algorithm
number theory
least common multiple
Euclidean algorithm

i started learning number theory yesterday with reading a book.
i discovered two new theorem myself. i dont know if this has been discovered before.
but i need a proof for the first theorem.

[color=blue]Theorem 1[/color]:
if $ a$ and $ b$ are two consecutive integers then $ lcm(a,b)\equal{}ab$

[color=red]Theorem 2[/color]: if $ a$ and $ b$ are two consecutive integers then $ gcd(a,b)\equal{}1$

[color=green]proof of theorem 2[/color]:
we know that $ gcd(a,b).lcm(a,b)\equal{}ab$
now assume that theorem 1 is true for all consecutive integers $ a$ and $ b$
$ lcm(a,b)\equal{}ab$
$ gcd(a,b).lcm(a,b)\equal{}ab$
$ gcd(a,b).lcm(a,b)\equal{}lcm(a,b)$
$ gcd(a,b)\equal{}\frac{lcm(a,b)}{lcm(a,b)}\equal{}1$
$ gcd(a,b)\equal{}1$

almost 18 why ?what about you?  we dont study too much higher stuff in our class, i always say why dont we, here students only look for passin n getting a good grade not for maths, but bbcoz i love maths i read books my self

[quote="binomial_4eva"]i started learning number theory yesterday with reading a book.
i discovered two new theorem myself. i dont know if this has been discovered before.
but i need a proof for the first theorem.

[color=blue]Theorem 1[/color]:
if $ a$ and $ b$ are two consecutive integers then $ lcm(a,b) \equal{} ab$

[color=red]Theorem 2[/color]: if $ a$ and $ b$ are two consecutive integers then $ gcd(a,b) \equal{} 1$

[color=green]proof of theorem 2[/color]:
we know that $ gcd(a,b).lcm(a,b) \equal{} ab$
now assume that theorem 1 is true for all consecutive integers $ a$ and $ b$
$ lcm(a,b) \equal{} ab$
$ gcd(a,b).lcm(a,b) \equal{} ab$
$ gcd(a,b).lcm(a,b) \equal{} lcm(a,b)$
$ gcd(a,b) \equal{}\frac{lcm(a,b)}{lcm(a,b)}\equal{} 1$
$ gcd(a,b) \equal{} 1$[/quote]

Do you know the euclidian algorithm? It says that for integers $ a > b$, we have $ \gcd(a,b) \equal{}\gcd(a\minus{}b,b)$. Now that makes theorem 2 trivial: let $ a \equal{} n$ and $ b \equal{} n\plus{}1$, then $ \gcd(a,b) \equal{}\gcd(n,n\plus{}1) \equal{}\gcd(n,(n\plus{}1)\minus{}n) \equal{}\gcd(n,1) \equal{} 1$. Now, indeed, $ \text{lcm}(a,b)\gcd(a,b) \equal{} ab$, so thereom 1 follows now easily from theorem 2.

so those theorems are right ?were they discovered b4? :lol:

This part of the post used to have what the poster 2 above said. No point in repeating myself. :P

To prove the Euclidean Algorithm:
- let $ gcd(a,b) \equal{} d$.
- show that $ gcd(a,b\minus{}a)$ is divisible by $ d$.
- suppose there exists a number $ m$ such that $ m\mid a$ and $ m\mid b\minus{}a$, but it does not divide $ d$.
- use the fact that $ d\mid a$ and $ d\mid b$ implies that $ d\mid a\plus{}b$ to get a contradiction to $ gcd(a,b) \equal{} d$, which finishes the proof

Yes, the theorems are well-known, and I've used the second one very often on olympiads.

let $ a,b$ be two consecutive integers. $ b \equal{} a\plus{}1$.

If $ gcd(a,b) \equal{} d$ then there are $ x,y$ integers such that $ ax\plus{}by \equal{} gcd(a,b)\equal{}d$.

Now, $ (1)\cdot (a\plus{}1)\plus{}(\minus{}1)\cdot a \equal{} 1$ then $ a,b$ are relative primes and $ gcd(a,b) \equal{} 1$

stop being sarcastic Elemennop, i am new to number theory, but may be i am the next euclid :lol: , coz he worked so hard for discovering n i discovered it in the first day. :P

[quote="t0rajir0u"]My patience with you is wearing thin.

You should be happy that you [b]rediscovered[/b] this result.  It suggests you might have something of an intuition for number theory.  But you should realize that it only gets harder, as the years progress, for any mathematician to make new discoveries.  When Euclid made his discoveries, he was operating in a far less advanced mathematical climate, and he did not have many great mathematicians before him to work from, like you do.  The fact that you have thousands of years of mathematical history and discoveries to work from makes it much easier for you to [b]rediscover[/b] results like this than it was for past mathematicians to discover them [b]for the first time.[/b]

And I don't care if statements like

[quote="binomial_4eva"]may be i am the next euclid :lol: , coz he worked so hard for discovering n i discovered it in the first day. :P[/quote]

are a joke.  Stop talking that way.  In the context of what I'm heard from you, I'm afraid that you might actually believe this.  And as much as I don't want to quash your enthusiasm for mathematics, if you really understood anything about mathematics, you would understand how [b]incredibly hard[/b] it is to make new discoveries, and the way you talk about discovering "new" theorems is [b]disrespectful[/b] to the effort of those who have done real mathematics before you.[/quote]

Quoted for truth.

This is in Pre-Olympiad? My first serious browse too... how depressing.

Can a moderator move this to classroom math, or whatever you consider fitting?

[quote="t0rajir0u"]
And I don't care if statements like ... are a joke.
[/quote]

This is a little related to what you said. Statements like,
[quote]
Any person today is smarter than Newton because he knows everything he knows.
[/quote]
are really disrespectful and wrong. It makes absolutely no difference that you (not you, but "you" in general) are 350 years ahead. The person (Newton in this case) is still far superior in his abilities than anyone else who says that. Though he might not have much mathematical tools available he still cant be replaced nor matched.

Wait if you don't know the proof for a statement ("theorem 1" from first post) you conjecture, it is not called a theorem, it is called a conjecture.

Yeah, and this is in the wrong forum.

[quote="binomial_4eva"]I agree with you t0rajir0u, what you say is right
sorry i was just ......what do you call it , joke+missing about=?
but is there anything wrong with rediscovering i wrote the theorems
for you load to check, if its wrong or right.
Elemennop started bein sarcastic first.[/quote]

Based on what I've read, it's clear that you were joking after the first post, but do realize that this is a very elementary result.  Moreover, you have posted other such elementary results in this forum.  People here are starting to get a little harsh though: they're losing patience (as t0r explicitly stated).  I think we appreciate your enthusiasm, but you maybe want to post in a lower level forum.  It's a little frustrating to see elementary, very well-known facts posted as though they are not well-known.  (though "well-known" is a matter of opinion here: I'm sure many of the problems posted in this forum would be considered "well-known" and "obvious" to a mathematician)

But don't be discouraged that you're rediscovering only elementary results.  Results that are considered "elementary" can at times be nonobvious before you get a feel for a subject.  Just keep studying mathematics. :)

[quote="Elemennop"][quote="Elemennop"][quote="binomial_4eva"]were they discovered b4? :lol:[/quote]

nope, you're the first  you should get it published in a mathematical paper or something[/quote]

Quoted for truth.[/quote]

Quoted for humor.


After getting my distance,

d = sqrt( (x1 - x2)² + (y1 - y2)² )

I would times it by 25.4 to get how many millimeters there is between those two points?

Only if $ x_1,x_2,y_1,y_2$ are measured in inches.

Okay so what If I don't have any kind of measurement in units with x1,x2,y1,y2.  How am I supposed to figure out the mm in between the points?

If $ x_1, x_2, y_1, y_2$ aren't given with units, you need to find their units, or alternatively how those values were gotten. If they never had units in the first place, nothing you do will give them units.

Can you give us a context?

Well, as in my other post, I'm referring to the In-Game Mini Map...
http://img209.imageshack.us/my.php?image=examplexw9.png

So, where ever I click, X2 and Y2 is that position, and X1 = 748, Y1 = 298 which is my current position.  So, what can I do ?

Is there anybody that can help or just tell me the steps on what to do so I can get how many mm is inbetween X1 and X2.

So with my length below, how would I put this into a unit of inches so I then can convert into mm ?

X1: 748, Y1: 298
X2: 630 Y2: 312

d = 236 + 28 = 264 Sqrt: 16.2480

d = (x1 - x2)² + (y1 - y2)² = total = Sqrt

What would I need to do though so I can convert into mm.  I don't need to be spoon fed it, I just would like to know what steps I need to take to get to my desired goal.  Thanks.

I'm going to try to summarize what most of us are trying to say: [i][b]you can't[/b][/i]
[u]Please[/u] refer to these following posts:

[quote="Zyloch"]What you want is difficult, because while you can find the distance in pixels, you will then need to convert to millimeters depending on a variety of factors, such as screen resolution. And the value you get will be different for different monitors with different settings, and so forth. You would be better off sticking with the game coordinates as they are given.[/quote]

[quote="MellowMelon"]If $ x_1, x_2, y_1, y_2$ aren't given with units, you need to find their units, or alternatively how those values were gotten. [b]If they never had units in the first place, nothing you do will give them units.[/b]

Can you give us a context?[/quote]

[quote="Patterns_34"]What you're seeking is a length described by a certain unit. The only question is: what are your given lengths for this scenario?

Unfortunately, this scenario does not have any given lengths to work with. All that's given are numbers.
It's vital to have units to describe these numbers in this situation because they help describe the given numbers as lengths, which is what we need to help answer.

Otherwise, we can't do much else to properly answer the question.[/quote]

Even if you could, how would this be meaningful or useful? (consider Zyloch's statement)

I recommend addressing our posts before posting the same question again.

If I understand, this is coming from a game of some sort. You need to figure out how many millimeters or inches correspond to each unit in your game's coordinates... that is, what is the distance (in mm or inches) between $ (x,y)$ and $ (x\plus{}1,y)$ if those are two are points in your game, or equivalently the distance between $ (x,y)$ and $ (x,y\plus{}1)$. We have no way of knowing or doing this. You're on your own, unless you are prepared to post a [i]very[/i] detailed description of what you have.


find $\sum_{k=1}^n \frac{(-1)^k*!k}{k!}$ in terms of $n$, and $!n$ denotes the subfactorial function
(look here) [url]http://mathworld.wolfram.com/Subfactorial.html[/url]

[hide]I think you can make the sum easier by using the identity $!k=k!\sum_{i=0}^k \frac{(-1)^i}{i!}$, but I think you're still going to be left with a summation symbol in there somewhere.  The simplest form I could get it into was $\sum_{k=1}^n \frac{(-1)^k*!k}{k!}=\sum_{k=1}^\frac{n}{2} \frac{1}{(2k)!}$ for even $n$ and $\sum_{k=1}^n \frac{(-1)^k*!k}{k!}=-1+\sum_{k=0}^\frac{n-1}{2} \frac{1}{(2k+1)!}$ for odd $n$.[/hide]
The series diverges, but the two series for even and odd $n$ converge.  If we look at the Maclaurin series for $\cosh{x}$ and $\sinh{x}$, we see that the series for even $n$ converges to $\cosh{(1)}-1$ and the series for odd $n$ converges to $\sinh{(1)}-1$.


Hi everyone. I am searching for the mathmeticians who are related to Gottingen, but it is hart for me to find a complete list of their names. So can anyone help me find a complete list of mathmeticians related to Gottingen?

Thanks.

[quote="yasiru89"]try
[url]http://www.ams.org/notices/199510/maclane.pdf[/url][/quote]
Thanks a lot for the article!  :D


When pirates go ashore to dig up buried treasure, each pirate in the digging party receives a different-sized share of the plunder. The treasure is always completely shared out with the captain getting the largest share, the first mate the next largest and so on down to the cabin boy, if he is with the digging party.

As pirates are not smart, the fractions they use for the sharing of treasure always have $1$ as the numerator.

 [b]a)[/b] Show that there is only one way to share out the treasure amongst a digging party of $3$.

[b]b)[/b] Find the six different ways which may be used to share out the treasure amongst a digging party of $4$.

[b]c)[/b] Show that there is at least one way to share out the treasure amongst a digging party of any size greater than $2$.

[hide="A."]Let's say that the Captain get's 1/2. Now we need two different fractions with numerator 1 that add up to 1/2.

Let's say that the next person get's one third of the treasure. Then the lasp person MUST get 1/6 of the treasure.

Let's say that the second guy get's 1/4 or 1/5 or 1/6 or so on of the treasure. Then the third guy must get more, but that's a contradiction.

We did when the fiirst guy get's 1/2 of the treasure, let's try 1/3.

Now the second and third guy must sum up to 2/3, but the maximum sum, (1/4)+(1/5), is 9/20, which is less than 2/3.

We can present the same case with the first guy getting 1/4, 1/5, 1/6, so on.[/hide]

[hide="B."]Let's say the the first get's 1/2 and the second get's 1/3.

$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{1}{6}$

$(x-6)(y-6)=36$

Since x is bigger than y, there are four possibilities. We make a table:

\begin{tabular}[t]{|c|c|}\multicolumn{7}{l}{{\bf{\large x and y values}}}\\\hline{\bf x}&{\bf y}\\\hline 42 & 7 \\\hline 30 & 8 \\\hline 18 & 9 \\\hline 15 & 10 \\\hline\end{tabular}

So we get four possibilities:

$\frac{1}{2}+\frac{1}{3}+\frac{1}{15}+\frac{1}{10}$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{18}+\frac{1}{9}$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{30}+\frac{1}{8}$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{42}+\frac{1}{7}$

There aren't any more for the first guy get's 1/2 and the second get's 1/3.

So let's say that the first guy get's 1/2 and the second guy get's 1/4.

$\frac{1}{2}+\frac{1}{4}+\frac{1}{x}+\frac{1}{y}$

$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{1}{4}$

$(x-4)(y-4)=16$

We make another table of x and y values:

\begin{tabular}[t]{|c|c|}\multicolumn{7}{l}{{\bf{\large x and y values}}}\\\hline{\bf x}&{\bf y}\\\hline 20 & 5 \\\hline 12 & 6 \\\hline\end{tabular}

There aren't any more for 1/2 and 1/4.

For 1/2 and 1/5, there are none. The same for 1/2 and 1/6 and so on.

for 1/3, the maximum sum is .95 and the rest of the sums are less than 1.

The sums are:

$\frac{1}{2}+\frac{1}{3}+\frac{1}{10}+\frac{1}{15}$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{9}+\frac{1}{18}$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{30}$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}$

$\frac{1}{2}\frac{1}{4}+\frac{1}{5}+\frac{1}{20}$

$\frac{1}{2}\frac{1}{4}+\frac{1}{6}+\frac{1}{12}$[/hide]

[hide="C"]Proceed by induction.

For $n=3$, $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ works.

Now, also, note that $\frac{1}{k}=\frac{1}{k+1}+\frac{1}{k(k+1)}$.  So by repeatedly stretching out the smallest term in a series of fractions, we can add one more term.  For example, when $n=4$, we can do $\frac{1}{2}+\frac{1}{3}+\mathbf{\frac{1}{7}+\frac{1}{42}}$, replacing the $\frac{1}{6}$ with the boldface fractions.  This can be continued on, and thus the induction is complete.[/hide]


There are 50 even numbers within every 100, so there are $ 50\times 8\equal{}400$ even numbers in between 199 and 1000.

[quote="GameBot"]How many three-digit even integers exist which are greater than 199?[/quote]
It's easy. $ (999\minus{}199)/2\equal{}400$


Jim gave $ \frac12$ of his baseball cards to Bruce. Bruce gave $ \frac13$ of the cards he received to Vince. Vince gave $ \frac14$ of the cards he received to Steve. Steve received 8 cards. With how many cards did Jim begin?

Working backwards yields $ 8 \times 4 \times 3 \times 2\equal{}\boxed{192}$.


Each dice has a total of 1+2+3+4+5+6=21 dots. So, all three together have 63 dots. We subtract the number of dots we see to get how many are hidden.
63 - (1+2+3+6+4+1+5)
63 - 22
$ \boxed{41}$


geometry
geometric transformation
rotation

Why do we not use [b]alcohols[/b] in place of[b]sugar[/b] in polarimetry experiment(where we calculate the specific rotation of the solution using polarimeter)?

you must know the true purpose, and in some substance there is too high specific rotation(about 5600 degrees) useless for interpretation.


yea ive been trying to write some problems lately.... so here's one that i made... even though there may be some flaws to it *heh*

but yea here it goes....

Given that $xyz = 100$, and that $\displaystyle\sum_{\text{cyc}} (\log x)^2 = 108$, show that the maximum value of:

$\sqrt{\log_y x} + \sqrt{\log_z y} + \sqrt{\log_x z}$ is $2\sqrt3$

err like $(\log x)^2 + (\log y)^2 + (\log z)^2$

or... at least i hope i used the cyclic sum right...

yea my friend taught me yesterday sooo im just trying to incorporate everything that iev learned haha

[quote="gotztahbeazn"]yea ive been trying to write some problems lately.... so here's one that i made... even though there may be some flaws to it *heh*

but yea here it goes....

Given that $xyz = 100$, and that $\displaystyle\sum_{\text{cyc}} (\log x)^2 = 108$, show that the maximum value of:

$\sqrt{\log_y x} + \sqrt{\log_z y} + \sqrt{\log_z x}$ is $2\sqrt3$[/quote]

Is there some inconsistency in the problem?

I worked out like this.

I assumed that, in $\displaystyle\sum_{\text{cyc}} (\log x)^2 = 108$, the logarithms are for base 10.

calling $logx$, $logy$, $logz$ as $a, b, c$ respectively,

we have $a+b+c=2$ and $a^2 +b^2+c^2=108$
this gives $ab+bc+ca= -52$

which makes at least one of $a, b$ or $c$ negative(of course, all three can't be negative), because of which there is a danger that the quantity $\sqrt{\log_y x} + \sqrt{\log_z y} + \sqrt{\log_z x}$  becomes unreal.

So I think this problem must be restated properly.

oh... well my intended solution was this:

by cauchy you have:

$(\log x + \log y + \log z)\Big(\frac{1}{\log y} + \frac{1}{\log z} + \frac{1}{\log x}\Big) \ge \Big( \sqrt{\log_y x} + \sqrt{\log_z y} + \sqrt{\log_x z} \Big)^2$

then by RMS-AM-HM we have

$\sqrt{\displaystyle\frac{(\log x)^2 + (\log y)^2 + (\log z)^2}{3}} \ge \displaystyle\frac{\log x + \log y + \log z}{3} \ge \displaystyle\frac{3}{\frac{1}{\log x} + \frac{1}{\log y} + \frac{1}{\log z}}$

So, we have:

$ 6 \ge \frac{1}{\log x} + \frac{1}{\log y} + \frac{1}{\log z} \ge \frac{9}{2}$

Thus from the original problem we have:

$(2)(6) \ge (\text{RHS})^2 \implies 2\sqrt3 \ge \text{RHS}$

Cauchy doesn't hold for negative numbers... so if any of $x,y,z$ is less than $10$ which at least one has to be for $xyz = 100$ the first step is false.


We are given set $S=1,2,3,...,36$ and we choose $n$ distinct 3-element subsets, no two of them disjoint such that no element in $S$ belongs to all $n$ chosen subsets. Prove that $n\leq 100$ and determine the number of families of these $n$ subsets when $n=100$


Hi,

I was wondering wether any AoPSers will be attending MathPath this year.

There will definitely be AoPSers at MathPath. A+MATH and nr1337 are going, and so are some others. There's an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22358]earlier thread about MathPath[/url] to check for more AoPSer screen names.


calculus
derivative
calculus computations

Hi Guys,

Please, I need some help with the following equation, any help is VERY highly appreciated. I am lost as to where I should start.

Thanks guys.
[img]http://img149.imageshack.us/img149/9736/problemkt7.jpg[/img]

[hide]
This only looks difficult.

By definition $\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}\cdot \frac{\partial u}{\partial x}+\frac{\partial w}{\partial v}\cdot \frac{\partial v}{\partial x}$

and likewise $\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}\cdot \frac{\partial u}{\partial y}+\frac{\partial w}{\partial v}\cdot \frac{\partial v}{\partial y}$.

Since $u = x+y \text{ and }v = x-y$,

$\frac{\partial u}{\partial x}= 1$

$\frac{\partial v}{\partial x}= 1$

$\frac{\partial u}{\partial y}= 1$

$\frac{\partial v}{\partial y}=-1$

Substituting these into the above equations:

$\frac{\partial w}{\partial x}=\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}$

$\frac{\partial w}{\partial y}=\frac{\partial w}{\partial u}-\frac{\partial w}{\partial v}$.

Applying the FOIL method from elementary algebra:

$\frac{\partial w}{\partial x}\frac{\partial w}{\partial y}= \left(\frac{\partial w}{\partial u}\right)^{2}-\frac{\partial w}{\partial u}\frac{\partial w}{\partial v}+\frac{\partial w}{\partial u}\frac{\partial w}{\partial v}-\left(\frac{\partial w}{\partial v}\right)^{2}$.

Cleaning up:

$\frac{\partial w}{\partial x}\frac{\partial w}{\partial y}= \left(\frac{\partial w}{\partial u}\right)^{2}-\left(\frac{\partial w}{\partial v}\right)^{2}$.

All that remains is to make the substitution $w = f$ on the right hand side.

$\boxed{\frac{\partial w}{\partial x}\frac{\partial w}{\partial y}= \left(\frac{\partial f}{\partial u}\right)^{2}-\left(\frac{\partial f}{\partial v}\right)^{2}}$.

[/hide]


The function $ f$ is defined on the non-negative integers. $ f(2^n \minus{} 1) \equal{} 0$ for $ n \equal{} 0, 1, 2, ...$ . If $ m$ is not of the form $ 2^n \minus{} 1$, then $ f(m) \equal{} f(m \plus{} 1) \plus{} 1$. Show that $ f(n) \plus{} n \equal{} 2^k \minus{} 1$ for some $ k$, and find $ f(2^{1990})$.

It is quite easy . Here is general solution .
$ f(2^n\minus{}k)\equal{}k\minus{}1$ $ \forall k\in N$  such that $ 2^n\minus{}k$ doesn't have the form $ 2^m\minus{}1$


MATHCOUNTS
search
geometry
FTW
number theory
prime factorization
prime numbers

I would also suggest Art Benjamin's book on mental math. (I forget the actual title, I believe it is somewhere along the lines "Secrets of mental math". I saw Art Benjamin show his "Mathemagics" one time, and it was incredible. Try squaring 5 digit numbers in you head. :) )

[quote="ak47"]help please, i need shortcuts for mathcounts help. you know like a list of useful formulas or something like the human calculator. plz respond[/quote]

The prime factorization of $ 1001$ which is $ 7\cdot11\cdot13$

However, as you approach higher level contests, just memorizing formulas is a bad idea because usually you can't just directly apply them. For example this is unlikely: Find $ 1\plus{}2\plus{}3\plus{}4\plus{}\cdots\plus{}n$ in terms of $ n$

If you want a formula sheet you should go to the website

education.ti.com

for help on mathcounts just learn algebra 1 and geometry as well as possible (not by your self but with a professor of mathematics) then you'll be fine.

Also do amc10/12 problems and sit down and work on them for days and increase your ability to do problems and your set.

You should also memorize the prime factorization of the current year and the prime numbers that are near the current year value. This will be useful in the countdown round.

Dude heres a totally crazy off-the-wall idea. Bear with me because its like just so crazy.

Theres still like 6 months left until mathcounts, right? Now heres where the idea gets really insane. Instead of just blindly memorizing formulas, you can use AoPS to your advantage by actually studying! Crazy, right? Sometimes even I surprise myself with my ideas.:P

Take a few tests, find your weaknesses, get the appropriate AoPS book (Intro to CP, Intro to Alg., whatever), and also go through AoPS volume 1.

You might find that you actually...enjoy it... *shudder* :P

Wow what a coincidence I use the same strategy!! :lol:

AoPS actually says NOT TO BLINDLY MEMORIZE formulas.

That's why people get AOPS, to help them study. Math is not memorization, it's insight

Actually I did

lets just say I asked my dad how to and then he said

"how about you do it,it'll make you get used to thinking unlike FTW where all these silly kids just memorize these stupid questions"

but deriving the formula was very tough.

[quote="athunder"]@AwesomeToad: easier said then done....no really...out of curiousity how many people learned how to derive the heron's formula immediately after they found it?[/quote]

You make a good point. However, what Im saying is memorize formulas that are hard to derive (herons, etc.)
but for easier formulas, its best to know how to derive them.

When you are taking a test and the question asks "what is the area of an equilateral triangle with height of 4?" your not gonna go like "oh thats easy, the formula is 1/3xh^2x\sqrt(3) when your given the height." Chances are you will have to actually derive it.

My dad has a Ph.D in math so he says"who cares about mathcounts, in later mathematics mathcounts will help but usually the kids that win rarely ever do well in later mathematics(as in masters and Ph.D)"

as you can see mathecounts is about who can do math the fastest.

PH.D is about who can spend the most time concentrated on a problem and find the shortest proof to your answer.

That is true. But the level math required for Mathcounts doesn't matter for college app. like AMC/AIME/Olympiad does

But Mathcounts is a good foundation for the HS math contests
Your dad probably meant the math itself may not actually be required for real life.
But as you pointed out, the problem solving skill is what comes out of that.


Prove that the length of the Fibonacci sequence $\mod n$ is periodic of length $p\cdot ord_{n}(F_{p-1})$ where p is the smallest positive integer such that $n|F_{p}$.

Hmm.  I've been wondering about this for some time.

Well, the Fibonacci sequence $\bmod n$ will have $F_{p}\equiv 0, F_{p+1}\equiv F_{p-1}, ... F_{p+k}\equiv F_{p-1}F_{k}$.

It follows inductively that

$F_{pd+k}\equiv F_{p-1}^{d}F_{k}$

And therefore that

$F_{p \cdot ord_{n}(F_{p-1})}\equiv F_{0}= 0$
$F_{p \cdot ord_{n}(F_{p-1})+1}\equiv \frac{1}{F_{p-1}}F_{p-1}= 1$


hello,

Prove that : $ \quad\large \forall x \in \left]0,\frac{\pi}{2} \right[ \quad,\quad \frac{1}{\sin^2 x} \;\leq \;\frac{1}{x^2} \plus{} 1 \minus{} \frac{4}{\pi^2}$


algebra
polynomial
limit
integration
LaTeX
function
real analysis

Assume $ f$ is Riemann integrable with respect to $ \alpha$ (i.e. f belongs to $ R(\alpha)$) on $ [a, b]$, and prove that there are polynomials $ P_n$ such that

$ \lim_{n\to\infty} \int_a^b |f \minus{} P_n|^2\,d\alpha \equal{} 0$.

(I'm so sorry I don't know Latex, I would really appreciate it if someone would translate this into code for me. If it helps, this is Rudin's problem 22, Chapter 7.)

Thanks.

Weierstrass' Approximation Theorem allows you to approximate a function $ f$ by a polynomial function and you get as accurate as you want.

It's not that simple. Weierstrass says that you can approximate a continuous function on a closed, bounded interval uniformly by polynomials, but this isn't a continuous function. We also need to approximate an arbitrary integrable function by a continuous function in the $ L^2$ sense.

There are other paths, and Lebesgue integration is more natural than Riemann integration in all of them.


The product of three consecutive integers is 120. That product divided by the mean of the three integers is 24. What is the largest of the three consecutive integers?

$ \frac{120}{\frac{x\plus{}x\plus{}1\plus{}x\plus{}2}{3}}\equal{}24$.  That is, $ \frac{360}{3x\plus{}3}\equal{}24\Rightarrow\frac{120}{x\plus{}1}\equal{}24\Rightarrow 120\equal{}24x\plus{}24\Rightarrow 24x\equal{}96\Rightarrow x\equal{}4$.  So 6 is the largest number.

In my opinion is easier if you call the three integers
$ a \minus{} 1$, $ a$, and $ a \plus{} 1$.

The mean of the three numbers is $ a$,

so

$ \frac {120}{a} \equal{} 24$, which leads $ a$=5.

So, the largest integer is $ a \plus{} 1 \equal{} 6$


Ne radi se baš o ozbiljnoj i konstruktivnoj temi, ali u trenucima odmora i dokolice, lutajući threadovima na MLu počesto naiđem na smiješne threadove... Vjerujem da je to slučaj i sa ostalim posjetiocima.

E pa podijelite linkove s nama, da se i mi nasmijemo!*

Za početak:

http://www.artofproblemsolving.com/viewtopic.php?t=236243

http://www.artofproblemsolving.com/viewtopic.php?t=84559

http://www.artofproblemsolving.com/viewtopic.php?t=253453

http://www.artofproblemsolving.com/viewtopic.php?t=179302

____
*Ovo je bio pokušaj aluzije na staru pedagošku mudrost naših učitelja i nastavnika još od starih vremena da u situacijama kad se dva učenika smiju nečemu u toku časa, glasno podviknu: "Recite šta je smiješno, da se cijeli razred smije!". Uvijek sam to mrzio.

Nisam mogao odoljeti da ne ubacim par svojih favorita, makar ih nisam sve uspio pronaći. Ovo nisu produkti šale ili ironije kao većina tvojih, Harune (treći je zakon!), nego uglavnom neznanja (koje i nije samo po sebi loše), ambicija i egoizma pomiješanih s izvjesnom dozom ludosti.

[url=http://www.mathlinks.ro/viewtopic.php?t=205286]The most important equation ever.[/url] [size=75]Definitivni pobjednik. Dotični se ne ograničava, na sreću, samo na postanje po Mathlinksu, pa ga ima svuda po Internetu. Samo ga potražite na Googleu. Ima i [url=http://donblazys.com/]svoju web stranicu[/url].[/size]

[url=http://www.mathlinks.ro/viewtopic.php?t=42157]INTERESTING FINDINGS[/url] [size=75]Nije loš, a primijetite i comeback nakon dvije godine sa rečenicom "Pretpostavljam da ste uzeli malo vremena da razmislite o mojim tvrdnjama...".[/size]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=202084]Two proofs of Fermat's last theorem[/url] [size=75]Sam topic i nije toliko impresivan, ali autor voli ubacivati rečenice, na primjer, o tome kako ide naučiti što više jezika da može čitati reference koje mu daju i kako se neće baviti ničim drugim od sada. I neke rečenice su mu kao iz nekog akcijskog filma (ili barem neke poslovice), samo primijenjene na matematiku. Npr. "Don't make fun of the boy who teaches himself number theory if he has spreaded in a wrong way. I am anxious enough to keep awake tonight." Ovaj mi je zapravo i simpatičan.[/size]

[quote="mornik"] (treći je zakon!)[/quote]

Još malo našeg genijalca:

http://www.mathlinks.ro/viewtopic.php?t=253763

Još kaže da će nam svaki dan objavljivati nove pojedinosti, te odgovarati na naša pitanja (već je par moderatora, uključujući i mene, zamolilo administratore da ga ne banuju, da se malo zabavimo  :lol:  )

[quote]neznanja (koje i nije samo po sebi loše), ambicija i egoizma pomiješanih s izvjesnom dozom ludosti. [/quote]

Tu je trenutno najaktuellniji slavni FTI  :lol:

http://www.mathlinks.ro/viewtopic.php?t=250886

[quote="hsiljak"]
Još malo našeg genijalca:
http://www.mathlinks.ro/viewtopic.php?t=253763[/quote]

Nee, obrisali su ga...

Vjerovatno jer su u njemu prepoznali starog i iskusnog trola... http://mathforum.org/t2t/discuss/message.taco?thread=1722

Nije s Mathlinksa, ali evo, jedan naš domaći... zanimljiv amalgam matematike i svih ostalih znanosti i filozofsko-religijskih promišljanja: http://www.teskera.com/ . Preporučujem posebno, što se matematike tiče, "Temelji matematike za buduća vremena (17.07.2008.)", ali cijela stranica je zbilja vrijedna čitanja. Čovjek smišlja i viceve!

Tu je i Taoli...

http://www.mathlinks.ro/viewtopic.php?t=266270

http://www.mathlinks.ro/viewtopic.php?t=287035

A i naslov jedne sve kaze :) Nije toliko smesan, ali uporno tvrdi svoje, kao i prethodnici ...

Post #7

http://www.mathlinks.ro/viewtopic.php?p=1592210#1592210


I want to type in nCr (only more fancy than that) on TeXnic center.  Unfortunately, neither the LaTeX tutorial nor the TeXnic center's help guide tells me how to do this.  I've been stuck doing this: \left(^{n}_{r}\right), which leaves me with something rather ugly: $\left(^{n}_{r}\right)$ (it gets even uglier if one of the terms in the binomial is longer than the other $\left(^{n+1}_{r}\right)$).  I've been advised to try the \binom command, which works just fine here $\binom{n}{r}$, but TeXnic center doesn't seem to understand the command!  Am I doing something wrong here?

\binom is part of the AMS-LaTeX package so you need to add \usepackage{amsmath}. In fact, you might like to get the full benefit by using
[code]\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}[/code]

[quote="joml88"]I'm not sure.  I personally use {n\choose r}  for nCr.  It comes out like this:
\[{n\choose r}\]
\[{n+1\choose r}\][/quote]

Thanks!  I'll probably do that in the future.

you all use the brackets denotion? I keep stuck to the good old $C^r_n$.

it seems that Englishmen prefer $^nC_r$, right, eh?

I don't like $C_n^r$ very much because the size of subscript is small...sometimes I don't see it very clearly.  :D

[quote="dts"]I've been advised to try the \binom command, which works just fine here $\binom{n}{r}$, but TeXnic center doesn't seem to understand the command!  Am I doing something wrong here?[/quote]

Stevem is correct in advising you to use the amsmath package.  For the record, TeXnicCenter has nothing to do with building the file - that falls on the compiler, probably MiKTeX - so if a command isn't recognized, it's not TeXnicCenter's fault, but either the user's or the compiler's.


To build a set of book shelves, a carpenter needs 30 identical boards, each 2 feet 7 inches long.  For this type of wood, the lumber yard sells only 12 foot long boards.  How many 12 foot long boards must the carpenter purchase?

A.6   B.7  C 8 D. 9	E. More than 9

[hide] Keeping everything in feet, we have $ \frac{30(\frac{31}{12})}{12}\approx6.458$. Thus, the carpenter needs $ 7$ boards, so the answer is B.[/hide]


How many integer solutions does x/y + (y+1)/x = 4 have ?

[hide="hint"]
Expand, then use modular
[hide="which modular"]
4
[/hide][/hide]


geometry
rectangle
AoPSwiki
probability
number theory
relatively prime

[quote="kyyuanmathcount"]Moderator kyyuanmathcount must delete 7 spam replies in each topic involving number theory, 9 spam replies in algebra topics, and 8 in geometry topics.

He deletes 4 in topics with both number theory and algebra, 3 in geometry and algebra, but didn't count how many he deleted for number theory and geometry.  However, he knows that he deleted one reply in a topic with all three.  If he deleted 15 replies total, how many had number theory and geometry?

[hide="hint"]I like Pie[/hide][/quote]

Oh, I like  Pie too!  What kind of pie do you like? Apple? Cherry? Blueberry? I love apple pie.

[hide]Either way my answer is

$3$, by using Principle of Inclusion-Exclusion. :rotfl:

$3$[/hide]

[quote="mathgeniuse^ln(x)"]
Oh, I like  Pie too!  What kind of pie do you like? Apple? Cherry? Blueberry? I love apple pie.
[/quote]

key lime pie is rather awesome

oh btw answer is 3 using IEP.

Blueberry pie is clearly the best!

[hide="by blueberry pie"]Add up the number with at least one of the requirements to get $7+8+9=24$.

OH, NO, we overcounted, because those with 2 are counted twice, so we must subtract those.  Call number theory and geometry $x$.  $24-4-3-x$

WHAT SHALL WE DO?  We have oversubtracted this time.  The numbers with 3 were subtracted 2 twice, instead of once, so add back in the number with 3.  $17-x+1$

WE DON'T HAVE AN EQUATION!!!  This is the total number of replies, so set that equal to 15 and solve for x.

$18-x=15\Rightarrow\boxed{x=3}$  WOW, THE ANSWER!![/hide]

[hide="the lemon meringue pie says"] i got 3 because of that one thing where you go  7+8+9=24, and then 24-4-3-x, which goes next to 17-x= pie, i mean 15, and then 18-x=pie, i mean 15. so the answer is pie, i mean 3.[/hide]

I agree with Treething, Key Lime pie is the best.

[hide="Answer"]
24-(4+3+x)+1=15
25-(7+x)=15
18-x=15
x=3[/hide]

Our answers are correct, right kyyuanmathcount?

[hide="I like pumpkin pie"] I think we can all agree that $\boxed{3}$ is the answer. [/hide]

[quote="Ubemaya"]What's the inclusion-exclusion principle?[/quote]

I guess thats the most important part of this problem.

Principle of Inclusion-Exclusion, or PIE, is a fancy name for counting things by adding and overcorrecting until we get the right amount.  Usually used with items in multiple sets.

That's the name of it? I was wondering what all the talk about pie was, I knew how to use it, just not the name, thanks for that 10000th User, I just thought it was a method without a name.

[quote="10000th User"]This is ridiculous. Guys, please control yourself. If you like talking about real-life food pies, there is a forum for your digestion (try G&FF). Stop SPAMMING, [u]PLEASE[/u].

Not only is this topic full of ridiculous spams, daermon and Ubemaya asked what is PIE (Principle of Inclusion-Exclusion). To my disappointment, not a single one of you helped your fellow AoPSers. Is this the way you treat others?

[quote="kyyuanmathcount"]Principle of Inclusion-Exclusion, or PIE, is a fancy name for counting things by adding and overcorrecting until we get the right amount. Usually used with items in multiple sets.[/quote]

That's really helpful. I preferred you provided more information.

In set notation,

$N(A\cup B\cup C)=N(A)+N(B)+N(C)-N(A\cap B)-N(B\cap C)-N(C\cap A)+N(A\cap B\cap C)$

where $N(S)$ means the number of elements in set $S$.

If you need further explanation, please let me know.[/quote]

Yes thank you for helping us, 10000th User.  We are sorry for not helping our fellow friends.  Please forgive us.

[quote="life=tennis+math"]Same with me for Ignite, I didn't know it really had a name. As for the hint, I assumed that he was hungry...
Also, what are some other weirdly named theorems which are kind of common but their names aren't?[/quote]

Well I didn't know the largest number you cannot make from relatively prime m and n was mn-m-n was called "Chicken McNugget" until I found AoPS, or from point P in rectangle ABCD $(PA)^2+(PC)^2=(PB)^2+(PD)^2$ or something like that was called British Flag Theorem either. There also is something called Hockey Stick I heard mentioned a few times but all I remember is that it's complex and way above my level.

Hockey stick is actually very simple.  You take out the pascals triangle, and draw a hockey stick on it.  I can't really show you that, but I'll put it in algebra.

$\binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+....+\binom{k}{r} =\binom{k+1}{r+1}$

Sometimes useful.

Well it is useful if you forget the sum of the first n numbers and the first n triangular numbers and combinatorical patterns.

Yeah, I envy those who don't.

In some cases, the problem isn't even a counting problem, or it's hidden, so always be on the lookout ;) .

[hide]From PIE (I like pie), 7+9+8-4-3-x+1=15. Solving for x, x=3.[/hide]

For more info on PIE and the Hockey Stick Identity, see the AoPSWiki: [url]http://artofproblemsolving.com/Wiki/index.php/PIE[/url] and [url]http://artofproblemsolving.com/Wiki/index.php/Combinatorial_Identities[/url].

Similar problem split and put here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99466[/url]

[quote="mathgeniuse^ln(x)"]Thanks for the links.

Is AoPSWiki new?[/quote]Yes, it's part of the new server

I have a question.  Is the formula the same pretty much for four five or n items?

[quote="solafidefarms"]Essentially. If you want to learn more, I highly recommend the Intermediate Couting and Probability class, which has 2 weeks devoted to PIE - usage and generalization.[/quote]

Yeah, Intermediate Counting and Probability was cool.  I thought I knew how to use PIE until I took it  :blush: .


[i][b]Βάζω ένα ακόμα γεωμετρικό πρόβλημα. Δεν έχω τη λύση , αλλά θα ψάξω να τη βρω (ίσως και μόνος). Παλαίψτε την και νικήστε την !!![/b][/i]

Οι παράλληλες από το περίκεντρο $ O$ ενός τριγώνου $ ABC$ προς τις πλευρές  $ BC,CA,AB$  τέμνουν αντίστοιχα τις εφαπτομένες του κύκλου στα $ A,B,C$ στα σημεία $ K,L,M$.( δηλαδή, η εφαπτομένη στο $ A$ και η παράλληλη προς την $ BC$ από το $ O$ τέμνονται στο $ K$ , η εφαπτομένη στο $ B$ και η παράλληλη προς την $ AC$ από το $ O$ τέμνονται στο $ L$ ενώ   η εφαπτομένη στο $ C$ και η παράλληλη προς την $ AB$ από το $ O$ τέμνονται στο $ M$). Να αποδειχθεί ότι τα σημεία $ K,L,M$ είναι συνευθειακά .

 [u]Μπάμπης[/u]

Βάζω μια λύση με μιγαδικούς.

Θεωρούμε ότι ο περιγεγραμμένος κύκλος του τριγώνου είναι ο μοναδιαίος. Τότε $ |a| \equal{} |b| \equal{} |c| \equal{} 1$. Βρίσκουμε τώρα το Κ. Αφού $ AK\perp AO$ θα έχουμε ότι $ \frac{k\minus{}a}{a}\in I$, δηλαδή μετά από πράξεις $ \frac{k}{a}\plus{}a\overline{k}\equal{} 2$ (1). Ακόμη αφού $ KO\parallel BC$ θα είναι $ \frac{k}{b\minus{}c}\in R$, άρα $ \overline{k}\equal{}\minus{}\frac{k}{bc}$ (2).
Από τις σχέσεις (1), (2) παίρνουμε $ k \equal{}\frac{2abc}{bc\minus{}a^{2}}$ και αντίστοιχα βρίσκουμε τους $ l,m$. Τότε $ \frac{k\minus{}l}{l\minus{}m}\equal{}\frac{(a\minus{}b)(ab\minus{}c^{2})}{(b\minus{}c)(bc\minus{}a^{2})}\in R$ και άρα το ζητούμενο αποδείχτηκε.

[b]Moder.[/b]  Μιχάλη, προσθέτω το σχήμα .

[quote="savvas_michael"]Βάζω μια λύση με μιγαδικούς.

Θεωρούμε ότι ο περιγεγραμμένος κύκλος του τριγώνου είναι ο μοναδιαίος. Τότε $ |a| \equal{} |b| \equal{} |c| \equal{} 1$. Βρίσκουμε τώρα το Κ. Αφού $ AK\perp AO$ θα έχουμε ότι $ \frac{k\minus{}a}{a}\in I$, δηλαδή μετά από πράξεις $ \frac{k}{a}\plus{}a\overline{k}\equal{} 2$ (1). Ακόμη αφού $ KO\parallel BC$ θα είναι $ \frac{k}{b\minus{}c}\in R$, άρα $ \overline{k}\equal{}\minus{}\frac{k}{bc}$ (2).
Από τις σχέσεις (1), (2) παίρνουμε $ k \equal{}\frac{2abc}{bc\minus{}a^{2}}$ και αντίστοιχα βρίσκουμε τους $ l,m$. Τότε $ \frac{k\minus{}l}{l\minus{}m}\equal{}\frac{(a\minus{}b)(ab\minus{}c^{2})}{(b\minus{}c)(bc\minus{}a^{2})}\in R$ και άρα το ζητούμενο αποδείχτηκε.[/quote]

Μιχάλη ,
η  λύση αυτή είναι υπέροχη. Θα σε παρακαλέσω μόνο να απλώσεις λίγο τη λύση με τις απαραίτητες θεωρητικές εξηγήσεις ( όσο γίνεται βέβαια) , για να την έχουμε έτοιμη  για το βιβλίο.

 Μπάμπης

[b] ΣΧΟΛΙΟ [/b]

 Η άσκηση ειναι η

 [color=red][b]Problema 2760 din “Матматика в школе”-Rusia, nr.4/1984:[/b][/color]

και όχι από το Kvnat που έγραψα. Sorry.

Ωραία άσκηση .... :)
Λοιπόν νομίζω ότι βρήκα μια συνθετική λύση (dedicated to masters of geom Nick Rapanos and staretak)
Η άσκηση γίνεται σχεδόν προφανής(γιατί ;) μετά το ακόλουθο λήμμα ...
Έστω τρίγωνο ΑBC . An η εφαπτομένη στο Α τέμνει την ΒC sto X , η εφαπτομένη στο B τέμνει την CA sto Y, η εφαπτομένη στο C τέμνει την ΒA sto Z tote ta X,Y,Z είναι συνευθειακά .

Υπόδειξη για την απόδειξη του λήμματος :
Έστω Α' το ίχνος της συμμετροδιάμεσου από το Α στη BC (A' sthn BC) .Tότε το X είναι το συζηγές αρμονικό του Α' ως προς τα Β και C.
Όμοια τα Υ,Ζ είναι τα συζηγή αρμονικά των ιχνών των συμμετροδιαμέσων από τα B,C αντίστοιχα.
Τώρα επειδή οι συμμετροδιάμμεσοι συντρέχουν στο σημείο Lemoine από το θεώρημα Desarques ,τα X,Y,Z είναι συνευθειακά

[quote="silouan"]Ωραία άσκηση .... :)
Λοιπόν νομίζω ότι βρήκα μια συνθετική λύση (dedicated to masters of geom Nick Rapanos and staretak)
Η άσκηση γίνεται σχεδόν προφανής(γιατί ;) μετά το ακόλουθο λήμμα ...
[color=red][b]Έστω τρίγωνο ΑBC . An η εφαπτομένη στο Α τέμνει την ΒC sto X , η εφαπτομένη στο B τέμνει την CA sto Y, η εφαπτομένη στο C τέμνει την ΒA sto Z tote ta X,Y,Z είναι συνευθειακά .[/b][/color]

Υπόδειξη για την απόδειξη του λήμματος :
Έστω Α' το ίχνος της συμμετροδιάμεσου από το Α στη BC (A' sthn BC) .Tότε το X είναι το συζηγές αρμονικό του Α' ως προς τα Β και C.
Όμοια τα Υ,Ζ είναι τα συζηγή αρμονικά των ιχνών των συμμετροδιαμέσων από τα B,C αντίστοιχα.
Τώρα επειδή οι συμμετροδιάμμεσοι συντρέχουν στο σημείο Lemoine από το θεώρημα Desarques ,τα X,Y,Z είναι συνευθειακά[/quote]

[b]Σιλουανέ , [/b]

 αυτό το ενδιαφέρον θεώρημα βασίζεται , εκτός από τη λύση που έδωσες , και στο εξής επίσης βασικό συμπέρασμα :

  Για το σημείο $ X$ ισχύει η σχέση :

                $ \frac{XB}{XC}\equal{}\frac{\gamma^{2}}{\beta^{2}}$.

 Το συμπέρασμα ( η απόδειξη δηλαδή του λήμματος)είναι τώρα εφαρμογή του θεωρήματος Μενελάου στο τρίγωνο $ ABC$ , αν πάρουμε τις αντίστοιχες σχέσεις για τα δύο άλλα σημεία.

 [u]Μπάμπης[/u]

[quote="pontios"]Έστω $ H$ το ορθόκεντρο του τριγώνου. $ AH\perp BC\Rightarrow AH\perp OK$

Αλλά η $ KA$ είναι εφαπτομένη, επομένως η $ AH$ θα είναι η πολική ευθεία του $ K$

Το $ H$ λοιπόν ανήκει στην πολική ευθεία του $ K$, άρα και το $ K$ ανήκει στην πολική ευθεία του $ H$ (Θεώρημα La Hire)

Ομοίως και τα $ L,M$ θα ανήκουν στην πολική του $ H$[/quote]

[b] Λουκά, [/b]

 ψάχνοντας το πρωϊ στο φόρουμ βρήκα την ίδια απόδειξη με πολικές σε ένα παλιό μήνυμα στα Ρουμάνικα ! Η λύση σου αξίζει εύγε :lol:  , διότι είναι σύντομη και φυσική. Είμαι σίγουρος ότι αυτός που έφτιαξε την άσκηση , την έφτιαξε πατώντας στις πολικές και στη λύση σου.
 Θα κοιτάξω και για λύση με αντιστροφή . Έτσι και αλλιώς η αντιστροφή και η πολικές έχουν και κοινά χαρακτηριστικά.

Μπάμπης

Καταρχήν Λουκά εύγε εκπληκτική λύση  :)

[quote="stergiu"]
[color=red][b]Έστω τρίγωνο ΑBC . An η εφαπτομένη στο Α τέμνει την ΒC sto X , η εφαπτομένη στο B τέμνει την CA sto Y, η εφαπτομένη στο C τέμνει την ΒA sto Z tote ta X,Y,Z είναι συνευθειακά .[/b][/color]
[/quote]

Τώρα που το ξανακοιτάω βγαίνει και κατευθείαν από το θεώρημα Pascal για το εκφυλισμένο εξάγωνο AABBCC   :wink:
(όταν θα έχουμε ΑΑ ως τμήμα θα είναι η εφαπτομένη στο Α)

[quote="pontios"]Έστω $ H$ το ορθόκεντρο του τριγώνου. $ AH\perp BC\Rightarrow AH\perp OK$

Αλλά η $ KA$ είναι εφαπτομένη, επομένως η $ AH$ θα είναι η πολική ευθεία του $ K$

Το $ H$ λοιπόν ανήκει στην πολική ευθεία του $ K$, άρα και το $ K$ ανήκει στην πολική ευθεία του $ H$ (Θεώρημα La Hire)

Ομοίως και τα $ L,M$ θα ανήκουν στην πολική του $ H$[/quote]

Λουκά, πολύ όμορφη λύση.

$ \bullet$ Εναλλακτικά, αφού οι πολικές των $ K,$ $ L,$ $ M,$ διέρχονται δια του ορθοκέντρου $ H$ του $ \bigtriangleup ABC,$ συμπεραίνουμε ότι τα σημεία αυτά κείνται επί ευθείας κάθετης επί την $ OH,$ όπου $ O$ είναι το περίκεντρο του $ ABC,$ στο σημείο $ H',$ αρμονικό συζυγές του $ H,$ ως προς τα σημεία τομής του περιγεγεγραμμένου κύκλου $ (O),$ από την ευθεία $ OH$ $ ($ δηλαδή κείνται επί της πολικής του $ H$ ως προς τον κύκλο $ (O)$ $ ).$

Μια στοιχειώδης απόδειξη αυτής της βασικής ιδιότητας από τη θεωρία περί Πολικών, έχει δημοσιευτεί στο αγγλικό φόρουμ.

http://www.mathlinks.ro/Forum/viewtopic.php?t=139564

Κώστας Βήττας.

[quote="KMN"]να η λύση με αντιστροφή . Θεωρούμε αντιστροφή κέντρου Ο και ακτίνας ίσης με του περ κύκλου .έστω  $ K'$ η εικόνα του $ K$ μετά την αντιστροφή . Τότε το $ K'$ είναι το σημείο τομής της ΟΚ με την κάθετη από το Α στην ΟΚ . Όμοια ορίζονται και τα
$ L',\ M'$ . Τότε εύκολα βλέπουμε λόγω καθετότητας ότι το τετράπλευρο $ OL'K'M'$ είναι εγγράψιμο (και το Η (ορ του ΑΒC) ανήκει σε αυτό τον κύκλο) . Άρα αφού  $ OL'K'M'$ οι εικόνες των L,K,M πριν την αντιστροφή είναι συνευθειακά[/quote]

Αυτό ακριβώς είναι !!! Μπράβο Κωνσταντίνε!
 Πάμε τώρα στην επόμενη άσκηση - ακολουθεί νέο μήνυμα.


Is it possible to make commands which manipulate some data structures? For example, queue or stack.

Queue-like structure:
[code]\NEWQUEUE{myQueue}

\QPUSH{myQueue}{x}
\QPUSH{myQueue}{y}

...

\QPOP{myQueue} -> gives x
\QPOP{myQueue} -> gives y
[/code]

See [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/37985212f2dfb9d3/5e03fe6de1be785c]Help: implementation for stack, array, catenation Options[/url] but this is deep stuff.

[quote="stevem"]See [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/37985212f2dfb9d3/5e03fe6de1be785c]Help: implementation for stack, array, catenation Options[/url] [b]but this is deep stuff[/b].[/quote]

That's why I need it. Don't worry about my programming experience. ;)


geometry
parallelogram
geometry unsolved

I' m looking for proof of
$PA + PB + PC \geq  2*(m_a*a + m_b*b + m_c*c)/(a + b + c)$
P-any point in the trangle
m - median

I think I am a proposer of this problem.  ;) I posted this problem a long time ago. The proof is based on Ptolemei inequality.
 Call A' the fourth vertex of the parallelogram ABA'C. Then we apply Ptolemei to quadrilateral BPCA'. :D


modular arithmetic
number theory proposed
number theory

Find all positive integers $ x , y$ such that $ 5^{y} \plus{} 1 \equal{} 2\cdot 3^{x}$ .

use the fact that if $ 3^x | 5^y\plus{}1$ then we have $ 3^{x\minus{}1} | y$
now use inequality...

Claim:$ 5,(mod3^{k})$ is pirimitive root

$ k \equal{} 1$  $ \implies$  $ 5^{d}\equiv 1\pmod3$,$ ord_{3}5 \equal{} 2$

$ k \equal{} n$  $ \implies$ $ ord_{3^{n}}5 \equal{} 2.3^{n \minus{} 1}$

$ k \equal{} n \plus{} 1$  ,$ 5^{d}\equiv 1\pmod{3^{n \plus{} 1}}$  $ \implies$  $ 5^{d}\equiv 1\pmod{3^{n}}$

$ \implies$  $ 2.3^{n \minus{} 1}|d$ ,We know $ d \equal{} 2.3^{n}$ satisfies this equation

$ d \equal{} 2.3^{n \minus{} 1}.k$  $ \implies$ $ 5^{d} \minus{} 1 \equal{} 5^{2k.3^{n \minus{} 1}} \minus{} 1 \equal{} [5^{2k.3^{n \minus{} 2}} \minus{} 1][5^{4k.3^{n \minus{} 2}} \plus{} 5^{2k.3^{n \minus{} 2}} \plus{} 1]$

$ 3\parallel{}5^{4k.3^{n \minus{} 2}} \plus{} 5^{2k.3^{n \minus{} 2}} \plus{} 1$

$ \implies$  $ 3^{n}\parallel{}5^{2k.3^{n \minus{} 2}} \minus{} 1$  $ \implies$  $ 3|k$

$ \implies$  $ d \equal{} 3^{n}.2l$ ,We know $ d \equal{} 2.3^{n}$ satisfies this equation

$ \implies$  $ 3^{x \minus{} 1}|y$

$ y \equal{} 3^{x \minus{} 1}u$ , $ 5^{3^{x \minus{} 1}u} \plus{} 1 \equal{} 2.3^{x}$

if $ 5^{a} > 6a$ Contradiction!$ (a \equal{} 3^{x \minus{} 1})$

$ a\geq 2$  $ \implies$  $ 5^{a} > 6a$

$ \implies$  $ 3^{x \minus{} 1}\leq 1$  $ \implies$  $ 3^{x \minus{} 1} \equal{} 1$

$ \implies$  $ x \equal{} 1$  $ \implies$  $ 2.3^{1} \equal{} 5^{y} \plus{} 1$  $ \implies$  $ y \equal{} 1$  :wink:


group theory
abstract algebra
superior algebra
superior algebra solved

what does $[G: H]$ actually mean ?
and what links $G/H$ with $[G: H]$ ?
can one give some examples here too ?

$[G: H]$ is exactly : the number of cosets, thus the number of disjoint sets of form
$a H$ in a group $G$ with subgroup $H$

you can prove that there is a bijection with the number of right cosets $H a$

when $H$ is a normal subgroup, the quotient $G/H$ will have as number of elements : $[G: H]$

Now note that all this does not assume that the group $G$ or subgroup $H$ is finite

this example might be interesting : $G=\mathbb{R}\backslash\{0\}$ with multiplication
$H$ consists of all the positive real numbers
well both are infinite but $[G: H]$=$2$

but if G and thus also H are finite, then $[G: H]=\frac{|G|}{|H|}$

[quote="fredbel6"]$[G: H]$ is exactly : the number of cosets, thus the number of disjoint sets of form
$a H$ in a group $G$ with subgroup $H$
[/quote]

That is my definition of $[G: H]$ and as far as I know, $G: H$ has no meaning?


The sequence $ a_{1}, a_{2}, a_{3}, ...$ is defined by $ a_{1}= 0$, $ a_{n}= a_{[n/2]}+(-1)^{n(n+1)/2}$. For any positive integer $ k$  find all $ n$ in the range $ 2^{k}\leq n < 2^{k+1}$ such that $ a_{n}= 0$.

[quote="quangpbc"]The sequence $ a_{1}, a_{2}, a_{3}, ...$ is defined by $ a_{1}= 0$, $ a_{n}= a_{[n/2]}+(-1)^{n(n+1)/2}$. Show that for any positive integer $ k$ we can find $ n$ in the range $ 2^{k}\leq n < 2^{k+1}$ such that $ a_{n}= 0$.[/quote]
It is not true.
For example k=1 $ a_{2}=-1;a_{3}=1$
k=3 $ a_{8}=1;a_{9}=-1;a_{10}=-3;a_{11}==-1;a_{12}=1;a_{13}==-1;a_{14}=1;a_{15}=3.$
It is easy to show, that $ a_{n}=k(mod \ 2)$, when $ 2^{k}\le n<2^{k+1}$.
Therefore your statment is true only for even k.

Sorry , this problem is not true  :blush: .I have edited it .Can you slove it ?


Show that if a,b, are elements in R and $ a$ does not equal $ b$, then there exist $ \epsilon$-neighborhoods $ U$ of $ a$ and $ V$ of $ b$ such that $ U$ $ \cap$ $ V$ = $ \emptyset$

Any help would be greatly appreciated.

$ a\neq b \Rightarrow |a \minus{} b| > 0$. Let $ \epsilon \equal{} \frac {|a \minus{} b|}{2}$, now $ U \equal{} (a \minus{} \epsilon,a \plus{} \epsilon )$ and $ V \equal{} (b \minus{} \epsilon,b \plus{} \epsilon )$ are $ \epsilon$-neighborhoods of $ a$ and $ b$ and $ U\cap V \equal{} \emptyset$.

Note that Ognjen09's proof would work with very little change in any metric space. In short, a metric space is Hausdorff.


Solve:

1. Given a point A and two circles K1 and K2, construct a third circle K3 so that K3 passes through A and is tangent to K1 and K2.

2. Given two points A and B and circle K1, construct another circle K2 so that K2 passes through A and is tangent to K1.

Using inversion.....


In $ \triangle ABC$ with $ BC\equal{}a,CA\equal{}b,$ and $ AB\equal{}c$, given that $ 9a^2\plus{}9b^2\minus{}17c^2\equal{}0$, find the value of $ \frac{\cot A\plus{}\cot B}{\cot C}.$

$ \cot A\plus{}\cot B\equal{}\frac{\sin(A\plus{}B)}{\sin A\sin B}\equal{}\frac{\sin C}{\sin A\sin B}$

Then $ \frac{\cot A\plus{}\cot B}{\cot C}\equal{}\frac{\sin^2C}{\sin A\sin B\cos C}$

We have $ \sin A\equal{}\frac{a}{2R}, \ \sin B\equal{}\frac{b}{2R}, \ \sin C\equal{}\frac{c}{2R}, \ \cos C\equal{}\frac{a^2\plus{}b^2\minus{}c^2}{2ab}$

Then $ \frac{\sin^2C}{\sin A\sin B\cos C}\equal{}\frac{2c^2}{a^2\plus{}b^2\minus{}c^2}\equal{}\frac{2c^2}{\frac{17c^2}{9}\minus{}c^2}\equal{}\frac{18c^2}{8c^2}\equal{}\frac{9}{4}$


algebra
polynomial
inequalities
logarithms
least common multiple
number theory proposed
number theory

Here is a very very beautiful problem
   There exists an absolute constant $A>0$ such that for any non-constant polynomial $f$ with integer coefficients, the least common multiple of the orders of those roots of unity that are also roots of $f$ is at most $e^{A\cdot\sqrt{nlnn}}$.  I think this one is very hard (but it is not open!! :D ). An easier, but still nice question would be to prove that the above number is at most $e^{\frac{4n}{e}}$.

[hide="Hint"] Each root of unity of order exactly $k$ is a root of the cyclotomic polynomial $\Phi_k$ of degree $\varphi(k)$. Since the cyclotomic polynomials are irreducible over $\mathbb Z$, it suffices to prove that for any integers $n_1,\dots, n_m$, the inequality $\varphi(n_1)+\dots+\varphi(n_m)\leqslant n$ implies that the least common multiple of $n_1,\dots, n_m$ does not exceed $e^{A\sqrt{n\ln n}}$.[/hide]

Yes, that's the idea, but I do not think that proving the last assertion is an easy thing, is it?  :?

It is less or more easy using primes distribution (it suffices to use that $\pi(n)\le c\frac{n}{\ln n}$). The last estimate may be proved respectively easy: it follows from a fact that $\prod_{p<n} p <4^n$, which follows from $\prod_{n<p\le 2n} p | C_{2n}^n<4^n$.


Let $ n>1$ be a given integer and $ A$ be an infinite set of positive integers satisfying: for any prime $ p\not|n$, there exist infinitely many elements of $ A$ not divisible by $ p$. Prove that for any any $ m>1$, $ (m,n)\equal{}1$, there exists a finite subset of $ A$ whose sum of elements, say $ S$, satisfies $ S\equiv1\pmod{m}$ and $ S\equiv0\pmod{n}$.

nobody interested? this is not so hard..


function
inequalities
number theory unsolved
number theory

Find all solutions to the equation $ x^y\plus{}y\equal{}y^x\plus{}x$ in positive integers.

Function $ f(x,y)\equal{}y^x\plus{}x$ is strictly increesing. That is why equation $ f(x,y)\equal{}f(y,x)$ can have $ \leq 1$ solutions. But $ y\equal{}x$ is a solution. So, this will be the answer.

[quote="greenvert"][quote="Yuriy Solovyov"]Function $ f(x,y) \equal{} y^x \plus{} x$ is strictly increesing. That is why equation $ f(x,y) \equal{} f(y,x)$ can have $ \leq 1$ solutions. But $ y \equal{} x$ is a solution. So, this will be the answer.[/quote]
Wrong;$ 2^3 \plus{} 3 \equal{} 3^2 \plus{} 2$[/quote]

[color=red]Yuriy Solovyov[/color] said that the equation may have $ \leq 1$ solution.So it has a solution $ (x,y)\equal{}(2,3)$ and it is the solution.

For $ x\equal{}y$ the equation becomes a identity which has infinitely many solutions. (i.e. you may put any value in that identity and get the same value for $ L.H.S$ and $ R.H.S$)

[quote="greenvert"]If I misunderstood Yuriy Solovyov words, then I'm very sorry.[/quote]

Actually his/her solution was incomplete.You have rather completed it.So, I think that you needn't be sorry.
(And most importantly, we all do mistakes. To err is human  :) )

Obviously $ x \equal{} y$ is a solution.Now wlog assume that $ x \equal{} y \plus{} k$,where $ k\in\mathbb{N}$.
Hence our equation becomes:
$ (y \plus{} k)^{y} \minus{} y^{y \plus{} k} \equal{} k$.
We will prove that for all $ y\geq 3$:
$ (y \plus{} k)^{y}\leq y^{y \plus{} k}$.
The last inequality is equivalent to the following one:
$ \sqrt [y \plus{} k]{y \plus{} k}\leq \sqrt [y]{y}$.To prove it we must consider a function $ f(x) \equal{} \sqrt [x]{x}$.It reaches max. value at point $ x \equal{} e$,hence as it could be easily proved for all $ x\geq 3$,function $ f$ is strictly decreasing.Hence the inequality is proved.
If $ y \equal{} 1$ then $ x\in\mathbb{N}$ is a solution too.
If $ y \equal{} 2$ then $ \sqrt [k \plus{} 2]{k \plus{} 2}\leq\sqrt {2}$.Which is true for all $ k\geq 2$.Hence $ k \equal{} 1$.
But then $ x \equal{} 3$.So finally our solutions are:
$ 1.x \equal{} y\in\mathbb{N}$
$ 2.(x \equal{} 1,y\in\mathbb{N})$,$ (x\in\mathbb{N},y \equal{} 1)$.
$ 3.(x \equal{} 3,y \equal{} 2)$,$ (x \equal{} 2,y \equal{} 3)$.

[quote="Erken"]So finally our solutions are:
$ 1.x \equal{} y\in\mathbb{N}$
$ 2.(x \equal{} 1,y\in\mathbb{N})$,$ (x\in\mathbb{N},y \equal{} 1)$.
$ 3.(x \equal{} 3,y \equal{} 2)$,$ (x \equal{} 2,y \equal{} 3)$.[/quote]
And obviously, $ x \equal{} 1$ or $ y \equal{} 1$ makes it another identity.

sorry... i wrote for the equations of the type $ a^x \plus{} x \equal{} a^y \plus{} y$, but our is another type. This work for functions of one variable, but not for two. Maybe we should change it a little and will be ok.


linear algebra
matrix
function
algebra
polynomial

Let $ A$ and $ B$ be $ n$x$ n$ matrices which are invertible. Prove that there exist a constant $ c$ such that $ A+cB$ is not invertible.

A counterexample:
Let $ A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, $ B=\begin{bmatrix}0&1\\1&1\end{bmatrix}$, both over the field with two elements. Neither of the two possible choices for $ c$ make $ A+cB$ singular.

Over $ \mathbb{R}$, we could use $ A=I,B=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$.

You need another hypothesis; this requires an algebraically closed field.

ok here is the exact wording from the book

Let $ A$,$ B\in M_{nxn}(\mathbb{C})$
Prove that is $ B$ is invertible, then there exist a scalar $ c\in \mathbb{C}$ such that $ A+cB$ is not invertible.

A hint: What is the determinant as a function of $ c$?

The required $ c$ will simply be the negative of any eigenvalue of $ AB^{-1}.$ If $ AB^{-1}+cI$ is not invertible, then neither is $ A+cB.$

The point jmerry was making about an algebraically closed field (that is, a field in which the Fundamental Theorem of Algebra is true) was so that eigenvalues are guaranteed to exist.

i dont see how you could express $ det(A+cB)$ as a function of $ c$.

Oh, you can. The key: it's a [i]polynomial[/i] function of $ c,$ an $ n$-th degree polynomial. That's why both jmerry and I are stressing the idea of an algebraically closed field (the Fundamental Theorem of Algebra). In order to make the claim of this problem, you have to know that polynomials have roots. It looks like jmerry is content with the statement that it's a polynomial function of $ c,$ and indeed that suffices for the result. I go one step further and identify the polynomial in question as a constant multiple of the characteristic polynomial of $ AB^{-1}$ (with a sign change: $ -c$ instead of $ c.$)


Mulitply both sides by 0 to get 0=0.  Since this equation is true, 1+1 = 1 is also true.

Or, you could replace 0 with infinity.[/hide]

interesting answer

my solution

one drop of water + one drop of water = one drop of water

don't criticize my answer or me for saying these

because I did it just for fun ^>^
happy new year

one piece of cloth stitched on with another piece of cloth equals to one piece of cloth, :)

There is an old old joke  . . .

A farmer has 2 haystacks in one field and 3 haystacks in another.
If he puts them together, how many haystacks does he have?

5 :) . They magically stay separated because all of the hay piles are just hay glued to a magnet.

1 + 1 = 1
2 = 1

And we all know that's true.

You'd have to arrange them so that the magnets don't attract but repel.
But more reasonably, if the farmer uses machinery to make the haypiles, then they'll be neatly bundled up so that they stay seperate.

Or the old and boring solution    x=y      x^2 = xy       x^2 - y^2 = xy - y^2
(x-y)(x+y)=y(x-y)      (x+y)=y     x = 2y          y = 2y      1 = 2     1 = 1 + 1

On a little serious note  .. According to Einstein.  this is how "speeds" should be added ...

have you  ever  heard speed expressed as 'Mach'.. eg  Mach 2 means twice the speed of sound etc.. Well many Physicists  express  velocity as  unit-less quatntity as a fraction of the speed of light. (Being lazy, they  select a unit where c=velocity_of_light=1)
In this unit,   A speed of, say  a, fast runner is expressed as  (0.00000003) . (vs , say 10  meter/second or 22 miles/hour).

Now suppose, if there a particle going in one direction with the veloity x, and another particle going in opposit  direction with the velocity y and we define "+" (as in x +y) being  the relative velocity between the  two.

With this definition, if x=.00000003, and y=.00000003, x+y indeed is .00000006

But when x=1, and y=1 x"+" Y  is equal to 1 too!.

(Here "+" accoding to  Einstien is given by :

x "+" y   =    (Is defined as ) =  (x + y )/ sqrt( 1 + xy)

(On the RHS, + is our oridnary +)
and thus for small x and y the "+" is almost same as ordinary + but for speeds close to velocity of light ( or 1)  "+" behaves as 1"+"1 = 1

P.S for those who have read Veda's the formula also translates into:
"ultimate truth"+ ultimate truth" = "ultimate truth"

(Where  a number means "truth_ness" and 1 means  "ultimate truth".. etc...[/b]

Gyan, your answer is unbelivebly beautiful. The best solution I ever heard in my entire life!!

I really loved the way Einstein described addition

Cool

[quote="shinwoo"][size=167][color=red]Cool[/color][/size][/quote]

If any one of the admins is listening, one of my wishes is that we could be lazy and combine parameters like shinwoo tried to do.

That's nice....

I don't really get what you mean, but.....

to RC-7th

[code][size=20][color=red]Cool[/color][/size][/code]
That's the code I used. I used separate tags for size and color, while you put them in the same tag.

oh okay  I just did in the way I do

well, I thought we were supposed to put "font" and all those stuff..

we could all be lazy^.^


Ah, you already reposted it, I'm sorry :)

You know, there's an "edit" button on posts, you can correct your problem that way :)

Anyway here goes my solution:

We know that $a^{669}+c^{669} \geq a^nc^{669-n} + a^mc^{669-m}$ with $m,n \in \left[ 0,669 \right]$. (order inequality)
Newton's binomium tells us that $(a+c)^{669}$ contains $2^{669}$ such terms.
Applying this to each of the 3 separate things, we get that the LHS can be at most $(2^{668})^3$, which is $2^{2004}$.


If you are given the height and the width of a rectangle and have a point in the exact center of the rectangle and draw a line from the center to the width or height of the rectangle, what is the length of the line? I need a formula to calculate the length of this line for each the length and height.  Thanks in advance.[/img]


floor function
combinatorics proposed
combinatorics

The member of a committee were choosing a president, each member gave a vote to one of 27 candidate. For each candidate, the exact percentage of votes the candidate got was smaller by at least 1 than the number of votes for that candidate. What is the smallest possible number of members of the committee.

Answer: 134

Let the number of members be $s$. We will prove $s\ge134$. There exists a candidate who got $k$ votes where $k\le\lfloor\frac{s}{27}\rfloor$. Hence we have $\frac{k}{s}\cdot100\le k-1$. Obviously $k>1$, and note that $s\ge\frac{100k}{k-1}>100$. We have $k\ge\frac{s}{s-100}$, so $\lfloor\frac{s}{27}\rfloor\ge\frac{s}{s-100}$. Let $s=27a+b$ where $0\le b<27$. So $a\ge\frac{27a+b}{27a+b-100}$, $b\ge\frac{(127-27a)a}{a-1}$. Since $s>100$, then $a\ge3$. For $a\ge5$ we have $s\ge135>134$. If $a=3$ then $b\ge69>27$, a contradiction. Lastly, assume $a=4$, so $b\ge\frac{76}3$, $b\ge26$. Therefore $s\ge 27\cdot4+26=134$.

The case is possible for 134 members with number of votes 4,5,5,5,...,5. Since we have $\frac{4}{134}\cdot100<3$ and $\frac{5}{134}\cdot100<4$, then our proof is complete.


Given that f is a continuous, nondecreasing function, defined on$ [0,1]$, with$ f(0)=0$ and $f(1)=1$, prove:

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

Perhaps the condition needs to be 'increasing' rather than 'non-decreasing' - otherwise, how can you define $f^{-1}(x)$?  For instance, what if $f(x)=\left\{\begin{array}{ccc}2x & \mbox{for} & x<\frac{1}{4}  \\ \frac{1}{2} & \mbox{for} & \frac{1}{4}\le x\le\frac{3}{4} \\ 2x-1 & \mbox{for} & \frac{3}{4}<x \end{array} \right.$?  What value would you take for $f^{-1}\left(\frac{1}{2}\right)$?

The function injective.
There is no need to mention it ,when we use $f^{-1}$.
As for you question [quote]what will be the $f^{-1}(\frac{1}{2})$?[/quote]
$f$ is not injective so $f^{-1}$ is not defined.

The original problem is correct .
The function is Nondecreaing.


This one seems to be preety trivial, but I cant seem to find elegant solution.

x^2 + y^2= 25

x + y + xy=19

I've turned 2nd equation into (x+1)(y+1)=20, and with discussion solution is trivial, but I dislike that solution.  :D

I tryed to solve it with geometry, but couldnt do anything better.

Help ^^

For this problem, I'd recommend changin the equations around a bit and adding them. For the first one we can complete the square by $ x^2 \plus{} y^2 \plus{} 2xy \equal{} 25 \plus{} 2xy \implies (x \plus{} 2)^2 \minus{} 2xy \equal{} 25.$ Multiplying the second equation by two gives us $ 2x \plus{} 2y \plus{} 2xy \equal{} 38,$ adding them we get $ : (x \plus{} y)^2 \plus{} 2x \plus{} 2y \equal{} 63 \implies (x \plus{} y)^2 \plus{} 2(x \plus{} y) \equal{} 63 \implies (x \plus{} y \plus{} 2)(x \plus{} y) \equal{} 63.$ So the factors are $ 9$ and $ 7,$ and you can do the rest :)

Notice that $ x^2 \plus{} y^2 \equal{} (x \plus{} y)^2 \minus{} 2xy$. Let $ a \equal{} x \plus{} y$ and $ b \equal{} xy$. Solve the system

$ a^2 \minus{} 2b \equal{} 25 \\
a \plus{} b \equal{} 19.$

Then, find the roots of the quadratic $ x^2 \minus{} ax \plus{} b$ for each ordered pair $ (a,b)$ that satisfied the previous system and that will give you the ordered pair(s) $ (x,y)$ that satisfies the original system.

@limac

Tomekk's problems is not to find integer solutions ,you can not use factors of $ 63$ ,you have to substitue $ x\plus{}y\equal{}t$ and then to solve quadratic equation $ (t\plus{}2)t\equal{}63$

hello, solving the second equation for $ y$ we get $ y\equal{}\minus{}\frac{x\minus{}19}{1\plus{}x}$, plugging this in the first equation we have
$ \frac{2x^2\plus{}2x^3\plus{}x^4\minus{}38x\plus{}361}{(x\plus{}1)^2}\equal{}25$
simplifying this and factoring we get
$ \frac{(x\minus{}3)(x\minus{}4)(x^2\plus{}9x\plus{}28)}{(x\plus{}1)^2}\equal{}0$
from here you will get all solutions.
Sonnhard.


Find all continous functions $F,G,H_1,H_2$ from the real numbers to the same set which fulfill
$F(x+y)=G(x)+H_1(x)H_2(y)$ for all $x,y \in\mathbb{R}$

There are 4 families of solutions :
$F(x) = a$
$G(x) = a$
$H_{1}(x) = 0$
$H_{2}(x) = \text{arbitrary}$

$F(x) = a$
$G(x) = a-b H_{1}(x)$
$H_{1}(x) = \text{arbitrary}$
$H_{2}(x) = b$

$F(x) = aux+2av+b$
$G(x) = aux+av+b$
$H_{1}(x) = a$
$H_{2}(x) = ux+v$

$F(x) = a \exp(bx)+c$ with non-zero $a$ and $d$
$G(x) = c-ud \exp(bx)$
$H_{1}(x) = d \exp(bx)$
$H_{2}(x) = \frac{a}{d}\exp(bx)+u$

However, I won't post my proof since it is pretty long and boring (and anyway not so hard to find).


I decided to start a Pokemon discussion thread.

Does anyone know how to catch a Miltank on Pearl.

At serebii.net it says I need to use a "Poke Radar".  I think I have it, but how do I use it exactly to catch a Miltank?

route 209 i think

u have nat. dex right

Man... I used to be hooked. Like, a while back I had a Pokemon Diamond, and like I caught practically every pokemon... I beat the elite 10 times.... and I beat the whole thing, I caught all the legendaries..  then I lost it. Did I mention that my memory's worse than my sister's? (whose 6). Then I got Platinum. Then I still have to beat er... the Canalave Gym I think?

reminds me of my old gameboy color....

*sighs*

Hey does anyone know any good quote from Pokemon (like the TV series)? I need them for debate.

I have a quote from Pokemon the First movie;
It's Jesse when she speaks in a french accent to like.... trick Ash and the gang I think?
"Don't vorry. This ship von't let you down. Ve alvays sail in da rough veather. Don't get upset. "

"I need pants" - Slowking from The Power of One
xD?

and if u want a non-weird one,

inscription on Amity Square-"When every life meets another life, something will be born"

=)
its my all time favortie pokemon quote

I have a National Pokedex, but I used an action replay :P

I don't have ranger and I don't know how to use the GTS, and none of my friends have a Manaphy, so I caved in................

But I do have a lot of awesome Pokemon in Pearl:

Palkia, Dialga, Mesprit, Uxie, that other one where I forgot the name

Plus, I traded stuff from Kanto, so I have a nice diverse little collection of stuff.

And the Miltank seems IMPOSSIBLE to find.  I looked everywhere around Solaceon Town and I clicked on the Poke Radar again and again.............

Pokemon I use: Spiritomb Lv.100
Ability: Wonder Guard
Moves: Aura Sphere, Dark Pulse, Shadow Ball, Draco Meteor
Pokerus

I had Diamond, and my friend has this cartridge where I had Unlimited Rare Candies... it took a while but my most powerful pokemon were level 100;
I think I remember my team...
Empoleon - 100
Staraptor - 100
Roserade - 100
Dialga - 100
Vaporeon - 100
I don't remember my last one... T.T

So yeah that's my 'old' team, on my new one I have er... a Torterra? And a whole bunch of common pokemon. :P
[img]http://images.icanhascheezburger.com/completestore/2008/3/4/wannaseemaip128491323165308193.jpg[/img]

I only trained 1 Pokemon(prior to my action replay, now I have 2 Lv. 100s, and 14 Pokemon Lv. 65 and Higher), my Infernape, which was Lv. 85 when I first tried against the Elite Four.  I got killed because the PP levels of the awesome moves got low and I didn't have any Ethers.  I eventually beat the Elite Four, but if Ether's were available at the Poke Mart, I'd win easily on the first try.

[quote="EggyLv.999"][quote="Scamper"]Hi, I have Leaf Green Version. I am just about to battle the Elite Four. Is this a good team? I don't have the wireless adapter so I can't trade by the way.

Level 47 Golbat - Later evolved to Crobat once I get the National Pokedex.
Level 50 Raichu
Level 47 Lapras
Level 47 Charizard
Level 43 Hitmonchan
I plan to use this place for Espeon once I get the National Pokedex.

This team kinda sucks and I have a lot of training to do. I choose Mount Ember as my training spot right next to the spa. Please give me some suggestions.[/quote]

not even close- try 1 lv.80, 1 lv. 75 backup, 4 of whatever.[/quote]

I don't think you have to be that good..... In my silver, I defeated the Elite Four with all 40-50s, and one 63~. Now I only use one pokemon (Lvl. 80~), and can pwn the Elite Four, without potions, or elixers, and so on....

Do you guys have rescue team?(i played on my friends DS)
It's a much betta version of pokemon, im my opinion.

Rescue team? Oh yeah. I have the old version (blue something?) ... I think I'm a err.... Pikachu or a err... Torchic?

My friend let me check out his salamance; it was BEASTLY

Does anyone know the cloning trick in Pokemon Emerald?
My friend has like 59 Raquazas.

OH yeah.  You just have to use a Ditto(I got mine in FireRed) and put it with a Pokemon and viola, you get an egg of that Pokemon.

Unfortunately, it doesn't work for most legendaries and a couple others like Jigglypuff and for some(such as Snorlax) you'll get an identical copy of it being in Lv. 1 rather than it being before evolution.  And I want a Munchlax, so that's kinda sad..........

Oh, I want a Munchlax too...
I'm trying to find one of the old versions so I can transfer that to Platinum etc. etc...

No, not the egg. I mean the cloning trick.
You get the pokemon you want in the 1st 2nd 3rd 4th and 5th slot then go to the Battle Tower in Battle Frontier.
Then go to the PC and deposit all the pokemon that you want to clone (1,2,3,4,5) inside an empty box.(not the 6th because the PC won't let you.)
Then turn the PC off then get back all the pokemon you want put inside your party.
Then go to the lady that makes a game the one on the very left.
Enter a game.
Then when the power is saving "Saving... please don't turn off the power" just turn the game off.
Go back to the game then you will have all the pokemon you want in your party and inside the PC.

I might have left something or two out so don't blame me if it doesn't work.

I'm a Piplup with a Torchic as my partner (in Darkness). You can go ahead and guess my partner's name. :D

Awww...
That's sweet.
Oh right..
I have to guess (even though like I know)
Izzy?
 :)
On mine my partner's named Pookie.
I wasn't very creative... :P

[quote="meewhee009"]
On mine my partner's named Pookie.
I wasn't very creative... :P[/quote]

Are you kidding?? I'm even less creative...  :lol:

My partner's name is my name..... Except it gets confusing sometimes...  :)

I'm waiting for the English releases of Heart Gold and Soul Silver to come out. I'm excited about them, considering that Gold and Silver were my favorite games.

[quote="mz94"]inscription on Amity Square-"When every life meets another life, something will be born"[/quote]

It's in the Solaceon Ruins and the quote is "Friendship - All lives touch other lives to create something anew and alive."

[quote="cygao15"]my friends, so his name is alan right?
so he accidently typed a and then hit enter by accident. lolz
[/quote]

alan?

I have a friend named alan who does aops!

here is a pokemon secret. if you want to get Celebi on Diamond, Pearl, or Platinum, go to the basment of solaceon ruins, and don't pick up the four items. once you have your national pokedex, go back and you will find Celebi.

[color=red][size=84]Please don't revive. Thank you. [/size][/color]


function
limit
real analysis
real analysis unsolved

Consider the functions $ f,g : \mathbb{R} \to \mathbb{R}$. If $ g$ is continuous and $ \lim_{x \to \infty} g(x)\equal{} \plus{} \infty$ and $ \lim_{x \to \infty} {f (g(x))} \equal{}l$, prove that $ \lim_{x \to \infty} f(x)\equal{}l$.


$\text{Compute the value of }x\text{ when }f(x)\text{ is at its maximum and }f(x)=x^{\frac1x}\text{.}$
It's easy to do with a graphing calculator (if you know how to use it), but I want to know if it's solvable without one.

Oh, it's an easy calculus problem... I kinda had a feeling when I saw that $x=e$.

Out of curiosity, how would you do it?

By looking at it, i'd say use logarithmic differentiation (as one would to find the derivative of $x^x$) and set the derivative equal to zero.

This probably isn't the right forum for this type of problem, but here it goes.

$f(x) = x^{1/x}$
$\ln f(x) = \frac1x \ln{x}$
$\frac{f'(x)}{f(x)} = \frac1{x^2} - \frac{\ln x}{x^2}$
$f'(x) = \frac{x^{1/x}}{x^2}(1 - \ln x)$

We are looking for $x$ such that $f'(x)=0$.  It's easy to check that this is satisfied if and only if $x = e$.

I think this is the wrong place for a problem like this, now that I see how you do it, but I was wondering whether it could be done without calculus.  I guess not.

Actually it kind of can be solved wihtout calculus. Here's how I did it
[hide]

(x)^(x+a)   (x+a)^(x)
divide both sides by x^x
x^a            (1+a/x)^x
x^a            e^a
the left side is greater if x>e and the right side is greater if x<e
therefore x^1/x is greater than (x+a)^1/(x+a) if x=e and a is not 0.
therefore x=e is the maximum.

[/hide]

Okay, but how did you get $a$ involved in this?

if x is a maximum than f(x) should be greater than f(x+a) for all nonzero values of a.

[quote="0714446459923"]

(x)^(x+a)   (x+a)^(x)
divide both sides by x^x
x^a            (1+a/x)^x
x^a            e^a
the left side is greater if x>e and the right side is greater if x<e
[/quote]
how did you get from x^a and (1+a/x)^x to x^a    and e^a
(1+a/x)^x is only e^a iff x is infinite, which it is clearly not if x is e.

(1+a/x)^x < e^a for all positive real a
that only proves that the maximum is at most e though.

however if you plug x=e into that then you can see that it is the maximum
e^a  >   (1+a/e)^e

[quote="0714446459923"]
however if you plug x=e into that then you can see that it is the maximum
e^a  >   (1+a/e)^e[/quote]
so is there a cool non-calculus proof of it like you mentioned? (and i dont see.... why would that be so, without a calculator?)

[quote="0714446459923"]because (1+a/x)^x < e^a for any real x not equal to 0[/quote]
i know. This only proves that the value of x cannot be greater than e, but you never proved that x could not be less than e


MATHCOUNTS calculus geometry circumcircle perimeter

This problem is probably bad because I made it up but whatever. I'll try to come up with a better problem soon.

EDIT- this problem is really bad. I made a new one (see below)

Hmm... I'll try thinking about it... but could you only post around AMC 10/12 problems, because that's what I need to focus most on (besides Mathcounts) now.

I actually found a mistake in this... it's still doable, but as far as I know it requires calculus (it required calc in the first place, but now it needs more than it used to). There still may be a non-calc solution... I'm looking for it

Edit: non-calculus solution found (well you may have to use a LITTLE... told you this was a bad problem)

Edit 2: Okay just scrap this problem it's really bad

I have a better one (but still probably bad):

In $ \triangle ABC$ $ \angle A\equal{}60$. H and O are the orthocenter and circumcenter respectively. $ OH$ intersects $ AB$ at D and $ AC$ at E. Given that $ DE\equal{}7$, find the perimeter of $ \triangle ADE$


calculus
integration
number theory unsolved
number theory

Prove that the expressions $2x+3y$ and $9x+5y$ are divisible by 17 for the same set of integral values of $x$ and $y$.


hey guys! i need help in my homework....basically its solving simultaneous equation using Substitution or Elimination methods. Heres the questions:

1. Mr. Brown travels 30 kilometers in 3 hours. Part of the time he travels at 5km/h anf the rest of the time at 11km/h. How long does Mr. Brown travel at each speed?

2. A number is formed by two digits whose sum is 11. If the digits are reversed, the number is increased by 27. Find the number.

3. Find the fraction which reduces to 2/3 when one is added to both the numerator and the denaminator.

I really appreciate if you help me solve it guys!! Thanks

Do NOT and NEVER double post.

Locked.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72264


let $a, b, c$are measurements of dimensions of a triangle and checking :
$a+b+c=2$ prove that$a^{2}+b^{2}+c^{2}+2abc\prec2$

Putting $a = x+y, b = y+z, c = z+x$ gives $x+y+z = 1$. Subtracting $2(x+y+z)^{2}$ from LHS and 2 from RHS, we find we have to prove

$2\sum_{sym}x^{2}y+4xyz \le \sum_{sym}xy$

Multiplying LHS by 1 and RHS by $(x+y+z)$ gives $2xyz \ge 0$, which is clear. :D

with $a\le b+c$ then $a,b,c \le 1$
so $(1-a)(1-b)(1-c) \ge 0$
you will find it is equivalent to your inequality.

But I don't think this is the proper subforum you can post this problem :)


inequalities
Euler
inequalities proposed

Prove that

$ \frac{1}{r_ah_a}\plus{}\frac{1}{r_bh_b}\plus{}\frac{1}{r_ch_c}\geq\frac{1}{r(2R\minus{}r)}$

[quote="Ligouras"]Prove that

$ \frac {1}{r_ah_a} \plus{} \frac {1}{r_bh_b} \plus{} \frac {1}{r_ch_c}\geq\frac {1}{r(2R \minus{} r)}$[/quote]

we can prove that

$ \sum{\frac{1}{r_ah_a}}\equal{}\frac{4R\plus{}r}{{s}^{2}r},$

so

$ \frac {1}{r_ah_a} \plus{} \frac {1}{r_bh_b} \plus{} \frac {1}{r_ch_c}\geq\frac {1}{r(2R \minus{} r)}$

$ \Leftrightarrow \frac{4R\plus{}r}{{s}^{2}} \ge \frac {1}{2R \minus{} r}$

$ \Leftrightarrow \frac{8{R}^{2}\minus{}2rR\minus{}{r}^{2}\minus{}{s}^{2}}{{s}^{2}\left(2R\minus{}r\right)} \ge 0,$

by Gerretsen Inequality $ 4R^2\plus{}4Rr\plus{}3r^2\ge s^2$ and Euler Inequality $ R\ge 2r$,

we have

$ 8{R}^{2}\minus{}2rR\minus{}{r}^{2}\minus{}{s}^{2} \ge 8{R}^{2}\minus{}2rR\minus{}{r}^{2}\minus{}(4R^2\plus{}4Rr\plus{}3r^2)\equal{}2(r\plus{}2R)(R\minus{}2r)\ge0,$

so ineq holds.


geometry
3D geometry
HMMT
Harvard
college
MIT
algebra proposed

This was the HMMT (Harvard-MIT) #10 question from the 1998 Algebra Event.

"G. H. Hardy once went to visit Srinivasa Ramanujan in the hospital, and he started the conversation with: “I came here in taxi-cab number 1729. That number seems dull to me, which I hope isn’t a bad omen.” “Nonsense,” said Ramanujan. “The number isn’t dull at all. It’s quite interesting. It’s the smallest number that can be expressed as the sum of two cubes in two different ways.” Ramujan had immediately seen that 1729=12^3+1^3=10^3+9^3. What is the smallest positive integer representable as the sum of the cubes of three positive integers in two different ways?"

The answer is 251 (1^3+5^3+5^3 and 2^3+3^3+6^3) if that helps, and I know that brute force works, but I'm wondering if there is any other better way.

Thanks

hello, see also here
http://mathworld.wolfram.com/TaxicabNumber.html
Sonnhard.

That doesn't explain how to get the taxicab numbers without guess and check and it's only for the sum of 2 cubes, not 3 cubes :)

but thanks anyways.


Could you help me to find the determinant of this matrix :?: :

$ \left [\begin{array}{ccccc}1 & x & x^{2}& x^{3}& x^{4}\\ \noalign{\medskip}1 & 2x & 3x^{2}& 4x^{3}& 5x^{4}\\ \noalign{\medskip}1 & 4x & 9x^{2}& 16x^{3}& 25x^{4}\\ \noalign{\medskip}1 & y & y^{2}& y^{3}& y^{4}\\ \noalign{\medskip}1 & 2y & 3y^{2}& 4y^{3}& 5y^{4}\end{array}\right ]$

If you mean, "in a smart way," my personal answer is "no."  If you mean, "using Mathematica," my answer is $ 2x^{9}y\minus{}12 x^{8}y^{2}\plus{}30 x^{7}y^{3}\minus{}40 x^{6}y^{4}\plus{}30 x^{5}y^{5}\minus{}12 x^{4}y^{6}\plus{}2 x^{3}y^{7}$

Hmm, Mathematica's answer factors as $ 2 x^{3}y (x\minus{}y)^{6}$.  Maybe there's a smart way to see why.

There's no generalizable rule (ie. the matrix is too 'small' to find any pattern). It's very easily reduced though; try it, you'll find a whole bunch of zeros.


let $n$ is natural number prove that is available prime number such that all of them this numbers $p-1,p+1,p+2$minimum have $n$ different  prime divisor

For all $k\in N$ suppose that $p_k$ be the $k$th prime number. According to chinese remiander theorem(CRT) there exist $a\in N$ such that
$a\equiv 1(mod p_1...p_n)$
$a\equiv -1(mod p_{n+1}...p_{2n})$
$a\equiv -2(mod p_{2n+1}...p_{3n})$
and we know that $gcd(a,p_1p_2...p_{3n})=1$. Then there exist infinity prime number in this form :$a+p_1p_2...p_{3n}k$. then we can prove $p-1,p+1,p+2$ have the requested speciality. ;)


From the no.s $ 1,2,\dots\dots,(2n \plus{} 1)$  3 are selected at random.What is the probability that this no.s are in Arithmetic progression?

Does order matter (can 4,3,5 qualify as well as 3,4,5)?

there are $ \binom{2n\plus{}1}{3}\equal{}\frac{(2n\plus{}1)(2n)(2n\minus{}1)}{6}$ ways to pick 3 numbers.

if three numbers' common difference is 1
(1,2,3)(2,3,4)---(2n-1,2n,2n+1) = 2n-1 ways

if three numbers' common difference is 2
(1,3,5)(2,4,6)---(2n-3,2n-1,2n+1) = 2n-3 ways

...

if three numbers' common difference is n-1
(1,n,2n-1)(2,n+1,2n)(3,n+2,2n+1) = 3 ways

if three numbers' common difference is n
(1,n+1,2n+1) = 1 ways

so there are total $ 1\plus{}3\plus{}...\plus{}(2n\minus{}3)\plus{}(2n\minus{}1)\equal{}n^2$ ways.

its $ \frac{(2n\minus{}1)(2n)(2n\plus{}1)}{6n^2}$, which simplifies to $ \frac{(2n\minus{}1)(2n\plus{}1)}{3n}$

answer : $ \frac{(2n\minus{}1)(2n\plus{}1)}{3n}$

[quote="solveitall"]Nice solution! But I think you meant

$ \frac {3n}{(2n \minus{} 1)(2n \plus{} 1)}$

Also, i think we need to consider positive common difference series and negative as different ones. So,

My solution would be:

$ 2 . \frac {3n}{(2n \minus{} 1)(2n \plus{} 1)}$[/quote]
No, try it for $ n\equal{}1$.
A positive series (1,4,7) is the same as a negative one (7,4,1), when order doesn't matter.

oh oops, changed numerator and denominator  :D

yep, $ \frac{3n}{(2n\minus{}1)(2n\plus{}1)}$ would be the answer.


geometry
ratio
analytic geometry
symmetry
angle bisector
area of a triangle
geometry solved

[quote="romano"]Let ABC be a triangle with acute angle A and AC= 2 AB . The angle bisector AD cuts the altitude BH at K ( D lies on BC , H on AC) . The line CK cuts AB at E . Prove that ABC is right at B when and only when the areas of the triangles BDE and HDE are equal[/quote]

That's definitely the ugliest geometry problem I have ever seen (no offence to you, Romano, of course, just my attitude towards the problem). Anyway, I have a quite long and not really elementary solution:

We denote by [XYZ] the (non-directed) area of a triangle XYZ.

We have to prove that < ABC = 90° if and only if [BDE] = [HDE].

Let Q be the point of intersection of the lines BH and DE. Then, the triangles BDQ and HDQ have the same altitude from the vertex D; hence, the ratio of their areas equals the ratio of their bases BQ and HQ. In other words, [BDQ] : [HDQ] = BQ : HQ. Similarly, [BEQ] : [HEQ] = BQ : HQ. Now, a well-known algebraic rule says that for any five reals a, b, c, d, e, if a : b = c : d = e, then also (a + c) : (b + d) = e. Hence, from [BDQ] : [HDQ] = BQ : HQ and [BEQ] : [HEQ] = BQ : HQ, it follows that ([BDQ] + [BEQ]) : ([HDQ] + [HEQ]) = BQ : HQ, or, equivalently, [BDE] : [HDE] = BQ : HQ. Thus, [BDE] = [HDE] is equivalent to BQ = HQ, i. e. equivalent to the condition that the point Q is the midpoint of the segment BH. And it remains to show that < ABC = 90° if and only if the point Q is the midpoint of the segment BH.

Considering the complete quadrilateral formed by the lines AB, BC, AD and CE, we see that the lines DE, BK and CA are the diagonals of this complete quadrilateral. Hence, after the complete quadrilateral theorem, the points Q and H divide the segment BK harmonically. This means that, now using directed lengths, BQ : QK = - BH : HK. Introducing coordinates on the line BH, and calling b, q, k and h the coordinates of the points B, Q, K and H, respectively, we have BQ = q - b, QK = k - q, BH = h - b and HK = k - h. Thus, BQ : QK = - BH : HK yields (q - b) : (k - q) = - (h - b) : (k - h). Hence, $\left(q-b\right)\cdot\left(k-h\right) = -\left(k-q\right)\cdot\left(h-b\right)$. After some algebra, this can be rewritten as bh + bq + kh + kq = 2bk + 2hq. This, again, is equivalent to $\left(k-b\right)\cdot\left(h-q\right) = 2\left(q-b\right)\left(h-k\right)$, so that (k - b) : (h - k) = 2 (q - b) : (h - q). In other words, BK : KH = 2 BQ : QH. Hence, BQ : QH = 1 if and only if BK : KH = 2.

But the point Q is the midpoint of the segment BH if and only if BQ : QH = 1, and thus, this holds if and only if BK : KH = 2. Hence, it remains to prove that < ABC = 90° if and only if BK : KH = 2.

[[i]Note:[/i] The step above, where I proved BK : KH = 2 BQ : QH, could have been done synthetically, but this would require some more projective geometry.]

Now, let M be the midpoint of the side CA of triangle ABC. Since AC = 2 AB, we have AM = AC / 2 = AB, and thus, the triangle ABM is isosceles. Hence, the line AD, being the angle bisector of the angle CAB, i. e. of the angle BAM, is the symmetry axis of this isosceles triangle. Hence, $AD\perp BM$. In other words, the line AD is an altitude of triangle ABM. The line BH is another altitude of this triangle. The point K, being the point of intersection of these two altitudes, must therefore be the orthocenter of triangle ABM.

Since the line AD is the symmetry axis of the isosceles triangle ABM, it passes through the midpoint U of its side BM. In other words, the line AD is a median of triangle ABM.

Let the line MK meet the side AB of triangle ABM at S. Then, the van Aubel theorem (this theorem is called "Van Obel theorem" on http://www.cut-the-knot.org/triangle/VanObel.shtml ), applied to triangle ABM and the concurrent cevians AU, BH and MS, yields BK : KH = BS : SA + BU : UM. Since the point U is the midpoint of the segment BM, we have BU : UM = 1, and thus BK : KH = BS : SA + 1. Thus, BK : KH = 2 if and only if BS : SA = 1, i. e. if and only if the point S is the midpoint of the segment AB. Thus, it remains to show that < ABC = 90° if and only if the point S is the midpoint of the segment AB.

Now, if the point S is the midpoint of the segment AB, then the line MS is a median of triangle ABM; since the line AD is also a median of triangle ABM, it follows that the point of intersection K of these two lines is the centroid of triangle ABM. Conversely, if the point K is the centroid of triangle ABM, the line MS, passing through K, must be a median of triangle ABM, so that the point S is the midpoint of the segment AB. Hence, we have obtained that the point S is the midpoint of the segment AB if and only if the point K is the centroid of triangle ABM. But the point K is the orthocenter of triangle ABM; hence, the point K is the centroid of triangle ABM if and only if the triangle ABM is equilateral (remember: the centroid of a triangle coincides with its orthocenter only if the triangle is equilateral). Hence, what we have to prove now is that < ABC = 90° if and only if the triangle ABM is equilateral.

Well, this is very easy. If < ABC = 90°, then the point B lies on the circle with diameter AC. The center of this circle is clearly the midpoint M of the segment AC. So we have AM = BM. Since we also have AM = AB (triangle ABM is isosceles!), it follows that AB = AM = BM, and the triangle ABM is equilateral. Conversely, if the triangle ABM is equilateral, we have AM = BM, and thus, as in the above, we see that the point B lies on the circle with diameter AC, so that < ABC = 90°.

Solved!

  Darij


limit
absolute value
real analysis
real analysis unsolved

Hi folks,

I'm having trouble with the proof (using $ \epsilon-\delta$, mind you) that

$ \lim_{n\rightarrow\infty}(-1)^n\cdot{n}$

doesn't exist. I wanted to do a proof by contradiction, but I'm mired and confused by the numerous logical operators and the length of the definition of convergence. I don't really know where to begin.

I've assumed that the limit converges to an arbitrary number, call $ a$. This gives,

$ \forall{\epsilon>0}\exists{N>0}\ni{n>N\implies{|(-1)^n\cdot{n}-a|<\epsilon}}$.

I wanted to extract the $ n$ out of the absolute value but it turned out pretty tricky!

Thanks all.

Prove that the expression $ (\minus{}1)^n\cdot n\minus{}a$ with $ a\in\mathbb{R}$ and $ n\in\mathbb{N}$ is not bounded. This gives the contradiction with your assumption that the sequence $ \{ (\minus{}1)^{n}\cdot n\}_{n\in\mathbb{N}}$ converges to $ a\in\mathbb{R}$.

The second way is to prove that $ \lim_{n\to\infty}|(\minus{}1)^{n}\cdot n|\equal{}\plus{}\infty$. From the uniqueness of the limit you obtain that either $ \lim_{n\to\infty}(\minus{}1)^n\cdot n$ can't exist or is infinite.


When will they be up? I know they will be up next week, but has a more specific date been determined?


Kara rolls two six-sided die.  Given that at least one of the dice lands on an odd number, what is the probability that the two dice add up to 8?

Odd numbers are 1, 3, 5. Getting a 7 is impossible. So $ \frac{2}{3} \times \frac{1}{3} \equal{} \boxed{\frac{2}{9}}$.

copied solution:

There are $6\cdot 6=36$ possible ways to roll two dice. Of these, the dice can add up to 8 in five ways: $2+6,3+5,4+4,5+3,6+2$. Only 2 of these combinations include odd numbers. \\ \\ We need to count the number of rolls that include at least one odd die. There are a total of $3\cdot 3=9$ distinct rolls with both dice even, which leaves $36-9=27$ rolls that include at least one odd die. Of these 27, exactly two will add up to 8, as described above. So, the probability the two dice add up to 8 is $\boxed{\dfrac{2}{27}}$.

I need some help.  I picture the problem this way:  one die is limited to coming up 1,3,5 and the other can be any 1 through 6.  This gives 18 possible pairs of which only two add up to 8.  Then, I get the result 1/9 for the probability.  What am I doing wrong?  Thanks.


When doing problems involving energy or KE (the standard relativistic formulas), does anyone know if you are always supposed to use the $\frac{MeV}{c^{2}}$ form of the mass, or what about $kg$. It doesn't seem like these should be different if they are both measuring mass.

For example, prb. from book: find the momentum of an electron (in MeV/c) travelling at 0.600c.

My solution: $p=\gamma mv=\frac{0.511(.6c)}{\sqrt{1-.36}}\approx 1.44\times 10^{8}~\frac{MeV}{c}$. Books answer: $0.383~\frac{MeV}{c}$.

??

Nobody forces you to use units you don't like ;) You just need to be consistent -- do not switch between units in mid-problem. Using MeV's for mass is generally a smarter move if you're dealing with particles with a small mass, such as electrons; if you don't mind huge exponents, kg's are ok.

Btw, you've got a mistake in your calculation...

I know I made an error, that is why I was posting. I was assuming it had to do with the mass, but it doesn't sound that way from what you said. All of my problem are coming out way too big like that. (I know it isn't just a factor of $c$, although that does put me on the right order of magnitude).

The problem is that you just multiplied incorrectly, that's all. The quantities you have specified are ok. Check again with a calculator...

I see what my problem was with that extra factor of $c$. I was thinking about it wrong. In $p=\gamma mv$ you have $\gamma \left(\frac{MeV}{c^{2}}\right)(\%\text{of speed of light})c$, so you don't include $c$ in the multiplication, because it cancels the units:
$p=\gamma mv$ where $v$ is the fraction of the speed of light it is travelling, not the actual speed. My answer was in $MeV$, not $\frac{MeV}{c}$.


Let $ n$ be an integer greater than $ 3.$ Points $ V_{1},V_{2},...,V_{n},$ with no three collinear, lie on a plane. Some of the segments $ V_{i}V_{j},$ with $ 1 \le i < j \le n,$ are constructed. Points  $ V_{i}$ and $ V_{j}$ are [i]neighbors[/i] if  $ V_{i}V_{j}$ is constructed. Initially, chess pieces $ C_{1},C_{2},...,C_{n}$ are placed at points $ V_{1},V_{2},...,V_{n}$ (not necessarily in that order) with exactly one piece at each point. In a move, one can choose some of the $ n$ chess pieces, and simultaneously relocate each of the chosen piece from its current position to one of its neighboring positions such that after the move, exactly one chess piece is at each point and no two chess pieces have exchanged their positions. A set of constructed segments is called [i]harmonic[/i] if for any initial positions of the chess pieces, each chess piece $ C_{i}(1 \le i \le n)$ is at the point $ V_{i}$ after a finite number of moves. Determine the minimum number of segments in a harmonic set.

The answer is $ n\plus{}1$.

If we have $ n$ segments, we have only one cycle in the graph and so we can only reach at most $ n$ different permutations, but we need $ n!$.

So suppose we have $ n\plus{}1$. Construct the segments so that $ V_i V_{i\plus{}1}$ are connected (including $ V_n V_1$) and $ V_1 V_4$ are connected. That is, we have an $ n$-cycle and a $ 4$-cycle. In other words, we can basically pick any $ 4$ consecutive pieces and permute them in a $ 4$-cycle by shifting all pieces with the $ n$-cycle so those $ 4$ are on the $ 4$-cycle and then shifting back. Suppose we have $ 5$ adjacent pieces $ a,b,c,d,e$. Then we can get $ c,d,a,b,e$ :arrow: $ c,a,b,e,d$ :arrow: $ a,b,e,c,d$. Namely, $ c,d,e$ became $ e,c,d$. So we can generate any $ 3$-cycle, too.

Using the $ 3$ or $ 4$ cycles, simply move $ C_1$ to $ V_1$, then move $ C_2$ to $ V_2$ without changing $ C_1$'s place. We can guarantee we will be able to move $ C_1, C_2, \ldots C_{n\minus{}2}$ into the right positions in this way using the $ 3$-cycle. The only issue is if we end up with $ C_{n\minus{}3}, C_{n\minus{}2}, C_n, C_{n\minus{}1}$ in $ V_{n\minus{}3}, V_{n\minus{}2}, V_{n\minus{}1}, V_n$ respectively. This is easy to fix: use the $ 4$-cycle to get $ C_{n\minus{}1}, C_{n\minus{}3}, C_{n\minus{}2}, C_n$, and then the $ 3$-cycle to get $ C_{n\minus{}3}, C_{n\minus{}2}, C_{n\minus{}1}, C_n$, and that finishes it.