geometry
angle bisector
geometry unsolved

Let ABCD be a convex quadrilateral. Diagonal AC bisects angle BAD. Let E be a point on side CD. Segments BE and AC intersect at F. Extend segment DF trough F to intersect segment BC at G. prove that <GAC=<EAC.

Let be the point $ P\equiv BD\cap EG$ and from the complete quadrilateral $ CEFGBD,$ we have that the points $ P,\ B,\ M,\ D,$ where $ M\equiv AC\cap BD,$ are in harmonic conjugation.

So, the pencil $ A.PBMD,$ is also in harmonic conjugation and because of $ \angle BAC \equal{} \angle DAC,$ we conclude that $ PA\perp AC$ $ ,(1)$

The points $ P,\ G,\ N,\ E,$ where $ N\equiv AC\cap EG,$ are also in harmonic conjugation as well.

So, the pencil $ A.PGNE,$ is in harmonic conjugation and because of $ (1),$ we have that the the ray $ AN,$ bisects the angle $ \angle EAG.$

Hence, we conclude that $ \angle GAC \equal{} \angle EAC$ and the proof is completed.

Kostas Vittas.

Let BD intersect AC at N. By Ceva on BCD, (BG/GC)(CE/ED)(DN/NB)=1, but by angle bisector on ABD, this simplifies to (BG/GC)(CE/ED)(AD/AB)=1, hence BA/DA=(CE/DE)/(CG/BG)=([CEA]/[DEA])/([CAG]/[BAG]). If we assume $ \angle GAC \equal{} \angle CAE$, then $ \angle DAE\equal{}\angle GAB$. If we rewrite the areas of all four triangles using the formula 1/2absinC, where the included angle is at A, we get BA/DA after simplification. So if one angle was not the same as the other, that would throw off the proportion.

If you don't understand the end, think of the proof of the converse of Menelaus, how you show the converse must be true because if the point was any farther to the left or right, it would throw the proportion off.


Can someone create a wiki page describing how to whitelist email addresses from different providers

( :arrow: Put a notice for hotmail, yahoo, aol, etc. users to check their spam box for activation emails / notifications from wishlist )

The page should be similar to this one: http://www.ionemoney.com/whitelist.htm

Email addresses to be whitelisted: webmaster@artofproblemsolving.com and webmaster@mathlinks.ro

I've created a page at [[A:WL]] - [[AoPSWiki:Whitelisting]] (hopefully it suffices).


integration
trigonometry
calculus
real analysis
real analysis unsolved

This integral has been posted several times.


induction
modular arithmetic
algebra
polynomial
number theory proposed
number theory

Let $p$ be a prime number, and define a sequence by: $x_i=i$ for $i=,0,1,2...,p-1$ and $x_n=x_{n-1}+x_{n-p}$ for $n \geq p$
Find the remainder when $x_{p^3}$ is divided by $p$.

Please always add the year and if possible the round.  :)

A nasty solution :

A straightforward induction on $k$ shows that for all integers $n,k$ such that $n \geq kp \geq p$, we have :
$a_n = \sum_{i=0}^k C_k^i a_{n-ip-k+i} .$

It follows that for $n \geq p^2$ we have $a_n = \sum_{i=0}^p C_p^i a_{n-ip-p+i}.$
But, it is well-known that, since $p$ is prime, we have $C_p^i = 0 \mod [p]$ for $i=1, \cdots , p-1.$
Therefore  $a_n = a_{n-p}+a_{n-p^2} \mod [p].$

In another hand, using that $a_n = a_{n-1} + a_{n-p}$, we deduce that for $n \geq p^2$, we have $a_{n-1} = a_{n-p^2} \mod [p]$. It means that the sequence is $(p^2-1)$_ periodic modulo $p$. Since $p^3 = p(p^2-1) + p$, it follows that $a_{p^3} = a_p = a_{p-1}+a_0 = p-1 \mod [p] .$

Pierre.

Here's another interesting way (although not nicer, I think :() of showing the $p^2-1$-periodicity $\pmod p$:

The characteristic equation is $x^p-x^{p-1}-1$. We can write $x_n=\sum\alpha_i u_i^n$, where $\alpha_i$ are some constants and $u_i,\ i\in\overline{1,p}$ are the roots of $x^p-x^{p-1}-1$ in an extension of $\mathbb Z_p$ (the splitting field of the polynomial), which is clearly not $\mathbb Z_p$ itself. The next step would be showing that $x^p-x^{p-1}-1|x^{p^2}-x$ (I haven't figured out how to prove this in a nice, clean, quick way, even though I think I know the form of the quotient). From here we find that the roots $u_i$ also satisfy $u^{p^2}-u=0$, and we get the desired result.

I also proved mod p the sequnce was $p^2-1$ periodic, working with the sequence backwards. and using inductions.


Find a second degree polynomial with roots 1/2 and -4 and passes through the point (-1, 6) ?
Write your answer in the form f(x) = ax^2 + bx + c.

this looks suspiciously like a homework problem... [hide=":dry:"]instead, let $f(x) = k\left(x-r_{1}\right)\left(x-r_{2}\right)$  it is obvious that $f(r_{1}) = f(r_{2}) = 0$, so $r_{1},r_{2}$ are the roots of $f$.  so let $f(x) = k\left(x-\frac{1}{2}\right)\left(x+4\right)$  we now need only choose $k$ such that $f(-1) = 6$ which is very simple to do.  to get it into your form just expand.[/hide]

sorry kouichi click on the smiley in my post and out pops a partial solution (i left the finding $a$ to sharkman)

hiding stuff with smileys is fun, you should try it sometime.  it apparently really fools people.

[hide=" :lol: "]haha you thought there was something cool under this one, didn't you?  LOL[/hide]


Well, some starting idea was here on mathlinks (inspired by IMO2005/3), then I posted a long one post with many the key ideas at

http://www.mathlinks.ro/Forum/viewtopic.php?t=56369

at and Darij replied in solving ineq 1
http://www.mathlinks.ro/Forum/viewtopic.php?t=103182
and that was useful for me. I have to admit that Darij helped me with it. Honestly he did. So I named it Grinberg-aTToWe Inequality. In that order because abbreviation AG already exists... On the other hand aTToWe is my nickname. So GA as its abbreviation.

====================
Denotion is $(r,s,t)=\sum_{sym}x^{r}y^{s}z^{t}$
====================

[color=blue]Lemma 1[/color]
For $a,b,c\ge0$ we have for every $n\ge3$
$(n,0,0)+(n-3,2,1)\ge(n-2,2,0)+(n-1,1,0)$
We get the result as follows:

$(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}+a^{n-3}bc+ab^{n-3}c+abc^{n-3}-a^{n-2}b-ab^{n-2}-b^{n-2}c-bc^{n-2}-c^{n-2}a-ca^{n-2})+$
$+(a^{n}+b^{n}+c^{n}+a^{n-2}bc+ab^{n-2}c+abc^{n-2}-a^{n-1}b-ab^{n-1}-b^{n-1}c-bc^{n-1}-c^{n-1}a-ca^{n-1})\geq0$

Just add Schur's Ineqaulity for $n-1$ multiplied with $(a+b+c)$ with Schur's Inequality for $n$.

Lemma 1 is proven.

[color=blue]Lemma 2.[/color]
For $a,b,c\ge0$ we have for every $n\ge3$
$(n,0,0)+(n-2,1,1)\ge(n-1,1,0)+(n-1,1,0)$

That is exactly Schur's Inequality.

Now I will say something about permitted operations.

There are left and right operations. If we have that

$(p_{1},p_{2},p_{3})+(q_{1},q_{2},q_{3})\ge(r_{1},r_{2},r_{3})+(s_{1},s_{2},s_{3})$

we define the left operations like that:

$(p_{1},p_{2}+t,p_{3}-t)+(q_{1},q_{2},q_{3})\ge(p_{1},p_{2},p_{3})+(q_{1},q_{2},q_{3})\ge(r_{1},r_{2},r_{3})+(s_{1},s_{2},s_{3})$

for some permitted value of $t$. (permitted $t$ because definition of multiindex and because not make it muirheadable)

right operations is similar:

$(p_{1},p_{2},p_{3})+(q_{1},q_{2},q_{3})\ge(r_{1},r_{2},r_{3})+(s_{1},s_{2},s_{3})\ge(r_{1},r_{2}-t,r_{3}+t)+(s_{1},s_{2},s_{3})$

again for some permitted value of $t$.

Now using Lemma 1 and Lemma 2 and our right and left operations we can get what we want:

We can do... for example:

$(n,0,0)+(n-3,2,1)\ge(n-2,2,0)+(n-1,1,0)$ is Lemma 1

using left oparation we get that

$(n,0,0)+(n-3,3,0)\ge(n-2,2,0)+(n-1,1,0)$

or summing

$(n,0,0)+(n-3,2,1)\ge(n-2,2,0)+(n-1,1,0)$
with
$(n,0,0)+(n-2,2,0)\ge(n-1,1,0)+(n-1,1,0)$
we get
$2(n,0,0)+(n-3,2,1)\ge2(n-1,1,0)+(n-1,1,0)$

and doing similar left and right operarions we can get thones of inequalities, which are very strong. But the basics ones are Lemma 1 and Schur.

====================
So everything neccessary now is to remember these two ones:
$(n,0,0)+(n-3,2,1)\ge(n-2,2,0)+(n-1,1,0)$
$(n,0,0)+(n-2,1,1)\ge(n-1,1,0)+(n-1,1,0)$
and the permitted left and right operations.
====================
Further note that there will not be any nicer one look, because it is not possible to create it.

If you have bunched some ineq, and after that it has a form

$E_{1}(p_{1},p_{2},p_{3})+E_{2}(q_{1},q_{2},q_{3})\ge E_{3}(r_{1},r_{2},r_{3})+E_{4}(s_{1},s_{2},s_{3})$ (t)
($E_{i}\in\Re^{+}$)
than you can try to get using these two ineqs and left and right operations to get it. There is no way how to say immediatelly that some bunched ineq is right or not (if it has a shape like (t) does).
====================
Further I can say that it is very strong inequality.

By Muirhead we have that
$(n,0,0)\ge(n-2,2,0)$
and
$(n-1,1,0)\ge(n-3,2,1)$
Now by "super summing" these inequalities we get (I hope you understand what super summing means):
$(n,0,0)+(n-3,2,1)\ge(n-2,2,0)+(n-1,1,0)$
what is our lemma 1.

Super summing is not a method to prove some ineq, it is a way how to come up with a new idea, very strong ineq, what has to be afterwards proved. And that summing these particular ones is correct we can see by Lemma 1.
====================
I will write it again to end:

So everything neccessary now is to remember these two ones:
$(n,0,0)+(n-3,2,1)\ge(n-2,2,0)+(n-1,1,0)$
(Lemma 1)

$(n,0,0)+(n-2,1,1)\ge(n-1,1,0)+(n-1,1,0)$
what is Schur's Inequality.
and the permitted left and right operations.
====================

$E_{1}(p_{1},p_{2},p_{3})+E_{2}(q_{1},q_{2},q_{3})\ge E_{3}(r_{1},r_{2},r_{3})+E_{4}(s_{1},s_{2},s_{3})$ (t)

is called Grinberg-aTToWe's Inequality

1.9.2006, Czech Republic

I believe he means that he takes inequalities
$A \ge B$ (1)
$C \ge D$ (2)
and considers whether it holds:
$A+D \ge B+C$ (*)
If it is true, then it is a stronger inequality than either (1) or (2). Another way to think of it is this:
An inequality $A \ge B$ can instead be thought of as $A-B \ge 0$. We have shown that certain quantities are positive in this way. Now, what else can we say about them? It is natural to ask (especially in the case of similar things, like symmetric sums) how these differences compare. For example, is $(7, 1, 1)-(5, 2, 2) \ge (6, 2, 1)-(4, 3, 2)$? Both
$(7, 1, 1) \ge (5, 2, 2)$
$(6, 2, 1) \ge (4, 3, 2)$
follow from Muirhead and are not particularly strong (since Muirhead is basically AM-GM). But
$(7, 1, 1)+(4, 3, 2) \ge (6, 2, 1)+(5, 2, 2)$
(i.e., the first difference is the greater) is a significantly stronger ineq. Note (the obvious fact that) in general, $A-B \ge 0$ and $C-D \ge 0$ does not necessarily imply that ($A-B \ge C-D$ OR $A-B \le C-D$). It may happen that in some cases one expression is larger and in other cases the other expression is larger. That's why [b]conceive[/b] says "what has to be afterwards proved."  ;)
Hope that helped...

to be frank, I don't know what you are doing. Your lemma 1 is a consequence of schur, which means your whole method is based on schur. We know that muirhead and schur is not compatible, so of course, the ineq produced by this  way is not muirheadable.

but then, you are only proposing to use schur for bunching. But isn't that something known already?!

[quote="Soarer"]but then, you are only proposing to use schur for bunching. But isn't that something known already?![/quote]

As I have already said:

[quote="conceive"]... I mean of the form

$(t_{1},t_{2},t_{3})+(s_{1},s_{2},s_{3})\ge(r_{1},r_{2},r_{3})+(p_{1},p_{1},p_{3})$ (1)
¨
...[/quote]

GA Inequality is natural consequence os Schur as well as Muirhead is consequence of AM-GM. Just another (better form of writting some ineq, quicklier form and more understandable in hard cases).

GA is so strong as Schur is.

Every Muirhead is possible to solve with AM-GM, same here. Every GA can be proven with Schur.

I hope it is clear now.

EDIT: I and Darij are working on methods how to make stronger GA, and we have some key ideas.. but not many. We will see what can be done.

EDIT2:
[quote="me@home"]Sorry, what is "super summing"?[/quote]

Note that it is not new. It first appear here (third post):

http://www.mathlinks.ro/Forum/viewtopic.php?t=78857

not convinced.

Muirhead somehow hides amgm, because you don't need to know how to use amgm in a clever way to apply muirhead.

You are now only proposing a method, not a new ineq, if i am not mistaken. I don't think this is an ineq unless you propose a way of adding up ur lemma 1 and schur so that it can be useful in a large number of cases.

Otherwise, I don't need muirhead. Why don't I just tell others to divide an ineq ingeniously by themselves and use amgm?

[quote="Soarer"]

You are now only proposing a method, not a new ineq, if i am not mistaken.[/quote]

That is quite right. I am proposing a method how to prove ineqs in form

$E_{1}(t_{1},t_{2},t_{3})+E_{2}(s_{1},s_{2},s_{3})\ge E_{3}(r_{1},r_{2},r_{3})+E_{4}(p_{1},p_{1},p_{3})$

That ineqs we called GA, and we show the METHOD how to prove MANY of them using Lemma 1, Schur and left and right operations.

Honestly I do not believe in that there there exists any explicit exact conditions for

$E_{1}(n-w_{1},w_{1}-w_{2},w_{2})+E_{2}(n-v_{1},v_{1}-v_{2},v_{2})\ge E_{3}(n-t_{1},t_{1}-t_{2},t_{2})+E_{4}(n-u_{1},u_{1}-u_{2},u_{2})$

which is my ineq10 I proposed in Open Question.

Note that these inequalities which we obtained are very strong.

[quote="Soarer"]I don't think this is an ineq unless you propose a way of adding up ur lemma 1 and schur so that it can be useful in a large number of cases.[/quote]

It can be useful in many number of cases if you realize what everything you can do with left and right operations.

[quote="Soarer"]Otherwise, I don't need muirhead. Why don't I just tell others to divide an ineq ingeniously by themselves and use amgm?[/quote]

Well, I have something to say here. It is quite wellknown, that the more "super", more general, more strong theorems and statement we create in mathematics, the more finite and the more narrow they are.

We all know that every Muirhead is AM-GM doable, and Muirhead is very nice and convenient, but I am not sure there will be anything similar here for Schur, because Schur is very strong. That is why GA also is. And maybe that is why GA could not have any better conditions. As I said, it is very intuitive. You'll have to feel it.

Anyway thanks for your replies.

you don't get my point.
Why do many consider Muirhead useful? It is because we need to deal with symmetric sums very often, and it provides an useful criterion of comparing it. It doesn't tell you how to break it into amgm, it only tells u the result, that's why it is useful. That's why it is nice and convenient.

Usually when you develop a tool at your aid, you want to equip them to solve inequalities. If it is not convenient to use, what is its use?

Your idea is absolutely fine -- in fact many people have used schur in their steps of bunching, and very often schur and muirhead are used in mix. The problem is, if you do not give an ineq that can help us CONVENIENTLY, i don't know why it can be useful.

[quote="Soarer"]you don't get my point.The problem is, if you do not give an ineq that can help us CONVENIENTLY, i don't know why it can be useful.[/quote]

The point is that u didnt get my point.

[color=orange]The more stronger inequalities we have the more they are limited.[/color]

I said I think there will probably not be any conditions just like in Muirhead here in GA. There will be just a method, but such one to prove very strong ineq, while Muirhead tells you everything immediatelly, but the ineqs it produces are very weak. That is the price.

[quote]I said I think there will probably not be any conditions just like in Muirhead here in GA.[/quote]
Then I think Soarer's point is "What actually is GA?" It's not an inequality like Muirhead, because, as you say, there will probably not be any conditions like Muirhead for such strong inequalities. Therefore, what actually does GA do? Can I not just learn Schur, and solve inequalities with the same ease?

I cannot understand what the GA means. If it means that inequalities with degree n can be proved by the special case of Schur's ineq and Muirhead's ineq, (left and right application is actually just application of Muirhead's inequality.) I do not think that this way can brighten all of the kind of inequalities. First, we do not know whether Schur's ineq is the sharpest ineq, and actually there exist some inequalities which cannot be solved with Schur. Moreover, left and right application, namely Muirhead, makes the inequality easier, as it expands the "gap" of the original inequality.

[quote="Ilthigore"][quote]I said I think there will probably not be any conditions just like in Muirhead here in GA.[/quote]
Then I think Soarer's point is "What actually is GA?" It's not an inequality like Muirhead, because, as you say, there will probably not be any conditions like Muirhead for such strong inequalities.[/quote]

Well, at least GA is an abbreviated form of written something (like congruences - you don't need them necessary, just to work more effectively). GA in the current form we have it now can us help in proving inequalities after bunching. It is more convenient to work with multiindexes, then write $ x^{2}y \plus{} xy^{2} \plus{} y^2z \plus{} yz^{2} \plus{} z^{2}x \plus{} zx^{2} \plus{} \ldots\geq \ldots$. But it is not only that. GA is more. We can have $ (n,0,0) \plus{} (n \minus{} 3,3,0)\ge(n \minus{} 1,1,0) \plus{} (n \minus{} 2,2,0)$.

[quote="Ilthigore"] Therefore, what actually does GA do? Can I not just learn Schur, and solve inequalities with the same ease?[/quote]

With multitindexes you can do so many things.. e.g. you can multiply it by $ abc$ to increase every number in there of 1.. to obtain what would you need to have to prove some ineq after bunching. You can add GAs to obtain more like the weight version of GA I mentioned. This is a revolution. You cannot see it?

Note that this what we have now is the beginning. We can even think about GA such that will have $ n$ tuples on both sides.. Could you imagine it? Such inequalities will be very strong.

[quote="Sung-yoon Kim"]I cannot understand what the GA means. If it means that inequalities with degree n can be proved by the special case of Schur's ineq and Muirhead's ineq, (left and right application is actually just application of Muirhead's inequality.) I do not think that this way can brighten all of the kind of inequalities. First, we do not know whether Schur's ineq is the sharpest ineq, and actually there exist some inequalities which cannot be solved with Schur.[/quote]

Sure I personally know infinitely many inequalites stronger than Schur. But for me is Schur sufficient for now. GA is generalization of Schur.

[quote="Sung-yoon Kim"]Moreover, left and right application, namely Muirhead, makes the inequality easier, as it expands the "gap" of the original inequality.[/quote]

Left and right operations are there to obtain something more similar to the ineq you want to prove. It is right that they make GA more weaker.

An example:

http://www.mathlinks.ro/Forum/viewtopic.php?p=619874

Why is GA a generalisation of schur ?
i think it is just an application of it. :(

I guess according to your signature, the actual generalization is the Zaimi-Marku... is this is a joke or is it really an inequality?

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103737]here[/url]
but there is not a unique way to generalise schur.although Zaimi-Marku is the only with n-variables :)


pigeonhole principle
floor function
search
inequalities
algebra
polynomial
number theory unsolved

$ n$ integers are given: $ a_{n}>a_{n\minus{}1}>....>a_{1}$ (I think that this condition is unnecessary).
Prove that:
$ \displaystyle\prod_{2\leq i\leq n}(\displaystyle\prod_{1\leq j\leq i\minus{}1} (a_{i}\minus{}a_{j}))$ is divisible by $ \displaystyle\prod_{2\leq i\leq n}(\displaystyle\prod_{1\leq j\leq i\minus{}1} (i\minus{}j))$

This question appeared in the first stage of getting into the Israeli IMO team (actually, it was a much easier one: it was the special case of n=7, and then you can use the pigeonhole principle). I have proved this generalized form by proving a not-pretty-at-all combinatorial identity (using induction). What are your solutions? and what is the origin of this beautiful problem (which looks almost trivial)? (my guess - there's a trivial solutions with polynomials...)

[hide="hint-"]For every prime P not bigger that n, split $ a_{1},....a_{n}$ into modulo classes of $ P^{k} (p^{k}<\equal{}n)$ . It is sufficient to prove that:
$ min \{\sum_{1<\equal{}i<\equal{}p^k} \binom{b_{i}}{2} | \sum_{1<\equal{}i<\equal{}p^k} b_{i}\equal{}n\} >\equal{} \sum_{1<\equal{}j<n\minus{}1} \lfloor\frac{j}{p^{k}}\rfloor$

[/hide]

you can search schur polynomial (this is my proof but is in italian (here i'm jordan, using PMI, while piever combinatorial :wink: ) http://olimpiadi.dm.unipi.it/oliForum/viewtopic.php?t=9815 )

bye

Yes, the condition $ a_{n} > a_{n \minus{} 1} > .... > a_{1}$ is unnecessary.
You can find the problem discussed at http://www.mathlinks.ro/viewtopic.php?t=4858 .

Bboypa, how can we use Schur polynomials?

  darij


[quote="FieryHydra"][quote="kimby_102"][quote="maokid7"]Find all non-negative integral solutions $(n_{1},n_{2},\dots,n_{14})$ to
$n_{1}^{4}+n_{2}^{4}+\dots+n_{14}^{4}=1599$[/quote]
I think this one was posted before, but im lazy to search...
[hide="hint"]
consider mod 16

[/hide][/quote]
Thanks, [b]kimby_102[/b], for the suggestion.

[hide]I found out for all $n\geq 1$, either $n^{4}\equiv 0(\mod16)$ or $n^{4}\equiv 1(\mod 16)$ holds.
$\therefore n_{1}^{4}+n_{2}^{4}+\dots+n_{14}^{4}\equiv k (\mod16)$, where $0\leq k \leq 14$, whereas $1599 \equiv-1(\mod16)$.
$\therefore$ No nonnegative integral solution exists.[/hide][/quote]
lol cool! but how did you find that for all $n\geq 1$, either $n^{4}\equiv 0(\mod16)$ or $n^{4}\equiv 1(\mod 16)$ holds? does you just check the numbers $1^{4},2^{4}...15^{4}$ mod 16? or there is a trick?

To avoid any checking approach give something general:

When $a$ is even, then $a^{k}\equiv 0 \mod 2^{k}$, but not necessary $\mod 2^{k+1}$.
When $a$ is odd, then $a^{2^{k}}\equiv 1 \mod 2^{k+2}$, but not necessary $\mod 2^{k+3}$ for $k \geq 1$.

Setting $k=4$ into the first and $k=2$ into the second, the result follows.


"Mirobolant" is a Romanian word for "wonderful".

I like to keep it close to where my heart used to be.

That's why I don't dare to drive in Bucharest !
Very actuall situation ! I hope it will be better in the future !

I think you posted this picture twice, the first was the "Ugly Geometry"  :lol:

Enough is enough!!! I've had it with these...

lol What's with the random comment?
good song tho

And in the future Physics exam bind them


[i] prove that./[/i]
$\large\sqrt{\frac{a}{b}+\frac{b}{c}}+\sqrt{\frac{b}{c}+\frac{c}{a}}+\sqrt{\frac{c}{a}+\frac{a}{b}}\ge\sqrt{\frac{a+b}{b+c}+1}+\sqrt{\frac{b+c}{c+a}+1}+\sqrt{\frac{c+a}{a+b}+1}$
[i]where[/i] $\large a,b,c$ [i]positive are real numbers./[/i] :D
[i]@hungkhtn: If you think this problems is interesting,i'll post its answer to you./[/i]

Let $\frac{a}{b}=x.\frac{b}{c}=y \frac{c}{a}=z$->xyz=1'
 We prove:
    $\sum\sqrt{x+y}\geq \sqrt{\frac{x+1}{1+\frac{1}{y}}+1}+\sqrt{\frac{y+1}{1+\frac{1}{z}}+1}+\sqrt{\frac{z+1}{1+\frac{1}{x}}+1}$.
    $\Leftrightarrow\sum\sqrt{x+y}\geq \sqrt{\frac{y(x+1)}{y+1}+1}+\sqrt{\frac{z(y+1)}{z+1}+1}+\sqrt{\frac{x(z+1)}{x+1}+1}$.
    Let$A=\sum\sqrt{x+y}B=\sqrt{\frac{y(x+1)}{y+1}+1}+\sqrt{\frac{z(y+1)}{z+1}+1}+\sqrt{\frac{x(z+1)}{x+1}+1}$
    We have:$A^{2}\geq 4(x+y+z)+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$
                  $\Rightarrow Leftrightarrow A^{2}\geq 4(x+y+z)+6$.
      Hence.$B^{2}\leq 3(\sum(\frac{y(x+1)}{y+1}+1))$.
     But it's easy to prove:
      ${\sum x \geq \sum(\frac{y(x+1)}{y+1}}$
  then->qed


trigonometry
function
number theory proposed
number theory

Prove that if $ c$ is a real number, then

a. $ c$ is rational if and only if $ f(x) \equal{} cos(x) \plus{} cos(cx)$ is periodic. (easyish)
b. $ c$ is rational if and only if $ g(x) \equal{} sin(x) \plus{} sin(cx)$ is periodic. (harder)

The function $ g(x)$ is periodic iff there exist $ h$ such that:
$ sin(x\plus{}h)\plus{}sin(cx\plus{}ch) \equal{} sin(x)\plus{}sin(cx)$, for all $ x$.
($ h$ is not 0)
$ \leftrightarrow sin(x\plus{}h)\minus{}sin(x) \equal{} sin(cx)\minus{}sin(cx\plus{}ch)$
$ \leftrightarrow 2sin(x\plus{}h/2)cos(h/2) \equal{}2sin(c(x\plus{}h/2))cos(ch/2)$
Like this is for all $ x$:
$ \leftrightarrow cos(h/2) \equal{}cos(ch/2)\equal{}0$
Then: $ h/2\pi\equal{}rational$ and $ ch/2\pi\equal{}rational$ and $ c\equal{}rational$
The other part is the same thing.


The product of two consecutive odd whole numbers is 255. What is the greater number?

If we square root 255, we get about 16. Thus the two numbers we are looking for are 15 and 17, and our answer is $ \boxed{17}$.


Can I ask someone to help me with these questions please ?

1)Use Gauss's Lemma to solve ((a+5)/37).

2)Let p be any prime congrent to 1 (mod 8) and let g be a generator for Z*p.

a)Explain why there is an element x of order 8 in Zp and why x^2= -x^(-2).

b)Now show that (x-x^(-1))^2= -2.

c)Use part b to find all elements y=squareroot(-2) for all elements of Zp.

[quote="natelli"]Can I ask someone to help me with these questions please ?

1)Use Gauss's Lemma to solve $ (\frac{a\plus{}5}{37})$

2)Let p be any prime congrent to 1 (mod 8) and let g be a generator for Z*p.

a)Explain why there is an element x of order 8 in Zp and why$ x^{2}\equal{}\minus{}x^{\minus{}2}$.

b)Now show that$ (x\minus{}x^{\minus{}1})^{2}\equal{}\minus{}2.$

c)Use part b to find all elements $ y \equal{}\sqrt\minus{}2$ for all elements of Zp.[/quote]
E have edit it.


What is the perimeter of the right triangle whose exradius of the hypotenuse is $ 30$ ?

$\textbf{(A)}\ 40  \qquad\textbf{(B)}\ 45  \qquad\textbf{(C)}\ 50  \qquad\textbf{(D)}\ 60  \qquad\textbf{(E)}\ 75$

[hide]$ r_c\equal{}{a\plus{}b\plus{}c\over 2}\iff p\equal{}60$, so the answer is $ D$.[/hide]


Find :
${ \lim_{n \rightarrow \infty}\frac{ ( n!)  ^{\frac{1}{n}}} {n}}$

Well, it's related with your Stirling post:

$ n!\sim n^ne^{ \minus{} n}\sqrt{2\pi n}$ (Stirling)

using it in your limit yields the solution $ \frac {1}{e}$.

[quote="Carcul"]$ \frac {(n!)^{\frac1n}}{n} \equal{} \left(\frac {n!}{n^n}\right)^{\frac1n} \equal{} (a_n)^{\frac1n}$

Observe now that

$ \lim \frac {a_{n \plus{} 1}}{a_n} \equal{} \lim \left(\frac {n}{n \plus{} 1}\right)^{n} \equal{} \lim \left[\left(1 \plus{} \frac { \minus{} 1}{n \plus{} 1}\right)^{n \plus{} 1}\right]^{\frac {n}{n \plus{} 1}} \equal{} (e^{ \minus{} 1})^1 \equal{} \frac1e$

Since the limit $ \lim \frac {a_{n \plus{} 1}}{a_n}$ exists, we can therefore conclude that

$ \lim (a_n)^{\frac1n} \equal{} \lim \frac {a_{n \plus{} 1}}{a_n} \equal{} \frac1e$.[/quote]

Could you explain why?
$ \boxed{\lim (a_n)^{\frac1n} \equal{} \lim \frac {a_{n \plus{} 1}}{a_n} \equal{} \frac1e}$

$ \left(a_n\right)^{\frac {1}{n}} \equal{} \left(a_1\prod_{k \equal{} 1}^{n \minus{} 1}\frac {a_{k \plus{} 1}}{a_k}\right)^{\frac {1}{n}}$, so if $ n$ is large and $ \frac {a_{k \plus{} 1}}{a_k}\to L$ then $ \left(a_n\right)^{\frac {1}{n}}$ is the geometric mean of $ n$ numbers all but a few of which are very close to $ L$.  Make this estimate precise.


The Application has been posted on the NYSML web site.  If any of you are too lazy to go there, it's right here, as well.

So, who's going?

Yay Upstate NY! I'll be there as usual... 4 years baby!

I'll give a little report on Westchester.  Michael and I tied for first, and we're both going.  Two people tied for third; one can't go (but I recruited him for our HMMT etc. team anyway), the other has already applied.  There's two other people I'm trying to convince, including the guy who won NYSML.  So yeah.  From Reuter's latest email, it looks like we're missing a few USAMOers:
Steph (huh?)
Anthony (aw, you guys failed)
Joseph Lee (the 9th guy that Chris found outside the breakfast place--oh well, I wasn't sure he was convinced anyway)

Anyway, I think we're pretty much okay.

[quote="perfectnumber628"][quote="TimReyn"]Steph (huh?)[/quote]
I'm officially in now, don't worry.  Seriously, who mails in their application on time?[/quote]

Well, judging by the email, 32 people.

sooooo, according to the email, looks like i got in, somehow.  :maybe:

in other news, i found the step i was missing on the bus from nysml to solve the 5^3 cube


Compute :

\[ arccos(sin(5)) \equal{} ?\]

[hide="solution"]$ \sin{5}\equal{}\cos{90\minus{}5}\equal{}\cos{85}$

$ \arccos{\cos{85}}\equal{}\boxed{85}$[/hide]

I didn't mean $ 5^\circ$ degree . When I type $ 5$  I mean $ 5$ radians .

[quote="1=2"]*jaw drops to floor and shatters*[/quote]It's not that hard, check on which of the quadrants does 5 radians lie in.

[hide]Uh, $ \sin 5\equal{}\cos (2\pi\minus{}5)$, so $ \arccos(\sin 5)$ is just $ 2\pi\minus{}5$.[/hide]

[quote="Temperal"][hide]Uh, $ \sin 5 \equal{} \cos (2\pi \minus{} 5)$, so $ \arccos(\sin 5)$ is just $ 2\pi \minus{} 5$.[/hide][/quote]That's not correct because if you apply sum angle formula for cosine,
$ \cos(2\pi\minus{}5)\equal{}\cos 2\pi\cos 5\plus{}\sin 2\pi\sin 5\equal{}\cos 5\neq\sin 5$

With $ arc\cos$ defined to give values in $ [0,\pi]$:

$ \sin(5)\equal{}\cos(5\minus{}5\frac{\pi}{2})\equal{}\cos(5\frac{\pi}{2}\minus{}5)$ and as $ 0<5\frac{\pi}{2}\minus{}5<\pi$

$ arc\cos(\sin(5))\equal{}5\frac{\pi}{2}\minus{}5$

 :)

[quote="10000th User"][quote="Temperal"][hide]Uh, $ \sin 5 \equal{} \cos (2\pi \minus{} 5)$, so $ \arccos(\sin 5)$ is just $ 2\pi \minus{} 5$.[/hide][/quote]That's not correct because if you apply sum angle formula for cosine,
$ \cos(2\pi \minus{} 5) \equal{} \cos 2\pi\cos 5 \plus{} \sin 2\pi\sin 5 \equal{} \cos 5\neq\sin 5$[/quote]

Whoops. Is that because the angle is in the fourth quadrant? Or did I just remember the mutiple of $ \pi$ incorrectly?


In $ \triangle ABC$, let $ O,H$ be the circumcenter and orthocenter respectively. Let $ A',B',C'$ be points on the circumcircles of triangles $ BHC,CHA,AHB$ respectively, such that each of the sets $ \{A',C,B'\},\{B',A,C'\},\{A',B,C'\}$ have points collinear. Let $ H'$ be the orthocenter of triangle $ A'B'C'$. Find $ \angle HOH'$, in terms of angles $ A,B,C$ of triangle $ ABC$


Solve the following equation:

Find all primes p, q and even n > 2 such that
pn + pn-1 + ... + p + 1 = q2 + q + 1.

This was a problem in INMO 2003 paper.

I have seen this one like 5 times in the forum


[quote="jli"]MATHCOUNTS 2004-2004 Warm-up 13, problem #10

Let a and b be two integers for which 7a + 12b = 1.  What is the largest possible positive value of a+b which is less than 2005?

[hide="my solution"]Here is how I did it, but I am wondering if there is a better way to solve this problem.

$12(a+b)=1+5a$

$(a+b)=\frac{1+5a}{12}<2005$

$a<4811.8$

Since b has to be an integer, a=4811 and 4809 do not work, but a=4807 works.

EDIT NOTES: fixed errors from 1-5a to 1+5a
So, the answer is 2003.[/hide][/quote]

Same,  but  may be a little easier (or straight forward) way to think about this :

[hide]$7(a+b)+5b = 1$ --- as you said.
Take mod 5 we get
$2(a+b) \equiv 1 \equiv 6  \bmod 5 $
or $(a+b) \equiv 3 \bmod 5$

(In other words  $(2(a+b) -1)$ is divisible by 5,  you know  (6-1) is divisible by 5 and 6/2 is 3 so $(a+b)$ has to be of the form  $5k+3$)
So the required answer is $2003$..[/hide]


Solve the equation:
$ (x \plus{} 2)^4 \plus{} 98cosx^2\equal{} x^8 \plus{} 98cos(x \plus{} 2)$

[quote="imedeen"]Solve the equation:
$ (x \plus{} 2)^4 \plus{} 98cosx^2 \equal{} x^8 \plus{} 98cos(x \plus{} 2)$[/quote]

Let $ f(x) \equal{} x^4 \minus{} 98\cos(x)$. We have to solve $ f(x^2) \equal{} f(x \plus{} 2)$

$ f'(x) \equal{} 4x^3 \plus{} 98\sin(x)$ and so :
$ \forall x\in(0,\pi]$ :$ f'(x) > 0$
$ \forall x > \pi$ : $ f'(x) > 4\pi^3 \minus{} 98 > 4\cdot 3^3 \minus{} 98 \equal{} 10 > 0$

So $ f'x) > 0$ $ \forall x > 0$ and so $ f(x)$ is an even function strictly decreasing on $ \mathbb R^ \minus{}$ and strictly increasing on $ \mathbb R^ \plus{}$.

So $ f(u) \equal{} f(v)$ $ \iff$ $ u \equal{} v$ or $ u \equal{} \minus{} v$
So $ f(x^2) \equal{} f(x \plus{} 2)$ $ \iff$ $ x^2 \equal{} x \plus{} 2$ or $ x^2 \equal{} \minus{}x \minus{} 2$

Hence the two unique solutions : $ x \equal{} \minus{} 1$ and $ x \equal{} 2$


James,

   UC Santa Cruz - late application  :P

   Korean American, entering 10th grade this fall, 5' 10'' tall, runs track, "likes" cross country / basketball / swimming...

   Social and outgoing, so I'm hoping to make great new friends at AMSP.

   Also hoping that the program will hone my USAMO skills..... :lol:

   See you on July 31th!

  [b] BTW, can ANYONE let me know how to get to San Jose Int'l Airport from LAX or SFO??[/b]

   No international flights from Seoul go to SJ, and I need transportation...... :rotfl:


Mudlark Airlines has a 15-seater commuter turboprop that is used for short
flights. Their data suggest their on average about 8% of the customers who
buy tickets are no-shows. Wanting to avoid empty seats, they decide to sell
17 tickets for each flight. Ticketed customers who can't be seated on the
plane will be accommodated on another flight and will receive a certificate
good for a free flight at another time. Determine if this particular
overbooking is a sound strategy.

Doesn't look like a particularly good strategy to me, but we'll see.

$ 60\%$ of the time, the airline will be bumping (and compensating) at least one person from the flight, and that includes $ 24\%$ of the time in which they will bump two people from the flight.

On average, there will be $ 0.84$ displaced and compensated fliers per flight.

So in terms of full fares: the airline has the income from selling $ 17$ full fares. Call that $ \plus{}17.$ From that, they need to accommodate $ .84$ passengers on other flights, dropping the capacities (and ability to sell tickets) for future flights by $ .84.$ and then there's the free certificate - that costs the airline another $ .84$ fares. Hence the airline's average income per flight is $ 17\minus{}.84\minus{}.84\equal{}15.32$ fares. (Some of the certificates will go unused, but I didn't account for that.)

On the other hand, if they never overbook at all, but sell out every flight, they will fly with an average of $ .08\cdot 15\equal{} 1.2$ empty seats. But they'll pocket all $ 15$ full fares.

So they do earn a net $ .32$ extra fares per flight with using the overbook and bump strategy. That seems like a pretty small gain for the massive PR hit. ("Have you heard about Mudlark Airlines? I swear that every time I've ever flown with them, someone's been bumped from the flight.")

What if they sell $ 16$ tickets per flight, not $ 17?$ then they'll bump at most one person from each flight, and that will happen $ 26\%$ of the time. Doing the calculation the same way, I get an average income per flight of $ 16\minus{}.26\minus{}.26\equal{}15.48.$ (Well, it rounds to $ 15.47.$) That's a higher average net income than the book-17 strategy, and with much less PR damage - bumping people a quarter of the time isn't the same as bumping them the majority of the time.

(All calculations performed using the binomial distribution with $ p\equal{}0.08$ and $ n\equal{}15,16,$ or $ 17,$ depending on which part of this solution.)


trigonometry
complex numbers
algebra unsolved
algebra

What is the value of:

$ \tan^21^\circ\plus{}\tan^23^\circ\plus{}\cdots\plus{}\tan^289^\circ$

[quote="triplebig"]Tell where? Can't find[/quote]

maybe here : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=63539[/url]

It isn't the same problem...

Does anyone have any hints or ideas to solve this one?

Well, maybe we can somehow solve this by grouping all $ \tan^2 \alpha$ with $ \tan^2 \left( \frac{\pi}{2}\minus{}\alpha \right) \equal{} \cot ^2 \alpha$.

[quote="triplebig"]It isn't the same problem...[/quote]

Maybe you could just have looked at the beautiful demo of Zetax, tried to understand it and  simply adapt it to your problem ...  :( :

Let $ (1 \plus{} i\cdot\tan(x))^{n} \equal{} \frac {1}{\cos(x)^n}(\cos(nx) \plus{} i\cdot\sin(nx))$, as in Zetax demo.

But choose $ n \equal{} 90$ instead of $ 180$ and the real part instead of the imaginary one :

$ P(z) \equal{} (1 \plus{} iz)^{90} \equal{} \sum_{k \equal{} 0}^{90}\binom{90}{k}i^kz^k$

Real part is $ Re(P(z)) \equal{} \sum_{k \equal{} 0}^{45}\binom{90}{2k}( \minus{} 1)^k(z^2)^k$

But $ P(\tan((2p \plus{} 1)\frac {\pi}{180})) \equal{}$ $ \frac {1}{\cos((2p \plus{} 1)\frac {\pi}{180})^{90}}(\cos(90(2p \plus{} 1)\frac {\pi}{180}) \plus{}$ $ i\cdot\sin(90(2p \plus{} 1)\frac {\pi}{180}))$

So $ Re(P(\tan((2p \plus{} 1)\frac {\pi}{180}))) \equal{}$ $ \frac {1}{\cos((2p \plus{} 1)\frac {\pi}{180})^{90}}(\cos((2p \plus{} 1)\frac {\pi}{2}) \equal{} 0$

So, the $ 45$ numbers $ (\tan((2p \plus{} 1)\frac {\pi}{180}))^2$ ($ p \equal{} 0\to 44$) are the roots of $ Q(x) \equal{} \sum_{k \equal{} 0}^{45}\binom{90}{2k}( \minus{} 1)^kx^k$.

So $ \sum_{p \equal{} 0}^{44}(\tan((2p \plus{} 1)\frac {\pi}{180}))^2$ $ \equal{} \minus{} \frac {\binom{90}{88}( \minus{} 1)^{44}}{\binom{90}{90}( \minus{} 1)^{45}}$ $ \equal{} 4005$

Congrats, Zetax !!!

Patrick, I did look at it and I did try to adapt without success. Now I see it's quite trivial, thanks for the solution, appreciate it

I, however, was also wondering if there was a solution without the use of complex numbers. Does anyone know of such?

I give another solution in these forum, when calculate sums:
\[ \sum_{i\equal{}1}^{89} \tan^2 i\] and
\[ \sum_{i\equal{}1}^{45}\tan^2 (2i\minus{}1).\]


[hide="answer"]
100/5=20, so the average is 20.
So the answer is 8, 14, 20, 26, 32
$\framebox{8}$[/hide]

[quote="math92"]A boy ate 100 cookies in five days. Each day he ate 6 more than the day before. How many cookies did he eat on the first day?[/quote]
[hide]
I'm TOO attached to Algebra so...

$x+x+6+x+12+x+18+x+24=100$

$5x+60=100$

$5x=40$

$x=8$[/hide]

[quote="math92"]A boy ate 100 cookies in five days. Each day he ate 6 more than the day before. How many cookies did he eat on the first day?[/quote]
[hide]
5x+6+12+18+24=5x+60
x=8

[/hide]


A cylindrical closed toxic waste container eighteen inches high with an outer diameter of 12 inches is made of stainless steel with a uniform thickness of $ \frac{1}{4}$ inch. Assuming the cylinder is full, to the nearest percent what is the volume of the liquid to the volume of the entire cylinder?

Firstly, note that a diameter of $ 12$ inches means that the radius, used in our volume formulas, is $ \frac{12}{2}\equal{}6$ inches.
The percent of the volume occupied by the liquid is simply the volume of the liquid over the volume of the entire container. Given the $ \frac{1}{4}$ inch uniform thickness of the container, the volume the liquid covers is $ \pi(6\minus{}\frac{1}{4})^2(18\minus{}2(\frac{1}{4}))\equal{}\frac{18515}{32}$. The volume of the container is much easier to calculate, simply $ \pi(6)^2(18)\equal{}648$. There percent is $ \frac{\frac{18515}{32}}{648}$, which comes out to be approximately $ \fbox{89}$ percent.


If $ x,y,z$ are positive real numbers with $ xyz\equal{}1,$ prove that
\[ \frac{x}{y^3\plus{}2}\plus{}\frac{y}{z^3\plus{}2}\plus{}\frac{z}{x^3\plus{}2}\ \geqslant\ 1\]

y^3+2=y^3+1+1
am:gm y^3+1+1>3y
(y^3+1+1)^-1<(3y)^-1
now we use am-gm for 3 fractions
then we use taht xyz=1
(sorry but i downt know how shoud i type formulas)

[quote="Mahan73"]
$ (y^3\plus{}1\plus{}1)^{\minus{}1}\leq(3y)^{\minus{}1}$
[/quote]
I think it does not help here. :wink:


inequalities
calculus
geometry
geometry solved

Here is an other problem that I think is easier than the first. It was given to me by the same student-R Necula:

Let $ABC$ be a triangle with sides $a$, $b$, $c$, area $S$ and semiperimeter $p$ (these are standard notations). Prove that

$S\cdot\sqrt{\frac{4p^{2}}{27}+OG^{2}}\leq\frac{abc}{4}$.

here is a solution:

sqrt(4p^2/27+OG^2)<=abc/(4S) which is equivalent to   4p^2/27+OG^2<=R^2, because R=abc/(4S);
now: 3OG=OH =>OG^2=OH^2/9=R^2(1-8cosA*cosB*cosC)/9
so the inequality becomes:
4p^2/27+R^2(1-8cosA*cosB*cosC)/9<=R^2
p=R*(sinA+sinB+sinC) and the last inequality is equivalent to:
4R^2(sinA+sinB+sinC)^2/27+R^2(1-8cosA*cosB*cosC)/9<=R^2
we divide it by R^2 and we know that:
sinA*sinA+sinB*sinB+sinC*sinC=2+2cosA*cosB*cosC and we now have the inequality like so:
4(sinA+sinB+sinC)^2/27+(9-4(sinA*sinA+sinB*sinB+sinC*sinC))/9<=1 which becomes (after some calculus)
(sinA+sinB)^2+(sinB+sinC)^2+(sinA+sinC)^2>=0 which is obious!

hope is correct! :D :D

Congratulations! It's a good solution. I myself though of another but I'm not going to post it yet. Try to solve an inequality posted by RNecula(that's me). I registered later an I asked Maverick to post it for me.(this one)


Let $a,b\in\mathbb{N}$, $(a,b)=1$ and $\frac{a+b}{a-b}\in\mathbb{N}$, show that either $ab+1$ or $4ab+1$ is a square number.

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

Instead of $4ab-1$ and $ab-1$ it should be respectively $4ab+1$ and $ab+1$

If $gcd(a,b)=1$, then $gcd(a, a-b)=1$

$a-b|a+b \Rightarrow a-b|a+b+a-b \Rightarrow a-b|2a \Rightarrow a-b=1 \vee a-b=2$
In the first case we have $4ab+1=(2a-1)^2$, in the second $ab+1=(a-1)^2$

I found that $ab+1$ or $4ab+1$ are squares... can you show me the mistake?

[size=75][color=green]no, you're right -mod[/color][/size]

[hide]

$\frac{a+b}{a-b}=k$, $k \in \mathbb{N}$ We have to assume $a>b$

$a(1-k)=b(-1-k)$ but $gcd(a,b)=1 \implies a|-1-k$ and $b|1-k$
then, $-1-k=ax$ and $1-k=by \implies aby=abx \implies x=y$

So, $k=-ax-1$
      $k=1-bx$

$x(a-b)=-2$ but $a>b \implies$ $(a-b=2 \wedge x=-1)$ [case 1] $\vee$ $(a-b=1 \wedge x=-2)$ [case 2]

[case 1]

$a-b=2$
$a^2-ab=2a$ $a^2-2a+1=ab+1$ $ab+1=(a-1)^2$

[case 2]
$a-b=1$ $4a^2-4ab=4a$ $4a^2-4a+1=4ab+1$ $4ab+1=(2a-1)^2$
[/hide]

[quote="Hokkage"]Oh, so those should have been + signs rather than -?[/quote]

Sorry...they're "+" signs. :blush:

[quote="hydro"][quote="Meghrabi"]How did you came up with 'abx=aby' ?

Thanks[/quote]

because if $a(1-k)=b(-1-k)$ and $gcd(a,b)=1$, then $-1-k$ must be a multiple of $a$, multiple that we can write in the form $ax$.
The same thing for $1-k$, that must be a multiple of $b$. Call it $by$. Now substitute $-1-k$ with $ax$ and $1-k$ with $by$ and you have
$abx=aby$[/quote]

Thanks :)

[quote="shadysaysurspammed"]What do the $\vee$ and  $\wedge$  signs mean?[/quote]

$\wedge$ means "and", $\vee$ means "or" in math logic language


Another problem I got a different answer for, #27:

27. A collection of 20 coins made up of only nickels, dimes and
quarters has a total value of $\$$3.35. If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of$\$$2.75. How
many quarters are in the original collection?

This time, I got 5 nickels, 26 dimes, and 2 quarters.  2 was my answer.

dude that's messed up a little. fix the tex by putting [code]$\$$[/code] for the dollar sign

[quote="Zmastr"]Another problem I got a different answer for, #27:

27. A collection of 20 coins made up of only nickels, dimes and
quarters has a total value of $3.35. If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of$2.75. How
many quarters are in the original collection?

This time, I got 5 nickels, 26 dimes, and 2 quarters.  2 was my answer.[/quote]

They said 20 coins in the problem.

[hide]
Try 8 quarters, 3 dimes, 9 nickels.
x= nickels
y=dimes
z=quarters
5x+10y+25z=275
x+2y+5z=55
x+y+z=20
y+4z=35
Guess and check gives you the answer.
[/hide]

That one was long and annoying.

Maybe my guess and check wasn't narrowed enough?

I probably had 20 minutes going into #25 and wasted 10 minutes on 25 & 27.

It's three equations with three variables each.  That's enough to have only one possible solution.

Also, I think 5x+10y+25z=335, not 275.

Heres #27:

A collection of $20$ coins made up of only nickels, dimes and
quarters has a total value of $3.35$. If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75$. How
many quarters are in the original collection?

Here's what I did:

[hide]Guess and Check!
i put in some random numbers until i cam eto having $11$ quarters.  From here i played around with the nickels and dimes, only to come to the conclusion that regardless the amount of dimes or nuckels, the answer asked for quarters.  So i put down $11$, and got it right.[/hide]

[hide]Um:
5n + 10d + 25q = 335
5d + 25n + 10q = 275

Multiply the second equation by two and subtract the equations
9n-q = 43
guess and check, n = 5, 6 only because 63-43 = 20, which is too many!

6 nickels works, so there are 54-43 = [b]11[/b] quarters, (3 dimes).[/hide]

You have to realize that they said that the number of coins is 20.

lol! i remember this one

i guessed two since it was obviously 1-3, and it wasn't 1 or 3

i still believe i should've gotten a 25 on the sprint

i don't remember getting 11 wrong

jorian

[quote="superwizard188"]
9n-q = 43
[/quote]

Sidenote on that:
Yeah, you should have taken into account that there were 20 coins and stuff, but if you wanted to solve that equation over the positive integers, you would use the following process...

[hide]
$9n-q=43$
We can take the modulo 9 values of both sides to get
$-q\equiv7\pmod{9}$
$q\equiv2\pmod{9}$
Hence $q\in\{2,11,20,\ldots\}$, generating the solutions $(p,q)=(k+4,9k-2)$ for $k\in\mathbb{Z}^{+}$.[/hide]


limit
invariant
real analysis
real analysis unsolved

[quote="kunny"]By $A.M.\geq G.M.\geq H.M.$ we have that $a_{n+1}\geq b_{n+1}\geq c_{n+1}.$ Maybe answer is $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=\lim_{n\to\infty}c_{n}=1$, but I can't show that.[/quote]

I'm not think answer is $1$. But lima_n=limb_n=limc_n are true!  :wink:

On my class , problem has condition $c_{1}a_{1}=b_{1}^{2}$  :D I soved it!

But if remove this conditon , I'm can't!  :(

I'm not sure there is a simple formula for the limit. Just try the (apparently?) simpler sequence:
$a_{n+1}=\frac{a_{n}+b_{n}}{2}$, $b_{n+1}= \sqrt{a_{n}b_{n}}$
It has already been discussed, and it is well known that the sequence converges towards the arithmetic-geometric mean of $a$ and $b$, but this limit is not that simple. Take a look at http://en.wikipedia.org/wiki/Arithmetic-geometric_mean

This just can't be true for that simple reason that, if you start with $a_{2}, b_{2}, c_{2}$, you will get a different quantity. We should look for something that is invariant under the given transformation.

[quote="fermat3"]we have $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=\lim_{n\to\infty}c_{n}=\sqrt[3]{a_{1}b_{1}c_{1}}$ i will post the solution tonight.[/quote]

post your solution, Please! :P

Don't bother waiting- it's false.

If it were true, the geometric mean of $a_{2},b_{2},c_{2}$ would be equal to the geometric mean of $a_{1},b_{1},c_{1}$. It's not.

The answer is a mean which is preserved by the transformation. That can't be any of the familiar ones.


function
induction
algebra unsolved
algebra

Find all functions$f: Z\rightarrow{Z}$ such that
           $f(x^{2}+y^{2})=f(x)^{2}+f(y)^{2} for all x,y\in{Z}$
:)

Well, there was also : http://www.mathlinks.ro/Forum/viewtopic.php?t=14908 .

So, I wonder : can we generalize? :) ($f(x_1^n + \ldots + x_n^n) = f(x_1)^n + \ldots + f(x_n)^n$)


find all integers solutions to the equation:
 $x^2 =y^7+7$


[quote="LydianRain"]Who do you keep quoting?[/quote]Certainly not a native English speaker.

The problem has $ m$ and $ n$ hopelessly confused, and doesn't make sense in its current form.

[quote="LydianRain"][quote="studen of jean"][b]"..A nice of math ..."[/b][/quote]
Who do you keep quoting?  :roll:[/quote]
".... everyone ...."
"... Love math....."

Use $ Lim_{m \to \plus{}\infty}U_{m}\equal{}L <\plus{}\infty$ then
 $ Lim_{m \to \plus{}\infty}\frac {U_{1}\plus{}U_{2}\plus{}...\plus{}U_{m}}{m}\equal{}L$
 :oops: [b](Casino)[/b]

No, the question is nonsense. You started with a fixed $ n$ and some $ n\times n$ matrices; you can't change $ n$ later.

[quote="studen of jean"]Use $ Lim_{m \to \plus{} \infty}U_{m} \equal{} L < \plus{} \infty$ then
 $ Lim_{m \to \plus{} \infty}\frac {U_{1} \plus{} U_{2} \plus{} ... \plus{} U_{m}}{m} \equal{} L$
 :oops: [b](Casino)[/b][/quote]

casino, cesaro, what's the difference?


Joe was in charge of lights for a dance. The red light blinks every two seconds, the yellow light every three seconds, and the blue light every five seconds. If we include the very beginning and very end of the dance, how many times during a seven minute dance will all the lights come on at the same time?

They come on together every 30 seconds, so twice a minute. If we count the beginning AND the end, it will be $ 7\cdot2\plus{}1\equal{}\boxed{15}$ times.


Prove that for arbitary integer $ n > 16$, there exists the set $ S$ that contains $ n$ positive integers and has the following property:if the subset $ A$ of $ S$ satisfies for arbitary $ a,a'\in A, a\neq a', a \plus{} a'\notin S$ holds, then $ |A|\leq4\sqrt n.$

Consider the set $ S\equal{}\{1,\dots,65\}$ and the odd number subset of it $ A\equal{}\{1,3,\dots,63,65\}$. $ |A|\equal{}33$ and $ 4\sqrt {65}\approx 32.25$

From this [url=http://www.mathlinks.ro/viewtopic.php?t=1491]post[/url] we know the maximum length of the sum free sequence. Hence $ (k\plus{}1)^2 \leq 16(2k\plus{}1)$ according to the problem statement. Clearly this does not hold as LHS is a higher degree polynomial.

[quote="Fang-jh"]Prove that for arbitary integer $ n > 16$, there exists the set $ S$ that contains $ n$ positive integers and has the following property:if the subset $ A$ of $ S$ satisfies for arbitary $ a,a'\in A, a\neq a', a + a'\notin S$ holds, then $ |A|\leq4\sqrt n.$[/quote]
we want say that?: (it's what i understand)

$ \forall n > 16,\exists set\ S: \ |S| = n:$
$ (\forall A\subset S): \ |A| > 4\sqrt {n}\Rightarrow \exists (a,a')\in A: \ a\neq a'\ and \ a + a'\in S$

but $ \exists N\in%Error. "mathh" is a bad command.
{N}:$ such that $ N > 16$ and $ N\ge 2(4\sqrt {N} + 1)$
for $ S$ such that $ |S| = N$ let $ S=\{M,M-1,M-2,.......,M-N+1\}$
we take $ A = \{M,M - 1,M - 2,...,M - ([4\sqrt {N}])\}$ here we have $ |A| > 4\sqrt {N}$
and $ \forall a,a'\in A$ such that $ a\neq a'$ : $ a + a'\ge 2M - 2[4\sqrt {N}] + 1\ge (M + 1) + (N - 2[4\sqrt {N}])\ge M + 1$ (because $ M\ge N$)
then $ a + a'\not\in S$

$ A\subset S$ cheks the property but $ |A| > 4\sqrt {N}$
so the probleme is fault or i don't understand  :ninja:

[quote="zhaobin"][quote="aviateurpilot"]$ A \equal{} \{M,M \minus{} 1,M \minus{} 2,...,M \minus{} ([4\sqrt {N}])\}$ here we have $ |A| > 4\sqrt {N}$
[/quote]
Hello,But I think $ A$ should be the subset of $ S$.[/quote]
$ S\equal{}\{M,M\minus{}1,..,M\minus{}N\plus{}1\}$

[quote="aviateurpilot"][quote="zhaobin"][quote="aviateurpilot"]$ A \equal{} \{M,M \minus{} 1,M \minus{} 2,...,M \minus{} ([4\sqrt {N}])\}$ here we have $ |A| > 4\sqrt {N}$
[/quote]
Hello,But I think $ A$ should be the subset of $ S$.[/quote]
$ S \equal{} \{M,M \minus{} 1,..,M \minus{} N \plus{} 1\}$[/quote]
But in the problem,we should prove there exist a set $ S$,not for all $ S$.
You just prove that for some $ S$ the conclusion doesn't hold.

[quote="zhaobin"][/quote][quote="aviateurpilot"][quote="zhaobin"][quote="aviateurpilot"]$ A = \{M,M - 1,M - 2,...,M - ([4\sqrt {N}])\}$ here we have $ |A| > 4\sqrt {N}$
[/quote]
Hello,But I think $ A$ should be the subset of $ S$.[/quote]
$ S = \{M,M - 1,..,M - N + 1\}$[/quote][quote="zhaobin"]
But in the problem,we should prove there exist a set $ S$,not for all $ S$.
You just prove that for some $ S$ the conclusion doesn't hold.[/quote]

Zhaobin is right. The most difficult part in this problem is to construct a set $ S$ which satisfies the property mentioned in the question. So please stop searching for a certain $ S$ to negate the problem  :) .

[quote="zhaobin"][/quote][quote="aviateurpilot"][quote="zhaobin"][quote="aviateurpilot"]$ A = \{M,M - 1,M - 2,...,M - ([4\sqrt {N}])\}$ here we have $ |A| > 4\sqrt {N}$
[/quote]
Hello,But I think $ A$ should be the subset of $ S$.[/quote]
$ S = \{M,M - 1,..,M - N + 1\}$[/quote][quote="zhaobin"]
But in the problem,we should prove there exist a set $ S$,not for all $ S$.
You just prove that for some $ S$ the conclusion doesn't hold.[/quote]
sorry  :blush:

I have an idea of such a set S, i have not computed quantitatively

Let K is set of integers such that its representation in base 4 contains only the digit 1,2.  Arrange this set in increasing order and for each n, take the first n elements of K

Anybody solutions? I think this is a very famous problem but I don;t know solution. :(

I thought this is too hard for a 2008TSTP2. Here's a hint:
We construct a graph with vtx set $S$
$a$~$b$ iff $a+b$ belongs to $S$
We need to ensure the independent set if the graph isn't too large
Recall what we have learnt in Graph theory, how to ensure there isn't a large indep. set with little edges?
That's Turan's theorem! Thus the graph  roughly look like a combination of cliques and a indep set(the largest few numbers)
The next question is how to add numbers to the vertices.
Actually, between different cliques, the numbers on their vertices should have different order,
To illustrate, we can let their power of 2 is different and their odd part is the same.

Hope you can work out for the details! (It's not hard but it would be helpful if you work out yourselves)

Is there a way to do it without graph theory ?

Actually we don't need to use graph theory, turan's theorem is just a motivation lead to the construction!


calculus
derivative
calculus computations

Hi, sorry just stuck on this simple question, I would just like to know the basic idea how to get it:
Let f be a polynumial of degree n such that f is greater or equal to 0 on the whole real. Then prove that f + d(f) + d^2(f)+...+d^n(f) is greater or equal to 0 ( here d^k(f) means the k th derivative of f).Thanks

Let

$g=\sum^n_{i=0} f^{[i]}$

It is clear that $f=g-g'$.  Say that $g(x_0)<0$.  Then there exist $x_1,x_2$ such that $g(x_1)=g(x_2)=0$ and $g(x)<0$ for every $x\in(x_1,x_2)$ by continuity, since clearly $f\to\infty\implies g\to\infty$.  Also, by Rolle's theorem, there exists $x_3\in (x_1,x_2)$ such that $g'(x_3)=0$.

But then $f(x_3)=g(x_3)-g'(x_3)\implies f(x_3)=g(x_3)<0$, which contradicts $f\geq 0$.

The result follows.

[quote="blahblahblah"]Let

$g=\sum^n_{i=0} f^{[i]}$

It is clear that $f=g-g'$.  Say that $g'(x_0)<0$.  Then there exist $x_1,x_2$ such that $g(x_1)=g(x_2)=0$ and $g(x)<0$ for every $x\in(x_1,x_2)$ by continuity, since clearly $f\to\infty\implies g\to\infty$.  Also, by Rolle's theorem, there exists $x_3\in (x_1,x_2)$ such that $g'(x_3)=0$.

But then $f(x_3)=g(x_3)-g'(x_3)\implies f(x_3)=g(x_3)<0$, which contradicts $f\geq 0$.

The result follows.[/quote]
Maybe you mean $g(x_0)<0$? And $g(x_3)$ is, in fact, the minima of $g$ on $[x_1,x_2]$. Hence $g'(x_3) = 0$(Fermat).

A problem is that why can we find $x_1,x_2$ such that $g(x_1)=g(x_2)=0$?

Hi,
I'll prove it using extreme value theorem.
Let g(x)=f(x)+f'(x)+....+f_n(x).Suppose min{g}occurs at x=c ,then
g(c)-f(c)=g'(c)=0.Thus g(x)>=g(c)=f(c)>=0

[quote="blahblahblah"]Yes, I meant $g(x_0)<0$.

Since $f\geq 0$, it is clear that $f$ has even degree.  Since $g$ has the same degree as $f$, it goes to infinity as $x$ becomes arbitrarily large or small.  The existence of $x_1,x_2$ follows.  I don't understand what your point about $x_3$ being the minima of $g$ is, because Rolle's implies nothing about the uniqueness of the critical points whose existence it yields.[/quote]

Oh, thanks. I didn't see the existence of $x_1,x_2$ (now I see it), so my argument goes like this:

If $g(x_0)<x_0$ for some $x_0$, then $g(x)<0$ for $x\in [x_1,x_2]$, where $x_0\in [x_1,x_2]$. Suppose $g(x)$ attains its minimum value at $x=\xi\in[x_1,x_2]$ then $g(\xi)<0$ and $g'(\xi)=0$. Thus $f(\xi) = g(\xi) - g'(\xi) = g(\xi) < 0$, which contradicts with the assumption of $f\geq 0$.


I just wonder, are you allowed to bring Graph paper into Mathcounts?

I checked everywhere but I can't find it.

no, graph paper isn't allowed on mathcounts.

Why would you need graph paper? I surmise that all the problems there can be done without graph paper. (Or if it's not Sprint you can graph it on your calculator  :P ).

You must use their paper.  I've seen someone try it, but Mathcounts said "No."
As D3m0n was saying, you can use your calculator for the target round, most TIs above 7X have graphing abilities.  Just don't bring your PDA in there!


Suppose that the domain of the function $f : \mapsto f(x)$ is the set $D$, and the range is the set $R$, where $D$ and $R$ are subsets of the real numbers. Consider the functions:

$f(2x), \ f(x+2), \ 2f(x), \ f(\frac{x}{2}), \ \frac{f(x)}{2}, \ f(x+2)-2$

If $m$ is the number of the functions listed above that must have the same domain as $f$, and $n$ is the number that must have the same range as $f$, then the ordered pair $(m,n)$ is:

$A. (0,5) \ B. (1,5)\ C. (2,3) \ D. (3,2) \ E. (3,3)$

[hide]C: Only 2f(x) and f(x)/2 have the same domain as f(x).  f(2x), f(x+2) and f(x/2) have the same range.[/hide]


inequalities
search
trigonometry
geometry proposed
geometry

[quote="Virgil Nicula"]Prove that in any triangle $ ABC$ there is the relation $ \boxed {\ OH\ge OI\cdot\sqrt {1 \plus{} \frac {2r}{R}}\ }$ ([b]standard notations[/b]).[/quote]

$ OI^2 \equal{} R^2 \minus{} 2Rr,OH^2 \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2)$

$ OI\sqrt {1 \plus{} \frac {2r}{R}} \equal{} \sqrt {OI^2\frac {R \plus{} 2r}{R}} \equal{} \sqrt {R(R \minus{} 2r)\frac {R \plus{} 2r}{R}} \equal{} \sqrt {R^2 \minus{} 4r^2}$

We have to prove

$ 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2)\ge R^2 \minus{} 4r^2\Leftrightarrow 8R^2 \plus{} 4r^2\ge a^2 \plus{} b^2 \plus{} c^2$ which is true and quite strong.

I can show a trigonometric proof,it was posted many times but I can't search

by law of sine and $ \frac {r}{R} \equal{} \cos A \plus{} \cos B \plus{} \cos C \minus{} 1$ we need to prove

$ 8 \plus{} 4(\cos A \plus{} \cos B \plus{} \cos C \minus{} 1)^2\ge 4(\sin^2A \plus{} \sin^2B \plus{} \sin^2C)$
Using $ \cos^{2}A \plus{} \cos^2B \plus{} \cos^2C \equal{} 1 \minus{} 2\cos A\cos B\cos C$ and $ \sin ^2A \equal{} 1 \minus{} \cos^2A$ the inequality equivalent to $ (1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C) \geq \cos A\cos B\cos C\Leftrightarrow \tan A\tan B\tan C \geq \cot{\frac {A}{2}}\cot{\frac {B}{2}}\cot{\frac {C}{2}}\Leftrightarrow \tan A \plus{} \tan B \plus{} \tan C \geq \cot{\frac {A}{2}} \plus{} \cot{\frac {B}{2}} \plus{} \cot{\frac {C}{2}}$ which is true by $ \tan B \plus{} \tan C\ge\cot\frac {A}{2}$ e.t.c

[quote="Erken"][quote="encyclopedia"]
$ 8R^2 \plus{} 4r^2\ge a^2 \plus{} b^2 \plus{} c^2$ which is true and quite strong.[/quote]
Can I see the proof of this inequality?
I know one,which follow from some algebraic inequality given at Poland Olympiad,if I am not mistaken,so I'm interested in other solutions.[/quote]
it's equivalent to:

$ HI^2\geq 0$

There is a link [url=http://www.mathlinks.ro/viewtopic.php?search_id=117899440&t=216506]here[/url] of course you mean
$ IH^2 \equal{} 4R^2 \plus{} 4Rr \plus{} 3r^2 \minus{} p^2\ge 0\Rightarrow 8R^2\plus{}4r^2\ge a^2\plus{}b^2\plus{}c^2$ but how to prove it?

[quote=encyclopedia]I can show a trigonometric proof,it was posted many times but I can't search

by law of sine and $ \frac {r}{R} \equal{} \cos A \plus{} \cos B \plus{} \cos C \minus{} 1$ we need to prove

$ 8 \plus{} 4(\cos A \plus{} \cos B \plus{} \cos C \minus{} 1)^2\ge 4(\sin^2A \plus{} \sin^2B \plus{} \sin^2C)$
Using $ \cos^{2}A \plus{} \cos^2B \plus{} \cos^2C \equal{} 1 \minus{} 2\cos A\cos B\cos C$ and $ \sin ^2A \equal{} 1 \minus{} \cos^2A$ the inequality equivalent to $$ (1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C) \geq \cos A\cos B\cos C$$
$\Leftrightarrow \tan A\tan B\tan C \geq \cot{\frac {A}{2}}\cot{\frac {B}{2}}\cot{\frac {C}{2}}\Leftrightarrow \tan A \plus{} \tan B \plus{} \tan C \geq \cot{\frac {A}{2}} \plus{} \cot{\frac {B}{2}} \plus{} \cot{\frac {C}{2}}$ which is true by $ \tan B \plus{} \tan C\ge\cot\frac {A}{2}$ e.t.c[/quote]
$$ (1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C) \geq \cos A\cos B\cos C$$
$$\implies  (1 \minus{} \sin\frac{A}{2})(1 \minus{} \sin\frac{B}{2})(1 \minus{} \sin\frac{C}{2}) \geq \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$

 George Apostolopoulos:
$$3\sqrt{3} (1 \minus{} \sin\frac{A}{2})(1 \minus{} \sin\frac{B}{2})(1 \minus{} \sin\frac{C}{2}) \leq \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$
We have
$$3\sqrt{3}\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq 3\sqrt{3} (1 - \sin\frac{A}{2})(1 - \sin\frac{B}{2})(1 - \sin\frac{C}{2}) \leq \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$


Let $ a,b,c$ be real numbers such that $ ab,bc,ca$ are rational numbers. Prove that there are integers $ x,y,z$, not all of them are $ 0$, such that $ ax\plus{}by\plus{}cz\equal{}0$.

$ \frac {ab \cdot bc}{ac} \equal{} b^2$ and so forth give $ a^2, b^2, c^2$ rational.  Then $ \mathbb{Q}[a, b, c] : \mathbb{Q}$ has dimension at most $ 4$, so it must be the case that one of $ a, b, c$ is contained in the extension field generated by the other two.

Edit:  In particular, we have the following cases.  If $ \mathbb{Q}[a] : \mathbb{Q} \equal{} 1$, we are done; $ a, b, c$ are rational.  Otherwise, if $ \mathbb{Q}[a, b] : \mathbb{Q}[a] \equal{} 1$, we are done; $ b$ is some rational linear combination of rationals and $ a$ and if the rational part is nonzero, square the relation to conclude that $ a$ is rational, contradiction..  Otherwise, $ \mathbb{Q}[a, b, c] : \mathbb{Q}[a, b] \equal{} 1$ and $ c$ is some rational linear combination of rationals and $ a, b$.  Again, if the rational part is nonzero, square the relation to conclude that a linear combination of $ a, b$ is rational, contradiction.

Or just make x = bct, y = act and z = -2abt, for some integer t, multiple of de denominators of ab, bc and ca.

Since $ ab,ac$ and $ bc$ are rational $ ab/bc\equal{}a/c$ and $ ab/ac\equal{}b/c$ must be rational as well. Thus $ a\equal{}mc,b\equal{}nc$ where $ m,n$ are rational numbers. Then $ ax\plus{}by\plus{}cz\equal{}0\Leftrightarrow mx\plus{}ny\plus{}z\equal{}0$. There obviously exist integers $ x,y,z$ satisfying the last equality, for example let $ x,y$ be the denominators of $ m$ and $ n$ respectively and $ \minus{}z$ be the sum of their numerators.


Or, what about the shorter version:

$ \minus{}1 \equal{} (\minus{}1)^{(2/2)} \equal{} \sqrt{(\minus{}1)^2} \equal{} \sqrt{1} \equal{} 1$

The answer is that the identity $ a^{bc} \equal{} (a^b)^c$ is only valid if $ a$ is positive.

I hope you understand that a square root and a principle square root are different. A positive number has 2 square roots, but only 1 principle square root.
ex: $ n^{\frac12}\equal{}\pm\sqrt n$


adding and subtracting the two equation gives us:
2=x^5+5y^4x+10y^2x^3
and 1=5x^4y+y^5+10x^2y^3
adding and subtracting yields:
(x+y)^5=3 hence x+y=5^(1/3)=a
(x-y)^5=1 hence x-y=1
its a putnam problem.

the same way I did it.
I found it in an italian forum, could you, please, provide exact and complete reference of this problem?

It was problem B2 of Putnam 2001. You can find at address:
http://www.kalva.demon.uk or forumgeom.fau.edu/POLYA/ProblemCenter/POLYA018.html

Have joy!


calculus
integration
trigonometry
function
calculus computations

I'm having a bit of trouble with $ \int{\sqrt{tan(x)}dx}$. How would you do it? Thanks.

Substitute $ t\equal{}\sqrt{\tan x}, x\equal{}\arctan(t^2)$. That gives you a rational function; apply partial fractions.

[hide="Answer"]Make the substitution $ \tan x \equal{} \frac {1}{t^2} \equal{} > x \equal{} \arctan\left(\frac {1}{t^2}\right) \equal{} > dx \equal{} \frac { \minus{} 2t}{1 \plus{} t^4} dt$, and so

$ P \equal{} \int \sqrt {\tan x}\,dx \equal{} \int \frac {1}{t} \cdot \frac { \minus{} 2t}{1 \plus{} t^4}\,dt \equal{} \minus{} 2 \int \frac {1}{1 \plus{} t^4}\,dt$,

and now you are to factorize $ 1 \plus{} t^4$.

[hide="Hint"]$ 1\plus{}t^4 \equal{} (1\plus{}t^2)^2 \minus{} 2t^2$[/hide][/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

are all the irreducible polynomials over $ \mathbb{Z}[X]$ which have coefficients in $ \{0,..,9\}$ described by Cohn's irreducibility criterion ?
I mean does the converse of that criterion work for polynomials in $ \mathbb{Z}[X]$ with all coefficients in $ \{0,..,9\}$ ?

http://en.wikipedia.org/wiki/Cohn's_irreducibility_criterion

[quote="billwgd"]See the reference to the the Bunyakovsky conjecture in said Wiki page, and see my post
http://google.com/groups?selm=y8z4r81qhab.fsf%40nestle.ai.mit.edu[/quote]

you're right.


how many way are there to express 100000 as the product of 3 positive integers? not that $a*b*c$ is the same as $b*a*c$ for numbers $a,b,c$.

This was a pretty tricky problem...
Think about the paired factors of 100000.  Now, if you pick one of those factors as one of your numbers for the product, you have to split the other paired factor into two numbers.  However, you can pick any factor first, meaning that you are looking for the number of factors for each of the factors of 100000.  Then you are looking to evaluate $(m+1)(n+1)$ for $2^{m}* 5^{n}$ for all $0 \le m \le 5$,and $0 \le n \le 5$...this is easily changed into $(1+2+3+4+5+6)(1+2+3+4+5+6)$, which equals $21^{2}=441.$
Sorry if I didn't explain this too well, but it's kind of abstract.


geometry
parallelogram
geometric transformation
reflection
geometry proposed

$P$ and $\mathcal{C}$ are a fixed point and a fixed circle in the plane. A pencil of chords $AA',$ $BB',$ $CC',$ $DD',$ etc, pass through $P.$ Circles $C_1$ and $ C'_1$  passing through $ P$ are tangent to $ \mathcal{C}$ at  $ A$ and  $ A',$ respectively. These circles intersect at $P$ and $ Q_1.$ Analogously, we defined the points $Q_2,Q_3,Q_4,Q_i.$ Prove that all points $Q_i$ lie on a same circle and find its diameter and center.

$ O, O_1, O_1'$ are centers of the circles $ \mathcal C, \mathcal C_1, \mathcal C_1'.$ Isosceles $ \triangle AO_1P \sim \triangle AOA'$ and $ \triangle A'O_1'P \sim \triangle A'OA$ are centrally similar with similarity centers $ A, A',$ respectively $ \Longrightarrow$ $ PO_1OO_1'$ is a parallelogram, its diagonals cutting each other in half at $ M.$ $ \mathcal M$ is a circle with diameter OP. $ \mathcal C_1, \mathcal C_1', \mathcal M$ have collinear centers $ O_1, M, O_1'$ and all go through P, therefore they are coaxal and all go through the reflection $ Q_1$ of P in the center line $ O_1O_1'.$

Inversion with pole $P$ and power equal to the power of $P$ WRT the circle $\mathcal{C}$  transforms $ C_1$ into the tangent line $ p$ to $\mathcal{ C}$ through $ A'$ and $ C_1'$ into the tangent line $ p'$ to $\mathcal{ C}$ through $A.$ Thus, lines $p$ and $ p'$ meet at the inverse $ Q'$ of $Q,$ but $Q'$ lies on  the polar $\tau$ of $ P$  with respect to $\mathcal{C}.$ So, if $Q'$ describes $\tau,$ then $ Q_1$ describe the circle with diameter $\overline{OP}.$


Given two circles $(O;R)$;$(O';R')$ with $OO' >R+R'$,from $O$ ,a tangent $OT'$ is constructed to $(O')$,from $O'$  a tangent $O'T$ to $(O)$ is drawn.Line $TT'$ meets $(O)$ and $(O')$ at $S$ and $S'$,respectively(distinct from $T$ and $T'$).Prove thatr $ST=S'T'$

Let M, M' be midpoints of ST, S'T'. Let OT' meet (O) at P and let O'T meet (O') at P'. Let F, F' be feet of perpendiculars from P, P' to OO'. $OT' \perp O'T',$ $O'T \perp OT,$ OO'T'T is cyclic, $\angle MT'O \equiv \angle TT'O = \angle TO'O$ and $\angle M'TO' \equiv \angle T'TO' = \angle T'OO'.$ Since

$\angle MOT+\angle TOT'+\angle TT'O = 90^\circ = \angle TOP+\angle POO'+\angle TO'O$

$\angle M'O'T'+\angle T'O'T+\angle T'TO' = 90^\circ = \angle T'O'P'+\angle P'O'O+\angle T'OO'$

it follows that $\angle MOT = \angle POO' \equiv \angle POF,$ $MT = FP$ and $\angle M'O'T' = \angle P'O'O \equiv \angle P'O'F',$ $F'P' = M'T'.$ By the [url=http://agutie.homestead.com/files/Eyeball0.htm]eyeball theorem[/url], $FP = F'P',$ hence $MT = M'T'$ and $ST = 2MT = 2M'T' = S'T'.$

first solution
Let $M,M'$ be the midpoint of $ST, S'T'$
we have triangle OMT is similar with $TM'O'$ and triangle $OMT'$ is similar with $T'M'O'$
so $O'M'*OM=MT*TM'$ and $O'M'*OM=T'M'*MT'$
then $MT/MT'=M'T'/M'T$.
So $MT/TT'=M'T'/TT'$. Then $MT=M'T'$
Thus $ST=S'T'$ $q.e.d$

second solution
Let $M,M',K,N$ be the midpoint of $ST, S'T',OO',TT'$
we have $KT=KT'$ so triangle $TKT'$ is isosceles
Then $KN\bot TT$'. So $KN//OM//O'M'$
then $NM=NM'$
so $MT=M'T'$.Thus $ST=S'T'$ $q.e.d$

We have:
$\left\{\begin{array}{c}T'O^{2}-R^{2}=T'S\cdot T'T\\ TO'^{2}-R'^{2}=TS'\cdot TT'\\ T'O^{2}+R'^{2}=TO'^{2}+R^{2}\end{array}\right\|\Longrightarrow T'S=TS'\Longrightarrow TS=T'S'$


superior algebra
superior algebra unsolved

assume that we have a ring$R$ and there exist $a\in R$  such that for any $x\in R$ we have $x^3=a$ prove that $R$ is field or for any $r\in R$ there exist $n$ such that $r^n=0$.

I think there are some things you need to clarify. For instance, I assume $x^3=a$ for all $x$ except $x=0$, right? Second of all, you don't assume $R$ to have a multiplicative unit, right (otherwise we could take $a=1^3=1$, and the problem would be easy).

There's another thing: do you assume $R$ to be commutative? If not, do we have to prove that it's commutative? (i.e. do we have to show it's a field, which is by definition commutative, or just a division ring, which has all the properties of a field, except maybe for commutativity?).


function
linear algebra
linear algebra unsolved

For nonempty sets A,B, and C, let f:A->B and g:B->C be functions. Prove if g o f is one-to-one then f is one-to-one using as many of the following proof techniques as possible: direct proof, proof by contrapositive, and proof by contradiction.
I don't know how to prove it. Can anyone help?

Contrapositive is really nice. Suppose $f$ is not 1-1. Show: $g\circ f$ is not 1-1. Since $f$ is not 1-1, then there exists $x,y\in A$ such that $x\neq y$ and $f(x)=f(y)$. Take $g$ of both sides to get $g\circ f(x)=g\circ f(y)$, and since $x\neq y$ we have that $g\circ f$ is not 1-1.

You can do contradiction almost exactly the same way: Suppose $g\circ f$ is 1-1 and that $f$ is not. Try to get the contradiction using the same technique. (Don't want to give too much away).


geometry
geometric transformation
reflection
trigonometry

Two mirrors form an angle of 30 degrees at the point V. A ray of light, coming from a source S, is issued in parallel with one of the mirrors and is reflected by another mirror at point A, as shown. After a certain amount of reflection, the beam returns to S. If AS and AV are 1 meter in length, which is the distance traveled by the ray of light in meters?

[img]http://img27.imageshack.us/img27/1703/obmt.jpg[/img]

Portuguese question:

Dois espelhos formam um ângulo de 30o no ponto V. Um raio de luz, vindo de uma fonte S, é emitido paralelamente a um dos espelhos e é refletido pelo outro espelho no ponto A, como mostra a figura. Depois de uma certa quantidade de reflexões, o raio retorna a S. Se AS e AV têm 1 metro de comprimento, qual é a distância percorrida pelo raio de luz, em metros ?

Thanks. How to do ?

So do I.

The ray strikes the bottom mirror at $ B$, say, where angle$ VAB\equal{} 30$ degrees and so angle $ ABV\equal{} 120$ degrees. It then strikes $ AV$ at $ C$, where angle $ CVB \equal{} 60$ and so angle $ BCV \equal{} 90$ degrees.

Then, by the cosine rule, $ AB \equal{} BV\equal{} \frac{1}{\sqrt{3}}$ and so $ BC  \equal{} \frac{1}{2 \sqrt{3}}$.

Therefore the total distance travelled is $ 2 \plus{} \frac{2 \sqrt{3}}{3} \plus{} \frac{1}{\sqrt{3}} \equal{} 2 \plus{} \sqrt{3}$.

I hope this makes sense, luis.

Yes Andrew
Your sense is correct  :lol:
The answer is it.
Thanks.


inequalities function logarithms integration calculus geometry trapezoid

Let $ 0 < a < b$. Prove that \[ a < \frac{1}{e} \cdot \left( \frac{b^b}{a^a} \right)^{\frac{1}{b\minus{}a }}<b .\]

[quote="turcas_c"]Let $ 0 < a < b$. Prove that
\[ a < \frac {1}{e} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a }} < b .
\]
[/quote]

The first half says to prove that $ (\frac{b}{a})^{b} > e$ which is not always true. e.g. take b=1,a=0.9999

[quote="turcas_c"][quote="ith_power"]
The first half says to prove that $ (\frac {b}{a})^{b} > e$ which is not always true. e.g. take b=1,a=0.9999[/quote]

Your example is irrelevant. The first half it is equivalent to:

$ e < \frac {1}{a} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a}}$. For $ b \equal{} 1$ and $ a \equal{} 0.9999$ you get :
\[ RHS \equal{} 2.71842
\]
( I used Mathematica ) which is greater than $ e \equal{} 2.7182...$ so the inequality remains true.

I have a nice proof using analysis, but I'm waiting for something else.[/quote]

Sorry my mistake. :oops:  :P

Anyway,

We start with the inequality $ ln x \ge 1 \minus{} \frac {1}{x}$ for $ x \ge 1$.
Then, if $ x \equal{} b/a$, then
$ lnb \minus{} lna > 1 \minus{} a/b \\
\rightarrow a \plus{} blnb > b \plus{} blna \\
\rightarrow (\frac {b}{a})^b > e^{b \minus{} a}$

For second part,
Use
 $ ln x \ge 1 \minus{} \frac {1}{x}$ for $ 0 \le x \le 1$ put $ x \equal{} a/b$.

Another solution :

Consider the function $ f: \mathbb{R} \to \mathbb{R},$ with $ f(x)\equal{}1\plus{} x \cdot \ln(x)$. Just apply Lagrange for this one:

$ f^{\prime}(a) \leq \frac{f(b)\minus{}f(a)}{b\minus{}a} \leq f^{\prime}(b)$ and you will get the answer.

Btw: Nice solution ith_power :D

[quote="turcas_c"]Let $ 0 < a < b$. Prove that
\[ a < \frac {1}{e} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a }} < b .
\]
[/quote]

$ a < \frac {1}{e} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a }} < b$

$ \Longleftrightarrow \ln a < \minus{} 1 \plus{} \frac {1}{b \minus{} a}\ln \frac {b^b}{a^a} < \ln b$

$ \Longleftrightarrow \ln a < \frac {b(\ln b \minus{} 1) \minus{} a(\ln a \minus{} 1)}{b \minus{} a} < \ln b$

$ \Longleftrightarrow \ln a < \frac {1}{b \minus{} a}\int_a^b \ln x \ dx < \ln b$, which can be proved by the mean value of thereom of Integration, or by the comparison of the areas of two rectangulars and some region.

More harder problem created by kunny

$ \boxed{\sqrt {ab} < \frac {1}{e} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a }}\ (0 < a < b)}$

[quote="kunny"]

More harder problem created by kunny

$ \boxed{\sqrt {ab} < \frac {1}{e} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a }}\ (0 < a < b)}$[/quote]

Use $ ln x< 2\frac{x\minus{}1}{x\plus{}1} \forall 0<x<1$. let x=a/b

[quote="ith_power"][quote="kunny"]

More harder problem created by kunny

$ \boxed{\sqrt {ab} < \frac {1}{e} \cdot \left( \frac {b^b}{a^a} \right)^{\frac {1}{b \minus{} a }}\ (0 < a < b)}$[/quote]

Use $ ln x < 2\frac {x \minus{} 1}{x \plus{} 1} \forall 0 < x < 1$. let x=a/b[/quote]

That's right  :lol: , my solution is the comparison of the area of a trapzoid and two rectangulars.  :D


Let $ a,b,c$ be nonnegative real numbers. Consider the following inequality
\[ \frac{a}{a\plus{}kb}\plus{}\frac{b}{b\plus{}kc}\plus{}\frac{c}{c\plus{}ka} \plus{}\frac{2k\minus{}1}{k\plus{}1}\cdot \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2} \le 2.\]
1. Prove that the inequality holds for all $ k \ge \frac{1}{3}$.
2. $ \bigstar$ Determine the least value of $ k>0$ so that the inequality holds.

Hello can_hang2007! I liked ur nice problem. It doesn't seem very easy to me! :)
Just one query, in the 1st question u told to proof the ineq for all $ k \geq \frac{1}{3}$ .....and in the 2nd u asked again for the least possible $ k$? That $ k$ will also be $ \frac{1}{3}$ for sure.
Did u meant to proof the two parts, by treating them as two isolated problems? :maybe:

[quote="Sunkern_sunflora"]Hello can_hang2007! I liked ur nice problem. It doesn't seem very easy to me! :)
Just one query, in the 1st question u told to proof the ineq for all $ k \geq \frac {1}{3}$ .....and in the 2nd u asked again for the least possible $ k$? That $ k$ will also be $ \frac {1}{3}$ for sure.
Did u meant to proof the two parts, by treating them as two isolated problems? :maybe:[/quote]
Dear Sunkern_sunflora,
I proposed the problem into 2 parts since I only have solution for 1).
For 2), it is still openned for me, so I include $ \bigstar$ as my openned question.
Thanks for your interest,
:)

Thank u, can_hang2007.
I still get no clue how to prove that for all $ k \geq \frac{1}{3}$ the ineq holds! Could u post your solution please, I'm eager to see it.
One more thing, I can't handle the Schur & Muirhead ineq. Could u please help me out! :(
[P.S. please tell me about "majorisation" also.]
Thank you. :)


I'm having trouble understanding the problem.  The numbers on the bottom - do you mean 124 drivers had 0 accidents, 234 drivers had 1 accident, etc.?  If so, $ 346\plus{}176\plus{}120\equal{}642$, or the number of drivers with at least two accidents.  This is out of $ 1000$ drivers.  $ \frac{642}{1000}\equal{}.642\equal{}$$ \boxed{64.2}$ (percent).

[quote="sri340"]An automobile insurance company has compiled data from a survey of 1000 16-year-old drivers during the year 2003. According to the results below, what percent of them have had at least two accidents? Express your answer to the nearest tenth.

Total #                                                        # of  16-Year-Old
of Accidents                                                       Drivers

0                                                                   124
1                                                                   234
2                                                                   346
3                                                                   176
4                                                                   120[/quote]

Just add up the ones that had 2 or more accidents which would be 642 then put that over 1000 then that would be 64.2 percent.


[i]Let a,b,c be nonnegative real numbers
 Prove that
        $ (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2) \geq \frac {5}{16}(1 \plus{} a \plus{} b \plus{} c)^2$[/i]

              Y An Forever
    Wait for a nice solution

This can be killed just by expanding + homogenizing via AM-GM. (The AM-GM's are motivated by noting that equality holds for $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$.)

After expanding, it's equivalent to
\[ 11\plus{}11(a^2\plus{}b^2\plus{}c^2)\plus{}16(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}16a^2b^2c^2\]
\[ \geq 10(a\plus{}b\plus{}c)\plus{}10(ab\plus{}bc\plus{}ca)\]
But this can be rewritten as
\[ 0\leq \left( 16a^2b^2c^2\plus{}\frac{1}{2}\minus{}3(abc)^\frac{2}{3}\right)\]
\[ \plus{}\sum_\text{cyc} \left( 16a^2b^2\plus{}1\minus{}8ab\right)\]
\[ \plus{}\sum_\text{cyc} \left( 10a^2\plus{}\frac{5}{2}\minus{}10a\right)\]
\[ \plus{}\left( a^2\plus{}b^2\plus{}c^2\plus{}3(abc)^\frac{2}{3}\minus{}2(ab\plus{}bc\plus{}ca)\right)\]
AM-GM, or just factoring, can be applied to all the lines except the last. The last is homogenous and can be done by Schur (on $ a^\frac{2}{3},b^\frac{2}{3},c^\frac{2}{3}$) and AM-GM.

[quote="nkht-tk14"][i]Let a,b,c be nonnegative real numbers
 Prove that
        $ (1 \plus{} a^2)(1 \plus{} b^2)(1 \plus{} c^2) \geq \frac {5}{16}(1 \plus{} a \plus{} b \plus{} c)^2$[/i]

[/quote]
Your inequality is true for all reals $ a,$ $ b$ and $ c.$ :wink:

Thank you for dedicating your problem to me. I saw this just now. I am really lost in words. I'm not sure if I'm worthy of your dedication.. But anyhow I'll try to solve it.

In Algebraic Inequalities (page 389), is proved the following inequality (with a solution of Harazi):

If $ a_1,a_2,...,a_n$ are nonnegative numbers and $ m\ge n$, then

$ (1\plus{}a_1^2)(1\plus{}a_2^2)...(1\plus{}a_n^2) \ge \frac {m^{n\minus{}2}}{(m\minus{}1)^{n\minus{}1}}(\frac{m\minus{}n}{\sqrt{m\minus{}1}}\plus{}a_1\plus{}a_2\plus{}...\plus{}a_n)^2$.

For $ n\equal{}3$ and $ m\equal{}5$, we get the proposed inequality.


There are 100 groups of pebbles on a table containing 1, 2, 3, ..., 99, 100 pebbles, respectively. In one step you may reduce the number of pebbles of any number of groups as long as you take the same number of pebbles from each group you selected. What is the least number of steps you can take all the pebbles off of the table?

This is a disturbing problem.
Take 1 from each of the first 10 pebble groups (1-10), 11 from each of the second groups (11-20), 21 from the third groups (21-30), etc. this is 10 steps. Now there are 10 groups of 1-9 pebble groups. Taking one from each group at a time will result in 9 steps, making the total 19 steps.

Does this make sense?

Step 1:
Remove 50 pebbles from all groups with at least 50 pebbles.
This leaves groups with 1,2,...,50 pebbles.

Step 2:
Remove 25 pebbles from all groups with at least 25 pebbles.
This leaves groups with 1,2,...,25 pebbles.

Step 3:
Remove 13 pebbles from all groups with at least 13 pebbles.
This leaves groups with 1,2,...,12 pebbles.

Step 4:
Remove 6 pebbles from all groups with at least 6 pebbles.
This leaves groups with 1,2,...,6 pebbles.

Step 5:
Remove 3 pebbles from all groups with at least 3 pebbles.
This leaves groups with 1,2,...,3 pebbles.

Step 6:
Remove 2 pebbles from all groups with at least 2 pebbles.
This leaves groups with 1 pebbles.

Step 7:
Remove all remaining pebbles.

wow that was a smart way... but how did you come up with the fact that you remove half all the time? I'm not sure i'll be able to do similar problems because the solution doesn't really tell me why it's half of it.


bueno un problema en el cual quiero ver varias soluciones  :D

Dos cirfunferencias secante $C_{1}$ y $C_{2}$  de radis $r_{1}$ y $r_{2}$ se cortan en los puntos $A$ y $B$ por $B$ se traza una recta variable que corta de nuevo a $C_{1}$ y $C_{2}$  en dos puntos que llamarenos $P_{r}$ y $Q_{r}$ respectivamente.
Demuestre la siguiente propiedad : Existe un punto $M$ (sugerencia : que depende solo de $C_{1}$ y $C_{2}$ ) , tal que la mediatriz del segmento  $P_{r}Q_{r}$ pasa por $M$.

Oye, yo tengouna solucion de ese ejercicio, fue una tarea que nos marcaron aki en Mexico. No tengo tiempo aora para poner toda la solucion. Apenas tenga tiempo yo lo pongo!!   :lol:

[b]Proposición:[/b] Sean $ C_1$ y $ C_2$ dos circunferencias secantes en $ A,B$ cuyos centros son $ O_1,O_2.$ Una recta variable que pasa por $ A$ corta a $ C_1$ y $ C_2$ respectivamente en $ P,Q.$ Demostrar la mediatriz de $ PQ$ pasa por un punto fijo $ M.$

Sea $ N$ el punto medio de $ PQ.$ Las potencias de $ N$ respecto a $ C_1$ y $ C_2$ son respectivamente $ NA \cdot NP \equal{} \frac {_1}{^2}AN \cdot PQ$ y $NA \cdot NQ \equal{} \frac {_1}{^2}AN \cdot PQ.$  Es decir que $ N$ tiene igual potencia respecto a ambas. En otras palabras el lugar geométrico de $ M$ es una circunferencia $ \mathcal{K}$ coaxial con $ C_1$ y $ C_2$ cuyo centro es el punto medio de $ O_1O_2.$ La perpendicular a $ PQ$ por $ N$ (mediatriz de PQ) cortará pues a $ \mathcal{K}$ en el antipodal $ M$ de $ A$ en ella, obviamente  fijo.


Why is it true that

$ \sum_{n\in S;10^{k\minus{}1}\leq n < 10^k}\frac{1}{n}\leq\frac{9^k\minus{}1}{10^{k\minus{}1}}$, where $ k\leq 1$? I think it should be the same argument for showing that the harmonic sum on the larger set is unbounded, no?

Thanks in advance,

los

Well, very strange question. If $ k\le 1$ and $ k$ is not necessary integer then it's just brute force. If $ k\in\mathbb{Z}$ then you have $ k \equal{} 1$ and the problem is trivial (just note that $ \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{6} \equal{} 1$ and all other termes are not greater then $ 1$). If you meant that $ k\ge 1$ then the statement is wrong (your sum is approximately $ ln(10^k) \minus{} ln(10^{k \minus{} 1}) \equal{} ln(10)$ and your fraction $ \frac {9^k \minus{} 1}{10^{k \minus{} 1}}\to 0$ when $ k\to\infty$).


calculus
integration
calculus computations

let Q be the solid that is outside both [tex]x^2 +y^2 +z^2=1[/tex] and [tex]z=2x^2 +2y^2 +2[/tex], yet inside [tex]z=4\sqrt{x^2 + y^2}[/tex]

Could someone give me a little jump start on this problem? I would love a complete solution but anyhelp will do.

Hint: notice that the only way that either $x$ or $y$ appears is as $x^2+y^2.$ In other words, this is rotationally symmetric about the $z$ axis. It's a solid of revolution.

I suggest that you replace $x^2+y^2$ by $r^2$ everywhere it occurs, then sketch the curves you have on an $r$-$z$ plane. That way you have some chance of understanding the figure.


function
integration
calculus
real analysis
real analysis unsolved

Let $ r$ be a positive real number. We are looking for a family of
non-constant functions  $ \lambda_{r,n}(x)$, where $ n\in \mathbb{Z}$ and $ x\in \mathbb{R}$,  such that
\[ (1)\;\;\; \int_0^r \lambda_{r,n}(x)e^{\frac {2\pi inx}{r}}dx \equal{} 0,\;\;\; \mbox{for all integers}\; n\neq 0
\]
and satisfy the following condition:
For a given $ \epsilon > 0$ and  complex-valued continuous function
$ f$ with period $ r$, there exists a function of the form: $ \Psi (x) \equal{} \sum_{n \equal{} \minus{} N}^N c_n\lambda_{r,n}(x)e^{\frac {2\pi inx}{r}}$  ($ c_n\in \mathbb{C}$), such that
\[ (2)\;\;\; \sup_{0\leq x\leq r} |f(x) \minus{} \Psi (x)|\leq \epsilon.
\]
Notice that if $ r \equal{} 1$, then the Weierstrass theorem implies that the family of $ 1$-constant functions: $ \lambda_{r,n}\equiv 1$ satisfy (1) and (2).

Erm... How can you expect to approximate an [b]arbitrary[/b] continuous function on an interval [b]uniformly[/b] by a function [b]with $ 0$ integral[/b] over that interval? :what?: Or, maybe, I'm misunderstanding something? :?


vector
linear algebra
matrix
linear algebra unsolved

Hi everyone. I need some help with this problem.

Define L(R^3,R^2) as the set of all linear mappings from R^3 into R^2. Let T belong to L(R^3,R^2) and let T' belong to L(R^2,R^3). Prove that T'T belongs to (R^3,R^3) is not invertible. Thanks for any help!

If $T'T$ were invertible, $T$ would be injective and $T'$ would be surjective. Both are impossible...

[quote="OwlMan"]If  T'T were invertible,  T would be injective and  T' would be surjective. Both are impossible... why exactly is this impossible? Thank you for your help![/quote]Show that you can find a non zero vector $x$ such that $Tx=0$. Then watch what happens to $T'Tx$.

I see what you are getting at, show that T'Tx=0 for some non-zero vector x. I can show it for a specific matrix T and T' where x is a non-zero vector. Hence T'Tx=0 implies T'T is not ivertible. However, how do I show in general that T'Tx=0 without using actual numbers in matrices? Thanks!


I'll write up here since that's so long.  If your USAMTS is higher than usual--and a better index than the AMC, don't worry about it.  But otherwise, you need to contact the AMC directly.  I tried much the same thing to get in to take the B test:  AMC wants a $32 registration for you to take the test legally (not throwing the results for the other school--team, SLIFFE improvement scores).  By now, you'd have to do some major jumping, including getting the money in, to get it all figured out.  Otherwise, you can just continue in the competition at the host school, hoping AMC doesn't catch on and disqualify you...  Also, teacher doesn't matter.  As long as they teach math, you should be fine.  (I think.)

[quote="Ultimatemathgeek"]When are we supposed to get the AMC A score back?

I took AMC A at another school, and received no response so far.  I had to enter AMC A and take it under the name of another school, since my school didn't participate.

I also got USAMTS notice for qualifying AIME.  It says I must provide CEEB number for my school unless I know I qualify through AMC.

Well, I am pretty sure I qualified through AMC, but I don't have any response from the other school that I took the test at, so I am really perplexed here:  if I enter my school CEEB for USAMTS, then I will have to take it at my school.  But if I am also qualified through AMC, then I would have to take AIME at the other school.  Since I have no contact with the other school, I have no way of finding out if I have to go to that school again or not.  Thus, I have two AIME available for me:  one at my school for USAMTS and one at the other school for AMC.

Who can solve this problem?  I hope I explained my problem straight foward.

Also, the name of the teacher I provide--I understand AIME will be sent to that teacher.  Does this mean that teacher has to be the proctor of AIME for me?  I don't think i have a teacher that could do that for me.  (Most of them are also college professors and are very busy.)[/quote]

I think USAMTS already sent their stuff to AMC, and I didn't not provide my teachers name, so I think I am late.

I am just going to take AIME at the other school.  Can AMC really disqualify me because I don't enter under my school's name?  hmm........that is quite surprising.

The AMC A scores have already been emailed.

Virtually all AMC A score reports have also been postal mailed and should either be in the hands of the contest manager or will be this week (at least in the continental United States)

The school and the adminstering teacher or contest manager where you took the AMC A already has your score.
Because you did not register as a visitor, it is incumbent upon you to get your scores from the teacher at the school where you took the contest.  The AMC has no way to redirect your scores to you, or to redirect an AIME (assuming you qualified)  to you at your everyday school.

This is precisely the reason we ask students taking the contest at another school to register as a visitor.
The fee that we charge covers the additional handling that a visitor requires.

Steve Dunbar
AMC Director

I didn't know anything about visitor policy.

I don't mind going to another school.


What is the smallest prime the sum of whose digits is $ 19$?

when i first read the problem i started thinking of primes
but then i realized, that this way would be pretty slow

the next thing i thought of is that the sum of the digits is 19
therefore, it cannot be a two-digit number
the greatest sum possible for a two-digit number is 18 (99)
therefore, the number has three or more digits

i then thought about the smallest possible value for the hundreds place
if the hundreds place is 1, then the last two digits have to be both 9's
so we make the number 199

199 has a digit sum of 19
with a little examination, we find that 199 is also prime
thus, our answer is $ \boxed{199}$


ARML geometry calculus integration trigonometry

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b) : (b+c) : (c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

The American Regions Mathematics League.  There's a forum at AoPS for it and a lot of information about it in the AoPSWiki.

What are the ages for the ARML? Like who is allowed. Where are the centers. :huh:

[quote="Ignite168"]Bump seeing as it's almost time for ARML (last practice is this week)[/quote]

bumb* great time I had with your mom

[hide="8?"]
\[\frac{a}{\sin{A}}= \frac{b}{\sin{B}}= \frac{c}{\sin{C}}\]

\[\frac{a^{3}+b^{3}+c^{3}}{\sin^{3}{A}+\sin^{3}{B}+\sin^{3}{C}}= 7 \]

\[a^{3}\left(\frac{\frac{\sin^{3}{A}}{\sin^{3}{A}}+\frac{\sin^{3}{B}}{\sin^{3}{A}}+\frac{\sin^{3}{C}}{\sin^{3}{A}}}{\sin^{3}{A}+\sin^{3}{B}+\sin^{3}{C}}\right) = 7 \]

\[\left(\frac{a}{\sin{A}}\right)^{3}= 7 \]

\[a \leq \sqrt[3]{7}\]
I'm too lazy to check if equality is possible, meh. Also there might be trouble if the triangle has complex angle measures :ninja:
[/hide]


inequalities
trigonometry
function
logarithms
inequalities unsolved

Please help me with this problem:

 :| Given that a,b and c are positive real numbers and
\[ ab \plus{} bc \plus{} ca \equal{} 1
\]
Prove that:
\[ (1 \plus{} \frac {1} {\sqrt {1 \plus{} a^2}})(1 \plus{} \frac {1} {\sqrt {1 \plus{} b^2}})(1 \plus{} \frac {1} {\sqrt {1 \plus{} c^2}}) > 4
\]
Equivalently, prove that:
\[ (1 \plus{} \cos(\frac {A}{2}))(1 \plus{} \cos(\frac {B}{2}))(1 \plus{} \cos(\frac {C}{2})) > 4
\]
where ABC is an triangle

$ f(x) \equal{} ln(1 \plus{} cos x)$
$ f'(x) \equal{} \frac { \minus{} sin x}{1 \plus{} cos x}$
$ f''(x) \equal{} \frac { \minus{} cosx(1 \plus{} cosx) \minus{} sin^2 x}{(1 \plus{} cos x)^2} \equal{} \frac { \minus{} cosx \minus{} 1}{(1 \plus{} cos x)^2} < 0$  for  $ x\in(0;\frac {\Pi }{2})$
$ \Rightarrow f(x)$ is a concave function in $ (0;\frac {\Pi }{2})$.

$ (\frac {A}{2},\frac {B}{2},\frac {C}{2})\prec (\frac {\Pi }{2},0,0)$ .

 $ \Rightarrow ln((1 \plus{} cos\frac {A}{2})(1 \plus{} cos \frac {B}{2})(1 \plus{} cos\frac {C}{2}))\geq ln((1 \plus{} cos\frac {\Pi }{2})(1 \plus{} cos0)^2) \equal{} ln(4)$ from Karamata's inequality.

For the first part we can donate $ a \equal{} tg\frac {A}{2},b \equal{} tg\frac {B}{2},c \equal{} tg\frac {C}{2}$ because of the trigonometric identity $ tg\frac {\alpha}{2}tg\frac {\beta}{2} \plus{} tg\frac {\beta}{2}tg\frac {\gamma}{2} \plus{} tg\frac {\gamma}{2}tg\frac {\alpha}{2} \equal{} 1$ and so on.

:( Sorry, but I don't know the ln() function and Karamata's inequality. Any more help?

ln means the [url=http://en.wikipedia.org/wiki/Natural_logarithm]natural logarithm[/url] and [url=http://en.wikipedia.org/wiki/Karamata's_inequality]Karamata's inequality[/url] is a powerful inequality.

Sorry, but I'm a Vietnamese, so it's not easy for me to understand difficult things written in English, especially when the concepts are all new to me. :(

I got a hint, prove that:
\[ \cos(\frac {A}{2}) \plus{} \cos(\frac {B}{2}) \plus{} \cos(\frac {C}{2}) \plus{} \cos(\frac {A}{2})\cos(\frac {B}{2}) \plus{} \cos(\frac {B}{2})\cos(\frac {C}{2}) \plus{} \cos(\frac {C}{2})\cos(\frac {A}{2})\plus{}\cos(\frac {A}{2})\cos(\frac {B}{2})\cos(\frac {C}{2})>\sin^2(A) \plus{} cos^2(A) \plus{} \sin^2(B) \plus{} cos^2(B) \plus{} \sin^2(C) \plus{} cos^2(C)
\]
But it's still no use. Please help me to solve this problem in a simple and easy-to-understand way. :|


If $ \$2,400.00$ was borrowed at an annual interest rate of $ 11\%$, how much simple interest in dollars was charged for $ 8$ months?

In one year $ 264$ dollars interest.
Since it is only for 2/3 of a year, $ 264*(2/3)\equal{}176$ dollars interest.


function
limit
real analysis
real analysis unsolved

Let $ f_n: [0,1] \rightarrow \mathbb{R}$ be a sequence of functions such that $ |f_n(x) \minus{} f_n(y)| \leq |x\minus{}y|$ for any $ x,y \in [0,1],$ and any positive integer n. Let  $ f: [0,1] \rightarrow \mathbb{R}$ be a continuous function such that for any rational number $ x \in [0,1],$ we have
\[ \lim_{n\rightarrow \infty}f_n(x) \equal{} f(x).\]
Prove that $ \lim_{n\rightarrow \infty}f_n(x) \equal{} f(x)$ for any $ x \in [0,1].$

It is enough to assume that $ |f_n(x)\minus{}f_n(y)|\le h(x\minus{}y)$ with $ h(t)\to 0$, $ t\to 0\plus{}$ (Arzelà–Ascoli theorem).
And, moreover, we will get a uniform convergence.


modular arithmetic
number theory proposed
number theory

solve the equation $ x!\plus{}y!\plus{}z! \equal{}2^{v!}$

[quote="Victory.US"]solve the equation $ x! \plus{} y! \plus{} z! \equal{} 2^{v!}$[/quote]

Clearly $ v \ge 2$, so $ 2^{v!} \equal{} 4 \pmod{6}$

WLOG, $ x \le y \le z$, and $ 1! \equal{} 1 \pmod{6}, 2! \equal{} 2 \pmod{6}, n! \equal{} 0 \pmod{6}$ for $ n \ge 3$
$ 1 \plus{} 1 \plus{} 2 \equal{} 4 \pmod{6}..........(1)$
$ 2 \plus{} 2 \plus{} 0 \equal{} 4 \pmod{6}..........(2)$

[b]Case (1):[/b] $ (x,y,z,v) \equal{} (1,1,2,2),(1,2,1,2),(2,1,1,2)$ are the solutions.

[b]Case (2):[/b] we have $ 2! \plus{} 2! \plus{} z! \equal{} 2^{v!} \Leftrightarrow 4(2^{v! \minus{} 2} \minus{} 1) \equal{} z!$,
$ v \equal{} 2$ is not a solution so $ z! \equal{} 4 \pmod{8} \Rightarrow$ there are no solutions in this case as $ 4 \nmid z!$ for $ z < 4$ and $ 8|z!$ for $ z \ge 4$.


induction
algebra
polynomial
inequalities
vector
inequalities proposed

How many proofs of Cauchy-Schwarz can we have?

I know three:

-Induction (the easiest)
-Considering a polynomial of 2nd degree
-$ L \minus{} R \equal{} \sum_{i < j} (a_{i}b_{j} \minus{} a_{j}b_{i})^2$

The last 2 ones are basically the same...
 :)

[hide]Nice Avatar  :lol: [/hide]

One of the most kown proofs (it can be easily generalised to a proof of the Holder inequality) is to transform the inequality
\[ \sqrt {\sum\limits_{i = 1}^{n}a_{i}^{2}}\cdot \sqrt {\sum\limits_{i = 1}^{n}b_{i}^{2}}\geqslant \left|\sum a_{i}b_{i}\right|
\]
into the form
\[ \sum\limits_{i = 1}^{n}\sqrt {\frac {a_{i}^{2}}{\sum\limits_{j = 1}^{n}a_{j}^{2}}}\cdot \sqrt {\frac {b_{i}^{2}}{\sum\limits_{j = 1}^{n}b_{j}^{2}}}\leqslant 1,
\]
which follows instantly from the am-gm inequality applied to each constituent of the sum.

There is also a proof that uses only some properties of inner product in $ \mathbb{R}^{n}$.

[b]Theorem.[/b] Let $ \left(\cdot,\cdot\right)$ be inner product in $ \mathbb{R}^{n}$ that satisfies conditions $ \left(\vec{x},\vec{y}\right) = \left(\vec{y},\vec{x}\right)$, $ \left(a\vec{x} + b\vec{y},\vec{z}\right) = a\left(\vec{x},\vec{z}\right) + b\left(\vec{y},\vec{z}\right)$ for any $ \vec{x},\vec{y}\in\mathbb{R}^{n}$, $ a,b\in\mathbb{R}$, and $ \left(\vec{x},\vec{x}\right) > 0$ for any $ \vec{x}\in\mathbb{R}\setminus\left\{\vec{0}\right\}$. We define a notation $ \left|\vec{x}\right| = \sqrt {\left(\vec{x},\vec{x}\right)}$. Then the inequality
\[ \left|\vec{x}\right|\cdot\left|\vec{y}\right|\geqslant \left|\left(\vec{x},\vec{y}\right)\right|
\]
holds for any two vectors $ \vec{x},\vec{y}\in\mathbb{R}^{n}$.

[i]Proof.[/i] If any of the considered vectors is the zero vector, the inequality becomes trivial, so we can assume $ \vec{x},\vec{y}\neq \vec{0}$. Then there exists a positive real number $ k$ such that $ \left|\vec{y}\right| = k\cdot \left|\vec{x}\right|$. Hence,
\begin{eqnarray*}0 & \leqslant & \left(k\vec{x} + \pm \vec{y}, k\vec{x} + \pm \vec{y}\right) \\
& = & \left(k\vec{x}, k\vec{x}\right) + \left(\pm\vec{y},\pm\vec{y}\right) + 2\left(k\vec{x},\pm\vec{y}\right) \\
& = & k^{2}\left|\vec{x}\right|^{2} + \left|\vec{y}\right|^{2}\pm 2k\left(\vec{x},\vec{y}\right).\end{eqnarray*}
Therefore
\begin{eqnarray*} \pm 2k\left(\vec{x},\vec{y}\right) & \leqslant & k^{2}\left|\vec{x}\right|^{2} + \left|\vec{y}\right|^{2} \\
& = & k\left|\vec{x}\right|\left|\vec{y}\right| + k\left|\vec{x}\right|\left|\vec{y}\right| \\
& = & 2k\left|\vec{x}\right|\left|\vec{y}\right|,\end{eqnarray*}
and since $ 2k > 0$, we get $ \pm \left(\vec{x},\vec{y}\right)\leqslant \left|\vec{x}\right|\left|\vec{y}\right|$, so obviously $ \left|\left(\vec{x},\vec{y}\right)\right|\leqslant \left|\vec{x}\right|\left|\vec{y}\right|$, which ends the proof.

In particular, we can take $ \left(\vec{x},\vec{y}\right) = \sum_{i = 1}^{n}x_{i}y_{i}$ and obtain the Cauchy-Schwarz inequality.