Euler
AMC
AIME
LaTeX
limit
geometric series
real analysis

For this post, $ .x^ \minus{}$ means $ .xxx...$, a.k.a. infinitely many $ x$ digits in the decimal expansion.  For example, I believe (and so will most or all of you) that $ .1^ \minus{} \equal{} \frac {1}{9}$.

The question that I am about to ask is related to the fact that $ .\displaystyle{9^ \minus{} } \equal{} 1$.
First of all, I do believe that $ .\displaystyle{9^ \minus{} } \equal{} 1$, so don't think that I don't believe so.

Is $ .1^ \minus{}$ [b]defined[/b] to be $ \frac {1}{9}$?

I've been trying to (unsuccessfully) convince someone that $ .9^ \minus{} \equal{} 1$, but that person refuses to believe that $ .1^ \minus{} \equal{} \frac {1}{9}$.
I've already scanned through the past AoPS topics about this, and all of the proofs that I've found either relied on $ .1^ \minus{} \equal{} \frac {1}{9}$, $ .3^ \minus{} \equal{} \frac {1}{3}$, $ 10(.9^ \minus{} ) \equal{} 9.9^ \minus{}$, or $ .9^ \minus{} \equal{} \sum_{k \equal{} 1}^\infty\frac {9}{10^k}$; all of these facts are rejected by the person (according to him, infinitely repeating decimals are only an estimate of whole numbers such as $ 1/3$, and are not defined to be equal to those whole numbers).  If my question is answered (with some sort of proof from a reliable source), then I would have a better chance of convincing the person.

you just won't let it go will you? :P

i'm not the person he was trying to convince don't worry.

Then don't use a fraction-based proof:

Let $ x\equal{}.\bar{9}$.  Then obviously $ 100x\equal{}99.\bar{9}$.  So $ 99x\equal{}100x\minus{}x\equal{}99.\bar{9}\minus{}.\bar{9}\equal{}99$.  Then $ x\equal{}1$, as we wanted.

[quote="JRav"]Then don't use a fraction-based proof:

Let $ x \equal{} .\bar{9}$.  Then obviously $ 100x \equal{} 99.\bar{9}$.  So $ 99x \equal{} 100x \minus{} x \equal{} 99.\bar{9} \minus{} .\bar{9} \equal{} 99$.  Then $ x \equal{} 1$, as we wanted.[/quote]
Umm, I said:
[quote="MathAndKnowledge"]I've already scanned through the past AoPS topics about this, and all of the proofs that I've found either relied on $ .1^ \minus{} \equal{} \frac {1}{9}$, $ .3^ \minus{} \equal{} \frac {1}{3}$, $ 10(.9^ \minus{} ) \equal{} 9.9^ \minus{}$, or $ .9^ \minus{} \equal{} \sum_{k \equal{} 1}^\infty\frac {9}{10^k}$; all of these facts are rejected by the person (according to him, infinitely repeating decimals are only an estimate of whole numbers such as $ \frac{1}{3}$, and are not defined to be equal to those whole numbers).[/quote]
One of the things that I mentioned which the person doesn't believe to be true is that $ 10(.9^ \minus{} ) \equal{} 9.9^ \minus{}$, so he certainly won't believe that $ 100(.9^ \minus{} ) \equal{} 99.9^ \minus{}$.  Your proof will not convince him (and I've already tried it).

To cognos599: Yes, I can't let this go as of yet.  It quite frustrates me when someone rejects mathematical proof that is clearly true.  It has been widely believed that $ .9^ \minus{} \equal{} 1$ (in fact, Euler proved it in 1770).  Yet this arrogant person thinks that he is right.  It really is not wise to question (unless you are extremely well-knowledgable in the subject in question yourself) centuries of proven knowledge that is believed by professionals today. (By the way, this person is an 11th grader that couldn't make it to AIME.  While this fact isn't something to laugh at, seeing as many people don't make it to AIME, it certainly shows that the person can't just assume that he knows more about this than I or millions of other people who agree with me, because he is not very knowledgable about mathematics.)
I need some convincing way to prove him wrong.

A LaTeX tip:  [code]$\overline{xx}$[/code] produces $ \overline{xx}$.

If this person doesn't believe that $ 100(.\overline{9})\equal{}99.\overline{9}$, then what does he believe it's equal to?

[quote="cf249"]If this person doesn't believe that $ 100(.\overline{9}) \equal{} 99.\overline{9}$, then what does he believe it's equal to?[/quote]

Most likely a number that's infinitesimally close to $ 1$ (from $ 0.\overline{9}$). We can see why a person might believe this - it's not exactly one, and there's always that infinitesimal amount missing.

I don't know what he believes it does equal to, but I think that 7h3.D3m0n.117 is correct.  Either way, if I can say that $ \frac{1}{9}$ is defined to be $ .\overline{1}$, then I think I'll be able to convince him.

[quote="MathAndKnowledge"]Either way, if I can say that $ \frac {1}{9}$ is defined to be $ .\overline{1}$, then I think I'll be able to convince him.[/quote]

If he doesn't believe it, then he's saying that $ \frac{9}{9} \neq 1$.  :D

How can this person reject that

$ .\overline{9} \equal{} \sum_{k \equal{} 1}^\infty\frac {9}{10^k}$?

That is the very definition of $ .\overline{9}$.

It is a fact that the sum of this infinite series is $ 1$.

[quote]Is  $ .\overline{1}$ defined to be $ \frac19$? [/quote]

The answer to this is no.

$ .\overline{1}$ is defined to be $ \sum_{k\equal{}1}^\infty \frac{1}{10^k}$.

It is a theorem that the sum of this infinite series is $ \frac19$.

Your friend has a misunderstanding. An infinite decimal is not an approximation.

.9 repeating is not an infinitesimal amount less than 1, and therefore equal to 1. It [i]is[/i] 1.

An infinite decimal defines a unique real number by producing a Cauchy sequence that converges on that number. It is much like saying $ \lim_{x \to \infty} \left( 1 \plus{} \frac{1}{x} \right)^x$ [i]is[/i] $ e$, not an approximation to $ e$.

Recall that the real numbers is the smallest set R such that any convergent Cauchy sequence of rational numbers converges to some element of R.

No infinite geometric series, eh?

[quote="MathAndKnowledge"](By the way, this person is an 11th grader that couldn't make it to AIME. [/quote]

Simply having heard of AIME is an comparably impressive mathematical feat for any person of any age, the way it is now...  :lol:

I hope my post will contain some new ideas.

How about this argument: if $ 0.\overline{9}$ and $ 1$ are not the same, then there has to be some number in between them. (Hopefully your friend will accept THIS fact... see below) Can you name it?

If you need to convince your friend that there is always a number between two different numbers: the average (the mean $ \frac{a\plus{}b}{2}$) should be one.

Or how about this:

$ x \equal{} 0.999999999999999\ldots \\
1 \minus{} x \equal{} 0.00000000000000\ldots\ldots1$ (the "1" is after infinitely many zeros)
$ 10x \equal{} 9.99999999999999\ldots \\
10 \minus{} 10x \equal{} 0.0000000000000\ldots\ldots1\\
1\minus{}x \equal{} 10 \minus{} 10x \\
1\minus{}x \equal{} 10(1\minus{}x) \\
1 \equal{} 10$ ??????
Contradiction, thus you must assume you divided by $ 0$ somewhere. Where? At the second to last line, you divided by $ 1\minus{}x$. So $ 1\minus{}x \equal{} 0 \Rightarrow x\equal{}1$.

[quote="MathAndKnowledge"]
Is $ .1^ \minus{}$ [b]defined[/b] to be $ \frac {1}{9}$?

I've been trying to (unsuccessfully) convince someone that $ .9^ \minus{} \equal{} 1$, but that person refuses to believe that $ .1^ \minus{} \equal{} \frac {1}{9}$.
I've already scanned through the past AoPS topics about this, and all of the proofs that I've found either relied on $ .1^ \minus{} \equal{} \frac {1}{9}$, $ .3^ \minus{} \equal{} \frac {1}{3}$, $ 10(.9^ \minus{} ) \equal{} 9.9^ \minus{}$, or $ .9^ \minus{} \equal{} \sum_{k \equal{} 1}^\infty\frac {9}{10^k}$; all of these facts are rejected by the person (according to him, infinitely repeating decimals are only an estimate of whole numbers such as $ 1/3$, and are not defined to be equal to those whole numbers).  .[/quote]
people here have given enough arguments to convince yre friend.
he doesnt believe $ .1^ \minus{} \equal{} \frac {1}{9}$
prov like this: let x=0.111111.....
10x=1.11111111........
10x=1+0.1111111111......
10x=1+x
9x=1
so, $ x\equal{}\frac{1}{9}$

As the last poster pointed, we've already constructed several mathematically sound arguments.  If you're friend doesn't want to believe them, then he's the ignorant one because any mathematician will agree with it, and, if you get around to real analysis and the construction of the real numbers, you learn that this is indeed true.

Here's a couple more:

$ x\equal{}.\overline{9}$

$ 10x\equal{}9.\overline{9}$

$ 10x\minus{}x\equal{}9.\overline{9}\minus{}.\overline{9}$

$ 9x\equal{}9$

$ x\equal{}1$

$ .\overline{9}\equal{}1$

$ \frac{1}{3}\equal{}.\overline{3}$

$ \frac{2}{3}\equal{}.\overline{6}$

$ \frac{1}{3}\plus{}\frac{2}{3}\equal{}.\overline{3}\plus{}.\overline{6}$

$ 1\equal{}.\overline{9}$


induction
inequalities unsolved
inequalities

$x\in \mathbb{R}\ ,\ n\in \mathbb{Z}^-$.prove that:
$(1+\frac{1}{x})^n\geq 1+\frac{n}{1+x}$


induction
LaTeX
combinatorics proposed
combinatorics

In the following, a [i]word[/i] will mean a finite sequence of letters "$a$" and "$b$". The [i]length[/i] of a word will mean the number of the letters of the word. For instance, $abaab$ is a word of length $5$. There exists exactly one word of length $0$, namely the empty word.

A word $w$ of length $\ell$ consisting of the letters $x_1$, $x_2$, ..., $x_{\ell}$ in this order is called a [i]palindrome[/i] if and only if $x_j=x_{\ell+1-j}$ holds for every $j$ such that $1\leq j\leq\ell$. For instance, $baaab$ is a palindrome; so is the empty word.

For two words $w_1$ and $w_2$, let $w_1w_2$ denote the word formed by writing the word $w_2$ directly after the word $w_1$. For instance, if $w_1=baa$ and $w_2=bb$, then $w_1w_2=baabb$.

Let $r$, $s$, $t$ be nonnegative integers satisfying $r + s = t + 2$. Prove that there exist palindromes $A$, $B$, $C$ with lengths $r$, $s$, $t$, respectively, such that $AB=Cab$, if and only if the integers $r + 2$ and $s - 2$ are coprime.

All of this would be a lot easier with a diagram to look at :).

We can perform some sort of induction, going from smaller numbers to larger numbers. We may assume WLOG that $s\ge 4$ (if $s=3$ it’s easy: we simply take $A=bb\ldots b$, and $B=bab$). I’ll also consider only the case $r\ge s$ (I’m not even sure, but I think somehing similar can be done otherwise).

The word consisting of the first $s-2$ leters of $B$ must be found at the beginning of $A$ as well, only that it’s reversed. This word begins with $ba$, so $A$ consists of a word of $s-2$ letters ending in $ab$, followed by a palindrome $T$ (because $T$ is equidistant from the endpoints of $C$). The first $s-4$ letters of $A$ also form a alindrome, so $Aba=XY$, where $X$ is the $s-4$-letter palindrome at the beginning of $A$, and $Y=abTba$, which has $r-s+6$ letters. Conversely, it’s clear hat if we can establish this situation, we can reconstruct $AB=Cab$. According to the induction hypothesis, this is true iff $(s-4+2,r-s+6-2)=1$, which is equivalent to $(s-2,(r+2)-(s-2))=1$, which, in turn, is equivalent to $(r+2,s-2)=1$ because, in general, $(u,v)=(v,u-v)$.

Like I said, I didn’t check what happens if $s>r$, but I imagine something very will turn up.

I think it isn't a big problem to prove that :
  r+2 and s-2 are consecutive numbers -->this will give us the solution

I'd post my own solution but I have very big problems with LaTeX

[quote="grobber"]All of this would be a lot easier with a diagram to look at :).

We can perform some sort of induction, going from smaller numbers to larger numbers. We may assume WLOG that $ s\ge 4$ (if $ s \equal{} 3$ it’s easy: we simply take $ A \equal{} bb\ldots b$, and $ B \equal{} bab$). I’ll also consider only the case $ r\ge s$ (I’m not even sure, but I think somehing similar can be done otherwise).[/quote]

I think you should consider just the case when : r <= s


calculus
integration
trigonometry
calculus computations

${\int_{\sqrt{\frac{\pi}{2}}}}^{\sqrt\pi} \theta^3 \cos(\theta^2) d\theta$

the book sats to "First make a substitution and then use integration by parts to evaluate the integral."

What I did was to try to substitute $t = \theta^2$ but that just complicated things when I tried to do integration by parts....

Any help is appreciated, I am pretty stuck...

Go ahead and make that $t=\theta^2$ substitution. We have that $dt=2\theta\,d\theta.$

Write $\theta^3\,d\theta=\frac12\theta^2\cdot2\theta\,d\theta=\frac12t\,dt.$

And go ahead and change the limits of integration.

When $\theta=\sqrt{\frac{\pi}2},\,t=\frac{\pi}2$ and when $\theta=\sqrt{\pi},\,t=\pi.$

The integral becomes

$\frac12\int_{\frac{\pi}2}^{\pi}t\cos t\,dt.$

You should be able to handle that.

By $\theta ^ 2=t,$ we have $\frac{dt}{d\theta}=2\theta.\ \therefore 2\theta d\theta=dt.$ Thus

$\int \theta ^3\cos \theta ^2\ d\theta=\int \theta ^2 \cos \theta ^2\cdot \theta d\theta=\frac{1}{2}\int t\cos t dt.$

It was the step to rewrite

$\theta^3\,d\theta=\frac12\theta^2\cdot2\theta\,d\theta=\frac12t\,dt.$

that I was not getting. Now it's clear...

Thanks Kent and Kunny


There are only red marbles and green marbles in a bag. The ratio of red marbles to green marbles in the bag is 4:7. Julia then adds 90 red marbles and 36 green marbles to the bag, which makes the probability of selecting a red marble from the bag on a random draw equal to $ \frac{1}{2}$. How many total marbles are in the bag after Julia has added the 126 marbles?

set up proportions

$ \frac{r}{r\plus{}g}\equal{}\frac{4}{4\plus{}7}\equal{}\frac{4}{11}$
$ \frac{r\plus{}90}{r\plus{}g\plus{}126}\equal{}\frac{1}{2}$
cros-multiplying the second equation and solving gives us
g=r+54
plug it into the 1st equation
$ frac{r}{2r\plus{}54}\equal{}\frac{4}{11}$
cross-multiplying and solving yields r=72
we could plug it into one of the longer equations, but that would take too long...
What havent we used...

oh right!

$ \frac{r}{g}\equal{}\frac{4}{7}\equal{}\frac{72}{7x}$
$ x\equal{}72/4\equal{}18$
$ 7x\equal{}126$
$ 126\plus{}126\plus{}72\equal{}\boxed{324}$ marbles after the girl adds the 126


ABC triangle. 2p=1 Prove that:
a 2 +b 2 +c 2  +4abc <1/2

really ez.
(2a-1)(2b-1)(2c-1)<0
Expand and get the desired inequality


A single cable supports an 837-kg elevator car. What is the tension in the cable when the car is moving with a) constant speed, b) an upward acceleration of 3.21 m/s^2 and c) a downward acceleration of 3.21m/s^2

I'm not totally lost. Since the T is directed downward, I can derive from

F=ma
T+mg=ma
T=ma+mg (since g is negative)

so I thought T would be for
a)873(1)+837(9.8)
b) 873(3.21)+(873)(9.8)
c) 873(-3.21)+873(9.8)

is this correct?? if not someone please help me!

In part (a), you have constant speed so $a=0$ $m/s^{2}$. This means $T$ is balanced with $mg$, so $T=mg=837(9.8)$, the others look correct. I would always draw your force diagram to make sure about signs and any possible extra forces instead of completely relying on formulas (which you might have done, I don't really know).


If eggs had cost x cents less a dozen, it would have cost 3 cents less for x+3 eggs than if they had cost x cents more a dozen. Find x.

(from Math Team team round)

[hide="write an algebraic equation"]Let the price of the eggs (per dozen) be $ z$.  Then, we know that $ \frac{(z\minus{}x)(x\plus{}3)}{12}\plus{}3\equal{}\frac{(x\plus{}3)(z\plus{}x)}{12}$.

This conveniently simplifies to:

$ 3\equal{}\frac{\left[(z\plus{}x)\minus{}(z\minus{}x)\right](x\plus{}3)}{12}$
$ (x\plus{}3)2x\equal{}36$
$ x^2\plus{}3x\minus{}18\equal{}0$
$ (x\plus{}6)(x\minus{}3)\equal{}0$
$ x\equal{}3$[/hide]


Consider a regular $ n$-gon inscribed inside a circle of radius 1.  Each of the vertices of the $ n$-gon lie on the circle.  Draw a line between each pair of the $ n$ vertices.  Prove that the sum of the squares of the lengths of these lines is $ n^2$.

The lengths of the diagonals are given by $ L \equal{} 2\sin(\frac {k\pi}{n})$ for $ 0 < k \le \frac {n}{2}$.
There are $ n$ of each length except when $ n$ is even and $ k \equal{} n$, then there is $ n/2$.
We can add these lengths for $ 0 < k < n$ to give the correct double counting for both even and odd $ n$.

$ 2S \equal{} n\sum_{k \equal{} 0}^{n \minus{} 1} 4\sin^2\frac {k\pi}{n}$

$ S \equal{} n\sum_{k \equal{} 0}^{n \minus{} 1} (1 \minus{} \cos\frac {2k\pi}{n}) \equal{} n^2 \minus{} \Re(\sum_{k \equal{} 0}^{n \minus{} 1} e^{\frac {2k{\pi}i}{n}} ) \equal{} n^2$


calculus
derivative
function
calculus computations

Find a function U: R+ x R+ -> R+ satisfying the following conditions

1. Ux >0 , Uy>0 where Ux, Uy are partial derivatives of U

2. Uxx>0 Uyy<0

3. dy/dx<0 d^2y/dx^2<0

In condition (3), differentiate U(x,y)=u where u is fixed

If the function satisfying three conditions doesn't exist, prove that

Condition 3 of your post says $ \frac{d^2y}{dx^2}<0$, so it is supposed to be minus.

Don't double (or triple) post, please! You have been answered in this thread, so continue the discussion here. I removed your two identical posts.

What does mean " \sqrt{ x^2 + 4y}??

If it means sqrt(x^2+4y), d^2y/dx^2=0

  :huh:

Try $ \sqrt{(x\plus{}1)^4\plus{}y\plus{}1}$ then :). What do you need this function for, by the way?

I'm majoring in economics. so I have to study economical mathematics

Thank you for your answer!!!!!! :lol:


Find all function f:$ R_{+}\mapsto R_{+}$ :
i) $ f(f(f(x)))+2x= f(3x)$
ii)$ \lim_{x}\mapsto\infty(f(x)-x)=0$
for all x $ \in{R_{+}}$

[hide="Hint"]1) Prove  $ f(x)\geq x\forall x>0$ by method of Problem 3 from VMO 2002.
2) Prove $ f(x)\leq x\forall x>0$ by 1).[/hide]


I was trying to find an elegant solution to a problem that was about tangent lines, circle, and graph.

It said that a line was tangent to a circle of some radius. Another line was also tangent to that circle and no points on the circle were above that line.

They wanted to know the center of the circle in coordinates. How is this done? I did it brute-force using distance equations and equation of tangent lines, but it took such a long time and we only get 5 minutes per problem.

Give us the problem - there might be an elegant solution only applicable to that problem.


A polynomial P with integer coefficients is such that $ P( \minus{} n) < P(n) < n$ for some integer $ n$. Prove that $ P(\minus{}n) < \minus{} n$ .

It is false. For examplt $ P(x)\equal{}2x\minus{}5,n\equal{}2.$  $ P(\minus{}2)\equal{}\minus{}9<P(2)\equal{}1<2$.

Sorry, my terrible mistake :/ , I just corrected the content

HI Sylwek,
$ a\minus{}b|f(a)\minus{}f(b)$ , $ n\minus{}(\minus{}n)|P(n)\minus{}P(\minus{}n)$, -> $ 2n|P(n)\minus{}P(\minus{}n)$ -> $ 2n\equal{}<P(n)\minus{}P(\minus{}n)$ ->
$ P(\minus{}n)<P(n)\minus{}2n<\minus{}n$

I remember i did this once, it's probably from Pawłowski, greetings


Okay, I don't want to sound annoying, but I have a few other problems that I have been having trouble with.
But if you guys are busy, I can ask someone else, too.

#1. The product of four consecutive positive integers is 17160. What is their sum?
      (Is there a way to do it beside guess and check)(again, no calculator)
#2. If a+b=12 and a+b=250, then what does a+b=?
#3. How many of the divisors of 8! are larger than 7! ?

For number 1, factor it (but not necessarily prime factorize it.

17160 = 10 * 1716

The rest of the solution in spoiler:

[hide]= 10 * 4 * 429

= 10 * 4 * 3 * 143

= 10 * 4 * 3 * 13 * 11

by that you can tell the numbers are 10, 11, 12, 13.[/hide]

For number 2.

Keep in mind that

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

You can solve for ab by squaring (a+b) and subtracting A^2 + b^2 from that, and then dividing by two.

a[hide]+b = 12

a^2 + b^2 = 250

a^2 + 2ab + b^2 = 144

Subtract a^2 + b^2 from that to get

2ab = 144 - 250 = -106

ab = -53

a^3 + b^3 = 12(250 - (-53)) = 12(303) = 3636 ... I believe.[/hide]

For problem 3,

[hide]8!/1, 8!/2, 8!/3, 8!/4, 8!/5, 8!/6, 8!/7 are all factors of 8!, but greater than 7!.[/hide]

Hope this helps.

[quote="k.lee"]#1. The product of four consecutive positive integers is 17160. What is their sum?[/quote]

rep123's factoring solution is excellent.  Here's another approach.  Estimate the geometric mean of the 4 consecutive positive integers.  For this problem we take the 4th root of 17160.  [This is equivalent to the square root of the square root of 17160.]

By looking at the first 3 digits (171), I know the square root of 17160 is between 130 and 140, because 13^2=16900 and 140^2=19600.

The square root of a number between 130 and 140 is between 11 and 12, because 11^2=121 and 12^2 = 144.  Thus I am certain the geometric mean is between 11 and 12.  I suspect the arithmetic mean is close to the geometric mean.

For the countdown round, I would buzz-in with 11.5x4=46.  For a written round, you should check your work:  10x11x12x13=17160.


search
combinatorics unsolved
combinatorics

A country has 2001 towns. Each town has a road to at least one other town. If a subset of the towns is such that any other town has a road to at least one member of the subset, then it has at least k > 1 towns. Show that the country may be partitioned into 2001 - k republics so that no two towns in the same republic are joined by a road.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=787095433&t=4998

Pierre.


Well, I guess poles and polars do the job again... But this is way too well-known.

The problem trivially follows from the fact that the line FG is the polar of the point E with respect to the circumcircle of the quadrilateral ABCD. This fact was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=38871 (problem 2), http://www.mathlinks.ro/Forum/viewtopic.php?t=25036 , http://www.mathlinks.ro/Forum/viewtopic.php?t=5326 , http://www.mathlinks.ro/Forum/viewtopic.php?t=550 .

  darij

[quote="M@re"]What are and where I can learn a little more about [b] poles and polars[/b]?[/quote]Check http://www.mathlinks.ro/Forum/viewtopic.php?t=31812 for this.

It's not much but thank you anyway :) :) :)

its clear.EF is the polar of G.then OG will be prependicular to EF.

see  http://mathlinks.ro/forum/viewtopic.php?t=51865 :D


geometry
rectangle
trigonometry
analytic geometry
combinatorics proposed
combinatorics

Suppose an $a \times b$ rectangle $R$ can be dissected into finitely many triangles with the property that every angle $\alpha$ of a triangle in the dissection has $\tan{\alpha} \in \mathbb{Q}$. Show: $a/b \in \mathbb{Q}$.

It can be easily proven that if $A,B \in \mathbb{Q}^2$ and $C$ satisfies ${\tan{\angle{CAB}}, \tan{\angle{CBA}}} \in \mathbb{Q}$ then $C \in \mathbb{Q}^2$, by writing the rational equations of the lines $AC,BC$. Join two points $X,Y$ by an edge iff $XY$ is the side of some triangle of the partition. Now let $ABCD$ be our rectangle and let $X\in (AB)$ be connected with $A$. Choose a system of coordinates with $A(0,0),X(1,0)$. If $AXY$ is a triangle of the partition then we deduce that $X\in \mathbb{Q}^2$. We can repeat this procedure until we add all vertices of the partition, so all vertices of the partition have rational coordinates. Particularly $B(u,0), C(0,v)$ where $u,v$ are rational. Then $\frac ab=\frac uv$ is rational.

But the dissection need not be a triangulation; there need not be a triangle in the partition with side $AY$ or $XY$. So, you cannot repeat this procedure.


a,b,c are integers.
Find f(x): x^2 + ax + b so that we can find 3 integer in [1,9] : |f(m )|=|f(n)|=|f(p)|=7

if $ n < m < p$ then we must have $ f(n) \equal{} f(p) \equal{} \minus{} f(m)$

$ f(x) \minus{} 7$ or $ f(x) \plus{} 7$ have $ 2$ integres roots
1) if $ f(x) \minus{} 7$ have $ 2$ integres roots then $ x^2 \plus{} ax \plus{} b \minus{} 7 \equal{} (x \minus{} i)(x \minus{} j)$
so $ a \equal{} \minus{} i \minus{} j$ and $ b \equal{} 7 \plus{} ij$ and $ (i \minus{} k)(j \minus{} k) \equal{} \minus{} 14$ impossible because $ \{i,j,k\} \equal{} \{m,n,p\}\subset[1,9]$

2) if $ f(x) \plus{} 7$ have $ 2$ integres roots then $ x^2 \plus{} ax \plus{} b \plus{} 7 \equal{} (x \minus{} i)(x \minus{} j)$
so $ a \equal{} \minus{} i \minus{} j$ and $ b \equal{} 7 \plus{} ij$ and $ (i \minus{} k)(j \minus{} k) \equal{} 14$
then $ i \equal{} k \plus{} 2,j \equal{} k \plus{} 7$ where $ k\in\{1,2\}$
or $ i\equal{}k\minus{}2,j\equal{}k\minus{}7$ where $ k\in\{8,9\}$
then $ (a,b) \equal{} ( \minus{} 11,31),( \minus{} 14,52),(\minus{}9,21),(\minus{}7,13)$


inequalities
geometry
inequalities proposed

Prove that the following inequality holds in any triangle:

$ \frac{3\sqrt{3}abc}{2pS} \geq \sum \frac{a(b+c)}{r_{a}(r_{b}+r_{c})}$$.$
When does equality occur? (standard notations)

I hope this is original and even more: correct! :)

Enjoy!  :cool:

Its trivial  because  after some  boring computations  you will  find
$\Sigma\frac{a(b+c)}{r_a(r_b+r_c)}=4  $


I just lost $ 102$ points in $ 8$ minutes, a rate of $ 12 \frac{3}{4}$ points per minute. Is this a record for a period longer than $ 5$ minutes?

I think izzy has lost more than that before.

lol I think you're the only one who keeps track of this record. I'll keep this topic open. If it becomes spammy me or izzy will lock it.

This would not be an accurate description. Games tend to last longer than cd's, so people who play games more han cd's may not be able to beat this record. However, one who plays 2/5/ranking over and over and loses all teh time could easily beat this.

it's not that hard to beat it
say i started a cd and a 1200 rating person joined
say i start the game and leave immediately
say i keep doing that, and whoever joins the game finishes

that way, i could lose 100's of points in the times it takes two average people to finish two games while competing

PS it works even better if you have a VERY high rating

I think r15 might mean not on purpose. actually trying. I may be wrong

no duh i mean trying...

btw when ur rating is really high for some reason u actually don't lose at much as if it were slightly lower

Yeah, I once lost 170 points in about 15 minutes. The points per minute is lower, but I was playing normal games, which usually last longer.

none of you can beat me. joined a game against a 1100er, got flashies, thye didnt leave, forgot to refresh, made a new cd, lfet it bc of flashies, also an 1100er, and lost 60 pts in 2 minutes. 30 pts a minute, baby.

=D

[quote="r15s11z55y89w21"]Is this a record for a period longer than $ 5$ minutes?[/quote]

Even for $ 5$ minutes, that would be $ 12$ points a minute.

@Nikey - Yeah, but you got flashies. I failed for real. :P


Galois Theory
superior algebra
superior algebra unsolved

Let $ F \subseteq E$ be a Galois extension of fields. Let $ G$ be the Galois group $ \Gamma(E,F)$. For $ \theta \in E$, define \[ N_{E/F}(\theta) \equal{} \prod_{\tau \in G} \tau(\theta).\] It is easy to see that $ N_{E/F}(\theta) \in F$.

If $ E'$ is a Galois extension of $ E$ such that $ E \subseteq E'$, then show that for all $ \theta \in E$ \[ N_{E'/F}(\theta) \equal{} \left(N_{E/F}(\theta)\right)^{[E': E]}.\]

I'm sure this has to be easy, but alas, I don't manage to prove it.

Lemma 1:
If $ K \subseteq L \subseteq M$ are finite field extensions, then $ N_{M|K} \equal{} N_{L|K} \circ N_{M|L}$.

Lemma 2:
If $ K \subseteq L$ is a field extension, then $ N_{L|K} (x)\equal{}x^{[L: K]}$ for all $ x \in K$.

Both are easily checked from definitions.
Now apply both, the result is immediate.

The second lemma is obvious.

For the first lemma I need a bijection between $ \Gamma(E': K)$ and the set $ \left\{\varphi \psi : \varphi \in \Gamma(E: K),\psi \in \Gamma(E': E)\right\}$. Why does this bijection exist?

We fix some representants to be able to see $ \mathrm{Gal}(E|K)$ as a subset of $ \mathrm{Gal}(E^\prime|K)$.

Assume that $ \phi_1 \psi_1 \equal{} \phi_2 \psi_2$ on $ E^\prime$ with $ \phi_i \in \mathrm{Gal}(E|K) , \psi_i \in \mathrm{Gal}(E^\prime|E)$. Then as $ \psi_1, \psi_2$ fix $ E$, we get that $ \phi_1 \equal{} \phi_2$ (both act the same way on $ E$). Setting this into the first equation, we get $ \psi_1\equal{}\psi_2$.

So the mapping $ \mathrm{Gal}(E|K) \times \mathrm{Gal}(E^\prime|E) \to \mathrm{Gal}(E^\prime|K)$ with $ (\phi,\psi) \mapsto \phi \psi$ is injective. But both sets have same finite cardinality, so it's surjective as well.


Find the mean of the data in this stem-and-leaf plot.
\begin{tabular}{|cc|}
\hline
Stem&Leaf\\
1&4\\
2&3,4\\
3&0,4\\
\hline
\end{tabular}

$ \begin{tabular}{|cc|}
\hline
Stem&Leaf\\
1&4\\
2&3,4\\
3&0,4\\
\hline
\end{tabular}$

By definition, the answer is $ \frac{14\plus{}23\plus{}24\plus{}30\plus{}34}{5}\equal{}\boxed{25}$


A committee is to be made up from a group of 10 candidates. The committee must consist of a chairman and some (not necessarily nonzero) number of members-at-large. How many ways can the committee be chosen?

I didnt really get this problem. I separated it into ten cases where 1 case was just the chairman, another was just the chairman and 1 member, and a third was a chairman and 2 members.....up all the way to a chairman and 9 members. I calculated all the combinations up and got some absurd answer of  9864100. Did I do this wrong because the members aren't distinguishable? I dont know.

There are $ 10$ ways we can select the chairman. When we have picked a chairman, we have $ 9$ possibilities left for the committee. There can be any number of committee members from $ 0$ to $ 9$. We must then add up all cases together.

$ 10\binom{9}{0} \plus{} 10\binom{9}{1} \plus{} 10\binom{9}{2}... \plus{} 10\binom{9}{8} \plus{} 10\binom{9}{9}$

$ \equal{}10(\binom{9}{0} \plus{} \binom{9}{1} \plus{} \binom{9}{2}... \plus{} \binom{9}{8} \plus{} \binom{9}{9})$

$ \equal{}10\times{2^9}$

$ \equal{} \boxed{5120}$

I got this right but I don't understand the official solution.

@above [hide=Solution]You can pick the chairman $10$ ways. For the remaining $9$ members, you can either include or exclude them, giving us $2^9$. Thus, $10 \cdot 2^9 = \boxed{5120}$.[/hide]

who else bashed out the answer by computing all of the combinations? lol


Are you trying to say $2003^{2003}$?[hide] If you are then it would be a pattern. $3^{1}=3$ and $3^{2}=9$ and $3^{3}=27$ and $3^{4}=81$. The pattern is 3,9,7,1... So divide 2003 by 4 and the remainder will be 3. So it is 7.[/hide]

[quote="xxxxx"]what is the units digit of $2003^{2}003$?[/quote] Usually these kinds of problems are in Math Counts competitions and they always rely on patterns.
[hide="3"]Since the units digit is 3 in 2003, you find the pattern of the units digits. This pattern goes 3, 9, 7, 1, 3, 9, 7, 1.... repeating every four times. 2003/4=500 r 3 so then the digit is $\boxed{3}$[/hide]
Also, you should do this for the powers: 2003^{2003} or else it will turn out bad.

These patterns exist with all kinds of numbers. Try it out!  :)

[quote="ckck"]Note to Quevvy:The remainder isn't the unit digit. It will tell you which one of the four is the units digit. Since the remainder is 3 it will be the third digit in order starting from the left so it will be 7.[/quote] Sorry, I didn't mean 3 - I meant the [i]third number[/i] in the pattern.

[quote="xxxxx"]what is the units digit of $2003^{2003}$?[/quote]

Modular arithmetic  :lol:  I was learning it a couple weeks ago.

if you need more help chess64 posted some extra problems so I could practice for MC

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=124340[/url]


http://news.bbc.co.uk/2/hi/uk_news/england/6524495.stm

Basically talking cameras are being installed so they can remind people not to litter, etc. One Mr. Reid defending it says there will be some kook jobs in the minority who will complain about it being an invasion of civil rights, but the majority will be happy with the decrease in littering and other such abuses. Of course various minorities think it terrible and an abuse of civil rights. What do you think of this, and does the resemblance to 1984 occur to you?

[Yes, I know that lead-in was biased. I want a civil debate here.]

Now I must go work on my debate case along the lines of "this happens when we overvalue democracy, so we should be very careful not to overvalue it."

[quote="solafidefarms"]do you think of this, and does the resemblance to 1984 occur to you?
[/quote]

My class is currently reading the same book  :D .
I think that pretty soon, people are going to get annoyed and it became justified for them to litter more...

I don't know...but I'm pretty sure people don't have a civil right to privacy in public places like streets or squares. I don't necessarily think talking cameras are a good idea, but I'm not sure I see the connection with 1984.

I think that the nigthmare of 1984 is "What happens when people can't think what they want?". And that in this case, nobody is really messing whit anyone thougths, they are just saying "Don't litter, we are watching".
Now, I also think that there are surely lots of better ways to solve this problem. Cameras cannot solve this problem by their own, because
1.- People can just ignore them
2.- If someone can litter when there are people around, he/she can surely litter when there are cameras around
3.- Even if there are tickets, well, they have always been there.

The power of the talking cameras is the guilt- public humiliation is incredibly powerful. There's no more invasion of privacy than with ordinary cameras, and those are already common.
People ignore cameras, but they won't ignore a voice telling them what to do.

[quote="PenguinIntegral"]There's the quote about sacrificing freedom for liberty...[/quote]

Freedom for security, you mean.

DUDE
i didn't click that link or read anything much but it seems that as long as it's public we have to remember people can see us in the public
but i didn't click the link

[quote="bookaholic"][quote="PenguinIntegral"]The question is, "why should the government be doing this?", not "why not?". There's the quote about sacrificing freedom for liberty...[/quote]
But nothing is being sacrificed, because the freedom to behave as you wish in public places never existed to begin with. I agree that it seems like a silly policy, but that doesn't mean any rights have been taken away.[/quote]
You miss the point. It's so easily abused and so useless it's just laughable. Unless you instill some culture of fear, and then it's just idiotic and totally against American ideals.

So I'm getting the sense that you're like

"well they won't commit crimes if they are being watched"

and that's totally stupid, because a giant percentage of people incarcerated for real crimes get convicted, and they know that they are going to get caught, and they still do it.

So your cost/benefit ratio isn't great.

And then there's this whole you give the government extreme power thing. That's kinda like the main argument here. Because the government having anymore power than absolutely necessary is bad, and this is way more than necessary.

Also, if you don't agree, you are a Communist.

For the most part PenguinIntegral, I agree with your arguments, but please do not resort to name calling.  It does not make your points any stronger.

[quote="bookaholic"][quote="PenguinIntegral"]The question is, "why should the government be doing this?", not "why not?". There's the quote about sacrificing freedom for liberty...[/quote]
But nothing is being sacrificed, because the freedom to behave as you wish in public places never existed to begin with.[/quote]

Why doesn't that freedom exist? As long as you aren't invading other's rights to life, liberty, pursuit of happiness, property, etc. why aren't you free to do whatever you like in public? Certainly littering is infringing on the right of the person whose land is littered, but is it the government's place to correct them while they are doing it? Wouldn't it work almost as well and not raise such a stir if the cameras were simply used to record who littered and then the government or whomever's land was littered sued them for it?

PenguinIntegral--What? Just because it's a bad idea doesn't mean it violates civil rights. I wasn't arguing for or against the policy in any way--I was just pointing out that the cameras don't take away any specific rights or freedoms. The government already has the power to watch you when you're in public. I didn't say the cameras are especially useful, or that they don't have negative consequences, or anything remotely similar. Note the "...I agree that it seems like a silly policy..." part of my previous post.

Solafidefarms--you're probably right. I haven't taken a position on the actual policy in question yet. I reserve the right to comment on a policy without first taking a side on it. Or having a side forced on me.

So last time I checked it's pretty easy to distinguish between public streets and bathrooms. In one case there's an expectation of privacy, and in the other there isn't. I don't know enough about law to know exactly where the line is drawn, but we already have a system in place to determine whether someone has an expectation of privacy in any one area. They certainly don't when they're in a street.

[quote="bookaholic"]PenguinIntegral--What? Just because it's a bad idea doesn't mean it violates civil rights. I wasn't arguing for or against the policy in any way--I was just pointing out that the cameras don't take away any specific rights or freedoms. The government already has the power to watch you when you're in public. I didn't say the cameras are especially useful, or that they don't have negative consequences, or anything remotely similar. Note the "...I agree that it seems like a silly policy..." part of my previous post.
[/quote]
There's a difference between stalking someone every day and taking pictures of them...

[quote="tim1234133"]I think it is rather pathetic and that it will not actually achieve much - there are more cost-effective ways of achieving the same things - but I can't see how it is an invasion of privacy. Cameras already exist in public places, and have done for some time. You are not obliged to do what the camera tells you, it just puts extra onus on you to act sensibly in public.[/quote]
Hmm, invasion of privacy or rights is probably the wrong way to put it. Contrary to the principles of limited government would be better. Considering that many stupid laws have been passed and my corrupt politicians have surfaced, I do not want them tracking my every move in public. Once again, see the definition of stalking.

EDIT: Time to engage in link wars! Take ten paces back and [url=http://www.nyclu.org/pdfs/surveillance_cams_report_121306.pdf]fire![/url]

There was also the wiretapping controversy in the US. Although in that situation, the stakes are much higher (terrorism versus littering), it still boiled down to freedom versus security.

On a side note, you've got to feel sorry for the people monitoring the cameras, who have to sit in front of a screen every single day watching for people littering. :D

[quote="mathnerd314"]There was also the wiretapping controversy in the US. Although in that situation, the stakes are much higher (terrorism versus littering), it still boiled down to freedom versus security.[/quote]

You've conflated two extremely different kinds of privacy invasion.
The key element of this discussion is whether there is in fact a freedom being interefered with in this case, since all that's happening is pictures are being taken of you [i]in public[/i], i.e. in a place where you (correctly) have a lowered expectation of privacy.  In the case of wiretapping, we unquestionably have the invasion of a legitimate expectation of privacy, i.e. when I'm having a private conversation in the privacy of my home I have an authentic expectation and interest that no one be listening to me.  (There are dozens of other things one could say about how the eavesdropping was stupid, invasive and illegal, of course, but I don't want to spend the time right now.)

With regards to the wiretapping, they used to just wait two weeks to tell you, but then they started getting these annoying lawsuits from people who were (rightly) enraged at the invasion of privacy.  They solved that problem by simply making a policy of simply not telling people that their phones had been tapped :mad: .


Let ABC be a triangular surface of centroid G.
Divide this surface in  two surfaces $S_1, S_2$ such that
$\frac{S_1}{S_2}=\frac{m}{n}$,  $m,n\in$$N^*$, using only a ruler.

What is a "surface" for purposes of this problem?  One can construct arbitrary 2-dimensional figures with given centroid.  Even for a disc it is not clear how to solve this problem with ruler only (and some arguments why it might not be solvable).


Q) a non decreasing series of the form , where each odd integer k is repeated k times , ex. 1,3,3,3,5,5,5,5,5....where each term corresponds to the value a1 a2 a3...of the series , it is a known fact that
a(n)= b[ (n-c)^1/2] + d. where [] is greatest integer function. Find b,c,d as they belong to Z.

Since $ 1\plus{}3\plus{}5\plus{}...\plus{}(2n\minus{}1)\equal{}n^2$, if $ n^2<k\le{(n\plus{}1)^2}$, $ a_k\equal{}2n\plus{}1$.
We know that $ n^2\le{k\minus{}1}<(n\plus{}1)^2$, so $ \lfloor{\sqrt{k\minus{}1}}\rfloor\equal{}n$.  Thus $ a_k\equal{}2\lfloor{\sqrt{k\minus{}1}}\rfloor\plus{}1$.

[quote="TZF"]Just for fun, and just because this kind of thinking can score last minute points in timed contests.

[hide="Hand Wavy Approach"]The sequence values go up in steps of 2, but the [] term goes up in steps of 1, so [b]b=2[/b].

The indexes at which the sequence increments are 1, 2, 5, each one more than a square. We guess [b]c=1[/b].

[i](Alternately, c<=1 so that the function is defined for index 1, but the rate at which its growth slows down 'looks like' the 'beginning' of a squareroot graph, so most likely, [b]c=1[/b].)[/i]

d is odd since the 2[] term is even and the sequence elements are all odd. Looking at the first term, [b]d=1[/b].[/hide]

[b]Related Question:[/b] Suppose we remove the restraint that b, c, and d are integers. Are there any other ways of expressing the sequence in the suggested form?
- If so, what are they?
- If not, and you don't have to answer this here (just think about it), effect might the problem writer's choice (of b,c,d in Z or R) have on the method you might end up solving the problem with if you didn't have to show your work?[/quote]
@bpgbcg
The answer is correct , but TZF got the way i solved it. Nice question at the end though , hopefully ill get that done.
thanks!


ratio
geometry
trigonometry
perimeter
geometric transformation
reflection
function

I think this kinda has to do with the A=(1/2)aP thing, but here it is:

A regular octagon and an equilateral triangle have equal perimeters. What is the ratio of the area of the hexagon to the area of the triangle? Express your answer as a common fraction.

and how would u use this problem to solve an even harder problem:

A regular octagon and a regular decagon have equal perimeters. What is the ratio of the area of the hexagon to the area of the decagon? Express your answer as a common fraction.

Thanks! :)

I don't think this is MC level, but...

the side of the equilateral triangle is $ \frac13 p$.

Thus, the area of that is  $ \frac{\frac{p}9 \sqrt3}{2}$, or $ \frac{p\sqrt3}{18}$.

Now, we must use trig to find the apothem of the octagon.

The side is $ \frac18 p$.  Half of this is $ \frac1{16} p$.

Then, you divide $ 135^o$ by $ 2$, to get $ 67.5$.

Now, it's time for trig...

You have:

$ \tan{67.5} \equal{} \frac{a}{\frac{1}{16} p}$

$ \frac1{16} p \times \tan{67.5} \equal{} a$

Thus, you must multiply everything by $ p$ (the perimeter) and then halve it for the area.

$ A (\text{octagon}) \equal{} \frac{p^2}{16} \times \tan{67.5}$

Thus, the answer is $ \frac{\frac{p^2}{16} \times \tan{67.5}}{\frac{p\sqrt3}{18}}$.

$ \frac{p^2 \times \tan{67.5}}{16} \times \frac{18}{p\sqrt{3}}$

$ \frac{18p^2 \times \tan{67.5}}{16p\sqrt{3}}$

$ \frac{9p \times \tan{67.5}}{8\sqrt{3}}$

$ \frac{9p\sqrt{3} \times \tan{67.5}}{24}$

$ \boxed{\frac{3p\sqrt{3} \times \tan{67.5}}{8}}$

Same method for the second.

er, a similar question wuz on the state sprint once, so how do u evaluate that without a calc? (i'm newby at trig: only really know law of sines/cosines)

and idk about this geo problem 2:
"Point P is 9 units from the center of a circle of radius 15. How many different chords of the circle contain P and have integer lengths?" (Source: 2001 MC State Target)

Thx!  :)

Hmmm... for the first one, you stated octagon, then you stated hexagon.

If it meant hexagon, then it is [b]a lot[/b] easier!

As for the target question, notice that the longest possible length is the diameter, which is of length $ 30$. The shortest is $ 24$, which occurs when the chord is perpendicular to the segment containing the center of the circle and P. For the lengths $ 25\minus{}29$, though, you have $ 2$ possible chords, as you can reflect the chords about the diameter. This gives us a total of $ \boxed{12}$. Sorry if this doesn't make sense.

[quote="math154"]As for the target question, notice that the longest possible length is the diameter, which is of length $ 30$. The shortest is $ 24$, which occurs when the chord is perpendicular to the segment containing the center of the circle and P. For the lengths $ 25 \minus{} 29$, though, you have $ 2$ possible chords, as you can reflect the chords about the diameter. This gives us a total of $ \boxed{12}$. Sorry if this doesn't make sense.[/quote]

Wow!  I actually understand it!

lol I was gonna ask this prob too...

is there a general way to solve problems like
"The number 4 can be written as a sum of one of more natural numbers in exactly five ways: 4, 3+1, 2+1+1, 2+2, and 1+1+1+1; and so 4 is said to have five partitions. What is the number of partitions for the number of 7?" (2001 State Sprint)
or some problem like

Toads from Toadoncartville can either hop 4 hops forward or 3 hops backward. Mrs. Toad wants to travel to Funnytoadville to buy some funny groceries, and Funnytoadville is 24 hops away from Toadoncartville. How many ways can she reach Funnytoadville by using exactly 9 hops? :| (srry, math154, i just edited)

edit: generating functions? wut r those? could u show me how u would list them out? (i usually overcount or don't count all of them or something)

You can use generating functions for partitions, but for Mathcounts, just list them out.

As for your second question, there are an infinite number of ways. You didn't put any restraints on the backwards moving.

Remember: [url=http://en.wikipedia.org/wiki/Generating_function]Google is your friend.[/url]

[quote="toadoncart"]so how do u evaluate that without a calc?[/quote]

Assuming it was a [b]hexagon[/b] and not an octagon, it's actually much easier than you think.

Once again, assume that the perimeter is $ p$.  The side of a hexagon is $ \frac {p}6$ and the side of the triangle is $ \frac {p}3$.

The area of an equilateral triangle with side $ s$ is $ \frac {s^2\sqrt3}4$.  Thus, the area of this triangle is $ \frac {\frac {p^2}{9} \times \sqrt3}4$, or $ \frac {p^2\sqrt3}{36}$.

Now, we must find the area of the hexagon.

Half of the side is $ \frac {p}{12}$.

[asy]draw((0,0)--(1,0)--(1,sqrt(3))--cycle);
label("$\frac{p}{12}$",(0.5,0),S);[/asy]

The angle in the lower right is obviously $ 90^o$.

[asy]draw((0,0)--(1,0)--(1,sqrt(3))--cycle);
label("$\frac{p}{12}$",(0.5,0),S);
label("$90^o$",(1,0),NW);[/asy]

The angle in the lower left is obviously half of the interior angle of a hexagon, so it is $ 60^o$.

[asy]draw((0,0)--(1,0)--(1,sqrt(3))--cycle);
label("$\frac{p}{12}$",(0.5,0),S);
label("$90^o$",(1,0),NW);
label("$60^o$",(0,0),NE);[/asy]

Thus, the top angle is $ 30^o$.

[asy]draw((0,0)--(1,0)--(1,sqrt(3))--cycle);
label("$\frac{p}{12}$",(0.5,0),S);
label("$90^o$",(1,0),NW);
label("$60^o$",(0,0),NE);
label("$30^o$",(1,sqrt(3)-0.5),SW);[/asy]

Let's gasp!!!  It's a $ 30$-$ 60$-$ 90$ triangle!!!

Thus, the long side is $ \frac {p\sqrt3}{12}$.

[asy]draw((0,0)--(1,0)--(1,sqrt(3))--cycle);
label("$\frac{p}{12}$",(0.5,0),S);
label("$90^o$",(1,0),NW);
label("$60^o$",(0,0),NE);
label("$30^o$",(1,sqrt(3)-0.5),SW);
label("$\frac{p\sqrt3}{12}$",(1,sqrt(3)/2),E);[/asy]

And now, we use $ A \equal{} \frac {aP}{2}$.

$ A \equal{} \frac {p \frac {p\sqrt3}{12}}{2}$

$ A \equal{} \frac {p^2\sqrt3}{24}$

Now, we make a ratio with the area of the triangle.

$ \frac {\frac {p^2\sqrt3}{24}}{\frac {p^2\sqrt3}{36}}$

$ \frac {\frac1{24}}{\frac1{36}}$

$ \boxed{\frac32}$

Hmmm... the first time i did this, I got $ \frac{\sqrt6}{6}$... Did I goof up?

[geogebra]7c068e8cea17a5ad3372ecafb967d849b8c43669[/geogebra]

From this, you can see that $ \Delta BEH$ is an equilateral triangle with the same perimeter as the shaded hexagon. How many small triangles does $ \Delta BEH$ have? What about the hexagon? What's their ratio?


I am trying to find out how a step transformer works i get everything except how the current gets induced to the secondary coil(the main part) plz help! :D  :D

Eddy currents I think, induced by the alternating current (alternating current=changing magnetic field).


My file built correctly with no errors, yet when i try to open it i get an error

[DocOpen("%bm.pdf")]{FileOpen("%bm.pdf")]
Cannot execute the command

It was working fine, then I did some minor code change and It stopped working

any help would be appreciated

Are you absolutely sure that there are no errors after the change? What does the log say? Is the pdf file actually there? What happens if you undo the changes?
Just occasionally TexnicCenter stops communicatig with Adobe Reader. Closing and repopening TexnicCenter solves this.

Also, if you upgraded to the new version of Acrobat reader recently, you should edit the profiles. A good test for checking whether this is the case is to open Acrobat reader and then ask to view the output from the TeXnic center. If it works in this case and doesn't work with the closed AR, the probability that the cause is this mismatch is 90%.


Triangle ABC is isoceles with AB = AC. D is the midpoint of BC. E is on AC such that DE is perpendicular to AC. F is the midpoint of DE. Prove that AF is perpendicular to BE.

This is well known and posted many times.
Let K be the midpoint of EC, then F is the orthocenter of AKD, since $ KF \perp AD$ and $ DE \perp AC$, so since $ AF \perp KD$ and $ KD \parallel BE$, $ AF \perp BE$.
See [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110196[/url]


There are some points inside a unit square and there is no two points from these points, of which have a distance 0.001. Prove that these points cover an area less than a)0.34 b)0.287 .

I know I am stating the obvious but is not .287 less than .34 and therefore if you prove it for .287 then you don't have to prove the .34 cuz it is automatically less than that.

Yes, but there is a weaker approach to this problem that you can prove 0.34 but not 0.287, and it seems like the author of the problem didn't want to waste this approach either.


Given that $x,y,z$ are positive real numbers satisfying $xyz=32$, find the minimum value of
$x^{2}+4xy+4y^{2}+2z^{2}$.


Find lim
1) $ lim \frac { n^2 \plus{} 3n \plus{} 5}{2^n}$

[quote="bobbym"]Hi;

What is n approaching?[/quote]

$ n \geq 0$

Hi;

That is not clear enough. For each n>=0 you will have a different limit. You have to declare what n is approaching.

I believe he refers to $ \displaystyle\lim_{n \to{\plus{}}\infty}{}$

Hi;

Maybe, but how do you draw that conclusion from n>=0 ?

Hi,

I thought of the case that may have more difficulty. There is no reliable conclusion. :)

Hi pablo;

You might be right. So:

Because:

$ \lim_{n\to \infty}\frac{n^2\plus{}3n\plus{}5}{2^n} \equal{} \frac{\infty}{\infty}$

We use LHospitals rule.

$ \frac{1}{\log(2)}\lim_{n\to \infty}\frac{2n\plus{}3}{2^n} \equal{} \frac{\infty}{\infty}$

We use LHospitals rule again.

$ \left (\frac{1}{\log(2)}  \right )^2 \ \lim_{n\to \infty}\frac{2}{2^n} \equal{} 0$

So:

$ \lim_{n\to \infty}\frac{n^2\plus{}3n\plus{}5}{2^n} \equal{} 0$

WE have :

$ \lim_{n\to \infty}\frac {n^2 \plus{} 3n \plus{} 5}{2^n} \equal{} \frac {\infty}{\infty}$

We use LHospitals rule.

$ \lim_{n\to \infty}\frac {2n \plus{} 3}{2^n.ln2} \equal{} \frac {\infty}{\infty}$

We use LHospitals rule again.

$ \lim_{n\to \infty}\frac {2}{2^n.(ln2)^2} \equal{} 0$

So:

$ \lim_{n\to \infty}\frac {n^2 \plus{} 3n \plus{} 5}{2^n} \equal{} 0$[/quote]


geometry
geometric transformation
Ross Mathematics Program
3D geometry

state ur favorite band(s) and music genres.:)

Aargh (translation: if I must)

The Beatles
The Who
Nirvana
My Chemical Romance
Finger Eleven
Linkin Park
BTO
Nine Inch Nails
that dude who wrote the Star Wars music
The Doobie Brothers
Eric Clapton
Lynard Skynyrd (or however you spell it)
John Lennon (his solo stuff)
Alabama Rhythm Section
Blue Oyster Cult
Bob Marley and the Wailers
Green Day
The Offspring
The Police
Yes

That's all I can remember right now. I have a very eclectic taste in music (But i HATE country and rap)

The Beatles
The Red Hot Chile Peppers
Elvis Costello (band?)
Beck (band?)
The Pretenders
David Bowie (band?)

meh

Led Zeppelin
Linkin Park
Lifehhouse
Guns n' Roses
AC/DC

Mozart (yea i know he doesnt really count but who cares i luv classical music)

Soundtrack artists:

Hans Zimmer (The Rock, Pirates of the Carribean)
Marty O'Donnell (Halo)
John Williams (Star Wars, Jurassic Park, Harry Potter (?), etc.)

Bands:

The Killers
Green Day
Three Days Grace
Linkin Park
All American Rejects

etc.

In order:

The Who/Pete Townshend
Simon and Garfunkel
Led Zeppelin
Peter Paul and Mary (anyone else? =P)
Queen
Styx
Beatles
Rolling Stones
Paul Simon
Steve Miller Band
Kinks
Lovin' Spoonful
Mamas & Papas
U2
...
...
...

Ok well, not really in order I guess because it changes a lot, but you get the idea.

umm...I can't believe no one said The Strokes....

couple more:
Red Hot Chili Peppers
Electric Light Orchestra
Papa Roach
Disturb'd
Seether
Weezer
Limp Bizcit
Cream
...

I keep forgetting more...

The rest of my list would pretty much be the ones Erabel said, with the exception of Weezer, etc.

i change my mind: i like Beethoven more than Mozart.

lil wayne
chamillionaire
will smith
timbaland
kanye west
rick ross
lil john
birdman.
Eminem
ice cube
afroman
Papa Roach
Etc.
I like Rap if you havent noticed. my favorite songs are from lil wayne. my favorite songs from him are: fireman, 1000 degrees and The Sky is Da Limit. and my other favorite song is by Eminem, when im gone

meh....

in order:

queen
mama cass elliot
beatles
U2
eagles
dropkick murphys (anyone?)
crosby stills & nash
Nickelback
Rihanna (I know)
Guns n roses
pink floyd (only the brick in the wall)
a few rolling stones stuff
one group that has a name I cant think of

cant think of anymore right now...

anyways I love listening to music and my ipod has like 1200 songs!

ok fine i'll put them in order

Avril Lavigne (dont comment)
Linkin Park
AC/DC
Led Zeppelin
Metallica
Lifehouse
T-Pain
Sum 41
Green Day
OneRepublic
Immortal Technique(a bit)


the sum of two numbers is 10 and the sum of their reciprocals is 5/12 . find the numbers

[hide="not enough time for explanation"]4 and 6[/hide]

Let the two numbers be $ x$ and $ y$. The sum of their reciprocals is $ \frac{1}{x} \plus{} \frac{1}{y} \equal{} \frac{x \plus{} y}{xy}$. Their product is thus $ 24$, and their sum is given to be $ 10$. The two numbers are thus $ \boxed{4}$ and $ \boxed{6}$


If  r = 20/(r + 5) and  s = -5 + 20/s, where r and s are two distinct real numbers, compute r + s.

This problem looks really familiar. Where did it come from? Or did you make it up, in which case I'm just hallucinating?

Solution:

[hide]If you multiply out the fractions, you see that r^2+5r=20, and s^s+5s=20.

This means that r,s are roots to x^2+5x=20, and the sum of the roots of that quadratic is -b/a, or -5. [/hide]

[color=cyan]Well, honestly, I don't know . . . I assembled the problems I'm posting at the beginning of last year.  I think I made it up, but it may be ripped off a Math League or ARML/NYSML.  Possibly even an old AHSME, but I don't think so.  It's not of a horribly original form, so it might be just coincidence.[/color]


Find all functions f: R -> R such that:
$ f(n^2)\equal{}f(n\plus{}m).f(n\minus{}m)\plus{}m^2$ for all $ m,n \in R$.

Take $ n\equal{}m\equal{}0$, we get $ f(0)\equal{}f(0)^2$.

(i) $ f(0)\equal{}0$. Set $ n\equal{}m\equal{}\sqrt{x}$. So $ f(x)\equal{}x$ for positive $ x$. Set $ n\equal{}0$ and $ m\in\mathbb{R}^\plus{}$, we get $ f(\minus{}m)\equal{}\minus{}m$. So $ f(x)\equal{}x$ for all $ x$.

(ii) $ f(0)\equal{}1$
set $ m\equal{}n$, $ f(n^2)\equal{}f(2n)\plus{}n^2$ ... (1)
$ m\equal{}0$, $ f(n^2)\equal{}f(n)^2$ ...(2)
$ n\equal{}0,m\equal{}1$, $ f(1)f(\minus{}1)\equal{}0$. So $ f(1)\equal{}0$ or $ f(\minus{}1)\equal{}0$. Sub this to (1) ($ 2n\equal{}\pm1$), we get $ f(\frac14)\equal{}\frac14$.
$ n\equal{}\frac12\equal{}m$, $ f(1)\equal{}0$.
set $ m\equal{}n\minus{}1$, $ f(n^2)\equal{}(n\minus{}1)^2$. Combined with (2), we get $ f(n)\equal{}\pm(n\minus{}1)$ for all $ n$. Now assume that there exist $ a,b$ s.t. $ f(a)\equal{}a\minus{}1$ and $ f(b)\equal{}1\minus{}b$. Take $ a\equal{}n\plus{}m,b\equal{}n\minus{}m$. Sub this to the given equation, it is easy to get that $ a\equal{}1$ or $ b\equal{}1$. But for $ x\equal{}1$, $ \pm(1\minus{}1)\equal{}0$. So for all $ x$, we must have $ f(x)\equal{}x\minus{}1$ or $ f(x)\equal{}\minus{}x\plus{}1$. But these functions do not satisfy the given equation.

So the only function is $ f(x)\equal{}x,\forall x\in\mathbb{R}$.

 :D

Hey, it's problem T11/380 in MY magazine, and the deadline for this problem is 15th April 2009.

My proof is similar except for the part (ii)
Into $ f(n^2)\equal{}f(2n)\plus{}f(n^2)$ put $ \minus{}n$. Hence $ f(2n)\equal{}f(\minus{}2n)\Longrightarrow$ $ f$ is even function. So we can assume that $ f\geq 0$. In the main equation put $ n\equal{}0 \Longrightarrow f(0)\minus{}m^2\equal{}f(\minus{}m)f(m)\equal{}f^2(m)$. Hence $ f(x)\equal{}\sqrt{1\minus{}x^2}$ for $ x^2\leq 1$. But this is false because $ f^2(x)\equal{}f(x^2)$. Contradiction!


How do I add a header to my document? I couldn't find anything about itin the LaTeX section of AoPS.

Yes it does. [[LaTeX:Layout]].

\section{TITLE}
\subsection{TITLE}
\subsubsection{TITLE}

They're numbered. To take out the numebr from one of them, do \section*{title} \subsection*{TITLE} etc.

There's more, that are something like \paragraph{TITLE} \subparagraph{TITLE} etc. but I forget them.

No I mean like a little title at the top of the paper on every page.

Hi

I have some problems with fancyhdr (including a \leftmark)
I wan't to have a Chapter that I isn't listed in the TableOfContents and that does not have a chapter number, so I use the command: [b]\chapter*{Chaptername}[/b]
This is all good, except for one thing, the name in the Header (\leftmark) doesn't change to "Chaptername"
I want to have the chapter after TOC (or list of figures/tables for that matter) but having Chaptername in the header, not Contents! (or the name that was used before this Chapter)

If I instead use the command: [b]\chapter{Chaptername}[/b], it changes the header, but also sets it as "Chapter 1, Chaptername" and includes it in the TOC, and this is not the way I want it..
Is it possible to maybe redefine what is printed in the header (\leftmark)??

Best regards
 Perra

My pagestyle:
[code]
\fancypagestyle{plain}
{
    \fancyhf{} % clear all header and footer fields
    \fancyhead[RO]{\leftmark \hspace{5mm} \bfseries \thepage}
    \fancyhead[LE]{\bfseries \thepage \hspace{5mm} \normalfont \leftmark}
    \renewcommand{\headrulewidth}{0.4pt}    % Line under the header
    \renewcommand{\footrulewidth}{0pt}      % No line over the footer
    \renewcommand{\baselinestretch}{1.25}   % Stretches line spacing
}
[/code]

Have a look section 9, pages 9-11 of the [url=http://tug.ctan.org/get/macros/latex/contrib/fancyhdr/fancyhdr.pdf]fancyhdr manual[/url] which deals with this situation. In particular, the instructions at the top of page 11 looks like what you are asking for.


Let $ n\ge 3$ be a natural number. Find all nonconstant polynomials with real coeficcietns $ f_{1}\left(x\right),f_{2}\left(x\right),\ldots,f_{n}\left(x\right)$, for which
\[ f_{k}\left(x\right)f_{k+ 1}\left(x\right) = f_{k +1}\left(f_{k + 2}\left(x\right)\right), \quad  1\le k\le n,\]
for every real $ x$ (with $ f_{n +1}\left(x\right)\equiv f_{1}\left(x\right)$ and $ f_{n + 2}\left(x\right)\equiv f_{2}\left(x\right)$).

[quote="SAPOSTO"]Let $ n\ge 3$ be a natural number. Find all nonconstant polynomials with real coeficcietns $ f_{1}\left(x\right),f_{2}\left(x\right),\ldots,f_{n}\left(x\right)$, for which
$ f_{k}\left(x\right)f_{k \plus{} 1}\left(x\right) \equal{} f_{k \plus{} 1}\left(f_{k \plus{} 2}\left(x\right)\right)$,$ 1\le k\le n$,
for every real $ x$ (with $ f_{n \plus{} 1}\left(x\right)\equiv f_{1}\left(x\right)$ and $ f_{n \plus{} 2}\left(x\right)\equiv f_{2}\left(x\right)$).[/quote]

Let $ d_k>0$ the degree of $ f_k(x)$. The equation implies $ d_k\plus{}d_{k\plus{}1}\equal{}d_{k\plus{}1}d_{k\plus{}2}$ and so $ d_k\equal{}d_{k\plus{}1}(d_{k\plus{}2}\minus{}1)$
If $ d_{k\plus{}2}\equal{}1$, $ d_k\equal{}0$, which is impossible. So $ d_{k\plus{}2}>1$ and so $ d_k\geq d_{k\plus{}1}$
So, the cyclic equality implies $ d_k\equal{}$constant, and so $ d_k\equal{}2$ $ \forall k$

So $ f_k(x)\equal{}a_kx^2\plus{}b_kx\plus{}c_k$ and the equation becomes :
$ (a_kx^2\plus{}b_kx\plus{}c_k)(a_{k\plus{}1}x^2\plus{}b_{k\plus{}1}x\plus{}c_{k\plus{}1})$ $ \equal{}a_{k\plus{}1}(a_{k\plus{}2}x^2\plus{}b_{k\plus{}2}x\plus{}c_{k\plus{}2})^2\plus{}$ $ b_{k\plus{}1}(a_{k\plus{}2}x^2\plus{}b_{k\plus{}2}x\plus{}c_{k\plus{}2})$ $ \plus{}c_{k\plus{}1}$

Equating coefficients of $ x^4$ in this equality implies $ a_ka_{k\plus{}1}\equal{}a_{k\plus{}1}a_{k\plus{}2}^2$ and so $ a_k\equal{}a_{k\plus{}2}^2$ (since $ a_{k\plus{}1}\neq 0$, else $ d_k\equal{}0$). And so $ a_k\equal{}a_k^{2^n}$ and so $ a_k\equal{}1$ and the equation becomes :
$ (x^2\plus{}b_kx\plus{}c_k)(x^2\plus{}b_{k\plus{}1}x\plus{}c_{k\plus{}1})$ $ \equal{}(x^2\plus{}b_{k\plus{}2}x\plus{}c_{k\plus{}2})^2\plus{}$ $ b_{k\plus{}1}(x^2\plus{}b_{k\plus{}2}x\plus{}c_{k\plus{}2})$ $ \plus{}c_{k\plus{}1}$

Equating coefficients of $ x^3$ in this equality implies $ b_k\plus{}b_{k\plus{}1}\equal{}2b_{k\plus{}2}$ and so $ b_k\equal{}b_{k\plus{}1}$ else $ b_{n\plus{}1}\neq b_1$.
So $ b_k\equal{}b$ $ \forall k$ and the equation becomes :
$ (x^2\plus{}bx\plus{}c_k)(x^2\plus{}bx\plus{}c_{k\plus{}1})$ $ \equal{}(x^2\plus{}bx\plus{}c_{k\plus{}2})^2\plus{}$ $ b(x^2\plus{}bx\plus{}c_{k\plus{}2})$ $ \plus{}c_{k\plus{}1}$

Equating coefficients of $ x^2$ in this equality implies $ c_k\plus{}c_{k\plus{}1}\plus{}b^2\equal{}b^2\plus{}2c_{k\plus{}2}\plus{}b$ and so $ c_k\plus{}c_{k\plus{}1}\equal{}2c_{k\plus{}2}\plus{}b$
Adding the $ n$ equalities $ c_k\plus{}c_{k\plus{}1}\equal{}2c_{k\plus{}2}\plus{}b$ implies $ b\equal{}0$ and then $ c_k\equal{}c$ constant and the equation becomes :
$ (x^2\plus{}c)(x^2\plus{}c)$ $ \equal{}(x^2\plus{}c)^2\plus{}c$ and so $ c\equal{}0$

So, the only solution is $ f_k(x)\equal{}x^2$


[hide="Problem as I understood it"]Let $ a_1, a_2, \ldots, a_n$ be a sequence of real numbers. Suppose that for each positive integer $ k$, we have $ (1\plus{}a_1^k)(1\plus{}a_2^k)\cdots (1\plus{}a_n^k) \equal{} 1$. Prove that $ a_1 \equal{} a_2 \equal{} \ldots \equal{} a_n \equal{} 0$.

[hide="But this seems too easy"]Suppose $ a_m \neq 0$ for some $ m$. Pick any even positive integer $ k$. Then for all $ 1 \leq i \leq n$, $ (1\plus{}a_i^k) \geq 1$, and $ (1\plus{}a_m^k) > 1$. So the product of all $ (1\plus{}a_i^k)$ is greater than 1. Contradiction.[/hide][/hide]

Clarify?

For real numbers it followed from case $ k\equal{}2$. But it istrue for all complex numbers.

If all numbers are complex can you give a proof rust?

I think Virgil posted this problem in $ \mathbb{C}$

What about a solution?

 :)


probability
floor function
quadratics
Vieta
factorial
algebra
quadratic formula

It's been ages since I did this but COP-Shortlist is basically set of problems that will help you on problem solving and get you ready for the problems in COP (Contest of Opportunit). Post to this thread if you find solution to any of the problems.

None of the problems are mine. All problems are from various competitions.

1. The integers from 1 to 2001 are written in order around a circle. Starting at 1, every 6th number is marked (that is 1,7,13,19,etc). This process is continued until a number is reached that has already been marked. How many unmarked numbers remain?

A. None
B. 1668
C. 1669
D. 1004
E. 1334

2. Find the probability that an element picked at random from the set {1,2,3,...1001} is divisible by 7 or 11 (or both).

A. 1/13
B. 1/11
C. 1/77
D. 17/77
E. 18/77

3. If $a+b+c = 0$ and $a^3+b^3+c^3 = 27$, find $abc$.

A. -3
B. 0
C. 3
D. $3 \sqrt[3] {9}$
E. 9

4. Bottle A contains a quart of milk and bottle B contains a quart of black coffee. Pour a small amount from B into A, mix well, and then pour back from A into B until both bottles again each contain a quart of liquid. Then:

A. The fraction of coffee in A is equal to the fraction of milk in B.
B. The fraction of coffee in A is bigger than the fraction of milk in B.
C. The fraction of coffee in A is lesser than the fraction of milk in B.
D. The answer varies; it depends on how well you do the mixing.
E. The answer varies; it depends on how much you pour.

5. Suppose that there are some blabs, some blibs, and some blubs. Suppose also that all blabs are blibs and some blubs are blabs. On the basis of this information, which of the statements X,Y,Z must be true? X: All blabs are blubs; Y: Some blabs are not blubs; Z: Some blibs are blubs.

A. X only
B. Y only
C. Z only
D. X,Y only
E. Y,Z only

6. How many positive integers less than 10,000 are of the form $x^8+y^8$ for some integers $x>0$ and $y>0$?

A. 5
B. 6
C. 7
D. 8
E. > 8

7. Suppose that 100 more than the sum of $n$ consecutive integers is equal to the sum of the next $n$ consecutive integers. Then $n$ equals:

A. 10
B. 31
C. 50
D. 100
E. 200

8. If $17!$ is written as base 8 numeral, how many zeros will appear at the end of the numeral?

A. 0
B. 3
C. 5
D. 9
E. 17

9. In a class of 100, 40 students study Spanish, 35 study German, and 27 study French. Two study all three subjects, while 3 study Spanish and German only, and 40 do not study either Spanish or French. Finally, 20 students study no language at all. How many students who study German do not study Spanish?

A. 25
B. 30
C. 31
D. 33
E. 34

10. If $\Big(1+\frac{1}{x^2}\Big)\Big(1+\frac{1}{x^4}\Big)\Big(1+\frac{1}{x^8}\Big)\Big(1+\frac{1}{x^{16}}\Big)\Big(1+\frac{1}{x^{32}}\Big) = \frac{x^m - 1}{x^m - x^{m-2}}$

A. 32
B. 64
C. 70
D. 72
E. 128

11. The difference between the roots of $x^2+px+q = 0$ is the same as the difference between the roots of $x^2+qx+p$, and $p \neq q$. Then $p+q =$?

A. -4
B. -1
C. -2
D. 0
E. 2

12. Find the units digit of the sum $\sum^{2005}_{k=0} k^5$.

A. 1
B. 3
C. 5
D. 7
E. 9

[hide="Answer Keys"]
E
D
E
A
C
A
A
C
B
B
A
C [/hide]

[hide="2"]
143 of these numbers are divisible by 7, 91 are divisible by 11. (operating on 1001/7 and 1001/11).

However, we counted the numbers that are divisible by both 7 and 11 (which is 77) twice. We subtract the numbers that are divisible by 77 (or 1001/77 = 13) to get 221.

221/1001=17/77[/hide]

[hide="3"]
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$.

$27-3abc=0$

$abc=9$. [/hide]

What do you want in number 10?

[hide="Number 8"]
We can rewrite the question as: "Find the largest integer $n$ such that $8^n$ divides $17!$."

Note that $8^n=2^{3n}$.

$\lfloor\frac{17}{2}\rfloor +\lfloor\frac{17}{4}\rfloor +\lfloor\frac{17}{8}\rfloor +\lfloor\frac{17}{16}\rfloor =8+4+2+1=15$, so $3n=15$.

Therefore, $n=5$ :arrow: $\boxed{C}$ [/hide]

[quote="shadysaysurspammed"]In the 11th one,i can prove p-q=0...why can't we take it,the proof is right.

Anyway this is what i'm getting fir 11th one

Let A and B be roots of 1st and a and b be roots of second.Then,

A+B=-p
AB=q
a+b=-q
ab=p

A+B+a+b=-AB-ab
$p^2+q^2=p+q$ :? How can p+q be -2?[/quote]

The questions wants p+q, not p-q.  Also, the answer key gives A, not C, the answer is -4 not -2.

[hide="solution"]
The differences of the roots are equal, so let's find the difference and set them equal using quadratic formula
$\frac{2\sqrt{p^2-4q}}{2} = \frac{2\sqrt{q^2-4p}}{2}$
$p(p+4) = q(q+4)$
Since $p \not = q$, the only way that can happen is if both sides are 0.  So either p or q is -4, and the other must be 0.  Therefore the sum is -4.
[/hide]

[hide="1"]$2001\equiv 2\bmod{6}$, so we have $334$ numbers marked the first time around, and we end up on the $4$. $(2001-4)\equiv 5\bmod{6}$, so we're going to end up on the $1$ the next time around, which we've already marked, so our answer is $2001-(334+333)=1334\Rightarrow \boxed{E}$.[/hide]

[hide="5"]X is not necessarily true; some blubs are blabs, but there is nothing to stop us from having some blabs not be blubs. Y is not necessarily true; we do know that some blubs are blabs, but nothing stops us from having all blabs be blubs, while some blubs are something else. Z must be true; all blabs are blibs, and some blubs are blabs, which means that some blubs are blibs, and therefore vice-versa, so our answer is $\boxed{C}$.[/hide]

[hide="6"]We have $1^8=1<10,000$, $2^8=256<10,000$, $3^8=6,561<1000$, but $4^8=65,536>10,000$. Since $x$ and $y$ do not have to be distinct, our possibilities are $(1,1)$, $(1,2)$, $(1,3)$, $(2,2)$, $(2,3)$, $(3,3)$, so our answer is $6 \Rightarrow \boxed{B}$.[/hide]

[b]EDIT:[/b] [hide="4"]Pour $\frac{1}{100}$ of $B$ into $A$. $A$ is now $\frac{1}{101}$ coffee and $\frac{100}{101}$ milk. Now, pour $\frac{1}{101}$ of $A$ back into $B$. $B$ is now $\frac{100}{101}\cdot \frac{1}{101}=\frac{100}{10201}$ milk and $\frac{10101}{10201}$ coffee, while $A$ is $\frac{10101}{10201}$ milk and $\frac{100}{10201}$ coffee, so our answer is $\boxed{A}$.

This can also be proved just algebraically, but it's easier to explain why it works with a sample amount.[/hide]

[hide="10"]$(1+\frac{1}{x^2})(1+\frac{1}{x^4})...(1+\frac{1}{x^{32}})=\frac{x^{62}+x^{60}+...+x^2+1}{x^{62}}=[1]$
$[1]\cdot \frac{x^2-1}{x^2-1}=\frac{x^{64}-1}{x^{64}-x^{62}} \Rightarrow m=64 \Rightarrow \boxed{B}$.[/hide]

[hide="12"]The units digit of every $n^k$ repeats every 1, 2, or 4 consecutive values of $k$, so $n^5\equiv n\bmod{10}$. Therefore, $\sum_{k=0}^{2005}k^5\bmod{10}=\sum_{k=0}^{2005}k\bmod{10}$. Every set of $10$ integers will give a $5$ in the units place, and the sum of $5$ integers also gives a $5$, so we have $200\cdot 10\Rightarrow 0+5=5 \Rightarrow \boxed{C}$.[/hide]

What's wrong in my solution to the 11th one???

I have no idea what you even did in your solution... you didn't even go anywhere with what you wrote down.

He used vieta's fomulas and then did something I couldn't follow either.

I used vieta's formula(whatever that is) adn then try and understand.....then i did  $(a+b)^2$...What step can you not understand???

How did you get from $A+B+a+b=-AB-ab$ to $p^2+q^2=p+q$?

By using $(a+b)^2$...then i substituted all the values....don't you guys get it???Try it...

A+B+AB+a+b+ab=0
$(A+B)^2-AB+(a+b)^2-ab=0$
Substituting values....
$p^2-q+q^2-p=o$
$p^2+q^2=p+q$

What is wrong in this?

What are you doing with the squared terms? $(A+B)^2=A^2+B^2+2AB$...

Number 7 is nice.

[hide]

Let the sum of the $n$ consecutive integers equal $x$.

  We know 100 more than $x$, (100+x) is eual to the next $n$ consecutive integers.

  Now consider the very first term of the $n$ consecutive integers, 1 and the very first term of the next $n$ consecutive integers, $n+1$.

 Their difference is $n$ as each term can in the first and next $n$ consecutive integers can be paired up with 1 term that is $n$ greater.  There are $n$ terms so the sequence is $n \cdot n=n^2$ greater.

$100 + x = x + x^2$

$100=x^2$

$\boxed{x=10}$[/hide]

[quote="shadysaysurspammed"]A+B+AB+a+b+ab=0
$(A+B)^2-AB+(a+b)^2-ab=0$
[/quote]

Can you explain that? By the way, I think it's $(a+b)^2-2ab$.  ;)

:rotfl:  :rotfl: Sorry for wasting your time factorial and jesusfreak...I appreciate your concern towards my doubt...but now i've realised that i'm totally wrong after the step factorial mentioned :(  :( I have been acting pretty dumb lately,I'll think $thrice$ before i speak now.


Gauss
Euler
calculus
articles
Pythagorean Theorem
geometry
complex analysis

I've always wondered who's considered to be the most famous mathematician!
Please post your opinions here. :)

[b]Shouldn't this post be on the Fun&Games forum instead of here?[/b]

The question is somehow subjective. Nevertheless, even when you are going to have very different opinions here (because we are a mathematical community and we know many mathematicians) I would guess that, if you ask to any one in the street, you will hear once and again [b]Pitagoras[/b]. Actually, it is the only one most of the people know.

Best,

hm i believe the mathematician who has been the most neglected yet contributed the most to the mathematical community given his limited resources was [b]Archimedes[/b]. hehe, his mind was so brilliant that the romans even sent soldiers out to capture him. his accomplishments are way too many to list, but given the time period he lived in, he sure did know a LOooooot of math. *shrug, but that's just my opinion

Grothendieck!!!

And of course Gromov!!!!!!!!!!!!!!!!!

When I look how much is named after him: C. F. Gauss

[quote="tokenadult"]Definitely Pythagoras. He is the one mathematician that every schoolkid learns the name of, all over the world.[/quote]

Not by the Chinese. They know Pythagorean theorem by a different name.

Euler: I did an essay in fourth grade, and since then, I believed it.

[quote="hello"]This is slightly off topic, but the three greatest mathematicians are for some reason considered to be Archimedes, Newton, and Gauss. Not sure why them, but that's what most books say.[/quote]

A bit of a correction.

Those three men are considered the three best [u]scientists[/u], not the three greatest mathematicians.

In terms of the question at hand.  I do not think it is Pythagoras because people hear of the "Pythagorean Theorem," but they may not know the namesake.

While Gauss is almost certainly the best mathematician of all time [whatever that means], the most [u] famous [/u] is likely Sir Isaac Newton.  In the sense that he is the most famous man who happens to also have been a mathematician.  He was more famous for his science than his mathematics, but I would say he is more famous than fermat [whom I would say is the second most famous].

Remember, the vast majoirty of people wouldn't know a harmonic mean from a mean harmony  :lol: , so when you speak of "most famous" you have to look at whose name is most in the pool of common knowledge [and that pool can be mighty shallow!].

[quote="hello"]the three greatest mathematicians are for some reason considered to be Archimedes, Newton, and Gauss. Not sure why them, but that's what most books say.[/quote]

The contributions of Archimedes and Newton were on a seemingly ineluctable road to discovery by others.  Gauss' and Galois' discoveries have more of the "singular genius" aspect where the ideas could have gone for centuries without further progress.

I'd say Pythagoras.  I don't think he made the greatest contributions, but he's certainly the most famous.  Everyone knows about the Pythagorean Thm. Unless one studies advanced math to a great depth, not many hear of Gauss, Cauchy, etc.

Actually it is referenced to mathematicians, not scientists.

[quote="hello"]This is slightly off topic, but the three greatest mathematicians are for some reason considered to be Archimedes, Newton, and Gauss.[/quote]

I have read that Gauss had something to do with establishing that list. But I can't confirm that with a source citation today.

i note no east europe/asian mathematicians here
and i think reimann (spelling?) is most well known among economists, cryptography, mathematicians, etc..., even in the eastern hemisphere

[quote="1234567890"]Cauchy; half the theorems in complex analysis are named after him.[/quote]

You should study a little deeply complex analysis. There are relatively few things in complex analysis named after him.

[quote="EUCLA"][quote="tokenadult"]Definitely Pythagoras. He is the one mathematician that every schoolkid learns the name of, all over the world.[/quote]i think so :)[/quote]
Me too. I've actually read in a Japanese book the result of a poll of this sort...maybe I'll look for it. (Japanese people know him as "pitagorasu" though.)

[quote="pkothari13"]i suggest you make a poll with a bunch of famous mathematicians, and then have other as another choice[/quote]

Well, depending on your requisites to select [b]a bunch of famous mathematicians[/b], it could include a few hundreds (or thousands) and the polls here just accept up to 10 poll options! That means, I think your idea is not really practic!

Newton is quite famous, but more so for his lil falling apple than his calculus. Pythagoras would be famous if people actually knew that the Pythagorean Theorem was named after him, but no such luck exists. Everyone I know connects Pythagoras's theorem to the Egyptians. Fermat is quite famous for his last theorem and his work in number theory, among mathematicians that is. And on that note, even Andrew Wiles is famous among modern mathematicians for  proving Fermat's theorem. Archimedes is probably remembered more for his physics than his math as well. Euler would be good, imo, but very few people outside of the select group of extremely motivated students actually remember him after their test on mathematicians. Cauchy and Gauss are standard knowledge for AoPSers but definitely not the rest of the world. I personally think DeCartes may be a good contender. DaVinci may or may not be, depending on whether or not you consider him an actual mathematicians. Einstein would definitely be up there, but again, more physics than math in his case.

Actually, Einstein was only an average mathematician. Not only did he make little or no strictly mathematical discoveries, he just wasn't that extraordinarily good at math, nor did he explore lots of advanced math. I'm sure he was above average, but I don't even think he's worth even mentioning as a mathematician.

[quote="09husbri"]I say Newton. He simultaneously invented calculus! C'mon![/quote]

Well, this has been under discussion for centuries. There are people who believes that Liebtniz developed most of his ideas a few years before Newton, but when the Royal Academy of Sciences decided to give a veredict, well, they created a committe to review the facts, and you won't believe who was the head of the commitee....

On the other hand, some biographical references of Isaac Newton described him as a, let's say, not quite honorable person, to be euphemistic! Eitherway, there is no doubt that he is one of the greatest men in the history of science.

I know, Newton headed the committee, right? I wrote a paper on him a while back.

[quote="pkothari13"][quote="Jhyon8149"]The best mathematician is Gauss but I think the most famous is either Newton, Archimedes, or Fermat. Fermat especially because people were captivated by Fermat's Last Theorem for what, hundreds of years?[/quote]
ah, good point.  i was thinking of "best" mathematician. i suppose when it comes to fame, i'll have to go with newton.[/quote]

With respect to Mathematics, well, Newton is quite remarkable, but he is at the high of Gauss's ankle!, Gauss just did too much, more than what any other mathematician has done ever! If by that time there had been a scientific publishing system as we have now, probably Erdos would have been an amateur compared to Gauss.

By the way, I don't really know much about the moment where mathematics start being published in Jurnals instead of recopilated in sporadic books and mathematicians mail, but I am surprised to know that Erdos has have the most articles published ever, specially considering that Hilbert is not that far back in the past.

If you are going to name most famous...Think of what people you run into everyday may tell you...

I ran this experiment, and amazingly I got the answer "Albert Einstien" more than anything.  Although his ideas where great...I consider him more of a physicist than mathematician.

The other math people I know said things like Euler, Archemedies, Pythagoras, Newton, Fermat, Gauss... (Gauss would be my pick for best  :lol: )

IMHO, most famous modern mathematician : Ronald Rivest


I think it is good that President Bush tapped Ben Bernanke because of his solid qualifications as both the current Chairman of Economic Advisors, former Fed governor and former department head of economics at Princeton. Any other opinions?

LOL, I like your joke Mathew, but what is your opinion on Bernanke being chosen?

I haven't paid all that much attention, but there's a senator in Alabama who knows enough about economics to think he's a fine choice, so I guess that's a good enough endorsement.

Mathew, I dont understand why you base your decisions on some senator from Alabama, there are more talented people who endorsed him, you saying that surprised me. But I tjhink he is a safe choice, I studied by his principles of macro book.

I'm not saying he's a bad choice.  I admit to not being up to date on the qualifications of all the candidates -- particularly his.  For all I know, he's the best one yet!  For all I know, he's the devil.

I was just taking a moment to take a shot at the esteemed senator from my home state who likes to see his name in print.  He might be as good at anyone in the senate at finding a reporter to drop names to during historic events.  It's really a special talent.  I wish I had that.  I really wish I did.

lol, Mathew, you have good humor skills too. But I think academically at least, Bernanke, is the most qualified of previous chairman's, he did much more academics then Greenspan, horray to a good economy in the future!! And may the financial markets give more then historical returns on my future stock investments (I only want to start investing in the not too near future) and I will become a millionaire 10^n where n is a positive integer greater then 10. LOL. What is wrong with the dollar symbol?

Unfortunately the policy makers have much greater control over the economy.  If only it were as simple as having one smart person in a good position...

Yes, but one smart person probably cant get everything done themselves too. But the Fed Chairman is also a policymaker even though I think there are a lot of things that he will not get to influence anymore except in speeches which will be considered political as when Greenspan did.


calculus
integration
trigonometry
function
complex numbers
complex analysis
complex analysis unsolved

How would you intergrate the following from -inf to inf

(Cosx/(a^2 - x^2))dx for a>0

We started simple integrals with residues but then our prof just whipped this one out on us.

That's obviously divergent; we shouldn't have any zeros of the denominator on the $x$-axis.

Use that $\cos x$ is the real part of $e^{ix}$. Find an appropriate closed curve to integrate your function (with $e^{ix}$ replacing $\cos x$) around, and show that the part of the curve away from the axis does not contribute in the limit.

I had a feeling you were going to say something like that. argh

Perhaps you wish the Cauchy Principal Value:

$\text{pr.v.}\int_{-\infty}^{\infty}f(x)dx=2\pi i \sum_{\text{u}}\text{Res}\: f(z)+\pi i\sum_{\text{x}}\text{Res}\: f(z)$

Where the frist sum extends over all residues in the upper half plane and the second sum extend over all the poles on the x-axis

Edit:  I'll leave it here cus' sometimes going down a wrong avenue is helpful but I now see the above formula for calculating the  principal value is not valid for the integral you wish to calculate.  Know why?  However I believe the principal value still exists.


I was inspired by the exercise from [url=http://www.mathlinks.ro/viewtopic.php?t=294392][color=red][b]here[/b][/color][/url].

$ \sin x\plus{}\sin^2x\plus{}\sin^3x\equal{}a\in [\minus{}1,3]\ \implies\ \cos^6x$ $ \minus{}4\cos^4x\plus{}2(a\plus{}3)\cos^2x\equal{}(a\plus{}1)(3\minus{}a)$ .


Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. Prove that
$ \frac{ab}{(a\plus{}b\plus{}c)^2}\plus{}\frac{bc}{(b\plus{}c\plus{}d)^2}\plus{}\frac{cd}{(c\plus{}d\plus{}a)^2}\plus{}\frac{da}{(d\plus{}a\plus{}b)^2} \le \frac{4}{9}.$

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c,$ $ d$ be positive real numbers. Prove that
$ \frac {ab}{(a \plus{} b \plus{} c)^2} \plus{} \frac {bc}{(b \plus{} c \plus{} d)^2} \plus{} \frac {cd}{(c \plus{} d \plus{} a)^2} \plus{} \frac {da}{(d \plus{} a \plus{} b)^2} \le \frac {4}{9}.$[/quote]
Let me post my solution: http://can-hang2007.blogspot.com/2010/01/inequality-35-v-q-b-can.html


algebra
polynomial
real analysis
real analysis solved

We have a sequence of real numbers an.We know that 4a_(n+2) -4a_(n+1) +a_n is convergent to a.Prove that an is convergent to a, only by using the Stolz-Cesaro criteria!

Well,there is a general method for solving such problems, but in this particular case we can use the following fact:
If x_n is a sequence of real numbers such that 2x_(n+1)-x_n is convergent the x_n is convergent.
Proof by Stolz-Cesaro:
We have 2x_(n+1)-x_n = [2^(n+1)*x_(n+1) - 2^n*x_n]/(2^(n+1)-2^n)
Then we have by cesaro that x_n is convergent!
This is a bit simple but i like the trick.

Using the characteristic polinomial of the sequence!

I wanted to say if p is a polinomial of the form ax^2+bx+c with ac not 0 then if yn=p(xn) ,the roorts of the polinomial are in modulus smaller than 1 and yn is conv then xn is convergent.


Show that $ \forall n \in \mathbb{N}\;;\;n\geq 3$ then $ \exists$ two odd numbers $ x,y$ such that
\[ x^2\minus{}y^2\equal{}2^n\]
he he
it is hard !
who want to try?

am I missing anything? choose $ x\equal{}y\plus{}2\equal{}2^{n\minus{}2}\plus{}1.$  :lol:

[i]Additional assumption x>y>0. [/i] 2|x iff 2|y and 2|x-y (otherwise $ 2 \nmid x \plus{} y \implies 2 \nmid xy$) and surely x+y>x-y>0. So it is necessary and sufficient condition that for all integer i such that 1≤ i≤ [(n-1)/2] we have $ 2^i \equal{} x \minus{} y < x \plus{} y \equal{} 2^{n \minus{} i}$. So for all $ n \in \mathbb{N}_0$ there exist exactly [(n-1)/2] couple (x,y) of solutions. Now we can calculate number of solutions with x,y both even, setting x=2a,y=2b, so $ a^2 \minus{} b^2 \equal{} 2^{n \minus{} 2}$ that has exactly [(n-1)/2]-1 solutions. So with the additional assumption x>y>0 we have proved that exist exactly one couple of odd integers (x,y) such that $ x^2 \minus{} y^2 \equal{} 2^n$.


I will post some problems that appeared in the University Entrance Chemistry Exams in my country, from 2006 and 2007.
----------------------------------------------------

Problem 3. Consider mercury(II) oxide (HgO), and mercury(II) sulphide (HgS), the main constituent of cinnabar, mineral from which mercury is extracted.
Supposing that the corresponding crystals are purely ionic, explain the following true statement: mercury(II) oxide has a greater lattice energy (in absolute value) than mercury(II) sulphide.

since it's given they are ionic and since the charge on each ion of the respective compounds is the same we rae left to analyse the internuclear distance which in case of $ O$ is less than that of $ S$.
since $ \Delta U$ is inversely propotional to $ r$ we have the required ans


How would you predict the charges of ions in ionic compounds if the elements are transition metals, and note close to the left or right side of the Periodic Table as shown below?

[img]http://preparatorychemistry.com/images/ions_PT.jpg[/img]

Thanks in advance!

Transition metals from the middle of d block have several possible oxidation states, as you might know. However, given for example $ \ce{CrCl3}$ I think its easy to see what is the charge of chromium. Am I missing something?

So how would I name something like $ \text{Ag}_2 \text{S}$ and $ \text{ZnBr}_2 \cdot 7\text{H}_2 \text{O}$?  Also, what does the dot in $ \text{ZnBr}_2 \cdot 7\text{H}_2 \text{O}$ mean?

Thanks!

[quote="elcric"]So how would I name something like $ \text{Ag}_2 \text{S}$ and $ \text{ZnBr}_2 \cdot 7\text{H}_2 \text{O}$?  Also, what does the dot in $ \text{ZnBr}_2 \cdot 7\text{H}_2 \text{O}$ mean?[/quote]

I'm currently taking AP chemistry, so I don't know a lot about technical terms.

1.) Transition metals have several oxidation states (as Carcul stated earlier) and these [b]must be memorized[/b]. We were given a chart/packet of some of the common transition metal ions. This is also similar in memorizing polyatomic ions (how many oxygens in sulfate vs sulfite compared to nitrate/nitrite).

2.) The dot sign indicates that the compound before it ($ \text{ZnBr}_2$) is a solid that absorbs the water.


If $ a$, $ b$, $ c$ are nonnegative integers, then prove that

$ \sqrt {\left(a^2b \plus{} b^2c \plus{} c^2a\right)\left(ab^2 \plus{} bc^2 \plus{} ca^2\right)}$
$ \geq abc \plus{} \left(\left(a^3 \plus{} abc\right)\left(b^3 \plus{} abc\right)\left(c^3 \plus{} abc\right)\right)^{1/3}$.

This is exactly the same as the problem from the KMO problem topic and was solved.

[quote="Maverick"]If $ x, y, z$ are nonnegative real numbers, then prove that

$ \sqrt {\left(x^2y \plus{} y^2z \plus{} z^2x\right)\left(xy^2 \plus{} yz^2 \plus{} zx^2\right)}$

$ \geq xyz \plus{} \sqrt [3]{\left(x^3 \plus{} xyz\right)\left(y^3 \plus{} xyz\right)\left(z^3 \plus{} xyz\right)}.$[/quote]Using the arithmetic-mean - geometric-mean inequality, we have a stronger inequality :

$ xyz \plus{} \sqrt {\sum_{cyc}x^2y\sum_{cyc}xy^2} \equal{} xyz \plus{} \sqrt {x^2y^2z^2 \plus{} \prod_{sym}{\left(x^2 \plus{} yz\right)}}$

$ \geq xyz \plus{} 3(xyz)^{\frac {1}{9}}\left[\frac {1}{8}\prod_{sym}{\left(x^2 \plus{} yz\right)}\right]^{\frac {4}{9}}\geq 4(xyz^{\frac {1}{4}}(xyz)^{\frac {1}{12}}\left[\frac {1}{8}\prod_{sym}{\left(x^2 \plus{} yz\right)}\right]^{\frac {1}{3}}$

$ \equal{} 2\sqrt [3]{xyz\prod_{sym}{\left(x^2 \plus{} yz\right)}} \equal{} 2\sqrt [3]{\prod_{sym}{\left(x^3 \plus{} xyz\right)}}$.

By the way, the following inequality

$ \left(2k^2 \minus{} 3k\plus{} 2 \right)xyz \plus{}k \sqrt {\left(x^2y \plus{} y^2z \plus{} z^2x\right)\left(xy^2 \plus{} yz^2 \plus{} zx^2\right)}$

$ \geq 2\sqrt [3]{\left(k^2x^3 \plus{} xyz\right)\left(k^2y^3 \plus{} xyz\right)\left(k^2z^3 \plus{} xyz\right)}$

holds for $ k\geq0$.


geometry
circumcircle
parallelogram
geometric transformation
rotation
ratio
perpendicular bisector

Let $ ABC$ be an $ A$-isosceles triangle. Let $ M$ be a moving point on $ [BC]$. The line through $ M$ parallel to $ AB,AC$, respectively, intersects $ AC,AB$ at the point $ X,Y$. Let $ O_a$ be the circumcenter of triangle $ AXY$. Prove that $ O_a$ moves on the fixed line. :lol:

This is a very strange problem. I have found out the solution but it is now $ 12: 15\text{am}$ so my mom won't allow me to stay up and post the solution :D (something from Fermat :P)

[b]Hint.[/b] Let $ BH_b, CH_c$ be the altitudes of $ \triangle ABC$. Perpendicular bisectors of $ AB,AC$ intersects $ CH_c,BH_b$ at $ K_c,K_b$. The locus of $ O_a$ is segment $ K_bK_c$.

[b]Comment.[/b] After found, the locus $ O_a$ is not very hard to prove but it is really a strange way to find out the locus of it.

$ AXMY$ is a parallelogram with $ AX \plus{} AY \equal{} AB \equal{} k \equal{} \text{const}.$ Let $ M,N$ be the midpoints of segments $AB,AC.$ Since $ AM \equal{} AN \equal{} \frac {_1}{^2}k,$ it follows that $ XM \equal{} YN \equal{} \frac {_1}{^2}|AX \minus{} AY|.$ If $ O$ is circumcenter of $ \triangle ABC,$ then $ O$ is the antipode of $A$ in the circle $\odot(AMN)$ and it's the center of the rotation  taking the oriented segments $ MX$ and $ NY$ into each other $\Longrightarrow$ $ O \equiv{} \odot (AXY) \cap \odot (AMN).$ Hence, locus of the circumcenter $ O_a$ of $\triangle AXY$ is the perpendicular bisector of $\overline{AO}.$

$ M_0, X_0, Y_0$ are midpoints of $ BC, AC, AB.$ $ M' \in BC$ is also arbitrary and parallels to $ AB, AC$ through $ M'$ cut $ AC, AB$ at $ X', Y'.$ $ \mathcal P_0, \mathcal P, \mathcal P'$ are circumcircles of $ \triangle AX_0Y_0, \triangle AXY, \triangle AX'Y'.$ Powers of $ X_0, Y_0$ to $ \mathcal P, \mathcal P'$ are in the same ratio:

$ \frac {p(X_0, \mathcal P)}{p(X_0, \mathcal P')} \equal{} \frac {\overline{X_0X} \cdot \overline{X_0A}}{\overline{X_0X'} \cdot \overline{X_0A}} \equal{} \frac {\overline{X_0X}}{\overline{X_0X'}} \equal{} \frac {\overline{M_0M}}{\overline{M_0M'}} \equal{} \frac {\overline{Y_0Y}}{\overline{Y_0Y'}} \equal{}$ $ \frac {\overline{Y_0Y} \cdot \overline{Y_0A}}{\overline{Y_0Y'} \cdot \overline{Y_0A}} \equal{}$ $ \frac {p(Y_0, \mathcal P)}{p(Y_0, \mathcal P')}$

$ \Longrightarrow$ circles $ \mathcal P_0, \mathcal P, \mathcal P'$ are coaxal and their centers collinear.

I have another approach for yetti's generalization.

[b]Proposition:[/b] $ D$ is a variable point on the side $\overline{BC}$ of $\triangle ABC.$ Parallels through $ D$ to $ AC$ and $ AB$ meet $ AB$ and $ AC$ at $X,Y.$ Then,  locus of the circumcenter of $ \triangle AXY$ is a line perpendicular to the A-symmedian of $\triangle ABC.$

Let $ M,N$ be the midpoints of $ AB$ and $ AC.$ It's easy to prove that $ \frac {_{XM}}{^{NY}} \equal{} \frac {_{c}}{^{b}}$ $\Longrightarrow$ $\odot (AXY)$ and $\odot (AMN)$ meet in the center $ P$ of the spiral similarity that takes the oriented segments $\overline{XM}$ and $ \overline{NY}$  into each other $ \Longrightarrow$ $ \frac {_{PM}}{^{PN}} \equal{} \frac {_{c}}{^{b}}$ $\Longrightarrow$ $ P$ is fixed on the arc $ MN$ of $\odot(AMN)$ and  lies on the A-Apollonius circle of  $ \triangle AMN.$ Now, the conclusion follows.


function
calculus
integration
derivative
LaTeX
calculus computations

Can someone integrate this for me:

$xsinx dx$.....I cannot use the let u technique for this...

Are you familiar with integration by parts? This is a very typical function for that approach.

So u= x and dv = sinx; so the derivative of x is dx and integration of dv=sinx is v=-cos x.

Using the formula:

$\int udv= uv- \int vdu =x(-cosx) - \int (-cosx)dx =-xcosx - -(sinx) =-xcosx+sinx$

Is that the final answer?

Looks right to me. :)

You can always check whether you integrated correctly by taking the derivative of your answer and see if you get back what you started with. :D

Also, you can use \sin(x) and \cos(x) to make your functions look nicer in LaTeX.

edit: Oops! Don't forget the constant of integration!

I almost forgot the c...Thanks for the help and for the latex tips! :D


MATHCOUNTS
geometry
perimeter
inequalities
calculus
integration
triangle inequality

Question:

  Are the problems from Ellias Saab's "Mathcounts Drills"  from Mathcounts Nationals, States, or Chapter.  I would think state, but a few look more like National problems.

  Problem:

  Anyways, heres an interesting problem I came across:

  The perimetre of a triangle is 106.  If one side is 21, what is the smallest possible length of the other side.

[hide="Hint"] Triangle Inequality [/hide]

[hide]a+21>85-a
2a>64
a>32
a=33 (assuming you mean integers)[/hide]

[hide]
a+21>b
a+b=88

a>32
smallest a = 33. [/hide]

[quote="G-UNIT"]Im giving you guys only partial credit for your solutions, you didnt answer my question! :D[/quote]
What do you mean, yeah we answered it. The shortest possible integral length for the other side is 33

No, you only answered his problem :P

His question is/was :

[quote="G-UNIT"]Question:

  Are the problems from Ellias Saab's "Mathcounts Drills"  from Mathcounts Nationals, States, or Chapter.  I would think state, but a few look more like National problems.

[/quote]

Oh  :P

THe answer is nationals, i think. Many of those problems if not all were on the national tests, cause I've taken like all the nats tests, and all those problems are familiar


trigonometry
inequalities unsolved
inequalities

Use cauchy:
We have:
$(\sin a \sin b+\cos a \cos b)\cdot LHS \ge (\sin^{2}a+\cos^{2}a)^{2}=1$
But we aslo have $\sin a \sin b+\cos a \cos b=\cos(a-b) \le 1$


function
probability
parameterization
probability and stats

$ X_1,...,X_n$  sample from distribution with unknown cumulative distribution function, strictly increasing and continuous.
Also $ X_{1: n},...,X_{n: n}$ are order statistics. Create asymptotic confidence interval for $ p$-Quantile of this distribution, with confidence $ 1\minus{}\alpha$ that is in form $ [X_{k_1: n},X_{k_2}: n]$. $ 0\leq k_1 \leq k_2\leq n$.

I was trying with approximating Bernoulli distribution with normal distribution. But I have no idea how to make this from the begining to the end.

If I understand it right, you are asked to find $ k_1$ and $ k_2$ such that $ P\{X_{k_2: n}<Q_p\}<\frac \alpha 2$ and similarly for $ k_1$ (you may want to split the probabilities differently, of course, but there is no apparent reason why you should). The event in question is just the event that $ k_2$ of the $ n$ Bernoulli trials with probability $ p$ of the favorable outcome will yield $ k_2$ or more favorable outcomes. That probability is just the sum $ \sum_{k\equal{}k_2}^n {n\choose k}p^k(1\minus{}p)^{n\minus{}k}$ and if you want an exact answer that works in full range of $ p,\alpha$, you'll have to stop here and just compute. If you want an approximate answer that works for reasonable values of parameters and large $ n$, you can say that it is about the same as the probability that the Gaussian random variable with mean $ pn$ and variance $ p(1\minus{}p)n$ is above $ k_2$, i.e., that the standard Gaussian is above $ \frac{k_2\minus{}pn}{\sqrt{p(1\minus{}p)n}}$, which can be read from the tables. The answer will then be of the form $ k_{1,2}\equal{}pn\pm f(\alpha/2)\sqrt{p(1\minus{}p)n}$ with $ f$ defined by $ P\{N>f(t)\}\equal{}t$ where $ N$ is the standard Gaussian random variable.

Yes you understand it correct. I also solved it before looking at your answer and get the same result. Thanks for your help :)


geometry
geometric transformation
reflection
rotation

reflect $(5,4)$ over the line $y=3x$.  Do we have a generalize for a point reflected over $y=mx+b$ or $ax+by+c=0$?

The idea is to make up a formula that will create a perpendicular through the point to be reflected, and then use distance formula and the equation of the perpendicular to find the new desired point.

Also, you can define reflections in the complex plane by taking the conjugate, and then rotating it (multiply by cis(x) for angle of rotation x) to match the reflection.