the weather, do you think something like $123km$ or maybe just around $100km$ is realistic in case you've got a clear sight (flat ground and no buildings or other things in the way). Of course you can only see the tip of the antenna, but let's say $100km$ away from the tower. What do you think?

That seems like a conservative, fair estimate.

You sound like you would expect even more? How far would you guess and based on what? How many percent extra do you think we could gain from bending light in warm weather?

The formula (given above by other posts) = $\sqrt{(R+h)^{2}-R^{2}}\approx \sqrt{2Rh}\approx 3.57 \sqrt h$
(The answer is in  Km, the height is in meters - For 1200 meter tower it is about $124 km$

The effect due to earth being not a "perfect sphere" is quite negligible.  (The difference between  equatorial and polar  radius is  less than 1/2 a percent) The  uncertainty  (in about 100km)  in distance you can see would be at the most about 100 meters ... much smaller effect  than due to say seeing a person's toe or his head - remember just a few meters and your  horizon is already a few km's)

The effect due to refraction is certainly not negligible. I remember visiting Empire State Building with my father, and  the distance one can see from top was much different than what I (and my father) calculated at first..(About 4-5km more, a value much higher than our  uncertainty allowed). .. so this  got us back to dosome  more calculation using simple model to include refraction)

Some quick calculations (assuming homogeneous atmosphere etc) we found that adding about 8% to the theoretical-no-refraction value is quite ok if  weather conditions are normal.

(If you want to use this formula - multiply 3.9Km with square root of the height in meters to get to see how far you can see  ---  assuming calm air - For 1200 meter tower this is about $135 Km$)

Note that  for sunrise (or sunset), one can see the  sun when it is  almost a degree  below the horizon. (Refraction is about 34' and radius of the disk is about 16 giving the value as 50 minutes of arc)... and every degree the air is cold, the effect is almost 0.3% per degree... (You can see sunrise earlie -- colder it gets)

But this is all in normal weather conditions... Dubai, I assume , is too hot.(ground air can be hotter and lighter).. so actual values can vary a lot ( I assume by LOT --- may be extra 20-30 Km)

.. Dubai tower, I will say, if there are inversion layers, theoretically, one can see  VERY far away-  limited by how clear the air is... and  you can get conditions where the horizon is limit-less....

Related to this:

People see  San Clemente Island,  from San Diego ( about 125 km)  (At sunset - if they know where to look) ... and there have been reported cases of seeing objects hundreds of km away, that is many times more distant than the theoretical horizon distance. (The weather condition was unusual) ...

(I  read somewhere  one observed  a Japanese mountain which was about 1000Km away from an observatory.. amazing)

Edited later: I just tried to look and found 1923 reference of seeing mountain peeks almost 750 km away:
http://mintaka.sdsu.edu/GF/bibliog/bibliog.html#Korzenewsky1923p

Yes, this phenomen is amazing. I don't know the distance, maybe it isn't that much, but in the old times there was a legend that said there existed some land from Iceland. However, it was consider only a legand until a few years ago when scientists discovered that it theoretically (though never happened for a nowaday scientists AFAIK) was possible that the people of Iceland had actually seen Greenland. This legend eventually led Erik The Red to Greenland (who later, AFAIK, discovered America - or maybe that wasn't him).


What is the maximum value of x+2y given that sqrt{x+2}+ sqrt{y-5}]=6.

Are you sure you want maximum? Or rather minimum?

sqrt{x+2}+ sqrt{y-5}=6<=>sqrt{x+2}=6-sqrt{y-5}=>x+2=36-12sqrt{y-5}+y-5=>x=29+y-12sqrt{y-5}
So, x+2y=3y+29-12sqrt{y-5}
We also have that 36>=y-5>=0 => 41>=y>=5
d(3y+29-12sqrt{y-5})/dy= 3-6/sqrt{y-5}
Minimum and maximum at the ends of the interval [5,41] or where derivative=0 (which happens when y=9)
Evaluating at these points we get the maximum at y=41 (x+2y=80), the minimum at y=9 (x+2y=32)[/code]

At Intermediate, the problems should be solved without calculus.

Another method for minimum, based on AM-QM:

$ 2 \equal{} \frac {{1\over 2}\sqrt {x \plus{} 2} \plus{} {1\over 2}\sqrt {x \plus{} 2} \plus{} \sqrt {y \minus{} 5}}{3}\leqslant\sqrt {\frac {{x \plus{} 2\over 4} \plus{} {x \plus{} 2\over 4} \plus{} y \minus{} 5}{3}} \equal{} \sqrt {{x \plus{} 2y \minus{} 8\over 6}}$

Hence $ x \plus{} 2y\geqslant 6\cdot 2^2 \plus{} 8 \equal{} 32$, and the minimum is obtained for $ {1\over 2}\sqrt {x \plus{} 2} \equal{} \sqrt {y \minus{} 5} \equal{} 2\implies x \equal{} 14\land y \equal{} 9$

[quote="salonikios"]Sorry for that!I forgot where I was!

For the minimum:I find the same way as before that x+2y=3y+29-12sqrt{y-5}=3(sqrt{y-5}-2)^2+32>=32 with equality holding when sqrt{y-5}=2=>y=9
For maximum:3(sqrt{y-5}-2)^2+32 maximum when sqrt{y-5}-2 maximum. But 6>=sqrt{y-5}=> 41>=y which gives us x+2y=80[/quote]

Please salonikios put a $ \$$ in each side of your math code so it is easier and nicer to read your posts.

Sorry for that!I forgot where I was!

For the minimum:I find the same way as before that $ x\plus{}2y\equal{}3y\plus{}29\minus{}12\sqrt{y\minus{}5}\equal{}3(\sqrt{y\minus{}5}\minus{}2)^2\plus{}32>\equal{}32$ with equality holding when $ \sqrt{y\minus{}5}\equal{}2\equal{}>y\equal{}9$
For maximum:$ 3(\sqrt{y\minus{}5}\minus{}2)^2\plus{}32$ maximum when $ \sqrt{y\minus{}5}\minus{}2$ maximum. But $ 6>\equal{}\sqrt{y\minus{}5}\equal{}> 41>\equal{}y$ which gives us $ x\plus{}2y\equal{}80$

By the way for sqrt it is "\sqrt"

 :)

Thanx for the advice ElChapin!!!I really need to find some time to learn how to type latex...

It's the Arithmetic Means Quadratic Means Inequality.
It states that $ (a_1\plus{}a_2\plus{}...\plus{}a_n)/n<\equal{}\sqrt((a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)/n)$


abstract algebra
vector
superior algebra
superior algebra unsolved

I assume that rings have a unit but are not necessary commutative. What are modules here¿

If $ R$ is a skew field, the same proof as for normal fields and vextor spaces applies (needing the axiom of choice).
If $ R$ is not a skew field, take a nonzero nonunit $ m$ and consider $ R/(Rm)$: it's a left $ R$-module but not free.

[quote="ZetaX"]If $ R$ is not a skew field, take a nonzero nonunit $ m$ and consider $ R/(Rm)$: it's a left $ R$-module but not free.[/quote]

This is not true. There are rings $ R$ such that $ R\cong R\times R$ as left $ R$-modules, take $ m$ to be the preimage of $ (1,0)$. owk

Since all left $ R$-modules are projective, $ R$ must be semisimple, so in particular left Artinian. Now if $ R$ were to have a proper non-zero left ideal, that ideal would be free and hence would contain a summand isomorphic (as an $ R$-module) to $ R$, which would again have a proper $ R$-submodule, etc. This contradicts the fact that $ R$ is left Artinian, so $ R$ has no proper non-trivial left ideals; the conclusion follows.

Alternatively, one could use [url=http://planetmath.org/encyclopedia/WedderburnArtinTheorem.html]the structure of semisimple rings[/url], but I suppose this is overkill :).

[quote="Omid Hatami"]Prove that the ring $ R$ is a skew-field, if and only if every $ R$-module is free.[/quote]

A more elementary proof.
Assume that every left $ R$-module is free. We may retain only the "weaker" hypothesis : every non-null left $ R$-module contains a free vector.
By Krull's theorem, $ R$ has a maximal left ideal $ M$. By our hypothesis, the left $ R$-module $ R/M$ contains a free vector, i. e. there exists an element $ a$ in $ R$ with the following property :
the only "scalar" $ c$ such tat $ ca$ lies in $ M$ is $ c \equal{} 0$. (Property (1) of a.)
Let $ x$ a non-null element of $ R$. We will prove that there exists an element $ x'$ of $ R$ such that $ x'x \equal{} 1$.
By the property (1) of $ a$, $ xa$ is not in $ M$. By maximality of $ M$, there exists $ y$ in $ R$ such that $ yxa \minus{} 1$ is in $ M$.
Since $ M$ is a left ideal, $ a(yxa \minus{} 1)$ lies in $ M$. But $ a(yxa \minus{} 1) \equal{} (ayx \minus{} 1)a$, thus $ (ayx \minus{} 1)a$ is in $ M$. By the property (1) of $ a$, $ ayx \minus{} 1$ = 0, thus
$ (ay)x \equal{} 1$, which proves our thesis with $ x' \equal{} ay$.
Thus we proved that each non-null element of $ R$ has a "left inverse". It is known that, in this case, each non-null element has a "bilateral" inverse. (Let $ x'$ be a left inverse of $ x$. By hypothesis, $ x'$ has a left inverse, say $ x''$. Associativity gives :
$ (x''x')x \equal{} x''(x'x)$, $ x \equal{} x''$, thus, by definition of $ x''$, $ xx'$ = 1, so $ x'$ is a bilateral inverse of $ x$.)
P.


when will it come out?????

so slow.....

I think they posted the COMC results even sooner! They are probably marking by hand because their equipment broke down :rotfl:


Find all functions $ f: Q\to Q$
 $ f(x\plus{}f(y))\equal{}f(x)f(y)$

If there exists $ f(a)\equal{}0$, $ P(x,a): f(x)\equal{}0$.
So $ f(x)\equal{}0$ is one solution.

Now assume $ f(x)\ne0,\forall x$ and let $ f(0)\equal{}a$.
$ P(\minus{}a,0): f(\minus{}a)\equal{}1$
$ P(x,\minus{}a): f(x\plus{}1)\equal{}f(x)$. (*)
Suppose there exists $ f(b)\equal{}c\ne1$.
$ P(x,b): f(x\plus{}c)\equal{}cf(x)$. (**)
Assume $ c\equal{}\frac pq$ with $ (p,q)\equal{}1$.
$ f(x\plus{}p)\equal{}f(x)$ from (*)
$ f(x\plus{}p)\equal{}(\frac pq)^qf(x)$ from (**)
Hence, $ (\frac pq)^q\equal{}1\Rightarrow \frac pq\equal{}1$. Contradiction!

Conclusion: $ f(x)\equal{}0$ or $ f(x)\equal{}1$


Do there exist 16 three digit numbers, using only three different digits in all, so that the all numbers give different residues when divided by 16?

[i]Bulgaria[/i]

The answer is no.

Suppose the answer would be affirmative. Obviuously the three digits cannot be all of the same parity, because we would not be able to cover all the residues modulo 2 :)

Thus there exists at least one even and one odd digit. Suppose WLOG that we have one odd digit and two even digits (the other case can be treated in the same way, replacing the word even with the word odd everywhere). Thus the 8 odd residues modulo 16 must be covered by odd numbers, that is the last digit is the odd one. But there are 9 ways to choose the first two digits, and out of those at least 4 must give odd residue when divided by 8, because the number $\overline{xy}\cdot 10$ ( $\overline{xyz}=\overline{xy}\cdot 10 +z\ \Rightarrow\ \overline{xy}\cdot 10$ ) must pass through all the even residues modulo 16.

The last 4 numbers which give odd residue when divided by 8 must be formed having the last digit y, and odd number, but there are only 3 such numbers, which leads to a contradiction.

Thus the problem is solved.

I am not sure, but I belive this was my solution in the contest.

why there are at least 4 odd residues modulo 8 and why xy.10 must pass through all even residues modulo 16?
(any help is appreciated)

[quote=guesswho1729]why xy.10 must pass through all even residues modulo 16?[/quote]
According to the solution above, when there is exactly one odd digit, and the last digit is odd, 8 odd numbers $\overline{xyz}=\overline{xy}\cdot 10 +z$ gives the distinct odd residues 1, 3, 5, 7, 9, 11, 13, 15. So  $\overline{xy}\cdot 10$ gives the distinct even residues 0, 2, 4, 6, 8, 10, 12, 14.

[quote=guesswho1729]why there are at least 4 odd residues modulo 8[/quote]
From the 8 cases of distinct even residues mentioned above,
$\overline{xy}\cdot 10=16a+0\Rightarrow \overline{xy}\cdot 5=8a+0\Rightarrow \overline{xy}$ is even.
$\overline{xy}\cdot 10=16a+2\Rightarrow \overline{xy}\cdot 5=8a+1\Rightarrow \overline{xy}$ is odd.
$\overline{xy}\cdot 10=16a+4\Rightarrow \overline{xy}\cdot 5=8a+2\Rightarrow \overline{xy}$ is even.
$\overline{xy}\cdot 10=16a+6\Rightarrow \overline{xy}\cdot 5=8a+3\Rightarrow \overline{xy}$ is odd.
$\overline{xy}\cdot 10=16a+8\Rightarrow \overline{xy}\cdot 5=8a+4\Rightarrow \overline{xy}$ is even.
$\overline{xy}\cdot 10=16a+10\Rightarrow \overline{xy}\cdot 5=8a+5\Rightarrow \overline{xy}$ is odd.
$\overline{xy}\cdot 10=16a+12\Rightarrow \overline{xy}\cdot 5=8a+6\Rightarrow \overline{xy}$ is even.
$\overline{xy}\cdot 10=16a+14\Rightarrow \overline{xy}\cdot 5=8a+7\Rightarrow \overline{xy}$ is odd.

So there are exactly 4 odd numbers for $\overline{xy}$. That happens when $y$ is an odd digit. But there are only 3 choices for $x$, which leads to a contradiction.


Numbers $ a,b,c,d>0$ such that
\[ \frac1{1\minus{}ab}\plus{}\frac1{1\minus{}bc}\plus{}\frac1{1\minus{}cd}\plus{}\frac1{1\minus{}da}\plus{}\frac1{1\minus{}db}\plus{}\frac1{1\minus{}ac}\equal{}8,\]
prove that
\[ a^2\plus{}b^2\plus{}c^2\plus{}d^2\geq1.\]

$ \frac {1}{1\minus{}ab}>1\plus{}ab$
So, $ \sum_{cyc}ab<2$
$ \sum_{cyc}a^2\geq1 \Leftrightarrow (a\plus{}b\plus{}c\plus{}d)^2\minus{}2\sum_{cyc}ab\geq1 \Leftrightarrow (a\plus{}b\plus{}c\plus{}d)^2\geq5$
How to do that??

Suppose a,b are the smaller two (if applicable)
Then:
\[ 8\leq \frac{6}{1\minus{}ab}\iff ab\geq \frac14 \Rightarrow a^2\plus{}b^2\geq \frac12\]
Since $ c^2\plus{}d^2\geq a^2\plus{}b^2$ it implies the desired result.

[quote="hitek"]Numbers $ a,b,c,d > 0$ such that
\[ \frac1{1 \minus{} ab} \plus{} \frac1{1 \minus{} bc} \plus{} \frac1{1 \minus{} cd} \plus{} \frac1{1 \minus{} da} \plus{} \frac1{1 \minus{} db} \plus{} \frac1{1 \minus{} ac} \equal{} 8,
\]
prove that
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\geq1.
\]
[/quote]
Let $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 < 1,$  $ a \equal{} kx,$ $ b \equal{} ky,$ $ c \equal{} kz$ and $ d \equal{} kt,$
where $ x,$ $ y,$ $ z$ and $ t$ are non-negative such that $ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 1.$
Hence, $ k^2(x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2) < 1.$ Id est, $ 0\leq k < 1.$
Hence, $ 8 \equal{} \frac {1}{4}\sum_{sym}\frac {1}{1 \minus{} ab} \equal{} \frac {1}{4}\sum_{sym}\frac {1}{1 \minus{} k^2xy} < \frac {1}{4}\sum_{sym}\frac {1}{1 \minus{} xy}.$
But $ \frac {1}{4}\sum_{sym}\frac {1}{1 \minus{} xy}\leq8$ for non-negative $ x,$ $ y,$ $ z$ and $ t$ such that $ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \equal{} 1.$
For proof see the zhaobin's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109029


find the sum
\[ 1 \plus{} \left( \begin{gathered}
  n \hfill \\
  1 \hfill \\
\end{gathered}  \right)\cos \theta  \plus{} \left( \begin{gathered}
  n \hfill \\
  2 \hfill \\
\end{gathered}  \right)\cos 2\theta  \plus{} ........ \plus{} \left( \begin{gathered}
  n \hfill \\
  n \hfill \\
\end{gathered}  \right)\cos n\theta
\]

hello,prove by induction that $ \sum_{i\equal{}0}^n\binom{n}{i}\cos(i\theta)\equal{}2^n\cdot\cos(\frac{\theta}{2})^n\cos(\frac{n\theta}{2})$.
Sonnhard.

[hide="Alternately"] This is the real part of $ (1 \plus{} e^{i \theta})^n$, which we can write as (edited:)

$ e^{ i \frac {n \theta}{2} } \left( e^{ \minus{} i \frac {\theta}{2} } \plus{} e^{ i \frac {\theta}{2} } \right)^n \equal{} 2^n \left( \cos \frac {n \theta}{2} \plus{} i \sin \frac {n \theta}{2} \right) \cos^n \frac {\theta}{2}$. [/hide]

[hide="Hint and answer"]Complex numbers; $ S\equal{}2^n\cos^n{\theta\over 2}\cos{n\theta\over 2}$[/hide]

[quote="t0rajir0u"][hide="Alternately"] This is the real part of $ (1 \plus{} e^{i \theta})^n$, which we can write as

$ e^{ \minus{} i \frac {n \theta}{2} } \left( e^{ \minus{} i \frac {\theta}{2} } \plus{} e^{ i \frac {\theta}{2} } \right)^n \equal{} 2^n \left( \cos \frac {n \theta}{2} \minus{} i \sin \frac {n \theta}{2} \right) \cos^n \frac {\theta}{2}$. [/hide][/quote]
Doesn't this equal to $ (1\plus{}e^{\minus{}i\theta})^n$ instead?

I think he meant to say this:
$ (1\plus{}e^{i\theta})^n\equal{}e^{i\frac{n\theta}{2}}\left( e^{\minus{}i\frac{\theta}{2}}\plus{}e^{ i\frac{\theta}{2}}\right)^{n}\equal{} 2^{n}\left(\cos\frac{n\theta}{2}\plus{}i\sin\frac{n\theta}{2}\right)\cos^{n}\frac{\theta}{2}$


The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime number.  For such representations of the even number 126, the largest possible difference between the two primes is
\[ \text{(A)}\ 112 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 92 \qquad \text{(D)}\ 88 \qquad \text{(E)}\ 80 \]

[hide][color=indigo]starting with 125, I went through the odd numbers to find the primes.
113+13 = 126, and 13 is prime, so 100 is the biggest difference. B is the answer[/color][/hide]

[hide="my guess"]
A or B[/hide]
i don't know how to do this type of problem

[hide="Something"]
A would mean 7 and 119. You might be tempted to write A as your answer, but 119 = 7 * 17.
B would mean 13 and 113. Both are prime.
[/hide]


Find the geometric mean of the positive divisors of the natural number n.

[hide]Let $ a_{1},a_{2},...,a_{k}$ be all the positive divisors of the natural number n.

G.M. = $ \sqrt[k]{a_{1}a_{2}...a_{k}}$

Since $ a_{1}a_{k}=n$, $ a_{2}a_{k}=n$, ...

Therefore G.M. = $ \sqrt{n}$[/hide]

Just wondering. I fail to see why IrinaSharova's post was originally rated as spam.Except for a minor Latex error it is perfectly right. I did it the same way but I put a bit more detail that's all.

[hide=" My Solution"]

Let us denote the divisors as $ d_{1}< d_{2}< .......<d_{k}$ and let us consider two cases. Furthermore $ d_{1}=1, d_{n}=n$

1. If $ n$ is not a perfect square we have $ n=d_{2}d_{k-1}=......=d_{\frac{k}{2}}d_{\frac{k}{2}+1}$ and so their $ G.M. = \sqrt[k]{n.n^{\frac{k}{2}-1}}=\sqrt{n}$

2. If $ n$ is a perfect square we have $ n=d_{2}d_{k-1}=..........=d_{\frac{k+1}{2}-1}d_{\frac{k+1}{2}+1}$ and so their $ G.M. = \sqrt[k]{n.d_{\frac{k+1}{2}}.n^{\frac{k-3}{2}}}= \sqrt{n}$.[/hide]


Is this tournament holding this year?
Can anyone post some practice problems?


inequalities
limit
algebra
binomial theorem
real analysis
real analysis unsolved

Prove that $ \lim_{p\to\infty}(a^p\plus{}b^p)^{1/p}\equal{}\max(a,b)\forall a,b\geq 0$.

$ \left(a^p \plus{} b^p\right )^\frac {1}{p} \equal{} \max (a,b)\left(1 \plus{} x^p \right)^\frac {1}{p}$ where $ x \equal{} \frac {\min (a,b)}{\max (a,b)}\leq 1$.  Therefore for $ p\geq 1$ the binomial theorem yields
\[ \max (a,b) \left(1 \plus{} \frac {x^p}{p} \minus{} \frac {(p \minus{} 1)x^{2p}}{2p^2}\right) \leq \left(a^p \plus{} b^p \right )^\frac {1}{p}\leq \max (a,b)\left(1 \plus{} \frac {x^p}{p}\right)
\]
and the conclusion follows.

That's trivial if $ max(a,b)\equal{}0$. If $ max(a,b)>0$ we can write : $ (a^p\plus{}b^p)^{1/p} \equal{} max(a,b).(1\plus{}(\frac{min(a,b)}{max(a,b)})^p)^{1/p}$
But $ 1 \leq (1\plus{}(\frac{min(a,b)}{max(a,b)})^p)^{1/p} \leq 2^{1/p}$ , and  $ 2^{ 1/p} \rightarrow 1$, so  you can conclude ...
In the same way you can prove  that $ \lim_{p\to\infty}(a_1^p\plus{}...\plus{}a_m^p)^{1/p}\equal{}\max(a_1,...,a_m)$, where $ a_1,...,a_m\geq 0$ ..  :cool:

It was shown on this forum that if $ a_i  > 0$, $ x_i  \ge 0$, $ \sum\limits_{i \equal{} 1}^n {a_i  \equal{} 1}$, then $ \left( {\sum\limits_{i \equal{} 1}^n {x_i a_i ^p } } \right)^{\frac{1}{p}}$has interesting asymptotic properties in p.

;)

Consider $ a \equal{} max (a,b)$. Therefore,

$ \lim_{p \to \infty} (a^p \plus{} b^p)^{1/p} \equal{} \lim_{p \to \infty} a \left[1 \plus{} \left(\frac{b}{a}\right)^p\right]^{1/p} \equal{} \lim_{p \to \infty} a \left[1 \plus{} \frac{1}{p} \left(\frac{b}{a}\right)^p \plus{} O[(b/a)^{2p}]\right]$

$ \equal{} a$,

since b/a < 1.


Let's say we have a right triangle consisting a hypotenuse c that is 10 cm.  In this triangle, there is a 1cm*1cm square that fits exactly in the triangle by touching the sides of the triangle (only one square can exactly fit in each triangle).  Now, what are the lengths of the opposite and adjacent sides of the triangle?

Masoud Zargar


limit
trigonometry
calculus
calculus computations

find
$ \lim_{ n\to \infty } \sqrt[n]{ 1\plus{}\sin n}$

[quote="aktyw19"]find
$ \lim_{ n\to \infty } \sqrt [n]{ 1 \plus{} \sin n}$[/quote]
It doesn't exist. Consider $ n \equal{} 2\pi k, k \in \mathbb{N}$, then $ \sqrt[n]{1\plus{}\sin n} \equal{} 1$. And $ n \equal{} 2\pi k \plus{} \frac{3}{2} \pi, k \in \mathbb{N}$ then $ \sqrt[n]{1\plus{} \sin n} \equal{} 0$

and so is  well?
$ \lim_{n\to\infty} (1 + \sin n)^{ \frac{1}{n}} =\lim_{n\to\infty}((1+ \frac{n \sin n}{n})^n)^ { \frac{1}{n^2}} = e^{\lim_{n\to\infty} \frac{n \sin n}{n^2}} = e^{\lim_{n\to\infty} \frac{\sin n}{n}} = e^0=1$

[quote="aktyw19"]and so is  well?
$ \lim_{n\to\infty} (1 + \sin n)^{\frac {1}{n}} = \lim_{n\to\infty}((1 + \frac {n \sin n}{n})^n)^ {\frac {1}{n^2}} = e^{\lim_{n\to\infty} \frac {n \sin n}{n^2}} = e^{\lim_{n\to\infty} \frac {\sin n}{n}} = e^0 = 1$[/quote]
$ \frac{n\sin n}{n}$ is not goes to 0, when $ n$ goes to $ \infty$. So how  you know that $ (1+\frac{n\sin n}{n})^{\frac{n\sin n}{n}}$ goes to $ e$ ??

[quote="aktyw19"]find
$ \lim_{ n\to \infty } \sqrt [n]{ 1 \plus{} \sin n}$[/quote]

It's clear that if the limit exists, it is $ 1$. That is because $ \sin n > 1/2$ for infinitely many $ n$ (this is not hard to see) and $ (1/2)^{1/n} \rightarrow 1$. Now it is also true that there is a sequence of positive integers converging to $ 3\pi/2 \mod 2\pi$, so it all boils down to how fast that convergence can be. There are theorems about such things, but I am no expert in this area.

we have: $ 2\minus{}\epsilon \geq1\plus{} \sin  n \geq \epsilon$ because of $ \sin n \neq 0$
Hence, $ \sqrt[n]{ \epsilon} \leq L \leq \sqrt[n] {2\minus{}\epsilon}$
Therefore, $ L\equal{}1$

@ Hong Quy
[quote="Hong Quy"]we have: $ 2 \minus{} \epsilon \geq1 \plus{} \sin n \geq \epsilon$ because of $ \sin n \neq 0$
Hence, $ \sqrt [n]{\epsilon} \leq L \leq \sqrt [n] {2 \minus{} \epsilon}$
Therefore, $ L \equal{} 1$[/quote]
$ 1\plus{}\sin n \neq 0$, but $ \forall \varepsilon > 0\exists n\in N: \ 1 \plus{} \sin n < \varepsilon$.

hello, i also think that the searched limit is $ 1$.
Sonnhard.

[quote="anik_andrew"]$ 1 \plus{} \sin n \neq 0$, but $ \forall \varepsilon > 0\exists n\in N: \ 1 \plus{} \sin n < \varepsilon$.[/quote]

This is correct and proves the limit of the sequence (restricted to integer n) is not defined.

[quote="atomicwedgie"]
... and proves the limit of the sequence (restricted to integer n) is not defined.[/quote]
Can you write your proof?  :)


algebra
polynomial
Rational Root Theorem
number theory proposed
number theory

Let $k=\frac{xy}{x+y}$.  A rearrangement gives $x+y-3k-k^2\in\mathbf{N}\Rightarrow k^2+3k\in\mathbf{Z}$, or $k^2+3k+b=0$ for some $b\in\mathbf{Z}$.  But since $k$ must be rational, the rational root theorem means that $k$ must be an integer - i.e. $\frac{xy}{x+y}\in\mathbf{N}$.

From $x^3+y^3-x^2y^2\ge0$, and assuming $y\ge x$ I get $y\ge x^2$ (unless $x=y=2$). This can be seen by considering separately two cases - $x\le y\le x^\frac{3}{2}$ and $x^\frac{3}{2}\le y\le x^2$, which can both be shown to be contradictions. So we let $y=x^2+p$.

But now $k=\frac{x^3+px}{x^2+x+p}=x-1+\frac{x}{x^2+x+p}\not\in\mathbf{N}$.

So the only solution is $x=y=2$, which gives $\frac{x^3+y^3-x^2y^2}{(x+y)^2}=0$.

That's quite a nice problem - I hope I haven't messed up the solution...

[quote="Diarmuid"]$x\le y\le x^\frac{3}{2}$ and $x^\frac{3}{2}\le y\le x^2$, which can both be shown to be contradictions.[/quote]

I can't see why. Can you show?

Case 1:
(There was a small typo here - I said the result was true for $y\ge x^2$, and false for $y\le x^2$ - I meant $y<x^2$, or $y\le x^2-1$.)

$x^\frac{3}{2}\le y\le x^2-1\Rightarrow y^3\le x^2y^2-y^2\le x^2y^2-x^3$, with equality iff $x^\frac{3}{2}=y=x^2-1$, which has no solutions in positive integers.

Case 2:
$x\le y\le x^\frac{3}{2}$

I can't remember what I did for this one. (already...my mind is not what it used to be)

One (ugly) way tp do it is by breaking it into further cases.  If we separate out $x\le4$ (this is just a handful of cases, and can be checked by hand), and define a sequence $a_1=1.5$, $a_i=\frac{3a_{i-1}}{2}-0.775$, then $a_i$ decreases, and in particular $a_7<1$.  We consider the case $\max(x^{a_{i+1}},1)\le y< x^{a_i}$ for $1\le i\le 6$.

$x^3+y^3<x^3+x^{3a_i}\le 2x^{3a_i}\le x^{3a_i+0.45}=x^{2a_{i+1}+2}\le x^2x^{2a_{i+1}}\le x^2y^2$.

(The sequence defined is chosen to satisfy $3a_i+\delta=2a_{i+1}+2$ for the largest possible value of $\delta$ that will still give a decreasing sequence.  The cases for $x<5$ are excluded since these do not meet the condition $2\le x^\delta$).

Sorry about that - I'm sure I had a nicer solution before...

I haven't thought much about that yet - but it seems a good idea to break into 2 cases $(x\le y\le x^\frac{3}{2} $and $x^\frac{3}{2}\le y\le x^2)$ like you did!


inequalities
induction
inequalities unsolved

Those are primitive inequalities in th case $n=2,3$.The prove is by the method of mathematical induction.Now let's assume for first that $n=2k,k\in{N}$,then $\sum_{i=1}^{2k}{(-1)^{i-1}{{a_{i}}^{2}}}\geq(\sum_{i=1}^{2k-1}{(-1)^{i-1}{a_{i}}})^{2}-{a_{2k}}^{2}$(by the iduction assumption),then as $\sum_{i=1}^{2k-1}{(-1)^{i-1}{a_{i}}}\geq{a_{2k}}\geq0$ we can write $(\sum_{i=1}^{2k-1}{(-1)^{i-1}{a_{i}}})^{2}-{a_{2k}}^{2}\geq(\sum_{i=1}^{2k}{(-1)^{i-1}{a_{i}}})^{2}.$
In the same way for $n=2k+1.$

[quote="mecrazywong"]Let $a_1\geq a_2\geq ...\geq a_n\geq0$, prove that \[a_1^2-a_2^2+...-(-1)^na_n^2\geq (a_1-a_2+...-(-1)^na_n)^2\][/quote]

The general problem:

$ a_1^k-a_2^k+...-(-1)^na_n^k \geq (a_1-a_2+...-(-1)^na_n)^k$
For all real $k>1$

Solve with the same way.


Joey's strategy for his first marathon (26.2 miles) was to run 2 miles, walk 1 mile, run 2 miles, walk 1 mile, and continue this pattern until he completed the race. Joey's average running pace is 8 minutes per mile, and his average walking pace is 16 minutes per mile. How many minutes will it take Joey to complete the marathon? Express your answer as a decimal to the nearest tenth.

Every 3 miles takes him 32 minutes, making the first 24 miles take him 256 minutes. He runs the next 2 miles in 16 minutes, making your total 272. The last 0.2 miles at 16 minutes per mile takes $ \frac{16}{5}\equal{}3.2$ add this to the 272 gives you a grand total of $ \boxed{275.2}$


inequalities
function
limit
real analysis
real analysis unsolved

let f be a derivable function on $ [a,b]$ such that $ f^2\plus{}(1\plus{}f')^2<\equal{}1$
prove that f is the zero function

What's wrong with $ f(x)\equal{}\minus{}x$ on $ [\minus{}1,1]$?

This is false.

Let $ f(x)\equal{}e^{\minus{}x}$ on $ [0,\infty).$

$ f(x)^2\plus{}(1\plus{}f'(x))^2\equal{}e^{\minus{}2x}\plus{}(1\minus{}e^{\minus{}x})^2$

$ \equal{}2e^{\minus{}2x}\minus{}2e^{\minus{}x}\plus{}1\equal{}2e^{\minus{}x}(e^{\minus{}x}\minus{}1)\plus{}1\le 1.$

Since $ f'\le 0$, $ f$ is nondecreasing. If $ L: \equal{}\lim_{x\to \plus{}\infty}f(x)\ne 0$, then $ f'<\sqrt{1\minus{}L^2/2}\minus{}1<0$ for sufficiently large $ x$, contradicting the boundedness of $ f$. Same argument applies to $ \lim_{x\to \minus{}\infty}f(x)$.

See a generalization ,  http://www.mathlinks.ro/Forum/viewtopic.php?t=135933 ,  and try to solve it ..  :D
 :cool:..


[quote="Pain rinnegan"][quote="leedt26"][quote="leedt26"]1.Let $ a,b,c \ge0$ .Prove that
$ \sqrt {\frac {{{a^2} \plus{} bc}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \plus{} \sqrt {\frac {{{b^2} \plus{} ca}}{{{c^2} \plus{} ca \plus{} {a^2}}}} \plus{} \sqrt {\frac {{{c^2} \plus{} ab}}{{{a^2} \plus{} ab \plus{} {b^2}}}} \ge \sqrt 6$
2.Let $ a,b,c\ge0$ .Prove that
$ \sqrt {\frac {{{b^2} \plus{} bc \plus{} {c^2}}}{{{a^2} \plus{} bc}}} \plus{} \sqrt {\frac {{{c^2} \plus{} ca \plus{} {a^2}}}{{{b^2} \plus{} ca}}} \plus{} \sqrt {\frac {{{a^2} \plus{} ab \plus{} {b^2}}}{{{c^2} \plus{} ab}}} \ge \frac {{3\sqrt 6 }}{2}$[/quote]
Can someone help me with double inequality?[/quote]

I think the second inequality is by Nguyen Van Thach, see [url=http://www.mathlinks.ro/viewtopic.php?t=195427]here[/url][/quote]
Thank you,Pain rinnegan . Now I known that it's been unsolved a long time ago.

[quote="leedt26"]1.Let $ a,b,c \ge0$ .Prove that
$ \sqrt {\frac {{{a^2} \plus{} bc}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \plus{} \sqrt {\frac {{{b^2} \plus{} ca}}{{{c^2} \plus{} ca \plus{} {a^2}}}} \plus{} \sqrt {\frac {{{c^2} \plus{} ab}}{{{a^2} \plus{} ab \plus{} {b^2}}}} \ge \sqrt 6$
2.Let $ a,b,c\ge0$ .Prove that
$ \sqrt {\frac {{{b^2} \plus{} bc \plus{} {c^2}}}{{{a^2} \plus{} bc}}} \plus{} \sqrt {\frac {{{c^2} \plus{} ca \plus{} {a^2}}}{{{b^2} \plus{} ca}}} \plus{} \sqrt {\frac {{{a^2} \plus{} ab \plus{} {b^2}}}{{{c^2} \plus{} ab}}} \ge \frac {{3\sqrt 6 }}{2}$[/quote]
We can prove 1. by this inequality:
$ \sum\frac{\sqrt{a^2\plus{}bc}}{b\plus{}c}\ge\sqrt{\frac{3}{2}}.\frac{a\plus{}b\plus{}c}{\sqrt{ab\plus{}bc\plus{}ca}}$

The first is [url=http://www.mathlinks.ro/viewtopic.php?t=202843]here[/url] and [url=http://www.mathlinks.ro/viewtopic.php?t=219237]here[/url], but no nice solution up to now :(


algebra
polynomial
calculus
integration
modular arithmetic
algebra proposed

Let $n$ be an integer. Prove that the polynomial $f(x)$ has at most one zero, where \[ f(x) = x^4 - 1994 x^3 + (1993+n)x^2 - 11x + n .  \]
[i]Greece[/i]

Warning : at most one INTEGER root
:cool:

I have a weird brute force solution.
Let the (complex) roots of the original polynomial be a,b,c, and d, suppose a and b are the integral roots, then
$ a\plus{}b\plus{}c\plus{}d\equal{}1994$
$ ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\equal{}n\plus{}1993\equal{}ab\plus{}(a\plus{}b)(c\plus{}d)\plus{}cd$
$ abc\plus{}abd\plus{}acd\plus{}bcd\equal{}11\equal{}ab(c\plus{}d)\plus{}cd(a\plus{}b)$
$ abcd\equal{}n$
We can solve for $ c\plus{}d$ in the first equation and then for $ cd$ in the second after substituting for $ n$. Then substitute everything in the third equation to get
$ 11\equal{}a^3\plus{}b^3\minus{}a^3b\minus{}ab^3\minus{}2a^2b^2\minus{}1994a^2\minus{}1994b^2\minus{}1994ab\plus{}1993a\plus{}1993b\plus{}1995a^2b\plus{}1995ab^2$
A contradiction, as the right side is always even (check every possible parity of a and b).
Is this right?

This is a funny problem. So, we think about parity. If $n$ is odd, you can easily check that it is always odd, so no roots. If $n$ is even, reduce over $\mathbb{F}_2$ (pretty much just $\pmod{2}$). We get the new polynomial $x(x^3+x-1)$. But the inside polynomial is irreducible, since it is a cubic and neither $0$ or $1$ is a root. So we can't have another integer root, so we're done. (This can all be easily translated into dumb parity arguments)


[quote="sri340"]if  2/3 . 3/4 . 4/5 . .... . n/n+1  , what is the sum of the numerator and denominator of the largest fraction on the left side of the equation?[/quote]

Most of the numbers cancel out except for 2, and (n+1).
So at the end you get:
2/(n+1).

So the sum is (n+3)

[quote="sri340"]if  2/3 . 3/4 . 4/5 . .... . n/n+1  , what is the sum of the numerator and denominator of the largest fraction on the left side of the equation?[/quote]

sri340, again, you need grouping symbols.

It's n/(n + 1).  As is, the answer should be 2n + 1.
It seems you did not ask the right question.

[quote="percy314"]
Most of the numbers cancel out except for 2, and (n+1).
So at the end you get:
2/(n+1).
So the sum is (n+3)
[/quote]

percy314, your answer involves *multiplying* the fractions,
not adding quantities together.

When sri is putting a dot, he means multiplication, not addition.

I think the problem just needs to be cleared up or something. He says "the sum," but puts dots. And he says "left side of the equation", but there is no equation.

The sum refers to the numerator and denominator.

The left side part makes no sense at all.

When he says "the sum" he is referring to the sum of the numerator and denominator once all those fractions are multiplied into one fraction. By dots, he means multiplication, and by n/n+1, he means $ \frac{n}{n\plus{}1}$. So percy's answer is right.

[quote="Arrange your tan"]I private messaged him (again?) to stop using periods for multiplication.

With the lack of compliance as to typing things correctly (explained
over and over again to the OP), even without using Latex,
there is no blame on forum users who cannot tell what is meant.[/quote]
You know in some countries they actually use the period for multiplication...


[quote="kunny"]Date: February 11, 2009

Work Time : 4 hour

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=259936]Problem 1[/url]

Find all positive integers $ n$ such that $ 8^n \plus{} n$ is devisible by $ 2^n \plus{} n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=259937]Problem 2[/url]

Let $ N$ be postive integer. Some integers are written in a black board and those satisfy the following conditions.

1. Any numbers written are integers which are from 1 to $ N$.

2. More than one integer which is from 1 to $ N$ is written.

3. The sum of numbers written is even.

If we mark $ X$ to some numbers written and mark $ Y$  to all remaining numbers, then prove that we can set the sum of numbers marked $ X$ are equal to that of numbers marked $ Y$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=259938]Problem 3[/url]

Let $ k\geq 2$ be  integer, $ n_1,\ n_2,\ n_3$ be positive integers and $ a_1,\ a_2,\ a_3$ be integers from $ 1,\ 2,\ \cdots ,\ k \minus{} 1$.
Let $ b_i \equal{} a_i\sum_{j \equal{} 0}^{n_i} k^{j}\ (i \equal{} 1,\ 2,\ 3)$. Find all possible pairs of integers $ (n_1,\ n_2,\ n_3)$ such that $ b_1b_2 \equal{} b_3$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=259941]Problem 4[/url]

4. Let $ \Gamma$ be a circumcircle. A circle with center $ O$ touches to line segment $ BC$ at $ P$ and touches the arc $ BC$ of $ \Gamma$ which doesn't have $ A$ at $ Q$. If $ \angle {BAO} \equal{} \angle {CAO}$, then prove that $ \angle {PAO} \equal{} \angle {QAO}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=259945]Problem 5[/url]

Find all functions $ f$, defined on the non negative real numbers and taking non negative real numbers such that $ f(x^2) \plus{} f(y) \equal{} f(x^2 \plus{} y \plus{} xf(4y))$ for any non negative real numbers $ x,\ y$.[/quote]

1.Since (2^n+n)|(8^n+n^3), (2^n+n)|(n^3-n). if n≥10, 2^n+n>n^3-n, and n^3-n is not 0. So it is impossible.

If we try n=1~9, only n=1, 2, 4, 6 suffice the problem.


A punctiform sourse S of light that emits izotropic a flux PHI is thrown with velocity Vo from O in the vertical plane XOY;the angle between Vo and Ox is h ,sin h=2*sqr(2)/3. Fiind illumination(E=d PHI/dA,A-surface) in O(horizontal surface) when OS is maximum.

yes of course. otherwise the problem would be senseless

and do we have to count with repelling force?


We will be offering a special AIME Problem Seminar during the weekend before the 2006 AIME.

The seminar will take place on Saturday March 4 and Sunday March 5, from 3:30-6:30 PM Eastern both days.  These times include a 1/2-hour break each day, so there is a total of 5 hours of instruction.

The seminar will consist of many past AIME problems along with some general discussion of test-taking strategy.  Note: much of the material is from our AIME Problem Series A, which was last offered in summer 2005.  In particular, if you are currently taking the AIME Problem Series B, then the material in this seminar will be different.

The cost for this special seminar is $\$50$.

[url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class1]Click here to enroll![/url]

[quote="towersfreak2006"]Will there also be another mock AIME available?[/quote]No, we're not going to have one.  There are undoubtedly many mock AIME's on the message board.

what exactly is taught in the seminar?  is it like a more general version of the problem series?  how have people who have taken it before thought of it?  would it be just as worthwile to spend the time doing practice aime's or does this yield some sort of further advantage?

[quote="1234567890"]So if we've taken AIME PS A, it would not help us much to take this?[/quote]The material in the special seminar is mostly a subset of the material from the AIME A class.  If you took the AIME A class last summer, then most of the seminar material will be a repeat.  Although there is some value in seeing the same material twice, it is not something that we would specifically recommend.

[quote="ffdbzathf"]what exactly is taught in the seminar?[/quote]Mostly practice problems from past AIME (with detailed solutions and, more importantly, strategies at how you would arrive at the solution), along with some disucssion of more general test-taking strategy.
[quote]is it like a more general version of the problem series?[/quote]It's (mostly) a subset of the AIME Problem Series A course -- we will pick and choose selections from the AIME A lectures.
[quote]how have people who have taken it before thought of it?  would it be just as worthwile to spend the time doing practice aime's or does this yield some sort of further advantage?[/quote]Hopefully people who took it last year can address this.

[quote="Calculus"]Why not use the same mock aime as last year. It would be better than nothing. And it would help people like me that didn't take the seminar last year.[/quote]

Calculus, I don't know if last year's mock AIME will be helpful to you. I think highest person got like 7 and this includes people who get nearly 14-15 in real AIME. Administrators used some evil problems like #15 was from Chinese Math Olympiad. So, don't worry about last year's mock AIME.

[quote]how have people who have taken it before thought of it? would it be just as worthwile to spend the time doing practice aime's or does this yield some sort of further advantage?[/quote]

It's really difficult to answer this question. If you solved many AIME problems, you might recognize some problems so it might not be helpful to you much. On the other hand, I didn't know many AIME problems last year so it helped me a lot last year. Again, this is just like AIME problem series. Your score and ability will depend on how much time and effort you put INSIDE and OUTSIDE of the class.

[quote="lowest_reynolds"]Will there be transcripts of the classes?[/quote]

Yes.

[quote="Calculus"]For someone that has prepared and done all past aimes, how helpful would this class be?[/quote]

The focus, as in all our classes, is not the individual problems/solutions, but the methods and motivations that more generally lead us there.  There are some past AIME Math Jams that might be worth looking at in order to see how much this might be helpful to you.

[quote="MaThWhIz2004"]Is the material the same as that of the seminar given last year?

Thanks. =)[/quote]No -- the material is new (to the seminar).  The material is from the AIME Problem Series A class.


calculus
integration
calculus computations

[color=darkred]Prove that for any even natural number $ n$ and any real numbers $ \{\begin{array}{c} a < b < c \\
 \\
a + c = 2b\end{array}$,

$ \int_a^c(x - a)^n(x - b)^n(x - c)^n\ \mathrm {dx} =$ $ 2\cdot (\frac {c - a}{2})^{3n + 1}\cdot\int_0^1x^n(x^2 - 1)^n\ \mathrm {dx}\ .$[/color]

Sorry, I corrected : must the condition $ a \plus{} c \equal{} 2b\ !$


Given a triangle ABC, find max of:
\[ P \equal{} (1 \plus{} sin^2(A))(1 \plus{} sin^2(B))(1 \plus{} sin^2(C))
\]
Also, prove that P>4 when ABC is an acute or a right triangle.

I think $ P>4$ is only for acute and right triangles.

Thanks for pointing that out. I have revised the original question accordingly.


trigonometry
complex analysis
complex analysis unsolved

Find the residue at $ z \equal{} 0$ of $ f(z)\equal{} \frac{cos(1/z)}{sin z}$

I recommend you find the $ a_{\minus{}1}$ term of the Laurent series.  Assume $ f(z)\equal{}\cdots\plus{}a_{\minus{}2}z^{\minus{}2}\plus{}a_{\minus{}1}z^{\minus{}1}\plus{}a_{0}\plus{}a_{1}z\plus{}a_{2}z^{2}\plus{}\cdots$.  Now multiply through my the Laurent (Taylor) series for sine:
$ (\cdots\plus{}\frac{a_{\minus{}2}}{z^{2}}\plus{}\frac{a_{\minus{}1}}{z}\plus{}a_{0}\plus{}a_{1}z\plus{}a_{2}z^{2}\plus{}\cdots)(z\minus{}\frac{z^{3}}{3!}\plus{}\frac{z^{5}}{5!}\minus{}\cdots)\equal{}1\minus{}\frac{1}{2!z^{2}}\plus{}\frac{1}{4!z^{4}}\minus{}\cdots$
Now just carry through multiplication and match coefficients.


[quote="funcia"][quote="bpms"][quote="funcia"]find the sum:

1/(1x3)+1/(3x5)+1/(5x7)+...+1/(199x201)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7751

it is more important to explain the answer in this case then to state it[/quote]
are you familiar with telescoping/partial fraction decompositions?
[hide]
$\frac{1}{(n+1)(n-1)}=\frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})$[/hide][/quote]

No, explain if you want to...I have to go to bed now.  :([/quote]
What I posted is a hint, rewrite the question in terms of that and see what you get. It's really cool.

[quote="bpms"][quote="funcia"]find the sum:

1/(1x3)+1/(3x5)+1/(5x7)+...+1/(199x201)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7751

it is more important to explain the answer in this case then to state it[/quote]
are you familiar with telescoping/partial fraction decompositions?
[hide]
$\frac{1}{(n+1)(n-1)}=\frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})$[/hide][/quote]

so with this ummm

1/2(1/1-1/3)+1/2(1/3-1/5) so we would get
2/gcf((n-1),(n+1))and add all of those from value of n=2 to n=100 but I don't see how to do that

Your first step is right. Take the 1/2 out of all the fractions and you're left with $\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}$... almost all of the fractions cancel.

Geometry? What does that have to do with it?

In partial fraction decomposition, take the denominator, here $(n+1)(n-1)$ and try to split into $\frac{K}{n-1}-\frac{M}{n+2}$ for some integers $K$ and $M$.

[quote="myspacesucks"]Just knowing the formulas isn't enough, you need to know when to apply them to the real world and problems like these, just like in geometry.[/quote]
Also, did anyone even post a formula?
@leet I think you made a typo


logarithms
function
inequalities
inequalities unsolved

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=36610
http://www.mathlinks.ro/Forum/viewtopic.php?t=29109

  darij

[quote="t0rajir0u"]Edit:  Okay, Karamata's doesn't really work.  :D[/quote]

It works..... :D     . See at the second link that darij gave and see myth's solution :lol:

[quote="othman"]i m sorry but karamata doenst work because we shold have $b_{1} \ge b_{2},...,\ge b_{n}$

in this problem we have only $a_{1} \ge a_{2},...,\ge a _{n}$ :([/quote]

That was my problem with it, too.  However, I think we can show that $\sum b_k$ is minimized for $b_i$ properly sorted in order for Karamata's to work...

can u show it ?
i want  a proof in details please :)

Let's sort $b_i$ and call the new sequence $c_i$.

[b]Lemma:[/b]  $\sum_{i=1}^{k} c_i \ge \sum_{i=1}^{k} b_i \, \forall k$

Proof:  $c_i$ is sorted such that $c_1 \ge c_2 \ge c_3 \ge ...$.  $c_1$ is the maximum of $b_i$, hence $c_1 \ge b_1$.  $c_1 + c_2$ is the sum of the two maximal elements of $b_i$, hence $c_1 + c_2 \ge b_1 + b_2$, etc.  The result follows.

[b]Lemma:[/b]  $\prod_{i=1}^{k} c_i \ge \prod_{i=1}^{k} b_i \, \forall k$

Proof:  Take $p_i = \ln c_i, q_i = \ln b_i$, and apply the previous lemma.

By our lemmas, sorting $b_i$ does not change whether the sequence satisfies the second condition.  Hence we can WLOG sort $b_i$ and then Karamata's finishes the problem.  QED.  :)

[quote="Lovasz"]Let's define the function $y=\to e^x$. Since $\frac{d^2y}{d^2x}=e^x>0\ \ ,y=e^x$ is [b]convex[/b] function.
In first conditon, we replace $a_i$ with $e^{a'_i}$
$a_1 \ge a_2 \ge \cdots a_n \iff e^{a'_1} \ge e^{a'_2} \ge \cdots e^{a'_n} \iff \boxed{a'_1 \ge a'_2 \ge \cdots a'_n}$
Second one, make similar substitution:
$b_1b_2 \ldots b_k \ge a_1a_2 \ldots a_k \iff e^{b'_1+b'_2+\cdots+b'_k} \ge e^{a'_1+a'_2+\cdots+a'_k} \iff \boxed{b'_1+b'_2+\cdots+b'_k \ge a'_1+a'_2+\cdots+a'_k}$
Now we must prove:
$e^{b'_1}+e^{b'_2}+\cdots+e^{b'_n} \ge e^{a'_1}+e^{a'_2}+\cdots+e^{a'_n}$
This is immediately followed by Majorization inequality.

Is it good ?[/quote]

The second condition isn't guaranteed to hold when $b_k$ is sorted, this must be proven or else Majorization inequality is inapplicable!


function
search
number theory unsolved
number theory

Show that there exist function $ f: \mathbb N \to \mathbb N$ such that $ f(f(n)) = n^{2}$.

[quote="Leonhard Euler"]Show that there exist function $ f: n\to n$ such that $ f(f(n)) \equal{} n^{2}$[/quote]

Let $ A$ be the set of positive integers which are not perfect squares.
Let $ B$ and $ C$ two disjoint infinite subsets of $ A$ such that $ A\equal{}B\cup C$ and  $ h(n)$ a bijective function from $ B$ to $ C$.
Let $ k(n)$ the function which gives, for any integer $ n>1$, the littlest integer $ m$ such that it exists an integer $ p\geq 0$ such that $ n\equal{}m^{2^{p}}$.
Let $ e(n)$ the integer $ p\geq 0$ such that $ n\equal{}k(n)^{2^{p}}$
Obviously, $ k(n)$ is not a perfect square and either $ k(n)\in B$, either $ k(n)\in C$

Then we can define $ f(n)$ :
$ f(1)\equal{}1$
$ \forall n>1$ :
if $ k(n)\in B$, $ f(n)\equal{}(h(k(n)))^{2^{e(n)}}$
if $ k(n)\in C$, $ f(n)\equal{}(h^{\minus{}1}(k(n)))^{2^{e(n)\plus{}1}}$


$O=(0;0)$

"Connect ABC and give each corner the following coordinates:

$A(a_1;a_2)$, $B(b_1;b_2)$, $C(c_1;c_2)$

The middle between BC is named M
Find the coordinates for $\vec{OA}+\frac{2}{3}\vec{AM}$"

So i write:
$\vec{OA}+\frac{2}{3}\vec{AM}=\left(\frac{a_1}{a_2}\right)+\left(\frac{\frac{2}{3}\cdot (a_1-0.5(c_1-b_1))}{\frac{2}{3}\cdot (a_2-0.5(c_2-b_2))}\right)$

Does this sound right to you?

If yes, is it possible to show the following without drawings?

$\vec{OA}+\frac{2}{3}\vec{AM}=\vec{OB}+\frac{2}{3}\vec{BM}=\vec{OC}+\frac{2}{3}\vec{CM}$

...where $M$ is a new spot in the middle between the opporsite corners.

wait how exactly did you get:

$\vec{OA}+\frac{2}{3}\vec{AM}=\left(\frac{a_1}{a_2}\right)+\left(\frac{\frac{2}{3}\cdot (a_1-0.5(c_1-b_1))}{\frac{2}{3}\cdot (a_2-0.5(c_2-b_2))}\right)$

??

I thought the coordinates would look more like...

make a line that goes through A and that is parallel to the x-axis and make that angle $\theta$, then we have the x-component of $\vec{AM}$ is $AM \cdot \cos \theta = \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2} \cdot \cos \theta$

also, since $O$ is already on the x-axis let the angle between $\vec{OA}$ and the x-axis be $\gamma$ and we have the x-component of that is $OA \cdot \cos \gamma = \sqrt{a_1^2 + a_2^2} \cdot \cos\gamma$

so, adding these components we get:

$<< \cos \gamma \sqrt{a_1^2 + a_2^2} + \frac{2}{3} \cdot \cos \theta \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2} , \sin \gamma \sqrt{a_1^2 + a_2^2} + \frac{2}{3} \cdot \sin \theta \sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2}>>$

yea i have a feeling i could be hugely wrong... but yea

Hum, i am not really sure how you did that.

What does the number actually indicate when you do a $\vec{a}\cdot \vec{b}$. We call it the skalarproduct, but i don't know what this indicates. If it was $\sqrt{a^2+b^2}$ we get pythagoras for the lenght of $c$. But but does the number indicate when we say $\sqrt{a^2+b^2}\cdot \cos{v}$

You have a scalar (we use a "c," not a "k" for that word in English) value equal to a vector -- perhaps you should reconsider that?
In fact, for this problem you probably don't want to break it into coordinates until the very end -- try putting everything in terms of vectors of the form $\vec{OX}$ instead.

If you think about what the later equality means, by the way, you'll see it's equivalent to the fact that the medians are concurrent and trisect each other.

Well, if we have two vectors with a lenght of 5 and 6 and the angle between them is 20, then:

$|a|\cdot |b| \cdot \cos(v)=5\cdot 6 \cdot \cos(20)=12.24$ What does the 12.24 mean? Is it anything that can be used for anything of interest?

Ok, so now i drew it and i found it medians obviously always intersect each others at $\frac{2}{3}$. I guess it is ok to prove this with words rather than showing it with vectors.

So how would you prove by words that they will always intersect at $\frac{2}{3}$? First of, the $\vec{OA}$, $\vec{OB}$ and $\vec{OC}$, isn't important since it is just another way to explain that we start vector at $A$, $B$ and $C$. I get that, but i can't explain how it can be that they intersect at exactly $\frac{2}{3}$.

[hide]$\overrightarrow{OM} = \frac12 (\overrightarrow{OB} + \overrightarrow{OC})$ since $M$ is de midpoint of $\overline{BC}$. Then:

$\overrightarrow{OM} = \left(\frac12 (b_1+c_1); \frac12 (b_2+c_2) \right)$

And...

$\overrightarrow{AM} = \overrightarrow{OM} - \overrightarrow{OA}$
$\overrightarrow{AM} = \left(\frac12 (b_1+c_1)-a_1; \frac12 (b_2+c_2) -a_2 \right)$

Therefore $\frac23 \overrightarrow{AM} = \left(\frac13 (b_1+c_1)-\frac23 a_1; \frac13 (b_2+c_2) - \frac23 a_2 \right)$

Hence
$\overrightarrow{OA} + \frac23 \overrightarrow{AM} = \left(\frac13(a_1+b_1+c_1);\frac13(a_2+b_2+c_2)\right)$
$\overrightarrow{OA} + \frac23 \overrightarrow{AM} = \frac13 (\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC})$

The result is the same for the other points... It's the center of mass of triangle ABC.[/hide]

Ah finally. It took me some time to get your point (not that i have wondered all the time since i first read it the same day you posted it, but after a busy couple of weeks i now found time to take a look at it - still took quite a nice handful of minutes to figure out what you meant :) ). Thx for your help.


just askin'. if you do, which brand, operator? do you use wap?

I use a Samsung C100 with Orange, and I sometimes use wap.

Alcatel O512
and
Philips 9@9++ (it seems that this one doesn't sell in Europe, and it is said that the SAR value is 1.14W -- that's high -- but the stand-by time can reach up to about 13 days)

Ericsson T28, Connex. The standby time is really lame! It's a bit over 2 days :D
50 hours or so...

please specify also if you use or not (and if you do how much time) wap services.

grobber - my samsung c100 is supposed to have the stand-by time of 90 hours, that is almost 4 days. Untill now, two days is at most that he could take :)

I use GPRS instead of WAP. Pretty frequently to browse the web, chat etc.

Edit: Pretty frequently being daily, maybe 30 minutes a day.

I haven't apply for WAP or GPRS although my mobile telephone supports both of them and it also supports receiving/sending e-mail.

wait a moment. GPRS is a connection protocol. WAP is wireless application protocol. I access WAP through GPRS.

also normal internet can be accessed through GPRS if the phone has an internal browser (just like dial-up), or even connect your computer to it.

So GPRS vs Old mobile WAP is pretty much the same for WAP as cable vs dial-up is for computers.

I find it amusing, but yet not so usefull, as the information on the wap sites is less updated then on www.

Nokia 3410.
Connex.
No wap yet. I don't need it yet.

I use Eriksson R380S. I haven't apply for WAP or GPRS although my mobile telephone supports both of them. I dont like cell phone, but it is my younger brother gave me (to catching me easier!)

Namdung

[quote="Namdung"]I dont like cell phone, but it is my younger brother gave me (to catching me easier!)[/quote]

I know that feeling :D My dad's working overseas, so I get to keep his cellphone, and I'm trying to "dump" it as soon as possible because I don't like cellphones, but there's always some event which I have to use it for :(

The best time of the day: telling somebody (whom I don't want to let that person know my cellphone number) that it belongs to my dad actually. :)

great news everyone ... the wap version of the site is now available. Just type [code]http://www.mathlinks.ro/phpBB/wap/index.php[/code] in your WAP enabled phone and you will access the WAP portal.

you can read the posts / check your private messages / login and post new ones.

Out of curiosity...why does the WAP version have the text
[quote]MathLinks EveryOne
the Mathlinks Forum - Please try to use English language as much as possible - Thank You and Enjoy! Valentin Vornicu[/quote]
on the index page which the normal version doesn't have?

it's because you have accessed the wap site through a html browser, not by cell phone. That what you saw is the phpBB2 shell of the wap version, and it uses the database !SITENAME! variable there, which is set out to be what you quoted (that's basically how the site looked a year ago).

oh, and quote that should have been posted on the main page, but I forgot about it, nice of you to remember it to me :D:D

try to access it by wap, I did and it works fine.

[quote="Valiowk"]Do you actually think that I would have WAP? (Read above post :D )

That's pretty much what I figured, only when you've never used WAP before it's not wise to be guessing blindly ;)[/quote]

Funny! :D  :D  :D

I have Alcatel 715. This is the only phone I have, so I connect to INEt only by GPRS. :)

okay this is a message sent through wap. let's see if it reaches its destination ..

Nec 341i,E-Plus.I use i-mode (similar to wap)

i've used the wap module a lot since I have put it online. unfortunately as of this moment there is no way to read your pm-s :)

Nokia 3310.
Really old when everyone is using mobile phone to take photos now  :)

I can't take photos, however I can store and view them on my phone :D

[quote="blondu"]What about an old phone?

Sagem MC820: no WAP, GPRS, photos... :D
Connex[/quote] old? :D you mean antique :P  (just kiddin').

You're right, quite antique, but still workin'  :D  :D

But then those people back in 2004 don't know about iPhones.

I don't have a phone.
Yet


geometry
rectangle
3D geometry
geometry unsolved

A parallelepiped is cut by a plane along a 6-gon. Supposed this 6-gon can be put into a certain rectangle $ \pi$ (which means one can put the rectangle $ \pi$ on the parallelepiped's plane such that the 6-gon is completely covered by the rectangle). Show that one also can put one of the parallelepiped' faces into the rectangle $ \pi.$

I think the following lemma is correct:
If we have a hexagon with opposite parallel sides contained in a rectangle R,
there is a rectangle congruent to R containing it with one side parallel to one of hexagon's sides.
I'm 90% sure it's true, and if it is true, it solves the problem

I've got a book, all the problems of ARMO 2004 are in it, but, orl gave a parallelepiped, in my book, that's not a parallelepiped but a cuboid. I don't know orl and my book which is right, could anyone give any advice? thank you!


What did everyone else get??

I got 84, which is 6 less than what i was aiming for . . -_-

I got 92, an improvement from last year.
By the looks of this pole there will be a lot of perfects. Congratulations! :D

only 4 perfects on the pole.
BTW do you have any suggestions of what age you should be to write this contest?

If you are what your teacher considers to be good at math it really doesn' tmatter what grade you are in.

But usually it is 6th Graders or higher.

ok, thanks. I'm going to be in grade 9 next year so I'll try it.

BC published the results at:
http://www.math.ubc.ca/Schools/Euclid/B.C.Euclid_results_2008.pdf

I wonder if other provinces also did the same (published the results)

[quote="nirvanatear"]I got 86, not even as good as two years ago=([/quote]
86 is pretty good, many school champs scored much lower than that!
at least, you are probably the highest one in your school right?


algorithm
number theory unsolved
number theory

Consider the following game: determine if P1 or P2 has a winning strategy (i think P2 has the strategy in which the case the problem is difficult since P2 has to account for every possible number P1 originally plays. if P1 has the strategy, then it would be a simple case bash but i dont think P1 has the strategy).
Each player is allowed to write positive integers on a piece of paper, with P1 starting; with the following rule in place: If the numbers a and b have already been written on the paper then all numbers of the form ax+by cannot be written for x,y>0. obviously some player must be forced to write down 1 (thats left as an exercise to the reader ..). this player loses.

Actually I've thought about the problem bit more, solved it and it's pretty easy though i'll wait for somebody else to solve it [hint: When P1 plays a prime at least 5, he's guaranteed to win].
A more involved problem would be the following: for a given number n, find a criteria which decides if when P1 starts off with n, P1 has the winning strategy or P2 has the winning strategy. I proved that if n is prime,  P1 has a strategy .. a more general answer would be appreciated because its quite a difficult problem now.
Another more involved problem would be defining an algorithm which allows P1 to win for n prime .. i only proved existance of a strategy and not an algorithm. look below for some hints about such an algorithm. [hidden though].
[hide] say P1 plays a prime p. Say P2 plays q with (q,p)=1. Now P1 plays pq-p-q, and i havent proven it but I'm dead sure that this the only number that P1 could possibly have played second to guarantee a win for P1 is pq-p-q; help in proving that would also be appreciated. [/hide]


I registered for the WOOT program in July but hadn't seen any change when clicking the "classroom" button on the left hand side. Also, I don't know where the WOOT forum is, where I'm supposed to find POTD's?

We actually do not show you as registered for WOOT.  Please contact our office at 619-659-1612 or at classes@artofproblemsolving.com so that we can investigate further.  Please let us know what date you registered for WOOT (approximately if you don't recall the exact date), the form of payment (or the fact that you were a MOSP student), and the email address that you used at the time of registration.


Prove that, for every positive integer k, the set {1, 2, 3, ..., 2k} can be partitioned into k pairs such that the sum of each pair of numbers is prime.

For example, if k = 4 :
1 + 6 = 7
2 + 3 = 5
4 + 7 = 11
5 + 8 = 13.

Do you know a proof Belenos ? :) I just love this one.

Arne wrote:Do you know a proof Belenos ?

No. Not yet

Btw, I'm a bit confused.

[hide]

This result implies Bertrand's postulate (which should be named Bertrand-Tchebyshev theorem !) for even numbers.

So either it uses Bertrand's postulate (which is not really basic knowledge  ) either there is a short proof of Bertrand postulate for even numbers (which I doubt) ![/hide]

Any clue Arne ?

It uses indeed [hide]Bertrand's Postulate[/hide].

Well in fact it's not too difficult if you know [hide]Bertrand's postulate[/hide].

Maybe a further hint :

[hide]Why wouldn't you use induction [/hide]?

So what is [hide]Bertrand's Postulate?[/hide]

``Chebyshev said it, and I'll say it again:
There's always a prime between n and 2n.''

Now you'll never forget it.

[hide]Obviously pigeon-hole was not the right idea

Usually induction doesn't work well with prime numbers. It is an exception.

. The result is true for n = 1 : 3 is prime !

. Let's suppose it is true for all k < n (strong induction hypothesis).

Bertrand's postulate : it's possible to find a such as 2n+a is prime and a < 2n.

Then a is odd, and there is an even number of integers in [a+1 .. 2n-1] (eventually 0).

Those integers can be paired together to sum to 2n+a, which is prime.

Then apply induction hypothesis to [1 .. a-1] (a-1 is even <= 2(n-1)) to finish the proof.

The solution is not necessarily unique :

for n = 3 : 1,6 - 2,5 - 3,4  or 6,5 - 4,1 - 3,2

Yes it is not so difficult for such an amazing result. IF you think of

Bertrand's postulate, and if you use induction. Without these hints it's probably very difficult ....[/hide]


What is the reason for studying math? The amount of time needed to become a great problem solver is tremendous and the benefits are low. Colleges care about what your GPA and your community service and your SAT are but will they accept you because of your mediocre performance in AIME? Unless you make IMO or MOSP or something major, your math accomplishments will just be touched upon to serve as a "tie-breaker" or something very miniscule.
     I have studied math for about a year. I haven't found a purpose. I have gotten better but not enough to get me USAMO or maybe even AIME. Why should I continue to study math?  Should pursuit of mathematical knowledge come before what college accepts me? Most  of the things I study won't even come handy in my life! What leads you guys to continue to study math?
     I would especially like to hear the responses of admins and the more experienced AOPS'ers

Please AOPS, help me.

Good question.

  First I would like to say Mathematics or rather "problem solving" is very much indeed a transcending skill in life, which can be applied to almost all professions and careers.  So even if you end up moving toward a career that has absolutely nothing to do with math, chances are your problem solving skills will help you along the way.

  I personally pursue mathmematics because of the aesthetic beauty, and pleasure of improving and mastering topics I had once found difficult.  There are so many intriguing ideas in Mathematics its just amazing, I feel it's Man's greatest invention.  Although contest mathematics may seem trivial at some stages, they are just stepping stones for the enjoyment that lies ahead.

Wow, I have the same question in my mind.

Well, my elementary math teacher helped me to discover that I like math, the AoPS teachers here helped me to realize that I love math... and I do appreciate math for its "aesthetic beauty" like what G-unit said... although I only discover the beauty of elementary math (those in math circles and AoPS vol 1) so imagine what advanced math would be like... they would be gorgeous in my opinion. Not everyone can get use to doing math for fun.

Problem solving gives me an unusually elated happiness, feelings that I can only experience after having math sessions with joejia or wdragyn... and that makes it priceless, I think, after being depressed for a long time.

I would try to balance between college admission and math... afterall, they do complement each other well since they are not totally opposites. Follow your instincts... that's what I usually do, to survive almost all the pop quizzes :lol: .

For me, it's something more for pleasure than for personal gain...
Whats that feeling you get after you solve an especially tricky problem?

It's the greatest in the world.

math is everywhere! :D
why'd you join this forum anyway?

[quote="daveed"]
Whats that feeling you get after you solve an especially tricky problem?

It's the greatest in the world.[/quote]

absolutely. couldn't have said it better. it all depends on your reason for studying math. I, like many AoPSers, love math - there's no other way to explain it. we don't do math to improve our college application or because the math team was the popular club to be in (imagine that  :P )

Ah, it's fun. Some people watch TV, other people... do math. :D

What's more relaxing than a nice elegant math problem after working on tedious dull school work for 3 hours? IT's like a jacuzee...

I think that the purpose of mathematics is itself. Maybe we won't never find a satisfactory answer to a question like "Why math?", because I think it's the same kind of question of "What's the purpose of our life?" etc :)

I'm agree with white_horse_king88. Some people paints, other make sculpture, other make poetry,... and finally, other do mathematics. I think that math can be perfectly considered as a manifestation of art. Math looks for beauty, and Bertrand Russell reminds us this:

"Mathematics, rightly viewed, posseses not only truth, but supreme beauty-a beauty cold and austere, like that of sculpture."

My opinion is that don't stop to make this sort of questions-you may not find a good answer. Just keep going!!! :lol:

When I was in college, many years ago, I was told that I would only directly use about 2% of what I was learning.  The rest would only be useful by analogy or as the basis for further work.

This is also true for Math.  Math is a logic driven system and learning how to move about in a math world will help one logically examine many other issues.  Understanding the basis of statistics, calculus and other mathematical modeling techniques will allow you to understand the types of limits that are likely and the significance one should attach to something that is "proven" or how can two studies "show" exactly the opposite thing.

I'll tell you years after finishing college, I've not regretted the time I've invested in math.  It has served me well, in many unexpected ways.


For each positive integer n, let $a_{n}$=$\frac{(n+9)!}{(n-1)!}$ Let k denote the smallest positive integer for which the rightmost nonzero digit of $a_{k}$ is odd. The rightmost nonzero digit of $a_{k}$ is

A. 1
B. 3
C. 5
D. 7
E. 9

I wouldn't be surprised if I messed up somewhere in here, but this is my attempt.

[hide="Answer"]For this to be possible, we must have all $2$'s cancel with $5$'s to make $0$'s. In ten consecutive integers, we must have at least 8 factors of $2$, so we must have at least 8 factors of $5$. We always have at least 2 factors of $5$, so we then take $5^7$ as one of the ten integers. Next, we choose $k$ so that we have only 8 factors of $2$ in the ten integers. Therefore, $k=78117$. We factor out each of the used $5$'s and $2$'s, and the units digits of the numbers that we are left with multiply to $7(9)(9)(3)(1)(1)(3)(1)(1)(3)\Rightarrow 9\Rightarrow \boxed{E}$.[/hide]


show that 1999*2*1000-1999 is a perfect square.


geometry
geometric transformation
reflection

$A$ is a point in plane $E$, $AB$ is a ray not lying in $E$, and $AC$ is a ray lying in $E$. Describe the locus of points that $C$ can occupy for $\angle BAC$ to be as large as possible. Do the same for the locus of points that make $\angle BAC$ as small as possible. Prove your answers.

[quote="srulikbd"]C on the line BA for the biggest value to the left of A, and to the right of A for the smallest value. is it right?  :maybe:[/quote]
?? Assuming $A$ is to the "right" of $B$.

Project $AB$ onto plane $\mathcal{E}$. Any point on that ray will give the largest angle. Reflect the line over point $A$ for the smallest angle.


So I was doing a Mildorf Mock AIME and get up to evaluating $ \dfrac{(\omega\minus{}1)^{2004}\plus{}(\omega^2\minus{}1)^{2004}}{3}$ where $ \omega\equal{}e^{2\pi i/3}$. But like how do I evaluate $ \omega\minus{}1$? The guy in the solution multiplies $ \sqrt{3}\cdot \dfrac{\minus{}\sqrt{3}\plus{}i}{2}$ and says WLOG. I neither have an idea where this expression comes from nor do I understand why this must be done WLOG.

Could someone like help me understand this?

Hmm...

So $ \omega\equal{}\cos \frac{2\pi}{3}\plus{}i\sin\frac{2\pi}{3}\equal{}\minus{}\frac{1}{2}\plus{}\frac{\sqrt{3}}{2}i$.

So $ \omega\minus{}1\equal{}\minus{}\frac{3}{2}\plus{}\frac{\sqrt{3}}{2}i$.

Drawing $ \omega\minus{}1$ in the complex plane and dilating by $ 2$ gives a right triangle with legs $ \sqrt{3}$ and $ 3$, so the argument is $ \arctan\frac{\sqrt{3}}{3}\equal{}\frac{5\pi}{6}$, and the magnitude is $ \sqrt{3\plus{}9}\equal{}2\sqrt{3}$.

So $ \omega\minus{}1\equal{}\minus{}\sqrt{3}e^{5\pi i/6}$ (I dilated back down by 2).

Now raising it to the 2004th power is significantly easier. I suspect the solution did something similar but skipped a few steps.

For $ \omega^2\minus{}1$, you can do something similar, or factor into $ (\omega\minus{}1)(\omega\plus{}1)$ to simplify things.

You could just note that $ \omega^3\equal{}1\implies(\omega^2\minus{}1)^{2004}\equal{}(\omega^2\minus{}\omega^3)^{2004}\equal{}\omega^{2\cdot3k}(1\minus{}\omega)^{2004}\equal{}(\omega\minus{}1)^{2004}$ so that we need to find $ \frac23(\omega\minus{}1)^{2004}$. Since $ \omega\minus{}1\equal{}\minus{}\sqrt3e^{\minus{}\pi i/6}$, $ (\omega\minus{}1)^{2004}\equal{}3^{1002}e^{\minus{}334\pi i}\equal{}3^{1002}$, and the value of the desired expression is $ 2\cdot3^{1001}$.


[quote="fredbel6"]Maybe he means this ?

[img]http://upload.wikimedia.org/wikipedia/en/f/fd/Winxtrix.jpg[/img]

I doubt anyone can convince me this topic is not  :spam:[/quote]

Please stay on topic and don't spam in my thread, thanks.

Calling me a spammer.... :|

I am serious, I have no idea what you are talking about then.

[quote="Ignite168"]Definitely not some cartoon of 3 girls or whatever.[/quote]
And would you please tell us what it is?  Or are we somehow not invited to that club?

The picture is from http://upload.wikimedia.org/wikipedia/en/f/fd/Winxtrix.jpg

:D

But I don't see how this topic isn't spam.

perhaps they transfer letters from the one word to the other?

No trix is a yogurt brand which has commericals constantly showing a rabbit trying to get his hand on the yogurt


logarithms
induction
real analysis
real analysis unsolved

Find two monotonically non increasing sequences $ (a_n)$, $ (b_n)$ such that the series $ \sum a_n$ and $ \sum b_n$ are divergent and $ \sum \min (a_n,b_n)$ is convergent.

at firts try to find  positive sequances $ a_{n}$ and $ b_{n}$ such that $ \sum a_{n} \equal{}\sum b_{n} \equal{} \infty$ and $ \sum \min (a_{n},b_{n}) <\infty$
Now the question is how to get monotonic sequace...
it's  easy:
 consider serie $ a_{1} \plus{} \underbrace{\frac{a_{2}}{n_{2}} \plus{} ... \plus{}\frac{a_{2}}{n_{2}}}_{n_{2}} \plus{} ... \plus{} \underbrace{a_{k}/n_{k}\plus{}...\plus{}a_{k}/n_{k}}_{n_{k}}\plus{}...$  (1).

similary for $ b_{n}$
 now consider new sequences
for example $ a'_{n} : \equal{}$ n-th term of the serie (1) (similary for $ b_{n}$)
note that we can choose $ n_{k}$ such that $ a'_{n}$ will be monotonically non increasing .

Thanks Extremal,

Can we characterize all positive non increasing sequence $ (a_n)$ with divergent $ \sum a_n$ such that:

For all positive non increasing sequence $ (b_n)$ with divergent $ \sum b_n$ we have $ \sum \min(a_n,b_n)$ diverges.
[hide=" Hint: "]$ a_n \equal{}1/n$ is one such example[/hide]

The Cauchy condensation trick takes care of this rather quickly. The question becomes: describe all sequences $ a_k$ of positive numbers satisfying $ a_{k+1}\le 2a_k$ such that for every sequence $ b_k$ with the same property, the divergence of $ \sum_k b_k$ implies the divergence of $ \sum_k\min\{a_k,b_k\}$. Now, if $ \inf_k a_k=0$, then we can choose a sequence $ n_k$ with $ n_{k+1}>n_k$ and $ a_{n_k}<2^{-2k}$. Now just put $ b_{n_k+k-\ell}=2^{-\ell}$ for $ 0\le \ell< n_k-n_{k-1}+1$. Then $ \sum_{k}b_k=+\infty$ (there are infinitely many ones in the sequence) but the sum of the minima is bounded by $ \sum_k [\sum_{\ell\ge k}2^{-\ell}+k2^{-2k}\cdot 2^k=\sum_k(k+2)2^{-k}<+\infty$.

On the other hand, every sequence with $ \inf_k a_k>0$ has the desired property. Indeed, if the sum of the minima is finite, then $ \min\{a_k,b_k\}\to 0$, whence $ \min\{a_k,b_k\}=b_k$ for all sufficiently large $ k$ and $ \sum_k b_k<+\infty$.

Thus, the description in the modified setting is $ \inf_k a_k>0$, which translates into $ \inf_k ka_k>0$ in the original setting.

Now something harder: describe all sequences $ a_k$ of positive numbers (not necessarily decreasing) such that if $ b_k$ are decreasing and $ \sum_k b_k=+\infty$, then $ \sum_k \min\{a_k,b_k\}=+\infty$. ;)

I asked this question a couple of years ago on one of the closed forums on this site. Here's an answer from that forum:

[quote="DrKevin"][quote="Kent Merryfield"]
(b) (Very difficult.) Find nonincreasing sequences $ a_n$ and $ b_n$ such that $ \sum_{n \equal{} 1}^{\infty}a_n \equal{} \infty$ and $ \sum_{n \equal{} 1}^{\infty}b_n \equal{} \infty$ but $ \sum_{n \equal{} 1}^{\infty}\min(a_n,b_n)$ converges.
[/quote]

Since $ \sum_{n \equal{} 1}^{\infty}\frac {1}{2^n}$ converges, the goal is to construct $ a_n$ and $ b_n$ so that $ \min(a_n, b_n) \equal{} \frac {1}{2^n}$.  We also want both $ \sum a_n$ and $ \sum b_n$ to diverge, so neither $ a_n$ nor $ b_n$ can be the minimum all the time.  Therefore they have to alternate: each assumes the smaller value (i.e. $ \frac {1}{2^n}$) for a certain number of times, while the other stays the same.  The length of each such "run" should be such that the larger value is repeated enough times to add up to 1.  Then, in each of the two sums $ \sum a_n$ and $ \sum b_n$, there will be infinitely many segments with sum 1, therefore they both diverge to $ \infty$, and yet the series of the minimums is convergent.

Thus we can set up the following:
\[ a_1 \equal{} a_2 \equal{} \frac {1}{2}, b_1 \equal{} \frac {1}{2}, b_2 \equal{} \frac {1}{4} \\
a_n \equal{} \frac {1}{2^n}, b_n \equal{} \frac {1}{8}, \quad n \equal{} 3,\ldots,10 \\
b_n \equal{} \frac {1}{2^n}, a_n \equal{} \frac {1}{2^{11}}, \quad n \equal{} 11,\ldots,2^{11} \plus{} 10 \\
a_n \equal{} \frac {1}{2^n}, b_n \equal{} \frac {1}{2^{2^{11} \plus{} 11}}, \quad n \equal{} 2^{11} \plus{} 11,\ldots,2^{2^{11} \plus{} 11} \plus{} 2^{11} \plus{} 10 \\
 \\
\cdots \cdots \cdots\]
You can see the length of each "run" increases rapidly.  Obviously the solution is not unique.  In fact, we can make $ \{a_n\}$ and $ \{b_n\}$ strictly decreasing ...
[/quote]

fedja we can use Cauchy condensation test in arbitrary $ n$-th times , thus it seems that  the sequance  $ \{ a_{k} \}$ if  for some $ n$ satisfy condition : ${ \inf_{k} a_{k}*k *\ln k *\ln \ln k ... \underbrace{\ln \ln ... \ln k}_{n}} > 0$ has the desired property. is it true ?  :)

Not if after the first application we get a sequence that is no longer decreasing ;)

[quote="fedja"]Now something harder: describe all sequences $ a_k$ of positive numbers (not necessarily decreasing) such that if $ b_k$ are decreasing and $ \sum_k b_k \equal{} \plus{} \infty$, then $ \sum_k \min\{a_k,b_k\} \equal{} \plus{} \infty$. ;)[/quote]
fedja,
Do you know the answer?

Thank you fedja.

[quote="fedja"]Now something harder: describe all sequences $ a_k$ of positive numbers (not necessarily decreasing) such that if $ b_k$ are decreasing and $ \sum_k b_k \equal{} \plus{} \infty$, then $ \sum_k \min\{a_k,b_k\} \equal{} \plus{} \infty$. ;)[/quote]
I think we need to make sequance $ a_{k}$ decreasing. I'm not sure but if we consider new sequance defineg by induction $ a'_{0} \equal{} a_{0}$ and suppose we have already defined it for $ j\leq k$ then if $ a'_{k} \geq a_{k \plus{} 1}$ we put $ a'_{k \plus{} 1} : \equal{} a_{k \plus{} 1}$ and if $ a'_{k} < a_{k \plus{} 1}$ we find mininal positive integer ((( "or..  real positive number  :?: " :ninja: , I mean just define $ a'_{k \plus{} 1}: \equal{} a_{k}$))) $ n_{k \plus{} 1}$ such that $ a_{k} \geq \frac {a_{k \plus{} 1}}{n_{k \plus{} 1}}$ and we put $ a'_{k \plus{} 1}: \equal{} \frac {a_{k \plus{} 1}}{n_{k \plus{} 1}}$  and so on.. Now if the sequance $ a'_{n}$ satisfy condition $ \inf_{n} n a'_{n} > 0$ then  if and only if this sequance have desired property .

If all of this is not true then Now I have no idea  what is the correct answer on this question..  :(

[quote="Fleury"]fedja, Do you know the answer?[/quote]
Yes. It is only slightly more complicated. In the monotone case, we need $ a_k>\frac ck$ with some $ c>0$ on the whole $ \mathbb N$. In the non-monotone case we need the same on a sufficiently large set (I leave it to you to figure out what exactly the word "large" should mean here ;)).

[quote="fedja"]In the monotone case, we need $ a_k > \frac ck$ with some $ c > 0$ on the whole $ \mathbb N$.[/quote]
Note that this fails for DrKevin's example.

[quote="fedja"]Now something harder: describe all sequences $ a_k$ of positive numbers (not necessarily decreasing) such that if $ b_k$ are decreasing and $ \sum_k b_k \equal{} \plus{} \infty$, then $ \sum_k \min\{a_k,b_k\} \equal{} \plus{} \infty$. ;)[/quote]

This is question Q661  in RMS Mai 2009 page 166


Find all solutions in positive integers $a,b,c$ of the equation

\[a^2+b^2+1=abc.\]

look for a problem $a|b^2+1$, $b|a^2+1$ which is posted somewhere in the forum (I remember vinoth giving a solution).

Try this

http://www.mathlinks.ro/Forum/viewtopic.php?t=46767&highlight=

Fair enough - perhaps a more tricky variant then.

If positive integers $a,b,c,d$ satisfy

\[a^2+b^2+c^2+1=abcd\]

determine all possible values of $d$.

Not quite - it's easy to get a sequence of solutions in the same way, but it's not so easy to show that they all terminate at the same value.

The same (or more accurately a similar) result does hold for this problem, but it's more difficult to prove.  After four variables, it breaks down and there is no longer a unique solution.

If you have a complete proof using the same method, I'd like to see it.

Oops, it seem you are right. One of my estimations was false. Sorry.


This is for mine:

I was the father of a son who kept pestering me into letting him enroll at Laval University in [i]Guitar Hero[/i], claiming it has the best undergraduate GH program in the world. I denied it time and time again, saying that many Guitar Hero (read music) graduates crash and burn as musicians and, even then, the R-score needed to get accepted at Laval University in [i]Guitar Hero[/i] was 31...

i dreamed that god gave me a lot of money, but made it come to me in 1000 tons of pennies dropping out of the sky at the speed of sound.... :alien:

I dreamed of sleeping. It was so cool. Too bad that isn't reality. :(

Have you ever dreamed of a dream?
Have you ever dreamed of dreaming of a dream?

[quote="crazychinesemaniac"][quote="isabella2296"]I once dreamed that I was 2nd place at MATHCOUNTS Nationals, and for some reason, that depressed me. Then a laptop bonked me on the head and I died.  :|[/quote]

second always sucks.[/quote]

5th is worse

2nd does not always suck. You can get second place once and first place the other three times, and yet get a 1-star ranking. :rotfl:  :rotfl:  :rotfl:

[quote="crazychinesemaniac"][quote="isabella2296"]I once dreamed that I was 2nd place at MATHCOUNTS Nationals, and for some reason, that depressed me. Then a laptop bonked me on the head and I died.  :|[/quote]

second always sucks.[/quote]

Seconded; being second is always worse than being 3rd. 4th is probably better than 3rd, since you barely made it and thus are proud of barely winning; 5th is again bad..

You know the best dream?  In one of my dreams, I woke up at 11:00 A.M., so I go to my friend's house but no one was there.  I went to all my neighbor's houses but no one was in my neighborhood.  I then called 911, but no one answered.  I then proceeded to call many different countries, but no one ever answered me.  I was pretty much the last guy on the planet.  But you know what?  I found out that I was probably dreaming, so I just took a bunch of food, video games, and Ipods from a nearby Wal-Mart and just went back home and watched TV all day. :thumbup:  :lol:  :thumbup:

I also dreamed that I lived in Florida and that Karlanna Lewis was elected governor - albeit forming a minority government (i.e. three or more candidates won a sizeable amount of counties.) Correct me if I'm wrong but such an occurence would be the first in Florida history as a State.

I have also dreamed about dreaming inside of a dream...

Well I dreamt that I was dreaming about this guy who went to bed and had a dream where everyone in his town was dreaming, and for some reason that startled him, and he ran around yelling "WAKE UP! WAKE UP!", and no one woke up, then he tripped into this black hole, where he was killed, then I woke up (in my dream) and figured that since no one was around and this was extremely weird, because a lot of weird things were happening, for example a tornado was flying upside down, and there was a ring of fire around me that was cold, and the floor was made of fire that didn't burn anything, so I figured I was dreaming, told myself to wake up 50 times, didn't wake up, so I went to sleep in my dream and then it got really weird.

 :P  This is actually true

I have had two weird dreams:

1. For some reason I was flying a plane for the first time and I had never flown before. There were like 50 pilots staring at me and about a million passengers on board. Somehow, we crash landed on the freeway near my local Costco... Then, FBI agents came for me in these police vans which then I climbed on a helicopter and watched them. Soon, there was a criminal breakout and there were people running all over the highway. Then I woke up...

The weird thing about the dream above is it repeats itself, except with different teachers. I even dreamed once that one of the FBI agents was my fifth grade teacher... scary.

2. I was walking along and suddenly, my mom knocked over a huge glass of orange juice, then BAM I stuck out my finger and somehow everything froze... I started slurping up the individual droplets of orange juice....

The extra weird thing was that it kept on repeating it self. I would walk, orange juice would fall then everything would rewind...

the weirdest dream i had was when i dreamed about math......

after a math tournament

and math practice......

i was counting everything, and putting them into equations, ...........it was scary. almost

and once, i played N too long, and dreamed i was a  :ninja:

[quote="Wickedestjr"][quote="crazychinesemaniac"][quote="isabella2296"]I once dreamed that I was 2nd place at MATHCOUNTS Nationals, and for some reason, that depressed me. Then a laptop bonked me on the head and I died.  :|[/quote]

second always sucks.[/quote]

Especially if there are only two people.

One of my wierdest dreams was probably when I was chased by pirates that were throwing knives at me.[/quote]

5th is worse if there are 2 people

And your weirdest dream is a bit typical

hahahahaha dragon,

[quote] Fifth always sucks if there are 2 peope[/quote]

 :rotfl:

and how do i make the "Dragon96 says" on top of my quote??

all i can make it say is "Quote:"

right after amsp 08 ended..
i dreamt about me& my roomates&friends from camp getting chased by brian basham because we didn't keep our curfew..[which sorta did happen in reality too..]

the dream was too realistic.[shivers..]


Someone observed that $6! = 8 \cdot 9 \cdot 10$.  Find the largest positive integer $n$ for which $n!$ can be expressed as the product of $n - 3$ consecutive positive integers.

[quote="ImagineKid"][quote="nsato"]expressed as the product of $n - 3$ consecutive positive integers.[/quote]

what does that mean?[/quote]

notice that $6!=8 \cdot 9 \cdot 10$. Since $6$ is $n$ in this case, it is expressed as $6-3$ or in other words, $n-3$ consecutive integers...

[quote="ImagineKid"]oic ok and how do u post those pictures? =\[/quote]

pictures? you mean latex? just add dollar signs to your text and go to latex forum to learn some codes. examples: \frac, \sqrt, \cdot, etc.

solution:

[hide="the words I want to put in??"]
Let $n!$ with $n$ as the maximum number we wish to find.  Since the product of $n-3$ consecutive integers have less numbers multiplied together, the greatest number, denote $k$, will be greater than $n$.  we will set $k=n$ initially and gradually increase $k$ to the point where the products are equal.  when $k = n$, it is $n*(n-1)*\dots*(4)$, because we took $3, 2, 1$ out of $n!$.  It is easy to see that the product is $6$ times less than $n!$.  To compensate, we have to increment $k$, i.e. shift $k$ 1 unit over on the number line.  This time it'll be $(n+1)*(n)*(n-1)*\dots * 5$.  This string of numbers had decreased 4 times and increased $n+1$ times.  Since $n$ is the maximum we seek, the multiplier $n+1$ certainly out did the effect of the decreasing by four times, and hopefully make up the factor of $6$ at the same time.  By wishful thinking we assume $k = n+1$.

Now we set up the equation $n! = (n+1)*n*\dots*5$, or $24 = n+1$.  As the result, we see that $n = 23$.

Since there is a solution to $k = n+1$, we're done.  Because the more units $k$ shifts to the right on the number line, the product of $n-3$ consecutive integers will increase faster, and as the consequence decreasing the value of $n$.  So we conclude that $n = 23$ is the maximum we seek.
[/hide]

btw, someone teach me how to change the [click to reveal hidden content] to the words I want to put in??

[size=75][color=green]edited by mod[/color][/size]

when you have [hide] (the first bracket) put "hide=xxxxxx" inside the brackets

[quote]btw, someone teach me how to change the [click to reveal hidden content] to the words I want to put in??[/quote]

[code][hide=nameurhidehere][/hide][/code]
that will change the name of ur hidden stuff

If the n-3 numbers are contained within (1,2,3...n), then they will multiply to less than n! So the least the product of the n-3 numbers could be is when you take the 1,2,3, and 4 out of (1,2,3...n) and tack on an n+1. This would work for n=23, since you take out the 1,2,3, and 4 and tack on the 24 to the end to make up for the lost product. For any n higher than 23, n+1 will be greater than 24, so you will be multiplying by more than the four numbers you took away, thus 23 is the highest possible.

[hide="Solution and my Generalization"]
Let the largest of the (n-3) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-3) consecutive positive integers will be less than n!.

[u]Key observation:[/u]
Now for n to be maximum the smallest number (or starting number) of the (n-3) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).

So the (n-3) consecutive positive integers are (5, 6, 7…, n+1)

So we have (n+1)! /4! = n!
=> n+1 = 24
=> n = 23

[u]Generalization:[/u]
Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = [b](a+1)! – 1[/b]

For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039

[u]Proof:[/u]
Reasoning the same way as above, let the largest of the (n-a) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-a) consecutive positive integers will be less than n!.

Now, observe that for n to be maximum the smallest number (or starting number) of the (n-a) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).

So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)

So we have (n+1)! / (a+1)! = n!
=> n+1 = (a+1)!
=> n = (a+1)! -1
[/hide]


A point moves so that the distance between the feet of the perpendiculars from it on the pair of non parallel straight lines given by bx^2 + 2hxy + by^2 = 0 is a constant 2 units. Show that the equation to its locus is x^2 + y^2 = 4h^2/(h^2 - b^2)

[hide="Start"]
We have $ bx^2\plus{}2hxy\plus{}by^2\equal{}0\implies x\equal{}\frac{\minus{}2hy\pm\sqrt{4h^2y^2\minus{}4b^2y^2}}{2b}$

$ \equal{}\frac{\minus{}2hy\pm2y\sqrt{h^2\minus{}b^2}}{2b}\equal{}\frac{\minus{}hy\pm y\sqrt{h^2\minus{}b^2}}b$.

Thus, our two lines are $ bx\plus{}(h\minus{}\sqrt{h^2\minus{}b^2})y\equal{}0$ and $ bx\plus{}(h\plus{}\sqrt{h^2\minus{}b^2})y\equal{}0$.

I don't have time to continue this right now.
[/hide]


[quote="balan razvan"]$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\geq a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2$ with $ a,b,c,d > 0$[/quote]

[hide]This is Tukervici inequality :wink: [/hide]

[quote="balan razvan"]$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd\geq a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2$ with $ a,b,c,d > 0$[/quote]
There is very many proofs of this inequality. Maybe the following proof is new.
$ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 2abcd \minus{} \left(a^2b^2 \plus{} a^2c^2 \plus{} a^2d^2 \plus{} b^2c^2 \plus{} b^2d^2 \plus{} c^2d^2\right) \equal{}$
$ \equal{} \frac {1}{48}\sum_{sym}(a \minus{} b)^2S_{ab},$ where $ S_{ab} \equal{} 4(a \plus{} b)^2 \minus{} (c \plus{}d)^2.$
Let assume $ a\geq b\geq c\geq d.$ Then $ a \minus{} c\geq b \minus{} c,$ $ a \minus{} d\geq b \minus{} d\geq c \minus{} d,$
$ S_{ab}\geq0,$ $ S_{ac}\geq0,$ $ S_{ad}\geq0,$ $ S_{ac} \plus{} S_{bc}\geq0,$ $ S_{ad} \plus{} S_{bd}\geq0$ and $ S_{ad} \plus{} S_{bd} \plus{} S_{cd}\geq0.$
Thus, $ \frac {1}{4}\sum_{sym}(a \minus{} b)^2S_{ab}\geq$
$ \geq(a \minus{} b)^2S_{ab} \plus{} (b \minus{} c)^2\left(S_{ac} \plus{} S_{bc}\right) \plus{} (c \minus{} d)^2\left(S_{ad} \plus{} S_{bd} \plus{} S_{cd}\right)\geq0.$

[hide="another way"]another way is to use SMV theorem :)[/hide]

We can use mixing variables to solve this as well. If a,b>c,d, then we get f(a,b,c,d)>=f((a+b)/2,(a+b)/2,c,d)

then f(x,x,c,d) can be proven by completing the square wrt. x.

SMV theorem? maybe I know it...but maybe not like SMV theorem

Suppose that $ a\ge b\ge c\ge d$.

Then observe that \[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{}
 2 a b c d \minus{} (a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 d^2 \plus{} d^2 a^2 \plus{} a^2 c^2 \plus{}
   b^2 d^2)\]\[ \equal{}   \frac{1}{2}\cdot (a\minus{}b)^2(a^2\plus{}ab\plus{}b^2\minus{}c^2\minus{}cd\minus{}d^2)\plus{}\frac{1}{16}\cdot ((a\plus{}b)^2\minus{}4(c^2\minus{}cd\plus{}d^2))^2\plus{}cd(c\minus{}d)^2\]\[ \plus{}\frac{1}{16}\cdot (a\minus{}b)^2(7a^2\plus{}18ab\plus{}b^2)\ge 0\]

I solve it this way:
$ a^4\plus{}2abcd\geq a^2bc\plus{}a^2cd\plus{}a^2db$ $ \implies$ $ a(a\minus{}b)(a\minus{}c)(a\plus{}2d)\plus{}a(a\minus{}c)(a\minus{}d)(a\plus{}2b)\plus{}a(a\minus{}d)(a\minus{}b)(a\plus{}2c)\geq 0$ and $ b^4\plus{}c^4\plus{}d^4\plus{}a^2bc\plus{}a^2cd\plus{}a^2db\geq a^2b^2\plus{}a^2c^2\plus{}a^2d^2\plus{}b^2c^2\plus{}b^2d^2\plus{}c^2d^2$ $ \implies$ $ (b\minus{}c)^2[(b\plus{}c)^2\minus{}a^2]\plus{}(c\minus{}d)^2[(c\plus{}d)^2\minus{}a^2]\plus{}(d\minus{}b)^2[(d\plus{}b)^2\minus{}a^2]\geq 0$


Let  $a$ is a (given) real number , not  equal  to $0$. Solve  the equation :$x^3-3ax^2-3x+a=0$

hello, one solution is $ x_1\equal{}\sqrt [3]{a\plus{}{a}^{3}\plus{}i\plus{}i{a}^{2}}\minus{}{\frac {\minus{}1\minus{}{a}^{2}}{\sqrt [3]{a\plus{}{a}^{3
}\plus{}i\plus{}i{a}^{2}}}}\plus{}a$
Sonnhard.

Are you sure the last two coefficients have the 3 in the right place? If you make the constant term $ 3a$ then the equation is easy to solve.


I just have a general question about proofs.  When writing them, is it ok to state assumed things like the product of 2 odd numbers will always be odd or should this be proved too?

Thanks.

I think you can assume something like that.

Well, it depends what you are trying to do.  In general, it is proper to use anything that anyone has proven before.  However, obviously, to make things readable it is frequently a good idea to use things that people have heard of.  In competitions, there is a preference that you not use results that trivialize problems.  Also, in classes, you will frequently be asked to prove known results.  In those cases, you are obviously not supposed to use a result in the proof of itself, and generally should assume that if you haven't proven something in class or seen it proved in a reading, you should not use it.  (There are exceptions, of course, but that's a pretty good rule of thumb.)


geometry
rectangle
combinatorics unsolved
combinatorics

Some identical napkins are arranged on a rectangle table (with overlappings), each in the shape of a square. Each napkin has its sides parallel to the sides of the table. Is it always possible to hammer some nails into the table such that every napkin is nailed and moreover, each napkin is nailed by only one nail?

I think its from russia...
any solutions??

This problem is so funny. Let $l$ be the sidelength of each napkin. Tile the plane with $l \times l$ squares parallel to the sides of the table, and hammer a nail into each lattice point of the grid (that is, each vertex of a square in the tiling). We can obviously place the tiling so that no nail falls on the edge of a napkin. Then it is easy to check that we're done.