Prove for positive integers $ n,m$ that
$ \sqrt [3m]{\sum_{i \equal{} 1}^{n^m}i^3}\ge n$

Are you sure you wrote this correctly?

[hide="Because"]$ \sqrt [3m]{\sum_{i \equal{} 1}^{n^m}i^3} \equal{} \sqrt [3m]{(m^n)^3 \plus{} \sum_{i \equal{} 1}^{n^m \minus{} 1}i^3} > \sqrt [3m] {(n^m)^3} \equal{} \sqrt [3m] {n^{3m}} \equal{} n$

where the inequality step is just removing the remaining summation from inside the root...  :maybe: [/hide]

Well, then I probably made a mistake somewhere

[hide="here"]
$ \sum_{i \equal{} 1}^{n^m}i \equal{} \frac {n^m(n^m \plus{} 1)}{2}\ge\sqrt {n^{3m}}$
Since $ \left(\sum_{i \equal{} 1}^{n^m}i\right)^2 \equal{} \sum_{i \equal{} 1}^{n^m}i^3$, rearranging, we have the desired inequality.[/hide]

I think you're both right, the only mistake is the strict inequality in the first solution, consider the equality case from the 2nd proof and it becomes obvious where the error is

Oh, I see. TZF's conversion of the summation into two pieces disregarded the case where $ n\equal{}m\equal{}1$.

[quote="Temperal"]Oh, I see. TZF's conversion of the summation into two pieces disregarded the case where $ n \equal{} m \equal{} 1$.[/quote]

Exactly


abstract algebra
Functional Analysis
advanced fields
advanced fields unsolved

Let $ \mathcal{H}$ be a Hilbert space with basis $ \phi_k$,  $ k \in \mathbb{N}$.  Let $ T$ be a bounded linear operator such that $ T(\phi_k) \equal{} 1/k \cdot \phi_{k \plus{} 1}$.  Show that $ T$ is compact but has no eigenvectors.

I assume that by a "basis" you actually meant an orthonormal basis.

It is easy to see that $ \|T^k\|\le \frac 1{k!}$, so the spectral radius is $ 0$ and the only possible eigenvalue is $ 0$. But the kernel of this operator is $ \{0\}$, so this case is excluded too.

Compactness follows immediately from approximability by the finite rank operators (just send $ \varphi_k$ to $ 0$ instead of $ \frac 1k\varphi_{k\plus{}1}$ for $ k\ge N$ to get the finite rank approximations).


linear algebra
matrix
algebra
polynomial
function
superior algebra
superior algebra solved

Let N be a $n \times n$ matrix over a field K. Show that $trN^j=0$ for all $j \ge 1$, hence N is nilpotent.

That needs to be a field of characteristic zero. For example, take the $p\times p$ identity matrix in a field of characteristic $p.$ Then this matrix an all of its powers have trace zero, but the matrix is not nilpotent.

If you do have a field of characteristic zero, we should be able to construct the characteristic polynomial from all of the traces of powers. The coefficients of the characteristic polynomial should be polynomial functions in the traces of the powers of $A$. Since all of those traces are zero, all the coefficients except the leading one should be zero. The characteristic polynomial should be $x^n.$ That's enough to make the matrix nilpotent.


The [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url] is where you try out writing mathematical stuff. But for those new to LaTeX you may need more help writing the code. There is help available on AoPWiki where you'll find a list of commands at [[LaTeX:Commands]] and symbols at [[LaTeX:Symbols]].

You may also like to use this [url=http://www.codecogs.com/latex/eqneditor.php]Equation Editor[/url] which will work out the commands for you which you can then copy and paste here. For example, if you click on the $ \int$ symbol it will put [b]\int_{}^{ }[/b] in the box below. Then add a and b between the {} to get [b]\int_{a}^{b}[/b]. You can see from the image it shows below that this is correct and then you can paste it here putting it between dollar signs to get $ \int_{a}^{b}$.

After a while you will learn the code and not need the Equation Editor.

I once accidentally revived a post that wasn't posted on for 3 years (!) and was informed that I shouldn't do that. But now, I need help with LaTeX but the help forum isn't posted on since 2005 or so. I'm just wondering if it's okay to receive help here.Help me!

Post your question with a new topic in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX forum[/url]. Click on the New Topic button.

[quote="stevem"]The [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url] is where you try out writing mathematical stuff. But for those new to LaTeX you may need more help writing the code. There is help available on AoPWiki where you'll find a list of commands at [[LaTeX:Commands]] and symbols at [[LaTeX:Symbols]].

You may also like to use this [url=http://www.codecogs.com/components/equationeditor/equationeditor.php]Equation Editor[/url] which will work out the commands for you which you can then copy and paste here. For example, if you click on the $ \int$ symbol it will put [b]\int_{}^{ }[/b] in the box below. Then add a and b between the {} to get [b]\int_{a}^{b}[/b]. You can see from the image it shows below that this is correct and then you can paste it here putting it between dollar signs to get $ \int_{a}^{b}$.

After a while you will learn the code and not need the Equation Editor.[/quote]But, if you are on the class, you need a semicolon before the $ \$$ sign.

How do you make a chart in LaTex? Also when I try to see how to make charts by other people in LaTex it showes that the page does not exist, is this normal?

[quote="AwesomeToad"]I once accidentally revived a post that wasn't posted on for 3 years (!) [/quote]
[hide="Anyways..."][quote="as2002"]How do you make a chart in LaTex?[/quote] See [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Layout#Tables]this[/url].[/hide]

$\text{\text\LaTeX}$ is a useful tool and Asymptote is too.  Check [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=519]this[/url] forum about that

The Equation Editor's link seems to be broken.

Thanks; it should now show the correct link.

How do you do C and P for combination and permutation?

Enclose them in dollar signs, like this:

[code]$C(n,r)$[/code]
It's only when you want a command that you preface it with a backslash.


For a quadratic equation , ax^2 + bx + b=0 , the roots are P and Q. If the signs of a and b are always same then find the value of :

(p/q)^1/2 + (q/p)^1/2 + (b/a)^1/2

[quote="Zed_darkx"]For a quadratic equation , $ ax^2 \plus{} bx \plus{} b\equal{}0$ , the roots are $ p$ and $ q$ . If the signs of $ a$ and $ b$ are always same then find the value of :

$ \sqrt{\frac{p}{q}}\plus{}\sqrt{\frac{q}{p}}\plus{}\sqrt{\frac{b}{a}}$[/quote]

$ \sqrt{\frac{p}{q}}\plus{}\sqrt{\frac{q}{p}}\plus{}\sqrt{\frac{b}{a}}\equal{}\frac{p\plus{}q}{\sqrt{pq}}\plus{}\sqrt{\frac{b}{a}}\equal{}\minus{}\sqrt{\frac{b}{a}}\plus{}\sqrt{\frac{b}{a}}\equal{}0$    :)

[quote="rachid"][quote="Zed_darkx"]For a quadratic equation , $ ax^2 \plus{} bx \plus{} b \equal{} 0$ , the roots are $ p$ and $ q$ . If the signs of $ a$ and $ b$ are always same then find the value of :

$ \sqrt {\frac {p}{q}} \plus{} \sqrt {\frac {q}{p}} \plus{} \sqrt {\frac {b}{a}}$[/quote]

$ \sqrt {\frac {p}{q}} \plus{} \sqrt {\frac {q}{p}} \plus{} \sqrt {\frac {b}{a}} \equal{} \frac {p \plus{} q}{\sqrt {pq}} \plus{} \sqrt {\frac {b}{a}} \equal{} \minus{} \sqrt {\frac {b}{a}} \plus{} \sqrt {\frac {b}{a}} \equal{} 0$    :)[/quote]
Typical solution , but not to be rude.
WRONG

If they are real, the roots are actually both negative. rachid assumed they were positive, so that's why s/he incorrectly had $ \sqrt{\frac{p}{q}}\plus{}\sqrt{\frac{q}{p}}\equal{}\minus{}\sqrt\frac{b}{a}$.

Right, Right. So the correct answer should be $ 2\sqrt{\frac{a}{b}}$

[quote="rofler"]Mkay, I guess that this is a trick question, since p,q have different signs... And since a square root isn't well-defined over the complex numbers then I guess your equation isn't well-defined, now is it? Don't be condescending, his solution is very nice asides from that minor oversight.[/quote]
His minor oversight is one of the many ways one can lose out on an IIT -JEE seat < Prime indian tech. institute>
We should always consider absolute values when taking roots or squaring.
2b/a is right. I myself got 0 when i 1st attempted the problem but i realised i would be stoopid to take postive roots.
Since some of 3 positive quantities can never be zero and P+q <0 and Pq>0 this is only possible when both are  negative.
So no point justifying the 0.


[quote="panos_lo"]Because p=3 mod4, we have that q,r are divisible by p[/quote]

Could you explain this part?

There is a very useful lemma, acording to which: If a prime p=3 mod4 divides the sum of two squares, then it must divide both of them.

[quote="panos_lo"][hide]If p=2, we have:q^2+r^2=4, which is impossible in primes. If r=2, we have: p^2(p-1)=q^2+4. The first part is even so the second must also be even. So q must be even which implies q=2. So p^3=p^2+8, impossible in primes. Similarly, q isn't equal to 1. So p,q,r are all odd and the second part of our first equation is congruent to 3 mod4. So p^3=3 mod4. From this we get p=3 mod4. Going back to the first equation, it's obvious that p dinides q^2+r^2. Because p=3 mod4, we have that q,r are divisible by p. So p=q=r, from which we easily get p=q=r=3.

Beautiful excercise!!! I hope my solution is correct. :)  [/hide][/quote]

I believe your solution is correct :) ,and I think it is essentially the same way of mine.
My solution is ,as [b]panos_lo[/b] did,we have shown $ p\equiv 3\pmod{4}$.then $ \minus{} q^2\equiv r^2\pmod{p}$,so if $ q$ is coprime to $ p$ then we have
$ 1\equal{}\left(\frac{\minus{}q^2}{p}\right)\equal{}\left(\frac{q^2}{p}\right)\left(\frac{\minus{}1}{p}\right)\equal{}\left(\frac{\minus{}1}{p}\right)$,
so $ \minus{} 1$ is a quadratic residue $ \mod p$ hence $ p\equiv 1\pmod{4}$ which is contradiction,thus $ p|q$ and $ p|r$ follows.

This just proved the lemma,am I right?

Exactly. The lemma could also be proven by Fermat's little theorem, without using
quadratic residues.

[quote=Anavel_Gato]Find all pairs $ (p,q,r)$ of primes such that $ p^3 \equal{} p^2 \plus{} q^2 \plus{} r^2$.[/quote]

$mod3$ gives that at least one of them equal to $3$.

First case: If $p=3$ then we have $18=q^2+r^2$ easy to see $p=q=r=3$.

Second case: if $r=3$ we have $p^3=p^2+q^2+9$ but then:
$p|q^2+9$ which means that $1=(-9/p)=(-1/p)$ so $p=3$ or $p=1(mod4)$
If $p=1(mod4)$ we have contradiction from $mod4$.

So $p=q=r=3$.


linear algebra
matrix
vector
linear algebra unsolved

[quote="miwalin"]Matrix $ A$ is of size $ 3\times 4$, obviously, the determinant is zero.  But I think you don't mean this. :)[/quote]

P,Q,S are $ 2 \times 1$ vector, so $ A$ is $ 4 \times 4$.


vector
function
analytic geometry
linear algebra

Let $ \mathcal{V}$ be a finite dimensional vector space over the field $ \mathcal{F}$. Let $ F \equal{} \{f_1, \dots, f_n\} \subset \mathcal{V}'$, where $ \mathcal{V}'$ is the dual of $ \mathcal{V}$, i.e. $ \mathcal{L}(\mathcal{V},\mathcal{F})$. Now define $ M : \equal{} \bigcap_{i \equal{} 1}^n \ker(f_i)$.
Show that $ < F > \equal{} M^\circ$.

Quite easily $ < F > \subset M^\circ$, as any linear combination of the $ f_i$ is also zero on $ M$, by linearity. However, how can we conclude the inclusion? Assuming wlog that the $ f_i$ are linearly independent, we can extend $ F$ to a basis of $ \mathcal{V}'$, but how to conclude that any functional zero on $ M$ must be a linear combination of the $ f_i$?

[quote="Tyl"]Assuming wlog that the $ f_i$ are linearly independent, we can extend $ F$ to a basis of $ \mathcal{V}'$, but how to conclude that any functional zero on $ M$ must be a linear combination of the $ f_i$?[/quote]Use a dual basis of $ \mathcal V ''\cong\mathcal V$ and count dimensions.

First, an example. Let $ V = \mathbb{R}^3$. Then, let $ f_1 (x,y,z) =y+z$, and $ f_2(x,y,z) = x+y$. $ M= \{(t,-t,t)|t \in \mathbb{R}\}$, which is of dimension 1, so $ M_0$ consists of all those functions of the form $ f(x) = ax+by+cz=0$ such that $ a-b+c=0$, which is of dimension 2. Also, $ <F>$ is of dimension 2, and consists of all those functions of the form $ f(x) = c_1x+(c_1+c_2)y+c_2z = 0$, which is the set of all functions such that $ f(x) = ax+by+cz$, where $ b=a+c$, which is identical to $ M^0$.

[hide="Now, my solution"]
We can assume $ f_1,...,f_n$ are linearly independent (ask if you do not see why). Now, let $ v_1,...,v_d$ be a basis for $ V$, and $ (e_1,...,e_d)$ be the coordinates of an arbitrary element $ v \in V$. Let (1) be the following system:

$ f_1(v)= a_{11} e_1+a_{12}e_2 + \cdots + a_{1d} e_d=0$
$ \vdots$
$ f_n (v)= a_{n1} e_1+a_{n2}e_2+ \cdots  + a_{nd} e_d = 0$

Note that the solution space of (1) is $ M$. Then, let $ \phi \in M^0$. Then, we may extend our system (1) to a system labeled by (2) by adding the equation:

$ \phi(v) = b_1e_1+...+ b_d e_{d}=0$.

Now, suppose (2) has a bigger rank than (1), for it is clearly impossible for it to have a smaller rank. Then, the solution space of (1) is of a bigger dimension than the solution space of (2), and hence the solution space of (2) is a proper subset of the solution space of (1), so $ \phi(m) \neq 0$  for some $ m \in M$. But this is a contradiction, for then $ \phi \notin M^0$. So, (2) and (1) have the same rank, so the added equation must be linearly dependent with the rest, so $ M^0 \subset <F>$. Also, it is obvious that $ <F> \subset M^0$, so $ M^0=<F>$, as desired.[/hide]

Third Edition, can you explain your idea of using the second dual space?

That should work. Thanks, Duelist. I was doing something similar but did not see this step [quote]the solution space of (2) is a proper subset of the solution space of (1), so $ \phi(m) \neq 0$ for some $ m \in M$.[/quote]

[quote="Duelist"]Third Edition, can you explain your idea of using the second dual space?[/quote]On order to unwind the argument I was having in mind a bit the following definition given a sub-vector space $ U$ of the vector space $ V$: $ U^\perp: \equal{}\{f\in V'\mid f(x)\equal{}0 \forall x\in U\}$ where $ V'$ denotes the dual vector space of $ V$ as above.

Under the additional assumption that $ V$ be finite dimensional, it holds $ \dim U \plus{} \dim U^\perp\equal{}\dim V$: [hide="Proof"]Extend a basis $ v_1,\ldots,v_r$ of $ U$ to a basis $ v_1,\ldots,v_n$ of $ V$. Then let $ f_1,\ldots,f_n$ a dual basis of $ V'$, i.e. $ f_i(v_j)\equal{}1$ if $ i\equal{}j$ and 0 otherwise. Then $ f\equal{}\sum_{i\equal{}1}^n\lambda_if_i\in V'$ is in $ U^\perp$ iff, for each $ j\equal{}1,\ldots,r$, $ 0\equal{}f(v_j)\equal{}\lambda_j$. Counting dimensions we get that $ \dim U^\perp\equal{}n\minus{}r$, hence the claim.[/hide]

Coming back to your assertion, we easily get $ \dim F\equal{}\dim F^{\perp\perp}$ from the lemma. Note that $ M^\circ$ is isomorphic to $ F^{\perp\perp}$ under the (natural) isomorphism $ \iota: (V')\to (V')''$, $ f\mapsto (\varphi\mapsto\varphi(f))$, thus $ \dim M^\circ\equal{}\dim F^{\perp\perp}\equal{}\dim F$. We have seen that then inclusion $ F\subseteq M^\circ$ you have already proven is an injection of [b]equidimensional[/b] vector spaces (of finite dimension), hence it is an isomorphism.

Hope that this makes my rather short comment a bit clearer. Note that the reasoning of your proof can be found in the construction or the claim of existence of a dual basis.


Euler
geometry
3D geometry
sphere
vector
advanced fields
advanced fields unsolved

Hi, why is the Euler class of a sphere bundle with even dimensional fibres 0?
Thanks~~

ps: this exercise appears in a very weird way: all through the section it talks about the sphere bundle whose dimension of fibre is 1 less than that of the base space, and this exercise suddenly turns out.

I think the Euler class of the tangent bundle to $S^2$ is not $0$.
When you say sphere, do you mean $S^2$?
Are you sure you mean even dimension for the fibres?

I mean the fiber, that is, the sphere is of even dimension.
For the tangent bundle of S2, its fiber, as a vector space is of dimension 2, but its corresponding sphere bundle is of dimension 1.

When I talk about sphere bundle, I refer to the fiber bundles with S^n, that is, sphere of any dimension as their fibers.

Sorry. I have only heard about an Euler class for vector bundles, so I thought you meant a vector bundle with fibres of even dimensions over the sphere.

What is the Euler class of a sphere bundle? Is it a characteristic class as it was for vector bundles?

[quote="Victor96"]For a fiber bundle there always be some characteristic classes, Euler class is one of them.  For sphere bundles things are simpler, so that in some cases you can view the Euler class as that of a vector bundle's.  In this problem, you can take it as vector bundles of odd dimensions.[/quote]

For vector bundles of odd dimension, the euler class has the propery $2e=0$. This may imply $2e=0$ if you take real coefficients, but as I know it to be defined, $e$ is an element of a cohomology group of the base space with coefficients in $\mathbb Z$ and then some torsion may appear when computing $H^*(B,\mathbb Z)$.

But why is 2e=0?  Or where can I find this conclusion?


Let $a$, $b$, $c$ be positive real numbers such that $abc=1$.  Prove that

$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right)\le 1$.

Let $a=x/y, b=y/z, c=z/x$, where $x,y,z$ are positive
Thus the inequality becomes
$\left(\frac{x}{y}-1+\frac{z}{y}\right)\left(\frac{y}{z}-1+\frac{x}{z}\right)\left(\frac{z}{x}-1+\frac{y}{x}\right)\leq 1$
or
$(x-y+z)(y-z+x)(z-x+y)\leq xyz$

Now I complete the proof.

There are three terms in the left-hand side.
If one of them is non-positive, the inequality holds obviously.
It is impossible that two or three of them are non-positive. (which can be easily proved by contradiction)
If all the three terms are positive, then
$(x+y-z)(y+z-x)=y^2-(x-z)^2\leq y^2$
similarly we have other two inequalities, multiply them and we get
$(x+y-z)(y+z-x)(x+z-y)\leq x^2y^2z^2$

This is an easy IMO problem.  Liyi's substitution reduces it to Schur, which is an inequality everyone should know.


Wow. We went from posting hard problems in Getting Started to posting easy problems in Pre-Olympiad. So for future reference, a problem like this belongs in Getting Started.

Square and factor. From inspection, -1 is a root, so we can factor this out. Simplify and dominate.

Hmm... I have:

$(x-1)(x^{5}-x^{4}-3x^{3}+3x^{2}+x-2)=0$

I can't seem to get anything interesting out of this apart from the fact that there must be a root between $-1$ and $2$.  :huh:

EDIT: Perhaps a calculus based solution?

[quote="i_like_pie"][quote="manunited"]Solve equation: $x^{3}-2x=\sqrt{x+2}$[/quote]
$x^{6}-4x^{4}+4x^{2}=x+2$

$x(x^{5}-4x^{3}+4x-1)=2$

There's a start for you guys...[/quote]
I think the answer must be just a real so this does not help at all.

[quote="lotrgreengrapes7926"][quote="Karth"]Hmm... I have:

$(x-1)(x^{5}-x^{4}-3x^{3}+3x^{2}+x-2)=0$

I can't seem to get anything interesting out of this apart from the fact that there must be a root between $-1$ and $2$.  :huh:[/quote]
Typo? $(x+1)...$

Isn't it in the interval $(1,2)$?[/quote]

He's talking about a different root.  Specifically, a root of the quintic part.

Phelpedo:  What made you call this problem easy?  It doesn't seem all that easy to me  :huh:

[hide="Hint"]Just square both sides, move the $x+2$ to the other side and then factor :wink: [/hide]

I misunderestimated the problem  :blush:  :wink:  I thought it was just one of the standard high school curriculum problems that have been spammed around here recently (see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111017). I thought it was just squaring and factoring.

[quote="Karth"][quote="seamusoboyle"][quote="manunited"]Solve equation: $x^{3}-2x=\sqrt{x+2}$[/quote]

Seems to me like $x^{3}-2x-\sqrt{x+2}$ in monotinically increasing[/quote]

What does monotonically increasing mean?

[quote="t0rajir0u"]
He's talking about a different root. Specifically, a root of the quintic part.
[/quote]

Yes, this is what I meant. Sorry  :blush:[/quote]
A function $f(x)$ is monotonically increasing if for all $x,y$ such that $x > y$, we have $f(x) > f(y)$.
For more information see [url=http://planetmath.org/encyclopedia/MonotonicallyIncreasing.html]here[/url].

[quote="seamusoboyle"][quote="manunited"]Solve equation: $x^{3}-2x=\sqrt{x+2}$[/quote]

Seems to me like $x^{3}-2x-\sqrt{x+2}$ in monotinically increasing[/quote]

It's not.  :P I graphed it. :D So, there are 3 roots.  :(

Decimal Approximations of the other 2:
$1.76103629148$ and $-0.814829624647$

In that case i need to do less maths with more sleep!


inequalities
Cauchy Inequality
inequalities unsolved

[quote="mathlink"]Let $ a, b, c \geq 0$ and $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 > 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 4$. Prove that
$ \frac {1}{(a \minus{} b)^2} \plus{} \frac {1}{(b \minus{} c)^2} \plus{} \frac {1}{(c \minus{} a)^2}\geq 1$[/quote]
Hello , this problem is VMO 2008 .
Since , $ \sum \frac{1}{(a\minus{}b)(b\minus{}c)} \equal{} 0$
It is suffice to show that $ ( \sum \frac{1}{(a\minus{}b)} )^2 \geq 1$
Now suppose $ c\equal{}min(a,b,c)$ and applying Cauchy inequality we have
$ (\sum \frac{1}{(a\minus{}b)} )^2 \geq \frac{4}{a\minus{}b}(\frac{1}{c\minus{}a}\plus{}\frac{1}{b\minus{}c}) \equal{} \frac{4}{(b\minus{}c)(a\minus{}c)} \geq \frac{4}{\sum ab} \equal{} 1$


What scores do you need in state Mathcounts to qualify for national? FYI, I live in Oregon. Also, I usually score around 38 to 40 on state practice tests. Is that enough to qualify?

[quote="gauss1181"]I guess so. Usually in most states you need something in the high to mid thirties to qualify.[/quote]

Is that true? I thought only 4 per state/territory. (i.e. probably a perfect paper? There's bound to be at least four perfect scorers in each state)

No.  For example, last year, CA cutoff was 40.

[quote="vallon22"]i thought last year was pretty difficult
the highest score for last year was a 41, the cutoff was around a 38

i live in MA, and the cutoff here is usually around 43 or so[/quote]

I just [b]happened[/b] to be chatting with a Massachusetts person who would know. He agreed that the cutoff is not even close to that usually.

Does it really matter though? Suppose you live in a state where the cutoff is usually around 35. So, on this year's state test, you try to make sure you get at least 35, aiming for a 36 or 37 for good measure. You manage to answer as many as you need to get a 37 and then decide to double check every single question to make sure it's correct. After the test, you see that you did get a 37, but the test was relatively easy and the cutoff was 38. Oh snap.

On the other hand, say you're struggling to get a 35, and, with a few minutes left, you realize you need at least two or three more sprints to get past a 35. You know that you're not sure about the ones you've answered, but you reason that you'll need at least a 35. So you spend the rest of your time trying to solve some of the remaining problems. At the end of the test, you find that you got one of those sprints that you spent the last few minutes on correct, but you made a couple of stupid errors that would've been easy to spot. Also, the test was [i]relatively hard[/i] and you barely missed the cutoff. If you had managed to fix those errors, you would've made it. Sucks for you.

Plus, even if you know the cutoff for something was 38, would that really stop you from aiming to get more?
[color=green]
[size=75]Edited by solafidefarms: delete potential difficulty information.[/size][/color]

last year, and vcez, i live in oregon too, yeah go beaver state, anyways, last year the cut off for oregon was 38 i think and guess what i got 37 yeah ik now 6th place...failed by a freakin point....cause bnoth the 4th//5th place dopes had low target scores anyways

i think above 40 should secure u in oregon

Guys, too much score discussion. Locked.


In a triangle $ ABC$ right angled at $ C$, the median through $ B$ bisects the angle between $ BA$ and the bisector of $ \angle B$.Prove that $ \frac 52<\frac {AB}{BC}<3$

[quote="hell_ever"]In a triangle $ ABC$ right angled at $ C$, the median through $ B$ bisects the angle between $ BA$ and the bisector of $ \angle B$. Prove that $ 1 \plus{} \sqrt 2 < \frac {AB}{BC} < 3$ . [size=134][color=darkred][b]Only so much ![/b][/color][/size] [/quote]
[b]Proof (?).[/b] Denote the midpoint $ M$ of the side $ [AC]$ , the point $ D\in (AC)$ for which $ \widehat {DBC}\equiv\widehat {DBA}$ and $ m(\widehat {ABM}) \equal{} x\ .$ Thus, $ B \equal{} 4x$ , $ m(\widehat {CBM}) \equal{} 3x$ and $ a\sin 3x \equal{} c\sin x$ $ \implies$ $ \frac {1}{\cos 4x} \equal{}$ $ \frac ca \equal{}$ $ \frac {\sin 3x}{\sin x} \equal{}$ $ 1 \plus{} 2\cos 2x$ , where $ 2x\in \left(0,45^{\circ}\right)\ .$ In conclusion, $ \frac ca\in \left(1 \plus{} \sqrt 2,3\right)\ .$
[b]Remark.[/b] Observe that $ \sin 7x \equal{} 3\sin x$ . But $ 1 \plus{} \sqrt 2 < \frac 52$ ?!
[color=darkred][b]Exists a mistake in the upper enunciation ?[/b][/color]


group theory
abstract algebra
email
limit
geometry
geometric transformation
superior algebra

[img]http://www.chatrefil.com/algebra/HungerfordEx7.JPG[/img]

is any one know this solve ?

please send me if you have this solutions...

 :|    thats important for me  :!:

I am not familiar with the Hungerford book, so I am not aware of the properties/representations of the Pruefer group $ \mathbb{Z}(p^{\infty})$ which might be known to the reader at the stage of this exercise. However, if you know the identification of $ \mathbb{Z}(p^{\infty})$ as a subgroup of the unit circle (with multiplication) given by $ \left\{\left.\exp\left(\frac{2\pi i n}{p^m}\right)\,\right|\,n,m\in\mathbb{Z}^\plus{}\right\}$, you will be capable to derive the results of this exercise. If not, pls let me know (via post or email)

yes it is exactly how they should know it
(see ex. 1.10)

okay, ..., although you will have to tranlate it into the textbook's terminology/notation

with the subgroup identification of the unit circle we can make it slightly more precise, that every element in the Pruefer-group is given by some $ g_{m,n}=\exp(\2\pi in}{p^m}$ with $ m,n\in\mathbb{Z}^+$ and actually, we can choose $ n\in (\mathbb{Z}/p^m\mathbb{Z})^*$. Then we observe the following simple properties:

(i) $ g_{m+1,n}^p=g_{m,n}$ and for $ n'$ co-prime to $ p$ we have $ g_{m,n}^{n'}=g_{m,nn'}$ and $ g_{m,n}=1\Longleftrightarrow m=0$
(ii) $ ord(g_{m,n})=p^m$ using (i)
(iii) $ <g_{m,n}> = <g_{m,1}>$ using (i) and that $ (\mathbb{Z}/p^m\mathbb{Z})^*$ is a multiplicative group

As for the exercise:
a) is directly implied by (ii)

b) From the description and using (ii) we have (*) $ g_{m,n}\in H\Longrightarrow m\leq k$. From (ii),(iii) we have $ <g_{k,1}> \subseteq H$ and using (1) and (*) this inclusion must be an equality.

c) Choose a sequence of group elements $ g_{m_k,n_k}\in H$ with $ \lim_{k\rightarrow\infty}m_k=\infty$. According to (iii) we can assume $ n_k=1$. Now let $ g_{m,n}\in\mathbb{Z}(p^{\infty})$ be arbitrary. We have to show that it is contained in $ H$. Choose $ m_k\geq m$, then using (i) we get $ g_{m_k,1}^{np^{m_k-m}}=g_{m,n}\in H$ as claimed.

d) If there is no bound on the orders of the subgroup we can apply c) and if there is a bound, we choose an element of highest order and apply b) to get all potential subgroups. The strict inclusion of the subgroups follows from (i) and (ii).

e) The isomorphism is actually the required translation into the textbook's terminology/notation with the identification $ x_i=g_{i,1}$ and is obvious from the defining group relations.

P.S. More or less a) to e) are just "one-liner" properties

small typo correction for the introductory paragraph of my last post:

instead of $ g_{m,n}\equal{}exp(\pi inp^m$ it shoud obviously read $ g_{m,n}\equal{}\exp\left(\frac{2\pi i n}{p^m}\right)$

sorry for the slippage  :blush:


hello, i think this is not true, setting
$ x\equal{}\frac{\pi}{8}$ in
$ \tan(x)\plus{}\tan(2x)\plus{}\tan(3x)\minus{}\tan(2x)\tan(3x)$
we get
$ 1\plus{}\tan(\frac{\pi}{8})$ and this is not zero.
Sonnhard.

I've edited it ,  it's  tan x -tan2x -tan3x=tanx tan 2x tan3x

[quote="enndb0x"]Prove that the following identity holds true :
\[ \tan x \minus{} \tan 2x \minus{} \tan 3x \equal{} \tan x \tan2x \tan3x\]
[/quote]

That's still wrong. Choose again $ x \equal{} \frac {\pi}8$, as Dr Sonnhard suggested :

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

$ LHS \equal{} \minus{} 3$
$ RHS \equal{} 1$

What is true is : $ \tan(3x)\equal{}\tan(2x\plus{}x)\equal{}\frac{\tan(x)\plus{}\tan(2x)}{1\minus{}\tan(x)\tan(2x)}$ and so $ \boxed{\minus{}\tan(x)\minus{}\tan(2x)\plus{}\tan(3x)\equal{}\tan(x)\tan(2x)\tan(3x)}$

oh ,yes this should be the correct problem , there was a typo in the book .

Thank you both !


[quote="galois1"]the incircle of a triangle $ ABC$  touch $ AC$ at  $ K$ and let $ M$ the midpoint of $ AC$ and $ N$ is the midpoint of $ BK$ prove that $ I, N$ and $ M$ are on the same line .where $ I$ is the incenter of the circle.[/quote]

nice! actually it is a special case of a well known theorem invoking miquels quadrilateral.

[quote="galois1"]the incircle of a triangle $ ABC$  touch $ AC$ at  $ K$ and let $ M$ the midpoint of $ AC$ and $ N$ is the midpoint of $ BK$ prove that $ I, N$ and $ M$ are on the same line .where $ I$ is the incenter of the circle.[/quote]
Very famous and easy.If i am not mistaken it was at First Zimo, :wink: .
Solution:
Let $ KL$ be the diameter of the incircle.And $ BL\cap AC\equal{}N$.It is well known that $ N$ is the tangency point of $ B$-excircle with $ AC$,hence $ M$ is the midpoint of $ KN$.Now the problem is obvious,because the line,that connected any two of them is parallel to $ BN$.


calculus
derivative
integration
abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $G$ be a group, then $G'$ is defind as usual i.e the group generated by the commutators. $\int G=H$ if and only if $H'=G$. For instance $\int\{e\}$= the set of all abelian groups.
[b]a[/b])Does there exist a group that is not integrable?
[b]b[/b])Find ${\int\mathbb{Z}_{2}}$ in the case it exists.

interesting problem, certainly challenging.

b) It exists,  Take the dihedral group G with 8 elements. G' is a flip.  Thats about all I know for sure.. there probably are others.

You can always think of these as extending a normal subgroup $C_{2}$ by some other group H.
Since you want $C_{2}$ to be the derived subgroup of the result $G$ you will have $G/C_{2}\cong H$ abealian. Furthermore you need $G'=C_{2}$ which seems to be a tricky little condition to pin down especially since in necessitates $G$ be non-abelian. So you need a non-abelian extension of two abelian groups. This still may not be sufficient there are probably some conditions I do not yet understand on H.

I'm not sure how often this will happen but I suspect non-split extensions may have to be involved.

cheers
-Q


LaTeX
number theory unsolved
number theory

[quote="greatestmaths"]find all  x in Q. such that:

f(x) = (5x-3)/(2x^2-2x+3)  and f(x) in  Z[/quote]
It is trivial problem.
$ m \equal{} \frac {2x^2 \minus{} 2x \plus{} 3}{5x \minus{} 3}$
Then
$ 2x^2 \minus{} x(5m \plus{} 2) \plus{} 3m \plus{} 2$
$ \Delta \equal{} (5m \plus{} 2)^2 \minus{} 8(3m \plus{} 3) \equal{} 25m^2 \minus{} 4m \minus{} 20$
$ x\in Q$ then $ \Delta \equal{} n^2$

you have a mistake we have  (5x-3)/(2x^2-2x+3)  not (2x^2-2x+3)/(5x-3)

Same method
$ m \equal{} \frac {5x \minus{} 3}{2x^2 \minus{} 2x \plus{} 3}$
then
$ 2mx^2 \minus{} x(2m \plus{} 5) \plus{} 3m \plus{} 3$
$ \Delta \equal{} (2m \plus{} 5)^2 \minus{} 8m(3m \plus{} 3) \equal{} \minus{} 20m^2 \minus{} 4m \plus{} 25$
Then it is quite easy.
Pleasy write it in Latex I think you problem is $ 5x \minus{} 3|2x^2 \minus{} 2x \plus{} 3$


For side lengths 11,15, and x, how many integer values of x form an acute triangle with distinct side lengths.

[hide]if it is a right triangle, the hypotenuse would be $ \sqrt{11^2\plus{}15^2}\equal{}\sqrt{346}$ that rounds down to 18. we know that x must be more than 4. for the triangle to be acute, x must be less than $ \sqrt{346}$ since we are dealing with integers, x is less than or equal to 18. there are $ \boxed{14}$ integers between 4 and 18 excluding 4.[/hide]

uh first of all i double posted (i think i hit the back button and then refresh so i think it double posted)

Sorry about that, but abacadea you forgot to consider the other case.

someone can lock one of these topics.

[hide]

well according to the the Pythagorean Inequality Theorem, a triangle is acute if a^2 + b^2 > x^2 and a^2 + x^2 > b^2 when b is the biggest of the known integers. ( if x isn't known)

so with side lengths 11, 15 and x,

11^2 + 15^2 > x^2 and 11^2 + x^2 > 15 ^2

or 346 > x^2 > 104
 or 19 > x > 10 since it has to all be integers.

So it can be x is 18- 11 so 8 possible answers.

[/hide]


Is it like a booklet or couple pages stapled together?  Can you take the question booklet after the test to estimate your own score?

its a booklet that actually a long, (but short vertically) page folded many times. You can take it home to estimate your score. Often times, your teacher will let you see a solution manual soon after (or right after). If not, I do not think it hurts to discuss among your classmates what the right answers are if you know that they have all taken it.

There is no official rule that forbids you to take the test home. If your teacher has ordered solution pamphlets, you may even check your answers, provided your school is able to afford the extra 6 dollars.

Our school lets us take it home...and we even check our answers like a little later during the day

The test comes as a folded booklet. Many times, schools let you take the booklet and/or show you the answers and solutions soon after the test. If not, you can probably just copy down your answers on a piece of paper, and then check them to find out your score. For example, the AMC8 answers were up on AoPS two days after the exam, and on the official website three days afterward.

An image of the front cover of the 2009 AMC 10 and the 2009 AMC 12 is on page 20, and 21 respectively of the Teachers' Manual, at:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-AMC_1012A,2009.pdf

The rule concerning taking the contest booklet home is in Section II, Item 16, parts a. b. on page 5 of the
Teachers' Manual.  The rule is repeated in the first paragraph on page 4.

Steve Dunbar
MAA Director, American Mathematics Competitions


Vieta
quadratics
algebra
quadratic formula

There are two solutions for the equation $ x^2 \minus{} x \minus{} 6 \equal{} 0$. What is the product of these two solutions?

Vieta's formulas state that in a quadratic $ ax^2 \plus{} bx \plus{} c$, the sum of the two roots is $ \minus{}\frac{b}{a}$ and the product is $ \frac{c}{a}$. That means the product is therefore $ \frac{\minus{}6}{1} \equal{} \boxed{\minus{}6}$.

As a sidenote, it is easy to derive these formulas from the quadratic formula, $ \frac{\minus{}b \pm \sqrt{b^2 \minus{} 4ac}}{2a}$. When you add $ \frac{\minus{}b \plus{} \sqrt{b^2 \minus{} 4ac}}{2a}$ to $ \frac{\minus{}b \minus{} \sqrt{b^2 \minus{} 2ac}}{2a}$, terms conveniently cancel out to leave us with $ \minus{}\frac{b}{a}$. Similarly, when we multiply the two general forms of roots, we are left with $ \frac{c}{a}$.


modular arithmetic
inequalities
number theory proposed
number theory

hxy09, Could you upload solution of problem 4?(China TST 2009)

This is a question like last years's third question of IMO, it is not hard enough for 3 but it is nice. I got the same solution as hxy09 and it seems like another proof for this appearently weak bound could be hard to find, maybe there can be a different approach resulting with a better bound. If somebody has a different approach, I would be pleased to see it.

[quote="birzhan"]Dear [b]hxy09[/b]
Sorry for my stupidity,but could you you explain me how you got:
$ x_1 \plus{} 2x_2 \plus{} ... \plus{} kx_k \equal{} a_r \minus{} a_1 \plus{} 1$[/quote]
Sorry for my late reply :oops:
and I had a typo for $ x_1 \plus{} 2x_2 \plus{} ... \plus{} kx_k \equal{} a_r \minus{} a_1 \plus{} 1$
the correct version is  $ x_1 \plus{} 2x_2 \plus{} ... \plus{} kx_k \equal{} a_r \minus{} a_1$
If $ p|a_i(a_i\plus{}1)...a_{i\plus{}1}\minus{}1$ the length is $ a_{i\plus{}1}\minus{}a_i$
then the sum of the length is $ a_r\minus{}a_1$
p.s. It made no difference whether  $ x_1 \plus{} 2x_2 \plus{} ... \plus{} kx_k$ is $ a_r \minus{} a_1 \plus{} 1$or $ a_r \minus{} a_1$
and thank you for pointing out my mistake :)

Assume otherwise. Like the previous solution, we will use the fact that $x(x+1)...(x+m-1) \equiv 1 \pmod p$ has at most $m$ roots. Partition into intervals $[0,p^{\frac{1}{3}}],[p^{\frac{1}{3}} , 2p^{\frac{1}{3}}] ..., [p-p^{\frac{1}{3}} , p]$ Then there are $p^{\frac{2}{3}}$ such intervals.  Let $a_i$ be the no. of solutions to $p|n!+1$ in $[ip^{\frac{1}{3}},(i+1)p^{\frac{1}{3}}]$ then any $a,b$ in the same interval gives us a distinct solution to $x(x+1)...(x+m-1) \equiv 1 \pmod p$ where $0 \leq m < p^{\frac{1}{3}}$.
$\implies \binom{p^\frac{1}{3}}{2} \geq \sum \binom{a_i}{2} \geq p^{\frac{2}{3}}\binom{c}{2}$ where the last inequality follows from the fact that $\binom{n}{2}$ is convex. This is a contradiction for large enough $c$ and thus we are done.

Assume that the requisite number of such integers is $r$ with the integers being $a_1<a_2<\dots <a_r=p-1$. Suppose $x_j= \{\# \text{times} \quad a_{n+1}-a_n=j\}$. Note that by Lagrange's theorem we have that there are atmost $j$ solutions $\pmod p$  to $$(x+1)(x+2)\dots (x+j) =1 \pmod p$$
This means that $x_j \leq j$. Because if $p\mid a_n!+1$ and $p\mid a_{n+1}!+1$ , then $p \vert (a_n+1)(a_n+2) \dots (a_n+(a_{n+1}-a_n)-1$.

Note that we have $:=$

$$ \sum_{i=1}^{p-2} x_i = r-1 \quad \text{and} \quad \sum_{i=1}^{p-2} ix_i= a_r-a_1 \leq p-2$$

Note that $ \sum_{i=1}^{p-2} x_i = r-1$ when the terms with the smallest index are maximal . Hence the second sum looks like $1^2+2^2 +\dots k^2 + (k+1)\cdot \ell$ for some $0\leq \ell \leq k+1$. We then have the following estimate :

$$p-2 > 1^2+2^2+\dots +k^2 \implies p-2 > \frac {k(k+1)(2k+1)}{6}$$
$$ \implies p > \frac {k^3}3 \implies k< (3p)^{\frac 13}$$

Then we have  $$ \sum_{i=1}^{p-2} x_i +1 \leq 1+2+\dots k + \ell+1= \frac {k(k+1)}2 + \ell \leq \frac {(k+1)(k+2)+2}{2} < 4k^2 < 4\cdot (3p)^{\frac 23}$$

This shows that $c= 4 \cdot 3^{\frac 23}$ is enough.

Fastest I  could ever solve a China TST!! Thanks to @Aryan-23 :love:
[quote=Fang-jh]Prove that for any odd prime number $ p,$ the number of positive integer $ n$ satisfying $ p|n! \plus{} 1$ is less than or equal to $ cp^\frac{2}{3}.$ where $ c$ is a constant independent of $ p.$[/quote]

[color=red][b]Claim[/b][/color](Lagrange Theoem):Let $p$ be a prime and $f\in\mathbb{Z[X]}$.If $f\neq 0$ in $\mathbb{F}_p$ then the number of solutions of $ \#\{f(x)\equiv 0\pmod p\}\leq \deg{f}$

[i]Proof[/i]:     We induct on the degree $d$ of $f$.          Base case $d=0$ is obvious now assume it holds for $d$ we will prove it for $d+1$.If $f(x)\equiv 0\pmod p$ with degree $d+1$ has no solutions then we are done by inductive hypothesis.           So assume there exists $g\in\mathbb{Z[X]}$ such that $$f(x)=(x-a)g(x)\pmod p$$ and $\deg(g)\leq d$.         But then this has atmost $\deg(g)+1\leq \deg(f)$ solutions so we are done.$\square$

Let $n_1<n_2\cdots <n_m=p-1$ be the set of solutions to the assertion.We have to show $m=\mathcal{O}(n^{\tfrac{2}{3}})$.
Define $j=\left\lfloor\dfrac{\sqrt{1+8m}-1}{2}\right\rfloor$ and note that $\tfrac{j(j+1)}{2}\leq m\leq \tfrac{(j+1)(j+2)}{2}$.

[color=red][b]Key claim[/b][/color] : For each $1\leq k<p$ there are atmost $k$ indices such that $n_{i+1}-n_i=k$
[i]Proof[/i]We have $$(n_i)!\equiv -1\equiv n_{i+1}!\pmod p\Longleftrightarrow (n_i+1)(n_i+2)\cdots (n_{i+1})\equiv 1\pmod p $$ and so using Lagrange's Theorm this has atmost $n_{i+1}-n_i=k$ solutions to this.$\square$
To finish see that $$p>(p-1)-n_1=\sum_{i=1}^{i=m-1}(n_{i+1}-n_i)\geq 1^2+2^2+\cdots j^2=\dfrac{j(j+1)(2j+1)}{6}>\dfrac{j^3}{3}\Longleftrightarrow j<(3p)^{\tfrac{1}{3}}$$
Thus $$m\leq \dfrac{(j+1)(j+2)}{2}<(j+1)^2\leq 4j^2<(576)^{\tfrac{1}{3}}p^{\tfrac{2}{3}}=\mathcal{O}(n^{\tfrac{2}{3}})\blacksquare$$

We claim $c=4$ works, and we will bound the number of $(i, j)$ such that $i(i+1)(i+2)\cdots(i+j-1) \equiv 1 \pmod{p},$ with $0 \le i < p, 0 < j < p^{1/3}.$
First, for any fixed $j,$ there are at maximum $j$ solutions for $i$ by Lagrange's theorem, so summing over all $j$ gives an upper bound of $p^{2/3}.$
Now, assume for the sake of contradiction there are more than $4 p^{2/3}$ solutions. We split the residues mod $p$ into $p^{2/3}$ blocks of size $p^{1/3}.$ If the $i$th block has $a_i$ solutions, then note that by taking any two solutions $n! \equiv m! \equiv -1 \pmod{p},$ and dividing, we get $\frac{n!}{m!} \equiv 1 \pmod{p},$ which gives a solution since the value of $j$ is less than $p^{1/3}$ by the condition that both solutions are in the same block. Therefore, we get at least $\binom{a_i}{2}$ solutions per block. Summing and using Jensen's, this tells us that we have at least $\binom{4}{2} p^{2/3}$ solutions, which is larger than the upper bound given before, so we're done.

Suppose that all the possible solutions to $n$ include $a_1<a_2<a_3<..........<a_X=p-1$
Our goal is to prove the existence of an constant $c$ such that $X \le c p^{\frac{2}{3}}$
Since $n_i! \equiv -1 \mod p$ and $n_{i+1}! \equiv -1 \pmod p$ hence $(n_i+1).........n_{i+1} \equiv 1 \pmod p$.
By [hide=Lagrange's theorem]Let $p$ be a prime and $f\in\mathbb{Z[X]}$.If $f\neq 0$ in $\mathbb{F}_p$ then the number of solutions of $ \#\{f(x)\equiv 0\pmod p\}\leq \deg{f}=k$[/hide],atmost $k$ indices such that $n_{i+1}-n_i=k$
This implies that the differences between $n_i$ are written in a ascending order then we get a the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5...........,k,k,......,k$
Adding everything and comparing gives the result

[quote=Fang-jh]Prove that for any odd prime number $ p,$ the number of positive integer $ n$ satisfying $ p|n! \plus{} 1$ is less than or equal to $ cp^\frac{2}{3}.$ where $ c$ is a constant independent of $ p.$[/quote]

Just notice that sigma1/p^2 which p is a prime less than 1/2(about 0.78)then notice that p=1 mod 4 and you can finish your proof.

[quote=Fang-jh]Prove that for any odd prime number $ p,$ the number of positive integer $ n$ satisfying $ p|n! \plus{} 1$ is less than or equal to $ cp^\frac{2}{3}.$ where $ c$ is a constant independent of $ p.$[/quote]

I love China and Iran’s tst, especially number theory and combination!!!!!!

I might as well post the solution I wrote up a year ago on a different thread.  The start is similar to other solutions, but I think the finish is a bit different.

Call an integer $n$ [b][color=blue]good[/color][/b] if $p$ divides $n! + 1$; observe that $p-1$ is the largest good integer.  The key to this problem lies in the following lemma.

[b][color=red]Lemma 1:[/color][/b] For each positive integer $t$, there are at most $t$ good integers $k$ such that $k+t$ is also good.

[i]Proof.[/i] If $p$ divides both $(k+t)! + 1$ and $k! + 1$, then $p$ must also divide
\[
\big((k+t)! + 1\big) - \big(k! + 1\big) = k!\big((k+1)\cdots(k+t) - 1\big).
\]
Since $k$ is a good integer, $\gcd(p,k!) = 1$, so in fact
\[
p\mid (k+1)\cdots (k+t) - 1.
\]
But the right hand side is a polynomial of degree $t$, which has at most $t$ solutions in $\mathbb F_p$ by Lagrange. $\square$

Let
\[
n_1 < n_2 < \cdots < n_m = p-1
\]
denote the set of good integers. For each $1\leq j\leq p$, let $a_j$ denote the number of integers $1\leq k\leq m$ such that $n_k - n_{k-1} = j$.  By the lemma, $a_j\leq j$ for each $j$.  Furthermore, we are given that
\[
a_1 + 2a_2 + \cdots + pa_p = \sum_{k=1}^m(n_k - n_{k-1}) = n_m - n_1\leq p,
\]
and we wish to maximize $a_1 + \cdots + a_p$.

------
The rest of this problem is encoded in the following purely algebraic lemma.

[b][color=red]Lemma 2.[/color][/b] Suppose $b_1,\ldots, b_n$ are nonnegative real numbers such that
\[
b_1 + 2b_2 + \cdots + nb_n \leq n\qquad\qquad(*)
\]
and $b_i\leq i$ for all $i$.  Then there exists a constant $c > 0$ such that
\[
b_1 + b_2 + \cdots + b_n \leq cn^{2/3}.
\]

[i]Proof.[/i] Throughout this proof, we will use the notation $a\lesssim b$ to denote "less than or equal to up to a constant".

Without loss of generality we may replace $(*)$ with the stricter condition $b_1 + 2b_2 + \cdots + nb_n = n$.  Consider an arbitrary $n$-tuple $(b_1,\ldots, b_n)$ of nonnegative real numbers satisfying the given conditions.  If there exist integers $r < s$ with $b_r < r$ but $b_s > 0$, then we may replace $(b_1,\ldots, b_n)$ by the tuple
\[
(b_1',\ldots, b_n') = (b_1,\ldots, b_{r-1},b_r + \tfrac{s\varepsilon}r,b_{r+1},\ldots, b_{s-1},b_s - \varepsilon,b_{s+1},\ldots),
\]
where $\varepsilon$ is chosen small enough that $b_r + \tfrac{s\varepsilon}r < r$ and $b_s > \varepsilon$.  (In other words, we adjust the values of $b_r$ and $b_s$ slightly while fixing all other elements in the tuple.)  Then
\[
b_1' + 2b_2' + \cdots + nb_n' = b_1 + 2b_2 + \cdots + nb_n = n,
\]
but the sum $b_1 + b_2 + \cdots + b_n$ increases.  This means that $(b_1,\ldots,b_n)$ does not achieve our maximum.  In particular, our maximum is achieved when
\[
b_j = \begin{cases}j&\text{if }1\leq j\leq t,\\
0&\text{if }j\geq t+2,
\end{cases}
\]
where $t$ is some positive integer at most $n$.

Finally, observe that
\[
n = b_1 + 2b_2 + \cdots + nb_n \geq 1^2 + 2^2 + \cdots + t^2 \gtrsim t^3,
\]
and hence $t \lesssim n^{1/3}$.  Then
\[
b_1 + b_2 + \cdots + b_n = b_1 + \cdots + b_{t+1} \leq 1 + \cdots + (t+1) \lesssim t^2 \lesssim n^{2/3},
\]
and so $b_1 + \cdots + b_n\leq cn^{2/3}$ for some constant $c$. $\square$

The problem falls upon applying Lemma 2 to the sequence $(a_1,\ldots, a_p)$.  $\blacksquare$

The difficult part is definitely realizing the combinatorial approach.

Let $1<n_1<n_2<\cdots<n_m = p$ be all the numbers that work, so it suffices to show $m \leq cp^{2/3}$. For each $i$, we have $$(n_i+1)(n_i+2) \cdots (n_i+n_{i+1}-n_i) \equiv 1 \pmod p.$$ In other words, for each $k$, there are at most $k$ indices $i$ with $n_{i+1} - n_i = k$ by Lagrange theorem.

The rest is easy. Let $j$ be maximal with $j(j+1) \leq 2m$. Then, note that $$p = \sum_{i=1}^{m-1} (n_{i+1} - n_i) \geq 1^2+2^2+\cdots+j^2 = \frac{j(j+1)(2j+1)}6.$$ The result follows.

[quote=HamstPan38825]Then, note that $$p = \sum_{i=1}^{m-1} (n_{i+1} - n_i) \geq 1^2+2^2+\cdots+j^2 = \frac{j(j+1)(2j+1)}6.$$[/quote]
Maybe I'm missing something (and you're not the only person to mention this in the thread, so I probably am), but why does this follow so quickly?  Because there are at most $k$ terms such that $n_{i+1} - n_i = k$, the contribution of these terms to the overall sum is at [i]most[/i] $k^2$, which seems like it should yield an inequality going in the wrong direction.


geometry
projective geometry
geometry proposed

Given a triangle ABC and a point in the plane, O, let D, E, F be the points where the perpendiculars at O to OA, OB, OC intersect BC, CA, AB respectively. Prove that D, E, F are collinear.

It is Menelai's theorem.

Let AO intersects BC at p.A1 and so on.
It is easy to show that
BD/DC=-cos(BOA1)*sin(OCB) / cos(COA1)*sin(OBC)

Product of these expressions is -(sin-form of condition AO, BO and CO are concurrent), i.e. -1.

That's all.

It's similar to what I did. It's not that hard, is it?

I don't like triginometric solutions. Geometry is not trigonometry.

Maybe I could look it up in some books and post a soln involving projective geometry? Would that qualify as geometry? :D  :D  :D


I'm currently preparing for college placement test and I ran into a few fairly simple Rational Expressions that are giving me trouble. Any help would be greatly appreciated.

(x^2 + 3x -10) / (5x)    *     (x^2 - 3x)/ (x^2 -5x +6)

and

(x^2 - 9x + 14) / (x^2 - 5x + 6) * (x - 5x -14) / (x + 2)[/list]

[quote="davidlizeng"][hide]
the two expressions are:
$(\frac{x^{2}+3x+10}{5x})(\frac{x^{2}-3x}{x^{2}-5x+6})$ and $(\frac{x^{2}-9x+14}{x^{2}-5x+6})(\frac{x^{2}-5x-14}{x+2})$
the first can be factored to get:
$(\frac{(x+5)(x-2)}{5x})(\frac{(x)(x-3)}{(x-2)(x-3)})$
then you can cross cancel and get $\frac{x+5}{5}$
the second can be factored to get:
$(\frac{(x-7)(x-2)}{(x-2)(x-3)})(\frac{(x-7)(x+2)}{(x+2)})$
then you can cancel to get $\frac{x^{2}-14x+49}{x-3}$
 [/hide]
 aww. kunny beat me to it......[/quote]

For the first problem: Ahhh! I think I see what I was doing wrong. The X on the top right cancels out with the X in 5x, right? I was leaving the top right X in and it was messing up the answer.

For the second problem: I'm sorry, I accidently gave the wrong equation -_- (checked over it like 10 times too). The real equation is:

(x^2 - 9x +14 / x^2 -5x + 6) / (x - 5x -14 /  x+2)

So, sorry about that.

Anyways, I've been working on these and I'm starting to get them but, I did get stuck on the following one:

(3/4xy^3) + (1/6x^2y)

Please break down the problem when you explain it. I actually have all the answers, I'm trying to figure out how to do them.

Also, how exactly do I write in that text that you guys use for equations?

Thanks for the help, it is greatly appreciated.

[quote="SmilieCx"]Also, how exactly do I write in that text that you guys use for equations?[/quote][url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]

[hide]$\frac{\frac{x^{2}-9x+14}{x^{2}-5x+6}}{\frac{x^{2}-5x-14}{x^{2}-5x+6}}$.
To divide fractions, you flip the fraction and multiply so you get $\frac{x^{2}-9x+14}{x^{2}-5x+6}*\frac{x^{2}-5x+6}{x^{2}-5x-14}$
Factor and get
$\frac{(x-7)(x-2)}{(x-3)(x-2)}*\frac{(x-3)(x-2)}{(x-7)(x+2)}$.
Cross cancel to get
$\frac{(x-2)}{(x+2)}$ so your final answer is $\boxed{\frac{x-2}{x+2}}$.

$\frac{3}{4}xy^{3}+\frac{1}{6}x^{2}y$
First of all, when you have something like $\frac{1}{3}*37y$ you can make it $\frac{37y}{3}$ so we'll do that.
$\frac{3xy^{2}}{4}+\frac{x^{2}y}{6}$.
Find a common denominator of 12.
$\frac{9xy^{2}}{12}+\frac{2x^{2}y}{12}$.
Add.
$\boxed{\frac{9xy^{2}+2x^{2}y}{12}}$. If you want to factor this, you'd get $\boxed{\frac{xy(2x+9y)}{12}}$.
[/hide]

I'm sorry again, not using the text format makes it hard to get across what the expression is supposed to look like and I think you copied one of the problems wrong (my fault). Here's the two that I'm working on, using the text format:

$\frac{3}{4xy^{3}}+\frac{1}{6x^{2}y}$

$\frac{x^{2}-9x+14}{x^{2}-5x+6}/ \frac{x-5x-14}{x+2}$

and the answers:

$\frac{9x+2y^{2}}{12x^{2}y^{3}}$

$\frac{1}{x-3}$

I'm not sure how to solve for these answers.
Sorry for the misunderstanding, and thanks for the help.

[hide]
factor the second one to get:
$\frac{(x-7)(x-2)}{(x-2)(x-3)}\frac{(x+2)}{(x-7)(x+2)}$
then the x+2, x-2, and the x-7 cancel out and you are left with
$\frac{1}{x-3}$

for the first one find the common denominator, which $12x^{2}y^{3}$
to achieve commone= denominators, multiply each fraction by something/the same something that results in the denominator becoming the one you want. the first fraction we multiply by $\frac{3x}{3x}$ the second one we multiply by $\frac{2y^{2}}{2y^{2}}$ thus we get:
$\frac{9x}{12x^{2}y^{3}}+\frac{2y^{2}}{12x^{2}y^{3}}$ add then to get the correct answer of $\frac{9x+2y^{2}}{12x^{2}y^{3}}$
[/hide]


trigonometry
integration
limit
calculus
calculus computations

Set
$ f(x)\equal{}0$ if $ x \le 0$
$ f(x)\equal{}\sin(\frac{\pi}{x})$ if $ x > 0$
Prove that $ f$ has Antiderevative.

An antiderivative is $ F(x)\equal{}\int_0^x\sin\frac{\pi}{t}\,dt$; this clearly works away from zero by the FTC. We need only to show that $ F'(0)$ exists and is zero.

To do this, integrate by parts: $ F(x)\equal{}\int_0^x \frac{t^2}{\pi}\cdot\frac{\pi}{t^2}\sin\frac{\pi}{t}\,dt\equal{} \left[\frac{t^2}{\pi}\cdot\cos\frac{\pi}{t}\right]_0^x\minus{} \int_0^{\pi}\frac{2t}{\pi}\cdot\cos\frac{\pi}{t}\,dt$
$ F(x)\equal{}\frac{t^2}{\pi}\cos\frac{\pi}{x}\minus{} \int_0^{\pi}\frac{2t}{\pi}\cdot\cos\frac{\pi}{t}\,dt$
Now both terms in $ F$ are clearly differentiable at zero; $ \lim_{t\to 0}\frac1t\cdot \frac{t^2}{\pi}\cos\frac{\pi}{x}\equal{}\lim_{t\to 0}\frac1t\cdot int_0^{\pi}\frac{2t}{\pi}\cdot\cos\frac{\pi}{t}\,dt\equal{}0$ by a squeeze.


[size=59]Hi! From now on, this thread will have everything writen in small text. You can read what other people are saying by copy and pasting everything into Microsoft Word and enlarging. This forum will kinda be like random facts forum. So, I'll start:[/size]
[size=59]The average human eats 8 spiders in their lifetime at night.[/size]

[size=9]The average human loses about 90 inches of eyelashes in their lifetime.[/size]

[size=9]See...I wrote the first one bigger so you people could actually know how to read it.
Donald Duck comics were banned from Finland because he doesn't wear pants.[/size]

[size=59]Mass and inertia are the same thing.

Units: a = [m/sec2], F = [kg•m/sec2] (newton), work = pe= ke = [kg•m2/sec2] (joule)

Solenoids are stronger with more current or more wire turns or adding a soft iron core.

Physics rocks...[/size]

[quote="shaggy75"][size=59]if u can read this, you have way too much time on ur hands  [/size][/quote]
[size=59]Don't quote this post[/size]
[size=34]Muahahaha even smaller[/size]
[size=9]Can you still see me?[/size]

[quote="mz94"][size=9]watever.

know how to keep a moron busy? look down

know how to keep a moron busy? look up

if u went thru tat, u have no life[/size][/quote]

[size=0]ROFL THIS IS SO FREKIN FUNNY HAHAHAHA[/size]

[size=34]If you can read this, your eyesight is obviously 20-20 (which means my eyesight is 20-20, because i can read this).
Even smaller![/size]
[size=9]The eating of insects is called entomophagy[/size]

[size=200]this is the smallest size ever[/size]

[size=200][b]Yes, I totally agree[/b][/size]

[size=200]why yes it is! small rules~[/size]

[size=200][b]I love writing small[/b][/size]

[size=17]This is the biggest size EVER!!! Go big![/size]

who would like to summarize this topic for me?

[size=200]How[/size] [size=9]dare [/size] [size=200]you[/size] [size=9]break [/size] [size=200]the[/size] [size=9]rules! [/size]

dont need to copy in word. press ctrl and scroll!

[quote="Nexmus"][quote="1=2"][size=9]This is small text. I like typing in small text. I love typing in small text. I'm obsessed with typing in small text. I do Yoga in small text. Wait, how can you do yoga in small text? Oh well. I like strawberry jelly. I also like blueberry jelly. I like a lot of jellies. My family makes jelly. We usually eat jelly with toast. We make the toast from the rolls we make. We make good rolls. I like spaghetti with tomato sauce. We call tomato sauce red sauce. I can eat spaghetti with red sauce for breakfast, unch, or dinner. I also drink milk. How I love milk. I also like lemonade, iced tea, and water. I like to eat pancakes, waffles, and french toast. French toast is warm and toasty. I like to eat things that are warm and toasty. I also eat bean dip with chips. The bean dip is made with beans and cheese. The chips are tortilla chips. I like watching some TV shows. My dad's favorite is Battlestar Galactica. My favorite is Eureka. My little brother's favorite is the Simpsons. My mom doesn't have a a favorite. I have a dog. His name is Barkley. Do you know why we named him that? Because when strangers came, he always barked. I used to have 4 fish, but they all died. The zebra fish went first. Don't know what happened to them. Then the cat fish went belly-up, but it was at the bottom of the tank. Then my Betta fish went. It was sitting on a fake coral reef. My mom and dad are making salsa. I like homemade salsa. We have it with tortilla chips. This is a long block of text, isn't it? Well, it all started with "This is small text. I like typing in small text. I love typing in small text. I'm obsessed with typing in small text. I do Yoga in small text. Wait, how can you do yoga in small text? Oh well. I like strawberry jelly. I also like blueberry jelly. I like a lot of jellies. My family makes jelly. We usually eat jelly with toast. We make the toast from the rolls we make. We make good rolls. I like spaghetti with tomato sauce. We call tomato sauce red sauce. I can eat spaghetti with red sauce for breakfast, unch, or dinner. I also drink milk. How I love milk. I also like lemonade, iced tea, and water. I like to eat pancakes, waffles, and french toast. French toast is warm and toasty. I like to eat things that are warm and toasty. I also eat bean dip with chips. The bean dip is made with beans and cheese. The chips are tortilla chips. I like watching some TV shows. My dad's favorite is Battlestar Galactica. My favorite is Eureka. My little brother's favorite is the Simpsons. My mom doesn't have a a favorite. I have a dog. His name is Barkley. Do you know why we named him that? Because when strangers came, he always barked. I used to have 4 fish, but they all died. The zebra fish went first. Don't know what happened to them. Then the cat fish went belly-up, but it was at the bottom of the tank. Then my Betta fish went. It was sitting on a fake coral reef. My mom and dad are making salsa. I like homemade salsa. We have it with tortilla chips." That was a long block of text, all right.[/size][/quote]

[size=9]Thanks for telling us that even though we didn't need to know. LOL[/size][/quote]

[size=59]Your welcome.[/size]

[quote="Nexmus"][color=white][size=9]LOL[/size][/color][/quote]

[color=white][size=200]Thank you.[/size][/color]

[size=1]OMG this is real eye strain[/size]

[size=100]Do not read the next sentence.[/size]

[size=1]You [hide=Insert Epic Roast Here][/hide][/size]

[size=50][pywindow]
nouns=['Your mom','My Pikachu Vmax','Joe Mama','Joe Biden','Bob the Builder','Elmo',' A cringe girl']
action = ['is sus', 'ate Donald Trump', 'was roasting me', 'was causing global warming', 'was listening to cringe music', 'was postfarming', 'was rickrolling people', 'declared war on Mickey Mouse', 'was building the Great Wall of China', 'was burning your house down', 'was robbing a bank', 'became really chunky', 'was walking your homework', 'was eating 500 cakes', 'was generating excuses', 'was watching TV', 'was working on a top secret military project', 'just realized the earth is round', 'was calculating pi']
soiwas = input("so i  _____(don't include so i, eg ' was late for school','forgot my homework')")
but=['but I love you','but remember 1+1=3','but your crush is kissing you tomorrow so get ready','so i will have to kiss you in an hour']
from random import choice
print('Sorry,', choice(nouns) , choice(action),'so I', soiwas , choice(but))
[/pywindow][/size]


$a_{0}=2,a_{1}=1$ and for $n\geq 1$ we know that : $a_{n+1}=a_{n}+a_{n-1}$
$m$ is an even number and $p$ is prime number such that $p$ divides $a_{m}-2$. Prove that $p$ divides $a_{m+1}-1$.

In order to make sense, I think that the correct statement is :
"Prove that p divides a_(n+1) - 1."

And, in fact, i'm quite sure of that, because it's true :D .
Let's prove it :

Let U_n = a_(n-1)*a_(n+1) - (a_n) 2 .
We have U_1 = 5, and for each n > 0 :
U_(n+1) = a_n*a_(n+2) - (a_(n+1)) 2
= a_n*a_(n+1) + (a_n) 2  - (a_(n+1)) 2
= a_(n+1)(a_n - a_(n+1)) + (a_n) 2
= -U_n.

It follows that U_n = 5*(-1) n-1  for each n > 0, that is
a_(n-1)*a_(n+1) = (a_n) 2  + 5*(-1) n-1 . (1)

Let p be a prime number which divides a_(2k) - 2.
Thus a_(2k) = 2 mod[p].
From (1), it follows that
(a_(2k)) 2  - 5 = (a_(2k+1) - a_(2k))*a_(2k+1)
that gives
-1 = (a_(2k+1)) 2  - 2*a_(2k+1) mod[p]
thus p divides (a_(2k+1) - 1) 2  and since p is prime we are done.

Pierre.

[quote="Omid Hatami"]$a_{0}=2,a_{1}=1$ and for $n\geq 1$ we know that : $a_{n+1}=a_{n}+a_{n-1}$
$m$ is an even number and $p$ is prime number such that $p$ divides $a_{m}-2$. Prove that $p$ divides $a_{n+1}-1$.[/quote]

Sorry, what is "n", in the last part?


For a given positive integer $a_0$ define the sequence $(a_n)$ by
$a_{i+1}$ $=$ $a_i/2$ if $a_i$ is even, and $3a_i-1$ if $a_i$ is odd.
 Prove that if $n$ is a natural number such that $a_n=a_0$, then $2^n>a_0$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=64610 (no complete solution there)


Find the four-digit number, which when multiplied by 4 gives the starting four-digit number with its digits reversed from right to left i.e $ 4385$ x $ 4 \equal{} 5834$
(the above is obviously not true but you get the point).

Start working digit by digit. And don't be so impatient!

[quote="Papalakis"]Sorry for sounding impatient, but i've been working on this the last 2-3 hours + half of my school hours.. I tryied heaps of stuff but nothing worked, thats why i was so impatient.. anyway i guess i'll just take a break and wait for someone brighter than me to respond :P[/quote]
What have you got so far?

My first thoughts:
- The number(Let's call it $ n \equal{} \overline{abcd}$) has $ n \le \frac{10000}{4}\equal{}2500$.
- The first digit $ a$ is even. So $ a \equal{} 2$.
- Since $ a\equal{}2$ we get $ d\equal{}8$ or $ d\equal{}9$. Only $ d\equal{}8$ works.
So we got the equation:
$ \overline{2bc8} \cdot 4 \equal{} \overline{8cb2}$
- $ b$ is odd and $ c$ is even. Since $ b < 3$ we get $ b\equal{}1$. So
$ \overline{21c8} \cdot 4 \equal{} \overline{8c12}$.
So $ c \equal{} 2$ or $ c \equal{} 7$. Only $ c\equal{}7$ works. So the number is $ 2178 * 4 \equal{} 8712$

Or when we have $ n\equal{}\overline{2bc8}$ we can write $ 4(200\plus{}100b\plus{}10c\plus{}8)\equal{}8000\plus{}100c\plus{}10b\plus{}2\iff 2c\equal{}13b\plus{}1$ => b=1 and c=7 ;)

Ah nice.. i'll show you how i solved it but can you explain your solution a bit more please? i cant seem to understand it.
Anyway here's how i thought about it;

let $ n \equal{} \overline{abcd}$ be our number, then $ 4 (\overline{abcd}) \equal{} \overline{dcba}$ $ \Rightarrow$
                    $ 4(1000a \plus{} 100b \plus{} 10c \plus{} d) \equal{} 1000d \plus{} 100c \plus{} 10b \plus{} a$

From the above we see that $ \overline{dcba}$ is even therefor its last digit $ a$ must be even too. Since $ a$ can only take values from $ 0 \minus{} > 9$, the only even numbers are $ 0,2,4,6,8$ but only $ a \equal{} 2$ satisfies our condition (a greater value of $ a$ would result in a number higher than 4-digits).
So                $ 4 (\overline{2bcd}) \equal{} \overline{dcb2}$.

The last digit $ d$ can take values from $ 1 \minus{} > 9$ only $ d \equal{} 8$ satisfies because after $ n$ is multiplied with $ 4$, the resulting product must have its first and last digits swapped with $ \overline{dcb2}$.
So                $ 4 (\overline{2bc8}) \equal{} \overline{8cb2}$.

Getting this far wasnt that hard, what follows made me go berserk.
I used (with the help of my dad, who's an elementary teacher) a mirror image of elementary multiplication. Here's what we did:

                                                     2 (1) (7) 8
                                                    x          4
                                                    -------------
                                                     8 (7) (1) 2

Basicly i put 2...8 which is the number to be multiplied by 4, with the 2 middle digits missing. Then on the bottom i put 8...2 which were the two standard digits i was sure to get after multiplying the first and last digits of $ n$.
Then in the first parenthesis i tryied all numbers from 1->9, and whatever was the outcome from the first top parenthesis x4, i put it on the first lower parenthesis, and then on the second top parenthesis. This sound wierd, i hope u understand what i'm trying to say. I just tryied to find 2 pairs of numbers -putting them in opposite order in the top & bottom- which when multiplied by 4 ( plus the 3 remaider from 4x8)  give an opposite pair. After trying 1,2,3,4,5,6,7 only the last satisfied the above conditions.

Therefor $ b\equal{}1, c\equal{}7$ and our number is $ n\equal{} 2178$

[quote="Papalakis"]Ah nice.. i'll show you how i solved it but can you explain your solution a bit more please? i cant seem to understand it.
Anyway here's how i thought about it;

let $ n \equal{} \overline{abcd}$ be our number, then $ 4 (\overline{abcd}) \equal{} \overline{dcba}$ $ \Rightarrow$
                    $ 4(1000a \plus{} 100b \plus{} 10c \plus{} d) \equal{} 1000d \plus{} 100c \plus{} 10b \plus{} a$

From the above we see that $ \overline{dcba}$ is even therefor its last digit $ a$ must be even too. Since $ a$ can only take values from $ 0 \minus{} > 9$, the only even numbers are $ 0,2,4,6,8$ but only $ a \equal{} 2$ satisfies our condition (a greater value of $ a$ would result in a number higher than 4-digits).
So                $ 4 (\overline{2bcd}) \equal{} \overline{dcb2}$.

The last digit $ d$ can take values from $ 1 \minus{} > 9$ only $ d \equal{} 8$ satisfies because after $ n$ is multiplied with $ 4$, the resulting product must have its first and last digits swapped with $ \overline{dcb2}$.
So                $ 4 (\overline{2bc8}) \equal{} \overline{8cb2}$.

Getting this far wasnt that hard, what follows made me go berserk.
I used (with the help of my dad, who's an elementary teacher) a mirror image of elementary multiplication. Here's what we did:

                                                     2 (1) (7) 8
                                                    x          4
                                                    -------------
                                                     8 (7) (1) 2

Basicly i put 2...8 which is the number to be multiplied by 4, with the 2 middle digits missing. Then on the bottom i put 8...2 which were the two standard digits i was sure to get after multiplying the first and last digits of $ n$.
Then in the first parenthesis i tryied all numbers from 1->9, and whatever was the outcome from the first top parenthesis x4, i put it on the first lower parenthesis, and then on the second top parenthesis. This sound wierd, i hope u understand what i'm trying to say. I just tryied to find 2 pairs of numbers -putting them in opposite order in the top & bottom- which when multiplied by 4 ( plus the 3 remaider from 4x8)  give an opposite pair. After trying 1,2,3,4,5,6,7 only the last satisfied the above conditions.

Therefor $ b \equal{} 1, c \equal{} 7$ and our number is $ n \equal{} 2178$[/quote]
You was not supposed to understand my solution! I did only write the facts which was used to get the solution, because I thought you might enjoy solving it yourself.
Your way to solve it was quite clever! Trying the combinations one after another is a great way to solve problems like these. What about $ 3\overline{abcd} \equal{} \overline{dcba}$?

[quote="Mathias_DK"] What about $ 3\overline{abcd} \equal{} \overline{dcba}$?[/quote]

Does such a number exist? i tryied it out but nothing matches.. in this case $ a$ is odd so $ a \equal{} 1$ or $ a \equal{} 3$. Well if $ a \equal{} 3$ then no such $ d$ can exist. So $ a \equal{} 1$ and $ d \equal{} 7$ are the only satisfying digits.

Then we have $ 3 (\overline{1bc7}) \equal{} \overline{7cb1}$...
With my mirror multiplication  technique nothing works.. and neither does [b]limes123[/b] method ;
                                  $ 3(1000 \plus{} 100b \plus{} 10c \plus{} 7) \equal{} 7000 \plus{} 100c \plus{} 10b \plus{} 1\iff 7c \equal{} 29b \minus{} 398$..

Here's another problem my teacher just gave me today :
Find a number with the following property:  $ \overline{xyzw} \plus{} x \plus{} y \plus{} z \plus{} w \equal{} 2003$

Haven't really started working on it yet, but hopefully i'll have it solved by the end of the day :D
But he did mention that its pretty hard :P

Ok i finally sold the second problem but i dont like my solution, i mean its more 'guess and try' than mathematics. Anyway here's my solution but please, someone post a better solution for maths' sake :)

We have: $ \overline{xyzw} \plus{}x \plus{} y\plus{} z\plus{} w\equal{} 2003 \Rightarrow$ $ \overline{xyzw}\equal{} 2003 \minus{} (x\plus{} y \plus{} z\plus{} w)$
But we know that $ 9\geq x \geq1$ and $ 9\geq y,z,w \geq0$

So we will find the values between which $ \overline{xyzw}$ is.

1st case:
max{$ x,y,z,w$}$ \equal{}${$ x\equal{}y\equal{}z\equal{}w\equal{}9$} $ \Rightarrow$ $ \overline{xyzw}>2003\minus{}36$ $ \Rightarrow$ $ \overline{xyzw}>1967$

2nd case:
min{$ x,y,z,w$}$ \equal{}${$ x\equal{}1$},{$ y\equal{}z\equal{}w\equal{}0$}  $ \Rightarrow$ $ \overline{xyzw}<2003\minus{}1$ $ \Rightarrow$ $ \overline{xyzw}<2001$

In general $ 1967<\overline{xyzw}<2001$

So now we try cases, making 3 columns of $ 1970 \minus{}> 1979$, $ 1980 \minus{}> 1989$, $ 1990 \minus{}> 1999$

[color=red]$ 1970 \minus{}> 17$||||[/color]     [color=blue]$ 1980 \minus{}> 18$||||[/color]    $ 1990 \minus{}> 19$
[color=red]$ .....1 \minus{}> 18$||||[/color]     [color=blue]$ .....1 \minus{}>  19$||||[/color]    $ .....1 \minus{}> 20$
[color=red]$ .....2 \minus{}> 19$||||[/color]     [color=blue]$ .....2 \minus{}>  20$||||[/color]    $ .....2 \minus{}> 21$
[color=red]$ .....3 \minus{}> 20$||||[/color]     [color=blue]$ .....3 \minus{}>  21$||||[/color]    $ .....3 \minus{}> 22$
[color=red]$ .....4 \minus{}> 21$||||[/color]     [color=blue]$ .....4 \minus{}>  22$||||[/color]    $ .....4 \minus{}> 23$
[color=red]$ .....5 \minus{}> 22$||||[/color]     [color=blue]$ .....5 \minus{}>  23$||||[/color]    $ .....5 \minus{}> 24$
[color=red]$ .....6 \minus{}> 23$||||[/color]     [color=blue]$ .....6 \minus{}>  24$||||[/color]    $ .....6 \minus{}> 25$
[color=red]$ .....7 \minus{}> 24$||||[/color]     [color=blue]$ .....7 \minus{}>  25$||||[/color]    $ .....7 \minus{}> 26$
[color=red]$ .....8 \minus{}> 25$||||[/color]     [color=blue]$ .....8 \minus{}>  26$||||[/color]    $ .....8 \minus{}> 27$
[color=red]$ .....9 \minus{}> 26$||||[/color]     [color=blue]$ .....9 \minus{}>  27$||||[/color]    $ .....9 \minus{}> 28$

The suitable number is the one that when multipliying their digit sum by their last digit we get $ 3$.. there are only three such numbers in these three colums; $ 1978, 1992, 1997$

Only $ 1978$ satisfies $ 1978 \plus{} 1 \plus{} 9 \plus{} 7 \plus{} 8 \equal{} 2003$.. So our number is 1978 :)

That was how i thought about it, please if you have something more fast/clever than this, please feel free to post.
Thanks in advance.

Another solution.
We see that x=1 (it can't be >1 which is easy to show and obviously it can't be 0). Next we see that y must be 9 (assuming y<9 we get a contradiction). And now he have to find numbers zw such $ \overline{zw}\plus{}z\plus{}w\equal{}93\iff 11z\plus{}2w\equal{}93$. We see that z<9 and z>6. We also see that z must be odd => z=7 and w=8 :)


function
calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Find the equation of the line tangent to the function at the given point.

$ f(x)\equal{}x$ at $ x\equal{}28$.

In general: $ f'(x)|_{x \equal{} x_0}$ (the derivative of f at x=28) gives you the slope of the line at the particular point $ x\equal{}x_0$

But here f(x) is a straight line itself, so the tangent to the function is pretty trivial.


geometry
3D geometry
number theory proposed
number theory

Let be $ m,n\in \mathbb{N}^*$ such that $ n^3\plus{}5m^2\plus{}5m$ is a perfect cube . Show that $ 2m>3n$ .

Let $ n^3\plus{}5m^2\plus{}5m\equal{}x^3$. Obviously $ x>n$

If $ n\equal{}x\plus{}1$, $ 5m^2\plus{}5m\equal{}3n(n\plus{}1)\plus{}1$, $ L.H.S.$ is even but $ R.H.S.$ odd , impossible.
If $ n\equal{}x\plus{}2$, $ 5m^2\plus{}5m\equal{}6n^2\plus{}12n\plus{}8$, $ 5|L.H.S.$ but $ 5 \nmid R.H.S.$, impossible.
If $ n\equal{}x\plus{}3$, $ 5m^2\plus{}5m\equal{}9n^2\plus{}27n\plus{}27$, $ L.H.S.$ is even but $ R.H.S.$ odd , impossible.

So $ x \ge x\plus{}4$, assume that $ 2m \le 3n$, then
$ n^3\plus{}\frac{45n^2}{4}\plus{}\frac{15n}{2} \ge n^3\plus{}5m^2\plus{}5m\equal{}x^3 \ge (n\plus{}4)^3\equal{}n^3\plus{}12n^2\plus{}48n\plus{}64$, impossible, done.

you are telling ovbiously $ X.>n$
then how are you assuming $ n\equal{}x\plus{}1$ etc?

He can assume what he wants as long as he handels all cases.


Eastlake is 50 miles due west of Northtown on Highway 7. Central City is 20 miles due south of Northtown on Route 19. If a road were built directly from Eastlake to Central City, how many miles, to the nearest mile, would be saved?

This is simply a right triangle with legs 50 and 20, and we want to find the hypotenuse. Thus, we have $ \sqrt{50^2 \plus{} 20^2} \approx 53.9$. The long way would take $ 50 \plus{} 20 \equal{} 70$ miles, so about $ 70 \minus{} 54 \equal{} 16$ miles are saved.


ratio
geometry
angle bisector
geometry solved

I have a homework assignment where I have to
1) Prove the concurrency of angle bisectors in $\tr ABC$ at point $I$, which lies on the bisector $AP$ using centers of mass
2) $\frac{AI}{AP}=\frac{b+c}{a+b+c}$ where a, b, c are the lengths of the sides

The hint in the book tells us to put point masses a, b, c at the points A, B, C

I could easily prove the the concerrency without point masses, but I'm not sure what process is required to get the second part completed as well.  We have not yet learned Ceva's thereom.  Anyone have a good proof that uses centers of mass for both parts?

[quote="Myth"]The hint devoted to part b), not for a).[/quote]

Myth, that's not quite correct - the idea of the center of mass can be used to prove both parts [b]1)[/b] and [b]2)[/b] together:

In the following, for every point T and every real number m, we will denote by T(m) the point-mass formed by the point T with the mass m.

Denote by I the center of mass of the point-masses A(a), B(b), C(c).

Let the angle bisector of the angle CAB meet the side BC of triangle ABC at a point P. Then, since the angle bisector of an angle in a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{BP}{PC}=\frac{c}{b}$; in other words, the point P is the center of mass of the point-masses B(b) and C(c).

Therefore, by the law of the redeployment of masses (I don't know how it's correctly called in English, sorry  :blush: ), the point I, being the center of mass of the point-masses A(a), B(b), C(c), can be considered as the center of mass of the point-masses A(a) and P(b + c). Hence, the point I lies on the line AP, and we have $\frac{AI}{IP}=\frac{b+c}{a}$.

Since the point P is the point where the angle bisector of the angle CAB meets the side BC of triangle ABC, the line AP is the angle bisector of the angle CAB. Hence, as the point I lies on the line AP, we can say that the point I lies on the angle bisector of the angle CAB. Similarly, we can show that the point I also lies on the angle bisectors of the angles ABC and BCA, and we conclude that the angle bisectors of the three angles of triangle ABC concur. This proves part [b]1)[/b] of the homework.

Now, from $\frac{AI}{IP}=\frac{b+c}{a}$, it follows that $\frac{IP}{AI}=\frac{a}{b+c}$, and thus

$\frac{AP}{AI}=\frac{AI+IP}{AI}=1+\frac{IP}{AI}=1+\frac{a}{b+c}=\frac{\left(b+c\right)+a}{b+c}=\frac{a+b+c}{b+c}$,

and thus, $\frac{AI}{AP}=\frac{b+c}{a+b+c}$, what solves part [b]2)[/b] of the homework.

  darij

You're wrong again Myth! :D
I think you are good in Geometry! (You've proved it many times)


Let $ a,b,c>0$ and $ abc\le 1$.Prove that $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\ge a\plus{}b\plus{}c$

We use $ \sum\frac{a}{b}\geq\frac{\sum a}{\sqrt[3]{abc}}\geq\sum a$

How can solve this?
[quote="karis"]We use $ \sum\frac {a}{b}\geq\frac {\sum a}{\sqrt [3]{abc}}$[/quote]


I think must a condition in hypothesis : for any $i\in\overline{1,4}\ ,\ A_{i}\in O_{i}\ .$

[quote="hunter_cvp_ic"]Prove that cyclic equilateral $A_{1}A_{2}A_{3}A_{4}$is harmonic(i.e $A_{1}A_{2}.A_{3}A_{4}=A_{1}A_{4}.A_{2}A_{3}$)iff there exist four circle$(O_{1})(O_{2})(O_{3})(O_{4})$such:
   1.$(O_{i})$ intanget to $(O)$(circumcircle of $A_{1}A_{2}A_{3}A_{4})at A_{i}(i=1,2,3,4)$
   2. $(O_{i})$ extangent to $(O_{i+1}) ,i=1,2,3,4$
   3. $(O_{1})$ tangent to $(O_{3})$[/quote]
OHHHHH!!!
 I'm sorry
I edited it now


Hi, I'd like to teach serious olympic math my younger brother who is 10 and knows almost nothing in math. I know the stuff I want to teach but I am a really poor teacher. What I would like is some advice on how to start really from nothing. He is willing to study.
I have no idea how to begin.
Please help,

Thanks
 :D

Hi, this should be in other problem solving topics.  I wrote up a nice little list of topics on what should be in that forum, and one of them is "methods of teaching math."  Unfortunately I can only move things out of Other Problem Solving Topics (I think, at least).

JS1527 is correct that this kind of question fits best in Other Problem Solving Topics, so I will move this thread there.

Welcome to the forum. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=33140&highlight=#33140]My very first post to AoPS[/url] includes advice for teaching younger math learners. Feel free to ask follow-up questions about that advice here in this thread.

Hope this helps!

The first,you should teach him some interesting questions,so that he will be interestded in mathematics.

I like that interesting questions one.

A few examples: Euclid's proof that there are infinitely many primes, and the fact that it is possible to construct a sequence of n consecutive integers in which none of them are prime (n!+k as k goes from 0 to n).

Start slowly.  Try to do (at least at first) problems that don't require a whole lot of additional knowledge beyond what he might know.  So, don't jump immediately into trigonometry or inversion or even simpler things (I don't know how much he already knows).  You might look in the Getting Started forum for ideas of things to use.  Build up an appreciation for problems and problem solving before you try to start teaching more curricular stuff.

Well one of the most important things is: make him do as much as possible by himself. Never talk an hour bout a problem while he's just listening.

Like you can make him start with simple olympiads like the round 1 multiple choices from most national olympiads. When he got the basics, you can let him advance, and so on and so on. The only things needed are his intelligence and his hard work.

There's no much 'teaching' required, unless you really want to get on high levels with him. Self-study is the best method, combined with you as a super-solver when he's got a problem he can't solve alone.

[quote="Peter VDD"]Like you can make him start with simple olympiads like the round 1 multiple choices from most national olympiads. When he got the basics, you can let him advance, and so on and so on. The only things needed are his intelligence and his hard work. [/quote]

ilyasu1's brother is only 10.  I doubt those types of problems would be easy enough so that he could answer them.  I would start out by having him try some medium/hard SAT questions.  Those require some problem solving, but no too much for a 10 year-old.  Some chapter level MATHCOUNTS problems would be good too.


limit
integration
logarithms
real analysis
real analysis unsolved

Evalute the limit

$\lim_{t\to1^{-}}(1-t)\sum_{k=1}^{\infty}\frac{t^{k}}{1+t^{k}}$

Well, just to go ahead and say what's cheap to say:

$\frac{t}2=(1-t)\sum_{k=1}^{\infty}\frac{t^{k}}2\le (1-t)\sum_{k=1}^{\infty}\frac{t^{k}}{1+t^{k}}\le (1-t)\sum_{k=1}^{\infty}t^{k}=t$

for $0<t<1.$

Hence, if the limit exists, it lies in the interval $\left[\frac12,1\right].$

Hint: this is a Riemann sum.

[hide="Solution"]Make it a sum from zero to $\infty$ instead (the $k=0$ term tends to zero, of course), and consider $\int_{0}^{1}\frac1{1+x}\,dx$ with the geometric partition $x_{j}=t^{n-j}$ for $1\le j\le n$ and $x_{0}=0$. The right-endpoint Riemann sum using this partition is $\frac{t^{n}}{1+t^{n}}+\sum_{j=1}^{n}\frac{t^{n-j}-t^{n-j+1}}{1+t^{n-j}}= \frac{t^{n}}{1+t^{n}}+(1-t)\sum_{k=0}^{n-1}\frac{t^{k}}{1+t^{k}}$. As $n$ goes to $\infty$ and $t$ goes to 1, this sum converges to $\int_{0}^{1}\frac1{1+x}\,dx=\ln 2$. The first term $\frac{t^{n}}{1+t^{n}}$ goes to zero as $n\to\infty$, so $\lim_{t\to 1^{-}}(1-t)\sum_{k=1}^\infty \frac{t^{k}}{1+t^{k}}=\ln 2$.[/hide]


geometry
circumcircle
trigonometry
trapezoid
geometry proposed

Let $ ABC$ be a triangle with the circumcircle  $ w$ and two points  $ \{P,Q\}\subset w$

so that $ PQ\parallel BC$ and the line $ BC$ separates $ A$ , $ P$ . Prove that

$ \left\|\ \begin{array}{c} PA\cdot QA \equal{} AB\cdot AC \plus{} PB\cdot PC \\
 \\
PA^2 \equal{} AB^\cdot AC \plus{} PB\cdot PC\ \Longleftrightarrow\ AB \equal{} AC\ \ \vee\ \ \widehat {PAB}\equiv\widehat {PAC}\end{array}\ \right\|\ .$

See [url=http://www.mathlinks.ro/viewtopic.php?t=264578][color=red][b]here[/b][/color][/url] .

Let $h_a,x,y$ be  the distances from $ A$ to $ BC,$ $ A$ to $ PQ$ and $ P$ to $ BC.$ If $PQ \parallel BC,$ it follows that $ x\equal{}h_a\plus{}y.$ But since $ \triangle ABC,$ $\triangle APQ$ and $\triangle BPC$ have the same circumcircle $(O,R),$ we deduce that

$ AB \cdot AC\equal{}2R \cdot h_a \ , \  AP \cdot AQ\equal{}2R \cdot x \ , \ PB \cdot PC\equal{}2R \cdot y$

$\Longrightarrow \ \frac{AP \cdot AQ}{2R}\equal{} \frac{AB \cdot AC}{2R}\plus{}\frac{PB \cdot PC}{2R}  \Longrightarrow  AP \cdot AQ\equal{}AB \cdot AC\plus{}PB \cdot PC$

[size=117][color=darkblue][u]Nice proof[/u], [b]Luisgeometra ![/b][/color][/size]

[quote="Virgil Nicula"] [color=darkred]Let $ ABC$ be a triangle with the circumcircle  $ w$ and two points  $ \{P,Q\}\subset w$

so that $ PQ\parallel BC$ and the line $ BC$ separates $ A$ , $ P$ . Prove that

$ \boxed{\ \begin{array}{c} PA\cdot QA \equal{} AB\cdot AC \plus{} PB\cdot PC \\
 \\
PA^2 \equal{} AB^\cdot AC \plus{} PB\cdot PC\ \Longleftrightarrow\ AB \equal{} AC\ \ \vee\ \ \widehat {PAB}\equiv\widehat {PAC}\end{array}\ }\ .$[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Suppose w.l.o.g. that  $ PB\ \le\ PC\ .$ Denote $ m(\angle PAB) \equal{} m(\angle QAC) \equal{} x\ .$

Observe that  $ \left\|\begin{array}{ccc} m(\angle APQ) & \equal{} & B \plus{} x \\
\ m(\angle AQP) & \equal{} & C \plus{} x\end{array}\right\|$ and $ AP\cdot AQ \equal{} AB\cdot AC \plus{} PB\cdot PC\ \Longleftrightarrow$

$ \sin (C \plus{} x)\cdot\sin (B \plus{} x) \equal{} \sin C\cdot\sin B \plus{} \sin x\cdot\sin (A \minus{} x)\ \Longleftrightarrow$

$ \cos(C \minus{} B) \minus{} \cos (B \plus{} C \plus{} 2x) \equal{} \cos (C \minus{} B) \minus{} \cos (C \plus{} B) \plus{} \cos (2x \minus{} A) \minus{} \cos A\ \Longleftrightarrow$

$ \cos(C \minus{} B) \plus{} \cos (2x \minus{} A) \equal{} \cos (C \minus{} B) \plus{} \cos A \plus{} \cos (2x \minus{} A) \minus{} \cos A$ , what is evidently.

The mentionned equivalence from the conclusion is also evidently.[/color]

See that BCQP is an isosceles trapezoid, apply twice Ptolemy, then cosine theorem, ready. For the 2nd relation see that, if AB = AC then obviously AP = AQ and, from 1st relation, done.

If $ \angle PAB \equal{} \angle PAC$, then P≡Q and, from 1st relation, done again.

Best regards
sunken rock


If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac{(b\plus{}c\minus{}a)^2}{b^2\plus{}c^2}\plus{}\frac{(c\plus{}a\minus{}b)^2}{c^2\plus{}a^2}\plus{}\frac{(a\plus{}b\minus{}c)^2}{a^2\plus{}b^2} \ge \frac 3{2}$.

Very nice ineq, Vasc. And it's trues with all the reals $ a, b, c$ because:
$ (a\minus{}b)^4\plus{}(b\minus{}c)^4\plus{}(c\minus{}a)^4 \geq\ 0$ (always trues)  :lol:

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {(b \plus{} c \minus{} a)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{c^2 \plus{} a^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{a^2 \plus{} b^2} \ge \frac 3{2}$.[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$
Hence, $ \sum_{cyc}\frac {(b \plus{} c \minus{} a)^2}{b^2 \plus{} c^2}\geq\frac {3}{2}\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(a^6 \minus{} 4a^5b \plus{} 5a^4b^2 \plus{} 2a^4bc \minus{} 2a^3b^3 \minus{} 4a^3b^2c \plus{} 2a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow w^6 \minus{} 2(6uv^2 \minus{} 5u^3)w^3 \plus{} 54u^6 \minus{} 144u^4v^2 \plus{} 117u^2v^4 \minus{} 26v^6\geq0,$
which is true for all reals $ a,$ $ b$ and $ c$ because $ (6uv^2 \minus{} 5u^3)^2 \minus{} (54u^6 \minus{} 144u^4v^2 \plus{} 117u^2v^4 \minus{} 26v^6)\leq0\Leftrightarrow$
$ \Leftrightarrow \minus{} (u^2 \minus{} v^2)^2(29u^2 \minus{} 26v^2)\leq0.$
What is your proof, nguoivn? Thank you!
I tried to use SOS but without success. :(

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {(b \plus{} c \minus{} a)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{c^2 \plus{} a^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{a^2 \plus{} b^2} \ge \frac 3{2}$.[/quote]
I think nguoivn's proof is my first proof:
Applying Cauchy Schwarz Inequality, we see that
\[ \left( \sum \frac {(b \plus{} c \minus{} a)^2}{b^2 \plus{} c^2}\right) \left( \sum (2b \plus{} 2c \minus{} a)^2(b^2 \plus{} c^2)\right) \ge (5x \minus{} 2y)^2
\]
where $ x \equal{} a^2 \plus{} b^2 \plus{} c^2$ and $ y \equal{} ab \plus{} bc \plus{} ca.$
Then we have to prove that
\[ 2(5x \minus{} 2y)^2 \ge 3\sum (2b \plus{} 2c \minus{} a)^2(b^2 \plus{} c^2).
\]
As $ \sum (2b \plus{} 2c \minus{} a)^2(b^2 \plus{} c^2) \equal{} 8x^2 \plus{} 4xy \minus{} 6y^2$, the last inequality is equivalent to
\[ (5x \minus{} 2y)^2 \ge 12x^2 \plus{} 6xy \minus{} 9y^2,
\]
or equivalently,
\[ 13(x \minus{} y)^2 \ge 0.
\]
which is trivial.

And for the SOS proof, see my secon proof :)
Without loss generality, we can assume that $ a,b,c$ are the sidelengths of a triangle. Now, by multiplying both sides for $ 2abc$, we can easily rewrite our inequality as
\[ \sum \frac {(b \plus{} c \minus{} a)(b \minus{} c)^2[(a \minus{} b)^2 \plus{} (a \minus{} c)^2]}{b^2 \plus{} c^2} \ge 0,
\]
which is trivial. :)

Here is my proof, my friends:
Using Cauchy Schwarts:
$ LHS \geq\ \frac{(2\sum\ ab \minus{} \sum\ a^2)^2}{2\sum\ a^2b^2} \geq\ RHS$
<=> $ \sum\ a^4 \plus{} 3\sum\ a^2b^2 \geq\ 2\sum\ ab(a^2\plus{}b^2)$
<=> $ (a\minus{}b)^4\plus{}(b\minus{}c)^4\plus{}(c\minus{}a)^4 \geq\ 0$
 :)

[quote="rahul1993"][quote="quykhtn-qa1"][quote="rahul1993"]my solution[/quote]
rahul1993 wrote:
$ \frac {(b + c - a)^2}{b^2 + c^2} + \frac {(c + a - b)^2}{c^2 + a^2} + \frac {(a + b - c)^2}{c^2 + a^2}\geq\ \frac {3}{2}$
sustituing:
$ \frac {3(a^2 + b^2 + c^2)}{b^2 + c^2} + \frac {3(a^2 + b^2 + c^2)}{c^2 + a^2) + \frac {3(a^2 + b^2 + c^2)}{a^2 + b^2} \geq \\
frac {3}{2}$
 This step is wrong :oops:[/quote]
Please can you explain why is it wrong :(

Edit: I got it :blush: Thanks for pointing it out[/quote]

Since $ (b+c-a)^2\le\ 3(a^2+b^2+c^2),b^2+c^2>0$
     =>$ \frac{(b+c-a)^2}{b^2+c^2}\le\ \frac{3(a^2+b^2+c^2)}{b^2+c^2}$  :oops:
i think you understood :oops:

[quote="nguoivn"]
$ LHS \geq\ \frac {(2\sum\ ab \minus{} \sum\ a^2)^2}{2\sum\ a^2b^2} \geq\ RHS$
<=> $ \sum\ a^4 \plus{} 3\sum\ a^2b^2 \geq\ 2\sum\ ab(a^2 \plus{} b^2)$
 :)[/quote]
The equivalence is wrong. :P

Can_hang'solution is nice. :wink:

I think this is very interesting.

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. For any real $ m$, the inequality holds

$ \frac {(mb \plus{}m c \minus{} a)^2}{b^2 \plus{}4bc\plus{} c^2} \plus{} \frac {(mc \plus{}m a \minus{} b)^2}{c^2 \plus{}4ca\plus{} a^2} \plus{} \frac {(ma \plus{}m b \minus{} c)^2}{a^2 \plus{}4ab\plus{} b^2} \ge \frac {(2m\minus{}1)^2}{2}$,

with equality for $ a\equal{}b\equal{}c$, and again for $ \frac a{2m\minus{}3}\equal{}b\equal{}c$ or any cyclic permutation, if $ m>\frac 3{2}$.

[quote="Vasc"]I think this is very interesting.

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. For any real $ m$, the inequality holds

$ \frac {(mb \plus{} m c \minus{} a)^2}{b^2 \plus{} 4bc \plus{} c^2} \plus{} \frac {(mc \plus{} m a \minus{} b)^2}{c^2 \plus{} 4ca \plus{} a^2} \plus{} \frac {(ma \plus{} m b \minus{} c)^2}{a^2 \plus{} 4ab \plus{} b^2} \ge \frac {(2m \minus{} 1)^2}{2}$,

with equality for $ a \equal{} b \equal{} c$, and again for $ \frac a{2m \minus{} 3} \equal{} b \equal{} c$ or any cyclic permutation, if $ m > \frac 3{2}$.[/quote]
Very nice inequality, but it is trivial, VasC. ;) Because by Cauchy Schwarz, we have
\[ \sum \frac{(mb\plus{}mc\minus{}a)^2}{b^2\plus{}4bc\plus{}c^2} \ge \frac{\left( \sum (mb\plus{}mc\minus{}a)\right)^2}{\sum (b^2\plus{}4bc\plus{}c^2)} \equal{}\frac{(2m\minus{}1)^2}{2}.\]
;) :)

[quote="Vasc"][quote="can_hang2007"][quote="Vasc"]I think this is very interesting.

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. For any real $ m$, the inequality holds

$ \frac {(mb \plus{} m c \minus{} a)^2}{b^2 \plus{} 4bc \plus{} c^2} \plus{} \frac {(mc \plus{} m a \minus{} b)^2}{c^2 \plus{} 4ca \plus{} a^2} \plus{} \frac {(ma \plus{} m b \minus{} c)^2}{a^2 \plus{} 4ab \plus{} b^2} \ge \frac {(2m \minus{} 1)^2}{2}$,

with equality for $ a \equal{} b \equal{} c$, and again for $ \frac a{2m \minus{} 3} \equal{} b \equal{} c$ or any cyclic permutation, if $ m > \frac 3{2}$.[/quote]
Very nice inequality, but it is trivial, VasC. ;) Because by Cauchy Schwarz, we have
\[ \sum \frac {(mb \plus{} mc \minus{} a)^2}{b^2 \plus{} 4bc \plus{} c^2} \ge \frac {\left( \sum (mb \plus{} mc \minus{} a)\right)^2}{\sum (b^2 \plus{} 4bc \plus{} c^2)} \equal{} \frac {(2m \minus{} 1)^2}{2}.
\]
;) :)[/quote]
Indeed, $ k \equal{} 4$ is the trivial case of this inequality type. Below there are 2 nontrivial inequalities for $ k \equal{} \frac { \minus{} 3}{2}$:

Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that

(a) $ \sum\frac {(b \plus{} c \minus{} 4a)^2}{2b^2 \minus{} 3bc \plus{} 2c^2} \ge 12$,

with equality for $ a \equal{} b \equal{} c$ and $ 4a \equal{} b \equal{} c$;

(b) $ \sum\frac {(3b \plus{} 3c \minus{} 4a)^2}{2b^2 \minus{} 3bc \plus{} 2c^2} \ge 12$,

with equality for $ a \equal{} b \equal{} c$ and $ 4a \equal{} 5b \equal{} 5c$.[/quote]

I have proved the following more general, VasC :)

If $ a,b,c,k \ge 0$, then
\[ \sum \frac {(b \plus{} c \minus{} ka)^2}{b^2 \minus{} \frac {8(k \minus{} 1)}{k^2}bc \plus{} c^2} \ge \frac {3k^2}{2}.
\]
The inequality holds for $ (a,b,c) \sim (1,1,1)$ and again for $ (a,b,c) \sim\left( \minus{} \frac {k^2 \minus{} 6k \plus{} 4}{k^2},1,1\right)$ if $ 3 \minus{} \sqrt {5} \le k \le 3 \plus{} \sqrt {5}.$

Moreover, if $ 0 \le k \le 2\left( \sqrt {2} \minus{} 1\right)$, the inequality holds for any[b] real [/b]numbers $ a,b,c.$ :)

[quote="can_hang2007"] I have proved the following more general, VasC :)

If $ a,b,c,k \ge 0$, then
\[ \sum \frac {(b \plus{} c \minus{} ka)^2}{b^2 \minus{} \frac {8(k \minus{} 1)}{k^2}bc \plus{} c^2} \ge \frac {3k^2}{2}.
\][/quote]
More general is the following.

If $ a,b,c \ge 0$, $ k>\minus{}2$ and $ \frac {2\minus{}\sqrt {4\plus{}2k}}{4}\le m \le \frac {2\plus{}\sqrt{4\plus{}2k}}{4}$, then

$ \sum \frac {(mb\plus{}mc\minus{}a)^2}{b^2\plus{}kbc\plus{}c^2} \ge \frac {3(2m\minus{}1)^2}{k\plus{}2}$.

[quote="Vasc"]If $ a,b,c$ are nonnegative real numbers, no two of which are zero, then

$ \frac {(b \plus{} c \minus{} a)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{c^2 \plus{} a^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{a^2 \plus{} b^2} \ge \frac 3{2}$.[/quote]

Natural extension

$ \frac {(b \plus{} c \plus{} ka)^2}{b^2 \plus{} (k \plus{} 1)bc \plus{} c^2} \plus{} \frac {(c \plus{} a \plus{} k b)^2}{c^2 \plus{} (k \plus{} 1)ca \plus{} a^2} \plus{} \frac {(a \plus{} b \plus{} kc)^2}{a^2 \plus{} (k \plus{} 1)ab \plus{} b^2} \ge \frac {3}{2}$.

holds for all $ k\ge \minus{} 3$

And

$ \frac {(b \plus{} c \plus{} ka)^2}{b^2 \minus{} (k \plus{} 1)bc \plus{} c^2} \plus{} \frac {(c \plus{} a \plus{} k b)^2}{c^2 \minus{} (k \plus{} 1)ca \plus{} a^2} \plus{} \frac {(a \plus{} b \plus{} kc)^2}{a^2 \minus{} (k \plus{} 1)ab \plus{} b^2} \ge \frac {3}{2}$.

holds for all $ \minus{}1\le k\le 1$

Sorry I edited some problems.

[quote="can_hang2007"]And for the SOS proof, see my secon proof :)
Without loss generality, we can assume that $ a,b,c$ are the sidelengths of a triangle. Now, by multiplying both sides for $ 2abc$, we can easily rewrite our inequality as
\[ \sum \frac {(b \plus{} c \minus{} a)(b \minus{} c)^2[(a \minus{} b)^2 \plus{} (a \minus{} c)^2]}{b^2 \plus{} c^2} \ge 0,
\]
which is trivial. :)[/quote]

I think this formula is wrong. It should be

\[ \sum \frac {(a\minus{}b)^2[(a \minus{} c)^2 \plus{} (b \minus{} c)^2]}{(b^2 \plus{} c^2)(c^2\plus{}a^2)} \ge 0,
\]

Please, tell me if I am wrong, my dear friend. :wink:

[quote="manlio"][quote="can_hang2007"]And for the SOS proof, see my secon proof :)
Without loss generality, we can assume that $ a,b,c$ are the sidelengths of a triangle. Now, by multiplying both sides for $ 2abc$, we can easily rewrite our inequality as
\[ \sum \frac {(b \plus{} c \minus{} a)(b \minus{} c)^2[(a \minus{} b)^2 \plus{} (a \minus{} c)^2]}{b^2 \plus{} c^2} \ge 0,
\]
which is trivial. :)[/quote]

I think this formula is wrong. It should be
\[ \sum \frac {(a \minus{} b)^2[(a \minus{} c)^2 \plus{} (b \minus{} c)^2]}{(b^2 \plus{} c^2)(c^2 \plus{} a^2)} \ge 0,
\]
Please, tell me if I am wrong, my dear friend. :wink:[/quote]
Dear my friend, I am sure about my formula. Please check it again ;)

OK I will try again, but please can you tell me why we can assume WLOG that a,b,c are sidelength of a triangle. Thank you very much.

[quote="can_hang2007"]And for the SOS proof, see my secon proof :)
Without loss generality, we can assume that $ a,b,c$ are the sidelengths of a triangle. Now, by multiplying both sides for $ 2abc$, we can easily rewrite our inequality as
\[ \sum \frac {(b \plus{} c \minus{} a)(b \minus{} c)^2[(a \minus{} b)^2 \plus{} (a \minus{} c)^2]}{b^2 \plus{} c^2} \ge 0,
\]
which is trivial. :)[/quote]

My dear friend, I made these computations :blush:

$ [(b\minus{}c)^2(a\minus{}b)^2 \plus{} (b\minus{}c)^2(a\minus{}c)^2]\equal{} (ab\minus{}b^2\minus{}ac\plus{}bc)^2 \plus{}(ba\minus{}bc\minus{}ac\plus{}c^2)^2\equal{} 2abc(b\plus{}c\minus{}a)\minus{}2a^2bc\plus{}(b^2\plus{}c^2)^2 \plus{}2a^2(b^2\plus{}c^2)\minus{}2a(b^3\plus{}c^3) \minus{}2bc(b^2\plus{}c^2)$

so we have

$ \sum \frac {(b \plus{} c \minus{} a)(b \minus{} c)^2[(a \minus{} b)^2 \plus{} (a \minus{} c)^2]}{b^2 \plus{} c^2} \equal{} 2abc\sum \frac{(b\plus{}c\minus{}a)^2}{(b^2\plus{}c^2)} \minus{}2abc\sum \frac{a(b\plus{}c\minus{}a)}{(b^2\plus{}c^2)}\plus{}\sum (b\plus{}c\minus{}a)(b^2\plus{}c^2)\plus{}2\sum (b\plus{}c\minus{}a)a^2\minus{}2\sum \frac{(b\plus{}c\minus{}a)a(b^3\plus{}c^3)}{(b^2\plus{}c^2)}\minus{}2\sum (b\plus{}c\minus{}a)bc\equal{} 2abc\sum \frac{(b\plus{}c\minus{}a)^2}{(b^2\plus{}c^2)} \minus{}2abc\sum \frac{a(b\plus{}c\minus{}a)}{(b^2\plus{}c^2)}\minus{}2\sum \frac{(b\plus{}c\minus{}a)a(b^3\plus{}c^3)}{(b^2\plus{}c^2)}\plus{}\sum a^2\sum(b\plus{}c\minus{}a) \plus{}\sum (b\plus{}c\minus{}a)a^2\minus{}2\sum(b\plus{}c\minus{}a)bc\equal{}2abc\sum \frac{(b\plus{}c\minus{}a)^2}{(b^2\plus{}c^2)} \minus{}2abc\sum \frac{a(b\plus{}c\minus{}a)}{(b^2\plus{}c^2)}\minus{}2\sum \frac{(b\plus{}c\minus{}a)a(b^3\plus{}c^3)}{(b^2\plus{}c^2)}\plus{}6abc$

So according your SOS form it should be

$ 2abc\sum \frac{a(b\plus{}c\minus{}a)}{(b^2\plus{}c^2)}\plus{}2\sum \frac{(b\plus{}c\minus{}a)a(b^3\plus{}c^3)}{(b^2\plus{}c^2)}\equal{}9abc$

Sorry, but I don't see why this is true. :blush:

P.S. Why you can assume WLOG that a,b,c are sidelength of a triangle. Thank you very much. :)


inequalities
trigonometry
inequalities unsolved

[quote="thanhnam2902"][color=darkblue][size=150]Let $ ABC$ is right triangle, with $ \angle A \equal{} 90^0$. Let $ BC \equal{} a,CA \equal{} b,AB \equal{} c$.

Prove that: $ b^n \plus{} c^n < a^n$, $ \forall n \in N, n > 2$.[/size][/color][/quote]

$ b^n \plus{} c^n\equal{}b^{2}*b^{n\minus{}2}\plus{}c^{2}*c^{n\minus{}2}<b^{2}*a^{n\minus{}2}\plus{}c^{2}*a^{n\minus{}2}\equal{}(b^{2}\plus{}c^{2})a^{n\minus{}2}\equal{}a^{2}*a^{n\minus{}2}\equal{}a^{n}$

${ \frac{b}{a} = \sin A}$ and $ \frac{c}{a} = \cos b$

We can now recast the problem as proving that $ \left (\frac{b}{a}\right ) ^n + \left (\frac{c}{a}\right )^n < 1$

which is equivalent to:

$ \sin^n A + \cos^n A < \sin^2 A + \cos^2 A = 1$

This is my solution:
$ \bullet$ Aplly theorem Pitago we have: $ a^2\equal{}b^2\plus{}c^2 \Longleftrightarrow a^n\equal{}b^2.a^{n\minus{}2}\plus{}c^2.a^{n\minus{}2}, (1)$.

$ \bullet$ But $ a>b$ $ \Longrightarrow a^{n\minus{}2}>b^{n\minus{}2} \Longrightarrow b^2.a^{n\minus{}2}>b^n$.

Same we have $ c^2.a^{n\minus{}2}>b^n$.

$ \Longrightarrow$ $ b^2.a^{n\minus{}2}\plus{}c^2.a^{n\minus{}2}>b^n\plus{}c^n, (2)$

$ \bullet$ Since $ (1)$ and $ (2)$ we have $ a^n>b^n\plus{}c^n$.

[quote="hsbhatt"]${ \frac {b}{a} = \sin A}$ and $ \frac {c}{a} = \cos b$

We can now recast the problem as proving that $ \left (\frac {b}{a}\right ) ^n + \left (\frac {c}{a}\right )^n < 1$

which is equivalent to:

$ \sin^n A + \cos^n A < \sin^2 A + \cos^2 A = 1$[/quote]

How are you going to prove the inequality $ \sin^n A + \cos^n A < \sin^2 A + \cos^2 A$?

Since $ \sin A < 1$, $ \sin^n A < \sin^2 A$ for $ n > 2$ Ditto for $ \cos^n A$

Right.  Just note that in this case $ A \neq \pi/2$, which is different from the previous one.


limit
topology
group theory
abstract algebra
geometry
geometric transformation
advanced fields

Let $ G$ be compact Lie group, $ e$ - unit in $ G$. Prove that for any $ g\in G$ $ e$ is the limit point of $ \{g^n, \ n\in \mathbb{N}\}$.

The sequence $ g^n$ has a limit point since $ G$ is compact. Therefore, there exist a convergent subsequence $ g^{a_n}$. Then
\[ \lim_{n\to \infty} g^{a_{n\plus{}1} \minus{} a_n} \equal{} \left( \lim_{n\to\infty}  g^{a_{n\plus{}1}}\right) \left(\lim_{n\to\infty} g^{a_n}\right)^{\minus{}1} \equal{} e,\]
since group operations are continuous.

[quote="Yury"]The sequence $ g^n$ has a limit point since $ G$ is compact. [/quote]
I think it is not true in general, only if topology is generated by some metric.

[quote="Lippilus"]Let $ G$ be [b]compact[/b] Lie group...[/quote]

anyway, Yuri, who tells you that $ a_{n\plus{}1}\minus{}a_n\to \infty$? :)
of course, fixing this is not that difficult, since you can just choose $ 2n$ instead of $ n\plus{}1$.

one different way could be this one, though: the closure of the orbit of $ g$ is a closed Lie subgroup of $ G$, so it's either finite or it intersects every neighbourhood around $ e$ (since it's homogeneous).

[quote="ma_go"][
one different way could be this one, though: the closure of the orbit of $ g$ is a closed Lie subgroup of $ G$, so it's either finite or it intersects every neighbourhood around $ e$ (since it's homogeneous).[/quote]
Can you give a bit more comment?

it's a standard fact that a closed subgroup of a lie group is itself a lie group (also wikipedia quotes it).
the closure of the set $ \{g^n\}$ is a subgroup $ H<G$, hence it's a lie group itself.

now either $ \{g^n\}$ is finite, or, by compactness, it has an accumulation point, which by definition lies in $ H$. but translations by elements of $ H$ (which give homogeneity of $ H$) bring this point around, so every point of $ H$ is an accumulation point.

after all, this is just a different wording for what Yuri wrote.

But can we extract a converging subsequence? We are not in a metric space, where we can do this using compactness.

A Lie group is a manifold, and the topology of any manifold is metrizable. It looks locally like $ \mathbb{R}^n$; patch together copies of that metric.