function
inequalities
induction
triangle inequality
algebra proposed
algebra

we know that for function $f: R^+\rightarrow R^+$
if $x,y,z$ are sides of a triangle ,then $f(a),f(b),f(c)$ are also sides of e triangle.
prove there exist positive reals $A,B$ such  for all $x$: $f(x)\leq Ax+B$

This is not strict but shows the main idea:

Suppose to the contrary that $f(x)$ grows faster than linear function, so we can assume that $f(x)=O(x^{1+\epsilon}+b)$ for some $\epsilon>0$. But then for sufficiently large $x$:  $f(2x-\delta)>2f(x)$ ($\delta$ is sufficiently small) which is contradiction (consider triangle with sides $x,x,2x-\delta$ - according to the statement we must have $2f(x)>f(2x-\delta)$)

Anyway I think it was discussed before

[quote="Megus"]Suppose to the contrary that $f(x)$ grows faster than linear function, so we can assume that $f(x)=O(x^{1+\epsilon}+b)$ for some $\epsilon>0$.[/quote]

I doubt you can assume this...

Anyway, here is my solution. I show an actual upper bound (very unsharp, but it works...):

[color=blue][b]Theorem 1.[/b] Let $f: \mathbb{R}^+\to\mathbb{R}^+$ be a function such that if three positive reals a, b, c are the sidelengths of a triangle, then f(a), f(b), f(c) are also the sidelengths of a triangle.

Then, $f\left(x\right)<2\left(x+1\right)\cdot f\left(1\right)$ holds for every positive real x.[/color]

[i]Proof of Theorem 1.[/i] If x < 2, then the numbers 1, 1, x are the sidelengths of a triangle (in fact, 1 < 1 + x, 1 < x + 1 and x < 2 = 1 + 1, so the triangle inequalities are fulfilled); thus, f(1), f(1), f(x) are also the sidelengths of a triangle, and thus

$f\left(x\right)<f\left(1\right)+f\left(1\right)=2\cdot f\left(1\right)<2\left(x+1\right)\cdot f\left(1\right)$.

So Theorem 1 is proven for x < 2. Also, as we showed that $f\left(x\right)<2\cdot f\left(1\right)$ for every x < 2, by taking $x=\frac32$ we can conclude that $f\left(\frac32\right)<2\cdot f\left(1\right)$; now, since $\frac 32$, $\frac 32$, 2 are the sidelengths of a triangle, we see that $f\left(\frac32\right)$, $f\left(\frac32\right)$, f(2) must also be the sidelengths of a triangle, and thus

$f\left(2\right)<f\left(\frac32\right)+f\left(\frac32\right)=2f\left(\frac32\right)<2\cdot 2\cdot f\left(1\right)<2\left(2+1\right)\cdot f\left(1\right)$.

Hence, Theorem 1 is proven for x = 2, too. Thus, Theorem 1 is proven for all $x\leq 2$. It remains to prove Theorem 1 for x > 2. In order to do this, we denote by n the greatest integer $\leq x$; then, $n\leq x<n+1$. Now, we prove Theorem 1 for all $x\geq 1$ by induction over n:

For n = 1, we have $1\leq x<1+1=2$, and thus Theorem 1 is already proven (since we proved Theorem 1 for all x < 2).

Now, let n be some positive integer, and assume that Theorem 1 holds for all x with $n\leq x<n+1$. We have to show that Theorem 1 also holds for all x with $n+1\leq x<n+2$. In order to prove this, we consider some x with $n+1\leq x<n+2$. If $x\leq 2$, then Theorem 1 was already proven above; thus, we assume that x > 2, so that x - 1 > 1. Hence, there exists a positive real t < 2 such that x - 1 > t > 1. Now, the numbers t, x - 1, x are the sidelengths of a triangle (since the greatest of these numbers is x, because x > x - 1 > t, and the triangle inequality t + (x - 1) > x is fulfilled, since t > 1 yields t + (x - 1) > 1 + (x - 1) = x). Hence, f(t), f(x - 1), f(x) are the sidelengths of a triangle, as well. Thus, f(x) < f(t) + f(x - 1). Since t < 2, we have $f\left(t\right)<2\cdot f\left(1\right)$ as shown above. On the other hand, we have x - 1 > 1, and since $n+1\leq x<n+2$, we have $n\leq x-1<n+1$; hence, by the induction assumption, Theorem 1 holds for x - 1, i. e. we have $f\left(x-1\right)<2\left(\left(x-1\right)+1\right)\cdot f\left(1\right)$. Thus,

$f\left(x\right)<f\left(t\right)+f\left(x-1\right)<2\cdot f\left(1\right)+2\left(\left(x-1\right)+1\right)\cdot f\left(1\right)=2\left(x+1\right)\cdot f\left(1\right)$.

Hence, Theorem 1 is proven for our number x. Thus, the induction step is complete, and Theorem 1 is proven.

  Darij


Hey all, since I didn't make nats, I was wondering if anyone knew if they are going to have nats on tv like they have had for a few years in the past.

Last year they didn't show it, don't count on it this year, but hopefully they do show it.

[quote="kevinchenxi"]what channel?[/quote]
In past (05/04) it has been on ESPN.

it was also shown '03.  if i remember correctly...

yah, i think rusczyk confirmed this once some time ago? i vaguely remember, but i do remember.

jorian

i vaguely remember reading something about mathcounts trying to get a deal with the discovery channel to show it for future years.

EDIT: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=channel+discovery&t=88847[/url]

Did espn think not enough ppl were watching it?  I know i was...it's really a sad story.  Hopefully Discovery Channel may cooperate.  :maybe:

on this years info frms it clearly sez that nats will b shown on espn, but i dont no when.


induction
combinatorics proposed
combinatorics

Given positive integer $ n$ with $ n \geq 2$,determine the minimum number of elements in set $ X$ such that for any n 2-element subsets $ S_1,S_2,...S_n$ of $ X$,there exists an n-element subset $ Y$ of $ X$ with $ Y \cap S_i$ has at most one element for every integer $ i \equal{} 1,2,...,n$

[hide="Looks like Darij was right.  My apologies"]

We must prove that $ |X|$ is at least $ 2n \minus{} 1$.  Suppose contrary that $ X$ is a set with less than $ 2n \minus{} 1$ elements.  Let $ S_1$, $ S_2$, $ \dots$, $ S_n$ be $ 2$-element subsets of $ X$ such that the union of $ S_1$, $ S_2$, $ \dots$, $ S_{n \minus{} 1}$ is all of $ X$.  Let $ Y$ be any $ n$-element subset of $ X$.  Thus, there are two elements of $ Y$ falling in the same $ S_i$ for some $ i \equal{} 1,2,\dots,n \minus{} 1$, a contradiction.

Hence, $ |X| \geq 2n \minus{} 1$.  We shall prove that this is a sufficient condition by induction on $ n$.  For $ n \equal{} 2$, there is nothing to say.  For convenience, a set $ X$ is said to be $ n$-cool if for any collection of $ n$ $ 2$-subsets $ S_1$, $ S_2$, $ \dots$, $ S_n$, there exists an $ n$-subset $ Y$ of $ X$ for which $ Y\cap{S_i}$ has at most one element for each $ i \equal{} 1,2,\dots,n$.

Suppose that $ n > 2$ and $ |X| \equal{} \{1,2,\dots,m\}$, where $ m \geq 2n \minus{} 1$.  Let $ S_1$, $ S_2$, $ \dots$, $ S_n$ be $ n$ arbitrary $ 2$-subsets of $ X$.  Since $ 2|X| > 2n$, then at least one element of $ X$, say $ m$, belong to at most one of the $ S_i$'s.  If $ m$ belongs to none of the $ S_i$'s, we can replace $ X$ by $ X' : \equal{} X\setminus\{m\}$, and prove that $ |X'| \geq 2n \minus{} 1$ is required.  So, we shall assume that $ m \in S_n$ and that $ S_n \equal{} \{m,m \minus{} 1\}$.

Suppose $ m \minus{} 1$ belongs to $ S_i$, says, $ S_i \equal{} \left\{m \minus{} 1,a_i\right\}$.  Then, replace $ S_i$ by either $ \left\{1,a_i\right\}$ or $ \left\{2,a_i\right\}$, and call this new set $ S'_i$.  For a set $ S_i$ that does not contain $ m \minus{} 1$, $ S'_i$ is equal to $ S_i$.

That is, $ S'_i$, $ i \equal{} 1,2,\dots,n \minus{} 1$, are $ 2$-subsets of $ X': \equal{} \{1,2,\dots,m \minus{} 2\}$.  As $ \left|X'\right|\geq 2(n \minus{} 1) \minus{} 1$, then $ X'$ is $ (n \minus{} 1)$-cool.  Thus, there exists $ Y'$ with the desired property for $ S_1$, $ S_2$, $ \dots$, $ S_{n \minus{} 1}$.  Then, $ Y: \equal{} Y'\cup\{m\}$ works for $ X$ with $ S_1$, $ S_2$, $ \dots$, $ S_n$, and the induction is complete.
[/hide]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299325 for a generalization of a problem.

@Batominovski:

[quote="Batominovski"]Suppose that $ n > 2$ and $ |X| \equal{} \{1,2,\dots,m\}$, where $ m \geq 2n \minus{} 1$.  Let $ S_1$, $ S_2$, $ \dots$, $ S_n$ be $ n$ arbitrary $ 2$-subsets of $ X$.  Since $ 2|X| > 2n$, then at least one element of $ X$, say $ m$, belong to at most one of the $ S_i$'s.  If $ m$ belongs to none of the $ S_i$'s, we can replace $ X$ by $ X' : \equal{} X\setminus\{m\}$, and prove that $ |X'| \geq 2n \minus{} 1$ is required.[/quote]

I have no idea what you mean by "$ |X'| \geq 2n \minus{} 1$ is required" here. In the generalized version of the problem, the case when $m$ belongs to none of the $ S_i$'s is the hardest case - so, yes, your solution might be somewhat too simple to be true. :(

[quote="Batominovski"] $ Y'$ with the desired property for $ S_1$, $ S_2$, $ \dots$, $ S_{n \minus{} 1}$.[/quote]

You might be trying to say "for $ S'_1$, $ S'_2$, $ \dots$, $ S'_{n \minus{} 1}$" here (although your version makes sense as well).

[quote="Batominovski"]  Then, $ Y: \equal{} Y'\cup\{m\}$ works for $ X$ with $ S_1$, $ S_2$, $ \dots$, $ S_n$, and the induction is completed.[/quote]

I wonder if they credited me with the authorship of the problem ... the should have had  :)


I'm Physics and very very scheptical about anything, So, dont call me idiot.
If you think this is a stupid think, just dont post here.

I'm asking if someone here in this forum belives in OBE, or Out-of-body experience.

Astral projection, lucid dreams, the effect of binaural beats. etc. etc.

we can talk about this, sending experiences etc.

I know that have a lot of forums about this, but here is best to talke, because have a lot of mathematicians and physics, a lot of inteligent peoples and schepticals about anything too.

thanks

Er, I'm not sure why you're grouping lucid dreams together with astral projections because lucid dreams are a scientifically verified fact.

http://en.wikipedia.org/wiki/Lucid_dream

because is impossible to talk about atral projection without talk about lucid dreams.

I'm ateist, so i did not belive that in a astral travel we leave the body and others things.

Because in my opinion astral projection and lucid dreams are the same thing. a Dream that we have control.

I dont belive that in a astral projection we see god, angels, etc... I just think that a astral projection is a induced dream.

In a Robert Bruce Book he talk about lucid dreams etc.

I dont think that the Moderator of this forum will be happy that I creat 17 posts talking about on thing in each post.
It's more simply just one ;)

Lucid dreams are possible. I have had them before and I think it is a really interesting experience. According to his autobiography, Feynman taught himself to do that at will. I haven't been able to do that, but I haven't tried with as much diligence as he did....

I don't really know what you are saying but do you mean something like this:

http://www.youtube.com/watch?v=gEDaCIDvj6I

?

I don't undersatnd what exactly is meant by "lucid dream," "astral projection," "binaural beats," etc., but I have had therapy from people who use "Eastern" medicine and therapies... one is a Shiatsu therapist who has had out of body experiences... another "therapist" is someone who can help one visualize the energies in one's body.  It's pretty cool stuff; like, there are "energy strands" that can connect people (not all of them are good to have because some can drain people's energy), and there are different "colors" and "names" to certain energies (not all of the energy is good).  It has helped me A LOT; it works.

I've lucid dreamed quite a bit (induced it). It's hard to say whether your 'soul' leaves your body when projecting, but it does seem to occur. I believe this is what your body goes through to reach sleep, whether or not you realise.

the thing is:
If someone say to you, "When you have a OBE you really leave your body". And so on, you try it and finally it happens.
But, if the person do not tells you that you leaves the body, you still wont have the sensation?

Because, when I try, i have the sensation of this. but i cannot the if this is influentiable by what i heard.

And my dear Zakary. Astral projections, etc etc... these so many names etc etc...

In my opinion, the meaning of this is just one thing.

Like God, he have a lot of names, but, for me is just one person.

I'm ateist, but this does not matter.

But OBE and astral projections are the same thing, people that like to use a lot of names and put religion on everything.

I'm the one who thinks that all this "stuff" can be explained by physics.

Just think one thing.

A Lot of years ago, people never imagined that have a eletromagnetic fields, forces, etc. never imaginated that this fields are in everywhere in the space.

I think that physics is not complete yet, so, is completelly ration to think that exists others "forces" "fields" that we cannot detect yet.

But someday, maybe.

We can understant the universe, but we cannot understand yourselves, hehe... sacarms

adie

Believing that you (whatever that means) have left your body (whatever that means) is not the same thing as you actually having left your body.  For a nice exposition on the falliability of human observational abilities, see http://zenoferox.blogspot.com/2006/04/miracles-on-interstate-80.html

Believing that YOU are Alive, does not mean that you are alive.

Plato says interesting things like that.

You can see the real word, you can do everything, so, IT's Real.

In a dream you can't see very well, you are not under the control, so, it's not real.

A day have 24 for hours, we dream 8 hours per day. but, if we dream 23 hours per day, the definition of reallity would be completelly different.

The question is that you believe in what you see. and this is completelly wrong.

To see the true you need to understant that the "real world" and the dreams are the same thing, and both are not reals.
uhahua

I can't parse everything you wrote, but it all sounds like a standard sort of new-age gibberish.  If you're going to say things like, "Reality isn't real," you aren't speaking the same language as I am.

the thing is like plato Sayed, you can trust in yours eyes.

you can see only a distorced world, what you see cannot or can be real, but... I'm not wrong.

Can you proove the Reality?

We don't have a proof of the real or non-real world.

[quote]but it all sounds like a standard sort of new-age gibberish[/quote]
Hey, this is a insult for me.

this is not new-age etc...

my thougth is very vert old.

in the ancient greek they thougth like me, don't trust in anything. in egipt too.
i've reading soem ocultists books etc

and this is not new-age

christianist is new-age

tsk tsk


geometry incenter perimeter rectangle ratio probability

Following is a problem set of ten problems that I made in my free time.  Most of them are pretty simple, but some require some thinking.  I know the answers to all but 5 and 10, and I know the answers, but can't prove, 4 or 6.  Have fun!  (Remember to hide all answers.)

Problem set #1

1.  ΔABC with ∠A = 90° has equal area and perimeter.  I is the incenter of ΔABC and IB = √13.  ΔBCD with ∠BCD = 90° has 18.2 times the area of ΔABC and has 91/30 the incenter of ΔABC.  Compute the length of BD.

2.  Point E is placed on the perimeter of rectangle ABCD so that ∠AED is 90°.  F is drawn on AD so that BF and EA intersect at P to make an equilateral ΔBAP of area 2√3.  What is the area of quadrilateral ABED?

3.  A bored mathematician writes down in his notebook 1, 2, 2, 3, 3, 3, and so on (the nth term is repeated n times).  What is the sum of the first 2007 terms in this sequence?

4.  A number called a perfect product if it can be written as n more than the product of three consecutive integers, where n is the middle of the three integers.  No perfect product can be rewritten as the sum of k consecutive positive integers.  Find the smallest possible value of k.

5.  Every number has a “P-index”. The P-index of the nth prime is n.  A palindrome is a number that when its numbers are reversed you get the same number.  A palindrome prime is a prime that is a palindrome and its P-index is also a palindrome.  Find the smallest palindrome prime whose P-index is at least two digits.

6.  Convex hexagon ABCDEF has a common intersection of AD, BE, and CF at O.  6 triangles are drawn (AOB, BOC, COD, DOE, EOF, and FOA).  The incenters of the triangles are G, H, I, J, K, and L.  If the smallest possible ratio of areas of GHIJKL:ABCDEF is 1:x, compute x.

7.  a/b is said to approximate a number if it is closer to the number than a/b+1 or a/b-1.  x/y approximates π and x/z approximates e.  The ratio (z-y):x = k:1.  Compute k to 4 decimal places.

8.  3 sides of quadrilateral ABCD have length 5.  The diagonals of ABCD intersect at P.  If AB, BC, CD, DA, AP, BP, CP, and DP are of integer length, compute the area of ABCD.

9.  Set S is composed of k distinct integers that sum to 2007.  The maximum of S is 2007 times the mean of S.  The minimum of S is p.  What is the maximum value of p?

10.  Mickey plays a solitaire game with a pack of cards.  On each turn, he flips the top card over.  The game ends when there is at least one of every number out.  Mickey wins if he deals himself a 4-of-a-kind before the game is over and if he doesn’t, he loses.  What is the probability of winning this game?

The  :mad:  means "over x".  Sorry, I had a smiley mishap.  Answers to any problems are fine.


logarithms
integration
calculus
real analysis
real analysis unsolved

I know that $\sum \frac{1}{p}$ is disvergent.
And what about $\sum \frac{1}{p\ln p}$ ?

More generally:If a positive decreasing sequences $a_{n}$ such that  $\sum a_{n}$ disvergent,
can we always have: $\sum a_{p}$ disvergent,
where the sum $\sum a_{p}$  means $p$ runs for all prime numbers.

[quote="N.T.TUAN"]$\sum_{i=1}^{\infty}\frac{1}{p_{i}}$ is disvergent, here $2=p_{1}<p_{2}<...$ are all prime numbers (well-know).[/quote]
What?I have said I know that :lol:

The Prime Number Theorem says that if $p_{k}=n,$ then $k$ is approximately $\frac{n}{\ln n}.$ This can be turned around to say that $p_{k}$ is comparable to $k\ln k.$

Given that, $\sum_{k}\frac1{p_{k}\ln p_{k}}$ converges by comparison to $\sum_{k}\frac1{k\ln^{2}k}.$

Oh,Thank you,you are right.It converges.
I also find a more elementary solution used $\sum_{p \le X}\frac{\ln p}{p}=\ln X+O(1)$
More generally:
I think this will be true:
For a nonnegative sequence $a_{n}$ such that:$\sum_{n=2}^{N}a_{n}=\ln N+O(1)$
Then we have:$\sum_{n=2}^{N}\frac{a_{n}}{\ln n}=\ln \ln N+O(1)$
and:$\sum_{n=2}^{N}\frac{a_{n}}{(\ln n)^{2}}$ converges.

Hi Zhaobin. I would appreciate to see the proofs of your 2 last assertions. Do you assume that $a_{n}$ is a decreasing sequence ? Otherwise $a_{n}\nsim\frac{1}{n}$. Moreover if $a_{n}$ is a decreasing sequence i don't think that $a_{n}\sim\frac{1}{n}$.
   We assume only $a_{n}\geq0$; if $\sum_{n}^{N}a_{n}\sim\frac{1}{N}$ then $\sum_{n=0}^{\infty}a_{n}x^{n}\sim{-ln(1-x)}$ when $x\rightarrow1-$; thus $\sum\frac{a_{n}}{n}$ is a convergent series. The case $\sum\frac{a_{n}}{ln(n)^{2}}$ is much more difficult.

Hi,Loup Blanc,Maybe I don't get your meaning very well,I don't what do you mean by $\sum_{n}^{N}a_{n}\sim \frac{1}{N}$,I also didn't understand what you what to say.

Here is my proof for the assertions:
$\sum_{n=2}^{N}\frac{a_{n}}{\ln n}=\sum_{n=2}^{N}a_{n}\left(\frac{1}{\ln N}+\int_{n}^{N}\frac{dy}{y\ln^{2}y}\right) =\frac{\sum_{n=2}^{N}a_{n}}{\ln N}+\int_{2}^{N}\frac{\sum_{1}^{[y]}a_{n}dy}{y\ln^{2}y}=\frac{\sum_{n=2}^{N}a_{n}}{\ln N}+\int_{2}^{N}\frac{(\sum_{1}^{[y]}a_{n}-\ln y)dy}{y\ln^{2}y}+\int_{2}^{N}\frac{dy}{y\ln y}$
$=\ln \ln N+O(1)$
The second will be proved similar.

Ok zhaobin; your integration by parts works very well.


On a $4\times4\times3$ rectangular parallelepiped, vertices $A$, $B$, $C$ are adjacent to vertex $D$. Find the perpendicular distance from $D$ to the plane containing $A$, $B$ and $C$.

[quote="13375P34K43V312"][quote="paladin8"]Just for your information, there is a resources section for a reason. Try not to post problems that have already been posted and have solutions.

http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=1996[/quote]

Where are the solutions?

Thanks[/quote]
click on the problem numbers, it links you to a thread with the problem+solution


What is the value of ......

(X-A)(X-B)(X-C)(X-D)........(X-Y)(X-Z)?

I want a real answer...

A big $0$.
 $x - x$ is the 24th term.
 old one, man.

No, it's been posted in the Puzzles & Brainteasers section about three times.

Interesting problem, though.

:rotfl: this is kind of spam.

[color=blue][size=75]Kyyuanmathcount: Yeah, thats spam.  Locked.[/size][/color]


conics hyperbola ellipse ratio trigonometry function

Is there a nice and simple way to remember graphs equation?

such as the equations of hyperbolas, ellipses, circles (-.-"'), and even graphs of inverse trigonometric ratios (arc sine, arc cos and arc tangent)?

my concern here is that i can't remember equations n graphs of rectangular hyperbolas, or rather, all hyperbolas, ellipses and circles.. i remember them for the sake of tests, and after a few days, i forget them all.. and i have many other subjects to revise as well.. some help here maybe?

If you understand the fundamentals of those curves, then there's no need to memorize; the formulas become pretty common sensish if you know how they were formed and the ground roots about conic sections.

Those fundamentals, however, are not basic middle school concepts so please go to the high school grounds  :D .

THis should be in High School math forum.

Just to note...it's spelled ellipse, not eclipse.

I think it is very easy to "remember" the inverse trig function graphs.
You just sketch the trig functions and then $f(x)=x$, and relect it...


inequalities
function
inequalities proposed

Prove that for any positive real numbers $ a_1,a_2,...a_n,x_1,x_2,...x_n$such that $ \sum_{1\le i < j \le n}x_ix_j = \frac {n(n - 1)}{2}$
the following ineq holds:
${ \frac {a_1}{a_2 + ... + a_n}(x_2 + ... + x_n) + ... + \frac {a_n}{a_1 + ... + a_{n - 1}}}(x_1 + ... + x_{n - 1}) \ge n$

p.s. I only had a proof for the case $ n=3$,and I am waiting for your nice solution :)

Any ideas?I am waiting for your solutions :)

My idea:Firstly,we can easily apply Cauchy to kill $ a_i$
if we assume $ y_1 \equal{} \frac {x_2 \plus{} x_3 \plus{} ... \plus{} x_n}{n \minus{} 1}$ similarly $ y_2...$
We can reduce the original to
Given $ 2\sum_{i < j} y_iy_j \equal{} n \plus{} (n \minus{} 2)(\sum y_i^2)$ Then $ 2\sum_{i < j} \sqrt {y_iy_j} \equal{} n \plus{} (n \minus{} 2)(\sum y_i)$ holds.
I didn't know if the subsitution made it easier or tougher,anyhow I think the above ineq is beautiful too.
Can anybody continue? :lol:

I do not like the solution using EV,function  convex...
have you other solution?

[quote="Vasc"]Write the inequality as
$ \sum (a_2 \plus{} ... \plus{} a_n)\cdot\sum\frac {x_2 \plus{} ... \plus{} x_n}{a_2 \plus{} ... \plus{} a_n}\ge (n \minus{} 1)^2\sum x_1 \plus{} n(n \minus{} 1)$.
By the Cauchy-Schwarz inequality, it suffices to prove that
\[ (1) \ \ \ \ (\sum\sqrt {x_2 \plus{} ... \plus{} x_n})^2 \ge (n \minus{} 1)^2\sum x_1 \plus{} \sqrt {n(n \minus{} 1)[(\sum x_1)^2 \minus{} \sum x_1^2]}.
\]
For $ \sum x_1 \equal{} 1$ and $ \sum x_1^2 \equal{} S$, the inequality becomes
\[ (\sum\sqrt {1 \minus{} x_1})^2 \ge (n \minus{} 1)^2 \plus{} \sqrt {n(n \minus{} 1)(1 \minus{} S)}.
\]
Since $ f(u) \equal{} \minus{} \sqrt {1 \minus{} u}$ has $ f'''(u) > 0$, by EV-Theorem it suffices to prove the inequality for $ x_2 \equal{} x_3 \equal{} ... \equal{} x_n$. Setting $ x_2 \equal{} x_3 \equal{} ... \equal{} x_n \equal{} 1$ in the homogeneous inequality (1), we find an inequality in $ x_1$, which is equivalent to $ (x_1 \minus{} 1)^2\ge 0$.[/quote]

Thank you for your nice solution,Vasc :lol:


Give A,B,C: A+B+C= \pi .Prove that:
3(cosA+cosB+cosC)  \geq  2(sinA*sinB+sinB*sinC+sinC*sinA).

I believe that this can be proved by this inequality:
    (cosA+cosB+cosC) 2  \geq sinAsinB+sinBsinC+sinCsinA
<=>(1-cos A)(a-cosB)(1-cos C) \geq cosAcosBcosC which can be easlyproved by let cosA=(1-x 2 )(1+x 2 ) where x=tan(A/2)

when i find the sol of a problem i find this interesting problem
my problem:
3(R+r) \geq h(a)+h(b)+h(c)
with a,b,c are sides of triangle, h(a)*a=h(b)*c=h(c)*c=a*b*c/(4R)=(1/2)(a+b+c)*r.
can you give me a short sol for this problem.

I think it can be srticter:
4(R+r) 2  \geq ab+ac+bc

I have a result same with my problem but to solve it i must have a long time.
(4(R+r) 2 )/R \geq h(a)+h(b)+h(c).


First of all we are speaking about positive natural numbers a and b right?
if so then I think the minimum is achieved by [b]a=23[/b] and [b]b=5.[/b]
Actually we have a+2b is divisible by 11 and a-2b is divisible by 13.
if [b]a-2b=0[/b] then [b]a=2b[/b] and [b]3b[/b] is divisible by 11 so[b] $ b\geq 11$[/b]and [b]a+b[/b] is at least 33.if[b] a-2b=13[/b] then [b]a+2b=13+4b[/b] is divisible by 11 so from here $ b\geq 5$ and a+b is at least 28. And the last case if [b]$ a \minus{} 2b\geq 26$ [/b]is easy now $ a\geq 28$
So the minimal sum is[b] 28.[/b]

[quote="Aranya"]Hi! I get answer as 24.

My method: will you please tell if I am wrong?

$a+13b=11m$ and $a+11b=13n$
Subtracting, $2b=11m-13n$ and also, $2a=169n-121m$

So, $a+b=78n-55m=55(n-m)+23n$

Now, we need to minimise $a+b$ so there is no harm in taking $n-m<0$ because that is the only possibility.

We now check the different multiples of 55 (mod 23) just to know the distance of each multiple of 55 from a multiple of 23.

We find that $55.5=-1(mod23)$ so if we select $n-m=-5$ then we may get close to a multiple of 23. We find that $23n=23.12=276$ which is just at distance $1$ away from $55.5=275$. But, no sum of two natural numbers give answer 1. So, we may select the next higher multiple of 23 i.e. $23.13=299$ so that the final answer is $a+b=299-275=24$ and that is less than your answer.

Can you please tell me where I am going wrong? Thanks!![/quote]

Hello Aranya! Here you go:

       According to you $n=13$, so $a+11b = 169$ and if $a+b=24$ , then $10b=125$. This implies that $b=12.5$.
Obviously this is not an integer. A contradiction. So there is something wrong in your equation, if you could specify the values of $a$ and $b$........

       Moreover, go to this link:
      http://olympiads.hbcse.tifr.res.in/uploads/rmo-2006
       It's the official solution. :D  :lol:

$11|a + 13b \implies a + 13b = 11x$  for some $x \in \mathbb{N}$.\\
Again $13|a + 11b \implies a + 11b = 13y$  for some $y \in \mathbb{N}$.\\
Therefore $11x - 13b = 13y - 11b \implies 11(x+b) = 13(y + b).$\\
For minimum values, we can take $x + b = 13$ and $y + b = 11$. Thus $x = y + 2.$\\
Since $2b = 11x - 13y \implies b = 11 - y.$\\
Thus $a + 11(11-y) = 13 y \implies a = 24 y - 121 > 0 \implies y \geq 6.$\\
For minimum value, choose $y = 6.$ Then $b = 5$ and $a = 23.$ Thus $a+b = 28.$

[quote=weiquan]$11|(a+13b) \iff 11|(a+24b)$
$13|(a+11b) \iff 13|(a+24b)$
Thus $143|(a+24b)$
$a,b>0 \Rightarrow a+24b\geq 143$
Therefore $a+b=(a+24b)-23b\geq 143-23b\geq 143-23(5)$ with equality at $b=5, a=23$[/quote]
This solution is better and shorter than the official solution...(NOTE THAT THE THREAD IS ALREADY REVIVED)

[quote=amar_04][quote=weiquan]Therefore $a+b=(a+24b)-23b\geq 143-23b\geq 143-23(5)$ with equality at $b=5, a=23$[/quote]

Can anyone tell how he got from $143-23b\geq 143-23(5)$. It could have been $1,2,3,4$ instead of $5$.[/quote]
$143-23(5) < 143 - 23(4)$, so your bound doesn't bold at all.

[quote=amar_04][quote=weiquan]Therefore $a+b=(a+24b)-23b\geq 143-23b\geq 143-23(5)$ with equality at $b=5, a=23$[/quote]

Can anyone tell how he got from $143-23b\geq 143-23(5)$. It could have been $1,2,3,4$ instead of $5$.[/quote]

Because you want to go as closer as possible to 143 with using maximum b value otherwise a will be too large and thus the sum will be large. With b=1,2,3,4 a will be really big.

[quote=amar_04][quote=weiquan]Therefore $a+b=(a+24b)-23b\geq 143-23b\geq 143-23(5)$ with equality at $b=5, a=23$[/quote]

Can anyone tell how he got from $143-23b\geq 143-23(5)$. It could have been $1,2,3,4$ instead of $5$.[/quote]
For the minimum value of a and b,
divide 143 by 24
Put b=quotient=5
        a=remainder=23

i think this question can have a much shorter and much clearer solution. So here goes my solution to it:
11 divides a+13b implies 11 divides 6a+b, a+24b.
13 divides a+11b implies 13 divides 6a+b, a+24b.
thus, 143 divides 6a+b, a+24b. which implies,
6a+b >= 143    ... (1) ,
a+24b >= 143   ... (2).
23*(1) + 5*(2) gives
a+b >= 28.
Also, for a=23, b=5 we see that a+b=28 and our required condition are also satisfies.
Thus, the least possible value for a+b is [b]28[/b].

[quote=shrayagarwal]If $ a$ and $ b$ are natural numbers such that $ a\plus{}13b$ is divisible by $ 11$ and $ a\plus{}11b$ is divisible by $ 13$, then find the least possible value of $ a\plus{}b$.[/quote]

$a+b=23+5=28$

[quote=Maths_1729]We can rewrite as
$2b=11x-13y$ for $(x, y) \in N$  and also we must have $\frac{11x}{13}>b, \frac{13y}{11}>b$
 $x, y$ is either both odd or both even and least value of $x, y$ following this is ${8, 6}$ so least $a, b$ is $23, 5$[/quote]
Is this solution true?

I don't think this is correct... if $x=4, y=2, b=9,$ then $\frac{44}{13} \not > 9.$

[hide=Solution]$11|(a+13b), 13|(a+11b)$.
These give $11|(a+2b), 13|(a-2b)$.
So $a^2-4b^2=143q=(a-2b)(a+2b)$.
For minimum value of $a+b$, we can optimize $q$ to equal $4$ because $q=1, 2, 3$ do not work.
This gives $a=24, b=1$.
Therefore $a+b=25$. [/hide]
Will this be sufficient for RMO? Also, is the answer correct?

[hide=Solution]$13\left|\left(a+11b\right)\Rightarrow13\right|\left(a-2b\right),11\left|\left(a+13b\right)\Rightarrow11\right|\left(a+2b\right)$. Let $a-2b=13x, a+2b=11y\Rightarrow4b=11y-13x,\ 2a=11y+13x$. $11y\equiv13x$ (mod 4) $\Rightarrow x\equiv-y$ (mod 4) $\Rightarrow4|(x+y)$.
The minimum value of $x,\ y$ we can take so that $11y>13x$ is $x=1,\ y=3\Rightarrow4b=33-13=20\Rightarrow b=5,\ 2a=33+13=46\Rightarrow a=23\Rightarrow a+b=23+5=28$ which is our answer.[/hide]
Hope it helps :)

My earlier solution was incorrect. Refer my latest solution above.

Since 13 divides a + 11b, we see that 13 divides a — 2b and hence it also divides
6a —12b. This in turn implies that 13|(6a +b). Similarly 11|(a + 136) = 11|(a + 2b) =>
11(6a + 125) = 11](6a + b). Since gcd(11,13) = 1, we conclude that 143|(6a + b). Thus
we may write 6a + b= 143k for some natural number k. Hence
6a + 6b = 143k + 5b = 144k + 6b — (k +b).

This shows that 6 divides + b and hence k + b > 6. We therefore obtain

6(a +b) = 143k + 5b = 138k + 5(k +b) > 138 +5 x 6 = 168.
It follows that a +b > 25. Taking a = 23 and b = 5, we see that the conditions of the
problem are satisfied. Thus the minimum value of a + b is 28.


[quote="math92"]If one wiggle equals two waggles and one waggle equals five woggles, how many wiggles are there in a wiggle plus a waggle plus a woggle?[/quote]

wiggle-$A$
waggle-$B$
woggle-$C$

$A=2B$, $B=5C$, so $A=10C$. We're trying to find out $A+B+C$, or $A+\frac{A}{2}+\frac{A}{10}=\frac{10A+5A+A}{10}=\frac{16A}{10}=1.6A$, which is $1.6$ wiggles. Terminator, you should show your solutions.  ;)

[quote="math92"]If one wiggle equals two waggles and one waggle equals five woggles, how many wiggles are there in a wiggle plus a waggle plus a woggle?[/quote]

wi= 2 wa
wa = 5 wo

wi + wa + wo = 1 wi + 6 wo

[quote="math92"]If one wiggle equals two waggles and one waggle equals five woggles, how many wiggles are there in a wiggle plus a waggle plus a woggle?[/quote]

Let wiggle equal A, waggle equal B, and woggle equal C.

A=2B
B=5C

Question
How many A's are in a A + B + C?

Rephrased question;
How many 10C's are in 10C + 5C + C?

16C/10C equals 16/10 which equals 8/5, which equals 1.6

wiggle=a
waggle=b
woggle=c
a=2b
b=5c
a=10c
a+1/2a+1/10a=1 6/10a=1.6a
 So the answer is 1.6 wiggles. :D


$ z \equal{} 2(x \plus{} y \plus{} z) \minus{} (2x \plus{} 2y \plus{} z) \equal{} 2(23) \minus{} 43 \equal{} 3$

$ \implies x \plus{} y \equal{} 20,x \plus{} 2y \equal{} 34\implies y \equal{} 14\implies x \equal{} 6\implies \frac xy \equal{} \frac6{14} \equal{} \boxed{\frac37}$.

I'm pretty sure that math154 has already listed how you do that problem professionally, but I'm just going to support him. I got 3/7 too. Or 6/14.

Hmmm i'm thinking this could be one of those "keep your eye on the ball" problems (i.e. solving directly instead of finding values and plugging in). does anyone else have a method that uses that?

There probably is one of those methods that utilizes proportion manipulations, but frankly, I tried longer looking for that clever solution (without any success) than I did solving it directly. :P


The side length of each of the five squares shown is 2cm. If a square is formed whose perimeter is equal to the total perimeter of the figure shown, what is the ratio of the area of the five squares below to the area of the new square? Express your answer as a common fraction.

[asy]path p = (0,0) --(0,1)--(1,1)--(1,0)--cycle;
draw(shift((1,1))*p); draw(shift((2,0))*p); draw(shift((0,2))*p); draw(shift((2,2))*p);
draw(p);[/asy]

[quote=ernie]The perimeter of the figure is $ 40$.  Thus, we would need a square having a side of $ 10$ and area of $ 100$.  Thus, the answer is $ 2 \times 2 \times 5 \div 100$, or $ \boxed{\frac25}$.[/quote]

shouldn't 2 times 2 times 5 divided by 100 equal 1/5?...


AMC
USA(J)MO
USAMO
inequalities
number theory

Is it USAMO 2000 Q6, the monster inequality?

Is it USAMO 1999 Q3, the moster number theory?

Which problem was solved by the lowest number of contestants, and which one has the lower average score, and which one has the most number of contestants scoring 0 points?

In my opinion USAMO 1999/3 is extremely difficult, even more difficult than USAMO 2000/6. John Scholes writes about USAMO 2000/6 that "no one made any headway with this in the USAMO exam. As an olympiad problem, it is exceptionally (and unreasonably) difficult." I guess it is more a matter of personal taste - if you are experienced with number theory and not with inequalities, you will find the 1999 problem the more difficult one, and vice versa. I am not really a number - theorist (well, I know quite a lot of number theory but still...) so USAMO 1999/3 really scares me ;) It is also funny that in 1999, Q3 was such an impossible problem while Q6 was completely trivial :)


4th time =\

The message is too small. Please make the message longer before submitting.

Back in my early days (300ish games) I averaged low 1300s so I was at 1337 a TON. Now if it hit 1337 i'll shoot myself. My new goal? 1337x2!!

[quote="AIME_is_hard"]THIS IS MY 9TH TIME AT ACHIEVING A RATING OF 1337! :rotfl:[/img][/quote]

good job.

except now ur bombing xD

mewto, it'll be impossible to get 2674, or wutever 1337*2 is.

[quote="math154"]Dude hell no.

You'll skip school and get $ 1337^2$ cding that [b]brother[/b] of yours. :wink:[/quote]

mewto cds his brother?

[quote="superllamel"]Mewto cds his brother a ton. Every time I'm about to catch up, his brother comes on and they cd. And mewto always wins and gets about +4 each time.[/quote]

Woah woah, more evidence that mewto = RomanianGenius. Come on, no one can deny it now.  It's almost as if I have magic powers where I'm unable to be deceived....

Shouldn't their ratings not change if they are brothers or the same person?

Different IPs.

[i]Ideally[/i] they wouldn't...

[quote="mewto55555"]Oh cmon, I get highest rating and everyone is crying that there was a conspiracy! Not everyone cheats to be #1 [hide="cough"]bobbymathking123[/hide] Just because I play my brother does not make him myself![/quote]

It's actually only one person "crying", and I was crying about it a very, very long time ago, not just since you got the highest rating.  Now I just have further evidence to support my theory.

Mewto, your highest rating would be more impressive if you didn't get all of it playing RG.

i do agree with iin. i mean why do romanian and mewto have differnent ip adresses. and whyy does rg always play mewto when super is catching up.

you guys honestly cannot be serious than you think mew and rg are the same people. mew still pwns anyone else he plays...so even if they are the same people, which they couldnt be, because when they play eachother, they glitch and stuff, he'd still be amazing.

stop being ridiculous...

[quote]Dude hell no.

You'll skip school and get $ 1337^2$ cding that brother of yours.[/quote]
He'll more likely get $ 1337!$.

Why do you people care so much about having a 1337 rating? I, personally, would rather have a 1338 rating.

I don't know what this 1337 craze is... I had never heard of it before I joined this site. But I have had a 1337 rating at least 12 times.

P.S. I do NOT think mewto55555 and RomanianGenius are the same person.

If you see:

LEET ---> 1337

You may find in on Wikipedia.  Not sure.

Yes, it is on Wikipedia.

The article is actually quite fascinating. I recommend. I read the whole thing. (No sarcasm.)

ROFL, i am 1337 AGAIN! :rotfl:  :rotfl:  :rotfl:

1337 is total leetness, but it gets boring after a while. gotta rejoin the n00bz0rs


geometry
angle bisector
geometry unsolved

Let $ ABC$ be atriangle and $ AD$ be angle bisector of angle $ BAC$.Let $ (W_1) and (W_2)$ be incircle of $ \Delta {ABD}$and$ \Delta{ACD}$.Let $ MN$ be the extangent of $ (W_1) and (W_2)$.Prove that the cicunference of trangle $ PQR$ tangent to $ MN$ wehre $ P.Q,R$ are midpoint of $ BD,CD,MN$.

can you explain me please what means "extangent"???
 :(  help please

He must have meant "external tangent of $ (W_1)$ and $ (W_2)$"

[quote="hoangclub"]Let $ ABC$ be a triangle and $ AD$ be angle bisector of angle $ BAC$.Let $ (W_1) and (W_2)$ be incircle of triangles ABD and ACD.Let $ MN$ be the external tangent (other than BC)of $ (W_1)$ and $ (W_2)$.Prove that the circumcircle of trangle PQR tangent to MN where P.Q,R are midpoints of BD,CD,MN. [/quote]
Dear hoangclub, this problem seems rather nice, unfortunately, it is incorrect for every triangles ABC.
You should edit your post.  :)


probability
linear algebra
matrix
vector
quadratics
function
calculus

Hi, I wonder if anyone here could help me with a little math problem I've ran into for a personal project... First, apologies to the moderators for almost certainly posting it in the wrong forum, but I had no idea where else to put it.

I have two variables X and Y whose value are randomly generated between -1 and 1. I have a variable Z which is equal to X*a + Y*b, where a and b are coefficients whose sum is one. Obviously, Z's value is between -1 and 1. My problem however is that I need its distribution to be uniform between -1 and 1, as if it had been randomly generated between -1 and 1, instead of favouring values close to 0, as the "averaging" of X and Y implies.
Is there a way to modify Z to make its distribution uniform again? Any solution would have to scale to variables generated from more than two other variables to be really useful to me.

If anyone knew how to solve this, I would be very grateful. I'm a CS major, not a math one, and this is stretching my probability capacities ;)

What is the actual problem?  If you have a means of generating a uniform random variable on an interval, you can use that directly to generate Z...

You didn't specify whether $X$ and $Y$ are supposed to be independent. Let's assume that they are.

Let's make this the question: if $X$ and $Y$ are independent identically distributed random variables, is there any way that $Z=aX+bY$ can have the same distribution as $X$ and $Y$?

Let's check the mean and the variance:

$E(Z)=aE(X)+bE(Y)=(a+b)E(X)$ so if we want to have $Z$ have the same mean as $X$, we must have $a+b=1.$ (You already noted this condition.)

$\text{Var}(Z)=a^2\text{Var}(X)+b^2\text{Var}(Y)=(a^2+b^2)\text{Var}(X).$

Thus in order for $Z$ to have the same variance as $X$, we would also need to have $a^2+b^2=1.$ But if $a+b=1$ and $0<a<1,$ then $a^2+b^2<1.$ Thus the only way to have the same distribution would be the trivial cases $a=0$ or $a=1.$

That's assuming independence. What if we scrap that assumption? We still must have $a+b=1$ for the expected value to turn out right.

Now, $\text{Var}(Z)=a^2\text{Var}(X)+b^2\text{Var}(Y)+2ab\,\text{Cov}(X,Y).$

Let $\text{Cov}(X,Y)=\rho\,\sigma_X\sigma_Y=\rho\,\text{Var}(X).$

Here, $ \sigma_X=\sqrt{\text{Var}(X)}$ is the standard deviation and $\rho$ is the correlation coefficient. Then

$\text{Var}(Z)=(a^2+2\rho ab+b^2)=(a+b)^2-(1-\rho)ab\le(a+b)^2=1$ with equality only if $\rho=1$ or $a=0$ or $b=0.$

Thus, $Z$ can have the same variance as $X$ (and hence have a chance at having the same distribution) only if $\rho=1$ which is perfect correlation between $X$ and $Y$ : $X=Y$ except possibly for a set of probability zero.

[quote="K"]First, apologies to the moderators for almost certainly posting it in the wrong forum, but I had no idea where else to put it.[/quote]
Let me invent the justifcation for keeping it right here, in "Matrix - Linear Algebra." Well, almost here; I'll move it to the "Computations and Tutorials" section.

I was quoting the two dimensional version of the following:

Suppose $X$ is a random $n$-vector. That is, let $X$ be a random variable taking values in $\mathbb{R}^n.$ In all of this, let all vectors be written as column vectors.

Define the covariance matrix of $X$ as

$\text{Cov}(X)=E(XX^T)-E(X)E(X^T).$

This is an $n\times n$ symmetric matrix whose $i$ th diagonal element is $\text{Var}(X_i)$ and whose $(i,j)$ element is $\text{Cov}(X_i,X_j).$

Let $a\in\mathbb{R}^n.$ Suppose $Z=a^TX.$

Then $E(Z)=a^TE(X)$ (just because the expected value is linear) and

$\text{Var}(Z)=a^T(\text{Cov}(X))a.$

Since the variance of any random variable is nonnegative, this quadratic form is nonnegative and the matrix $\text{Cov}(X)$ is (at least) positive semidefinite.

Thanks for the help :)

I see I should have been a bit more precise regarding the exact problem. My aim is to generate uniform variables correlated with other variables. As an exemple, I might have to generate R1 and R2, with R1 = 0.75 X + 0.25 Y and R2 = 0.5 X + 0.5 Z (X,Y,Z being random variables between -1 and 1). The "entry" variables (X, Y and Z in the exemple above) are independent.
I am not looking for conditions on the coefficients (apart from their sum being 1); those are imposed. What I am looking for is a function (which would depend on the coefficients, I guess) which applied to R1 and R2 would make their distribution uniform, "correcting" the averaging effect.

And it's not a pure math problem - I need it for a project and I will have to be able to code the result.

If nobody can find me the "correcting function", maybe someone could tell me what the probability density function of the resulting variable is, depending on the coefficients used? With that, I should be able to do my adjustment.

Thanks :)

If what you want is to assign a probability density to $(R_1, R_2)$ so that integrals in $R_i$ equal the corresponding $XY$-integrals, the answer is the (inverse) Jacobian determinant, i.e. the formula for change of variables in an integral.   Search for "pushforward of a measure" or "change of variables formula".

Thanks, I'll have a look at that. It might be beyond my mathematic abilities though ;)


When you reverse the digits of a particular two-digit number, the resulting number is 20% more than the original number. What is the original number?

How do you solve this problem without using guess and check?

call the 1st # 10a+b and its reversed # is then 10b+a

10b+a=1.2(10a+b)
10b+a=12a+1.2b
8.8b=11a
b=5a/4
4b=5a
so; b=5, and a=4

$ \boxed{45}$

A much faster way would be to see that for the 2nd number to be 20% more than the 1st, the 1st must be a multiple of 5, so units digit is either 0 or 5.

Cannot be 0 since then the reversed would have 0 for tens digit.................. so units digit is 5.

From there you can easily see that the tens digit would have to be 4.

Not a very good way of proof but it's definitely faster if you narrow down 1st.

Wait -- what do you mean it would have the tens digit has to be four? How did you get that?

If the original number is a5, then reversed it would be 5a which is 6/5 of a5.

basically a can be any number less than 5 (so, 1, 2, 3, or 4)
since 6/5 is not very large, a should not be much less than 5 so  should try 4  first.  If you want, trying all 4 would not take very much time either because 15, 25, and 35 all definitely cannot make 50-something when multiplied by 6/5 ------------------- 45 is the only reasonable-looking value of a.


ratio search function videos MATHCOUNTS ARML

I'm interested what is the ratio between the boys that do math and the girls that do math.

I wonder how many of these polls we have had.

the last one was 7 monthes ago  :blush:  moderators if you think that was asked to many times you can lock or delete this

sorry   :( .

[quote]Yeah, we did have a topic like this a while ago, and people started messing up the polls and voted for their opposite gender.[/quote]

I've noitced that  :) .

When i put 'girls boys' in the serarch function nothing senisble fell out. when i put just 'girl boy' some sensible things fell out.

Not trying to be sexist or start a flame war or anything, but in the AMC state scores (and I haven't checked them all) Boys seem to beat girls pretty much everywhere?

Dude you are missing some categories in the poll.

Well, Otter of course for all the xkcd-ers

I doubt there is a high proportion of hermaphrodites on AoPS though

[quote="SimonM"]Boys seem to beat girls pretty much everywhere?[/quote]

Well, if you put half the standard class as boys and the other as girls, no matter what subject, girls would prevail. Not to be mean but normally girls would skyrocket in theirs studies while boys would try only for average or a little above average.

Girls may perform better generally (but from what I can tell it is because they are generally less lazy), I was more commenting on the statistics I saw on the AMC website.

[quote="7h3.D3m0n.117"][quote="D3m0n Shad0w"]Well, if you put half the standard class as boys and the other as girls, no matter what subject, girls would prevail. Not to be mean but normally girls would skyrocket in theirs studies while boys would try only for average or a little above average.[/quote]

I read somewhere that there were statistics about that. Of course, you're a girl so you have no problem saying that  :P .[/quote]

Well the only people failing in my math class are currently boys...and one girl but thats a 2:1 ratio! That [b]has[/b] to count for something!

Right  :huh:

[quote="SimonM"]I was more commenting on the statistics I saw on the AMC website.[/quote]

Oh. What I've seen from middle schools is more boys participate in math programs than girls do. So their average "smartness" would have to be higher because of the more sheer numbers than the girls.

Sorry to be going so off topic.

EDIT:

[quote="cognos599"]why isn't there an option for both/neither, not.....that......um......well....yaknow :roll:[/quote]

That would be a waste and inappropriate...Also, there is no such thing and it would mess up poll results.

This is pretty much a dead thread until a mod feels like locking it.

Anyway, being smart isn't defined as being profficient in math, you can be skilled at other things too.

More $ \not \equal{}$ Better

I was also considering average statistics

Finally, your class is a small sample size and accepts the hard working hypothesis.

Usually  boys are smart and girls are hard-working.

hmm....  I'm a girl and I'm not a hard-working at all!!!  Can anybody tell me why did I say that stuff  blow???

we all are geting off topic. the question was if here are more boys on AOPS  or girls  and not who is smarter, a girl or a boy.

But I'll say that I think boys are often smarter than girls(it's not easy for me to say that as I'm a girl, ). I just think so.  maybe girls are more organized or more hard -working than boys, but we are not smarter then boys.
And if we have better results in anything then boys do, that's not because of our smartness, that's because of our hard work. if we beat boys in something, that's why they are lazy and do not do their best.

[quote="hunter34"]I guess to stay on topic, my gender is soccerball.

Getting sexist now are we?

I have a wii. I find it odd because my mom wants me to play it instead of doing math (Which I prefer). dreambold22, have you seen the make-up of a male teenager? most of it is about women.[/quote]

Wow, same here.  I'm like enjoying myself doing math, and my mom takes pity on me for working too hard and wants me to play the wii instead. :rotfl:

[quote="tinytim"]My mom says I'm addicted to math.... :o  :([/quote]

same... they think aops gets me high lol

[quote="RussianRocket"]im a boy (last time i checked)[/quote]

you sure? :D

i'm a boy for sure

That poll is obviously being cheated I doubt we have a 3:1 ratio, I thought it might have been 1:1.5 in favor of boys.

I'm a boy from 1 to 25 years old!!  :rotfl:  :rotfl:  :rotfl:

OK this irks me. You people are not very sensitive. There is no neither or both option.

This is weird, but I AM a girl. (haha suma_milli always laughs about that)

Cuz im a tomboy.

And I hate Miley Cyrus like ****.

And I hate girly gossip.

ANd I hate wearing high heels, dresses, and tank tops.

And I hate dating too.

Thank you.
That is all.
 :D

Then you are a lifeless being?
OMG 1=2 IS A BOT :D  HE'S BEEN MASQUERADING AS A HUMAN
I'm a guy.

AS for the topic which is so hotly debated on this thread, I say it would fluctuate.
The best person on my math team is a girl.
My sister is one of the four best people in her math team (who always make up the lead team at competitions). Three of them are girls.

But I have a theory, and one I made up just now. Studies show that men tend to give directions (I will come back to the relevant discussion in a while) that are based on cardinal directions and numerical distance (e.g. one mile north and turn west) while women tend to give directions that are based on landmarks (go past the park and turn left). Therefore my idea -and please forgive me, it's a stretch- is that boys tend to be more adept at concrete subjects (memorizing formulas) while girls are better at spatial and abstract ideas. AoPS is very dry and boring (in my experience) so more guys read it and thus see the website than girls.
Please take no offense. Feel free to completely and thoroughly dismantle my argument with a series of well-written replies.

I also noticed many differences between boys and girls besides something that many already know, and I'm not going to say. I noticed girls like keeping organized and balance out in their grades. They always seem to work much harder than some of the boys at my school. Girls also seem to take their time (not [i]that[/i] slow) when packing bags. For example, if you compare many girls' handwriting with boys', girls' always seem to be neater. Feminine also seem to "dominate" the top ten grades in about any classes. Girls also seem more social than boys, unless they are the very shy types. Girls like to think of themselves as the prettier and smarter ones, so many think of themselves better than guys. Females also don't seem to be nerds, because they usually don't specialize in anything but just get good grades. I'm not a girl, so I don't know that much about girls.

Boys on the other hand tend to focus on one thing and work towards it. It also seems that guys tend to be rushing and have messier handwritings than girls. I'm not sure if this is true about all boys, but I'm one, and I am very messy, just like some of my other friends. Last year in my class, there was a girl who had way too many things in her desk, and I think her desk was the second messiest after a boy who had to use 2 desks. In the grades, boys tend to be number 1 and 2, while girls seem to be numbers 3-10 in the class. Guys can survive without having any friends around in summer, but they still talk a lot, like girls. Boys tend to think of themselves as tougher than girls (which I don't think since I'm one of the shortest person. Ahh... I'm even shorter than izzy!) just because in the old days, men used to rule. At the age of 7 or 8, an average boy would rather have some kind of game system instead of Mp3 players, which girls tend to "need" more.

I'm not going to list the main difference between girls and boys, or what differenciates a boy from a girl, as it would be disgusting and inappropriate to discuss here.


[quote]Researchers at the UC Irvine Reeve-Irvine Research Center have used adult human neural stem cells to successfully regenerate damaged spinal cord tissue and improve mobility in mice. ...http://www.hypeandhope.com/wt/page/index/it_1127238553[/quote]

Your opinion on this?


[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ $ \in {\mathbb R}^{ \plus{} }$ , prove that :

$ \frac {bc}{\sqrt {(b \plus{} a)(a \plus{} c)}}$ $ \plus{}$  $ \frac {ca}{\sqrt {(c \plus{} b)(b \plus{} a)}}$ $ \plus{}$ $ \frac {ab}{\sqrt {(a \plus{} c)(c \plus{} b)}}$ $ \geq$ $ \frac {2(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab})^2}{3(a \plus{} b)(b \plus{} c)(c \plus{} a)}$[/quote]
$ \sum_{cyc}\frac {bc}{\sqrt {(b \plus{} a)(a \plus{} c)}}\geq\frac {4(a\sqrt {bc} \plus{} b\sqrt {ca} \plus{} c\sqrt {ab})^2}{3(a \plus{} b)(b \plus{} c)(c \plus{} a)}$ is true too. :wink:

thank's arqady

so your inequality is the best , can some one prove it ?


let m be a positive integer and M be a set containing $n^{2}+1$ positvie integers and having the following property: in each n+1 numbers from the set M there are two numbers for which one is divisible by the other. prove that the set M contains different numbers a_1, ,,, a_(n+1) for which
a_(i+1) l  a_(i) for i=1,2,    ,n


Help, please!
I'm enrolled in Introductory Counting & Probability...
I've been looking for the class forum and can't find it.

If you just enrolled, then you won't see it.  We have to cycle each class Message Board as new students join the class.  I'll go ahead and cycle it now.


When you go to school, it seems like homework is a big topic in the classroom, since most teachers will have a place to write homework assignments. But is homework really for its purpose? Teachers say that homework is to "reinforce the skills that you learned in class". That may be true, but doesn't homework add stress to both the students and the teacher? I know some teachers that leave a lot of homework because they feel that teachers should leave homework, when most of the things can be covered in class.

One of my teachers also say that homework is meant to keep you busy, and to try to keep the class important. If you're a student in China, chances are you have a lot of homework, and the same goes with other countries. I've never done any of that homework, but is that homework really to strengthen your mind, or is it just to make the teacher look like he/she is responsible (because the teacher has to grade it)? Is it that necessary to have [i]that[/i] much practice?

Without homework, there is more time to play sports and concentrate on other activities. The main question is: [b]Is it possible for students to achieve success without homework?[/b] I know that's probably impossible for math, but how about classes like history and english?

Thanks for reading my essay that goes barely anywhere.

homework does have a purpose. many people, probably not many of us, need to actually do multiple practices to understand and be able to perform the task. homework is the way to get the extra practice. for those of us who pick up on things quicker than the majority, howework seems boring, tedius, and pointless. however, the education system is not catoring to us but to the majority (remember: no child left behind) who need the extra practice whether they like doing it or not. and another thing, many people say there homework doesn't help them but they do it anyway to get good grades. they are actually solidifing the information mentally guarenteeing a full understanding of the concept whereas by not doing the howework you are relying more on memory than knowledge (thinking "this is how the teacher showed us" as oppposed to thinking "what i need to do to answer the question").
o yea, as to the bold question. yes, sucesses without homework is possible but would there be more sucess with homework? probably.

also homework is a way to improve your responsibility and work ethic, how much you are willing to work for school.  Plus a lot of things in life are boring and tedious.

probably not, unless you are a genius. for english, there are lots of things that need practice, like writing essays, POS, etc. and history too, like critical thinking, revisiting the facts, etc. maybe in some classes we could have less homework..

theres a big difference between homework and busywork, though. for example, "homework" in math is when your teacher goes through and gives you a few of every type problem to try in a practice exercise in your book. "busywork" is when he/she gives you the entire practice section to do.

my classes usually have this policy:

any work unfinished is homework

[quote="236factorial"]I sure don't get too much homework, but this is what my brother sometime has, in social studies:

for one chapter, he has to type an outline, a complete term sheet (vocab, people, places), some 20-30 pages to read, review sheets (5-8), and a test, in about a week.

It's strange how all the social studies teachers in my school leave the most homework. I have to turn in homework packets every week, when these concepts can easily be learned in just 10 minutes giving us a straightforward list of concepts. Instead, we go for 4 weeks of daily homework, revolving around about 20 vocab words. Is this busywork?[/quote]
I would say that's too much. It's a waste of time. Although I believe homework generally helps us understand the subject, too much homework ("busywork") is pointless.

In Japan, teachers don't give a lot of homework (as far as I know). Rather than giving an enormous number of practice problems or worksheets, teachers usually give us a few problems that require genuine critical thinking skills. In other words, they focus on quality, not quantity.

[b]Homework[/b] is a very broad term as far as I'm concerned.  All it means is work (school work) done at home.  As bleumoose points out, there is a critical difference between just 'homework' and 'busywork'.

[quote]
Is it possible for students to achieve success without homework?[/quote]

Yes, and no.  The answer depends on your definition of success.  If you think an A in the class is success, then you might not need the homework.  If you are aspiring to go beyond the limited scope of your high school (perhaps even college) curriculum, then you might actually learn something from your homework (depending on its content).

Now, if you change the topic of this thread from 'homework' to 'busywork' then I will agree whole-heartedly with you and say that it is highly over rated (to the advanced/gifted student that is).

Homework does have the purpose of reinforcing class material and learning new things. But it's really just because they don't have enough time in class for those things.

More importantly, teachers go way overboard on it. Homework tends to be boring. People could spend more time on here if they didn't have so much homework. I tend to find, [u]homework is the least favorite part of my life.[/u]

[quote="calc rulz"], [u]homework is the least favorite part of my life.[/u][/quote]

What a surprise  :roll:  same with some others I bet  :P

I think the main purpose of homework, contrary to what is often proclaimed to be its purpose, is to create and test a work ethic, and create a large effort-based grade, allowing hard work and dedication to triumph over actual understanding.

Of course there are many other purposes, and diferent teachers assign homework for different purposes.  Doing homework WILL help you remember, and hopefully understand material, but that doesn't mean it is necissary or even useful if you already understand and remember everything.

I hate homework, but I think in this year (Junior in high school,) some of it has actually taken on a purpose for me, and has helped me learn, although I still advocate homework free learning.

My physics and math teacher say, respectively,

"These are problems to help you work with and understand the concepts.  If you're going through them and find that it's circuitwork, then go ahead and skip them."

"At least try them to make sure you understand how to do them.  I care more that you understand how they work and can construct these things than that you can memorize them."

As for my history teacher...he doesn't check reading homework, nor does he really care if we haven't read it...
He just assigns and assumes that we read it.  There is no real consequence to not doing reading...except for the fact that we're completely lost in class and have no idea of what's going on.

I'm taking five aps this year and whats interesting is the way the different teachers view homework.  In calc she assigns problems but doesn't check although there are mandatory problem sets.  In comp sci theres the occasional reading (we've had one so far) and projects are due after five lab days but if you get it done in class there's no work.  In bio we have study guides that are optional and a review sheet we have to make for each unit that's required anything else is purely optional and up to us to do, he doesn't even tell us to read the chapters.  In history on the other hand we have tons of useless work.  We have to read nad outline the chapters and complete lots of worksheets but it doesn't seem to help much and she checks everything although she doens't always read it so in one case i only passed in half my assignment and got full credit.  I definately prefer the optional homework and I wind up reading and outlining bio without it being assigned because thats what i need to get good grades.

Homework is good. It builds character. ;)

Remember Gauss's assignment--homework can lead to new ways to figure things out. And it's practice. Especially math homework...

They say that "homework" is meant to be done at "home", hence the name. I define "home" as all the time between when the homework was assigned and when it was collected. ;)

[quote="mathnerd314"]They say that "homework" is meant to be done at "home", hence the name. I define "home" as all the time between when the homework was assigned and when it was collected. ;)[/quote]

How true  :D  I do nearly all of it at school. tHE math teacher doesn't care about what I do in class.

I think i understand what 236factorial was trying to get at.

I agree with everyone else that homework is indeed important in the learning process. I do not believe students can suceed without it. Sure a student might be able to get an A in a class without doing homework. That means 1 of 2 things. Either the kid is a freaking genious, or the class was a total kick back class.

But there is a limit to how much homework should be given. Too much homework causes alot more stress on a person. This can frustrate the student so severely that it turns the student off the subject which inherently slows down the learning process (I know because this has happened to me in both English and History).

Like everyone else has said, homework is important for reinforcing the conepts learned in class, but too much homework can result in unnecessary stress and often times, teachers who teach remedial or bullshit classes give homework daily when it is not needed to make them feel like the class is important to the students, when they know it is not.

So in conclusion, homework is good, but there needs to be reasonable limits otherwise (like what 236factorial said), the purpose is lost.

I know some teachers make the students do long note summaries (and are nearly required to be 10 pages long) because the teacher is too lazy to make up note sheets and copy them. Copying a whole chapter in 2 days doesn't teach the student anything since the student is so busy in getting it done.

Any of your teachers do this? (if they say that it's a good thing to study before your final exams, dont' believe him. If your textbook has 150 sections, and each section makes 2 note pages, you will have 300 pages on a 8 1/2 by 11" paper. good luck.  :) )

Some teachers also make you copy down the definitions and things verbatim, and therefore hinder true understanding of the words. My regular math teacher did this all the time (this, of course, was in the dark ages, before I was enlightened with honors math).

[quote="K81o7"]I've been told that homwork is meant to prepare you for college...[/quote]

I've also heard that homework is given so that you don't play video games or football the entire afternoon  :P

Now with the two packets and a chart...

In my AP physics class, the curve is insane - literally 65% curves to an A. Homework is optional in this class, and I haven't done it a single time. I am pulling like a 92 average, which is probably like 98-99 curved. Homework is not necessary for all classes and for all people - I think students should be able to choose which subjects they practice the most, because I could definitely use more time to practice English and history, but studying for AP physics would just be a waste of time for me.

Homework is given for the less gifted students to help them study and the teachers don't think it is fair that some people do homework and some don't.


Given any n + 2 integers, show that, for some pair of them, either their sum or their difference is divisible by 2n.

Do i use pigeon hole principle?

yes, but it's trickier finding out how than finding out if.

[quote="mathcrazed"]the problem can be reduced to the set S={1, 2, ..., 2n} and only worrying about the sum.  Now consider each pair (k, 2n-k).  See what you come up with.[/quote]

I don't get your solution. How can you reduce it to the set S? btw, it isn't stated that the integers are consecutive. Besides, for any k, how sure are you that there is a 2n-k in the set, assuming that the integers are not consecutive? Thanks

mathcrazed is simply looking at this problem $ \mod 2n$.  You choose the $ n\plus{}2$ integers from $ S$.

[hide]
Every term in a set $ S \equal{} {a_0,a_1...a_{n \plus{} 2}}$ that is not dividisble by 2n can be expressed as $ b_n\mod{2n}$ where $ 0 < b_n < 2n$
When adding or subracting two terms from the set, there equivalent values $ \mod{2n}$ cannot equal 0 or 2n. Therefore no two terms have equivalent values $ \mod{2n}$
therefore the largest possible set $ S$ contains terms with consecutive interget equivalents $ b\mod{2n}$ where $ 0 < b < n$ or $ n < b < 2n \minus{} 1$
Therefore the largest set contains $ n$ initgers

Due to the pigeonhole principal and set with $ n \plus{} 1$ intigers or indeed $ n \plus{} 2$ initgers must contain a pair of terms whose sum or difference is divisible by $ 2n$[/hide]


Write the formulas for the binary compounds corresponding to the following combination of elements:

$ P$ and $ Cl$

My textbook says $ PCl_{3}$ and $ PCl_{5}$.

Can't you use the oxidation numbers in the Periodic Table of Elements and have some more compounds such as:

$ P_{7}Cl_{3}$
(7 Molecules of $ P^{3\minus{}}$ and 3 Molecules of $ Cl^{7\plus{}}$)

Did I over look a rule?

Thanks!

Instead of "molecules", you should have written "ions". Also, do $ Cl^{7\plus{}}$ ions exist? And what would be the bonding in $ P_7Cl_3$?

[quote="Carcul"]Instead of "molecules", you should have written "ions". Also, do $ Cl^{7 \plus{} }$ ions exist? And what would be the bonding in $ P_7Cl_3$?[/quote]

According to my Periodic Table of Elements, $ Cl^{7 \plus{} }$ exists.

I don't get what you mean by

[quote="Carcul"]And what would be the bonding in $ P_7Cl_3$?[/quote]

by bonding he/she meant how the atoms bond.  For example, PCl_3 has a trigonal pyramidal shape exhibiting only sigma bonds, whereas PCl_5 has a tigronal bipyramidal exhibiting only sigma bonds as well.  If you think P_3Cl_7 exists, how many single and double bonds will it have?  What type of VSPER shape would exist?  Does it satisfy the octet rule?  etc.  You have to ask all these questions and make sure each element obtains 8 electrons if it was to satisfy the octet rule.

Then your Periodic Table is wrong. However, I bet what you are really trying to say is "the +7 oxidation state of chlorine exists", which is true (just consider the perchlorate anion). However, that does not mean, and is false, that the ion $ Cl^{7\plus{}}$ exists. In fact, with such a strong charge, it could never survive freely in solution.

[b]Iironiic[/b], phosphorus and chlorine are not obligated to obey octet rule, since they are elements from the 3rd period.

In general, the most electronegative element in a bond will have a negative oxidation number.  In this case, $ \text{Cl}$ is more electronegative than $ \text{P}$, and so it takes it negative oxidation state $ \text{Cl}^\minus{}$.  Then $ \text{P}$ takes on either of its positive oxidation states, $ \text{P}^{3\plus{}}$ or $ \text{P}^{5\plus{}}$.

[quote="Carcul"]However, that does not mean, and is false, that the ion $ \text{Cl}^{7\plus{}}$ exists. In fact, with such a strong charge, it could never survive freely in solution.[/quote]
I might argue that the $ \text{Cl}$ species in $ \text{ClO}_4^\minus{}$ is a $ \text{Cl}^{7\plus{}}$ ion.  I don't know that I've encountered a standard convention about this.[/quote]

It is common practice to say instead "the oxidation state of chlorine in ClO4- is +7".


inequalities
triangle inequality
combinatorics proposed
combinatorics

Let $ X_1,X_2,...,X_n$ be $ n$ points in plane. We denote
\[ S_i \equal{} \{X_j \in S; j\not \equal{} i; X_jX_i \equal{} \min\{X_kX_i; k\not \equal{} i\}\}\]
Prove that
\[ \sum_{i \equal{} 1}^n |S_i|\le 12n.
\]

Hmmm...are you sure about the bound? Because I think we can easily improve it to $ 6n\minus{}12$....

I assume that $ n \geq 3$ otherwise it is obvious.

For each $ i$ draw the line segment from $ X_i$ to each $ X_j$ which are nearest from $ X_i$. If some line segment is already drawn, you do not have to draw it twice.
Then, you have a representation of a simple finite undirected graph $ G$ with $ n$ vertices.
Note that $ G$ is planar because if the segments $ AB$ and $ CD$ intersect at $ O$ the triangle inequality leads to $ AC\plus{}BD < AB\plus{}CD$ so that $ AC<AB$ or $ BD<CD$, contradicting the definition of $ G$.

It is a well-known result (a direct consequence of Euler's formula) that since $ G$ is planar with $ n \geq 3$ vertices then it has at most $ 3n\minus{}6$ edges.
Now, for each $ i$, the number of $ X_j$ which are nearest from $ X_i$ is at most the degree $ d_i$ of $ X_i$ in the graph $ G$. This means that $ |S_i| \leq d_i$ for each $ i$.
Summing over $ i$, and using that the sum of the degree is twice the number of edges, we deduce that $ \sum_i^n |S_i| \leq \sum_i^n d_i \leq 2(3n\minus{}6)$.

Pierre.

Yes, you are absolutely right! The problem becomes really hard if some one doesn't think of planar graph.


Some stats for IMO 2005 :)

Beware, they might be somewhat inaccurate because of counting mistakes, but they should be relatively okay, and at least it gives you an idea!

Right...

[b]QUESTION 1[/b]

Question 1 was a bit tougher than expected this year. The average score for this very nice question is 2.61/7.

Number of students scoring 7/7: 140
Number of students scoring 6/7: 5
Number of students scoring 5/7: 10
Number of students scoring 4/7: 5
Number of students scoring 3/7: 19
Number of students scoring 2/7: 65
Number of students scoring 1/7: 57
Number of students scoring 0/7: 205

[b]QUESTION 2[/b]

This one was quite easy for a Q2 but again very nice. The average score for this question was 3.11/7.

Number of students scoring 7/7: 176
Number of students scoring 6/7: 2
Number of students scoring 5/7: 7
Number of students scoring 4/7: 12
Number of students scoring 3/7: 7
Number of students scoring 2/7: 12
Number of students scoring 1/7: 200
Number of students scoring 0/7: 96

[b]QUESTION 3[/b]

Again, a very nice algebra question by Hojoo Lee. This is his sixth question which makes the paper in 5 years, impressive! This question is not that hard for a Q3 but a lot of people didn't get anywhere with it. The average score is 0.92/7.

Number of students scoring 7/7: 55
Number of students scoring 6/7: 9
Number of students scoring 5/7: 0
Number of students scoring 4/7: 0
Number of students scoring 3/7: 0
Number of students scoring 2/7: 3
Number of students scoring 1/7: 22
Number of students scoring 0/7: 417

[b]QUESTION 4[/b]

Funny little question this one. I think this is the right level for a Q4 or Q1. Very nice, again.

Number of students scoring 7/7: 239
Number of students scoring 6/7: 7
Number of students scoring 5/7: 3
Number of students scoring 4/7: 0
Number of students scoring 3/7: 0
Number of students scoring 2/7: 30
Number of students scoring 1/7: 134
Number of students scoring 0/7: 87

[b]QUESTION 5[/b]

Beautiful geometry question, somewhat easyish for a Q5 (compared to last year's Q5 for example). The average for this question is 2.18/7 which is certainly not low.

Number of students scoring 7/7: 124
Number of students scoring 6/7: 8
Number of students scoring 5/7: 6
Number of students scoring 4/7: 4
Number of students scoring 3/7: 15
Number of students scoring 2/7: 30
Number of students scoring 1/7: 35
Number of students scoring 0/7: 284

[b]QUESTION 6[/b]

Strange. I am personally not very fond of this question. It is certainly not extremely hard for a Q6, with an average score of 1.35/7.

Number of students scoring 7/7: 56
Number of students scoring 6/7: 6
Number of students scoring 5/7: 2
Number of students scoring 4/7: 15
Number of students scoring 3/7: 12
Number of students scoring 2/7: 54
Number of students scoring 1/7: 39
Number of students scoring 0/7: 322

==============================

The average for the competition is 13.97/42. The paper seems easier than last year's paper to me so the bronze cut of 12/42 is surprisingly low, in my opinion. An obvious explanation for this is the fact that there are so many new countries participating (these countries score pretty low and bring the bronze cut down). The paper was definitely quite nice this year.

According to the official report (stats by Luis Cruz Romo), this should be corrected to:

QUESTION 1

Average score 2.61/7.
Number of students scoring 7/7: 141
Number of students scoring 6/7: 5
Number of students scoring 5/7: 11
Number of students scoring 4/7: 5
Number of students scoring 3/7: 20
Number of students scoring 2/7: 65
Number of students scoring 1/7: 59
Number of students scoring 0/7: 207

QUESTION 2

Average score 3.05/7.
Number of students scoring 7/7: 175
Number of students scoring 6/7: 2
Number of students scoring 5/7: 7
Number of students scoring 4/7: 11
Number of students scoring 3/7: 7
Number of students scoring 2/7: 13
Number of students scoring 1/7: 202
Number of students scoring 0/7: 96

QUESTION 3

Average score 0.91/7.
Number of students scoring 7/7: 55
Number of students scoring 6/7: 9
Number of students scoring 5/7: 0
Number of students scoring 4/7: 0
Number of students scoring 3/7: 0
Number of students scoring 2/7: 3
Number of students scoring 1/7: 23
Number of students scoring 0/7: 423

QUESTION 4

Average score 3.78
Number of students scoring 7/7: 240
Number of students scoring 6/7: 8
Number of students scoring 5/7: 6
Number of students scoring 4/7: 4
Number of students scoring 3/7: 16
Number of students scoring 2/7: 30
Number of students scoring 1/7: 145
Number of students scoring 0/7: 88

QUESTION 5

Average score 2.16/7
Number of students scoring 7/7: 125
Number of students scoring 6/7: 8
Number of students scoring 5/7: 6
Number of students scoring 4/7: 4
Number of students scoring 3/7: 16
Number of students scoring 2/7: 30
Number of students scoring 1/7: 36
Number of students scoring 0/7: 288

QUESTION 6

Average score 1.34/7.
Number of students scoring 7/7: 56
Number of students scoring 6/7: 6
Number of students scoring 5/7: 2
Number of students scoring 4/7: 15
Number of students scoring 3/7: 13
Number of students scoring 2/7: 57
Number of students scoring 1/7: 39
Number of students scoring 0/7: 325

TOTAL SCORES

Average score: 13.88/44

(Score - Number of participants - cumultative sum - rank)
Gold:
42 - 16 - 16 - 1
41 - 2 - 18 - 17
40 - 1 - 19 - 19
39 - 1 -20 - 10
38 - 2 - 22 - 21
37 - 6 - 28 - 23
36 - 6 - 34 - 29
35 - 8 - 42 - 35
Silver:
34 - 3 - 45 - 43
33 - 1 - 46 - 46
32 - 6 - 52 - 47
31 - 4 - 56 - 53
30 - 8 - 64 - 57
28 - 7 - 71 - 65
27 - 4 - 92 - 89
26 - 7 - 99 - 93
25 - 8 - 107 - 100
24 - 6 - 113 - 108
23 - 8 - 121 - 114
Bronze:
22 - 9 - 130 - 122
21 - 8 - 138 - 131
20 - 5 - 143 - 139
19 - 12 - 155 - 144
18 - 19 - 174 - 156
17 - 16 - 190 - 175
16 - 16 - 206 - 191
15 - 19 - 225 - 207
14 - 13 - 238 - 226
13 - 5 - 243 - 239
12 - 6 - 249 - 244
No medal:
11 - 12 - 261 - 250
10 - 17 - 278 - 262
9 - 19 - 297 - 279
8 - 25 - 322 - 298
7 - 6 - 328 - 323
6 - 13 - 341 - 329
5 - 15 - 356 - 342
4 - 12 - 368 - 357
3 - 27 - 395 - 369
2 - 49 - 444 - 396
1 - 41 - 485 - 445
0 - 28 - 513 - 486


number theory
greatest common divisor
number theory proposed

find all pairs of $x$ and $y$ of integers such that:

$x^y-y^x=1$

1) y<0 have not solution.
2)y=0 all $x\not=0$ is solution.
3) y=1 give x=2.
4) y=2 give x=3.
5)$y\ge 3$ give not solutions.

Rust: very good and complete solution...  :roll:

So what about posting the complete one¿

That's not really a (helpful) hint

Here's an elementary solution:

Assume $x, y \geq 3$.  (We have the solutions $(3,2), (2,1), (k, 0), k \neq 0$) We have $y$ even and $x$ odd (or else RHS is 3 mod 4).  Let $y = 2^m n$ with $n$ odd and $m \geq 1$.  Let $z = x^n$ so,

$2^{mx}n^x = x^{2^m n} - 1 = (z - 1)(z+1)\cdots (z^{2^{m-1}}+1)$

and note that all factors except $(z-1)$ and $(z+1)$ are necessarily 2 mod 4.  Now, the GCD of these two factors is 2.  Note that

$z - 1 = x^n - 1 = (x-1)(x^{n-1}+ \cdots + 1)$ has at most $x-2$ factors of 2 since $2^{x-1} > x$ for $x \geq 3$
Similarly factor $z+1$ so $x^{2^m n} - 1$ has at most $m+x-2$ factors of 2 while $2^{mx}n^x$ has a lot more -- $mx$ factors of 2, contradiction.

the solution i had in mind was a bit diffrent
but the idea is still the same (deriving contradictions)

ThAzN1 solution works :)


Find all $f: R+\to R+$ satisfying the condition:
$f(x^2+f(y))=y+(f(x))^2$
(R+ does not contain 0)
Another pro:solve it with f:Q+->Q+

Yes, it would be nice to see any solutions of it.  Unfortunately, I couldn't solve it.
The difficulty here comes from the fact that we can't use $f(0)=0$


Given that the positive integers $ a$, $ b$, $ c$, and $ d$ satisfy $ \frac {a}{b} < \frac {c}{d} < 1$, arrange the following in order of increasing magnitude:

$ \frac {b}{a}, \ \frac {d}{c}, \ \frac {bd}{ac}, \ \frac {b \plus{} d}{a \plus{} c}, \ 1$.

Thanks.

Use these numbers to boost up your confidence and then proceed to prove the statement: $ \frac45$ and $ \frac23$

[quote="10000th User"]Use these numbers to boost up your confidence and then proceed to prove the statement: $ \frac45$ and $ \frac32$[/quote]
*ahem*
$ \frac{3}{2}<1$???

[quote="Urc"]Maybe he means 2/3 instead.[/quote]Yeah, that's what I meant  :oops:  Edited.

Yeah, but you could be confusing someone since $ \frac{4}{5}\not<\frac{2}{3}$. =P


If you really really want to, I believe the admins can at least deactivate it.


geometry
3D geometry
trigonometry
ARML
rotation
limit
AoPSwiki

The Lassiter Invitational is this Saturday!

Who is coming?  I hope to see you there.  Even if I don't see you, you will probably see me there, as I'm helping to organize it.

Good luck on the tests!

I'll be there as an individual. I'm very excited as it's my first in-state tournament since Cobb. Also, the [varsity] written test last year was very good, so hopefully it will be this year too.

On a side note, my sister just told me that a kid in her math class lost points on his quiz for calling a hypotenuse a hippopotamus. :rotfl:

[quote]On a side note, my sister just told me that a kid in her math class lost points on his quiz for calling a hypotenuse a hippopotamus. Rotfl[/quote]
That's funny.
I will definitely be going, with the rest of my team.

I will be there. I'm assumuing that it starts at around 9, and registration starts at around 8:15. is this right?

I doubt that anybody from Lassiter will be there before 8:30.  I believe that registration is supposed to start at 9:00, so the competition should probably begin around 10:00.

Grr... I cannot sleep and thus will most likely be tired tomorrow at Lassiter. But I am very pleased that the Lassiter people made the tournament start at 10 so at least I will get to sleep in a little. Lack of sleep has been effecting me all week. On my last history quiz, I mistakenly implied that Abraham Lincoln was pro-slavery. :huh:

I wasn't able to stay for awards or to see scores, so how did you all do?
I personally scored a 123 on the written and a 40 on the ciphering. Disappointed in my ciphering score, but I understand what I didn't now, so overall I'm fine with it.

On the written I'm not sure how to go about doing 11, 13, 19, and 30 though - atleast without wasting too much time, any hints for those?

Jeff from Etowah High School by the way, pleasure to meet you all.

Hey Jeff,
For number 11, notice that once we choose any five digits, there is only one ordering that fulfils the given conditions (unless we choose five zeroes in which case, we can't fulfil the given conditions). After that, use the balls and urns arguement. If you don't know what that is, try searching for it on this forum.
For #13, notice that for a number to be a perfect square and a perfect cube, it must be a perfect sixth power. $ 10^{30}$ is the sixth power of $ 10^5$, so all the numbers less than $ 10^5$ have sixth powers less than $ 10^{30}$. There are $ 10^5\minus{}1$ such integers.
I don't have any great way to do number 19 and 30, so I'll leave them to someone else.

Anyways, congradulations to Santhosh, Howard, Harrison, and everyone else who did well. The written test had some interesting problems, but I would have liked to see a couple more harder problems (especially on the free response).

Although this did not really effect the problem, I see a mistake on #23: $ CE$ has a length of $ 11$, a length that cannot be contained within a unit circle - thus the situation presented is impossible. It really doesn't matter much though.

By the way, nice side-step on Billy's "what if $ k\equal{}0$" question, Miles. :rotfl:

Here is my hint for number 30 :  try to find something that doesn't change.

Here is the problem, by the way :
[hide]We define some sequences of numbers as follows :  $ x_0 = 17$, $ y_0 = 3$,
and $ z_0 = 1$, and for any positive integer $ n$,
\begin{align*} x_n & = \frac {4x_{n - 1} + y_{n - 1} + z_{n - 1}}{6} \\
y_n & = \frac {x_{n - 1} + 4y_{n - 1} + z_{n - 1}}{6} \\
z_n & = \frac {x_{n - 1} + y_{n - 1} + 4z_{n - 1}}{6} . \end{align*}
As $ n$ becomes larger, $ x_n$ always approaches a certain number.
What is this number?

[hide](I guess my only complaint about that problem is that it is kind of trivialized by considering an advanced concept (Markov chains).  But I actually thought of the problem independently of Markov chains; I later realized the implications for matrices, and then when we were talking about diffusion in chemistry, I remembered that my dad preparing a talk on osmosis and Markov chains, and realized that my problem was essentially an example of those.)[/hide][/hide]

I see nothing wrong with 23.  Who said that $ CE$ had to be contained in a unit circle?   :wink:
[asy]size(300); defaultpen(1); pair A=expi(1), B=expi(2), C=expi(5.2), D=expi(4.5); pair X=A+20(A-B), Y = C+20(C-D); path p=(A--X), q=(D--Y); pair E=intersectionpoint(p,q); draw(circle((0,0),abs(A))); draw(B--E); draw(D--E); label("$A$",A,NE); label("$B$",B,NW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,.5*E); dot(A,linewidth(3)); dot(B,linewidth(3)); dot(C,linewidth(3)); dot(D,linewidth(3)); dot(E,linewidth(3));[/asy]
(You're not the first person who thought that was impossible, though.)

[hide="Problem 23"]
Let $ \alpha$, $ \beta$, $ \gamma$, and $ \delta$ be angles.  Suppose $ A = (\cos{\alpha}, \sin{\alpha})$, $ B = (\cos{\beta}, \sin{\beta})$, $ C = (\cos{\gamma},\sin{\gamma})$, and $ D = (\cos{\delta}, \sin{\delta})$, and let $ E$ be the intersection of $ AB$ and $ CD$.  If $ AE\cdot EB = 132$ and $ CE = 11$, what is the length of $ ED$?[/hide]
Yes, we were trying not to murder everybody, and apparently we succeeded—the score distribution was about where it should have been.  To be honest, I would have been surprised if Harrison hadn't scored a perfect on that power question, but I thought it would be a cool thing to show everybody.  I admit I was slightly disappointed with Howard's perfect ciphering, although I did kind of expect somebody to score a perfect on ciphering.  It is important, though, that we not put all of the hard questions in the free response, as they stagger scores, making ties less likely.  This is obviously ineffective in general if non-ARML people can't solve any of them to begin with.  Thus you saw only two problems on the free response that are especially challenging by the standards of a local math tournament.  Thanks for your comments, though.

Anyhow, there were some typos in the JV test; I'm sorry about those.

[hide="Problem 19"]Either a domino is symmetric about its center, or it isn't. Since there are $ 4$ pairs of opposite dots, there are $ 2^4 \equal{} 16$ symmetric dominoes. Thus there are $ 2^8 \minus{} 2^4 \equal{} 240$ asymmetric dominoes. Each asymmetric domino can be rotated to form "different" asymmetric domino, while each symmetric domino rotated about its center forms the same domino. Thus there are $ \frac {240}{2} \plus{} 16 \equal{} 136$ distinct dominoes. [/hide]

[hide="Problem 30"]
30. Adding these equations together yields $ x_n \plus{} y_n \plus{} z_n \equal{} x_{n \minus{} 1} \plus{} y_{n \minus{} 1} \plus{} z_{n \minus{} 1}$
Thus $ x_n \plus{} y_n \plus{} z_n \equal{} x_0 \plus{} y_0 \plus{} z_0 \equal{} 21$ for every term in the sequence.
The equations show that $ x_n$, $ y_n$, and $ z_n$ are just different weighted averages of $ x_{n \minus{} 1}$, $ y_{n \minus{} 1}$, and $ z_{n \minus{} 1}$.  As $ n$ increases, $ x_n$, $ y_n$, and $ z_n$ get closer to each other.
Thus $ \lim_{n\to\infty}x_n \equal{} \lim_{n\to\infty}y_n \equal{} \lim_{n\to\infty}z_n \equal{} \frac {21}{3} \equal{} 7$[/hide]

[quote="Boy Soprano II"]I see nothing wrong with 23.  Who said that $ CE$ had to be contained in a unit circle?   :wink: [/quote]
Ah yes. No mistakes in the varsity competition then. That's pretty impressive.

i did terrible today
im a freshman and took the varsity
got like 75 mc points and 10 ciphering..
but my team placed 13th, and got the most improved trophy.
my school (Parkview) only had 1 team-and it consisted of a freshman(me) and 3 sophomores.
and our math team teacher is so stupid, we had to register ourselves..and she didn't even come.
and she had no excuse.

and what [i]if[/i] k=0?
that was really funny.
and miles just ignored it.
 :rotfl:

well because it's inconsequential, as the question wouldnt exist if k=0. anyway very good test miles. especially the ciphering. i think i liked that part the most ;)

by the way, on number 7, don't complex numbers require the imaginary part to be non-zero? i mean i got it anyway (by process of elimination, which was a pain, but still worked) but i did spend a lot more time on that than i would have liked.

[quote="captcha000"]by the way, on number 7, don't complex numbers require the imaginary part to be non-zero? i mean i got it anyway (by process of elimination, which was a pain, but still worked) but i did spend a lot more time on that than i would have liked.[/quote]

Yes, I did the same thing but got it wrong. I did all this searching for roots and synthetic division and wasn't very happy when my answer wasn't in the answer choices. According to AOPSwiki, the imaginary part can be zero (but I still think the question was slightly misleading).

Number 7 : [hide] ax + b = 0 The roots equal -b/a.
(x-a)(x-b)=x^2-(a+b)x+ab
(x-a)(x-b)(x-c) = x^3-(a+b+c)x^2...
And so forth for all polynomials.
So a polynomial of the form (x^y+bx^(y-1)...)
The sum of the roots is equal to -b.
However this only applies if the first coefficient is 1. So the sum of the roots is really equal to -b/a.[/hide]

To Miles:

Hey could you post the PDF version or TEX version of the test?

I still don't understand why the case k=0 doesn't matter; then x = 0 + undef so x is undefined, which is not equal to k...

[quote="Ziggurat"]Karth, if you'd like I could post scans of my test. You'd have to deal with the pages being filled with scribble mostly unrelated to the problems as well as some answers (and non-answers circled.) I bet they have an electronic copy somewhere though.[/quote]

Sure that'd be great :)

http://img521.imageshack.us/my.php?image=lassitervarsity1zi4.gif
http://img132.imageshack.us/my.php?image=lassitervarsity2vv5.gif
http://img204.imageshack.us/my.php?image=lassitervarsity3ag3.gif
http://img338.imageshack.us/my.php?image=lassitervarsity4sg7.gif
http://img151.imageshack.us/my.php?image=lassitervarsity5dn8.gif
http://img204.imageshack.us/my.php?image=lassitervarsity6yu2.gif
http://img204.imageshack.us/my.php?image=lassitervarsity7ev2.gif
The rules were: 80 minutes, no calculator, 5 points each on multiple choice, 7 for the free response. No penalty for guessing. Figures not drawn to scale.

I tried to crop as much of my chicken scratch as I could out, but there's still a lot there. And I apologize for the graininess, my scanner's rather old.

:mad:  :chief:  :wallbash_red:

I got 11th!!! Grrrr!!

So close to getting an award...no...I shall now go cringe in anger.

BTW who are you ziggurat?

Thanks for the kind comments, everybody.  I'm glad we were able to help some of you overcome your confusion about the definition of complex numbers.

If you want a higher-resolution, smaller version of the varsity written test, here it is.  I've also included the varsity power question.  Including the TEX version of the varsity written test would be rather difficult, as the test was not contained in a single file, but rather more than thirty.

I'm Jeff Powell from Etowah High School, I'm not at all surprised if nobody knows of me. This is only my second year on a math team (My school in California for both elementary and Freshman/Sophomore year didn't have anything like this). I'm a senior this year, and I want to atleast make it to the AIME before I graduate.
Edit: Plus I'm brand new to AOPS

It's ironic. I placed highest in my school yet there were still 8 people who all did better than I did.
Good thing Northview nabbed first place in both JV and varsity teams. If Karthik had gone, Heng wouldn't have placed.
Overall, the competition was nice. I unfortunately forgot to bring lunch and had to borrow Jimmy's.  :(

[quote="Benjamin Hu"]It's ironic. I placed highest in my school yet there were still 8 people who all did better than I did.[/quote]
Then I guess it was even more ironic at UGA 2006 when I placed highest in my school yet there were still 44 people who beat me. :wink:

@Jeff: Do you have at least AoPS2 and possibly also AoPS1? Those are really great books to start with in your quest for the AIME :).

I just ordered both of them recently actually, thanks for the suggestion. I don't suppose there's anything like college math competitions? (Short of things like the Clay problems) I'll probably end up donating them to our math team (it could definitely use the help) when I graduate. I probably need AOPS1 more than 2, my Geometry is really weak.

[quote="Ziggurat"]I don't suppose there's anything like college math competitions?[/quote]
The Putnam is a contest for undergraduates. I don't know much about it, but I gather that it's pretty popular. You should look it up.


[quote="silouan"]If $a,b,c>0$ prove or disprove the following inequality

$\frac{a^k}{(b+c)^{-1}}+\frac{b^k}{(a+c)^{-1}}+\frac{c^k}{(a+b)^{-1}}\leq\frac{(a+b+c)^k}{3^{k+1-1}[2(a+b+c)]^{-1}}$[/quote]
Why did you write your inequality in such a strange manner? :huh:
$a^k(b+c)+b^k(c+a)+c^k(a+b)\leq 6\left(\frac{a+b+c}{3}\right)^{k+1}$

It is generally not true. Since when $a=b=1$ and $c$ is very small it becomes
$2\le 6(\frac{2}{3})^{k+1}$

Or I am being stupid and in fact you're asking that for which k does the inequality hold?

It's not true for $a=b$ and $c\rightarrow 0$.

P.S. Just 1 minute later than Darkseer. :(


Suppose 3n + 1 points in the plane are given such that for any four of the given points, it is possible to select two points (among these four points) such that the distance between these points does not exceed 1. Show that there exists a disk of radius 1 which contains n + 1 of these points.

Isn't it straightforward? :wink:
I mean, assume for a contradiction that any disk of the plan contains at most $n$ of the given points.
Let $A$ be one of the points. Then the disk $D_{A}$ with center $A$ and radius $1$ contains at most $n-1$ of the other points, so that we may find at least $2n+1$ of the given points which are outside $D_{A}$.
Let $B$ be one of these points, and $D_{B}$ be the disk with center $B$ and radius $1$. As above, we may find at least $n+1$ points which are outside $D_{A}$ and $D_{B}$.
Let $C$ be one of these points and $D_{C}$ be blablabla...We may find at least one point, say $E$ which is outside each of the three disks $D_{A},D_{B},D_{C}.$
Now, $A,B,C,E$ form a set of four points with all mutual distances greater than $1$. A contradiction.

Pierre.

also it can be solved with graphs I think...
construct a graph with vertices the points we have,and connect two of them if the distance between the points didnt exceed $1$...

Straightforward indeed, is there anything wrong with that? :) I think this is a cute problem.


AD and BE are altitudes of triangle ABC. If AC=10, BC=8, and AD=5, then what is BE?

[hide="My sorution"]
The area of $\bigtriangleup \text{ABC}$ is \[\frac{1}{2}\text{(BC)(AD)}=\frac{1}{2}(8)(5)=20\]
This area may also be expressed as \[20=\frac{1}{2}\text(10)(BE) \Rightarrow \text{BE}=4\]
[/hide]


The graph below shows a portion of the curve defined by the quartic polynomial $ P(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$. Which of the following is the smallest?

$ \textbf{(A)}\ P( \minus{} 1)$
$ \textbf{(B)}\ \text{The product of the zeros of }P$
$ \textbf{(C)}\ \text{The product of the non \minus{} real zeros of }P$
$ \textbf{(D)}\ \text{The sum of the coefficients of }P$
$ \textbf{(E)}\ \text{The sum of the real zeros of }P$

[img]13164[/img]

[hide="Diagram Abuse"]P(-1) is between 4 and 5. The product of the zeros is d or the y-intercept which is between 5 and 6. The product of the non-real zeros of P is the product of the zeros of P divided by the product of the real zeros of P. The real zeros of P are between 1.5 and 2 and 3.5 and 4, summing to between 5 and 6, and multiplying to between 5.25 and 8. At most, the product of the non-real zeros of P is 6/5.25. The sum of the coefficients of P is P(1) which is between 3 and 4. Clearly, answer choice C is the least.  [/hide]

[quote="123456789"]How did you find the maximum bound on the product of the non-real roots?[/quote]
Using Vieta's theorem and the fact from the figure that f(0) is approximately 2.9, what can you say about the product of all roots of f(x)(note that the figure already shows that f(x) has at least two positive real-roots whose product is approximate 1.8*3.8>2.9, so what can you infer about the product of the other two(if f(x) has 4 real roots) :wink:

[b]EDIT: Oopps, the figure is not scale like I think it is. f(0) should be 5.8 instead of 2.9. Similar to the product of the two given positive real-roots[/b]


Prove or disprove that for positive real numbers $a,b,c$
\[ \dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{c^2+a^2}\geq\dfrac{a^3}{a^3+b^3}+\dfrac{b^3}{b^3+c^3}+\dfrac{c^3}{c^3+a^3}, \]
and it is curious to know that equality cases are when $a=b=c$ or any two of the three numbers equal.

It is not true. Let $a \rightarrow \infty$.

Yeah... Suppose that It's true. Then:

$\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{b^2+c^2}+\dfrac{a^2}{c^2+a^2} \ge \dfrac{b^3}{a^3+b^3}+\dfrac{c^3}{b^3+c^3}+\dfrac{a^3}{c^3+a^3}$

Sum of two inquality we have:
$1\ge 1$

Then:

$\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{c^2+a^2} = \dfrac{a^3}{a^3+b^3}+\dfrac{b^3}{b^3+c^3}+\dfrac{c^3}{c^3+a^3}$

Which is a contraction.