[hide]$2y$[/hide]
Same for an odd number of consecutive even integers.

[quote="anirudh"]The arithmetic mean of an odd number of consecutive odd integers is y. Find the sum of the smallest and largest of the integers[/quote]
[hide] The sum of an arithmetic sequence can be calculated by the formula $n(\frac{a_{1}+a_{n}}{2})$, so the arithmetic mean is $\frac{n(\frac{a_{1}+a_{n}}{2})}{n}=\frac{a_{1}+a_{n}}{2}=y$
Thus, $a_{1}+a_{n}=\boxed{2y}$ [/hide]

[quote="bpms"]One of the easiest national targets ever...[/quote]

was this the year that Mathew Crawford got a perfect score at nats?


Does anyone have solutions maybe already written out for numbers 21 through 30? I can't believe I got everyone wrong ... :(   I'm SO depressed - and now, very worried, as tomorrow is my state test -- ARRRGGHHHH!!!

i didn't do too well on those either... I think I only got 1 or 2 of those...kinda ran out of time.

Anyways, just relax...it's ok if you don't make nats. There is always USAMO and MOSP to shoot for.


modular arithmetic
quadratics
Diophantine equation
number theory proposed
number theory

Prove that $x^2-py^2=-1$ has infinitely many solutions when p is a prime in form 4k+1 and has none solution when p is prime in form 4k+3.

Can you please tell me what KMS means. I think it is much better to just use abbreviations if there are really known as IMO etc.  :)

For instance CMO can mean different things:

- Canadian Mathematical Olympiad
- Chinese Mathematical Olympiad etc.

To quickly answer the second assertion, if $p=4k+3$, then $p$ can't divide $x^2+1$, because the Legendre symbol of $-1\pmod p$ is $(-1)^{\frac{p-1}2}=-1$ (so $-1$ is not a quadratic residue $\pmod p$).

Another approach. It is impossible $x^2\equiv -1\pmod{p}$ for $p=4k+3$. Ohterwise considerig $2k+1$-th power of this equality we see that $x^{p-1}\equiv -1\pmod{p}$, but due to Fermat theorem $x^{p-1}\equiv 1\pmod{p}$ - cntradiction.

I thinl for $p=4k+1$ the following idea works.
We know that $p=a^2+b^2$. Suppose $y=u^2+v^2$, then $py^2=(au+bv)^2+(av-bu)^2$, so we just need to provide $av-bu=1$ (obviously, we can do it), then $x=au+bv$.

Edit: here is terrible typo, so solution is completely wrong :(

KMS is a competition in Slovakia in which you solve problems at home and then send it. It's solved just by people from Czech and Slovak Republic.

2Myth: I was thinking about using that $p=a^2+b^2$ but I didn't find that brilliant idea, which you've found.

My solution was: Let x,y be smallest non-trivial solution of $x^2-py^2=1$ (which exists). Transforming this Pell equation you get $(x+1)(x-1)=py^2$. You also know that x is odd, so you can replace x with 2z+1 and you know that y is even so you can replace it with w. So you know that $z(z+1)=pw^2$ and becouse (z,z+1)=1 arethere are 2 cases:

1) There exists positive integers m,n such that $z=m^2, z+1=pn^2$ => $x-1=2m^2,x+1=2pn^2$. Now try to compute $m^2-pn^2=\frac{x-1}{2}-p\frac{x+1}{2p}=-1$. So we have solution .

2) There exists positive integers m,n such that $z=pn^2, z+1=m^2$ => $x-1=2pn^2, x+1=2m^2$. Now if you try to compute $m^2-pn^2=\frac{x+1}{2}-p\frac{x-1}{2p}=1$ but it can't be because x,y is smallest non-trivial solution to the Pell equation.

[quote="pluricomplex"]So the equation $x^2-dy^2=-1$ has many infinity solutions if $d$ has no prime divisor $4k+3$ in which its exponent in $d$ is odd? :D[/quote]
Hmm... At least, it is not conclusion from my solution ;) I used fact, that $p=a^2+b^2$ and $(a,b)=1$.

No, it's not true, for example d=34 has no solutions. But I don't know all $d$ for which this equation has infinitely many solutions.

I found a mistake in you solution Myth. If your solution was good it would prove it also for 34, cause $34=5^2+3^2$. Your mistake is that from $y=u^2+v^2$ doesn't folw that $py^2=(au+bv)^2+(av-bu)^2$ but that $py=(au+bv)^2+(av-bu)^2$. Maybe it can be repaired by y,u,v being Pythagorean triple.

ah i see, i will check the last question later! :)

Thank you, ondrob. I don't know how to fix my mistake.


Hi guys,

Sorry for not speaking in Spanish, but I'm not fluent, so to speak :D
Anyway, for some time now it seems that Djmenez is gone, so we need a new moderator for this section. Since this is one of the most active national communities (of course the latin one :D) I am asking you first who would want to be a moderator.

Requirements: spanish speaking, visits the forums at least once every couple of days, and being older than 6 months on the forum.

Regards,
Valentin.

[quote="orl"]Well, I don't speak Spanish.  :D  But I would recommend Manuel who also wants to contribute to the Resources section, e.g. Ibero-American ShortList etc.  :)[/quote]Pascual2005 has been named already :) He fulfills all the criterions, in plus I know him ;)

Of course if the spanish forums continue to grow as they have, a new moderator might be required.

And to contribute to the Resources sections there is no need to be a moderator. I will detail this later ;)

In my opinion:

This community is growing quickly.......
and a person is not enought to organize this .........

Pascual and I are the more appropriate people to take this role in this moment .....

Carlos Bravo
Lima -Peru

:roll:

I know manuel is doing a great job but i think he should focus in research and getting ready for the next IMO.

I am not a participant anymore so I have more time.

Thanks, I think I can do a great Job.

OK, carlos is a moderator also. Since the traffic is high in these forums I decided two moderators are needed. Topic is now officially closed!


linear algebra
matrix
linear algebra unsolved

Let $ a_{1},\ldots,a_{n}\in\mathbb{R}$, and $ b_{0},\ldots, b_{n}\in\mathbb{C}$, s.t. $ b_{0}\equal{} b_{n}\equal{} 0$ and the other $ b_{i}\not \equal{} 0$.
$ A \equal{}\begin{pmatrix}a_{1}& b_{1}& 0 &\cdots & 0\\ \bar{b_{1}}& a_{2}& b_{2}&\ddots &\vdots\\ 0 &\ddots &\ddots &\ddots & 0\\ \vdots &\ddots &\bar{b}_{n\minus{}2}& a_{n\minus{}1}& b_{n\minus{}1}\\ 0 &\cdots & 0 &\bar{b}_{n\minus{}1}& a_{n}\\ \end{pmatrix}$
Let $ M \equal{}\max_{j \equal{} 1,\ldots ,n}(a_{j}\plus{}|b_{j\minus{}1}|\plus{}|b_{j}|)$ and $ m \equal{}\min_{j \equal{} 1,\ldots ,n}(a_{j}\minus{}|b_{j\minus{}1}|\minus{}|b_{j}|)$.
Show that the eigenvalues of $ A$ belong to $ [m,M]$.

Use the following:
1. a hermitian matrix has real eigenvalues;
2. http://en.wikipedia.org/wiki/Gershgorin_circle_theorem.

That's all.


calculus
integration
real analysis
real analysis solved

Let f :R--->R, C^2 (f" exist and is continous) such that
f(x) + f"(x) >= 0 for any x real
Show that, for any x we have f(x) + f(x+pi) >= 0
Oral examination Grandes Ecoles: Centrale

k(x) = f(x+ \pi )+f(x) =\int_[0; \pi ](f(x+t)sint - f'(x+t)cost) dt

integrate by part give

k(x)= \int_[0; \pi ](f(x+t)+f"(x+t))sint dt >=0

Great proof. I found it hard to understand, so I'll take some time to explain.
k(x)=f(x+pi)+f(x)=-f(x+pi)cospi+f(x)cos(0)= -f(x+t)cost | pi _{0} = the first integeral Moubinool used.

Fore the second integeral use f'(x+t)sint to integrate by part.

this problem was also used in Vietnamese Undergrad. Contest in 2001. :D nice soln. moubi ..


analytic geometry
HMMT
number theory
relatively prime
combinatorics theorems
combinatorics

Here's a proof I came up with for the fact that the number of paths from $ (0,0)$ to $ (n,n)$ consisting of unit steps right and up and not crossing the line $ y\equal{}x$ is $ \binom{2n}{n}/(n\plus{}1)$.  It's different from all the proofs I've seen, and I think this proof gives more insight into where the $ n\plus{}1$ comes from.  I'm curious whether this proof is already well-known or not.  Let me know what you think:

The number of paths from $ (0,0)$ to $ (n,n)$ that do not go above the line $ y \equal{} x$ is the same as the number of paths from $ (0,0)$ to $ (n\plus{}1,n)$ that do not go above the line through $ (0,0)$ and $ (n\plus{}1,n)$, so we'll count these instead.  The first step of any such path must be to the right.  We claim that the paths from $ (0,0)$ to $ (n\plus{}1,n)$ with first move to the right can be arranged in groups of $ n\plus{}1$ such that exactly one path in each group does not go above the line.

Represent these paths as sequences of $ n\plus{}1$ R's and $ n$ U's, such that the first character is R.  Then for each path, we form its group of $ n\plus{}1$ paths by taking the $ n\plus{}1$ cyclic permutations of its corresponding sequence that start with R.  Later we will establish that no two of these permutations are the same path.

Consider one group, and take an arbitrary path from the group.  Now concatenate the path with itself, so that we now have a path from $ (0,0)$ through $ (n\plus{}1,n)$ to $ (2n\plus{}2,2n)$.  This path contains our $ n\plus{}1$ paths, specifically for $ 0\le k\le n$, there is a path that starts at the highest point with x-coordinate $ k$, namely $ (k, y_k)$, and ends at the highest point with x-coordinate $ n\plus{}k\plus{}1$, namely $ (n\plus{}k\plus{}1, y_k \plus{}n)$.  For each path, draw the line from the start point to the end point.  These lines are all mutually parallel.

Suppose for contradiction that the line corresponding to the kth path coincides with another path.  Then there is a lattice point on the segment connecting $ (k, y_k)$ and $ (n\plus{}k\plus{}1, y_k \plus{}n)$, a contradiction since $ n$ and $ n\plus{}1$ are relatively prime, and thus the fraction $ n/(n\plus{}1)$ is irreducible.  Therefore all $ n\plus{}1$ lines are distinct.  Then each path in the group can be uniquely characterized by the number of lines above its line, so no two paths are the same.

Since the lines go through the highest points at each x-coordinate, and every path contains a point on each of the $ n\plus{}1$ lines, a path stays below the diagonal if and only if its line is above every other line.  Therefore there is exactly one path in the group that stays below the diagonal.

Therefore, the number of paths from $ (0,0)$ to $ (n,n)$ that do not go above the line $ y\equal{}x$ is $ \binom{2n}{n}/(n\plus{}1)$, as desired.

The crucial lemma in this proof was posted on AoPS a year or two ago as a question (with no reference to the Catalan numbers), so I assume it's in a textbook somewhere.  I answered it and, a few months later, realized it implied this result and its generalization to $ k$-ary trees.  I then saw Stanley present essentially this proof in a lecture some time later.

Ah, cool.  If you're interested, this was inspired by the following problem from one of Zuming's MOCK tests last year, similar to a problem that showed up on an HMMT test several years ago:

Given positive integers $ p,q$, find the number of subsets $ S$ of the non-negative integers such that

i)  $ 0\in S$
ii)  If $ n\in S$, then $ n\plus{}p, n\plus{}q\in S$

If you want a citation, apparently the lemma is known as Raney's lemma and the geometric proof is given on page 360 of Graham, Knuth, and Patashnik's [i]Concrete Mathematics.[/i]


Prove that if $ a,b$ are positive reals such that $ a^{2} \plus{} b^{2} \equal{} 1$ , then

$ a \plus{} b \plus{} \frac {1}{ab} \ge 2 \plus{} \sqrt{2}$

[hide="Solution"]Square both sides to get
$ a^2 \plus{} b^2 \plus{} \frac {1}{a^2b^2} \plus{} 2ab \plus{} \frac {2}{a} \plus{} \frac {2}{b} \ge 6 \plus{} 4\sqrt {2}$
Using our given condition,
$ \frac {1}{a^2b^2} \plus{} 2ab \plus{} \frac {2}{a} \plus{} \frac {2}{b} \ge 5 \plus{} 4\sqrt {2}$ (1)
We have
$ \frac {1}{a^2b^2} \plus{} 2ab \equal{} \frac {1}{(2ab)^2} \plus{} \frac {1}{(2ab)^2} \plus{} \frac {1}{(2ab)^2} \plus{} \frac {1}{(2ab)^2} \plus{} 2ab \ge 5 \sqrt [5] {\frac {1}{(2ab)^7}}$ by AM-GM
Now, $ 1 \equal{} a^2 \plus{} b^2 \ge 2ab \implies 2ab \le 1$
So $ \frac {1}{a^2b^2} \plus{} 2ab \ge 5$ by substitution.
Subtract this from both sides of (1) to get
$ \frac {1}{a} \plus{} \frac {1}{b} \ge 2\sqrt {2}$
But this is true, as
$ \frac {1}{a} \plus{} \frac {1}{b} \ge 2 \sqrt {\frac {1}{ab}} \ge 2 \sqrt {2}$ by AM-GM[/hide]

[hide="Easier one "]
Put $ a \equal{} \sin \theta$ and $ b \equal{} \cos\theta$.
To prove $ \sin\theta \plus{} \cos\theta \plus{} \frac {2}{\sin2\theta}\ge 2 \plus{} \sqrt {2}$
Use $ \sqrt {2}\ge \sin\theta \plus{} \cos\theta$

[/hide]

Edit:yes i made a mistake what mathwizarddude said is absolutely right but now my solution is incorrect

[quote="RAGHAV GROVER"]
Use $ \sin\theta \plus{} \cos\theta\ge\sqrt {2}$
[/quote]
This is wrong. What if $ \theta\equal{}0$?

In fact it should be $ \sin\theta\plus{}\cos\theta\le\sqrt2$.
But degreenb80b we cannot subtract two inequality as in your solution...

yes i made the mistake,so how to prove it using my substitutions

Note that when we put $ a\equal{}\frac{1}{\sqrt{2}}$ and $ b\equal{}\frac{1}{\sqrt{2}}$

It does not yield RHS but in most of the inequalities this condition satisfies

$ a\plus{}b\minus{}\sqrt{2(a^2\plus{}b^2)}\equal{}\frac{(a\plus{}b)^2\minus{}2(a^2\plus{}b^2)}{  a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)}}\equal{}\minus{}\frac{(a\minus{}b)^2}{  a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)}}$

and $ \frac{a^2\plus{}b^2}{ab}\minus{}2\equal{}\frac{(a\minus{}b)^2}{ab}$

So the inequality is equivalent to $ (a\minus{}b)^2 S_c\ge 0$ where

$ S_c\equal{}\frac{1}{ab}\minus{}\frac{1}{  a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)}}$

We must show that $ S_c\ge 0$. We can clear the nonnegative denominators to get

$ a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)}\ge ab$

$ \sqrt{a^2\plus{}b^2}(a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)})\ge ab$

which is obvious by AM-GM

$ \sqrt{a^2\plus{}b^2}(a\plus{}b\plus{}\sqrt{2(a^2\plus{}b^2)})\ge \sqrt{2ab}(2\sqrt{ab}\plus{}2\sqrt{ab})\equal{}4\sqrt{2} ab>ab$

So the inequality could be sharpened to

$ (a\plus{}b\minus{}\sqrt{2})(4\sqrt{2})\plus{}\left( \frac{1}{ab}\minus{}2\right)\ge 0$

[quote="mathwizarddude"]
But degreenb80b we cannot subtract two inequality as in your solution...[/quote]
Why not? Don't think of it as subtracting but rather substituting.
$ \frac {1}{a^{2}b^{2}} \plus{} 2ab \plus{} \frac {2}{a} \plus{} \frac {2}{b}\ge 5 \plus{} \frac {2}{a} \plus{} \frac {2}{b}$
So if we can show  $ 5 \plus{} \frac {2}{a} \plus{} \frac {2}{b} \ge 5 \plus{} 4\sqrt {2}$ we are done.
Remember with inequalities we are working backwards, so the chain of implications can be confusing.


algorithm
combinatorics unsolved
combinatorics

Given is a set
S={$1^{2},2^{2},3^{2},...,n^{2}$}
Find partition of set S into four subsets such that all subsets have equal sums (and their intersection is an empty set) if:
(i)    n=15
(ii)   n=16
(iii)  n=23
(iiii) n=24

Hvala ti brate.

This is similar to one Crux problem Actually these are some special cases of that problem.

[quote="mornik"]
Now, the last two were rather tough and I can only hope there is some kind of smart strategy for this and that it doesn't all have to come down to what I was doing.[/quote]

Yes, there is a pretty simple strategy - programming. Use your computer :)

[quote]Yes, there is a pretty simple strategy - programming. Use your computer[/quote]

Ye Of course you are right I also think about that but I am not so good at computers so...
(I know how to use Pascal but this is a litttle bit difficult for me...

[quote="delegat"][quote]Yes, there is a pretty simple strategy - programming. Use your computer [/quote]

Ok Then can you find me partition for
[b]n=31 n=39 and n=40.[/b]

Hajd molim te Hitno mi to treba da zavrsim jedan zadatak.[/quote]

Sorry, I cannot find algorithm which is fast enough to check all those partitions. Al jos cu probat.

[quote="delegat"]Hajd molim te Samo mi jos ovo treba.[/quote]

Pitao sam jednog informaticara i on kaze da se to da rijesiti nekim algoritmom "meet in the middle" kojeg ja na zalost ne znam, a on nema vremena. Probat cu ipak nesto smisliti.

Hajd dobro Ako sta skuzis javi mi Vozdra

Can anybody finish this problem (find partitions for 31, 39 and 40).

I ended up approaching this problem by hand (with the help of excel) using what was essentially a greedy algorithm: I'd fill up the sets one by one, filling each set by taking in turn the largest number remaining that would not cause the sum to exceed a fourth of the total.  If this failed, I would backtrack partway through the process and try the next largest square instead of the largest)

In the $ n=39$ case this worked easily, yielding
$ S_{1}=\{39^{2},38^{2},37^{2},28^{2},4^{2},1^{2}\}$
$ S_{2}=\{36^{2},35^{2},34^{2},33^{2},18^{2},6^{2},3^{2}\}$ (I had to go through twice, picking $ 18$ instead of $ 19$)
${ S_{3}=\{32^{2}, 31^{2},30^{2},29^{2},27^{2},26^{2},2^{2}]}$
$ S_{4}=$everything else

In the $ n=40$ case it worked, though the third set took a fair amount of trial and error

$ S_{1}=\{40^{2},39^{2},38^{2},31^{2},3^{2}\}$
$ S_{2}=\{37^{2},36^{2},35^{2},34^{2},20^{2},8^{2},5^{2}\}$ (had to use $ 20$ instead of $ 22$)
$ S_{3}=\{33^{2},32^{2},30^{2},29^{2},28^{2},26^{2},11^{2},10^{2}\}$
$ S_{4}=$everything else.

Unfortunately, I don't think this method generalizes well to larger $ n$.


calculus
integration
function
trigonometry
inequalities
real analysis
real analysis unsolved

(Ya.Mykytyuk) Let real-valued function $ f\in C^{2}[0;\pi]$ be convex. Prove that
\[ \int_{0}^{\pi}{f(t)\sin t dt}\le\int_{0}^{\pi}{f(t)\lvert\cos t\rvert dt.}\]

can u plz tell me what does f(belongs to)C^2(0,pi) mean?does it mean that it belongs to the set of all complex integrable functions on(0,pi)?

$ C^{2}$ means twice continuously differentiable. Since we're also told that $ f$ is convex, we can say that $ f''$ is continuous and that $ f''(t)\ge 0.$ That would certainly make me think about integrating by parts, twice.

Another idea, incomplete. Let $ g(t)\equal{}\frac{f(t)\plus{}f(\pi\minus{}t)}{2}.$ We have that $ g''(t)\ge 0$ and $ g(0)\ge g\left(\frac{\pi}{2}\right).$

Our proposed inequality can be written as:

$ \int_{0}^{\frac{\pi}{2}}g(t)\sin t\,dt\stackrel{?}{\le}\int_{0}^{\frac{\pi}{2}}g(t)\cos t\,dt$

Not sure what to do from there.

Here's half of an argument. I'm going to temporarily write up this half while I finish the arguments on the other half. Or maybe it's a quarter of an argument. Right now I'm just thinking aloud.

Suppose $ \frac{\pi}{4}\le x\le\frac{\pi}{2}.$ Then by convexity,

$ f(x)\le\frac{\pi}{4x}f\left(\frac{\pi}{2}\minus{}x\right)\plus{}\left(1\minus{}\frac{\pi}{4x}\right)f\left(\frac{\pi}{2}\plus{}x\right)$

Hence

$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}f(x)\sin x\,dx\le\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{\pi}{4x}f\left(\frac{\pi}{2}\minus{}x\right)\plus{}\left(1\minus{}\frac{\pi}{4x}\right)f\left(\frac{\pi}{2}\plus{}x\right)\right]\sin x\,dx$

Make the substitution $ x\equal{}\frac{\pi}{2}\minus{}u$ to get this integral to equal

$ \int_{0}^{\frac{\pi}{4}}\frac{\pi}{2\pi\minus{}4u}f(u)\cos u\,du\plus{}\int_{0}^{\frac{\pi}{4}}\frac{\pi\minus{}4u}{2\pi\minus{}4u}f(\pi\minus{}u)\cos u\,du$

Now replace $ u$ in that second integral by $ \pi\minus{}u:$

$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}f(x)\sin x\,dx\le\int_{0}^{\frac{\pi}{4}}\frac{\pi}{2\pi\minus{}4u}f(u)\cos u\,du\plus{}\int_{\frac{3\pi}{4}}^{\pi}\frac{3\pi\minus{}4u}{2\pi\minus{}4u}f(u)|\cos u|\,du$

Continuing in the same vein: for $ \frac{\pi}{2}\le x\le\frac{3\pi}{4},$

$ f(x)\le\frac{3\pi\minus{}4x}{4\pi\minus{}4x}f\left(x\minus{}\frac{\pi}{2}\right)\plus{}\frac{\pi}{4\pi\minus{}4x}f\left(\frac{3\pi}{2}\minus{}x\right)$

Integrating that and making a change of variables, I get:

$ \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}f(x)\sin x\,dx\le\int_{0}^{\frac{\pi}{4}}\frac{\pi\minus{}4u}{2\pi\minus{}4u}f(u)\cos u\,du\plus{}\int_{\frac{3\pi}{4}}^{\pi}\frac{\pi}{4u\minus{}2\pi}f(u)|\cos u|\,du$

---

OK, at this point, I have two issues:

(1) I must have a sign error. I think these are supposed to add up to

$ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}f(x)\sin x\,dx\le\int_{0}^{\frac{\pi}{4}}f(x)\cos x\,dx\plus{}\int_{\frac{3\pi}{4}}^{\pi}f(x)|\cos x|\,dx$

At the moment, I can't get it to do that.

(2) Even if I can make that work, so what?

Anyone who can see where this is going, feel free to jump in.

I don't know if I can help you with your argument, but here's something natural:

First of all, WLOG assume that $ f > C$, where $ C$ is very big (else we add a constant).

Write the conclusion as $ \int_{0}^{\pi}f(t)\left[\left|\cos t\right|\minus{}\sin t\right]\, dt\geq 0$.

Let's try to modify a given convex $ f$ so that the integral gets as small as possible.
On some intervals, $ h(t) \equal{}\left|\cos t\right|\minus{}\sin t$ is positive and on others is negative.
Now use the fact the graph of $ f$ lies above any tangent to it and below any chord (it's not true as I phrased it, but you get what I mean). For example, on the interval on which $ h$ is negative we need to make $ f$ as big as possible. That means we have to draw the corresponding chord. You should get a linear function in the end. Hence, it suffices to check the conclusion only for linear functions. (Actually, only for $ f(t) \equal{} t$, since I have already used the fact that the integral is $ 0$ for constants in the very beginning.)

Ah, I've been trying to make it too hard.

Suppose we have a continuous function $ g(x)$ defined on $ [a,b]$ with the following properties:

$ \int_{a}^{b}g(x)\,dx\equal{}0$ and,

if $ H(x)\equal{}\int_{a}^{x}\int_{a}^{t}g(u)\,du\,dt,$ then $ H(x)\ge0$ on $ (a,b)$ and $ H(b)\equal{}0.$

Then if $ f$ is $ C^{2}$ and convex on $ [a,b],$ we have

$ \int_{a}^{b}f(x)g(x)\,dx\ge0.$

We prove this by integration by parts. Let $ G(x)\equal{}\int_{a}^{x}g(t)\,dt$ and note that $ G(a)\equal{}G(b)\equal{}0.$ Then

$ \int_{a}^{b}f(x)g(x)\,dx\equal{}f(b)G(b)\minus{}f(a)G(a)\minus{}\int_{a}^{b}f'(x)G(x)\,dx\equal{}\minus{}\int_{a}^{b}f'(x)G(x)\,dx$

$ \equal{}\minus{}f'(b)H(b)\plus{}f'(a)H(a)\plus{}\int_{a}^{b}f''(x)H(x)\,dx\equal{}\int_{a}^{b}f''(x)H(x)\,dx$

But both $ f''$ and $ H$ are nonnegative, so the integral is nonnegative.

In this particular circumstance, $ g(x)\equal{}|\cos x|\minus{}\sin x.$ Integrating, we get that

$ G(x)\equal{}\begin{cases}\sin x\plus{}\cos x\minus{}1,&x\in\left[0,\frac{\pi}{2}\right]\\ \minus{}\sin x\plus{}\cos x\plus{}1,&x\in\left[\frac{\pi}{2},\pi\right]\end{cases}$

$ H(x)\equal{}\begin{cases}\minus{}\cos x\plus{}\sin x\minus{}x\plus{}1,&x\in\left[0,\frac{\pi}{2}\right]\\ \cos x\plus{}\sin x\plus{}x\plus{}1\minus{}\pi,&x\in\left[\frac{\pi}{2},\pi\right]\end{cases}$

We can explicitly check that this meets the conditions.

Hello, there!

I thought some days about this problem and I cannot find the mistake in my solution.

We can write $ \int_{0}^{\pi}f(x)\sin x\, dx$ as $ \int_{0}^{\frac{\pi}{2}}f(x)\sin x\, dx\plus{}\int_{\frac{\pi}{2}}^{\pi}f(x)\sin x\, dx$. Using the substitutions $ u \equal{} x\plus{}\frac{\pi}{2}$ and $ u \equal{} x\minus{}\frac{\pi}{2}$ we get
$ \int_{\frac{\pi}{2}}^{\pi}f(u\minus{}\frac{\pi}{2})\sin(u\minus{}\frac{\pi}{2})\, du\plus{}\int_{0}^{\frac{\pi}{2}}f(u\plus{}\frac{\pi}{2})\sin(u\plus{}\frac{\pi}{2})\, du \equal{}$
$ \int_{\frac{\pi}{2}}^{\pi}f(u\minus{}\frac{\pi}{2}) (\minus{}\cos(u))\, du\plus{}\int_{0}^{\frac{\pi}{2}}f(u\plus{}\frac{\pi}{2})\cos(u)\, du \equal{}$
$ \int_{\frac{\pi}{2}}^{\pi}f(u\minus{}\frac{\pi}{2}) |\cos u|\, du\plus{}\int_{0}^{\frac{\pi}{2}}f(u\plus{}\frac{\pi}{2}) |\cos(u)|\, du$ because $ \cos u$ is negative in the first sum and positive in the second one.

Using Taylor's forumula we get (and because $ f$ is convex)

$ f(u\plus{}\frac{\pi}{2}) \equal{} f(u)\plus{}\frac{\pi}{2}f'(u)\plus{}\frac{1}{2}(\frac{\pi}{2})^{2}f''(\xi_{1})\ge f(u)\plus{}\frac{\pi}{2}f'(u)$
$ f(u\minus{}\frac{\pi}{2}) \equal{} f(u)\minus{}\frac{\pi}{2}f'(u)\plus{}\frac{1}{2}(\minus{}\frac{\pi}{2})^{2}f''(\xi_{2})\ge f(u)\minus{}\frac{\pi}{2}f'(u)$

So we can write

$ \int_{\frac{\pi}{2}}^{\pi}f(u\minus{}\frac{\pi}{2}) |\cos u|\, du\plus{}\int_{0}^{\frac{\pi}{2}}f(u\plus{}\frac{\pi}{2}) |\cos(u)|\, du\ge$
$ \int_{\frac{\pi}{2}}^{\pi}(f(u)\plus{}\frac{\pi}{2}f'(u)) |\cos u|\, du\plus{}\int_{0}^{\frac{\pi}{2}}(f(u)\minus{}\frac{\pi}{2}f'(u)) |\cos(u)|\, du \equal{}$
$ \int_{0}^{\pi}f(u) |\cos u|\, du\plus{}\int_{\frac{\pi}{2}}^{\pi}\frac{\pi}{2}f'(u) |\cos u|\, du\minus{}\int_{0}^{\frac{\pi}{2}}\frac{\pi}{2}f'(u) |\cos(u)|\, du\ge$
$ \int_{0}^{\pi}f(u) |\cos u|\, du\plus{}\int_{\frac{\pi}{2}}^{\pi}\frac{\pi}{2}f'(\frac{\pi}{2}) |\cos u|\, du\minus{}\int_{0}^{\frac{\pi}{2}}\frac{\pi}{2}f'(\frac{\pi}{2}) |\cos(u)|\, du \equal{}$
$ \int_{0}^{\pi}f(u) |\cos u|\, du$ because $ f'$ is monontically increasing, giving us the wrong result :-(

The problem is that you have two inequalities, and they point in opposite directions. You proved two things:

(1) $ \int_{0}^{\pi}f(x)\sin x\,dx\le\int_{0}^{\pi}f(x)|\cos x|\,dx\plus{}A$

where $ A\equal{}\minus{}\frac{\pi}{2}\int_{0}^{\pi}f'(u)\cos u\,du$

(You wrote $ A$ a little differently, splitting it into two halves, but this is what it amounts to.)

And then you proved:

(2) $ A\ge0.$

Both proofs are correct. The problem is that the combination of the two results, with the inequality signs pointing in opposite directions, doesn't yield any result.

You could also have estimated the monotone $ f'$ in the opposite direction to get

$ A\le\frac{\pi}{2}(f'(\pi)\minus{}f'(0)).$

That also does not lead to the result we want.


Thanks Kouichi.

I had

$ x(x \plus{} 1)(x \plus{} 2)(x \plus{} 3) \minus{} 24$

$ \equal{} x(x \plus{} 3)(x \plus{} 1)(x \plus{} 2) \minus{} 24$

$ \equal{} (x^{2} \plus{} 3x)(x^{2} \plus{} 3x \plus{} 2) \minus{} 24$

$ \equal{} y(y \plus{} 2) \minus{} 24$, where $ y \equal{} x^{2} \plus{} 3x$

$ \equal{} y^{2} \plus{} 2y \minus{} 24$

$ \equal{} (y \plus{} 6)(y \minus{} 4)$

$ \equal{} (x^{2} \plus{} 3x \plus{} 6)(x^{2} \plus{} 3x \minus{} 4)$

$ \equal{} (x^{2} \plus{} 3x \plus{} 6)(x \minus{} 1)(x \plus{} 4)$

which is almost exactly what you had.

Alternatively, if you just expand the whole thing you get $ x^4 \plus{} 6x^3 \plus{} 11x^2 \plus{} 6x \minus{} 24$. You can synthetic divide to find that 1 is a zero of the equation $ x^4 \plus{} 6x^3 \plus{} 11x^2 \plus{} 6x \minus{} 24 \equal{} 0$ so the factor of the expression is $ x \minus{} 1$. If you synthetic divide again, you find that $ x \plus{} 4$ is also a factor. So the expression becomes $ (x \minus{} 1)(x \plus{} 4)(x^2 \plus{} 3x \plus{} 6)$, and by the discriminant of $ (x^2 \plus{} 3x \plus{} 6)$, we cannot factor anymore.

But this is more of a guess and check solution, so it's definitely not optimal.


Given three fixed integers a, b, c > 0; x, y, z are three positive variables so that:

$\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 1$

Prove that: $\sqrt[n+1]{a^n} + \sqrt[n+1]{b^n} + \sqrt[n+1]{c^n} \leq (\frac{a}{x} + \frac{b}{y} + \frac{c}{z})(x^n+y^n+z^n)$

Note: $n = 1, 2, 3...$

I know that we can write: $\sqrt[n+1]{a^n} = (\sqrt[n+1]{\frac{a}{x}})^n * \sqrt[n+1]{x^n}$, so can I use Cauchy Schwars inequality on this problem?  I think it only works with exponent of $2$, not $n+1$.

[quote="Laputa"]What's holder inequality?  I tried to look it up but can't understand what they mean.  Can you try to prove it using Cauchy?[/quote]

Well basically it's what Cauchy "should be" if you were to have more terms. However, you can't prove it with Cauchy (except possibly if you used some limit arguments? I'm not sure...). I believe there is a proof using Jensen, but I don't remember it off-hand.

This is how I would do it with Cauchy but I wonder it has legal operations:

$(\sqrt[n+1]{a^n} + \sqrt[n+1]{b^n} + \sqrt[n+1]{c^n})^{n+1}$ $=$ $[(\sqrt[n+1]{\frac{a}{x}})^n * \sqrt[n+1]{x^n} + (\sqrt[n+1]{\frac{b}{y}})^n * \sqrt[n+1]{x^n} + (\sqrt[n+1]{\frac{c}{z}})^n * \sqrt[n+1]{x^n})]^{n+1}$ $\leq$ $(\frac{a}{x}+\frac{b}{y}+\frac{c}{z})^n(x^n+y^n+z^n)$

Is this legal because I think that Cauchy only works with power 2 rather than power n+1?

[quote="shyong"]Holder Inequality (generalise of Cauchy) ，we have

$(\frac{a}{x}+\frac{b}{y}+\frac{c}{z})^n(x^n+y^n+z^n)\geq  a^\frac{n}{n+1}+b^\frac{n}{n+1}+c^\frac{n}{n+1}$[/quote]

Doesn't Holder's Say:

$(\frac{a}{x}+\frac{b}{y}+\frac{c}{z})^{\frac{n}{n+1}}(x^n+y^n+z^n)^{\frac{1}{n+1}}\geq  \frac{a^{\frac{n}{n+1}}}{x^{\frac{n}{n+1}}}\cdot (x^n)^{\frac{1}{n+1}}+\frac{b^{\frac{n}{n+1}}}{y^{\frac{n}{n+1}}} \cdot (y^n)^{\frac{1}{n+1}} + \frac{c^{\frac{n}{n+1}}}{z^{\frac{n}{n+1}}} \cdot (z^n)^{\frac{1}{n+1}}$
$= a^\frac{n}{n+1}+b^\frac{n}{n+1}+c^\frac{n}{n+1}$

$\Rightarrow (\frac{a}{x}+\frac{b}{y}+\frac{c}{z})^n(x^n+y^n+z^n)\geq (a^\frac{n}{n+1}+b^\frac{n}{n+1}+c^\frac{n}{n+1})^{n+1}$

P.S. Which still makes the off-holder mentioned true if $a^\frac{n}{n+1}+b^\frac{n}{n+1}+c^\frac{n}{n+1} \geq 1$ and $n + 1 \geq 1$.

http://mathworld.wolfram.com/HoeldersInequalities.html

So, the sum version states:

$\sum_{k=1}^{n} | a_k b_k | \le \left( \sum_{k=1}^{n} | a_k |^p \right)^{ \frac{1}{p} \right) \left( \sum_{k=1}^{n} | b_k |^q \right)^{ \frac{1}{q} }}$, which becomes Cauchy when $p = q = 2$.

I see, I get it now.  Thank you to all of you.


function
Gauss
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algebra

Let $ p$ be a prime. Let $ f: \mathbb{N}\rightarrow \{\minus{}1,0,1\}$ be a nonconstant function such that

(i) for all natural numbers $ m,n$, $ f(mn)\equal{}f(m)f(n)$

(ii) for all natural numbers $ n$, $ f(n\plus{}p)\equal{}f(n)$

Let $ w$ be a complex number such that $ w^p\equal{}1,w\neq 1$, and let $ a\equal{}\sum_{n\equal{}1}^{p\minus{}1}f(n)w^n$

Find all possible values of $ a^2$.

See [url=http://en.wikipedia.org/wiki/Gauss_sum]Gauss sum[/url] and [url=http://en.wikipedia.org/wiki/Dirichlet_character]Dirichlet character[/url].  The theory behind such sums is rather well-known, although it is instructive to work from the given conditions alone.

[hide="Hints"] - What are the possible values of $ f(1)$ if $ f$ is nonconstant?
- Let $ g$ be a generator $ \bmod p$.  What are the possible values of $ f(g)$?
- What does the above tell you about the value of $ f(n)$ where $ n$ is a quadratic residue $ \bmod p$?  A quadratic non-residue $ \bmod p$?
- Instead of evaluating $ a$, consider a slightly more general function. [/hide]


calculus
integration
calculus computations

Substitution $ t \equal{} \frac{2\minus{}x}{2\plus{}x}$ will work, giving $ \int^{2}_{0} \frac {(2 \minus{} x)^{2n\minus{}1}}{(2 \plus{} x)^{2n\plus{}1}} \,dx \equal{} \frac{1}{8n}$.

We can consider a general case though:

Let $ a,n > 0$ then $ \int_0^a {\frac {{(a \minus{} x)^{an \minus{} 1} }}{{(a \plus{} x)^{an \plus{} 1} }}\,dx} \equal{} \frac {1}{{2a^2n }},$ which confirms sos440 answer.

The problem is easy. :lol:

For $ n\neq 0,$

$ \int^2_0 \frac {(2 \minus{} x)^{2n \minus{} 1}}{(2 \plus{} x)^{2n \plus{} 1}} \,dx \equal{} \minus{} \frac {1}{4}\int_0^2 \left( \minus{} 1 \plus{} \frac {4}{2 \plus{} x}\right)^{2n \minus{} 1}\left( \minus{} 1 \plus{} \frac {4}{2 \plus{} x}\right)'dx$

$ \equal{} \minus{} \frac {1}{4}\left[\frac {1}{2n}\left( \minus{} 1 \plus{} \frac {4}{2 \plus{} x}\right)^{2n}\right]_0^2 \equal{} \frac {1}{8n}$

can we substitute x=$ \ce2cos2y$........................

then we can simplify using the reduction formula......

:lol: it is easy but nice ... why is the sense to invent very difficult problems no one may solve... Indeed some of integrals posted by me are still not solved  :) they doesn't seem to be very difficult.


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Evaluate

$ \int_0^{\pi/2} \frac{dx}{\sqrt{\sin x}}$

This is a version of the Beta integral. The value is $ \frac{\sqrt{\pi}\,\Gamma\left(\frac14\right)}{2\Gamma\left(\frac34\right)}.$

I don't know of any simpler way to express it. The general formula is this:

For $ \alpha>\minus{}1,\beta>\minus{}1,$ $ \int_0^{\frac{\pi}2}\sin^{\alpha}x\cos^{\alpha}x\,dx\equal{} \frac{\Gamma\left(\frac{\alpha\plus{}1}2\right) \Gamma\left(\frac{\beta\plus{}1}2\right)} {2\Gamma\left(\frac{\alpha\plus{}\beta\plus{}2}2\right)}.$

A somewhat simpler form may be obtained for this specific solution using the Gamma function identity:

$ \Gamma(x) \Gamma(1\minus{}x)\equal{}\frac{\pi}{\sin \pi x}$

In other words:

$ \Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right)\equal{}\frac{\pi}{\sin {\frac{\pi}{4}}}\equal{}\pi \sqrt{2}$

So

$ \int^\infty_0 \frac{dx}{\sqrt{\sin x}} \equal{} \pi\sqrt{\frac{\pi}{2}}\Gamma^{\minus{}2}\left(\frac{3}{4}\right)\equal{}\frac{1}{2\sqrt{2 \pi}}\Gamma^2\left(\frac{1}{4}\right)$

[quote="Kent Merryfield"]This is a version of the Beta integral. The value is $ \frac {\sqrt {\pi}\,\Gamma\left(\frac14\right)}{2\Gamma\left(\frac34\right)}.$

I don't know of any simpler way to express it. The general formula is this:

For $ \alpha > \minus{} 1,\beta > \minus{} 1,$ $ \int_0^{\frac {\pi}2}\sin^{\alpha}x\cos^{\alpha}x\,dx \equal{} \frac {\Gamma\left(\frac {\alpha \plus{} 1}2\right) \Gamma\left(\frac {\beta \plus{} 1}2\right)} {2\Gamma\left(\frac {\alpha \plus{} \beta \plus{} 2}2\right)}.$[/quote]

How was this general formula derived?

First step:

$ \int_0^{\infty}x^{\alpha}e^{\minus{}x^2}\,dx\equal{} \frac12\int_0^{\infty}u^{\frac{\alpha\minus{}1}2}e^{\minus{}u}\,du\equal{} \frac12\Gamma\left(\frac{\alpha\plus{}1}2\right).$

Knowing that, now compute $ \int_0^{\infty}\int_0^{\infty}x^{\alpha}y^{\beta}e^{\minus{}x^2\minus{}y^2}\,dx\,dy$

in two ways: first in rectangular coordinates, as written, and then in polar coordinates.

I'll let you finish the details.

I seem to be getting

$ \frac{\Gamma(\frac{\alpha+1}2) \Gamma( \frac{\beta+1}2 )}{8 \Gamma(\frac{\alpha+\beta+2}2 )}$

instead.

[quote="mathwizarddude"]I seem to be getting

$ \frac {\Gamma(\frac {\alpha + 1}2) \Gamma( \frac {\beta + 1}2 )}{8 \Gamma(\frac {\alpha + \beta + 2}2 )}$

instead.[/quote]

Never mind.  I forgot to halve each integral by changing the limits from 0 to infinity to -infinity to infinity.

May I know the reason for $ \alpha, \beta > -1?$

That's just a simple matter of size and comparison. (The integral equivalent of the "limit comparison test" for series.)

$ \sin x$ is comparable to $ x$ as $ x\to 0,$ so $ \sin^{\alpha}x$ is comparable to $ x^{\alpha}$ near $ 0.$ The integral $ \int_0^cx^{\alpha}\,dx$ converges for $ \alpha>\minus{}1$ and diverges for $ \alpha\le\minus{}1.$

It's the same at the other end, only this time it's the cosine term that matters.

hello, Maple says $ \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin(x)}}dx\equal{}\sqrt{2}Elliptic\left(\frac{1}{2}\sqrt{2}\right)$.
Sonnhard.

I want to do a home-brew numerical integral for this, using no software fancier than a spreadsheet. The singularity of the integral at $ 0$ is an annoyance, so I'll make the substitution $ x\equal{}u^2$ to get

$ \int_0^{\frac{\pi}2}\frac{dx}{\sqrt{\sin x}}\,dx\equal{} \int_0^{\sqrt{\frac{\pi}2}}\frac{2u}{\sqrt{\sin u^2}}\,du$

I thought about letting $ \sin x\equal{}u^2,$ but that introduces a new singularity at the other end. Now the integrand (with the value $ 2$ for $ u\equal{}0$) is smooth.

Running a Romberg integration on that, I get $ 2.622057554295,$ to 12 decimal places.

How can $ \int_0^{\frac {\pi}2}\sin^{\alpha}x\cos^{\beta}x\,dx\equal{}\int_0^{\infty}\int_0^{\infty}x^{\alpha}y^{\beta}e^{\minus{}x^2\minus{}y^2}\,dx\,dy$
And can you show your solution of the above integral? Thank you in advance :)

I didn't say exactly that. I said that if you do that double integral in polar coordinates, you get the product of two integrals, one of which is the one we're interested in and the other of which give you a gamma function.

$ \int_{0}^{\infty}\int_{0}^{\infty}x^{\alpha}y^{\beta}e^{-x^{2}-y^{2}}\,dx\,dy$

${ =[\int_{0}^{\infty}x^{\alpha}e^{-x^{2}}\,dx}][\int_{0}^{\infty}y^{\beta}e^{-y^{2}}\,dy]=\frac{1}{4}\Gamma(\frac{\alpha +1}{2})\Gamma(\frac{\beta +1}{2})$

And also

$ \int_{0}^{\infty}\int_{0}^{\infty}x^{\alpha}y^{\beta}e^{-x^{2}-y^{2}}\,dx\,dy$

$ =\int_{0}^{\infty}\int_{0}^{\frac{\pi}{2}}r^{\alpha +\beta +\frac{1}{2}}e^{-r^{2}}\cos^{\alpha}\theta\sin^{\beta}\theta\,d\theta\,dr=[\int_{0}^{\infty}r^{\alpha +\beta +\frac{1}{2}}e^{-r^{2}}\,dr][\int_{0}^{\frac{\pi}{2}}\cos^{\alpha}\theta\sin^{\beta}\theta\,d\theta]$

$ =\frac{1}{2}\Gamma(\frac{\alpha +\beta +2}{2})[\int_{0}^{\frac{\pi}{2}}\cos^{\alpha}\theta\sin^{\beta}\theta\,d\theta]$

$ \therefore[\int_{0}^{\frac{\pi}{2}}\cos^{\alpha}\theta\sin^{\beta}\theta\,d\theta]=[\int_{0}^{\frac{\pi}{2}}\sin^{\alpha}\theta\cos^{\beta}\theta\,d\theta]=\frac{\Gamma(\frac{\alpha +1}{2})\Gamma(\frac{\beta +1}{2})}{2\Gamma(\frac{\alpha +\beta +2}{2})}$

I got it!!! Thank you very much :blush:
One last question, how could I think of your double integral when I want to solve the original problem?  :maybe:


Here is an easy problem for those who are interrested:

 Let $S(p)=1^{p}+2^{p}+...+n^{p}$.Prove that:

   $(n+1)^{p+1}=1+C_{p+1}^1\cdot S(p)+C_{p+1}^{2}\cdot S(p-1)+...+C_{p+1}^{p}\cdot S(1)+n $ ;)

Have you taken it from "Manual clasa a X-a"?  :P  :P

No, I  didn't.So, I suppose it's there.Well, this easy formula helped me to deduce interresting things about $S(p)$.
That's why I posted.Maybe it is the wrong place.I think this should go in the "Classroom Math" :?  ;)

Sincerely, me too. This is the olympiad forum.

Cezar what do you mean by $C_i^j$
Should we just defind our notations?
I think you sould read this:
[quote="darij grinberg"]

[b]Please don't write up your problems as follows:[/b]

[quote="A typical geometric inequality"]In a triangle, prove $am_a+bm_b+cm_c\leq \sqrt{bc}m_a+\sqrt{ca}m_b+\sqrt{ab}m_c$.[/quote]

[b]Rather try to write them up as follows:[/b]

[quote="A typical geometric inequality"]Let ABC be a triangle, a, b, c its sidelengths and $m_a$, $m_b$, $m_c$ the corresponding medians. Prove $am_a+bm_b+cm_c\leq \sqrt{bc}m_a+\sqrt{ca}m_b+\sqrt{ab}m_c$.[/quote]

[b]In other words, please explicitely define all notations used instead of hoping that the reader will be able to guess their meaning.[/b] In fact, different people from different countries use very different notations for one and the same thing, and also sometimes denote by one and the same letter different things [/quote]


Let $S(n)$ be the number of integer solutions $(x,y)$ to the equation:
$3y^4+4ny^3+2xy+48=0$ such that $|x|$ is a square and $y$ is square-free
(so there is no prime $p$ such that $p^2$ divides $y$.)
Let $s$ be the maximal value of $s(n)$ (as $n$ varies).
Find all $n$ such that $s(n)=s$.

I will assume that n is integer.
we have that y(3y^3+4ny^2+2x)=-48 since y is square free its only possible values are 2,-2,3,-3,6,-6.but clearly y cannot be odd. hence
y={2,-2,6,-6}
case 1. y=2
2x+16n=-48
case2. y=-2
2x+16n=48
cass3. y=6
144n+2x=-656
case4. y=-6
hence 144n+2x=656

consider 9A=-(216+72n) and 9B=216-72n
C=-(328+72n) D=328-72n
at most one of A and B is perfect square similarly at most one of C and D is perfect square.  so the maximum number of solutions is 2.
Case 1. A and C are squares .
216+72n=9a^2
328+72n=b^2
hence 112=b^2-9a^2 =(b-3a)(b+3a)
this equation results in no solution.
so the only possibility is that A and D be perfect or B and C
A,D:
216+72n=9a^2
328-72n=b^2
hence 9a^2+b^2= 544=4*136=2^5*17 =(16+16)(4^2+1)= 20^2+12^2
hence a=4 and b=20
hence (x,y)=(16,2) ,(20,-6)
the other case is similarly done.


Show with a counter example that $\sqrt{a^2-b^2}\ne{a-b}$.

What does it equal instead?

[quote="biffanddoc"][hide]$a=1 \text{  } b=0$.
 You can't simplify $ \sqrt{a^2-b^2}$[/hide][/quote]

Actually, the example you gave us WILL work...  :D

[hide]A counter example to the equation 236factorial gave us would be if $a=2$ and $b=1$

The expression can't really be simplified, but it can be changed to $\sqrt{(a+b)(a-b)}$[/hide]


geometry
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I'm working on a research project and I realize I need to understand tempered distributions in order to pursue my current line of thought. Can someone please recommend a book from which I might learn about distributions (in particular tempered distributions)? Are there translations of Schwartz' works to English?

Thanks in advance.  :)

I found some:
Introduction to The Theory of Distributions; Friedlander & Joshi
Methods of Generalized Functions; Vladimirov
Distributions, Integral Transforms, and Applications; Kierat & Sztaba

I only made my choices by finding the right section in the library and picking ones that had nice-looking tables of contents. I'd still be appreciative of a recommendation, especially one that has a gentle introduction but still has some actual content.

There are frequently chapters on distributions in larger books. Here's one from my shelf: [i]Linear Partial Differential Operators[/i], by Lars Hörmander. The first two chapters of this book are devoted to distribution theory.

I also think there's quite a bit in the Eli Stein and Guido Weiss harmonic analysis book, but I seem to have misplaced my copy of that one.

I'd also check out the vast multivolume Gelfand and Shilov book.

Of course, none of these are exactly "gentle" - certainly not Hörmander.


analytic geometry
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[quote="mecrazywong"]Anyone here has a good source of learning polar?
Thanks :)

edit: 200th post in mathlinks :D[/quote]
Do you mean the Polar coordinates? :?

No.
The polar of a point P wrt to a circle is the line passing through the inverse of P and perpendicular to OP.
This has nothing to do with polar coordinate.

i have a question about polars.

i found in a soln on the forum the following:

if line $a$ is the polar of $A$ and $b$ is the polar of $B$ (wrt to the same circle). If $a$ intersected with $b$ $=C$, then $AB$ is the polar of $C$.

this seems obvious but i can't prove it. can anyone help me?

hey,

a solution to the problem perfect_radio posted (pretty easy):
We prove for a fixed cirle, and a point C, draw any line l through C. Then the inverse of l wrt to the circle lies on a fixed line as l varies.

Consider the circle w with diameter OC, let it meet the fixed circle at points Z1 and Z2. Our 'fixed line' is Z1Z2. For any line l through C, let L1 be the point where it intersects w, apart from C. The polar of l is in fact the inverse of L1 wrt to the fixed circle, call this point S. . Let OC meet Z1Z2 at R. Since OR.OC=OL1.OS=r^2 (where r is the radius of the fixed circle), RCL1S is cyclic, and <ORS=<OL1C=90,
so S lies on Z1Z2 as required.
This proof only works when C is outside, but when C is inside you still use the same idea, and is very similar (i'll leave it for you to finish.)

We can see this is equivalant to the problem.

we have this theorem that: ABC is a triangle and if the polars ofthe points A,B,C to circle(O)be the sides of the triangle XYZ.then the polars of the points X,Y,Zto the circle (O)will be AB,BC,CA

dear perfect radio look at it plz.
there is another theorem about pole and polar that says:
we have circle(O).let A be a point in the circle and      a       be the polar of it.
then for each point like B in line    a     we have:
the polar of B will go through point A.
its a basic theorem and i think every one who want to learn pole and polar should learn this theorem at first.
then it doesnt need proof here ;) .

[quote="ashegh"]dear perfect radio ,ididnt read  vinoth solution and i dont wether my solution is like him or not :D .
here is my solution:
point C is in line  a   .then the polar of Cto the circle (O),should go thrugh point A.(as i said at first)
also do it for pont B.
we will find that the polar of pointCshould go thrugh AandB.
and it means that AB is the polar of pointC to the circle(O). ;)[/quote]
Indeed, it is a basic theorem in pole and polar, so why do you use this thorem to prove the theorem itself?
btw, there are several ways to prove it, of which (I think) the simplest one is radical axis.

ashegh, i'm not really sure whether we can consider my problem a theorem...

btw, your solution is identical to vinoth's

dear perfect radio be sure that u question isnt a theorem.
ok.suppose that its a theorem then how can we use theorem to prove another theorem :?:  :)

[quote="ashegh"]the question that perfect radio asked wasnt a theorem.
if it was theorem i couldent use any theorem to solve it. :mad:  :!:[/quote]
OK, you are right. Anyway, I still consider the two theorems the same.

dear mecrazywong i dont want to force u my opinion.
so tell me why do u think the question that perfect radio says is a theorem?
and consider that it doesnt mather wether its theorem or not ;)

If "theorem" gives too much credit to this geometric fact, call it lemma then.

[quote="ashegh"]i believe that itsnt lemma at all.the problem that perfect radio said is not a lemma or a theorem.
its justa question like the other questions in pole and polar.[/quote]
Kindly look up your mathematics dictionary to see the definition of theorem and lemma.

i know the definition of them.
then plz think about the different betwin lemma and problem.


trigonometry
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how do you find the graph of 2(cost)i + (3sint)j when t = pi/2 ?  I don't understand.  Wouldn't it just be a point?

Well, the graph of $<x, y> = <2\cos t, 3\sin t>$ is an ellipse centered at the origin with its major axis along the y-axis.  But if $t$ has a specific value, then ya I guess it is just a point.

I believe this is a vector problem, not a polar.  Are they asking for a directional vector or anything of that sort instead?

Yes, it is a position vector of a particle in the plane time t.  I'm supposed to draw a graph of the path of the particle.  But again, I'm pretty much stumped.

This sounds like they are asking you to graph the path and then evaluate when t=pi/2.  Is that right?  If so Jimmy has already told you how to graph and you simply need to evaluate at the time.


15 aren't married, 30 don't have telephone, 25 don't have car, 20 don't have home, add this up you get 90, its possible to distibute the 90 empty slots to 90 men, so that only 10 men will ahve all 4.

before entering the last store he obviously had 2 dollars and spent them. earlier he had 5. next ,11, next,23, next 47.

Rep, in the second problem, you're doubling the money each time and then adding one.  You need to add one and then double.  For example, if he started with $5, he would spend $2.50+$1=$3.50 and end with $1.50, not $2.  The correct sequence is 2, 6, 14, 30, 62.


Hi, everyone, I am new here. I have some trouble to solve the problem below. I would really appreciate it if anyone can help me with it. Thanks in advance!

An open box is to be made from a flat piece of material 15 inches long and 4 inches wide by cutting equal squares of length x from the corners and folding up the sides.
(1) Write the volume V of the box as a function of x. Leave it as a product of factors, do not multiply out the factors.
(2)  If you write the domain of V as an open interval in the form (a,b), then what is a and b?

Ok. I figure out the answer for #1 which is (4-2x)(15-2x), but how do you find a and b?

[hide]
The volume is $x(4-2x)(15-2x)$ instead  ;)

Since the measurement of the width, length and height must be positive , it follows

$4-2x>0$ and $15-2x>0$ and $x>0$

these three gives you $x<2$ ; $x<\frac{15}{2}$ ; $x>0$

So combine it you will get $0<x<2$ which mean the domain is $(0,2)$ . So $a=0$ ,$b=2$
[/hide]

Thanks everyone for the help, but I still don't get how you get the volume. Can anyone explain it to me?

never mind, I figured out the above problem.

Here is another problem I need help with.
4^(5x-9)=(1/16)^(7x-8) Solve for X.

Please show me the steps of how you get x, thanks.

4^(5x-9)=(4^(-2))^(7x-8)
Thus,
5x-9=-2(7x-8)
19x=9+16
x=25/19
I think this should be the answer unless I have done some calcu mistake. :P
Atleast you can know the way to solve such problems. [/quote]

Hey, I gave reply to your problem within 10 mintes, isn't it nice.

Volume = length*width*height

the height will be x (draw a diagram  :P ).
The length will be 15-2x, because It is 15 to start with, and 2 bits with lenght x are removed.
Similarily, the width is 4-2x.
So $f(x)=x(15-2x)(4-2x)$.

To keep the box a box, x<2, because if it were greater than or equal to 2, the width would be negative.

Secondly, to keep the box a box, x>0, so that it has a height.

Thus, the limits are (0,2)
(OPEN intervals)


email
combinatorics unsolved
combinatorics

Maybe no one is interested in this strange problem but anyway it is just a question that i came up with...

Let $G_1,...,G_n$ be a list of of graphs. To perform a "step" is defined as follows:
pick any graph $G_i$ and perform one of the following operations:
(a) add an  edge
(b) delete an edge
(c) add an isolated vertex
(d) delete an isolated vertex

What is the minimum number of steps we need to perform to make every graph in the list isomorphic.

(NOTE: this should be in unsolved problems since i haven't solved it so any moderator please move it and thanks)

This question surely lack some other information like number of vertices or edges of all these graphs - otherwise it is unsolvable in general I guess - minimum number of steps could be arbitrary

P.S. moved  :)

I dont think is unsolvable.. it can be studied for different classes of graphs and for smaller values of $n$ and then maybe obtain general bounds and start lowering them...
for example, let see what happens if $n=2$. So we have graphs $G_1$ and $G_2$. Let $n_i$ be the number of vertices in $G_i$ .. WLOG we can assume tha $n_1 \geq n_2$. Let $d_i$ be the maximum degree in $G_i$ and let $t$ be the minimum number of steps to make both graphs in the list isomorphic. So now we can have a bound for $t$ (though maybe not the best one but we can start from here):
\[t \leq (n_1-n_2)(d_1+1)+{n_1-n_2\choose2}\]

Moreover if $G_1,G_2$ are both eigher complete or paths we can drop the ${n_1-n_2\choose2}$ in the bound.

Ah... that is what you meant - it seems to me unsolvable because I thought you were asking for exact value of minimal steps not "just" bound. Anyway I like the problem - I'll try to think on it

Yeah i like it to, is part of a small research im doing, so we can discussit privately with more depth and discuss mine and yours results at my email...
(manuelor@gmail.com)


Hi,

I just got a private message from one of the users in this site who is parent of MC participant and voluntary MC coach. Taking the idea what he/she said, I decided to introduce COP (Contest of Opportunity) which is more for high school students to the middle schoolers as well.

The contest is NOT easy if you just think you can ace it by guessing. If you work, however, you'll do very well. From 1st - 3rd COP, all the second placers were middle schoolers.  :) This means that even it's for high school students mostly, middle school students can win:

[hide="WIN"]
 :first: [/hide]

More information is found in:

[url=http://artofproblemsolving.com/Forum/viewtopic.php?t=30829]For more information[/url]

I hope this thread will introduce more middle schoolers to the competition.

Thanks. :)

Follow this link and try problems 9, 10, 13 after clicking Problems in the first post.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31211

I am giving them to my students who are going to Nats.

Naga


The Round 1 problems for the 2009-10 USA Mathematical Talent Search are now available on the [url=http://www.usamts.org]USAMTS website[/url].

The deadline for submitting your Round 1 solutions is October 13.

Please register and read the all rules on the website.  In particular, please note that, although you may use books, calculators, computers, and the internet, you may not discuss or ask questions about the problems with any other person and/or on this (or any other) forum.

Enjoy the problems!

Wait so we get two months to do round one?(evidently...)

But is it usually like that?

yes. round 1 is usually longer i guess because some people have school that start around now...


Prove that $ 67|A$ for any $ n \in \mathbb{N}$, where

\[ A\equal{}3^{n\plus{}2} \cdot 5^{n\plus{}1}\minus{}3^{n\plus{}1}\cdot 5^n\plus{}3^n \cdot 5^{n\plus{}2}.\]

hello, after some elementary computations we get
$ A\equal{}67\cdot 27^n$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, after some elementary computations we get
$ A \equal{} 67\cdot 27^n$.
Sonnhard.[/quote]

Please show your work in explaining your reasoning. Also, you made a typo--it's $ 15^n$, not $ 27^n$.

Factoring out $ A$ gives

$ A\equal{}3^n*5^n(3^2*5\minus{}3\plus{}5^2)$
$ \equal{}15^n(45\minus{}3\plus{}25)$
$ \equal{}67*15^n$

Therefore $ 67|A$.

hello, thank you for your hint, it must be $ A\equal{}67\cdot 15^n$.
Sonnhard.


Euler
number theory
greatest common divisor
number theory solved

Prove that if p|x, equation x^2-1=y^p doesn't have positive integer solution, where p is a prime larger than 3.

Just the beginning (the easier part :)):

We have (x-1)(x+1)=yp. We have 2 cases

(1) y odd => x is even, so (x-1,x+1)=1, so x-1=ap, x+1=bp, with (a,b)=1, but that's impossible because bp-ap=2 has no solutions in natural numbers. We haven't used p prime or p|x so far.

(2) y even, so x-1=2ap and
x+1=2p-1bp or vice-versa. I haven't worked this ouyt, but I think this is where we use p prime and p|x.

The condition p|x is not an 'hypothesis'...it is a special case of a theorem of Cassels on Catalan's equation (and the given problem is a special case of this equation).

Pierre.

So you mean it works even if p doesn't divide x? I mean, is this the problem:

Show that the equation x2-1=yp has no solutions in positive integers, where p>=3 is a prime

?

I've seen the problem with p=3 before (Mathlinks contest). I know a Z[i] approach works, and I think it's also the case here, but I haven't had time to really think about it.

If p = 3, I think the problem is easy and can be soved using elementary modulo (mod[4],...).
I say that the problem is a special case of Catalan's equation x m  - y n  = 1, which has recently been solved.
First, it is easy to see that we may restrict ourselves to the cases where m,n are primes.
There are many papers on the subject giving partial results, and one of the main theorems on this subject is a result due to Cassels, which states that, if p,q are odd primes and x p  - y q  = 1, where x,y > 1 are integers, then p|y and q|x.

In another hand, these results are generally not elementary and are far from olympiad level. Thus, I was only giving some informations about the problem. I think the condition p|x is essential to make it solvable from us.

Pierre.

according to this: http://mathworld.wolfram.com/CatalansConjecture.html

the only non-trival solution to x^p-y^q=1 is (x,y,p,q)=(3,2,2,3)

[quote]I've seen the problem with p=3 before (Mathlinks contest). I know a Z[i] approach works, and I think it's also the case here, but I haven't had time to really think about it.[/quote]

I don't think you can use the arithmetic of Z[i] for this problem because it is x^2-1 and not x^2+1.  For the special case of x 2 -1=y^3, it was proved by euler that this elliptic curve only has the rational points (-1,0), (0, \pm 1), and (2, \pm 3)

You're right: I was thinking about x2+1=y3

Let's solve the original problem :

The case p = 3 can be solved using elementary methods, as done in Sierpinski's 'Elementary theory of numbers' (p.77-79), and probably in many other number theory books. Since the proof is long i will not give it here (unless necessary?). I've not succeeded in prove it as the general case below, but it may be done. The classical proof uses  result of Legendre about the solutions of x 3 + y 3  = 2z 3 .

Now, I suppose that p > 3 is a prime, and that x 2  - 1 = y 3 , (1)
where x,y are positive integers, and p|x.

As proved by Grobber, we may restrict ourselves to the case where y is even, and x is odd.

Lemma.
for any integers a,b such that a - b =/= 0,  and gcd(a,b) = 1, we have
gcd(a - b , (a p  - b p )/(a - b) = gcd(a - b, pbp-1)

Proof.
Since a p  = [(a-b) + b] p , and using the binomial expansion formula, we get that :
(a p  - b p )/(a - b) = (a-b)N + pb p-1 , where N = sum{k=2 to p} C(p,k)(a-b) k b p-k , hence result.

From (1), and since gcd(x+1,x-1) = 2, we deduce that
x+1 = 2a 2 and x-1 = 2 p-1 b p
or
x-1 = 2a 2 and x+1 = 2 p-1 b p
where a is odd, and gcd(a,b) = 1.

We will prove the desired result in the first case. The second one may be proved following exactly the same lines, with some changes of signs in the expressions like a 2 - 2b,...

From x+1 = 2a 2 and x-1 = 2 p-1 b , we deduce easily that a > b.
Moreover, we have a p  - 2 p-2 b p  = 1.
It follows that (a 2 ) p  - (2b) p  = (a p  - 2) 2  = [ (x-3)/2 ] 2 .
Since p|x and p > 3 we deduce that p does not divide (x-3)/2, since a 2  - 2b divides (a 2 ) p  - (2b) p , it follows that p does not divides a 2  - 2b.

Moreover, since gcd(a,b) = 1 and that a is odd, we deduce that gcd(a 2  - 2b, p(2b) p-1 ) = 1. From the lemma, it follows that :
gcd (a 2  - 2b, [(a 2) p  - (2b), p ]/(a 2  - 2b) ) = 1.
Thus a 2  - 2b divides (a 2) p  - (2b), p  which is a square, and is coprime with [(a 2) p  - (2b), p ]/(a 2  - 2b) ). Then a 2  - 2b = c 2  is a square.
Note that c is odd and coprime with b. Thus gcd(ac,b) = 1.
Moreover 2b = a 2 -  c 2 = 0 mod[4], then b is even.

In another hand :
(a 2  - b) 2 = a 2 (a 2 - 2b) + b 2 = (ac) 2 + b 2 .
Then b,ac, a 2 - b form a primitive pythagorean triplet. It is well-known that we may deduce that :
b = 2mn, ac = m 2  - n 2 , and a 2  - b = m 2 + n 2 for some positive integers m,n with m > n  \geq  1.

It follows that : (m+n) 2  = a 2 , that is m+n = a.
Thus a - b = m+n - mn = (2m-1)(1/2 - n) + 1/2  \leq  0, which contradicts that a > b, and we are done!

Pierre.


I dunno if this is Middle School Level... you can solve it anyway  :roll:

A truck is traveling from point A to point B, which is 1000 miles away, at a rate of 50 mph. At the same EXACT time, going from point B to point A, at a rate of 100 mph, is another car. After a while of driving along, they finally meet at some point, and without stopping at all, the 2nd, faster car swiftly turns around (wasting NO time) and drives back to point B at the same rate. Once the faster car gets back to point B, he decides to go back and taunt the first car again, and so he does the same thing and the same thing until finally, the first car and the second car meet at point B. (Pretend that statement is true, if it isn't.)

How far did the 2nd car drive?

Don't calculate the intersection points, its a waste.

The car travels twice as fast. In the time it takes the truck to go 1000 miles, the car goes 2000.

In total, the second car is traveling for 1000/50=20 hours.
It's rate is 100 mph.
Therefore, in total, it travels 20*100=[u]2000 miles[/u]

This is a variation of a problem solved
by famous mathematician John von Neumann,
who worked with Kurt Godel and Albert
Einstein in the early 1920s.  You can
read about it a little here:
http://thesciencepundit.blogspot.com/2006/07/john-von-neumann-and-mathematicians.html


Consider a fair coin and a fair 6-sided die.  The die begins with the number 1 face up.  A [i]step[/i] starts with a toss of the coin: if the coin comes out heads, we roll the die; otherwise (if the coin comes out tails), we do nothing else in this step.  After 5 such steps, what is the probability that the number 1 is face up on the die?

[hide]If the coin flips tails all 5 times, then the dice will always have 1 face up; otherwise, the dice has a probability of $ \frac{1}{6}$ of having 1 face up. Thus, the desired probability is $ \frac{1}{32}\plus{}\frac{31}{32}\cdot\frac{1}{6}\equal{}\frac{37}{192}$.[/hide]


I was wondering if you can solve for y in the following equation:

$0 = \sqrt[9]{\sqrt[4]{20y}}$

if you can solve for y then solve for y
thanks

[hide="My Solution"]I think the solution's kinda obvious: $0$.

By definition, it becomes $0=\sqrt[36]{20y}$

$x^{36}=0$ has only one solution, 0, so $20y=0$ and $y=0$.[/hide]

[quote="math92"]How bout this?

$\displaystyle 1 = \sqrt[9]{\sqrt[4]{20y}}$[/quote]
let me exhaust this for you
[hide="extending RC-7th solution"]if $k=\sqrt[9]{\sqrt[4]{20y}}$, then $y=\frac{k^{36}}{20}$, done[/hide]

Extending Palytoxin's solution...

[hide]
Suppose we know that:

$k=\sqrt[n_n]{\sqrt[...]{\sqrt[n_2]{\sqrt[n_1]{cy}}}}$.

Then:

$k=\sqrt[n_1\cdot n_2\cdot ...\cdot n_n]{cy}$,

or:

$y=\frac{k^{n_1\cdot n_2\cdot...\cdot n_n}}{c}$.

[/hide]

And actually giving the answer to the problem, :P

[hide]
(20y)^36 = 1
20y=1
y=1/20=.05
[/hide]

Simply put, 1 to any power is 1, and the same applies with 0.

why are there 36 solutions?  can you give one complex number that satisfies it?  its not ^36, but rather ^(1/36)..   :?:  :?:

[quote="Palytoxin"][quote="nat mc"]For the $\sqrt[36]{20y}=1$, there are 36 solutions, by definition, but lets not get into complex numbers[/quote]
sorry but thats not true. you maybe thinking of: find $x$ if $x^{36}=1$. in this case, yea there are 36 solutions.[/quote]
O yeah, duh, I don't know what happened to me, must have read it wrong

Wait, there are 36 solutions. How are there not 36 solutions?

it is important that you be able to keep fundamentals without getting side-tracked when you learn higher level stuff!

there aren't 36 "solutions" because it becomes 1^36 = 20y = 1.  you have always known since grade school that 1^36 = 1, and it is no different here.

you have 36 solutions for x if you had something like x^36 = 1, where an undefined variable is raised to the power of 36 and we are trying to solve for the variable.  here we have 1^36, and we are certainly not trying to "solve for 1" because we already know what 1 is (one :)).  here we are solving for a normal linear variable y that is not raised to any power.


analytic geometry
conics
ellipse
geometry
circumcircle
geometry proposed

Triangle $ ABC$ with bisector $ AD,BE,CF$ ($ D,E,F$ on $ BC,CA,AB$ respectively). Call $ D',E',F'$ are symetric point of $ D,E,F$ over midpoint of $ EF,FD,DE$ respectively. Prove that $ BE',CF',EF$ are concurrent at $ T_1$, $ CF',AD',FD$ are concurrent at $ T_2$, $ AD',BE',DE$ are concurrent at $ T_3$ and $ D'T_1,E'T_2,F'T_3$ are parallel!

We'll use barycentric coordinates with respect to $\triangle ABC.$ Coordinates of the vertices $D',E',F'$ of the antimedial triangle of $\triangle DEF$ are given by:

$D' \equiv -\left (0:\frac{b}{b+c}:\frac{c}{b+c} \right)+\left (\frac{a}{c+a}:0:\frac{c}{c+a} \right)+\left( \frac{a}{a+b}:\frac{b}{a+b}:0 \right)$

$D' \equiv (a(b+c)(2a+b+c):b(c^2-a^2):c(b^2-a^2))$

$E' \equiv (a(c^2-b^2):b(c+a)(2b+a+c):c(a^2-b^2))$

$F' \equiv (a(b^2-c^2):b(a^2-c^2):c(a+b)(2c+a+b))$

$T_1 \equiv BE' \cap CF' \equiv (a(b^2-c^2):b(a^2-c^2):c(b^2-a^2)) \Longrightarrow T_1 \in EF$

$D'T_1 : \left[\begin{array}{ccc} a(b+c)(2a+b+c) & b(c^2-a^2) & c(b^2-a^2)\\  a(b^2-c^2)& b(a^2-c^2)& c(b^2-a^2)\\ x & y & z\end{array}\right]=0$

Infinity point of $D'T_1$ is indeed $X_{512} \equiv (a^2(b^2-c^2):b^2(c^2-a^2):c^2(a^2-b^2))$

In other words, Steiner circum-ellipse of $\triangle ABC$ cuts its circumcircle at $A,B,C$ and the Steiner point $X_{99} \equiv S.$ Then direction $D'T_1 \parallel E'T_2 \parallel F'T_3$ coincides with the direction of the isogonals of $AS,BS,CS.$


find the greater prime divisor of

$ 27! \plus{} 28! \plus{} 29!$

This is not a problem, only a nice coincidence:

27! + 28! + 29! = 27! ( 1 + 28 + 28·29 ) = 27! · 841 = 27! · 29·29  Then the grather prime is 29

Umm, really is not a nice coincidence, because:

(p-2)! + (p-1)! + p! = (p-2)! (1 + (p-1) + (p-1)p ) = (p-2)! · p²  Then the grather prime divisor of this sum is p when p is prime


complex analysis
complex analysis unsolved

Why are you posting it again¿ http://www.mathlinks.ro/Forum/viewtopic.php?t=77464


Let $ n$ be a positive integer. $ a_1\leq a_2...\leq a_n$; $ b_1\leq b_2...\leq b_n$ are two non-decreasing real number sequences such that $ a_1\plus{}a_2\plus{}...\plus{}a_i\leq b_1\plus{}b_2\plus{}...\plus{}b_i$ ;$ i\equal{}1,2,..n\minus{}1$ and $ b_1\plus{}b_2\plus{}...\plus{}b_n\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n$.
It is known that for any real number $ m$, the number of pairs $ (k,l)$ where $ a_k\minus{}a_l\equal{}m$ is equal to the number of pairs
$ (i;j)$ where $ a_i\minus{}a_j\equal{}m$ .Prove that:$ a_i\equal{}b_i$ $ \forall i\equal{}1,2,..,n$

Recheck your sources; this is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=982016#982016]USA TST 2007/2[/url].

I had mistakenly .Khi read a document on the problems I mistake it for an MOSP 2001

This is problem MOSP 2001
Let $ a_0,a_1,a_2,a_3,...$ and $ b_0,b_1,b_2,b_3,...$ be strictly increasing sequences of real numbers with $ a_0\equal{}b_0$ and with the following properties:
$ (1)$ $ a_i\plus{}a_j\plus{}a_k\equal{}b_t\plus{}b_m\plus{}b_n$ then the triple $ (i,j,k)$ is a permutation of $ (t,m,n)$
$ (2)$ a positive number $ x$  can be expressed as $ x\equal{}a_s\minus{}a_t$ (for some $ s>t\leq 0$) if and only if it can be expressed as
$ x\equal{}b_u\minus{}b_v$ (for some $ u>v\leq 0$)
Prove that: $ a_n\equal{}b_n$ for all positive integers $ n$.


One comment about the statement: the fields should match. That is, if we have $ A,A^{ \minus{} 1}\in M_n(\mathbb{K}),$ then we should have $ \alpha,\beta\in\mathbb{K}^n.$ Else if we have $ \alpha,\beta\in\mathbb{R}^n$ then we should have $ A,A^{ \minus{} 1}\in M_n(\mathbb{R}).$

Now: we have a claim that $ C^{ \minus{} 1} \equal{} D,$ where $ C$ and $ D$ are complicated expressions in other objects. There's a very obvious first thing to try, and that's to multiply out $ CD$ and see what happens. Have you tried that? Or perhaps I should ask, have you [i]really[/i] tried that?

One note for use somewhere in the middle: You can write $ \alpha\beta^TA^{ \minus{} 1}\alpha\beta^TA^{ \minus{} 1} \equal{} (\beta^T A^{ \minus{} 1}\alpha)(\alpha\beta^TA^{ \minus{} 1}).$ That's because $ (\beta^T A^{ \minus{} 1}\alpha)$ is scalar and can hence be factored out of the middle of the long matrix product.

[quote="Kent Merryfield"]
One note for use somewhere in the middle: You can write $ \alpha\beta^TA^{ \minus{} 1}\alpha\beta^TA^{ \minus{} 1} \equal{} (\beta^T A^{ \minus{} 1}\alpha)(\alpha\beta^TA^{ \minus{} 1}).$ That's because $ (\beta^T A^{ \minus{} 1}\alpha)$ is scalar and can hence be factored out of the middle of the long matrix product.[/quote]

That's why I hadn't solved it.Thanks.


Support
geometry
AMC
AIME
analytic geometry

[quote="CatherineAsaro"][quote="7h3.D3m0n.117"]What did you get when you took the MML? I only got a 36 out of 40 because I was hyped up on flu medicine that made me sleepy  :blush: .[/quote]

Hello!  Do you mean the Maryland Math League?  It's still going on this year.  I'm not sure that's the one you mean, though.  In the MML we do, the tests are 6 points per test, with 6 rounds, so that's a total of 36 points.

Best -- Dr. A[/quote]

Yes I mean the Maryland Math League...my results were here I do not remember how the competition actually worked though (it's been two years since i took it)

[url]http://www.mathleague.com/reports/2004_05/grade678/MD_6.HTM[/url]

[quote="chess64"]In middle school, you get one test with like 40 problems on it (30 minutes). In high school, you get 6 tests with 6 problems each (30 minutes).[/quote]

oh im gonna take one next year?

6. 40/40
7. 40/40
8. 40/40
9. So far 24/24.

I love the MML, though I noticed it is easy to make stupid mistakes.

MML=MIchigan Maht League.

jeez nice!

ive only taken the MML (maryland one) in sixth grade our school is like anti-math and only supports MC but not AMC or future MMLs....

[quote="calc rulz"]Jeez you people...

[url=http://www.massmathleague.org/]MML[/url] is a Massachusetts trademark.! :lol:[/quote]

well since this is the maryland forum i think MML qualifies as maryland math league  :D

I have 21/24 too, because I suck at math league.

[quote="mathgeniuse^ln(x)"][quote="E^(pi*i)=-1"]I'm not sure, but I think the problems are the same in every state anyway.   :wink:

I have only 21/24 this year so far because of two stupid mistakes.  :([/quote]

And what about the other mistake. :maybe:[/quote]

The other one I got wrong was Round 2 #6, which I just couldn't figure out.  I'm not a geometry person.

Yeah!!!

That question was just like #10 AIME 2005 I.

People are saying to me, are you ever not going to get a perfect score on a math league.

It was? It was really easy, just put it on the coordinate plane.

Oh, well, I didn't use a coordinate plane, I used a radius.


inequalities
function
inequalities theorems

If you have an inequality in variables $a_1,a_2,\ldots,a_n$, and replacing $a_i$ with $ka_i$ gives you the original inequality, then you can assume

$a+b+c=k$

$abc=k$

$a^2+b^2+c^2=k$

.... et cetera.

[quote="blackcode"]I saw some probem , sometimes they add a+b+c= 1 or another number . I want to ask you when and how they add this number . Thank you[/quote]
You can do so when the inequality is homogeneous.

Could you explain how to check an inequality is homogeneous? :)

Man, just look at the degree of each term!

Look at Mathworld if you don't know!

[url]http://mathworld.wolfram.com/HomogeneousFunction.html[/url]

Oh, okay, I see.. :)

Well, thanks a lot Arne! :D

[quote="LionKing_0804"]For example : Firstly , we have to solve an inequality with a,b,c >0 . After that , we assume a+b+c=k or abc=k , ...etc and we solve it with more conditions we just haved . So we have lost the general of the beginning inequality , i think so. Can anyone explain to me ?
(Sorry for my English  :blush:  )[/quote]
No, we didn't lose the generality.
Assume we have to prove
$f(x_{1},\ldots,x_{n})\geqslant 0$,
where $f$ is a homogenous function of order $k$, i.e. $f(\lambda t_{1},\ldots,\lambda t_{n})= \lambda^{k}\cdot f(t_{1},\ldots,t_{k})$. Then if we divide both sides by $\left(\frac{s}{x_{1}+\ldots+x_{n}}\right)^{k}$, we get
$f\left(\frac{x_{1}\cdot s}{x_{1}+\ldots+x_{n}},\ldots, \frac{x_{n}\cdot s}{x_{1}+\ldots+x_{n}}\right)\geqslant 0$
and by substituting $\frac{x_{i}\cdot s}{x_{1}+\ldots+x_{n}}$ as new $x_{i}$ we get an equivalent inequality of the same form as at the beginning:
$f(x_{1},\ldots,x_{n})\geqslant 0$,
but we have $x_{1}+\ldots+x_{n}=s$.
This is the reason why you may assume WLOG ([b]w[/b]ithout [b]l[/b]oss [b]o[/b]f [b]g[/b]enerality) that $x_{1}+\ldots+x_{n}=s$: you can just multiply all $x_{i}$ by $\frac{s}{x_{1}+\ldots+x_{n}}$ to obtain that their sum is $s$, without changing the inequality.
I hope it's clear now. :)

[quote="blahblahblah"]If you have an inequality in variables $a_{1},a_{2},\ldots,a_{n}$, and replacing $a_{i}$ with $ka_{i}$ gives you the original inequality, then you can assume

$a+b+c=k$

$abc=k$

$a^{2}+b^{2}+c^{2}=k$

.... et cetera.[/quote]

well if the inequality is homogenous then this works ofr all $k$ how do you know which k to choose

I think u use whatever k u like :P That's its benefit :wink:


inequalities
number theory unsolved
number theory

For positive integers $ m,n$, prove that
$ lcm(m,n) \plus{} lcm(m\plus{}1,n\plus{}1) > \dfrac{2mn}{\sqrt{m\minus{}n}}$.

First, you must change the term $ m\minus{}n$ to $ |m\minus{}n|$ in order to make the formula valid in all cases.

Applying AM-GM, we have:
$ LHS \equal{} \frac{mn}{\gcd(m,n)} \plus{} \frac{(m\plus{}1)(n\plus{}1)}{\gcd(m\plus{}1,n\plus{}1)} \geq 2\frac{\sqrt{mn(m\plus{}1)(n\plus{}1)}}{\sqrt{\gcd(m,n) \gcd(m\plus{}1,n\plus{}1)}} > \frac{2mn}{\sqrt{\gcd(m,n) \gcd(m\plus{}1,n\plus{}1)}}$ (*)

Moreover, because $ \gcd(m,n) \gcd(m\plus{}1,n\plus{}1) \equal{} \gcd(m\minus{}n,n) \gcd(m\minus{}n,n\plus{}1)$ and $ \gcd(n,n\plus{}1) \equal{} 1$, $ \gcd(m,n) \gcd(m\plus{}1,n\plus{}1) \equal{} \gcd(m\minus{}n,n(n\plus{}1)) \leq |m\minus{}n|$ (**).

Combining (*) and (**) yields the inequality.

[quote="Aravind Srinivas L"]Well isnt this a strict inequality? Sorry for the doubt u could add rigority to the proof by saying that they cant be equal for equalty.[/quote]
Sure :)

Assume wlog $ m > n$. ($ m\equal{}n$ makes no sense). Let $ k \equal{} m \minus{} n > 0$.
Then notice $ lcm(a,b)\cdot \gcd(a,b) \equal{} ab$. So we have:
$ lcm(m,n) \plus{} lcm(m\plus{}1,n\plus{}1) > \dfrac{2mn}{\sqrt{m\minus{}n}}$ $ \iff$
$ \frac{mn}{\gcd(m,n)} \plus{} \frac{mn}{\gcd(m\plus{}1,n\plus{}1)} \plus{} \frac{m\plus{}n\plus{}1}{\gcd(m\plus{}1,n\plus{}1)} > \dfrac{2mn}{\sqrt{m\minus{}n}}$ $ \iff$
$ \frac{1}{\gcd(m,n)} \plus{} \frac{1}{\gcd(m\plus{}1,n\plus{}1)} \plus{} \frac{m\plus{}n\plus{}1}{mn\gcd(m\plus{}1,n\plus{}1)} > \dfrac{2}{\sqrt{m\minus{}n}}$ $ \iff$
$ \frac{1}{\gcd(n\plus{}k,n)} \plus{} \frac{1}{\gcd(n\plus{}k\plus{}1,n\plus{}1)} \plus{} \frac{m\plus{}n\plus{}1}{mn\gcd(m\plus{}1,n\plus{}1)} > \dfrac{2}{\sqrt{k}}$.
Since $ \frac{m\plus{}n\plus{}1}{mn\gcd(m\plus{}1,n\plus{}1)} > 0$ we just have to prove:

$ \frac{1}{\gcd(n\plus{}k,n)} \plus{} \frac{1}{\gcd(n\plus{}k\plus{}1,n\plus{}1)} \ge \frac{2}{\sqrt{k}}$.
Now let $ a \equal{} \gcd(n\plus{}k,n) \equal{} \gcd(n,k)$ and $ b \equal{} \gcd(n\plus{}k\plus{}1,n\plus{}1) \equal{} \gcd(k,n\plus{}1)$. Then $ ab \le k$. And we just have to prove:
$ \frac{1}{a} \plus{} \frac{1}{b} \ge \frac{2}{\sqrt{ab}} \ge \frac{2}{\sqrt{k}}$ which is obvious. So the inequality is sharp. QED :)


well its actually quite simple, when the pressure is decreased in the chest cavity (and as a result in the lungs) the pressure in the lungs is lower than that of the atmosphere outside. Due to this imbalance air rushes into your lungs to equalize the pressure (this about piercing a pressurized container, and all the contents rushing out). I hope that makes sense, i just did my test on breathing last week (as well as the circulatory and digestive systems), but I know my bio otherwise:)

I'm not sure about the water, my guess would be that its the same water that we drink in.  I haven't done any of the body systems in biology yet, but I think that water is the same water that we breath out.

The was should be the plasma (90% water) of your blood and also the cytoplasm of the cells since the exchange occurs between blood and cells, and blood and lungs which is why air in the lungs is moistened in the nasal cavity by the tiny bones (cant remember their names)

It's been a long time, but I can't resist.

Water is present 70% in the body.

Carbonic Acid:

CO[size=59]2[/size] + H[size=59]2[/size]O -> H[size=59]2[/size]CO[size=59]3[/size].

This is reversed when the CO[size=59]2[/size] reaches your lungs.

Understand?!?!

Good...
[hide]If you all know this, wow...[/hide]


[quote="onlylove_math"]Let $ x,y,z \ge 0$ and $ x \plus{} y \plus{} z \equal{} 3$ . Prove that :

$ \frac {x}{xy \plus{} p} \plus{} \frac {y}{yz \plus{} p} \plus{} \frac {z}{zx \plus{} p} \ge \frac{3}{p \plus{} 1} \forall p \in [0,1]$[/quote]
It was also proposed by VasC's ;)

[quote="onlylove_math"][quote="can_hang2007"][quote="onlylove_math"]Let $ x,y,z \ge 0$ and $ x \plus{} y \plus{} z \equal{} 3$ . Prove that :

$ \frac {x}{xy \plus{} p} \plus{} \frac {y}{yz \plus{} p} \plus{} \frac {z}{zx \plus{} p} \ge \frac {3}{p \plus{} 1} \forall p \in [0,1]$[/quote]
It was also proposed by VasC's ;)[/quote]
Oh, Can you post solution ? Thanks! :)[/quote]
Sorry, I dont want to post the solution [b]for you[/b]

[quote="can_hang2007"][/quote][quote="onlylove_math"][quote="can_hang2007"][quote="onlylove_math"]Let $ x,y,z \ge 0$ and $ x + y + z = 3$ . Prove that :

$ \frac {x}{xy + p} + \frac {y}{yz + p} + \frac {z}{zx + p} \ge \frac {3}{p + 1} \forall p \in [0,1]$[/quote]
It was also proposed by VasC's ;)[/quote]
Oh, Can you post solution ? Thanks! :)[/quote][quote="can_hang2007"]
Sorry, I dont want to post the solution [b]for you[/b][/quote]
Sorry for I disturbed you :oops:  . And who can help me to prove  ?Thanks :)


[quote="hunter34"]For what? You didn't define which variable.[/quote]Solve for $ x$ perhaps... and the first fraction has two cancelling terms in the denominator, namely $ \minus{}ac$ and $ \plus{}ac$, and I wonder if there is a typo there  :roll:

I believe he means for $ x$.

[hide="Outline of method"]
Move the RHS to the LHS, write with common denominator, then you pretty much have whatever's in the numerator equal to $ 0$, unless the denominator's equal to zero, in which case some of our previous writing with common denominators was illegal, but we can ignore that. We set the denominator equal to zero, then it's polynomial solving time. Plug back in to see if the solution in terms of $ c$ is extraneous.[/hide]

Well, it might get more complicated one the typo is fixed, though.

[quote="Elio (n)"]$ \frac {2}{a^{2} \minus{} ac \minus{} ax \plus{} \boxed{cx}} \plus{} \frac {1}{x^{2} \minus{} ax \minus{} cx \plus{} ac} \equal{} \frac {1}{c^{2} \minus{} cx \minus{} ac \plus{} ax}$[/quote]

I belive that's what he intended.

[hide="Hint"]
$ \frac {2}{(x\minus{}a)(c\minus{}a)} \plus{} \frac {1}{(x\minus{}a)(x\minus{}c)} \equal{} \frac {1}{(x\minus{}c)(a\minus{}c)}$

Now multiply out by $ (x\minus{}a)(x\minus{}c)(a\minus{}c)$ and solve.
[/hide]

[hide="Solution"]$ \frac {2}{a^{2} \minus{} ac \minus{} ax \plus{} cx} \plus{} \frac {1}{x^{2} \minus{} ax \minus{} cx \plus{} ac} \equal{} \frac {1}{c^{2} \minus{} cx \minus{} ac \plus{} ax}$
$ \iff \frac {2}{a(a \minus{} x) \minus{} c(a \minus{} x)} \plus{} \frac {1}{x(x \minus{} a) \minus{} c(x \minus{} a)} \equal{} \frac {1}{c(c \minus{} x) \minus{} a(c \minus{} x)}$
$ \iff \frac {2}{(x \minus{} a)(c \minus{} a)} \plus{} \frac {1}{(x \minus{} a)(x \minus{} c)} \equal{} \frac {1}{(x \minus{} c)(a \minus{} c)}$
$ \iff \frac {2(x \minus{} c) \plus{} 1(c \minus{} a)}{(x \minus{} a)(x \minus{} c)(c \minus{} a)} \equal{} \frac { \minus{} 1(x \minus{} a)}{(x \minus{} a)(x \minus{} c)(c \minus{} a)}$
$ \iff \frac {2x \minus{} c \minus{} a}{(x \minus{} a)(x \minus{} c)(c \minus{} a)} \equal{} \frac {a \minus{} x}{(x\minus{}a)(x \minus{} c)(c \minus{} a)}$
$ \implies 2x \minus{} c \minus{} a \equal{} a \minus{} x$
$ \iff 3x \equal{} 2a \plus{} c$
$ \iff x \equal{} \boxed{\frac {2a \plus{} c}{3}}$

As long as the algebra is right.[/hide]

Hmmm...
how did you go from $ \frac{2x\minus{}c\minus{}a}{(x\minus{}a)(x\minus{}c)(c\minus{}a)}\equal{} \frac{a\minus{}x}{(x\minus{}c)(c\minus{}a)}$
to $ 2x\minus{}c\minus{}a \equal{} a\minus{}x$?

It would actually be $ 2x\minus{}c\minus{}a\equal{}x^2\minus{}a^2$, methinks.

Isn't it $ (a\minus{}x)(x\minus{}a) \equal{} \minus{}x^2 \plus{} 2ax \minus{} a^2$?

I left $ x \minus{} a$ out of the denominator in the step before I multiplied it out.

Also, the TI-$ 89$ returns $ \frac { \minus{} 3x \plus{} 2a \plus{} c}{(x \minus{} a)(x \minus{} c)(a \minus{} c)} \equal{} 0$ when the original equation is typed in.


A can is in the shape of a right circular cylinder. The circumference of the base of the can is 12 inches. and the height of the can is 5 inches. A spiral strip is painted on the can in such a way that it winds around the can exactly once as it reaches from the bottom of the can to the top. It reaches the top of the can directly above the spot where it left the bottom. What is the length in inches of the stripe?
[asy]size(150);
import solids;
currentprojection=orthographic(0,100,25);
defaultpen(linewidth(0.8));
revolution c=cylinder((0,0,0),(6/pi),5,Z);
c.draw(backpen=dashed);[/asy]

Unraveling the "sides" of the cylinder, we see that it becomes a $ 5\times12$, since one side is the height, and one is the circumference.  The spiral is now the diagonal of our rectangle.  Conveniently, we have a $ 5, 12, 13$ right triangle, so our answer is $ \boxed{13}$

"Unrolling" the can makes it a rectangle. Doing so gives the picture
[asy]dot((0, 0));
dot((6, 0));
dot((6, 2.5));
dot((0, 2.5));

draw("$12$", (0, 0)--(6, 0));
draw("$5$", (6, 0)--(6, 2.5));
draw((6, 2.5)--(0, 2.5));
draw((0, 2.5)--(0, 0));
draw("$h$", (0, 0)--(6, 2.5), linewidth(1.5bp));[/asy]

Using the pythagorean theorem, we have $ h \equal{} \sqrt {5^2 \plus{} 12^2} \equal{} \sqrt {25 \plus{} 144} \equal{} \sqrt {169} \equal{} \boxed{13}$

EDIT: beaten by two people? :maybe:

guys, I don't know if this is an error or not, but the answer (supposedly) is 15.......


Let $a,b,c,d$ be positive real numbers which satisfy $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that
\[\underset{cyc}{\sum}\frac{1}{\sqrt{1-ab}}\le \frac{8\sqrt{3}}{3}\]
:)

[quote="VIETNAM"][quote="toanhocmuonmau"]Let $a,b,c,d$ be positive real numbers. Prove that
\[\underset{cyc}{\sum}\frac{1}{\sqrt{1-ab}}\le \frac{4\sqrt{3}}{3}\]
:)[/quote]
It's false in case a=0,1;b=0,2;c=0,3;d=0,4. :)
I think you forgot the condition for a,b,c,d. :wink:[/quote]
Yes, i have edited it. :) Sorry.

Can, this follows from an old one proposed by me as an inspiration of Vasc
\[\sum_{\rm cyclic}\frac1{1-ab}\leq8.\]
It is true for real numbers.


function
number theory
prime numbers
number theory proposed

[quote="stvs_f"]let $ P \equal{} [{P_1,P_2,...,P_{2005}}]$ that $ p_i$ is prime.let$ A \equal{} [P_{b_1}P_{b_2}...P_{b_{1002}}] and B \equal{} [P_{c_1}p_{c_2}...P_{c_{1003}}]$
($ b_i,c_i \in [1,2,...,2005]$)
find an One to One Correspondence $ f: A \to B$ such that $ a|f(a)$[/quote]

Since $ f(a)$ is prime $ \forall a$ and $ 1\notin A$, we get that, in order to have $ a|f(a)$, we must have $ f(a)\equal{}a$ and so $ A\subset B$

[quote="stvs_f"][quote] Since $ f(a)$ is prime $ \forall a$ and $ 1\notin A$, we get that, in order to have $ a|f(a)$, we must have $ f(a) \equal{} a$ and so $ A\subset B$[/quote]
Why f(a) is prime? I think you didnt understand this problem.(source:Korean 2005 i think)[/quote]

$ f(a)\in B$
$ B \equal{} \{P_{c_1}...P_{c_{1003}}\}$

$ c_i\in[1,2005]$

$ p_i$ is prime $ \forall i\in[1,2005]$

So all elements of B are prime numbers and so $ f(a)$ is prime.

[quote="stvs_f"][quote] Since $ f(a)$ is prime $ \forall a$ and $ 1\notin A$, we get that, in order to have $ a|f(a)$, we must have $ f(a) \equal{} a$ and so $ A\subset B$[/quote]
Why f(a) is prime? I think you didnt understand this problem.(source:Korean 2005 i think)[/quote]

maybe we don't understand the problem because you didn't state it clearly. the problem as it is looks like a joke and it's doubtful that it would have appeared on a Korean olympiad. I must profusely apologize though if I'm the one whose misinterpreting things ;]

let P is set of  2005 prime and different number.A is set of all product of each 1002 member of P and B is set of all product of each  1003 member .find an One-to-one correspondence $ f: A \to B$ such that a|f(a)

(A={$ p_1 \times p_2 \times ... \times p_{1002},...,p_{1004} \times p_{1005} \times ... \times p_{2005}$}
B={$ p_1 \times p_2 \times ... \times p_{1003},...,p_{1003} \times p_{1004} \times ... \times p_{2005}$}

(sorry for my writing) :blush:

i'm sorry, i still dont quite get it. a "one to one correspondence" usually means bijection, but $ A$ and $ B$ have different sizes, so there can't be a bijection between them. on the other hand if you just want a function $ f$ that satisfies $ a|f(a)$, that's easy to get: the way, you've listed the elements, map the first element of $ A$ to the first of $ B$, the second to the second, ..., the 1003rd element of $ A$ to the 1003rd element of $ B$, and the 1004th element of $ A$ to the 1003rd element of $ B$.

[quote="pleurestique"]i'm sorry, i still dont quite get it. a "one to one correspondence" usually means bijection, ...[/quote]

I think he's speaking about injection.


This one is very very nice. I remember that lots of people spent long time with this one. In fact it follows from sum ab/(c+1)=sum ab/(a+c+b+c)<=1/4*sum (a/(a+c)+b/(b+c))=1/4.


For all $ a,b,c$ be nonnegative real numbers, show that
\[ \frac{a(a^2\plus{}bc)}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b(b^2\plus{}ca)}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{c(c^2\plus{}ab)}{a^2\plus{}ab\plus{}b^2}\ge2\frac{a^3\plus{}b^3\plus{}c^3}{a^2\plus{}b^2\plus{}c^2}\]

Dear dduclam, can you please your proof for this nice inequality again unsolved?

Thank you very much. :)

Has anyone tried to solve it?  I tried but I failed,sorry? :(

[quote="dduclam"]For all $ a,b,c$ be nonnegative real numbers, show that
\[ \frac {a(a^2 \plus{} bc)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b(b^2 \plus{} ca)}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {c(c^2 \plus{} ab)}{a^2 \plus{} ab \plus{} b^2}\ge2\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}\]
[/quote]
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$ Hence, your inequality is equivalent to
$ 2u^3w^6\plus{}(72u^6\minus{}90u^4v^2\plus{}24u^2v^4\minus{}2v^6)w^3\plus{}f(u,v^2)\geq0.$
But $ \left(2u^3w^6\plus{}(72u^6\minus{}90u^4v^2\plus{}24u^2v^4\minus{}2v^6)w^3\plus{}f(u,v^2)\right)_{w^3}'\equal{}$
$ \equal{}4u^3w^3\plus{}72u^6\minus{}90u^4v^2\plus{}24u^2v^4\minus{}2v^6\geq0.$
Thus, for the proof it remains to check our inequality for $ b\equal{}c\equal{}1$ and for $ b\equal{}1,$ $ c\equal{}0,$
which is not hard.

Thank you very much arqady for your interest in this NICE inequality?

See it here: http://can-hang2007.blogspot.com/2010/01/inequalities-66-d-d-lam.html

Thank you very much, my dear friend. :)

Your proof is really beautiful  :)

P.S. You are a very kind people to solve for us so difficult inequalities :)


ABC is a triangle. The points A', B', C' are on sides BC, CA, AB and AA', BB', CC' meet at O. Also AO/A'O + BO/B'O + CO/C'O = 92. Find (AO/A'O)(BO/B'O)(CO/C'O).

Bummer, no solution on kalva for this one.

Yes, mass points probably would work.  You might also want to use the fact that A'O/AA' + B'O/BB' + C'O/CC' = 1, which might make it easier.

Sorry for the late response, but would someone mind briefly running through the solution to this one using barycentric coordinates? I'm not very proficient with those, so I don't really know what to do here...

I havent worked out a barycentric solution, although I did solve it without them...

[hide]Let A'O, B'O, C'O be a, b, and c respectively, and let AO, BO, CO be ka, mb, nc respectively. Then it is given that k+m+n = 92 (the a, b, and c cancel out). Using JBL's identity A'O/A'A+B'O/B'B+C'O/C'C = 1 (which by the way is VERY useful), we get 1/(k+1)+1/(m+1)+1/(n+1) = 1, which after a bit of algebra becomes 3 + 2(k+m+n) + (km+mn+nk) = 1+(k+m+n) + (km+mn+nk) + kmn. After simplifying, kmn = k+m+n+2 = 94, which is the answer.[/hide]


Is there a 5-tuple of positive integers $(a,b,c,d,n)$ such that $n>d>c>b>a$ and $n^2=d^2+a^2=b^2+c^2$?

All one needs is for $n$ to have a lot more divisors of the form $4k+1$ than those of form $4k+3$. The number of representations of $t$ as the sum of two squares, counting both order and sign, is $4(d_1-d_3)$, where $d_i$ is the number of divisors of $t$ of the form $4k+i$. What you want here is $4(d_1-d_3)>12$ ($d_1$ and $d_3$ computed for $n^2$), which is easy to achieve. You can even find $n$ s.t. $n^2=a_1^2+b_1^2=\ldots=a_k^2+b_k^2$ for arbitrarily large $k$, where the couples $(a_i,b_i)$ are essentially distinct.

[quote="grobber"]The number of representations of $t$ as the sum of two squares, counting both order and sign, is $4(d_1-d_3)$, where $d_i$ is the number of divisors of $t$ of the form $4k+i$. [/quote]

Grobber,

Do you have a proof of this statement that you can show me?

Best,


Let $ a,b,c,d,p,q$ be positive reals such that $ p\geq q$. Prove that

$ \frac{a}{pb\plus{}qc}\plus{}\frac{b}{pc\plus{}qd}\plus{}\frac{c}{pd\plus{}qa}\plus{}\frac{d}{pa\plus{}qb}\geq \frac{4}{p\plus{}q}$.

Does the inequality still hold if $ p<q$?

[quote="outback"]Let $ a,b,c,d,p,q$ be positive reals such that $ p\geq q$. Prove that

$ \frac {a}{pb \plus{} qc} \plus{} \frac {b}{pc \plus{} qd} \plus{} \frac {c}{pd \plus{} qa} \plus{} \frac {d}{pa \plus{} qb}\geq \frac {4}{p \plus{} q}$.

Does the inequality still hold if $ p < q$?[/quote]
$ \sum_{cyc}\frac {a}{pb \plus{} qc} \equal{}\sum_{cyc}\frac{a^2}{pab\plus{}qac}\geq\frac{(a\plus{}b\plus{}c\plus{}d)^2}{p(ab\plus{}bc\plus{}cd\plus{}da)\plus{}2q(ac\plus{}bd)}.$
Thus, it remains to prove that
$ (p\plus{}q)(a\plus{}b\plus{}c\plus{}d)^2\geq4p(ab\plus{}bc\plus{}cd\plus{}da)\plus{}8q(ac\plus{}bd).$
But $ (p\plus{}q)(a\plus{}b\plus{}c\plus{}d)^2\geq4p(ab\plus{}bc\plus{}cd\plus{}da)\plus{}8q(ac\plus{}bd)\Leftrightarrow$
$ \Leftrightarrow p(a\minus{}b\plus{}c\minus{}d)^2\plus{}q((a\plus{}b\plus{}c\plus{}d)^2\minus{}8(ac\plus{}bd))\geq0\Leftrightarrow$
$ \Leftrightarrow(p\minus{}q)(a\minus{}b\plus{}c\minus{}d)^2\plus{}q((a\plus{}b\plus{}c\plus{}d)^2\plus{}(a\minus{}b\plus{}c\minus{}d)^2\minus{}8(ac\plus{}bd))\geq0\Leftrightarrow$
$ \Leftrightarrow(p\minus{}q)(a\minus{}b\plus{}c\minus{}d)^2\plus{}2q((a\minus{}c)^2\plus{}(b\minus{}d)^2)\geq0.$


Can anyone help me prove the following languages are not context free using the pumping lemma?

L = {w: na(w) = nb(w) = nc(w)}	(a,b and c are subscripts)

L = {a^n b^n c^m: n <= m <= 2n}

I just DO NOT understand this stuff. I would be eternally grateful if someone could help me

Thanks,
 MJF431

Simple. Find an element $a$ such that $a$'s allowable expansions depend on its neighboor. Then show that $a$ has two different neighboors.


algebra
polynomial
inequalities proposed
inequalities

Prove that for positive reals x, y, z we have

$(x+y)(y+z)(z+x)(x+y+z)^{2}\ge 24xyz(x^{2}+y^{2}+z^{2})$

Thank you for your info. ML is so big! Everything can be found here!

It is remained to note that this ineq has nice geometric interpretation:

$R/r \ge 2+8((a-b)^{2}+(b-c)^{2}+(c-a)^{2})/(a+b+c)^{2}$

and 8 is the best constant.


Given a positive integer n. If A is the union of the set of the first n positive integer and the set of first n negative integer, find how may ways we can select two integers, a and b from the set A so that :
a^2 - b^2 = a-b

do you mean how you could solve this combinatorically? I don't think it's necessary. Instead, try factoring the what you need to find.

After you factor it, you can apply combinatorics to count the number of pairs (a,b) that work.
From factoring, a=b or a=1-b. In the first case, there are 2n pairs. In the second case, there are 2n-2 (I think).

[hide]Factoring, we get $ a+b=1$ or $ a=b$, which would  yield $ n-1$ solutions (given the fact that $ 0$ is not in $ A$), and $ n$ solutions, respectively. Thus we have $ 2n-1$ solutions.[/hide]


geometry
calculus
derivative
trigonometry
similar triangles

Here's the problem:
A building is to be braced by a beam that must rest on the ground and pass over a vertical wall 10 ft high and 8 ft from the building. Find the shortest length that the beam can be.

I can't attach my picture for some reason. I think it's too big, almost 23 MB. How can I make it not so big?

I have it kinda planned out, if only I could show my picture...

The first diagram at the bottom of the page is of the problem, with $ x$ being the angle between the ground and the beam. It only requires elementary geometry to show that the two $ x$s are indeed equal. The length $ L(x)$ of the beam is equal to $ A + B$. We must find $ A$ and $ B$ in terms of $ x$.
$ \sec x = A/8$, so $ A = 8\sec x$.
$ \csc x = B/10$, so $ B = 10\csc x$
Combining these equations yields $ L(x) = 8\sec x + 10\csc x$
We want to minimize $ L(x)$, so we take the derivative the right side and set it equal to 0. After solving for $ x$, this will tell us where a local minimum of $ L(x)$ is.
Since $ \frac{d}{dx} \sec x = \sec x \tan x$ and $ \frac{d}{dx} \csc x = -\csc x\cot x$,
$ L'(x) = 8\sec x\tan x - 10\csc x\cot x$
$ 0 = 8\sec x\tan x - 10\csc x\cot x$
$ 10\csc x\cot x = 8\sec x\tan x$
To simplify the equation, we make the equation purely in terms of $ \cos x$ and $ \sin x$
$ 10(\frac{1}{\sin x})(\frac{\cos x}{\sin x}) = 8(\frac{1}{\cos x})(\frac{\sin x}{\cos x})$
$ 10(\frac{\cos x}{\sin^2 x}) = 8(\frac{\sin x}{\cos^2 x})$
$ \frac{\sin^3 x}{\cos^3 x} = \frac{10}{8} = \frac{5}{4}$
$ (\frac{\sin x}{\cos x})^3 = \frac{5}{4}$
$ \tan^3 x = \frac{5}{4}$
$ \tan x = \sqrt[3]{\frac{5}{4}}$
$ x = \tan^{-1} \sqrt[3]{\frac{5}{4}}$
Plugging this value of x into L(x) yields
$ minL(x) = 8\sec (\tan^{-1} \sqrt[3]{\frac{5}{4}}) + 10\csc (\tan^{-1} \sqrt[3]{\frac{5}{4}})$
This can be simplified with some help from the second diagram at the bottom of the page. We see that $ \tan^{-1} a = y$, $ \sec y = \sqrt{a^2 + 1}$, and $ \csc y = \frac{\sqrt{a^2 + 1}}{a}$. Therefore, $ \sec(\tan^{-1} a) = \sqrt{a^2 + 1}$ and $ \csc(\tan^{-1} a) = \frac{\sqrt{a^2 + 1}}{a}$. Letting $ a =  \sqrt[3]{\frac{5}{4}}$, we obtain
$ minL(x) = 8(\sqrt{(\sqrt[3]{\frac{5}{4}})^2 + 1}) + 10(\frac{\sqrt{(\sqrt[3]{\frac{5}{4}})^2 + 1}}{\sqrt[3]{\frac{5}{4}}})$
${ minL(x) = 8(\sqrt{(\frac{5}{4}})^{\frac{2}{3}} + 1}) + 10(\frac{\sqrt{(\frac{5}{4}})^{\frac{2}{3}} + 1}{\sqrt[3]{\frac{5}{4}}})$
Plugging this value into a calculator yields that $ L(x) \approx 25.40331527$.
I tried to do it without derivatives, but I couldn't work anything out with similar triangles or the Pythagorean Theorem. Anyway, hope this helped!


geometric sequence
number theory proposed
number theory

Problem from http://www.mathlinks.ro/Forum/viewtopic.php?t=64838  :

consider seq. An=3^n-2^n,n natural no.
show no 3 terms of this seq. can be in geometric progression.

[mod: g.p. is not the best know shortcut]

generalize this!

My generalization is that the sequence $x_{n}=a^{n}-b^{n}$ for $a>b>0$ integers contains no (or at most one) nontrivial geometric progression (the one exception can only happen for $a=2b$).

But please continue in the first link.


function
induction
algebra proposed
algebra

The function $f : \mathbb{Z}_0 \mapsto \mathbb{Z}_0$ is defined as follows:

$f(1) = 1, f(3) = 3$ and then for every $n \geq 1$, $f(2n) = f(n)$, $f(4n + 1) = 2f(2n + 1) - f(n)$ and $f(4n + 3) = 3f(2n + 1) - 2f(n)$.

Find how numbers $n$ where $1 \leq n \leq 2006$ such that $f(n) = n$.

Consier binary system:
We can prove by induction that
if $n=(a_ma_{m-1}\ldots a_0)_2$, then $f(n)=(a_0a_1\ldots a_m)_2$
So that's very easy following. :P

An old problem!


Let $ ABC$ be an isosceles triangle, $ AB\equal{}AC$ . Denote the midpoint $ D$ of the side $ [BC]$ . Consider two mobile points $ M\in (AB)$ , $ N\in (AC)$ so that $ \widehat {MDN}\equiv\widehat {ABC}$ . Prove that the product $ BM\cdot CN$ is constant and ascertain the geometrical locus of the point $ L\in MN$ for which $ DL\perp MN$ .

Hello!

$ \angle MDN \equal{} \angle ABC \equal{} \angle ACB \equal{} x$,
$ \angle MDB \equal{} \angle CND \equal{} y$, $ \angle BMD \equal{} \angle CDN \equal{} 180^{\circ} \minus{} \left(x \plus{} y\right)$.

$ \Delta BMD \sim \Delta CDN \Rightarrow \frac {MD}{DN} \equal{} \frac {BM}{CD} \equal{} \frac {BD}{CN}$
$ \frac {BM}{CD} \equal{} \frac {BD}{CN}\Rightarrow BM\cdot CN \equal{} BD\cdot CD\Rightarrow BM\cdot CN \equal{} \frac {BC^{2}}{4}$

$ \frac {MD}{DN} \equal{} \frac {BM}{CD}\Rightarrow \frac {DM}{BM} \equal{} \frac {DN}{BD}$ and $ \angle MDN \equal{} \angle MBD \equal{} x \Rightarrow \Delta MDN \sim \Delta MBD \Rightarrow$
$ \Rightarrow \angle MND \equal{} \angle MDB \equal{} y$ and $ \angle DMN \equal{} \angle BMD \equal{} 180^{\circ} \minus{} \left(x \plus{} y\right)$

$ \angle DMN \equal{} \angle DMB \Rightarrow MD$ is the bisector of the angle $ \angle BMN$   (1)
$ \angle DNM \equal{} \angle DNC \Rightarrow ND$ is the bisector of the angle $ \angle CNM$   (2)
(1), (2) $ \Rightarrow D$ is the center of the A-excircle of the triangle $ \Delta AMN$

Construct $ DP\bot AB$, $ P\in\left(AB\right)$ and $ DQ\bot AC$, $ Q\in\left(AC\right)$
The geometrical locus of the point $ L$ is the arc $ \widehat{PQ} \left(L\in\widehat{PQ}\right)$ of the circle with center $ D$ and radius $ DP$.  :)


A quadrilateral has the following vertices: (2,0), (4,5), (5,2), (1,4). How many square units are in its area?

There is a method called the [url=http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem]Shoelace Theorem[/url].

2  0
4  5
5  2
1  4

so the answer would be $ |38\minus{}27|/2$ which is simplified to 5.5, the area is 5.5

Eh, you forgot to wrap it around, or add 2 0 again at the end.  Read the statement of the Shoelace Theorem more closely.

oh yea, wow, i did forget to add 2 0 again.

2  0
4  5
5  2
1  4
2  0

so the ans is rreally $ |38\minus{}35|$ $ /2$  and get $ 1.5$

VERY good and easy link to learn this:
[url]http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/url]

Uh, well, when I used my calculator (I have a program to find the area of a quadrilateral given coordinates), it gave me 12.  Because of this, I drew a diagram of the quadrilateral, drew a 5 x 4 rectangle surrounding it and subtracted the area of the 4 triangles that lie inside the rectangle but outside the quadrilateral in question.  I got 12 again, so I think the answer is 12, but I am notorious for being incorrect.  For some reason though, I don;t think the Shoelace Theorem works in this case, I thought the coordinates had to be placed in clockwise order or something?

Yes, they have to be in cyclical order (doesn't matter clockwise or counterclockwise), so if the quadrilateral is $ ABCD$, you can't do shoelace with $ ACBD$ (which some of the above posters tried to do).

So your shoelace is actually:

2 0
5 2
4 5
1 4
2 0

Alternatively, you can break the quadrilateral into two different triangles by drawing a diagonal. Then, just use the simplified version of shoelace with determinants.

All we have to do then is find the determinant of two 3 x 3 matrices. I get 12 as the answer as well.

yes, that is how the program i wrote works.   i find the distance between all 4 of the coordinates,( all 4 sides and a diagonal) and then plug in herons for the two triangles created.  the only problem i have seen so far with it in that it will reduce radicals...so the program will really be of use only when the area is an integer or fraction.  otherwise, its better to use shoelace or do the work by hand.


geometry
geometric transformation
rotation
complex numbers
geometry proposed

on the sides of an acute triangle ABC are constructed externally a square, a regular n-gon and a regualr m-gon (m,n \geq  5) whose centres form an equilateral triangle. prove that m=n=6 and then find the angles of triangle ABC .
enjoy!!

I can't reproduce my calculations (too long), but I put x=cis(2Pi/m) and y=cis(2Pi/n), w=cis(2Pi/6), a, b and c the complex numbers corresponding to A, B and C respectively, and the condition that the center of the n-gon is obtained from the center of the m-gon by a rotation of 60 deg with respect to the square, and I got that the whole thing is equivalent to

(1-i)(1-x-wx-ywx)a=
[(1-w)(1-x)(1-y)+(1-i)(1-x)y]b + [(1-i)(1-y)w-i(1-w)(1-x)(1-y)]c. You can check that (1-i)(1-x-wx-ywx)=(1-w)(1-x)(1-y)+(1-i)(1-x)y+(1-i)(1-y)w-i(1-w)(1-x)(1-y), so it can be written as a=tb+(1-t)c, iff ta+(1-t)a=tb+(1-t)c iff t(a-b)=(t-1)(a-c) iff (a-c)/(a-b)=t/(t-1).

We don't know m and n, so we don't know x and y, and t depends on x and y, so I don't see how one could actually find everything here (x and y).

This is probably useless, since I'm sure I'm making a mistake here somewhere.

Can the file which contains the problem be downloaded from Mathlinks? I've looked through all the files from 1998 (Bulgaria) which are available in the downloads section, but I couldn't find it.

Never mind, I got the file containing the problem and I glanced at the solution, which is, by the way, horrible. I'll try to figure out what I did wrong.

There appears to be enough information to complete the problem, though.