geometry
AMC
AMC 12
AIME
AMC 10
USA(J)MO
USAMO

DO PROBLEMS

Also look at the 10+ threads that have already appeared on this topic during the summer (I can see 6 threads about preparing for AMCs just on the first page of this forum).

If you think you've got the concepts down from the book and previous tests, then it's all about PRACTICE.

[quote="basketball9"]What should I practice. My weakest area is geometry and I already ran out of amc 10 problems.[/quote]

go here

http://www.artofproblemsolving.com/Wiki/index.php/AMC_10_Problems_and_Solutions

or look at some old AMC 12 problems (the first 2/3 is about the same level as AMC 10)

http://www.artofproblemsolving.com/Wiki/index.php/AMC_12_Problems_and_Solutions

[quote="basketball9"]I already ran out of amc 10 problems.[/quote]
When did you do them? I think if it has been more than 6 months, you can do them again. Practice AMC 12s if you can, because if you go back to the 10 from a harder one, it will look easier.

You can always find more problems. Buy a book of problems. Look around in the Resources section here - there are a million contests.

Look here: http://www.artofproblemsolving.com/Wiki/index.php/Category:Introductory_Geometry_Problems

A lot of the problems there aren't just from the AMCs.

Well as most people on the forum suggest, do problems more difficult than  the competition you're preparing for. Chances are, if you raise your AMC12 and AIME scores, you will raise your average AMC10 score.

how should i raise my aime score. I can probably get 2 or 3 max. :D

[quote="basketball9"]CAN YOU GUYS SUGGEST WAYS TO RAISE MY AMC 10 SCORE BY 20-30 POINTS.
I already raised it by twenty points in two months by doing aops volome 1 and contest problems.

Thanks[/quote]

1) Read the AoPS Introduction books to get a thourough understanding of the stuff. Actually, that raised my score by almost 50(!)
2) Do some MORE contest problems. Don't just take one and go on. Look at your incorrect and blank ones. Try to understand the solutions.
3)Pinpoint your weak points.
4) Maybe buy [u]First Steps for Math Olympians[/u]
5) If you really want to push yourself and get a jumpstart on AMC 12, start doing those problems.

If money is not an issue, the AoPS books and/or classes are a great resource. According to Mr. Rusczyk, doing the intro series, volume 1, the intermediate series (including precalculus which is coming out in December), and volume 2 would set you up to do good on the AIME and make USAMO. Then taking WOOT would help you do good at USAMO. This is my plan for the next two years.

If your seriously still preparing for AMC 10, remember not to get too far ahead of yourself, like don't prepare for AMC 10 and USAMO.

So most of the AIME and USAMO is out for right now.


function
inequalities unsolved
inequalities

[quote="bugzpodder"]Prove that $\forall x\in \left]0,\frac{\pi}{2}\right[: \tan(x)>x^3$[/quote]

This one is easy: by taylor $\tan(x) > x+\frac13 x^3 + \frac2{15} x^5$
which is by AM-GM: $ \tan(x) > \frac13 x^3 + \frac23 x^3 = x^3$
(as $2\sqrt{\frac{2}{15}}>\frac23$)

More difficult (this trick fails) would be the following: prove that $\forall x\in \left]0,\frac{\pi}{2}\right[: \tan(x)>x^4$

An even more difficult variant (but I don't know if it has an elementary solution - I'm afraid not) is to find the greatest constant $k$ for which $\tan(x) > x^k$ for all $x \in ]0,\frac\pi2[$. The answer is somewhere around 4.88, but I have no clue how to prove it.


function
logarithms
algebra unsolved
algebra

Determine the functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy:

\[ f(x\plus{}y)\equal{}f( \ln (x \cdot y ))\]

for any $ x,y \in (0,\infty).$

[quote="moldovan"]Determine the functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy:
\[ f(x \plus{} y) \equal{} f( \ln (x \cdot y ))\]
for any $ x,y \in (0,\infty).$[/quote]

Let $ u,v\in\mathbb R$ such that $ u\ge 2e^{\frac v2}$ and let $ a,b$ the two positive roots of $ x^2\minus{}ux\plus{}e^v\equal{}0$.

Then $ f(a\plus{}b)\equal{}f(\ln(ab))$ and so $ f(u)\equal{}f(v)$

And so $ f(u)\equal{}f(v)$ $ \forall u\ge 2e^{\frac v2}$

And so $ f(u)\equal{}f(v)$ and $ f(u)\equal{}f(w)$ $ \forall u\ge\max( 2e^{\frac v2}, 2e^{\frac w2})$

And so $ f(v)\equal{}f(w)$ $ \forall v,w$ and it is easy to see that constant functions indeed are solutions.

Hence the solutions : $ \boxed{f(x)\equal{}c}$

let's put $ y\equal{} \frac{1}{x}$. then $ \forall_{x>2} \ f(x)\equal{}f(0)\equal{}a$
for x=2 we have:
$ a\equal{}f(2\plus{}y)\equal{}f( \ln(2y))$
so $ f(x)\equal{}a$


geometry
circumcircle
inradius
trigonometry
search
inequalities
inequalities unsolved

Prove that in any triangle

$ 2p^2(2R-r) \leq R(4R+r)^2 $

$p = $ semiperimeter
$R = $ circumradius
$r  = $ inradius

From this follows

$ 2(\cos\alpha +\cos\beta +\cos\gamma ) + (\tan \alpha/2)^2 +(\tan \beta/2)^2 + (\tan \gamma/2)^2 \geq4 $

This used to be discussed a long time ago. If you are interested in it, I 'll search itfor you. But is tha J Garfulkel or O Kooi ?

Here it is : http://www.mathlinks.ro/viewtopic.php?t=4575.

To make you easily understand, your inequality is equivalant to
   $\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\geq 2$


notice the 3 integrals ... well they're suppose to be on the next line.  i'm not sure what's wrong

$ \int\frac {1}{\sin{x} + \cos{x}}\times\frac {\sin{x} - \cos{x}}{\sin{x} - \cos{x}}dx \\
 \\
 \\
- \int\frac {\sin{x} - \cos{x}}{\cos^{2}x - \sin^{2}x}dx \\
 \\
 \\
- \int{\frac {\sin{x}}{2cos^{2}x - 1}dx\left\{\begin{array}{cc}u = \cos{x} \\
du = - \sin{x}dx\end{array}\right\} + \int\frac {\cos{x}}{1 - 2\sin^{2}x}dx\left\{\begin{array}{cc}u = \sin{x} \\
du = \cos{x}dx\end{array}\right\} \\
 \\
\int\int\int}$

code:

\int\frac {1}{\sin{x} + \cos{x}}\times\frac {\sin{x} - \cos{x}}{\sin{x} - \cos{x}}dx \\
 \\
 \\
- \int\frac {\sin{x} - \cos{x}}{\cos^{2}x - \sin^{2}x}dx \\
 \\
 \\
- \int{\frac {\sin{x}}{2cos^{2}x - 1}dx\left\{\begin{array}{cc}u = \cos{x} \\
du = - \sin{x}dx\end{array}\right\} + \int\frac {\cos{x}}{1 - 2\sin^{2}x}dx\left\{\begin{array}{cc}u = \sin{x} \\
du = \cos{x}dx\end{array}\right\}\\
\\
\\
\int\int\int

GRR!!!

To answer your question, if you look at the errors tex tells you that it is missing a }. This is because on one line you have - \int{ but no matching }. Removing the { will get you a new line but the result is a mess.

So let's make it look nicer, which is what LaTeX does so well.
1. You should never have more than one \\ together. There is no such thing as a blank line in LaTeX but instead it uses the idea of vertical space. The integrals are too close together vertically so you want more vertical space between them. Try \\[3ex] which says put 3ex of vertical space before the next line where 1ex is the height of an x in the current font.

2. In order to make things more readable you need to lay the equations out in display mode with all the = signs lined up. The align environment is made for that.

3. Put a little space before the dx

4. Use \iiint for a triple integral - it gives a much nicer spaced result $ \iiint$

5. When explaining the substitutions put them at the end of a line not in the middle of an equation, it makes it easier to read

6. Don't use arrays and \left{ and \right} use \Bmatrix though I prefer the sustitutions on one line.

So try this (you must of course be using the amsmath package but you should do for all tex files)

\begin{align*}
-\int \frac{1}{\sin{x} + \cos{x}}
&= -\int\frac{1}{\sin{x} + \cos{x}} \times \frac{\sin{x} - \cos{x}}{\sin{x} - \cos{x}}\;dx \\[3ex]
& = - \int \frac {\sin{x} - \cos{x}}{\cos^{2}x - \sin^{2}x}\;dx \\[3ex]
&= - \int \frac{\sin{x}}{2\cos^{2}x - 1}\;dx & u = \cos{x},\  du = - \sin{x}\;dx\\[3ex]
&\qquad + \int \frac{\cos{x}}{1 - 2\sin^{2}x}\,dx & u = \sin{x},\ du = \cos{x}\,dx \\
\end{align*}

Doesn't that look so much nicer and easier to follow? Here is the code:
[code]\begin{align*}
-\int \frac{1}{\sin{x} + \cos{x}}
&= -\int\frac{1}{\sin{x} + \cos{x}} \times \frac{\sin{x} - \cos{x}}{\sin{x} - \cos{x}}\;dx \\[3ex]
& = - \int \frac {\sin{x} - \cos{x}}{\cos^{2}x - \sin^{2}x}\;dx \\[3ex]
&= - \int \frac{\sin{x}}{2\cos^{2}x - 1}dx & u = \cos{x},\  du = - \sin{x}\;dx\\[3ex]
&\qquad + \int \frac{\cos{x}}{1 - 2\sin^{2}x}\,dx & u = \sin{x},\ du = \cos{x}\,dx \\
\end{align*}[/code]

Thanks for the advice, I am now working towards making my TeX cleaner.  I will incorporate your style into mine :p


topology
Functional Analysis
real analysis
real analysis unsolved

I have known the von Neumann theorem for unitary operators: suppose that $ U$ is a unitary operator on a Hilbert space $ H$ and let $ S_n\equal{}\frac{1}{n} \sum_{j\equal{}0}^{n\minus{}1} U^j$. Then $ S_n$ converges to $ P$ in strong operator topology, where $ P$ is the projection to $ \ker (I\minus{}U)$. I wonder why this convergence is not necessarily in norm topology. Could someone provide me a counterexmple that $ S_n$ does not converge to $ P$ in operator norm? Thanx.

Just take $ H\equal{}L^2(\mathbb T)$ and the operator of multiplication by $ z$.

I don't get. If $ A$ is the operator of multiplication by $ z$, then $ A^*$ would be the multiplication operator by $ \bar{z}$, and so $ AA^*$ is the multiplication operator by $ z\bar{z}$. How is $ A$ a unitary operator?

$ \mathbb T$ is the unit circumference (this notation is so standard that I automatically assumed it known; I apologize for that), so $ z\bar z\equal{}1$ everywhere on $ \mathbb T$.


Sometimes when I look at a post I just won't be able to see the LaTeX stuff on my computer.  For example,

"The equation is $ x^{2} \plus{} 1 \equal{} 0$."

will show up as

"The equation is ."

If I hit the quote button I can see their code but I can't see the final LaTeX'd result.  This happens on Internet Explorer and Firefox, so I have no idea what's causing the problem.

Make sure that you close the expression with "$ \$$" or else it will come up as an error.

Is it true for other posts or just yours? I presume that you have images turned on? Can you post a screen shot of what you see and give the URL of what you are showing? Is it the same for both mathlinks and AoPS URLs, http://www.mathlinks.ro and http://www.artofproblemsolving.com?

I see this on partially loaded pages- the images load after the rest of the page, and may sometimes stall out completely. It's especially common when there are many small code blocks on the page.

I attached a screenshot of what happens (sorry about the size), and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180658]here[/url] is a link to the topic in my screenshot.  I have visited this topic several times today, and the same LaTeX equation has failed to show up every time, even if I refresh.  It seems like it usually happens on other people's posts, not my own, and it actually happens pretty often.  I'll try Mathlinks to see if that makes a difference.  jmerry, do you have a way to deal with this when it happens?

EDIT: Going through mathlinks.ro didn't help.

How I would deal with it:
Wait for the page to finish loading. The blanks for any images that failed to load are replaced by broken image icons at this point. For each important bit I want to see, open that image (forcing a partial refresh). If that fails, give up on the pretty version.
The main fix is to simply have a better connection, which I do most of the time.

If you're not seeing broken image markers when the page stops loading, there's something else wrong. Is that part the same across browsers?

Since there are no $ \text{\LaTeX}$ errors involved, this probably belongs in the support forum.

[quote="jmerry"]Since there are no $ \text{\LaTeX}$ errors involved, this probably belongs in the support forum.[/quote]I agree and will move it across.

Is it only in hidden text that the problem occurs?
If you go to View Page Source, near the hint should be the following text:
[code]
 Let  then show that .
  [/code]
The missing image is [code]. [/code]Is that image code there? If it is there but you aren't seeing the image then there could be a problem with Javascript in your browser.

[quote="minicon"]Make sure JavaScript is enabled in your browser.  I think its Tools-->Preferences-->General
Make sure the box next to " JavaScript Enabled" is checked off.  :)[/quote]

If Javascript was disabled, it shouldn't display anything; not just hide a particular part.

Weird bug. Is it still happening? Have you tried viewing that IMAGE url by itself (does it load on your computer)? I had this problem back in the old office that some images (and only on my computer, with any browser, all antiviruses turned off, etc.) would just not load.

Ok this issue seems to be connected to the hide bbcode, so it needs more investigation. Possible causes are: images are not being loaded when a hide link is clicked which completely stops the loading of the images.

Can you tell me (the ones of you that had this happening to them), what skin (aops/mathlinks) and what browser where you using?

Actually, I'm not noticing it anymore.  When I posted before I was using Firefox and the AoPS skin, and I'm still using both of those.

I think the best way to ensure that it doesn't happen is to let the page fully load before clicking any hide tags, but now it's not happening even when I don't wait for the page to fully load.  Weird...


[hide="Method 1"] Cauchy ;) [/hide]
[hide="Method 2"] Generally we can rewrite $A \sin x+B \cos x$ as a single trig function.  I forget the exact form, but it is a very good exercise - try it for yourself. [/hide]

[quote="Allan Z"]Find the maximum value of:

$5\sin X+2\cos X$[/quote]

[hide]$(5\sin X+2\cos X)^{2}\le(25+4)(\sin^{2}X+\cos^{2}X)=29$
$5\sin X+2\cos X\le\boxed{\sqrt{29}}$[/hide]

[hide]We can treat this as the dot product of $(5,2)$ and $(\sin x, \cos x)$

The expression translates to: $(\sqrt{5^{2}+2^{2}})(\sqrt{\sin^{2}+\cos^{2}})\cos\theta$

$= \sqrt{29}\cos\theta$

This value is maximum when cos $\theta = 1$ or when $\theta = \pi$.

So max is $\sqrt{29}$[/hide]


A good one for the binomial series, with the "factor out what's big" hint:

$(x+\sqrt{2})^{\sqrt{2}}=x^{\sqrt{2}}\left(1+\frac{\sqrt{2}}x\right)^{\sqrt{2}}$

$=x^{\sqrt{2}}\left(1+\frac2x+O(x^{-2})\right)=x^{\sqrt{2}}+2x^{\sqrt{2}-1} +O(x^{\sqrt{2}-2})$

Similarly, $(x-\sqrt{2})^{\sqrt{2}}=x^{\sqrt{2}}-2x^{\sqrt{2}-1} +O(x^{\sqrt{2}-2})$

So the orignial quotient becomes

$\frac{4x^{\sqrt{2}-1}+O(x^{\sqrt{2}-2})}{x^{\sqrt{2}-1}}$

The limit as $x\to\infty$ is $4.$

(We're using: as $u\to0,\,(1+u)^{\alpha}=1+\alpha u+O(u^2).)$


inequalities MATHCOUNTS algebra function domain calculus integration

[hide="answer"]15?I just solved the inequality by taking the expressions 0s, and testing #s in between. But, knowing your coach, there's gotta be some elegant solution I'm missing...[/hide]

[quote="frost13"]Not 15 :lol:  I do like my solution although Leet does claim it too high level, but I think that's just because he's jealous he didn't think of it.  His solution was very good too though.  I'll give it more time before there are any hints though.

And just for the record this is not an original problem of mine, its from an old MATHCOUNTS handbook and the Acces database software I have.  Just for the record.[/quote]

[hide]There are 15 integers that will make the given expression positive.

On a number line mark the values of $n$ for which the given expression is zero. These are $-\frac{15}{2}, \frac{5}{2}, \frac{31}{3},$ and $\frac{17}{3}$. Then plug in some numbers for each of the ranges on this number line. You will see that only the integers in the two ranges $-\frac{15}{2}\le n \le \frac{5}{2}$ and $\frac{17}{3}\le n \le \frac{31}{3}$ will make the given expression positive. There are 15 integers in these two ranges.

I am eager to see Mr. Frost's simpler solution. [/hide]

[quote="mathgeniuse^ln(x)"][quote="13375P34K43V312"]For how many integer values of $n$ is $(2n+15)(5-2n)(31-3n)(17-3n)$ a positive number?[/quote]

[hide]The equation can be written as $-(2n+15)(2n-5)(3n-31)(3n-17)$.

The roots are $n=-\frac{15}{2}, \frac{5}{2}, \frac{31}{3}, \frac{17}{3}$

So this means that with some simple sketches we get that the domains that give positive numbers are $\frac{17}{3}<n<\frac{31}{3}$ and $-\frac{15}{2}<n<\frac{5}{2}$

There are 5 integral values in the first set (6,7,8,9,10) and there are 10 integral values in the second set (-7, -6, -5, ..., 0, 1, 2).

So the total number is 15.

I don't see how it is not 15.[/hide]
And  mthyiaga, I think you and I did it the same way, but you counted wrong.[/quote]
You're right mathgenius, my initial count was wrong. But the method is right.

But then where did we go wrong if Mr. Frost says it is not 15?

Maybe he thought I meant the only solution was 15? I don't know.

[quote="13375P34K43V312"][quote="frost13"]Not 15 :lol:  I do like my solution although Leet does claim it too high level, but I think that's just because he's jealous he didn't think of it.  His solution was very good too though.  I'll give it more time before there are any hints though.

And just for the record this is not an original problem of mine, its from an old MATHCOUNTS handbook and the Acces database software I have.  Just for the record.[/quote]

Oh, for the second time, I stated the problem incorrectly.
The second term should read $5n-2$, not $2n-5$, so the intended answer was $13$. But for the misstated problem, 15 is indeed correct.

@mthiyaga: be careful of your signs, you got lucky here, but if the roots were integers, you would have to make the distinction between $\le$ and $<$. It should be $<$, but since the roots are all fractions, it doesn't have an effect on the answer.

Here's my solution, it's slightly different: more algebra, less calculations

[hide="read"]
Let's find the conditions for which the four factors are positive or negative.
$2n+15>0 \Rightarrow n>-\frac{15}{2}$
$5-2n>0 \Rightarrow n<\frac{5}{2}$
$31-3n>0 \Rightarrow n<\frac{31}{3}$
$17-3n>0 \Rightarrow n<\frac{17}{3}$

For the product to be negative, we either have all four positive, two positive and two negative, or all four negative. Let's say all four are negative. Then, $2n+15<0$. This implies that $n<-\frac{15}{2}$. This would mean that everything else would be positive as well (make sure you see why), so we can't have all four negative.
Let's say everything is positive. We saw that if the first factor is negative, everything else has to be positive, so the first factor has to be negative. $n<\frac{15}{2}$. Also, we note that if $n<\frac{5}{2}$, $n<\frac{31}{3}$ and $n<\frac{17}{3}$ clearly hold as well, so our range here is $-\frac{15}{2}<x<\frac{5}{2}$, giving the solutions $-7,-6,\ldots,1,2$.
Let's say two are positive and two are negative. We know that the first factor must be positive. So exactly ONE of the following must be true, $n<\frac{5}{2}$, $n<\frac{17}{3}$, and $n<\frac{31}{3}$. This implies $\frac{17}{3}<x<\frac{31}{3}$ (again, make sure you see why), giving the solutions $6,7,8,9,10$.
Therefore, there are a total of $5+10=15$ solutions
[/hide]

Yes, it's slightly longer, but I like the logic in my solution, and the aviodance of calculation (I can never remember my multiplication tables  :P )[/quote]

If the second factor was $(5n-2)$, then the answer would be 13. Probably that is why Mr. Frost rejected perfect628's answer.

Multiplication tables to how high? I knew through 30 in first grade...but seriously, how high?

Yeah, I said it wasn't 15 because we had just done that problem after school and Leet and I both got 13 different ways.  My solution was basically shown here.  I looked at it as the graph of the equation y =(2n+15)(2-5n)(31-3n)(17-3n)

Set y = 0 to get the x intercepts and then just checked in between each solution to find the sections that worked.

Funny that Leet changed (2-5n) to (5n -2) since if I remember correctly when he did this problem at practice se said

(2 - 5n) = 0 gave n = 5/2 anyway.  I think it must have been his subconscience desire to make the root 5/2 instead of the intended 2/5 :lol:

Good job getting the correct 15 right off, sorry I said no, I should have checked Leets actual given problem.  My bad.


[quote="3X.lich"]Just read up this article about a chem whiz who recently graduated from Berkeley

[url=http://www.berkeley.edu/news/media/releases/2005/05/02_chau.shtml]www.berkeley.edu/news/media/releases/2005/05/02_chau.shtml[/url][/quote]

Impressive.


abstract algebra
group theory
superior algebra
superior algebra solved

let $H$ be a non-abelian group with $\left|H\right|=2p$, $p$ is prime. Show $H$ has $p$ elements of order 2.

Let's try for a solution using as few and as low level of tools as we can get away with. Since there are no nonabelian groups of order 4, $p$ is an odd prime.

The order of an element must divide the order of the group. If there's any element of order $2p,$ then the group is cyclic, hence abelian. Since it is non-abelian, the only possible orders of elements are $1, 2,$ and $p.$

Since any group in which all non-identity elements are of order 2 is abelian, there must be an element of order $p.$ Call this element $x.$

The cyclic subgroup generated by $x$ has order $p.$ Each element of this subgroup other than the identity has order $p.$ So far, we have one element of order $1$ and $p-1$ elements of order $p.$

Let $y\in H\setminus\langle x\rangle.$

The subgroup $\langle x\rangle,$ being of index 2, is normal. Since $\langle x\rangle$ is normal, $y^2\in\langle x\rangle$ and $y^p\not\in\langle x\rangle.$ The order of $y$ must be either $2$ or $p,$ but we have just shown that it cannot be $p.$ So, all $p$ of the elements in $y\in H\setminus\langle x\rangle$ have order $2.$ We have a complete count and are finished.

The more you know, the quicker you can finish this. As I said, I was trying to be as elementary as possible.


AMC
AIME
AMC 10
AMC 12
trigonometry
function
Putnam

What are usefull formulas for AMC10 AMC12 and AIME?

Someone else will probably say this, but it isn't the formulas as much as the techniques that matter in solving problems.  Besides, if you don't know how to apply the formulas, then much of the time you won't be able to use them.

That said, there is a basic toolbox of formulas that will help you.  In solving problems, reading AoPS, etc., you may discover or take note of these.  I'll mention some of the ones I find useful.  But not right now.  I'm too tired.  zzzz...

yeah, i agree with mathfanatic.  But since there is quite a bit of time from now til the contest, you might as well get familiar with some of the formulas and concepts that you always get wrong when practicing theses tests.
I remember seeing something that gave a list of all the topics / formulas that are on math contests.  It was on the mu alpa theta website i think, and it was actually written by the owners of this site, i think.  I cant seem to find it though, so if i do, i'll post it here.  Some of it is probablly too advanced for AMC, but by looking through some past tests you should be able to decide which are too advanced and which are not.

i've practiced a couple of old AIME tests recently and on a few problems i noticed that you need a lot of trig identities to solve... and i find it difficult to memorize all of them so i couldn't do any of them, which made my scores much lower... so trig identities are good to know :lol:

uh, as far as trig, know sin :^2: + cos :^2: =1, know the half angle formulas for sin and cos, and from those three, u can basically derive all the other ones...

wait no...the trig sums are important too: sin(x+y), etc...

Half angle is a direct result of the sum identities
sin(x + y) = sinx*cosy + siny*cosx and
cos(x + y) = cosx*cosy - sinx*siny.
Just substitute x = y = a/2 and solve for sin(a/2).

Basically, all you really need to [i]know[/i] are addition formulas and that sin:^2: + cos:^2: = 1.  However, it is helpful if you've seen the other standard identities, such as double- and triple- and half-angle.  That way, it's quicker to derive them and they'll seem more familiar to you when you arrive at them.

Also, of course, remembering the basic properties of the trig functions, which I wouldn't describe as identities so much as defining characteristics, such as sin(:theta:) = cos(90 - :theta:) and vice-versa.

what about sum-to-product? (the ones i always seem to forget) (snort)

They aren't as useful -- there are far fewer opportunities to use them.  But, anyhow, you should be able to derive them using the sum forumlae and the half-angle formulae (which are results of the sum formulae) in order to derive it, if you've seen it and you remember roughly what the form of the formula is.

If I were you I'll also memorize the Laws of Sine/Cosine.

Basically you can either choose to memorize everything (which is easier for some people who's good at memorization) or derive them as you go (which is easier for people who's used to trig stuff)
I memorized most since I usually don't have enough time.

Sum of arithmetic series, sum of geometric series, combinatrics formulas, and other bits and pieces might also come in handy.

I'd say it's better to err on the side of memorization. You won't have to be so concerned about making silly errors in your derivations of well-known formulae if you simply memorize them. Also, if you find that time is a big factor for you (i.e. you can't usually finish the competition on time), then you'll save yourself a lot of trouble by knowing stuff rather than wasting time figuring it out on your own. The professor who teaches Putnam seminar here keeps telling us that we should get through as much as possible in the seminars so that we can be dumb on the Putnam and still do well.

[quote]You won't have to be so concerned about making silly errors in your derivations of well-known formulae if you simply memorize them. [/quote]

But then you woudl have to be concerned about silly memorization kinks :/ I'd prefer the derivation if its simple (such as substituting sin^2 x + cos^2 x = 1 into cos(2x) = cos^2 x - sin^2 x to get cos(2x) = 1 - 2sin^2x = 2cos^2 x - 1). Of course, you probably don't want to be deriving the sum of a geometric series on a competition. IMHO, there's a fine (or not so fine) line between what you should memorize and what you should just derive when necessary.

What is IMHO?

The best solution will be to put everything on the calculator (or get a program with all the formulas) but that wouldn't work for competitions which bans calculators...

I was quite surprised to see that you (ComplexZeta) memorize formulas...I thought you'd be able to derive every single geometric formula within a minute :)

I have a lot of formulae memorized, but most of them weren't memorized on purpose. More or less anything that I have seen that I don't think I can derive without risk of making a mistake I have memorized. I have made a ridiculously large number of arithmetic mistakes on math contests.


[quote="Wumbate"]Prove that for any integer $m>0$ there exists integers a, b such that the following is true:
$0<a+b\sqrt{2}\leq\frac{1+\sqrt{2}}{m+2}$

This problem also comes from the Art and Craft of Problem Solving, under the pigeon hole chapter.[/quote]
The correct problem statement (as given in "The Art and Craft of Problem Solving") is as follows.

Prove that for any integer $m>0$, there exist integers $a,\ b$ such that the following are true:
$0<a+b\sqrt{2}\leq\frac{1+\sqrt{2}}{m+2},\ |a|\leq m,\ |b|\leq m$.

[hide]

Lemma: There are at least  ordered pairs  such that  Proof : Suppose we have some fixed element . Then a shift from  to  will shift the value of that number by only . However, our good region is of length , so we will have at least one  such that the inequality is true. Thus, for any , there is at least one  such that  and since , we have  's, and thus at least  unique ordered pairs  that satisfy the inequality.  for , so for the case  we can manually check that  works. For all , our lemma is true.

Now, partition the region  into  regions of equal length, so we have the regions  We have  regions, and at least  elements within these regions. By the pigeonhole principle, at least two elements must be in one of these regions of length  . Let these two elements be  and . The maximum difference between these two is . If we take the positive difference between these two, we get a new element  which satisfies all the original inequalities, and thus we have found our desired ordered pair .

[/hide]


Well, I threw it into Mathematica, and the solutions are far from nice. You'd have to solve the equation $ x^3 \minus{} 2x^2 \minus{} 7x \plus{} 6 \equal{} 0$, which doesn't have a rational solution.

however it is interestesting to see when you use the approximation (I used Wolfram for iPhone), you get 3.5079, a number in R, and that fits as a solution
Here is an approx
-2.26354 -> 0.75564 or
>3.5079


Consider the equation $ y^2\minus{}k\equal{}x^3$, where $ k$ is an integer.

Prove that the equation cannot have five integer solutions of the form
$ (x_1,y_1),(x_2,y_1\minus{}1),(x_3,y_1\minus{}2),(x_4,y_1\minus{}3),(x_5,y_1\minus{}4)$.

Also show that if it has the first four of these pairs as solutions, then $ 63|k\minus{}17$.

Such equations are called [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curves[/url].

Thanks for the information!

But still, nothing quite related to the question.

I start with the second part of the solution, which is a casework.

Let the first four be the solutions. Note that $x^3\in\{-1,0,1\}\pmod{7}$, for every $x$. We analyze the cases based on $y_1\pmod{7}$. Now, let $y_1\equiv 0\pmod{7}$. Since $y_1-1\equiv -1\pmod{7}$, thus $k\in\{0,1,2\}\pmod{7}$. Using now $y_1-2$, we get neither of these cases are possible. Thus if there are two such pairs whose second coordinates are $-1,-2$ in modulo $7$, the system is impossible to have a solution. This automatically eliminates the cases other than $y_1\equiv 5\pmod{7}$. In that case, $y_1-3\equiv 2\pmod{7}$, thus $k\in\{3,4,5\}\pmod{7}$. Now, $y_1-2\equiv 3\pmod{7}$, thus $k\not\equiv 4,5\pmod{7}$. Hence, $k\equiv 3\pmod{7}$ is indeed the case. Modulo $9$ can also be handled analogously.

Now, suppose that the equation admits five such pairs as it's solutions. As we have established already, it follows that $y_1\equiv 5\pmod{7}$. But then, $y_1-4\equiv 1\pmod{7}$, and $k\equiv 3\pmod{7}$, which then yields $x^3\equiv -2\pmod{7}$, not possible. Thus, the given equation cannot admit five solutions whose second coordinates are consecutive, as requested.


logarithms
pigeonhole principle
number theory
relatively prime
algebra theorems
algebra

1)Prove that the sequence $ \{na_n\}$ for all $ n\ge 1$ is dense in $ [0;1]$ if $ a$ is an irrational humber .
2) Please give me some applications

1) Suffices to show that we can find an $ n$ for which $ 0 < \{ na \}$ is arbitrarily small, because we get all multiples of it. First, we show given $ a_n$ there is an $ n$ such that $ \{na\} < \{a\}/2$. Take the first $ m$ for which $ \{ma\} < \{(m\minus{}1)a\}$ (so we had a "carry"). Then $ \{ma\} < \{a\}$. If $ \{ma\} < \{a\}/2$ we're done with $ n \equal{} m$. Otherwise as long as $ \{(1\plus{}k(m\minus{}1))a\}$ is positive, $ \{(1\plus{}(k\minus{}1)(m\minus{}1))a\} \minus{} \{(1\plus{}k(m\minus{}1))a\} \equal{} \{a\} \minus{} \{ma\} < \{a\}/2$. Take the maximum $ k$ such that $ \{(1\plus{}k(m\minus{}1))a\}$ is positive, then that means $ \{(1\plus{}k(m\minus{}1))a\} < \{a\} \minus{} \{ma\}$ and $ n \equal{} 1\plus{}k(m\minus{}1)$ works. Now just replace $ a$ with $ na$ and repeat ad infinitum.

2) One of many: If $ a$ is relatively prime to $ 10$ and $ d_1d_2 \ldots d_k$ is a sequence of digits, there are infinitely many powers of $ a$ that begin with $ d_1d_2 \ldots d_k$.

[quote="MellowMelon"]
2) One of many: If $ a$ is relatively prime to $ 10$ and $ d_1d_2 \ldots d_k$ is a sequence of digits, there are infinitely many powers of $ a$ that begin with $ d_1d_2 \ldots d_k$.[/quote]

How could you prove that? :?:

Alternatively you can use the Pigeonhole Principle. To prove that for every $N$ there exists an $n$ for which $\{na\}<1/N$, we break the unit interval $[0,1)$ into $N$ equal intervals $[(i-1)/N,i/N)$ for $i=1,2,\dots,N$. Since $a$ is irrational, no two $\{na\}$ can be equal, so we get infinitely many of these in the unit interval. By the Pigeonhole Principle, there must be two, say $\{an\}$ and $\{am\}$ which lie in the same interval. Then $\{a(m-n)\}<1/N$. Since the choice of $N$ was arbitrary and all the multiples of $\{a(m-n)\}$ must all lie in the unit interval, these numbers must be dense in $[0,1)$.


logarithms
inequalities
inequalities open

Let, $ f(x)\equal{}\ln (x^2\plus{}x\plus{}1)$, and Jensen ineq.

[quote="Kouichi Nakagawa"]Let, $ f(x) \equal{} \ln (x^2 \plus{} x \plus{} 1)$, and Jensen ineq.[/quote]
I have obtained $ f''(x) \equal{} \frac {1 \minus{} 2x \minus{} 2x^2}{(x^2 \plus{} x \plus{} 1)^2}.$ :maybe:
But an estimation $ \sum_{cyc}\left(3\ln3\minus{}\ln(a^2\plus{}a\plus{}1)\plus{}a\minus{}1\right)\geq0$ gives the proof. :wink:

[quote="AngelHolder"]$ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ (a^2 \plus{} a \plus{} 1)(b^2 \plus{} b \plus{} 1)(c^2 \plus{} c \plus{} 1) \le 27$[/quote]
Well
$ a^2b^2c^2 \plus{} abc(ab \plus{} bc \plus{} ca \plus{} 4) \plus{} \sum a^2b^2 \plus{} \sum a^2(b \plus{} c)\plus{}\sum a^2 \plus{} \sum ab \plus{} 4 \leq 27$
We have
$ \sum a^2b^2 \plus{} 6abc \leq 9$
$ \sum a^2\plus{}2\sum ab\equal{}9$
at last we have
$ 5\plus{}abc \plus{}ab\plus{}bc\plus{}ca \geq 3(ab\plus{}bc\plus{}ca)$
may be true !  :)


search
real analysis
real analysis unsolved

Show that $ \hat{\mathbb{Q}}$ is a complete ordered field.
Note: $ \hat{\mathbb{Q}}$ is the Cauchy sequences in $ \mathbb{Q}$ modulo the equivalence relation of being co-Cauchy.

Are you in the same class as this guy?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=563369982&t=312516

:maybe:
Maybe, I am not sure.
But that article is not very clear. Do you mind posting a step by step solution.

Thanks.


Let $x,y,z$ be positive real numbers. Find the greatest $p$ such that
\[\frac23\Bigl(\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}\Bigl)\geq\sqrt[p]{\frac{x^{p}+y^{p}+z^{p}}3}.\]

The following inequality is true and known:
let $x,$ $y$ and $z$ are positive numbers such that $x^{4}+y^{4}+z^{4}=3.$ Prove that:
$\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}\geq\frac{3}{2}.$
Thus, $p\geq4.$

Could anyone send a solution for $ p\equal{}5$?


Is it just me who realizes it ? Or is it tradition to include 1-2 Kvant problems in the Romanian TSTs each year ? You can find the problems at: http://kvant.mccme.ru/ The problems marked with an asterix are the more challenging problems.  :)

Well, there were so much problems published at Kvant (remember the magazine was founded in 1970 and is still alive!) that I can hardly imagine anybody systematically solving all of them... anyway, if anybody would do this, then I guess he would have no problems on any TST...

  Darij

[quote="darij grinberg"]Well, there were so much problems published at Kvant (remember the magazine was founded in 1970 and is still alive!) [/quote] the mathematical gazzete in Romania was founded over in the 19th century, and it is still very much alive! and it had an un-interrupted existance. that is incredible, isn't it? I belive that only Komal is older than MG.

[quote="darij grinberg"]Well, here in Germany, mathematical magazines have real existence worries... "Alpha" is dead, "Wurzel" has financial problems, about "Monoid" I don't know but they have little publicity; it is for that reason I find it incredible when a math magazine survives a century. BTW, with "the mathematical gazzete in Romania" you mean the "Gazeta Matematica"?
Darij[/quote]

Hi Darij,

I cannot evaluate ALPHA but I want to state my opinion regarding WURZEL ( http://www.wurzel.org/ ) and MONOID ( http://www.mathematik.uni-mainz.de/monoid/ ). WURZEL has a long tradition (I have all volumes, copied just the interesting parts), it has mostly a medium level and it publishes some nice and interesting articles sometimes. MONOID is rather for beginners, meaning easy level. It rather addresses students from low grades. It can be used for gettting started...  :)

Some forum links for WURZEL problems (mostly I used those problems from the journal which are more challenging):

- http://www.mathlinks.ro/viewtopic.php?t=5009
- http://www.mathlinks.ro/viewtopic.php?t=4976
- http://www.mathlinks.ro/viewtopic.php?t=4895
- http://www.mathlinks.ro/viewtopic.php?t=4991
- http://www.mathlinks.ro/viewtopic.php?t=5225
- http://www.mathlinks.ro/viewtopic.php?t=4986

Valentin,

you mentioned the Romanian "Gazeta" going back to the 19th
century. I believe that there are/were several different
series of this journal.  One is a high school journal with problems,
solutions and articles. The other serie(s) is/are at a higher level
and seems to be for research or university-level
articles in mathematics and physics.

Can you say more about these journals?  Does
the older university-level version still exist?


Let $ A$ denote a subset of the set $ \{ 1,11,21,31, \dots ,541,551 \}$ having the property that no two elements of $ A$ add up to $ 552$. Prove that $ A$ can't have more than $ 28$ elements.

$ 1\plus{}(n\minus{}1)10\equal{}551\implies n\equal{} 56$
Consider the following pairs $ 1,551; 11,541; 21,531;\dots$
each pairs sum is 552;there are total $ 56/2\equal{}28$ such pairs .
Now to construct $ A$,we can take only one element from each pair this gives maximum number of element in $ A$ is $ 28$


Two complete graphs, each of $ n$ vertices, are given such that a natural number is written at each vertex. Write along each edge the sum of the numbers written at its two endpoints. Suppose that $ n$ is not a power of $ 2$. Prove that, if the same numbers, with the same multiplicity, appear along the edges in the two graphs, then a similar statement is true for the vertices of the two graphs, too.

See (the links posted at) http://www.mathlinks.ro/viewtopic.php?t=240569 .

  darij


The radius of a cylindrical box is $ 8$ inches and the height is $ 3$ inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:

$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 5\frac {1}{3} \qquad\textbf{(C)}\ \text{any number} \qquad\textbf{(D)}\ \text{non \minus{} existent} \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
Since $ V=\pi r^2h$, we have $ \pi(r+x)^2h=\pir^2(h+x) \implies x=\frac{r^2-2rh}{h}$.  Then $ f=8,h=3 \implies x=5\frac{1}{3}$, or $ \boxed{\textbf{(B)}}$.
[/hide]

[quote="AIME15"][hide="Click for solution"]
Since $ V = \pi r^2h$, we have $ \pi(r + x)^2h = \pir^2(h + x) \implies x = \frac {r^2 - 2rh}{h}$.  Then $ f = 8,h = 3 \implies x = 5\frac {1}{3}$, or $ \boxed{\textbf{(B)}}$.
[/hide][/quote]Solution is correct, just a minor latex syntax typo: it should be $ \pi(r+x)^2h=\pi r^2(h+x)$


Three parallel lines are such that one passes through each vertex of the triangle $ABC$ . The line through $A$ meets $BC$ (extended if necessary)in $X$ and the lines through $B$ and $C$ meet $CA$ and $AB$ (extended if necessary) in $Y$ and $Z$ respectively .Prove that $(XYZ)=2(ABC)$  where (KLM) means the area of the triangle KLM .

WLOG assume $X$ is between $B$ and $C$. Then

$[AXZ]=[AXC]\qquad (1)$
(they share the base $AX$, points $C$ and $Z$ are equidistant from $AX$) and similarly

$[AXY]=[AXB]\qquad (2)$

Also

\begin{eqnarray*}\triangle ABY\sim\triangle AZC &\implies&{AB\over AZ}={AY\over AC}\\ &\implies& AB\cdot AC=AY\cdot AZ\\ &\implies&\frac{AB\cdot AC}{2}\sin\angle CAB=\frac{AY\cdot AZ}{2}\sin\angle YAZ\end{eqnarray*}
($\angle CAB$ and $\angle YAZ$ are opposite), hence

$[AZY]=[ABC]\qquad (3)$

Summing up $(1), (2), (3)$ we get

$[XYZ]=2[ABC]$

wlog assume $Z\in\overline{AB}$. $ABXY$ ia a trapezoid, with $[CXY]=[ABC]\ (\ 1\ )$. $AZCY, BXCZ$ are trapezoids and $[CYZ]=[AZC]\ (\ 2\ ), [CZX]=[CZB]\ (\ 3\ )$. But $[AZC]+[ZBC]=[ABC]\ (\ 4\ )$. Adding side by side the relations $(1),(2),(3)$, with $(4)$ we get the required relation, because $[CXY]+[CYZ]+[CZX]=[XYZ]$.

Best regards,
sunken rock


Let N be the product of all integers from 1 through 10 that are
not divisible by 5. What is the units digit of N?

So you're finding the units digit of $ 2*3*4*6*7*8*9$.

One way to do it:
2*3=6, 6*4=24, take the units digit 4.
4*6=24, take the units digit 4.
4*7=28, take the units digit 8.
8*8=64, take the units digit 4.
4*9=36, so your answer is 6.

Another way to do it:
2*3=6
7*8=56
9*4=36
6=6
Multiply a bunch of 6's, and the answer will always end in 6, so the answer is 6.


Today I solved this beautiful problem .
Find all numbers $n$ which have the form
$n=\frac{a^{2}+1}{b^{2}-1}$

I'm not sure if it is correct, because it seems to easy...
If $b$ would be odd than $b^{2}-1$ would be divisible by $4$. But $a^{2}+1$ is never divisible by $4$, contradiction.
If $b$ would be even than $b^{2}-1$ would be of the form $4k+3$ and it would have the prime divisor of the form $4t+3$. But then we wouldn't have $p|a^{2}+1$ since $p$ is of the form $4t+3$...

[quote="TomciO"]But then we wouldn't have $p|a^{2}+1$ since $p$ is of the form $4t+3$...[/quote]

You mean that $a^{2}+1$  can't have divisors of the form $4t+3$ ??? :maybe:

[quote="TomciO"]Yes, that's what I mean.[/quote]
Ok then it's also my solution ..... :lol:
Thank's for trying this problem

Can u give me more explanation about the $4t+3$ form and its uniqueness ??  :)

$a^{2}+1$ can't have divisors of the form $4t+3$. Indeed, if it would have one it would have, a _prime_ divisor of the form, let say $p$ (because every number of the form $4t+3$ has at least one prime divisor of this form, it's not hard to figure out why). So suppose that $a^{2}\equiv-1 \mod{p}$. It implies that: $(a^{2})^{\frac{p-1}{2}}=a^{p-1}\equiv 1 \equiv (-1)^{\frac{p-1}{2}}\mod{p}$.
So $\frac{p-1}{2}$ must be even, but that's impossible since $p$ is of the form $4l+3$!


[quote="Twins"]Find all $(m,k)$, so that $2!+3!+...+m!=2^{k}$[/quote]

[hide]k=1, m=2
k=2 doesn't work
k=3 doesn't work
Now look at the equation mod 16.
$2!+3!+4!+5! \equiv 8 \mod 16$
So the value is never divisible by 16 for any greater values.
So our only solution is (2,1)[/hide]


IMO 2008 is coming closer. Why not creating an own sub-forum for the IMO 2008 in Spain?

There is nothing to debate. Greenland is going to win.

  darij

What¿¿¿
The united penguins of antarctica will own them completely. Like last year!


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ A\in M_{n}(Q)$ and $ det(A \plus{} 2^\frac {1}{n}I_{n}) \equal{} 0$.Prove that
$ det(A \minus{} I_{n}) \equal{} det(A) \plus{} (det(A \plus{} I_{n}))^n$

$ \lambda\equal{}\minus{}2^{1/n}$ is an $ A$-eigenvalue. $ x^n\minus{}(\minus{}1)^{n}2$ is irreducible over $ \mathbb{Q}$. Thus $ x^n\minus{}(\minus{}1)^{n}2$ is the $ A$-characteristic polynomial; $ det(A)\equal{}\minus{}2$ and $ spectrum(A)\equal{}\{\minus{}2^{1/n}exp(2ki\pi/n),k\equal{}0\cdots{n\minus{}1}\}$. $ (x\plus{}1)^n\minus{}(\minus{}1)^{n}2$ is the $ A\minus{}I$-characpoly; then $ det(A\minus{}I)\equal{}\minus{}2\plus{}(\minus{}1)^n$. By a similar way we obtain $ det(A\plus{}I)\equal{}\minus{}2\plus{}1\equal{}\minus{}1$.   QED.


Let $ A\in GL_n(F)$. Prove or disprove that there exists $ r\in F$ such that $ \text{Per}(A \plus{} rI)\neq 0$, where $ \text{Per}$ is used to denote the [url=http://en.wikipedia.org/wiki/Permanent]permanent[/url] of a matrix.

PS. Currently I'm considering this problem in another context as well, where the permanent is replaced with the immanant. (See the same link, for the definition of the immanant of a matrix. Please don't ask me the reason for the difference between my spelling of immanant and that of [url=http://planetmath.org/encyclopedia/Immanent.html]PlanetMath[/url], since I don't know the reason.  :D  )

PPS. I'm not sure whether this problem has been discussed before in an article or a book such as [url=http://www.amazon.com/exec/obidos/ASIN/0521302269/ref=nosim/weisstein-20]this[/url]. If you know a source for it please let me know.

$ Per(A\plus{}rI_n)\equal{}r^n\plus{}\cdots\equal{}p(r)$ (a polynomial) has at most $ n$ roots in $ F$.
If $ |F|>n$ then there exists $ r\in{F}$ s.t. $ Per(A\plus{}rI_n)\not\equal{}0$.
If $ |F|\equal{}q\leq{n}$ and $ p(r)\equal{}0$ for all $ r\in{F}$ then $ r^q\minus{}r$ divides $ p(r)$.
In this last case I didn't find any counterexample for the Captain's proposition.

[quote="loup blanc"]$ Per(A \plus{} rI_n) \equal{} r^n \plus{} \cdots \equal{} p(r)$ (a polynomial) has at most $ n$ roots in $ F$.
If $ |F| > n$ then there exists $ r\in{F}$ s.t. $ Per(A \plus{} rI_n)\not \equal{} 0$.
If $ |F| \equal{} q\leq{n}$ and $ p(r) \equal{} 0$ for all $ r\in{F}$ then $ r^q \minus{} r$ divides $ p(r)$.
In this last case I didn't find any counterexample for the Captain's proposition.[/quote]Counterexamples can be chosen dependent on the characteristic of $ F$ where fields $ F$ which are not algebraically closed can serve as candidates.

$ char(F)\neq 2$:
Given $ d\in F$ such that $ 1\minus{}d$ is no square in $ F$, take $ A\equal{}\left(\begin{array}{cc}1&d\\1&\minus{}1\end{array}\right)$. Then $ Per(A\plus{}XI_2)\equal{}X^2\plus{}d\minus{}1$ has no zero in $ F$. If $ F$ is finite, we have $ [F^*: (F^*)^2]\equal{}2$ guaranteeing the existence of $ d$. For $ char(F)\equal{}0$ take $ F\equal{}\mathbb Q$ and $ d\equal{}2$.

$ char(F)\equal{}2$:
$ Per(XI_n\plus{}A)$ is equal to the characteristic polynomial of $ A$. Taking  a non-trivial nilpotent matrix $ A$, we have $ Per(XI_n\plus{}A)\equal{}0$ inducing the zero function.

Actually, $ A$ was chosen invertible. In characteristic 2, this means that the permanent of $ 0\cdot I\plus{}A$ is always nonzero.

Here's an explicit example with the permanents all zero, in the field with three elements:
$ \begin{bmatrix}1&1&0\\2&1&0\\0&0&1\end{bmatrix}$

That's as small as possible.

Oups.  :maybe: Thanks. But then we're done anyway since almost all my examples are non-singular.  :)

For $ {\rm char} F\not\equal{}2$ take $ B\equal{}\begin{pmatrix}1&1\\ \minus{}1&1\end{pmatrix}$ and $ C$ some triangular matrix with all of $ F\setminus \{0,2\}$ on the diagonal, but no $ 0$.
Then $ A\equal{}\begin{pmatrix}B&*\\ 0&C\end{pmatrix}$ is invertible with $ {\rm Per}(A\plus{}rI)\equal{}{\rm Per}(B\plus{}rI){\rm Per}(C\plus{}rI)\equal{}0$ for all $ r\in F$.

[quote="jmerry"]Third Edition: Your examples don't work, because you seem not to understand the question. The interesting question is whether there exist $ A$ making all of these permanents zero. Examples with none of the permanents zero are irrelevant.[/quote]Hmm, quite right. Apparently I had something else in mind.  :whistling:

Moreover, orolin's nice post implies that the analogue of Laplace's formula for the determinant is valid for the permanent. I was wondering if this is still true for the immanant.

Getting any nice formulas for the permanent depends on having a lot of zeros in the matrix, as our examples do. There aren't really any shortcuts for the calculation.

This is not what I was aiming at. I was looking for some structural evidence that the permanent respects decompositions of certain kind. I was thinking, e.g., of reducible representations.


function
complex analysis
complex analysis unsolved

There exists a family of distinct entire functions {f_a}_{a \in A}, with |A|>alef_0, but such that for each z \in C the set {f_a(x)}_{a\in A} is countable?

Yes if and only if the "hypothesis of the continuum" (hypothèse du continu in french) is true, and it is so undecidable.

:D

You can check the book named "proof from the book" for more details


geometry
circumcircle
trigonometry
geometry proposed

Let $ABC$ be a triangle, and let $C^{\prime}$ and $A^{\prime}$ be the feet of its altitudes issuing from the vertices $C$ and $A$, respectively. Denote by $P$ the midpoint of the segment $C^{\prime}A^{\prime}$. The circumcircles of triangles $AC^{\prime}P$ and $CA^{\prime}P$ have a common point apart from $P$; denote this common point by $Q$. Prove that:

[b](a)[/b] The point $Q$ lies on the circumcircle of the triangle $ABC$.
[b](b)[/b] The line $PQ$ passes through the point $B$.
[b](c)[/b] We have $\frac{AQ}{CQ}=\frac{AB}{CB}$.

  Darij

a) $AQPC' cyclic \Longrightarrow \angle{ABP} = \angle{BC'A'}$.
Analogous, $\angle{CQP} = \angle{C'A'B}$.

this implies that $\angle{AQC} = \angle{BC'A'} + \angle{C'A'B} = 180 - \angle{A}$.

b)$\angle{AC'C} = \angle{AA'C} \Longrightarrow AC'A'C$ is cyclic, that implies $BC' . BA = BA' . BC$.
so, the power of $B$ with relaction to circuncircles of $AC'P$ and $A'CP$ is the same.
so, $B$ lies on the radical axes $PQ$

c) by the sine law on $ABQ$ we have $AQ/AB = sin(AQP)/sin(ABQ) = sin(BC'A')/sin(C'BP)$.
analogous, $QC/BC = sin(BA'C')/sin(A'BP)$.

we know that in any triangle $ABC$
with one cevian $AD$ we have the relaction $AB/AC = AD/CD . sin(BAD)/sin(CAD)$.

$P$ is the midpoint of $A'C'$, and the conclusion follows (because we will have $BC'/BA' . sin(C'BP)/sin(A'BP) = 1$)  :)

E.L


geometry
cyclic quadrilateral
geometry proposed

Find the radius $R$ of the circle of a cyclic quadrilateral of sides $1,2,3,4$ using only Pytagora theorem.

I'm not that smart on here :( Probably one of the lowest member since I'm getting starter but I'll give this a shot.

I don't have an answer so I won't put in spoiler. Since the sides of cyclic quadrilateral are 1,2,3,4, I added them and divide them by 10 from 360 to get the measures of four angles: 36,72,108,144. And I know that on the cyclic quadrilateral, opposite angles are supplementary so 36-144 and 72-108. (- denote the facing). But now I'm stuck. :(

Can you explain your answer?

I'd like to see it.


analytic geometry
geometry
circumcircle
inequalities
Pythagorean Theorem
algebra unsolved
algebra

Two sequences of integers, $ a_1, a_2, a_3, \ldots$ and $ b_1, b_2, b_3, \ldots$, satisfy the equation
\[ (a_n \minus{} a_{n \minus{} 1})(a_n \minus{} a_{n \minus{} 2}) \plus{} (b_n \minus{} b_{n \minus{} 1})(b_n \minus{} b_{n \minus{} 2}) \equal{} 0
\]
for each integer $ n$ greater than $ 2$. Prove that there is a positive integer $ k$ such that $ a_k \equal{} a_{k \plus{} 2008}$.

I'll just post my solution in short notes ;).

Let $ x_i\equal{}a_{i\plus{}1}\minus{}a_i$ and $ y_i\equal{}b_{i\plus{}1}\minus{}b_i$. Then the condition from the task is equivalent to: $ (x_n\plus{}x_{n\minus{}1})^2\plus{}(y_n\plus{}y_{n\minus{}1})^2\plus{}x_n^2\plus{}y_n^2\equal{}x_{n\minus{}1}^2\plus{}y_{n\minus{}1}^2$
From this we obtain $ x_n^2\plus{}y_n^2 \leq x_{n\minus{}1}^2\plus{}y_{n\minus{}1}^2$ which means the some of squares of the differences is decreasing or staying equal(at start it is equal to $ x_1^2\plus{}y_1^2$). If $ x_{n\minus{}1}^2\plus{}y_{n\minus{}1}^2\equal{}x_n^2\plus{}y_n^2$ then by combining it with the original condition we easily obtain $ x_n\equal{}\minus{}x_{n\minus{}1}$ and $ y_n\equal{}\minus{}y_{n\minus{}1}$. From which follows $ a_{n\minus{}1}\equal{}a_{n\plus{}1}$. So we take first 2008 $ a_i$s and we conclude: if the $ x_i^2\plus{}y_i^2$ is not decreasing then we have $ a_1\equal{}a_{2009}$; if not then $ x_i^2\plus{}y_i^2$ decreased atleast by 1($ x$ and $ y$ are integer rays). So we continue by taking next 2008,,,, etc. By final amount of steps($ x_1^2\plus{}y_1^2$ is final) we will come to a moment where $ x_i\equal{}y_i\equal{}0$, from that point on, all members of $ x$ and $ y$ are $ 0$, so all members of $ a$ and $ b$ are equal and we can get what we wanted.

The end

In coordinate plane let point $ P_n\equal{}(a_n,b_n)$ and the given condition implies that $ P_n$ is on the circumcircle whose diameter is segment $ P_{n\minus{}1}P_{n\minus{}2}$. Denote by $ d_n$ the square of the distance of $ P_n$ and $ P_{n\plus{}1}$, namely $ d_n\equal{}|P_{n}P_{n\plus{}1}|^2$. Clearly $ d_n$ is non-increasing and $ d_n$ is an non-negetive integer. So take $ n$ to infinite we get that there exists $ n$ such that either $ 0\equal{}d_n\equal{}d_{n\plus{}1}\equal{}\ldots$ or $ 0<d_n\equal{}d_{n\plus{}1}\equal{}\ldots$, that is, $ P_{n}\equal{}P_{n\plus{}1}$ or $ P_{n}\equal{}P_{n\plus{}2}$. Either implies $ P_k\equal{}P_{k\plus{}2008}$.

Slightly more detailed version of the above.
[hide="Solution"]Consider the points $X_n=(a_n,b_n)$ in the coordinate plane. The condition is equivalent to stating that $X_nX_{n-1}$ is perpendicular to $X_nX_{n-2}$. Geometrically, this means that $X_nX_{n-1}X_{n-2}$ is (possibly degenerate) right triangle with right angle at $X_n$. If the triangle is non-degenerate, then we have the inequality \[{X_nX_{n-1}}^2+1\le {X_{n-1}X_{n-2}}^2\] This follows from the Pythagorean theorem and the fact that the smallest possible value of $X_nX_{n-2}$ for distinct lattice points $X_n$, $X_{n-2}$ is $1$. Note that $X_kX_{k-1}$ is bounded below since it's nonnegative, so it must be the case that the triangle is degenerate infinitely many times. In this case, $X_n$ must remain constant for infinitely many consecutive $n$, which implies the desired.  [/hide]

Let $s_n=a_n-a_{n-1}$ and $t_n=b_n-b_{n-1}$, so the initial condition becomes
\[
s_n^2+s_ns_{n-1}+t_n^2+t_nt_{n-1}=0
\]
Then by Cauchy-Schwarz
\[
(s_n^2+t_n^2)(s_{n-1}^2+t_{n-1}^2)\ge  (s_ns_{n-1}+t_nt_{n-1})^2=(s_n^2+t_n^2)^2
\]
Thus either $s_n^2+t_n^2\le s_{n-1}^2+t_{n-1}^2$ or $s_n^2+t_n^2=0$, in which case the first inequality holds anyway. As $s_n^2+t_n^2$ is now a monovariant, it must eventually converge to some value. If $s_n^2+t_n^2=0$ for all sufficiently large $n$, then $s_n=0$ for sufficiently large $n$, easily implying the desired results. So now assume that this is not the case; as $s_n^2+t_n^2$ converges to some positive value, equality must hold in each Cauchy-Schwarz, and $s_n=ks_{n-1}$, $t_n=kt_{n-1}$ for some $k$ and each $n$. But then $s_n^2+t_n^2=s_{n-1}^2+t_{n-1}^2$ implies $k^2=1$, and the initial condition implies $k^2+k=0$, so $k=-1$. Thus $s_n=-s_{n-1}$ for all sufficiently large $n$, yielding that $a_k$ eventually periodic with period $2$.

The key observation is that

[b]Lemma:[/b] For $A(x_1, y_1)$ and $B(x_2, y_2)$ in the plane, the equation of a circle with diameter $AB$ is given by $$C(x, y) \overset{\text{def}}{:=} (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.$$

(Proof) Follows from the fact that $AC \perp CB$ so $m_{AC}\cdot m_{BD}=-1.$

Define the sequence of points $\left(P_n(a_n, b_n)\right)_{n \ge 1}$ and observe that $P_{n+2}$ lies on the circle with diameter $P_nP_{n+1}$. It follows that the sequence $d_n=|P_nP_{n+1}|^2$ of positive integers is non-increasing, so for all large $n$ we have $P_{n+2} \in \{P_n, P_{n+1}\} \Longrightarrow P_k=P_{k+2008}$ for all large $k$ since the sequence is either eventually constant, or periodic with period $2$.

[b]Remark:[/b] Studying for boards isn't all that bad after all :D

define $x_n=a_n-a_{n-1} $ and $y_n=b_n-b_{n-1}$
then the given condition equivalent to $x_n^2+y_n^2+x_n.x_{n-1}+y_n.y_{n-1}=0 $ which is equivalent to $(2x_n+1)^2+(2y_n+1)^2=2 \implies |2x_n+1|=1$ then $x_n=0$ or $x_n=-1$
then $a_n=a_{n-1}$ or $a_n=a_{n-1}-1$ which implies $a_n=a_{n+2008} $ for a large enough $n$
and we win :D


Let $ u$ and $ v$ be positives such that $ \frac{u}{v}\le\frac{1}{2}$.
Prove for any positive $ a$, $ b$ and $ c$ that

\[ \frac{c}{bu\plus{}cv}\plus{}\frac{a}{cu\plus{}av}\plus{}\frac{b}{au\plus{}bv}\le\frac{3}{u\plus{}v}\]

 :)

[quote="ElChapin"]Let $ u$ and $ v$ be positives such that $ \frac {u}{v}\le\frac {1}{2}$.
Prove for any positive $ a$, $ b$ and $ c$ that
\[ \frac {c}{bu \plus{} cv} \plus{} \frac {a}{cu \plus{} av} \plus{} \frac {b}{au \plus{} bv}\le\frac {3}{u \plus{} v}
\]
:)[/quote]
$ \sum_{cyc}\frac {a}{cu \plus{} av}\leq\frac {3}{u \plus{} v}\Leftrightarrow\sum_{cyc}\frac {av}{cu \plus{} av}\leq\frac {3v}{u \plus{} v}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(1\minus{}\frac {av}{cu \plus{} av}\right)\geq3\minus{}\frac {3v}{u \plus{} v}\Leftrightarrow\sum_{cyc}\frac{c}{cu \plus{} av}\geq\frac {3}{u \plus{} v}.$
But $ \sum_{cyc}\frac{c}{cu \plus{} av}\equal{}\sum_{cyc}\frac{c^2}{c^2u \plus{} acv}\geq\frac{(a\plus{}b\plus{}c)^2}{(a^2\plus{}b^2\plus{}c^2)u\plus{}(ab\plus{}ac\plus{}bc)v}.$
Thus, it remains to prove that $ \frac{(a\plus{}b\plus{}c)^2}{(a^2\plus{}b^2\plus{}c^2)u\plus{}(ab\plus{}ac\plus{}bc)v}\geq\frac {3}{u \plus{} v},$
which equivalent to $ (v\minus{}2u)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)\geq0.$


If for example, two people have 44, how do they determine who wins? I know it takes sprint score first, but if they have the same sprint score, what happens?

Also, what if there are two perfect scores?

Hey, that happened to me!  :D

We took a tiebraker test, which is basically a two question test that tests your speed and precision.  Whoever gets the most right is first priority, I think, and then it is who finished first.

In my chapter, the people I faced and me all got 1 out of 2, but I FINISHED FIRST!!! :first:
[color=blue]Moderator kyyuanmathcount says: Seems to be a descrepancy with different sources,eh?  Lets not get into any of this chapter stuff, you're both going to states  :) [/color]

At our chaper their to lazy for it, they pick a random question, whoever gets it right wins....... grrrrrrrrrrr i always lose

*shivers*
I hate tiebreakers.
At my chapters, there was a tie of 3 for first place (perfect), tie of 2 for 4th place, and tie of 3 for 6th place.
Each of the groups who didn't get perfect had missed the same question(s).
I missed both.
And if this is too much info... delete this, I guess.

Seirously though, like what if two people get 44? After taking the sprint score, what do they do?

i seriously think thats how they did it at our chapter............. stupid people they hate me

If I remember correctly, the tiebreaker procedure for individuals is:
1. Sprint score
2. Score on the last 10 sprint problems (I'm not sure about this one, it might go straight to #3)
3. The last 10 sprint problems, taken in reverse order (compare 1 at a time)
4. Target round problems in reverse order
5. Middle 10 problems on sprint in reverse order
6. First 10 problems on sprint in reverse order
7. Tiebreaker round

Thus if two people had 44, with the same score on sprint, but one answered #25 and #28 wrong, while the other answered #29 and #30 wrong, the first one would win on step 3.

As robinhe described, the tiebreaker test is basically another target round set, except I don't think calculators are allowed.  It's only supposed to be used if two people get exactly the same set of questions right.  robinhe's description of the scoring is correct; it's whoever can get the most in the fastest time.

I don't know what happens if no one gets either question right; presumably in this case they would do the same thing as ARML tiebreakers and go to another tiebreaker round.  There are two sets of tiebreaker questions.  If this is the case and no one got any of the four questions right, I have no idea what they would do.

By the way, the team tiebreaker is:
1. Team round score
2. Average of sprint scores
3. Team round problems in reverse order
4. Tiebreaker round

The procedure for the team tiebreaker round is essentially the same as that for individuals.


geometry perimeter calculus derivative function algebra domain

Farmer Brown has 180 feet of fence and wants to enclose a rectangular pen, which adjoins to a side AB of his barn. AB is 100 feet long. (The bold black lines total 180 feet.) What should the dimensions of the pen be for maximum area?

So here's what I did: I said the total perimeter of the pen is 100+180=280 feet, so the short side is x feet and the long side is 140-x feet. (And BC is 40-x feet.)

So the area is x(140-x)=140x-x^2
The derivative of the area is 140-2x which equals 0 when x=70
x=70 is a global max of the area function because the second deriv of the area is -2

But the problem is... BC=40-x and if x=70 then BC would be negative... What did I do wrong?

Thanks everyone.

[quote="jmerry"]The total perimeter is not 280- that's only true in one configuration, which manages to use the whole barn side. The maximum actually comes with only part of the barn's side used.[/quote]
This problem asks for the maximum in the case that all 100 feet of the barn are be used - hence the included diagram.

[quote="jmerry"]I didn't see the diagram earlier. OK, in this case the maximum comes at a boundary, where $ B \equal{} C$. You always have to watch for the boundaries in optimization problems.[/quote]

Oh, right, I had forgotten about the boundaries. I should have first noted restrictions on x... $ 0 \le x \le 40$, and because A''(x)=-2 and A(x)=0 when x=70, A(40) yields the largest value out of the domain, so the dimensions are 40 by 100.

I actually had the same issue in another problem, and now I see the importance of always keeping track of restrictions. Thanks Jmerry.


The six edges of tetrahedron $ ABCD$ measure $ 7$, $ 13$, $ 18$, $ 27$, $ 36$, and $ 41$ units.  If the length of edge $ AB$ is $ 41$, the the length of edge $ CD$ is

$ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 36$

So the point is that each triangle has to have a+b>c where c is the longest side. Each line segment is a part of two triangles, so you must find two ways to fulfill it. Starting with 7, the only two pairs that work with it are 7,13,18 and 7,36,41. So we have to find what CD is from AB. CD is the opposite of AB, so it will be the only line that doesn't share a triangle with AB. For example, the line opposite to 7 was the 27 one. So 27, 36, and 7 are not opposite of 41. The only number that goes with 27 and 41 that has not already been used is 18, giving you a 27-18-41 triangle along with the earlier 7-36-41 one. Since 13 is the only side not included in the 41 triangles, it is CD. The answer is B.


Are you really going to use that formula to help on competitions?  Not saying I wouldn't want to.

[quote="nutz_for2.718281828"]I have decided to make the cubic formula better! Now we can use $ax^3+bx^2+cx+d=0$! (of course, divide by a) But I have denoted u and v in the cubic equation as e and f, respectfuly. I have found out that
$e=\frac{\frac{b^2-3ad}{3a^2}\pm\sqrt{\frac{9a^2d^2-6ab^2d+b^4}{9a^4}-\frac{12ab^2-4d^3-36abd}{27a^3}}}{2}$
I can't get a lot farther without plunging into a lot of algebra.
My calculator also doesn't have a $\pm$ button, either. How can I program it???[/quote]

Where's the $f$?

If you want to have a plus or minus in a formula for a program, then all you need to do is have the calculator output two different statements. In the first, put + and in the second put -. That way it will output both answers, the one with plus and the one with minus.

[quote]Or, you know, you could graph the equation and find the zeroes...

/smack[/quote]

Haha, yeah. That formula looks like it would take much more time to punch out an answer than just solving the equation graphically or algebraically.


calculus
derivative
trigonometry
calculus computations

([b]4[/b]) Let $ f(x) \equal{} \sin^6\left(\frac {x}{4}\right) \plus{} \cos^6\left(\frac {x}{4}\right)$ for all real numbers $ x$. Determine $ f^{(2008)}(0)$ (i.e., $ f$ differentiated $ 2008$ times and then evaluated at $ x \equal{} 0$).

[hide]$ \cos^6x\plus{}\sin^6x\equal{}(\cos^2x\plus{}\sin^2x)(\cos^4x\minus{}\cos^2x\sin^2x\plus{}\sin^4x)$

$ \equal{}\cos^4x\plus{}2\cos^2x\sin^2x\plus{}\sin^4x\minus{}3\cos^2x\sin^2x$

$ \equal{}1\minus{}\frac34\sin^2(2x)$

$ \equal{}1\minus{}\frac38(1\minus{}\cos(4x))\equal{}\frac58\plus{}\frac38\cos(4x)$

So, $ \cos^6\left(\frac{x}4\right)\plus{}\sin^6\left(\frac{x}4\right)\equal{}
\frac58\plus{}\frac38\cos x.$

Given that identity, the $ 2008$th derivative at zero is $ \boxed{\frac38}.$[/hide]

[hide="Solution"]

$ \sin ^ 6 \theta \plus{} \cos ^ 6 \theta \equal{} (\sin ^ 2 \theta \plus{} \cos ^ 2 \theta)^3 \minus{} 3\sin ^ 2 \theta \cos ^ 2 \theta (\sin ^ 2 \theta \plus{} \cos ^ 2 \theta)$
$ \equal{} 1 \minus{} 3\sin ^ 2 \theta \cos ^ 2 \theta$
$ \equal{} 1 \minus{} \frac {3}{4}\sin ^ 2 2\theta$
$ \equal{} 1 \minus{} \frac {3}{4}\cdot \frac {1 \minus{} \cos 4\theta}{2}$
$ \equal{} \frac {1}{8}(5 \plus{} 3\cos 4\theta)$
Since $ \frac {d^n}{dx^n}\cos x \equal{} \cos \left(x \plus{} \frac {n\pi}{2}\right)\ (n \equal{} 0,\ 1,\ 2,\ \cdots)$,we have $ f^{(2008)}(x) \equal{} \frac {3}{8}\cos x$.

$ \therefore f^{(2008)}(0) \equal{} \frac {3}{8}$.[/hide]


Solve the system:
i) x 2 +y 2 +z 2 =1
ii)x^3+y^3+z^3-3xyz=-1
Generalize it!!! :)

Maybe if we let ii)x^3+y^3+z^3-3xyz=- \sqrt 2,my problem is more hard a little

Dear Thanhliem and thanhtrung {oh,it seem both of you are one}
Thanks for your solution. :)
I think you mean we must solve 3 grade equation.So, what about the second problem,you can't  guess solutions....

oh I don't see the second problem of you so I don't solve .I'm try  :?

O.K., guys, by solving the first equation, we get that a+b+c=-1 and ab+bc+ca=0, which is not enough to solve the problem. Could you help us, Anh?

Dear Voice_of_gods;that means our system have innumerable solution and
a+b+c=-1;ab+ac+bc=0 is the formula.Sorry about this mistake,maybe I must write down abc=1.But the second one doesn't need this condition. :)


abstract algebra
superior algebra
superior algebra theorems

If $A\rightarrow B\rightarrow C\rightarrow 0$ is an exact sequence of $R$-modules with maps $\alpha,\beta,z$ respectively, show that
$0\rightarrow Hom(C,M)\rightarrow Hom(B,M)\rightarrow Hom(A,M)$, with maps $z',\beta^{*},\alpha^{*}$ respectively, is exact.

I believe that I showed $\beta^{*}$ is injective and $im~ \beta^{*}\subset \ker\alpha^{*}$.
I am stuck on how to show the other containment.  Also, what about $\alpha^{*}$ being surjective.

Also, thank you, Amfulger and others, for all of the help!

Let $A\stackrel{\alpha}{\to}B\stackrel{\beta}{\to}C\to0$ be an exact sequence of left $R$-modules
For $M$ an arbitrary left $R$-module we have a sequence:
$0\to Hom_R(C,M)\stackrel{\beta^*}{\to}Hom_R(B,M)\stackrel{\alpha^*}{\to}Hom_R(C,M)$.
The arrows are $\beta^*(f)=f\circ\beta$ for any $f\in Hom_R(C,M)$ and $\alpha^*(f)=f\circ\alpha$ for any $f\in Hom_R(B,M)$.

1. We prove that $\beta^*$ is injective:
Let $f\in Hom_R(C,M)$. Then $\beta^*(f)=0\Leftrightarrow f\circ\beta=0\Leftrightarrow f(\beta(b))=0\ \forall b\in B$. Since $\beta$ is surjective by hypothesis, the last is equivalent to $f(c)=0\ \forall\ c\in C$ which is the same as $f=0$.

2. We now prove that the sequence is exact at $Hom_R(B,M)$:
For every $f\in Hom_R(C,M)$, we have $\alpha^*\beta^*(f)=\alpha^*(f\circ\beta)=f\circ\beta\circ\alpha=f\circ(\beta\circ\alpha)=f\circ0=0$. So $Im\beta^*\subset\ker\alpha^*$.
Let $f\in Hom_R(B,M)$. Then $f\in\ker\alpha^*\Leftrightarrow f\circ\alpha=0\Leftrightarrow Im\alpha=\alpha(A)\subset\ker f\Leftrightarrow \ker\beta=Im\alpha\subset\ker f$. Since $C\simeq B/\ker\beta$ and $f: B\to M$ vanishes on $\ker\beta$, by the universality property of the factor module, there exists an unique $\bar f: C\to M$ such that $f=\bar f\circ\beta$ i.e $f=\beta^*(\bar f)$. This means that $f\in Im\beta^*$, hence we also have a reverse inclusion to prove $Im\beta^*=\ker\alpha^*$.

3. $\alpha^*$ IS GENERALLY NOT SURJECTIVE. Not even if you ask $\alpha$ to be injective.


Given a circle $ C1$, and two non-intersecting circles, say C2 an C3,
internally tangent to C1. Let l1 and l2 be the internal tangents to C2
and C3. C1 and l1 intersect at A and B, C1 and l2 intersect at C and
D, such that A and C lies on the same side of the line joining the
centers of C2 and C3. Prove that AC is parallel to one of the external
common tangents to C2 and C3.

Take a look at: http://www.mathlinks.ro/viewtopic.php?t=15945

Thanks for the link. this question has bothered me for quite some time.

i used Thebault but did not chase angle...:D


algebra
function
domain
calculus
calculus computations


Uh....someone mind moving this thread ? :? It doesn't seem to fit in the games and fun factory.

[quote="Rahul Singh"]$ABCD$  is  a quadrilateral with no 2 sides parallel.
Let $AB$ and $CD$ meet at $E$  and $AD$ and $BC$ meet at $F$
Prove that the midpoints of $AC, BD , EF$ are  collinear.[/quote]

This is the famous Gauss theorem or Newton theorem or Theorem on the Complete Quadrilateral.

See http://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml , http://www.mathlinks.ro/Forum/viewtopic.php?t=15036 , http://www.mathlinks.ro/Forum/viewtopic.php?t=842 .

  darij

[quote="math92"]this should be in Getting Started i think not here[/quote]
I think they should be in the Geometry Theorem and Formula Section.

i am  sorry  for posting  the problem  in  the wrong place.
this  is my  first  time and  i wont  make  the  mistake  again
thank u


inequalities
calculus
integration
function
real analysis
real analysis unsolved

Jensen's inequality in integral form states that

$ \Phi \left( \int_D h(x) w(x) dx \right) \leq \int_D \Phi (h(x)) w(x) dx$

for each h: D \to I and each weight function $ w : D \to [0,\infty)$ such that

$ \int_D w \equal{} 1$.

My book says that

$ \int_b^{b \plus{} 1} \frac {dx}{x \plus{} y} > \frac {1}{b \plus{} 1/2 \plus{} y}$

and justifies that with Jensen's inequality.

Unfortunately I don't see that at all--i.e. what are \Phi, h and w here?

i think $ h(x)\equal{}x\plus{}y$, $ w(x)\equal{}1$, $ \phi (x) \equal{} x^{\minus{}1}$

?

[quote="kenn4000"]i think $ h(x) \equal{} x \plus{} y$, $ w(x) \equal{} 1$, $ \phi (x) \equal{} x^{ \minus{} 1}$

?[/quote]

Nice!


AMC AIME trigonometry algebra polynomial probability USA(J)MO

I have found it very difficult. What I do is to go through past contests and see if I can take the existing problems in new directions or at least find out which concepts I want to include in a problem. Also remember that it's usually easier to solve a problem you wrote since you can go through it in reverse.

[color=cyan]I similarly have found it difficult, but not impossibly so at all.  For one, do it with a group of people -- they will be able to give a better idea of how tough the problems really are.  I'm sure I could think of more to say, but nothing is really coming.  Maybe when other people say some more I'll have more constructive things to add.[/color]

Um, I'm not quite understanding what you mean...Do you mean a collection of problems, or guidebook to past problems? or other?

Was there a past discussion about this? Because "The Art of Problem Writing" isn't really that easy to understand...any help?

This thread is about how to write math problems.

Oh. It's easy for me since I make them up on my own, so I just write down what I got. Course, that may not be the best way since a lot of stuff can go wrong (not to mention you won't be able to post a lot at a time), but at least you'll be proud when someone solves your problem.

I'd say it depends on what level you're writing your problems.  AIME-level problems are especially difficult to write, in my opinion, because they generally have to include some hidden "trick" as opposed to straight plug-and-chug work.  I found writing algebra problems (when I was teaching my brother algebra over the summer) easy, but that sort of problem is [i]very[/i] different from more advanced problems.

i play around with math on my own a lot, especially identities (trig, combinatoric, polynomial) and also with inequalities. ocassionally i come up with a result that i find worthy to ask others to prove/deduce. however i guess this is not really problem writing, since i never know directly what im going to come up with. i wonder how people who actually write problems for contests, especially the olympiads, do this.

[color=cyan]I would bet it's not all that different, actually.[/color]

I've got +/- 10 pretty nice self - made problems.
I can post them if you're interested ...

Of course, we're interested.

Problem writing may be an even greater test of creativity than problem solving, and I think it is a great way to become a better problem solver.  In the course of writing 1 good problem, you'll probably think about 5-10 different problems (some too hard that you can't solve, some that turn out to be uninteresting and too easy).

Sam, Sandor and I would occasionally work in a group.  Some of our most difficult and interesting individual test problems came from Sandor making up a problem that none of us could solve, then Sam turning it into a slightly easier problem which could be solved.

I think many problem writers also start with an idea - they want to write a projection problem, a Cauchy problem, a Ceva problem, etc., and build the problem around the desired solution.  However, often the most interesting problems come from just playful investigation (a la tetrahedr0n).

I wonder how hard the problems Mr. Lehoczky (or is it "Ms."?) makes...
Can you post one in Advanced, just to see how difficult it is?

I'll post them tomorrow ... some are "intermediate", some are "advanced".  :)

:?: Are you Ms. Rubio? Just guessing...

If not, where can I get those problems?

uh...Arne I believe is a student...correct me if im wrong...

Here's one of Mr Lehoczky's problems

http://www.artofproblemsolving.com/wforum/viewtopic.php?t=41

Yup I'm a sixteen year old student ... :mrgreen:

Mmm...chalenging, yet possible...out of my hands, but maybe in a few years...

so you're in college, Arne?

I think I'm not ...
I'm not sure about your word "college".
But I'm still in secondary education -  I think that's called high school in America. So I'm NOT at university !

This is all a bit confusing. In Belgium a high school is called "college" and a college is called university I think - Am I right ? :?:

Oh, I get it now. You're a sophomore at something...

In Spanish, a high school is a "colegio" and a college is a "universidad," so it's something like that, I guess.

As someone who writes a lot of harder problems, I find that there are many different ways of coming up with them.  Sometimes you know what kind of problem you want to write, and so you just create one that is a little weird and see if it can be solved.  A couple of years ago, I needed a probability problem and came up with:

"Two people play a game in which they alternate turns and the probability of the player whose turn it is winning on turn n is n/(n+2).  What is the probability that the second player to take a turn ends up winning the game?  Note that the first player's first turn is turn 1, the second player's first turn is turn 2, etc."

which had a very interesting solution.  Other times, you might take an existing problem and extend it, perhaps by making a two-dimensional problem three-dimensional, or by changing the base in which calculations are done.  The way I like the best, however, is when problems simply occur to you when you're not even trying, such as while you're eating pizza, trying to think up a computer password, or driving through traffic signals.

Wow...AoPS sure has grown...

Anyways, welcome!

i thought i'd bring this topic back up..

lately, after having discovored TeX and realized there aren't any more USAMO chances left for me, i've spent a lot of time writing fairly difficult problems.. and it can definitely be satisfying when you write a solid, difficult problem.

it's so much easier to write a hard problem than to solve one, personally.. and that definitely improves problem solving abilities.

i'll probably post some later when i have more written


I need to make my ellipses go diagonally, but ddots makes them go the wrong way.  Is there a way to make the dots go diagonally the other way (slanting up)?

Would it be possible to include the package mathdots so that we can use \iddots for diagonal dots going the other way?

I don't suppose we can use \usepackage in here.

If the packages haven't been installed on the server, I don't see how we can do a \usepackage.

How do you get details about what each packet has, what commands come with them, and stuff like that?

[url=http://www.ctan.org/tex-archive/info/symbols/comprehensive/symbols-a4.pdf]Everything you ever wanted to know about [tex]\LaTeX[/tex] and then some can be found here.[/url]


find all n such that the number of numbers less than n and relatively prime to n
is equal to the number of divisors of n

[hide="My Attempt"]Since $ \phi$ and $ \tau$ are both multiplicative functions, it suffices to find the instances where a prime power satisfies the condition, that is, for a power $ p^e$, $ p^{e\minus{}1}(p\minus{}1)\equal{}e\plus{}1$. The exponent is much larger than $ e\plus{}1$, especially for large $ p$, so we need only consider the smaller cases. Trial and error leads to $ 2^3$ and $ 3^1$ as solutions. Another solution is then $ 24$. [/hide]


geometry
circumcircle
trigonometry
angle bisector
perpendicular bisector
geometry proposed

Let $ABC$ be a triangle at $A$ with $AB<AC;BC=2+2\sqrt{3}$ and its incircle has radius $1$.Determine the value of the angles $\hat{B}$ and $\hat{C}$

[hide]
Let $(O,R)$ be the circumcircle and $(I,r)$ the incircle

$AB<AC \Rightarrow  C<B \Rightarrow I$ is in the interior of $\triangle BOA$

$BI$ is the internal angle bisector of $\triangle BOA \ \ (1)$

$AI^{2}= r^{2}+r^{2}=1+1=2$

$R=2a \Rightarrow  R=\sqrt 3+1$

$OI^{2}= R(R-2r) = (\sqrt 3+1)(\sqrt 3-1)=(\sqrt 3)^{2}-1 = 2$

$AI=OI \Rightarrow I$ is on the perpendicular bisector of $OA$. From $(1)$ we get that $OB=AB$.

So $\triangle OAB$ is equilateral $\Rightarrow B=60^\circ \Rightarrow  C=30^\circ$[/hide]

[quote="friendlist"]$\left\{\begin{array}{c}\triangle ABC\ ,\ AB\perp AC\\\ a=2(1+\sqrt 3)\ ,\ r=1\end{array}\right\|$ $\implies$ $\{B,C\}=\left\{60^{\circ},30^{\circ}\right\}$.[/quote]
[color=darkblue][b][u]Proof.[/u][/b] $\left\{\begin{array}{c}b+c=a+2r\\\ b^{2}+c^{2}=a^{2}\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}b+c=2(2+\sqrt 3)\\\ b^{2}+c^{2}=8(2+\sqrt 3)\end{array}\right\|$ $\implies$ $\left\{\begin{array}{c}|b-c|=2\\\ b+c=2(2+\sqrt 3)\end{array}\right\|$ $\implies$

$\{b,c\}=\left\{3+\sqrt 3\ ,\ 1+\sqrt 3\right\}$ $\implies$ $\tan B=\frac{b}{c}\in \left\{\sqrt 3\ ,\frac{1}{\sqrt 3}\right\}$ $\implies$ $B\in\left\{60^{\circ},30^{\circ}\right\}$.

[b]Remarks.[/b]

$1\blacktriangleright$ From the well-known relation $\tan\frac{A}{2}=\frac{r}{p-a}$ obtain $A=90^{\circ}\Longleftrightarrow \tan\frac{A}{2}=1\Longleftrightarrow p-a=r\Longleftrightarrow b+c=a+2r$.

$2\blacktriangleright (b+c)^{2}+(b-c)^{2}=2\left(b^{2}+c^{2}\right)\Longrightarrow$ $|b-c|=\sqrt{2\left(b^{2}+c^{2}\right)-(b+c)^{2}}$ and $(b+c)^{2}\le 2\left(b^{2}+c^{2}\right)$.

$3\blacktriangleright\left\{\begin{array}{c}2\cdot\max \{b,c\}=b+c+|b-c|\\\ 2\cdot\min \{b,c\}=b+c-|b-c|\end{array}\right\|$ $\implies$ $\{b,c\}=\left\{\frac{b+c+|b-c|}{2}\ ,\ \frac{b+c-|b-c|}{2}\right\}$.[/color]


integration
real analysis
real analysis unsolved

I think $f'(x)$ sould be continuous ...  :?

$I=\int_{-1}^1 x^{2n + 1}f(x) dx =\int_0^1 x^{2n + 1}(f(x)-f(-x)) dx = \int_0^1 x^{2n + 2}\frac{f(x)-f(-x)}x dx$
Manifestly $\frac{f(x)-f(-x)}x$ is continuous , so $\exists a\in]0,1[$ such that $I = \frac{f(a)-f(-a)}a\frac 1{2n+3}$ ,
 so $\exists c\in]-a,a[\ \subset\ ]-1,1[$ such that $I = \frac 2{2n+3}f'(c)$
 :cool:  Do you remember mean theorem :cool:


Solve the equation $ x^{5} \plus{}8x^{4} \plus{}6x^{3} \minus{}42x^{2} \minus{}19x \minus{}2\equal{}0$.

[quote="AndrewTom"]Solve the equation $ x^{5} \plus{} 8x^{4} \plus{} 6x^{3} \minus{} 42x^{2} \minus{} 19x \minus{} 2 \equal{} 0$.[/quote]

My attempt:
[hide]Factor theorem implies that $ (x \minus{} 2)$ is a factor:

$ x^{5} \plus{} 8x^{4} \plus{} 6x^{3} \minus{} 42x^{2} \minus{} 19x \minus{} 2 \equal{} (x \minus{} 2)(x^4 \plus{} 10x^3 \plus{} 26x^2 \plus{} 10x \plus{} 1)$

So $ (x \minus{} 2)(x^2 \plus{} 4x \plus{} 1)(x^2 \plus{} 6x \plus{} 1) \equal{} 0$

$ x \equal{} 2$
$ x \equal{} \frac { \minus{} 4\pm\sqrt {16 \minus{} 4}}{2}$

so $ x \equal{} \minus{}2 \plus{} \sqrt {3}$ and $ x \equal{} \minus{} 2 \minus{} \sqrt {3}$

$ x \equal{} \frac { \minus{} 6\pm\sqrt {36 \minus{} 4}}{2}$

so $ x \equal{} \minus{}3 \plus{} 2\sqrt {2}$ and $ x \equal{} \minus{} 3 \minus{} 2\sqrt {2}$[/hide]

Thanks.

We can also see that $ x^{4} \plus{}10x^{3} \plus{}26x^{2} \plus{}10x \plus{} 1$ is a reciprocal equation. (I thinks that is what it is called - the coefficients are palindromic.) So we can substitute $ y\equal{}x \plus{} \frac{1}{x}$ to get $ y^{2} \plus{} 10y \plus{} 24 \equal{}0$ and so on.

[quote="AndrewTom"]Thanks.

We can also see that $ x^{4} \plus{} 10x^{3} \plus{} 26x^{2} \plus{} 10x \plus{} 1$ is a reciprocal equation. (I thinks that is what it is called - the coefficients are palindromic.) So we can substitute $ y \equal{} x \plus{} \frac {1}{x}$ to get $ y^{2} \plus{} 10y \plus{} 24 \equal{} 0$ and so on.[/quote]

Oh, that's neat. Never heard of those before!

I don't have elegent methods for these sorts of questions, my tactic is to first find any easy factors first, and then just go through the remaining polynomials term by term, trying to match up some more factors. It's ugly and probably time-wasting, but I haven't come across better methods. Polynomial division gives me nightmares.


I made state in California from the Berkely chapter, and I won the countdown round.  How did the rest of AoPS fare in the Mathcounts chapter competition?

[color=red]Moderator rcv's note:  Until further notice, you may not say anything more than Yes or No (Did you make State?), and "I ranked nth".[/color]

I sure did! I ranked 2nd in the Central Florida Chapter Competition. One of my teammates ranked 4th. My team overall ranked 2nd, but we would have made it in first IF THE OTHERS WOULD HAVE LISTENED TO ME!!! :mad: :mad: :mad: Anyway, we're going to the Florida State Competition on March 31st, 2006. I'm so psyched, lol! :lol: :lol:

I was 4th in the Seattle Chapter because I lost 2 stupid tiebreakers and one girl made everyone go down a spot since she went from 10th to 1st.

I got 1st!!! Both written and countdown :winner_first::winner_first::winner_first::winner_first:!!

I got first in written :D . I didn't do countdown.

I tied with 2 other people for first written, BUT I WON THE TIEBRAKER!!!!! :first: :first:

I won in both written and countdown :winner_first:  :winner_first:
And our team got first, which means I got 3 trophies, though we only had one person in the countdown(me).

i got first written and countdown with three easy questions

[quote="Naonao"][quote="Green Chicken"]I got 5th in written 1st in countdown and our team placed 3rd cuz we were stupido. The funny part is: our team won cyphering.[/quote]

pretty sure 99% of you guys out there don't know what cyphering is because its only done in the state of florida.[/quote]

Mathcounts ciphering ?! I've heard of ciphering competitions, but never in conjunction with Mathcounts.

[quote="Naonao"]Its just special in Florida. Its just like countdown round but for a team of 4 individuals. Each getting 4 questions.[/quote]

Sounds a bit like the special "Mathbee" round Louisiana State holds in addition to the Countdown Round.

I made state in the Berkely chapter, 5th in written and 1st in countdown.  (Seems like a lot of AoPsers are getting 1st in countdowns :D )

Why can't Delaware do anything cool :mad: :cursing:

im happy  i got 3rd in chapter countdown :winner_third:

I think my chapter graded something wrong

I clearly remembered every answers for my test and when I checked it with answer key after the Individual Round,

I got only 1 wrong (#30 ) in sprint and 1 wrong for Target (#8)

I was supposed to get first and called down to take Countdown but I didn't even get into top ten :(

their system was horrible anyway. they spent literally 3 hours assigning people to seats.

( they did the countdown on the overhead too)

so I didn't get in :mad:

[quote="shinwoo"]
I got only 1 wrong (#30 ) in sprint and 1 wrong for Target (#8)[/quote]
Ha, those are the EXACT ones I missed.  What a coincidence.

One of the kids in my school got those two and some more wrong but still managed to get 2nd place in Individual's while I wasn't even selected to top 10.... :(

That must have been really unfair.
btw- Do you get to go to states or not?

unless I sue them, I don't think I will

My chapter's director was a big idiot anyway

he was like 90 years old, and he was yelling at his workers right in front of us

Why don't you or your coach ask the chapter coordinator to regrade your test?  That's more constructive than complaining and calling people "big idiots".

[quote="sugarwergoingdown"]I got second in the individuals in our chapter, but i cant go to state cuz they had this internal tiebreaker system where they divided up the test into parts and "this part was worth more than that" etc junk. and our team somehow got 1ST even though i like beat alot of the members....they were so lucky
so yeah i cant go to state now because I was ranked 6th.... :wallbash_red:  :furious:[/quote]

If your team got 1st, shouldn't you be going with them?

He is probably an individual and didn't make it..... :( GOod luck nextx year

I was 3rd in Northern Nevada.  The stupid bell wouldn't ring.  My team placed first, though, and one of my teammates was 6th.

Yeah, I got first in the written round year this year and last year, but after hearing you guys, I can't believe I deserved it.  I thought that Virginia was extremely hard, but i won last year with a 37 and this year with a 43.  Our team got first and we're going to state.  How many people do they allow to go to state in your states? Only 6 in ours.  (Top team and top 2 individuals not on the team) :lol:


search calculus derivative function limit

How did the number e come about?
Please explain.
Thanks.

There is no point in copy-and-paste well-written materials. A search on Google using keywords "Euler's number" gives a lot of links, first being [url]http://en.wikipedia.org/wiki/E_(mathematical_constant)[/url]

Note: this story will make sense if you know the definition of the derivative.
Someone decided to take the derivative of the function $ f(x)\equal{}c^x$ where $ c$ is a constant, and got $ c^x*\lim_{x\rightarrow 0}\frac{e^x\minus{}1}{x}$. Note that $ \lim_{x\rightarrow 0}\frac{c^x\minus{}1}{x}$ is a constant given $ c$. So someone decided to prove that there existed a constant $ e$ such that $ \lim_{x\rightarrow 0}\frac{e^x\minus{}1}{x}\equal{}1$, which means $ \frac{d}{dx}[e^x]\equal{}e^x$. Then plugging in 1 into the Taylor Series of $ f(x)\equal{}e^x$ (with $ a\equal{}0$ [Wikipedia the Taylor Series]), you get $ \frac{1}{1}\plus{}\frac{1}{1!}\plus{}\frac{1}{2!}\plus{}\cdots$, which equals $ e$.

It's from Euler. It is used for interest problems that compound continuously.

ex.

ln 7.4 = 2

It is log natural

hello, see also here
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Exponential_Function.pdf
Sonnhard.


[quote="hash_include"]Problem 208:

Assertion: If a,b,c are all non-zero reals, then $ (b^2 \minus{} ac)^2 \plus{} (c^2 \minus{} bd)^2 \plus{} (ad \minus{} bc)^2 \equal{} 0$implies $ a,b,c,d$are in Geometric Progression.

Reason: $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 0$  implies x=y=z=0.

I would go for Assertion is right, reason is wrong, because nowhere have you mentioned that $ x,y,z$ are reals.  :wink:

But something tells me i'm wrong  :|[/quote]

Right you are harry. No post on sundays. I blindly ticked "A". just wanted to illustrate the shackles of familiarity that bind us.  :rotfl:  :rotfl:

:D
i too was "binded by the shackles of familiarity", till i recently learnt (the hard way of course :D), that for assertion reasoning questions, first check validity of both statements SEPARATELY, and then only the connection between them.  :oops:
In fiitjee speaking terms, check options C and D first, and then only come to A and B  :rotfl:

[quote="hash_include"]
that for assertion reasoning questions, first check validity of both statements SEPARATELY, and then only the connection between them.  :oops:
[/quote]

Harry, thank me for that :D
Rememer how much you argued?

PS 2200 posts yay! :D

[quote="achu_182"]i bet if it was in a FIITJEE AITS, then it would have been A.[/quote]

You are right. It wasn't. It came in Mathematics today. :)

[quote="madness"]
Harry, thank me for that :D
Rememer how much you argued?

PS 2200 posts yay! :D[/quote]

fine. thank you, your madness-ness, for your humble insights on the problems that the world faces today :D


Let $ABCDEF$ be a cyclic hexagon with $AB=CD=EF$. Prove that the intersections of $AC$ with $BD$, of $CE$ with $DF$, and of $EA$ with $FB$ form a triangle similar to $BDF$.

Let $M,N,P$ be those three points, in the order in which you have mentioned them, and let $O$ be the center of the circle.

We can now notice $\angle CMD=\angle CND=\angle COD$, so $D,M,N,O$ are concyclic. We do exactly the same for $F,N,P,O$ and $B,P,M,O$.

We have

$\angle MNP=\angle ONM+\angle ONP=\angle ODM+\angle OFP=\angle ODB+\angle OFB$
$=\angle OBD+\angle OBF=\angle FBD$.

In the same way we show that the angles of $MNP$ are equal to those of $FBD$, respectively.

[color=red][Moderator edit: Some minor typos corrected.][/color]


function
calculus
integration
real analysis
real analysis unsolved

Let $f:[0,1]\to\mathbb R$ be a derivable function s.t. $f'$ is continuous. It is known that there exists $x_0\in[0,1]$ s.t. $f(x_0)=0$. Prove that
$\int\limits_0^1f^2(x)dx\le4\int\limits_0^1\left(f'(x)\right)^2dx$.

[quote="Moubinool"]If $x_0<1$

For $x\geq x_0$

$f(x)=\int_{x_0}^{x}f'(t)dt$

With Cauchy-Schwarz

$f^2(x)\leq (x-x_0)\int_{x_0}^{x}f'^2(t)dt\leq x\int_{x_0}^{x}f'^2(t)dt\leq (\int_{0}^{1}f'^2(t)dt )x$
[/quote]

Did you noticed that is only true for $x \geq x_0$.

But this is true for all $x$:

$f^2(x) \leq \left| x - x_0\right| \int_0^1 (f'(t))^2\,dt \leq x \int_0^1  (f'(t))^2\,dt $


Consider a triangle that has a circumcircumle of radius $ 1$. Suppose that there are $ 11$ points inside that triangle. Prove that among these points, there are $ 2$ of them that are separated by a distance not greater than $ \dfrac{1}{2}$.

It's passed a month since this was posted. No ideas yet ?

There is a simple model that shows $ 10$ points are not enough: an equilateral triangle of side $ \sqrt{3}$, with the points at the vertices, at thirds of  sides, and at the center ($ 3\plus{}6\plus{}1\equal{}10$ in all).

Usually (and especially for such a large number of points), in similar problems, it is very hard to hit the right threshold, so I wonder of a simple method to show $ 11$ points are enough. Pigeon-hole and area methods are too coarse...


trigonometry
geometry
angle bisector
AMC

A circle centered at $O$ has radius 1 and contains the point $A$. Segment $AB$ is tangent to the circle at $A$ and $\angle AOB = \theta$. If point $C$ lies on $\overline{OA}$ and $\overline{BC}$ bisects $\angle AOB$, then $OC =$

$\text{(A)} \ \sec^2 \theta - \tan \theta \qquad \text{(B)} \ \frac{1}{2} \qquad \text{(C)} \ \frac{\cos^2 \theta}{1+\sin \theta} \qquad \text{(D)} \ \frac{1}{1 + \sin \theta} \qquad \text{(E)} \ \frac{\sin \theta}{\cos^2 \theta}$

[hide]$OA=1$, $AB=\tan \theta$, $OB=\sec \theta$. If $OC=x$, then by Angle Bisector Theorem,

$\frac{\sec \theta}{x}=\frac{\tan \theta}{1-x}\Rightarrow \frac{1}{x}=\frac{\sin \theta}{1-x}$.

Then $1-x=x\sin \theta\Rightarrow x=\frac{1}{1+\sin \theta}\Rightarrow \boxed{D}$.[/hide]


If
$ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}2$
then
$ \frac{a^2\minus{}bc\plus{}1}{\sqrt{a^2\plus{}bc}}\plus{}\frac{b^2\minus{}ca\plus{}1}{\sqrt{b^2\plus{}ca}}\plus{}\frac{c^2\minus{}ab\plus{}1}{\sqrt{c^2\plus{}ab}}\geq2$

[quote="filomen"]If
$ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 2$
then
$ \frac {a^2 \minus{} bc \plus{} 1}{\sqrt {a^2 \plus{} bc}} \plus{} \frac {b^2 \minus{} ca \plus{} 1}{\sqrt {b^2 \plus{} ca}} \plus{} \frac {c^2 \minus{} ab \plus{} 1}{\sqrt {c^2 \plus{} ab}}\geq2$[/quote]

$ \sum_{cyc}\left(\frac {a^2 \minus{} bc \plus{} 1}{\sqrt {a^2 \plus{} bc}}\right)\equal{}\sum_{cyc}\left(\sqrt{a^2\plus{}bc}\plus{}\frac{1}{\sqrt{a^2\plus{}bc}}\right)\minus{}\sum_{cyc}\frac{2bc}{\sqrt{a^2\plus{}bc}} \ge 6\minus{}\sum_{cyc}\frac{2bc}{\sqrt{a^2\plus{}bc}}$

hence it's sufficent to show that

$ \sum_{cyc}\frac{bc}{\sqrt{a^2\plus{}bc}} \le 2$

By caushy shwarz

$ \sum_{cyc}\frac{bc}{\sqrt{a^2\plus{}bc}} \le \sqrt{(ab\plus{}bc\plus{}ca)\left(\frac{ab}{ab\plus{}c^2}\plus{}\frac{bc}{bc\plus{}a^2}\plus{}\frac{ca}{ca\plus{}b^2}\right)}$

and Thus it remains to prove that

$ \left(\frac{ab}{ab\plus{}c^2}\plus{}\frac{bc}{bc\plus{}a^2}\plus{}\frac{ca}{ca\plus{}b^2}\right) \le 2$
$ \Longleftrightarrow \sum_{cyc}\frac{a^2}{bc\plus{}a^2} \ge 1$

or we have by caushy shwarz $ \sum_{cyc}\frac{a^2}{bc\plus{}a^2} \ge \frac{(a\plus{}b\plus{}c)^2}{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca} > 1$


number theory
modular arithmetic
superior algebra
superior algebra solved

solve in $Z_{144}$ this equation : $x^{2}=5$ . it solved it using trivial programming :P , and it sais it has no solutions. how do I demonstrate it in an elegant way ?

$x^{2}\not\equiv 5$ mod $3$ (test $3$ numbers :-)), so $x^{2}\not\equiv 5$ mod $144$.

what particular statement do u base this ? i have no real knoledge about $Z_{n}$'s and stuff, so please be more explicit

If we do arithmetic mod 3, $0^{2}\equiv 0,1^{2}\equiv 1,$ and $2^{2}\equiv 1$. None of these are equivalent to 5, and they clearly exhaust all possibilities; no square can be equivalent to $5$ mod $3$.
Since $3|144$, $n\equiv 5\mod 3$ whenever $n\equiv 5\mod 144$.

We could have worked mod 8; 5 is also not a square mod 8.

Modular arithmetic is essential to number theory. You need to understand it.

As an exercise: What are the possible values of $n^{3}$ mod $7$?

[quote="jmerry"]
Modular arithmetic is essential to number theory. You need to understand it.
[/quote]

does my life depend on it?  :P
thank u so much jmerry

[quote="pureblack2003"]what particular statement do u base this ?[/quote]

the definitions

[quote="pureblack2003"]i have no real knoledge about $Z_{n}$'s and stuff, so please be more explicit[/quote]

this is not how it goes. first, [i]you[/i] have to read the definition and properties of something on which you want to work. my post is clear to everyone who has done that.

yes chief, you are so right  :P believe it or not , the definitions you told me about weren't given to me by my teacher and i didn't have the curiosity to learn more about 'em . i should have known better  :idea: