AMC
AIME
MATHCOUNTS
floor function
USA(J)MO
USAMO
AMC 12

Would you consider this one as hard, medium, or easy?  Just wondering, since this is my first one, and I got a 5 :ewpu:.

I can't really say with much authority, but it seemed pretty easy to me.  I got 7 (I got 5 last year and 1 the year before), so I think it was relatively easy.

kyyuanmathcount, didnt you take the aime at northville high?

just wondering if u were the middle schooler who came to my school to take the aime.

Yes, who are you again and what did you get?

lol. I was the indian one. dont know how else to describe who i was.

I got a 10

btw, gj on mathcounts. I am coaching my brother for the mathcounts (currently in 5th grade) so i know how hard it is to get a perfect score (especially with the time constraint)

I thought that it was at least as easy as last years (based mostly on my scores- 3 last year, 8 this year with the same amount of practice (i.e. none))
The first, oh, 8 were quite simple.

this years AIME, in my opinion, was the same difficutly as last year, easy.
although maybe a bit harder, not that much though

ummm...

i personally thought this was much harder than AIME 1994 because I got 11 on that one but I only got $\lfloor 11/2\rfloor$ on this one...

or maybe i just had a really bad day. :(

I thought it was kind of harder than usual ...

I thought it was harder since the concepts tested were generally on my weaker sides of problem solving.

:O YES PEOPLE ARE THINKING IT"S MODERATELY HARD!!! (Or at least medium)

In all the practice AIME's I've taken, at most I had 3. This time I THOUGHT I had 6 but then realized I made a HORRIBLE mistake on #2 and I think I have either a five or a six now.

I thought 1-5 were relatively easy while 6-10 were harder than usual. Then 12-15 were easier than normal. However, since most people don't look at 11-15 I thinking that scores are going to be lower overall.

[quote="idupda"]I don't know if this comment is really appropriate here. But for sophomores and below how does the qualification process differ from that for juniors and seniors? Will an AIME score of 7 not be enough for qualifying for USAMO?[/quote]

So this is how it works (someone should make a sticky about this): the top 250 juniors and seniors will be chosen, based on their index score (10*AIME + AMC12 = Index). Then the lowest AIME score will be taken among these people, and this will become the floor score. So then all sophmores and below with that score or higher will also qualify for the USAMO.

The floor should be either a seven or eight. The general consensus seems to be 8, but if you get lucky you might make it! Floor is typically easier to qualify on than the index.

I thought it was easier... the first 7 anyway. #8 was just a bad problem, and #10 was really easy if you tried doing it without math and hard if you tried using math. #11-15 were very doable in comparison to some of the last 5 on the 90's tests, but my solutions to #12 and #14 were very sorta tedious (both of which I ended up getting wrong because of dumb mistakes)... all this is my opinion of course.

1-7 and 9 were pretty easy. 13-15 were more approachable than usual (as some people have already said). I thought on the whole that this was even easier than last year's AIME. Or maybe i'm just better  :lol:

First 5: easier than normal

#6 was normal
#7 was harder
#8 was normal
#9 was easy
#10 was easy

Last 5: normal

So a little on the easy side, I think.

[quote="The Zuton Force"]I thought it was orders of magnitudes harder. I got an 11 on the 1994 but only got a 7 or so on this one. This is most likely because of the distribution of problems by subject.[/quote]

I blame it on the distribution as well. Not enough algebra and NO COMPLEX NUMBERS!?!?!?! :mad:

I am tempted to say that it was hard, because I got a 7...considering that I got 12s, 13s, 14s on practices...
But then again, I have to consider that it was the actual thing, and also that I had a bad day... (made 5 careless mistakes!)

I thought this was only slightly harder than the 2005 exam. After all, #8 and #9 could probably pass for AMC problems, at least in my opinion.

The questions ranged from ridiculously easy to hard.  The first page was rather easy (especially 1) and (14) was vicious.

#4 and #5 would have been good AMC questions.

I didn't think this one was that hard, actually.  My score almost doubled from last year's AIME (from 7 to 12.)  I think a lot of that, though, might be due to the fact that I'm older, and the astonishing number of stupid mistakes I made on last year's AIME.

It was about as easy as last year's even though I didn't make the usamo

[quote="filletwho"]I think number two was a MATHCOUNTS Countdown type question...[/quote]

What?!!  That's such an insult!

[color=white]Though I must agree[/color]

What?? That's an insult to MATHCOUNTS!

Just kidding... I thought the problems were too similar in difficulty. At least last year #14 and #15 were slightly tricky. So basically this year the number of stupid mistakes you made determined how well you did...


Given that $ f(x) \equal{} x^{\frac23}\minus{}x^2$, what is the value of $ f( 8 ) \plus{}f(2\sqrt2)$?


[quote="Hellstar"]I need help with this question. The answer was different than mine and I don't know where I went wrong...

For what values of a and b is $p(x)=x^2+1$ a factor of $q(x)= x{}^1{}^0+ax+b$ ?

This is what I got and what the "answer" is. You might not want to look until you've fiddled around with it a bit.
[hide] I got $a=i$ and $b=2$  but the correct answer is: There is no solution... I don't understand that. My answer seems to work.  [/hide][/quote]
[hide="No solution is not correct"]Take $a=0$ and $b=1$. Clearly $x^2+1$ divides $x^{10}+1$. Are you sure that's how the problem is stated?[/hide]
Edit:Fixed the exponent 10

Yeah, you're right; 10 isn't odd. ;)

Maybe the problem writer meant real solutions?

[quote="Altheman"][hide]

if $r_1$ and $r_2$ are factors of a polynomial $f(x)$, then $f(r_i)=0$
so we have

$-1+ai+b=0=-1-ai+b$
$1=-1$

so our assumption that those were roots was false

[/hide][/quote]
Are you sure what you are writing in there? That's two equations and you can solve for $a,b$... Please check your work.

[quote="236factorial"][quote="Hellstar"]I don't think $x^2+1$ divides $x{}^1{}^0+1$ I tried it with long division and ended up with a remainder of $1-x$
Yes and I am sure that's how the problem is stated. :)[/quote]

Really? I divided it and got a remainder of 0  ;)  (quotient: $x^8-x^6+x^4-x^2+1$).

Try it again...[/quote]

That's what I get too. So the solution is $ax+b=1$?

Yeah, that's what I have. (The solution should be $a=0$ and $b=1$)

[hide]
$x^{10}+ax+b=(x^{10}+x^8)-(x^8+x^6)+(x^6+x^4)-(x^4+x^2)+x^2+ax+b$
$=(x^2+1)(x^8-x^6+x^4-x^2)+x^2+ax+b$

So $x^2+1$ must divide $x^2+ax+b$.
Comparing the coefficients, we get $a=0$ and $b=1$.[/hide]

[quote="10000th User"]
Are you sure what you are writing in there? That's two equations and you can solve for $a,b$... Please check your work.[/quote]

woops...i made a little error,

[hide]

i divided by zero, so i lost the answer

$-1+ai+b=-1-ai+b$
$ai+b=-ai+b$
$ai=-ai$
$2ai=0$
$a=0$, $b=1$

[/hide]


LaTeX
inequalities
combinatorics unsolved
combinatorics

Let be given integers k and n such that 1<=k<=n. Find the number of ordered k-tuples (a_1,a_2,...,a_n), where a_1, a_2, ..., a_k are different and in the set {1,2,...,n}, satisfying
1) There exist s, t such that 1<=s<t<=k and a_s>a_t.
2) There exists s such that 1<=s<=k and a_s is not congruent to s mod 2.

P.S. This is the only problem from VMO 1996 I cannot find a solution or I cannot solve. But I'm not good at all in combinatoric...

I have a solution to it,it is too long and if nobody will solve it,i'll post it tomorrow.
P.S.Do you remember that you promise to use LaTeX :wink:

1. Solve in reals the system:
$ |\begin{aligned}y^{3}+9x^{2}-27x-27 & = 0\\ z^{3}+9y^{2}-27y-27 & = 0\\ x^{3}+9z^{2}-27z-27 & = 0\end{aligned}$

is a proof I started learning and using Latex. I even tried to modify one of mine inequalities but the topic was locked and I don't want to disturb the admins. I'm not very good in using latex but I'll use it for inequalities and systems etc. but I decided it is not too difficult for reading without Latex - it is the reason.

[quote="borislav_mirchev"]It seems no one didn't solved this problem :-) So if you have time you may send your solution. It may only seems the problem is not difficult as someone has said: "If you want to go to the Moon - you need rocket, nothing less is not enough".[/quote]
As i promised here is solution:
Let $ A$ be the set of ordered $ k$-tuples $ (a_{1},a_{2}\dots,a_{k})$,$ a_{i}\in\{1,2,\dots,n\},k\leq n,i=1,2,\dots n$ and $ A_{1}$ be the set of all ordered $ k$-tuples satisfying either $ 1)$ or $ 2)$ of the problem,let
$ A_{2}=\{$ordered $ k$-tuples $ (a_{1},a_{2},\dots,a_{k})|a_{i}<a_{i+1},i=1,2,\dots,k-1$,and   $ a_{i}\equiv i(mod 2),i=1,2\dots,k\}$.
It is obvious that $ A_{2}\subset A$ and $ A_{1}=A\backslash A_{2}$.Therefore:$ |A_{1}|=|A|-|A_{2}|$.For each $ (a_{1},a_{2},\dots,a_{k})\in A_{2}$,we have $ a_{i}+i\neq a_{j}+j$ for all elements $ i\neq j\in\{2,3,\dots,k\}$,
$ a_{i}+i$ is even and ${ a_{i}+i\in\{1,2,\dots,n+k}$ for $ i=1,2,\dots k$.Let
$ B=\{(b_{1},b_{2}\dots,b_{k})|b_{i}\in\{2,\dots,n+k\}$ and $ b_{i}$ is even,$ (i=1,2,\dots,k)$,$ (1+b_{i}<b_{i+1},(i=1,2,\dots k-1)\}$.
Consider the mapping $ f$ from $ A_{2}$ to $ B$,that maps from $ (a_{1},a_{2},\dots,a_{k})$ of $ A_{2}$ to $ (b_{1},b_{2},\dots,b_{k})=(a_{1}+1,a_{2}+2,\dots,a_{k}+k)$ of $ B$.
It is obvious that $ f$ is injection from $ A_{2}$ to $ B$.  [color=blue](1)[/color]
Consider any $ (b_{1},b_{2},\dots,b_{k})$ in $ B$.Since $ b_{i}$ is even and $ b_{i}\in\{2,\dots,k\}$,we have $ 1\leq b_{i}-i< b_{i+1}-(i+1)\leq n(i=1,2,\dots,k)$,and $ b_{i}-i\equiv i(mod 2)$ for all $ i=1,2,\dots k$.
Therefore $ (b_{1}-1,b_{2}-2,\dots,b_{k}-k)\in A_{2}$.
Moreover,the image of $ (b_{1}-1,\dots,b_{k}-k)$ through $ f$ is $ (b_{1},\dots,b_{k})$.So $ f$ is the mapping from $ A_{2}$ to $ B$.  [color=darkblue](2)[/color]
From (1) and (2) we conclude that $ f$ is the bijection from $ A_{2}$ to $ B$.Thus
$ |A_{2}|=|B|=\binom{[\frac{n+k}{2}]}{k}$,this leads us to
$ |A_{1}|=|A|-|A_{2}|=\frac{n!}{(n-k)!}-\binom{m}{k}$,where $ m=[\frac{n+k}{2}]$.Thus we are done.

I had no chances with this problem. I'm wondering how long did it take you to solve this problem?

[quote="borislav_mirchev"]I've the following question about the solution.
1) Does your solution count all the k-tuples for which at least one of 1) and 2) is true?
2) Please, describe in more details how did you calculate: $ |A_{2}|\equal{}|B|\equal{}\binom{[\frac{n\plus{}k}{2}]}{k}$ ?[/quote]

$ X_{2}=\{ (a_{1}, a_{2}, \dots , a_{k}) | 1 \leq a_{1}<  \dots < a_{k}\leq n \}      $
Let $X_{1}$ be the set of all $(a_{1}, a_{2}, \dots , a_{k})$ satisfying condition[u][b](1)[/b][/u] in problem.
$ |X_{1}|=|A|-|X_{2}|=\binom{n}{k} k!-  \binom{n}{k}$

Let $Y_{1}$ be the set of all $(a_{1}, a_{2}, \dots , a_{k})$ satisfying condition[u][b](2)[/b][/u] in problem.
$ Y_{2}=\{ (a_{1}, a_{2}, \dots , a_{k}) |~ i=1, \dots , k, ~ a_{i}\equiv i(mod 2) \}      $
$ |Y_{1}|=|A|-|Y_{2}|=\binom{n}{k} k!-  [\frac{n}{2}]!  [\frac{n+1}{2}]!      $

Therefore,
$ \binom{n}{k}( k!-1)-  [\frac{n}{2}]!  [\frac{n+1}{2}]! -   \binom{   [\frac{n+k}{2}]       }{k}                $


There are 6 points in a plane. If a line connect 2 points it touchs a third, show that the points are collinear.

oh....... come on. No one post any solution.

err?? It connects two points and passes through another one? It's beyond obvious that they have to be collinear... Unless that's not the question


how do you find, with proof, all positive integers $x$ such that $x^2+2$ is prime?

Umm... by not finding them¿ (it's even open if there are $\infty$ many of them)

x= 1 ,3 ,9 ,81 ,6561.....
I don't have any proof now but I will post it later ;)

Math is my life :!:

[quote="ZetaX"]What's written there¿
In most books, there are some really hard problems given (but most times declared that way).
And a proof that shows at least that there are $\infty$ many of them without very brutal methods would be revolutional![/quote]

ZetaX: Is there a proof using any kind of methods?

[quote="drunner2007"]
it was in a well written problem solving book so its probably not a waste of time
[/quote]

Perhaps a acops typo?  There are certainly enough of them.

[quote="TRAN THAI HUNG"]x= 1 ,3 ,9 ,81 ,729.....
I don't have any proof now but I will post it later ;)
[/quote]

If this were true, which it isn't, this would make you very, very famous.

Like those Chineese people that proved Poincaré

[quote="blahblahblah"][quote="drunner2007"]
it was in a well written problem solving book so its probably not a waste of time
[/quote]

Perhaps a acops typo?  There are certainly enough of them.

[quote="TRAN THAI HUNG"]x= 1 ,3 ,9 ,81 ,6561.....
I don't have any proof now but I will post it later ;)
[/quote]

If this were true, which it isn't, this would make you very, very famous.[/quote] :blush: Yes, I'm wrong $\ 6561^2+2$divide by 19
Math is mylife :!:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Find&t=80736]Way ahead of u[/url]


Suppose that $ ABCD$ is a tetragon satisfies that $ \widehat{BAC}\equal{}50$,$ \widehat{ABD}\equal{}60$,$ \widehat{DBC}\equal{}20$.Prove that $ \widehat{ACD}\equal{}80$

No info provided regarding any angle or length on the right side of AC!
Was ABD equilateral? Under such assumption, there is too much info provided, and makes $ \angle BDC \equal{} 80^\circ$.

Best regards,
sunken rock


$x=0$ is not a soluiton.
if $x>0$, squaring we get $3^{x}+4^{x}+2(\sqrt{12})^{x}=5^{x}$. so in order to $(\sqrt{12})^{x}$ be an integer $x$ must be even. let $x=2k$. it's easy to check that $k=2$ i.e. $x=4$ is a solution. if $k>2$, by Fermat's Last Theorem the equation has no solutons.

if $x<0$, first let $x=-2k$. so after manipulations $5^{k}(4^{k}+3^{k})=12^{k}$ for $k\in\mathbb N$ which is not possible.

if $x=-2k+1, k\in\mathbb N$, we get after squaring and manipulations, $4^{2k}5^{2k}3+3^{2k}5^{2k}4+2\cdot 3^{k}4^{k}5^{2k}\sqrt{12}=12^{2k}5$, which is impossible since $\sqrt{12}$ is not an integer.


calculus
integration
function
trigonometry
real analysis
calculus computations

hi,
let $ a>1$ and a function $ f: [\frac{1}{a},a] \rightarrow \mathbb{R}$ s.t.
\[ f(x) \plus{} f(\frac{1}{x}) \equal{} k , \forall x \in [\frac{1}{a},a].\]
Calculate:
\[ \int_{\frac{1}{a}}^{a}{\frac{(x\plus{}1)f(x)}{x\sqrt{x^{2}\plus{}1}}} dx\].

ups i forgot to say that f is continous  :)

I think there is only one thing one could possibly do:
[hide]Let $ x = 1/y$, so $ dx = -\frac{dy}{y^2}$.  Then we have
\begin{align*}
I & : = \int_{\frac{1}{a}}^{a}{\frac{(x+1)f(x)}{x\sqrt{x^{2}+1}}} dx \\
& = \int_{a}^{\frac{1}{a}}{\frac{(\frac{1}{y} + 1)f(\frac{1}{y})}{\frac{1}{y}\sqrt{\frac{1}{y^{2}}+1}}} \frac{-dy}{y^2} \\
& = \int_{\frac{1}{a}}^{a}{\frac{(y+1)f(\frac{1}{y})}{y\sqrt{y^{2}+1}}} dy \\
& = k\int_{\frac{1}{a}}^{a}{\frac{(y+1)}{y\sqrt{y^{2}+1}}} dy - I.
\end{align*}
Now you just need to remember your inverse hyperbolic trig function integrals or what-have-you.
[/hide]

yes, i see ,but do you need f to be continous ?

No. We just need it to be sufficiently nice that the integral exists. Since we're multiplying by a smooth function, this condition is simply that $ f$ is integrable.


calculus
integration
trigonometry
logarithms
calculus computations

good day ppl  :D

evaluate $ \int_0^\infty\;\dfrac{\tan^{\minus{}1}\;\left(\dfrac{x}{2\pi e}\right)}{e^x\minus{}1}\;\;\;\textbf dx$

Using Stirling's approximation and substituting $ x \equal{} 2 \pi t$ we obtain
$ \int_0^\infty\;\dfrac{\tan^{ \minus{} 1}\;\left(\dfrac{x}{2\pi e}\right)}{e^x \minus{} 1}\;\textbf dx \equal{} \pi \left\{\ln \Gamma (\textbf e) \minus{} \left( \textbf e \minus{} \frac {1}{2}\right) \plus{} \textbf e \minus{} \frac {1}{2} \ln \pi \right\}$
if I ain't wrong  :maybe:


function
limit
real analysis
real analysis unsolved

[quote="didilica"]Something is wrong.
On one hand if $n=2$ we get, according to your formula, that $\xi(2)=0$ which is definitely false and on the other hand for all $n$ the second sum diverges, since by way of contradiction if it converges then its term should converge to 0, but

$\lim_{q\rightarrow \infty}q\frac{1}{q}\sum_{k=1}^{q}\frac{1}{k^{n-1}}=\sum_{1}^{\infty}\frac{1}{k^{n-1}}=\xi(n-1)\neq 0$.[/quote]

You were right, some thing was wrong: I posted the formula incorrectly, thanks Didi! --it's fixed now (per [url=http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/Zeta/06/05/]this reference[/url]).

In this case write the second sum as follows:

$\sum_{m=1}^{\infty}\frac{1}{m}\sum_{k=m+1}^{\infty}\frac{1}{k^{n-1}}=\sum_{m=1}^{\infty}\frac{1}{m}\left(\xi(n-1)-\sum_{k=1}^{m}\frac{1}{k^{n-1}}\right)$
$=\sum_{m=1}^{\infty}\frac{1}{m}\left(\xi(n-1)-\sum_{k=1}^{m-1}\frac{1}{k^{n-1}}\right)-\sum_{k=1}^{\infty}\frac{1}{k^{n}}$
$=\sum_{m=1}^{\infty}\frac{1}{m}\left(\xi(n-1)-\sum_{k=1}^{m-1}\frac{1}{k^{n-1}}\right)-\xi(n)$.

Now apply the Abel summation by parts formula for the preceding sum and you should get it.


Incidentally it can also be proven that $ S=\frac{\pi^{2}}{6}$ but the proof is a bit beyond the scope of this forum.


search
modular arithmetic
Duke
college
number theory
prime numbers

(1) Let $p$ be a prime number for $p\ge5$. Prove $p^2-1$ is divisible by $24$

(2) Let $q$ be a prime number for $q\ge7$. Prove $q^4-1$ is divisible by $240$

Search the forum. I've seen (1) posted at least three times before.

1)

All prime numbers $p$ greater than 3 can be written in the form $p=6k\pm1$, for $k$ an integer. Thus $p^2\pm1=36k^2\pm12k+1-1=36k^2\pm12k=6(6k^2\pm2k)=4(9k^2\pm3k)$. Since $6k^2\pm2k$ and $9k^2\pm3k$ are integers, $p^2-1$ is divisible by 4 and 6, or equivalently, $p^2-1$ is divisible by 24.

[quote="drunner2007"]All prime numbers $p$ greater than 3 can be written in the form $p=6k+1$, for $k$ an integer.[/quote]
I don't think $6k+1$ is the only possible form.
$p=6k+5$ is another possible case.

(You basically need to repeat the same thing for $6k+5$ and you're done.)

>>Hokkage

Oh, I know it,but I just think (1) is hint of (2)  :(

use factorization ,these are moderately easy :lol:

It's actually all primes >5 can be written in the form $6k \pm 1$.
Just look at 6k+0, 6k+2, 6k+3, and 6k+4. They have factors of 6, 2, 3, and 2, respectively, thus not prime.

yeah u guys are right about the $6k+/-1$ thing. I'll edit my posts.

[quote="mna851"]Can anybody prove that all primes greater than $\ 5$ can be written in the form $\ 6k + 1$ .[/quote]

of course, if $k \not \in \mathbb{Z}$  ;)

check for the prime numbers $11,17,23,29,\cdots$ (they are in the form $6k-1$)

All numbers can be covered by the six:
6k = 6(k)
6k + 1
6k + 2 = 2(3k +1)
6k + 3 = 3(2k + 1)
6k + 4 = 2(3k + 2)
6k + 5

Hence, 6k + 1 and 6k + 5 are the only 2 types of numbers that could possibly be prime.

[quote="porte"](1) Let $p$ be a prime number for $p\ge5$. Prove $p^2-1$ is divisible by $24$
[/quote]
[hide]

$p=2k+1, \ k\geq 2$

$p^2-1 = (p-1)(p+1) = 2k(2k+2) = 4k(k+1)$

and of course $2 \mid k(k+1)$

Hence $\boxed{8\mid p^2-1}$

From the numbers $p-1,p,p+1$ one exactly is divisible by 3.
But $p>3\Rightarrow 3 \not {|} p\Rightarrow 3\mid (p-1)(p+1)\Rightarrow \boxed{3\mid p^2-1}$

(or we can say $p\equiv \pm 1 \mod{3} \Rightarrow  p^2\equiv 1 \mod{3}\ \ {}$)

and $(3,8)=1$ so we get $24\mid p^2-1$[/hide]

[quote="porte"](2) Let $q$ be a prime number for $q\ge7$. Prove $q^4-1$ is divisible by $240$[/quote]

[hide]
$q^4-1 = (q^2-1)(q^2+1)$

$24 \mid q^2-1$
$2 \mid q^2+1$

Check in $\mod 5$

$q\geq 7 \Rightarrow  q \neq 5 \Rightarrow  q \not \equiv 0 \mod 5$

$q \equiv \pm 1 \mod{5} \Rightarrow  q^2 \equiv 1 \mod 5 \Rightarrow  5 \mid q^2-1$

$q \equiv \text{2 or 3} \mod{5} \Rightarrow  q^2 \equiv -1 \mod 5 \Rightarrow  5 \mid q^2+1$

At any case $5 \mid (q^2-1)(q^2+1)$

so we have $24\cdot 2\cdot 5 =240 \mid (q^2-1)(q^2+1)$[/hide]

[quote="mna851"]Can anybody prove that all primes greater than $\ 5$ can be written in the form $\ 6k + 1$ .[/quote]

To prove that every prime no. is of the form  $6k\pm1$
Not sure of this proof......

Evey no. is of the form ,$3n$,$3n\pm1$   (Using Fermats little theorem)

So for a prime no.,we will have to see them in the form $3n\pm1$.

 $n$  has to be even.
so we can write the possible  prime as $3(2k)\pm1$

Thm: Every odd prime, except 3, is congruent to 1 or -1 modulo 6.

Proof: Suppose $p$ is a prime not equal to 2 or 3 such that $p\not\equiv\pm1\pmod{6}$. Then $p$ is congruent to 0, 2, 3, or 4 modulo 6. We now proceed by cases.

Case 1: $p\equiv 0\pmod{6}$. Then $2|p$, a contradiction.

Case 2: $p\equiv 2\pmod{6}$. Then $6|(p-2)$ so $p-2=6k_1$, $k_1\in\mathbb{Z}$. This implies $p=2(1+3k_1)$, a contradiction.

Case 3: $p\equiv3\pmod{6}$. Then $6|(p-3)$ so $p-3=6k_2$, $k_2\in\mathbb{Z}$. However, this means $p=3(1+2k_2)$, a contradiction.

Case 4: $p\equiv4\pmod{6}$. Then $6|(p-4)$ so $p-4=6k_3$, $k_2\in\mathbb{Z}$. Thus $p=2(2+3k_3)$, a contradiction.

Since all cases led to contradictions, it must be true that $p\equiv\pm1\pmod{6}$.

I first saw this problem in the book Elementary Number Theory with Applications, by Thomas Koshy. Its a great book for problem solving number theory, written for an  introductory college class in number theory, but accessible to anyone willing to work through it. Its problem-solving oriented and very interesting, and it is one of the main books Duke uses for its TIP magnet program.

[quote="Altheman"]a cool way to simplify showing that it is 0 mod 5:
$q^4-1 \equiv (q+1)(q-1)(q^2+1) \equiv$
$(q+1)(q-1)(q^2-4) \equiv (q+1)(q-1)(q+2)(q-2) \mod 5$
someone can finish...[/quote]

That is cool...

So, for any $q$, $5|(q-2)(q-1)(q)(q+1)(q+2)$.  Since $q$ is prime and $q > 5$, $5|(q-2)(q-1)(q+1)(q+2) \Rightarrow 5|q^4-1$. QED.


Find nature n ( n$\neq$0) satisfy there exist a,b are integer such that the equation
          nx^2 + ax + b =0 has 2 root x1 $\neq$ x2 and in the inteval (0;1)

Let $f(x)=nx^2+ax+b$. It has a minimum at $x_0=\frac{-a}{2n}$.

To satisfy the requirements the following must held:
$0 < \frac{-a}{2n} < 1$, $f(x_0) = \frac{-a^2}{4n}+b < 0$, $f(0)=b>0$, and $f(1)=n+a+b>0$

Is follows that such $a,b$ exist for every $n\geq 5$ and it is enough to take $b=1$ and $a=-n$.


[quote="texas137"]What is the latest that team selection can be made for Mathcounts chapter? ie, when does the coach have to submit actual names as opposed to just the number of kids attending?

also, has the 2009-2010 School Handbook come out yet?

Thanks!
-texas137[/quote]

As far as I know:

1. The coach does at some point have to submit actual names, but not until potentially the day of the chapter competition (although I wouldnt advise it)

2. The handbook has been out for quite some time, you can read it on mathcounts.org

I may be trumped by something who knows more about this.


LaTeX
function
algebra
polynomial
number theory
prime factorization
relatively prime

1)Prove that  $ 2^{147} \minus{}1$ is divisble by 343.

2) Demonstate that $ n^5 \minus{}5n^3 \plus{}60n^2\minus{}56n$ is divisble by $ 120$ if $ n$ is a natural number.

[quote="Chalao"]1)Prove that  $ 2^{147} \minus{} 1$ is divisble by 343.

2) Demonstate that $ n^5 \minus{} 5n^3 \plus{} 60n^2 \minus{} 56n$ is divisble by $ 120$ if $ n$ is a natural number.[/quote]
[hide]This will be divisible by 120 if it is divisible by 5,8 and 3. Taking the equation in mod 3 we have $ n^{5}\plus{}n^{3}\plus{}n$. Thus we have $ n(n^{4}\plus{}n^{2}\plus{}1)$ If $ n\equal{}1\bmod{3}$, then we have that the expression is divisible by 3. If $ n\equiv 2\bmod{3}$ then we have $ 2(21)$ which is divisible by 3. Finally if $ n\equiv 0\bmod{3}$ then the expression is obviously divisible by 3.

Taking the equation in mod 5 we have $ n^{5}\plus{}4n$ or $ n(n^{4}\plus{}4)$. Checking for $ n\equal{}1,2,3,4\bmod{5}$, we see that they all work, thus this is divisible by 5. Since we have that $ n$ outside the parentheses the case where n=0 is obvious.

Taking the equation in mod 8 we have $ n^{5}\plus{}3n^{3}\plus{}4n^{2}$ or $ n^{2}(n^{3}\plus{}3n\plus{}4)$. If n is even, then we have 2 2's from the $ n^{2}$ term, and an even term inside the parentheses making this divisible by 8 for any even $ n$. If n were to be odd, then n can be represented as $ 2k\plus{}1$. Plugging this into the equation gives us $ (4k^{2}\plus{}4k\plus{}1)(8k^{3}\plus{}12k^{2}\plus{}12k\plus{}8)$. This also equals $ 4(4k^{2}\plus{}4k\plus{}1)(2k^{3}\plus{}3k^{2}\plus{}3k\plus{}2)$. Checking parity from the cubic, we see that it is even in both cases of k, thus proving that this is also divisible by 8.
We have proved the divisibility of 3,5, and 8, and thus we can conclude that this is divisible by 120 for any natural number $ n$.[/hide]

Part 1:
 Hello, for the first part, is my soln. correct:

By Euler's theorem,
                       phi(343)=294

So, 2^294 is congruent to 1 modulo 343

4 (2^147) is congruent to 1 modulo 343

so 2^147 is congruent to 1 modulo 343

i.e 2^147 - 1 is divisible by 343

Thanks

P.S. Sorry, am still learning latex  :)

Euler's theorem states that for any 2 coprime positive integers, $ a$ and $ n$,

$ a^{\varphi(n)} \equiv 1 \mod n$

where $ n \equal{} p_1^{e_1}p_2^{e_2}...p_r^{e_r}$ and $ \varphi(n) \equal{} n(1 \minus{} \frac {1}{p_1})(1 \minus{} \frac {1}{p_2})...(1 \minus{} \frac {1}{p_r}) \equal{} p_1^{e_1 \minus{} 1}p_2^{e_2 \minus{} 1}...p_r^{e_r \minus{} 1}(p_1 \minus{} 1)(p_2 \minus{} 2)...(p_r \minus{} 1)$.

A similar theorem is Carmicheal's theorem. Sometimes it works better than Euler's. See here:

[url]http://en.wikipedia.org/wiki/Carmichael_function[/url]

EDIT: I should have also stated that $ n \equal{} p_1^{e_1}p_2^{e_2}...p_r^{e_r}$ is the prime factorization of $ n$ e.i. $ p_1,p_2,...,p_r$ are all primes and $ e_1,e_2,...,e_r$ are all positive integers.

[quote="Chalao"]If the original expression was $ n^{5} \minus{} 5n^{3} \plus{} 60n^{2} \minus{} 56n$, why did you (ckck), turn it into $ n^{5} \plus{} n^{3} \plus{} n$?

Can someone explain me in simple words the Euler's theorem? I don'nt understand the meaning of the symbol (phi). I read it in AoPs wiki but i don't get it at all.[/quote]
For divisibility of 3, I considered the equation in mod 3.

Excuse me, i don´t understand very much the notation of the Euler's theorem since i am a begginer. I suppose that $ \varphi(n)$ means the multiplication of the divisors of $ n$ substracting 1,2,3.... for each one.

And about the polynomial equation in mod 3,i think is something like this

$ n^{5}\minus{}5n^{3}\plus{}60n^{2}\minus{}56n\equiv 0\bmod{3}$
$ n^{5}\plus{}60n^{2}\equiv 56n\plus{}5n^{3}\bmod{3}$

but then i don´t know how did you turn it into $ n^{5}\plus{}n^{3}\plus{}n$

[quote="Chalao"]Excuse me, i don´t understand very much the notation of the Euler's theorem since i am a begginer. I suppose that $ \varphi(n)$ means the multiplication of the divisors of $ n$ substracting 1,2,3.... for each one.

And about the polynomial equation in mod 3,i think is something like this

$ n^{5} \minus{} 5n^{3} \plus{} 60n^{2} \minus{} 56n\equiv 0\bmod{3}$
$ n^{5} \plus{} 60n^{2}\equiv 56n \plus{} 5n^{3}\bmod{3}$

but then i don´t know how did you turn it into $ n^{5} \plus{} n^{3} \plus{} n$[/quote]
$ n^{5} \minus{} 5n^{3} \plus{} 60n^{2} \minus{} 56n\equiv 0\bmod{3}$
also $ 6n^{3} \plus{} 57n^{2}\equiv 0\mod{3}$
add to get,
$ n^{5} \plus{} n^{3} \plus{} n\equiv 0\mod{3}$

[quote="Chalao"]Excuse me, i don´t understand very much the notation of the Euler's theorem since i am a begginer. I suppose that $ \varphi(n)$ means the multiplication of the divisors of $ n$ substracting 1,2,3.... for each one.

And about the polynomial equation in mod 3,i think is something like this

$ n^{5} \minus{} 5n^{3} \plus{} 60n^{2} \minus{} 56n\equiv 0\bmod{3}$
$ n^{5} \plus{} 60n^{2}\equiv 56n \plus{} 5n^{3}\bmod{3}$

but then i don´t know how did you turn it into $ n^{5} \plus{} n^{3} \plus{} n$[/quote]

No. $ \varphi(n)$ is equal to the number of positive integers less than or equal to $ n$ that are relatively prime (or coprime) to $ n$. What I wrote down in my last post was the formula for computing $ \varphi(n)$, knowing the prime factorization of $ n$.


function
combinatorics unsolved
combinatorics

Find the number f:A->A that satisfies (1) and (2).(A={1,2,3,4,5,6,7})

(1) For all $ j\in A$, there is only on $ k\in A$ that $ j \equal{} f(k)$ and $ j \neq k$.

(2) there is some function g:A->A that for all $ i\in A f(i) \equal{} g(g(i))$

Thanks.

Use \to to get a nice arrow in LaTeX.

From the first half of condition 1, $ f$ and $ g$ must both be permutations.  From the second half of condition 1, $ g$ may have neither fixed points nor cycles of length 2.  Thus, $ g$ must have one cycle of length 3 and one cycle of length 4, and so $ f$ must have a cycle of length 3 and two cycles of length 2.  Conversely, any permutation with this cycle structure is a perfect square.  Thus, we just need the number of permutations with this cycle structure, and that (using whatever your favorite method is of counting such a thing) is 210.


ratio
geometry
parallelogram
function
cyclic quadrilateral
geometry proposed

The opposite sides of a cyclic quadrilateral  $ABCD$ , produced ,  intersect at points $M,N$. Prove that the bisectors of angles  $M,N$  intersect at a point ,  which is collinear with the midpoints of the diagonals  $AC,BD$ .
    [u] Babis[/u]

Let $M \equiv AB \cap CD,\ N \equiv BC \cap DA.$ Let the bisector of $\angle ANB$ meet AB, CD at P, R, let the bisector of $\angle BMC$ meet BC, DA at Q, S, and let the 2 angle bisectors meet at X. ABCD is cyclic => $\triangle MAC \sim \triangle MDB$ are similar => $\frac{MD}{MA}= \frac{MC}{MB}= \frac{AC}{BD}.$ Similarly, $\frac{NC}{ND}= \frac{NA}{NB}= \frac{AC}{BD}.$ It follows that the bisectors of $\angle ANB, \angle BMC$ cut AB, BC in the same ratios, hence $PQ \parallel AC.$ Similarly, $RS \parallel AC$ and $QS \parallel BD \parallel SP,$ so that PQRS is a parallelogram with the diagonal intersection X. The parallelogram PQRS inscribed in ABCD and its sides are parallel to the diagonals of ABCD. For any quadrilateral ABCD, the position of the diagonal intersection X of such parallelograms is a linear function of the position of $P \in AB,$ hence the locus of X is a line. This line passes through the midpoints of its diagonals, because when $P \longrightarrow A$ resp. $P \longrightarrow B,$ PQRS degenerates into the diagonal seqments AC resp. BD.

[color=darkblue][b]Lemma 1.[/b] Let $ABCD$ be a convex quadrilateral and let $x\ne 1$, $y\ne 1$ be two real constants. Let $P\in AB$, $R\in CD$ be two mobile points so that $\overline{PA}=x\cdot \overline{PB}$ and $\overline{RC}=x\cdot \overline{RD}$. Prove that the geometrical locus of the point $L\in PR$ for which $\overline{LP}=y\cdot\overline{LR}$ is the line $MN$, where $M\in AC$, $N\in BD$ so that $\overline{MA}=y\cdot \overline{MC}$ and $\overline{NB}=y\cdot \overline{ND}$ and in this case, $\overline{LM}=x\cdot \overline{LN}\ .$
[b]Proof.[/b] I"ll use the vectorial calculus. Prove easily that $P=\frac{A-xB}{1-x}$ ; $R=\frac{C-xD}{1-x}$ ; $M=\frac{A-yC}{1-y}$ ; $N=\frac{B-yD}{1-y}\ .$ Therefore, $L=\frac{P-yR}{1-y}$ $\Longrightarrow$ $(1-y)L=\frac{A-xB}{1-x}-y\cdot\frac{C-xD}{1-x}$ $\Longrightarrow$ $(1-x)L=\frac{A-yC}{1-y}-x\cdot\frac{B-yD}{1-y}$ $\Longrightarrow$ $L=\frac{M-xN}{1-x}$, i.e. $L\in MN$ and $\overline{LM}=x\cdot \overline{LN}\ .$[/color]


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14822[/img]

 :(

Let $ P, Q$ be points on the segments $ AA', BB',$ such that $ \frac {AP}{PA'} \equal{} \frac {BQ}{QB'} \equal{} \frac {AB}{A'B'} \equal{} k.$ These are easy to construct (see below). Let the lines $ AB, A'B'$ meet at $ X.$ Then $ PQ$ is parallel to the internal bisector of the $ \angle AXA' \equal{} \angle BXB'.$ See

[url]http://www.mathlinks.ro/viewtopic.php?t=116208[/url] and
[url]http://www.mathlinks.ro/viewtopic.php?t=116208[/url] for 2 different proofs.

Let $ M, M'$ be midpoints of $ AB, A'B'.$ Then of course, the line $ PQ$ cuts $ MM'$ at a point $ R$, such that  $ \frac {MR}{RM'} \equal{} k.$ $ U, U'$ are feet of the perpendiculars to $ PQ$ from $ M, M'.$  The right $ \triangle MRU \sim \triangle M'RU'$ are directly similar with coeficcient $ k,$ having equal angles at the common vertex $ R,$ $ \frac {MU}{U'M'} \equal{} \frac {MR}{RM'} \equal{} k.$ $ V, V'$ are intersections of perpendiculars to $ AB, A'B'$ with $ PQ.$ The right $ \triangle MUV \sim \triangle M'U'V'$ are oppositely similar with coefficient $ k,$, having equal angles $ \angle UMV \equal{} \angle U'M'V' \equal{} \frac {\angle AXA'}{2}.$ $ \frac {MV}{V'M'} \equal{} \frac {MU}{U'M'} \equal{} k.$ Let $ C \in PQ$ be arbirary and $ (O)$ circumcircle of the $ \triangle ABC.$ Let $ (O')$ be a circle through $ A', B'$ similar to the circle $ (O)$ with coefficient $ k,$ such that $ O$ is on the same side of $ AB$ as $ P, Q$ $ \Longleftrightarrow$ $ O'$ is on the same side of $ A'B'$ as $ P, Q.$ Then $ (O')$ cuts the line $ PQ$ at a point $ C'$ corresponding to $ C,$ such that the $ \triangle A'B'C' \sim \triangle ABC$ are oppositely similar with similarity coefficient $ k.$. It follows that $ PQ$ is the required line.
______________________

To construct $ P,$ draw circles $ (A), (A')$ centered at $ A, A'$ and with radii $ AB, A'B',$ respectively. Parallel to $ A'B'$ through $ A$ cuts the circle $ (A)$ at $ D, E$ where $ D$ is on the same side of $ AA'$ as $ B$ and $ E$ is on the opposite side of $ AA'$ than $ B.$ Line $ BE$ cuts $ AA'$ at $ P.$ ($ P$ is the internal similarity center of the circles $ (A), (A').$) The point $ Q$ can be constructed similarly with the help of circles $ (B), (B')$ centered at $ B, B'$ and radii $ BA, B'A'.$ Alternately, construct the internal bisector $ x$ of the $ \angle AXA'$ and parallel to $ x$ through $ P$ is the required line $ PQ,$ cutting $ BB'$ at $ Q.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14860[/img]

 :(

isn't the axis

 :(

The construction works for me just fine - wanna see? The only typo is "Line $ BE$ cuts $ AA'$ at $ P$ ...", which should be "Line $ B'E$ cuts $ AA'$ at $ P$ ...". However, I did say that $ P$ is the internal similarity center of the circles $ (A), (A')$ ...

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14861[/img]

 :thumbup:


[hide=" :-) "]
Jon has 450 legos and since Sarah has ten more she has 460 and since Hubert has half of Sarah, Hubert has 230.
$230+450+460=\boxed{1140}$[/hide]click the smiley to see my answer

[quote="MCrawford"]Jon has 450 legos.  Sarah has 10 more than Jon and twice as many as Hubert.  how many legos do they have altogether?[/quote]

Click Mr Green for Answer!!!!

[hide=":D"]Jon: 450 legos.

Sarah: 450 + 10 = 460 legos.

Hubert: 460/2 = 230 legos.

450 + 460 + 230 = 1140[/hide]

[quote="Fish_Factorial"]Click the ninja for the answer!
[hide=" :ninja: "]450 + (450+10) + [(450+10)/2] = 1140 legos[/hide][/quote]

How do you do that making a smiley a [hide]thing[/hide]?

put hide thing first
[hide]this thing[/hide]

then in the the first hide thing, put

" = :| "

[hide=" :lol: "] ha [/hide]

[quote="MCrawford"]Jon has 450 legos.  Sarah has 10 more than Jon and twice as many as Hubert.  how many legos do they have altogether?[/quote]
[hide=" :spam: "]
450+(450+10)+(460/2)=450+460+230=1140[/hide]
Click spammer for answer

[quote="apusey"]you only need to post the answer once[/quote]
um, no you don't....If people want to practice their math skills then they are welcome to answer any problem they want to answer. :)

[quote=peelybonehead]Lego lego[/quote]

nice 17 year bump lol

[quote=yiz76][quote=Theavenger1945][quote=peelybonehead]Lego lego[/quote]

nice 13 year 7 month 15 day 16 hour 24 minute bump lol[/quote]

FTFY[/quote]

$2022 - 2005 \neq 13$ but ok

Fun problem! Answer is $j+(j+10)+(j+10)/2$. Letting $j=450$ gives $1140$. :)

[hide=:diablo:]
bash theorem go brrrrr

i got $\boxed{1140}$ after guess-and-check

for the math addicts, here is some math

$450+460+230=\boxed{1140} \text{ } \mathbb{Q}.\mathbb{E}.\mathbb{D}.$

[s]ez[/s]
[/hide]

Woah! A question from the legendary [url=https://www.quora.com/profile/Mathew-Crawford]Mathew Crawford[/url]. Here is a [hide=Sol]Since John has $450$ legos, Sarah must have $450 + 10 = 460$ legos and Hubert must have $\frac{460}{2} = 230$ legos. Altogether, they have $450 + 460 + 230 = \boxed{1140}$ legos.[/hide]


Prove that in any triangle the following inequality holds:

     $\frac{b+c}{b+c-a}+\frac{c+a}{c+a-b}+\frac{a+b}{a+b-c}\geq\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}$ ;)

[quote="Fermat -Euler"]i have anther solution ;)

the inégality it's equivaut has   a/b+c-a +b/c+a-b +c/a+b-c >= (b+c)/a  -1  +  (c+a)/b  -1  +  (a+b)/c  -1

equivaut has    a/b+c-a +b/c+a-b +c/a+b-c >=  b+c-a/a +  c+a-b/b  + a+b-c /c

and we know that           a/b+c-a>= b+c-a/a
                                     b/c+a-b>=c+a-b/b
                                     c/a+b-c >=  a+b-c /c

then the inégalité is proofed[/quote]
The inequality $\frac{a}{b+c-a}\geq\frac{b+c-a}{a}$ is wrong!

[quote="Fermat -Euler"]for a b c the sides of triangles  ;; why??[/quote]

why is it true?!
$a \ge b+c-a$
$2a \ge b+c$
not necessary to be true!!


topology
advanced fields
advanced fields unsolved

Let f:X->Y be a cts map from a topological space X to a Hausdorff space Y. Prove that the graph L={(x,y) in XxY: y=f(x)} is a closed subset of XxY.

If $(x,y)\notin L$, then $f(x)$ and $y$ have disjoint neighborhoods $U$ and $V$. Consider $f^{-1}(U)\times V$.


I want to do HMMT next year in 8th grade. Is it posisble?

If you can arrange for transportation, then yes.  I believe the San Diego Math Circle came last year, for example.

[quote="darkprince"]I don't live close to either Harvard or MIT; is it possible to still participate?
I'm isolated in California!
 :([/quote]

I also live in California and flew to Boston with my mom to participate last year.  I joined a semi-random team on AoPS and coincidentally we won (WOOT!!) :)


On the same day as HMMT, Rice is holding the annual Rice Math Tournament.  If we could get together a team of AoPSers, we could do very well at the contest.  Is anyone interested?  If so, post your name, state, and grade here, or PM me.  If we have enough interest, I'll administer some kind of test to find the top 20 AoPSers who want to come, but I doubt that will happen.  So start posting, and we'll pull something together.

[url=http://www.ruf.rice.edu/~eulers/RMT.html]Link[/url]

[quote="MysticTerminator"]oh my god we will just have to finish HMMT really quickly and fly to rice hopefully we can use the time zone difference to our advantage[/quote]

are you really serious about that?

[quote="MysticTerminator"]oh my god we will just have to finish HMMT really quickly and fly to rice hopefully we can use the time zone difference to our advantage[/quote]

OK, good so I'll put you down on our team.

nrz:

I don't know if Rice allows conglomerate teams (there's nothing in the rules against it) so I emailed them to find out.  If they do allow it, would you like to be on our team?  If so, PM me for more information.

[quote="nrz"]Well, I'm going to it (since I live in Texas and everything), but unfortunately I think you have to be from the same school  :P .  However, I wouldn't mind teaming up with some fellow AoPSers if I could since I'm the only one from my school going haha.[/quote]

ummm no you don't have to be from the same school. i went last year and we had a kid from another school on our team.

[u]Interested AoPSers:[/u]

mysmartmouth
nrz
PTynan89
aahmadi

That's only 4 people.  Don't make plans to go just yet, we need more interest than this.  Just assume we are not going, unless you hear otherwise, because you don't want to show up without a team...


For all positive integers $n$, find the remainder of $\dfrac{(7n)!}{7^n \cdot n!}$ upon division by 7.

Well by dividing, you get rid of every 7th term in (7n)!.  If n is an integer.  So in that case it would be 0.  I don't know the definition for n! when n is not a positive integer.

yes, that's the same solution as myth's :)

As for K81o7, no it's not 0. check myth/nr1337's answer.

[quote="Peter VDD"]yes, that's the same solution as myth's :)[/quote]

Whoops.  Didn't read the other solutions first...

Oh you divide it by 7 again...that changes things...


"Assume Theta lies in quadrant 3 and the terminal side of Theta is perpendicular to the line y=-x-5

find cos(theta) and cot(theta)"

I drew the line y=-x-5 and then at y=-5 i formed a 90degree angle with another line, since it says its perpendicular to it. I can assume Theta being 45 degree now. So am i correct to say the cos(theta) is negative squareroot of 2 divided by 2 and the cot(theta) is 1?


geometry
circumcircle
congruent triangles
angle bisector
geometry unsolved

Let $ ABC$ is a triangle, let $ H$ is orthocenter of $ \triangle ABC$, let $ M$ is midpoint of $ BC$. Let $ (d)$ is a line perpendicular with $ HM$ at point $ H$. Let $ (d)$ meet $ AB, AC$ at $ E, F$ respectively. Prove that $ HE \equal{}HF$.

http://www.cut-the-knot.org/pythagoras/Butterfly.shtml

K_Er_Tus's solution is correct !Thank you !
[hide]Let $ BH$ cut $ AC$ at $ X$ and $ CH$ cut $ AB$ at $ Y$. We have $ MX\equal{}MY\equal{}MB\equal{}MC$ so the circle from Butterfly Theorem is of center $ M$ and radius $ MB$. The line (d) cut the cercle at $ P$ and $ Q$. Hence $ MH$ perpendicular to (d), we have that $ H$ is the middle of $ PQ$. Using Butterfly theorem for chords $ YC$ and $ XB$, we have that $ HE\equal{}HF$.[/hide]

Let $ K,\ L$ be, the orthogonal projections of $ B,\ C$ respectively, on the line segment $ EF$ and we denote the points $ X\equiv BC\cap AH,\ Y\equiv AC\cap BH,\ Z\equiv AB\cap CH.$

It is easy to show that $ HK \equal{} HL,$ from $ MB \equal{} MC$ and $ BK\parallel MH\parallel CL$ and then, the points $ B'\equiv CL\cap BH$ and $ C'\equiv BK\cap CH,$ are the reflexions of $ B,\ C$ respectively, with respect to $ H.$

We have now, the configuration of two congruent triangles $ \bigtriangleup HBC',\ \bigtriangleup HB'C,$ with $ E,\ F$ as their orthocenters respectively and so, their distances from their homologous vertices, are equal.

Hence, we conclude that $ EH \equal{} FH$ and the proof is completed.

Kostas Vittas.

[quote="K_Er_Tus"]This is actually a particular case of the Butterfly Theorem.[/quote]
[quote="thanhnam2902"]Thank you very much. I like this problem. It's easy but we can solve it by many ways. Use Butter fly Theorem is a nice example.[/quote]
Can someone explain to me what Butterfly theorem is?

oh! It is very famous Theorem. I have posted it at this forum.

You see it at here:
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml

 :P  :lol:

Thank you so much!. I realized that it is a particular case of a more general result:

$AB$ is a chord of a circle $(O)$ and $P$ is a point on $AB.$ Two distintc chords $MN$ and $RS$ pass through $P.$ $MS$ and $RN$ cut $AB$ at $X,Y,$ respectively. Then we have
\[ \frac{1}{PA}-\frac{1}{PX}=\frac{1}{PB}-\frac{1}{PY} \]

Let $ M'$ the diametrically oposite point of $ A$ in the circumcircle of $ \triangle{ABC}$. $ H,M,M'$ are collinear since by angle chasing $ BHCM'$ is a parallelogram. Also, $ EHM'B$ and $ HFCM'$ are both cyclic (opposite angles equal $ \frac{\pi}{2}$, sum $ \pi$. Since $ \angle{HBE}\equal{}\angle{HCF}$ this implies $ HM$ is angle bisector of $ EM'F$ then $ EH\equal{}HF$.

[quote="Nemion"]Let $ M'$ the diametrically oposite point of $ A$ in the circumcircle of $ \triangle{ABC}$. $ H,M,M'$ are collinear since by angle chasing $ BHCM'$ is a parallelogram. Also, $ EHM'B$ and $ HFCM'$ are both cyclic (opposite angles equal $ \frac {\pi}{2}$, sum $ \pi$. Since $ \angle{HBE} \equal{} \angle{HCF}$ this implies $ HM$ is angle bisector of $ EM'F$ then $ EH \equal{} HF$.[/quote]
This is exactly what I am thinking about

Dear Mathlinkers,
for the butterfly theorem you can see a nice article written by Darij Grinberg
http://www.cip.ifi.lmu.de/~grinberg/     On cyclic quadrilaterals and the butterfly theorem
Sincerely
Jean-Louis

Two orthogonal lines through orthocenter is the signature of Droz-Farny configuration.

   The required equality of segments follows immediately.

   M.T.

If N=BH∩AC and P=CH∩AB, then BCNP is cyclic, M is it’s excenter and, from butterfly theorem, EH = FH.

Best regards,
sunken rock


calculus
integration
trigonometry
function
calculus computations

Exactly two roots.

one in $ \ 0 < a < \frac{\pi}{2} $ and the other one in $ \pi < a < \frac{3\pi}{2} $

[quote="kunny"]Find the number of root for $\int_0^{\frac{\pi}{2}} e^x\cos (x+a)\ dx=0$ at $0\leq a <2\pi$[/quote]
We will prove that this equation has exactly two roots.

Firstly, $a=0$ can not be a root since

${\int_0^{\frac{\pi}{2}} e^x\cos x\ dx = \left[\frac{e^x(\cos x + \sin x) }{2}\right]_{0}^{\frac{\pi}{2}} = \frac{e^{\frac{\pi}{2}}-1}{2}}\neq 0$

Now let $f$ be a function such that

$f(a)= \int_0^{\frac{\pi}{2}} e^x\cos (x+a)\ dx$

$f'(a)= -\int_0^{\frac{\pi}{2}} e^x\sin (x+a)\ dx$

$f''(a)= -\int_0^{\frac{\pi}{2}} e^x\cos (x+a)\ dx=-f(a)$

$f''(a)+f(a)=0 \: \Longrightarrow \: f(a)=c_1\sin a+c_2 \cos a$ for some constants $c_1$ and $c_2$

Since $a \neq 0 \: , \: c_2=f(0) \neq 0$

$f(a)=0 \: \Longleftrightarrow \: \cot a = -\frac{c_1}{c_2}$

We know that the last equation has two solutions on the interval $(0, 2\pi)$

one root only in the given interval!!!!!!!!!!!!!


Find the product
1.$ \ce{CH3MgBr} \plus{} \ce{C6H5 \sbond CH \dbond HC \sbond CHO} \plus{} \ce{H}^{ \plus{} }$
2.1,4 di oxo cyclohex-2-ene + $ \ce{HBr}$
i forgo this in the figure the double bond on the left is not with carbon but with oxygen that is it is a keto group

[hide="Answers"]1. C6H5-CH=CH-CHOH-CH3.

2. 2-bromocyclohexa-1,4-dione.[/hide]

The answers ar
1.$ \ce{C6H5 \sbond CH(Me) \minus{} CH2 \minus{} CHO}$
2.o-Bromo hydroQuinone
For 1 it says there is a 1,4 addition  :maybe:
and for two i beleive the starting product is Quinone  and not that given .Carcul what is ur thought over this

Surely the alpha-beta unsaturations lead to a reaction called Micheal reaction.
It is basically 1,4-addition. It occurs because the nucleophilicity of the carbonyl carbon atom is less than the beta carbon which has a higher positive charge on it due to resonance

                                                +
C6H5-CH=CH-CH  <-->   C6H5-CH-CH=CH
                      ||                                   |
                      O                                   O-

That's why when we reduce cinnamaldehyde by LiAlH4 we get,

C6H5-CH2-CH2-CH2OH

First 1,4-addition and then the H- attacks the C=O group.[/img]

[quote="Pardesi"]For 1 it says there is a 1,4 addition[/quote]

I don't agree with that, and so I think your book is wrong. Not only the Grignard reagent used is very small, but also with alpha,beta-unsaturated aldehydes 1,2-addition is the preferred pathway because the carbonyl group is stereochemically uncrowded. For other alpha,beta-unsaturated carbonyls, the product obtained depends on the structures of both carbonyl and Grignard reagent. Some years ago I have written a monograph about Grignard reagents, featuring their structure, preparation, reactions, and other aspects of their chemistry, with more than 100 pages - so I think I am in a good position to talk about Grignard reagents.

[quote="Pardesi"]o-Bromo hydroQuinone[/quote]

That product cannot be formed from the molecule you gave, unless air oxidation or other oxidizing agent is allowed.

[quote="Chemrock"]Surely the alpha-beta unsaturations lead to a reaction called Micheal reaction.[/quote]

Wrong. The Michael reaction (or addition) is the name given only to the [i]conjugate addition of enolate anions to alpha,beta unsaturated carbonyls[/i].

In general, strong nucleophiles like Grignard reagents and organo lithium compounds give mainly 1,2-addition, while weaker nucleophiles like enolates, Gilman reagents, amines, alcohols, and cyanide give mainly 1,4-addition.

[quote="Chemrock"]It occurs because the nucleophilicity of the carbonyl carbon atom is less than the beta carbon which has a higher positive charge on it due to resonance[/quote]

Wrong again, that depends on the structure of the carbonyl. Also, for strong nucleophiles stereochemical factors are more important than electronic ones.

also what about the question in the diagram

Somehow, I can never see the pictures you post on this forum. Can you describe with words what is on your diagram?

Since i could not name the compound so i drew it anyway
the compound is basically a six membered Carbon ring .C1 is attached by double bond to O
C2-C3 have a double bond between them. C4 is attached to two oxygen atoms of the group obtained by removing the hydrogens bonded to oxygen of glycol.
rest all are normal
and this is reacted with a acid

So we have a carbonyl group protected as a cyclic acetal: then, with acid the acetal is reconverted to the carbonyl.


inequalities
trigonometry
inequalities unsolved

i think  it isnot  a  hard  inequality.   it  is     important  for  me. please  interest  with  my  question  i am waiting your solutions to my question   :lol:

[hide]
First, we have
\[ 2\minus{}A\equal{}\cos^{2}\alpha\plus{}\cos^{2}\beta\plus{}\cos\gamma\]
Now use
\[ \cos^{2}\alpha\plus{}\cos^{2}\beta\plus{}\cos^{2}\gamma\plus{}2\cos\alpha\cos\beta\cos\gamma\equal{}1\]
to get
\[ 2\minus{}A\equal{}1\minus{}2\cos\alpha\cos\beta\cos\gamma\plus{}\cos\gamma\minus{}\cos^{2}\gamma\]
\[ 1\minus{}A\equal{}\cos\gamma (1\minus{}\cos\gamma\minus{}2\cos\alpha\cos\beta )\]
Now note that
\[ 2\cos\alpha\cos\beta\equal{}\cos (\alpha\plus{}\beta)\plus{}\cos (\alpha\minus{}\beta)\equal{}\cos (\alpha\minus{}\beta)\minus{}\cos\gamma\]
so that
\[ 1\minus{}A\equal{}\cos\gamma (1\minus{}\cos (\alpha\minus{}\beta))\]
We must minimize the above. If $ \gamma\leq\frac{\pi}{2}$, then it is nonnegative. Otherwise, $ \cos (\alpha\minus{}\beta)\geq\cos (\alpha\plus{}\beta)\equal{}\minus{}\cos\gamma$ so that
\[ \cos\gamma (1\minus{}\cos (\alpha\minus{}\beta))\geq\cos\gamma (1\plus{}\cos\gamma)\geq\minus{}\frac{1}{4}\]
Therefore the minimum for $ A$ is $ \boxed{\frac{5}{4}}$, true for the degenerate
\[ \alpha\equal{}\frac{\pi}{3},\beta\equal{}0,\gamma\equal{}\frac{2\pi}{3}\]
[/hide]

I'm sorry, but scorpius119, how did you get

$ \cos^{2}\alpha\plus{}\cos^{2}\beta\plus{}\cos^{2}\gamma\plus{}2\cos\alpha\cos\beta\cos\gamma\equal{}1$ ?

Is this a property?
Thanks for your input

It is well known equality for all triangles. May express cosA, cosB, cosC with Cosine law. And you will see it is true. It was propsed as a problem in Bulgarian Math Olympiad in 1963-III round. There are lots of proofs.


How many distinct, natural-number factors does $ 4^3 \cdot 5^4 \cdot 6^2$ have?

Does it mean to say what the overall product's number of factors are?

No.  The problem states how many factors $ 4^3 \cdot 5^4 \cdot 6^2$ has.

Hey, it works.  And I don't have LaTeX.

i mean do you have to simplify it first?

Not sure... never learned how to do this...

I do know, however, that the number is $ 1440000$.  Not sure if this will help, however.

What it is asking for is what are the positive integral divisors of $ 4^3\cdot5^4\cdot6^2$.

First completely prime factorize to get $ 2^6\cdot5^4\cdot2^2\cdot3^2\equal{}2^8\cdot3^2\cdot5^4$. Now add one to each exponents and multiply to get: $ 9\cdot3\cdot5\equal{}\boxed{135}$

Would this method be used for all the numbers, such as $ 2^8 \cdot 3^4 \cdot 5^7 \cdot 23{}^1{}^4$?

Cough Cough.

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

1) Prime factorization $ 2^8\cdot 3^2\cdot 5^4$
2) Add one to each exponent and multiply $ 9\cdot 3\cdot 5\equal{}135$

[quote="ernie"]
Hey, it works.  And I don't have LaTeX.[/quote]

On the forum, you don't need to "have" $ \text{\LaTeX}$. It's sort of "built in".

@dragon96:

I shall repeat myself for the umpteenth time. Kindly [b]do not[/b] post the same solution with slightly different wording.

Thanks, hunter34. I get it. (and yes, ernie too.)


Find the polynomial $ P$ of smallest degree with rational coeffcients and leading coefficient $ 1$ such that $ P(\sqrt[3]{49}\plus{}\sqrt[3]{7})\equal{}4$.

[hide]
Since $ \sqrt[3]{49}\plus{}\sqrt[3]{7}$ is irrational and the polynomial coeficients are rational the polynomial has a degree greater than $ 1$

For $ n\equal{}2$...
$ (\sqrt[3]{49}\plus{}\sqrt[3]{7})^2 \equal{} \sqrt[3]{49} \plus{}7\cdot \sqrt[3]{7} \plus{}14$

$ \Rightarrow P(\sqrt[3]{49}\plus{}\sqrt[3]{7})) \equal{} (\sqrt[3]{49}\plus{}\sqrt[3]{7})^2 \minus{} a(\sqrt[3]{49}\plus{}\sqrt[3]{7}) \minus{} b \equal{} 4$

$ \Rightarrow 6\cdot \sqrt[3]{7} \plus{}56 \minus{}b \equal{} 4 \Rightarrow 6\cdot \sqrt[3]{7}\equal{}4\plus{}b\minus{}14$ which is impossible...

For $ n\equal{}3$
$ (\sqrt[3]{49}\plus{}\sqrt[3]{7})^3 \equal{} 21\sqrt[3]{49}\plus{}21\sqrt[3]{7}) \plus{} 56$

$ \Rightarrow P(x) \equal{} x^3 \minus{} 21x \minus{}52$
[/hide]

[quote="ocha"][hide]
Since $ \sqrt [3]{49} \plus{} \sqrt [3]{7}$ is irrational and the polynomial coeficients are rational the polynomial has a degree greater than $ 1$

For $ n \equal{} 2$...
$ (\sqrt [3]{49} \plus{} \sqrt [3]{7})^2 \equal{} \sqrt [3]{49} \plus{} 7\cdot \sqrt [3]{7} \plus{} 14$

$ \Rightarrow P(\sqrt [3]{49} \plus{} \sqrt [3]{7})) \equal{} (\sqrt [3]{49} \plus{} \sqrt [3]{7})^2 \minus{} a(\sqrt [3]{49} \plus{} \sqrt [3]{7}) \minus{} b \equal{} 4$

$ \Rightarrow 6\cdot \sqrt [3]{7} \plus{} 56 \minus{} b \equal{} 4 \Rightarrow 6\cdot \sqrt [3]{7} \equal{} 4 \plus{} b \minus{} 14$ which is impossible...

For $ n \equal{} 3$
$ (\sqrt [3]{49} \plus{} \sqrt [3]{7})^3 \equal{} 21\sqrt [3]{49} \plus{} 21\sqrt [3]{7}) \plus{} 56$

$ \Rightarrow P(x) \equal{} x^3 \minus{} 21x \minus{} 52$
[/hide][/quote]

I don't understand where your $ n$'s came from, and how you got $ (\sqrt[3]{49}\plus{}\sqrt[3]{7})^2 \equal{} \sqrt[3]{49} \plus{}7\cdot \sqrt[3]{7} \plus{}14$.  Thanks!

How was $ \Rightarrow P(\sqrt[3]{49}\plus{}\sqrt[3]{7})) \equal{} (\sqrt[3]{49}\plus{}\sqrt[3]{7})^2 \minus{} a(\sqrt[3]{49}\plus{}\sqrt[3]{7}) \minus{} b \equal{} 4$ gotten?

Okay sorry I should have explained.

Suppose the polynomial was degree $ 2$ and had a leading coeffiecient of $ 1$. Then it would be in the form
$ P(x) \equal{} x^2 \plus{} ax \plus{} b$, where $ a,b$ are rational

Now we have $ P(\sqrt[3]{49} \plus{} \sqrt[3]{7}) \equal{} 4$. If we sub that in and move the equation arround we get $ a$ and $ b$ on one side in a rational expression and on the other side we have an irrational expression, contradiction...Hence the polynomail degree is greater than 2.


When he is asked for his phone numbers, professor X usually says that it is a 7-digits number which is included in first 10000 digits of $\pi$ and $\sqrt 2$ (for example, number 346 is included in numbers 3.12[color=red]346[/color]234 and 4.12345677[color=red]346[/color]87). Most people consider him weird and ask no further questions. But, his student Y said (after thinking for a while) that he didn't have enough informations. The professor replied that the last two digits of his phone number are equal. After that student knew professor's phone number. Do you know it?

my computer says there is no such number

i am sorry...i didn't read well the proble, and i used my comp only for the first 1000 digits.  10,000 is far too much and i really don't know where to get from the first 10,000 digits of sqrt(2). :) so, i won't try it :)


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

we have more than $1000000$ points in space prove that we have at least 79 different distance.

I would like to know more about the source "Iran 2005". What I know is that the Iran TST 2005 has already taken place, and the Iranian team for IMO 2005 has been announced. So what's that ??  :?

mr erdos

this question is for the second round

one of the exams for selcting the team for next year 2006

there are 4 rounds in iran to select the team.

In general let there be N points in a d-dimensional space. What is the minimum number of distinct distances?

Answer. let P1,...,PN be the points and let m be the number of distinct distances.
Then m>= [(N-d)/2]^1/d , in case of N=10^6 and d=3 result follows.  m>=80.

Now to prove we do the following:
Let the distances be d1<d2<...<dm
Let us consider P1: draw spheres of radii d1,d2,...,dm.
Draw spheres of radii d1,...,dm about point P1.
Do the same thing for P2,P3,...,Pd-1.
Now consider Pd. any point other than Pd should either be among {P1,...,Pd-1} or should lie on the intersection of a sphere of radius dj (1<=j<=m) with all the spheres of P1,...,Pd-1. This means that the number of points with distances di1,di2,...,did-1 from P1,...,Pd-1 respectively are at most 2m. so since the number of tuples is m^(d-1)
we have that 2d^m + (d) >= N thus getting the result.
-Ali

See http://www.mathlinks.ro/Forum/viewtopic.php?t=34948
I posted it yesterday


One jar contains $ 40\%$ purple $ n\&n$'s. A second jar contains $ 60\%$ purple $ n\&n$'s. The second jar contains $ 50\%$ as many $ n\&n$'s as the first jar. When the contents of the two jars are mixed together, exactly what percent, of the $ n\&n$'s in the mixture are purple? Express your answer as a mixed number.

Let's say there are 20 $ n\&n$'s in the first jar.
Then, 8 are purple.
In the second jar, there are ten $ n\&n$'s,
so there are 6 purple ones.
So the entire percentage of purple $ n\&n$'s is $ \frac{8\plus{}6}{20\plus{}10} \equal{} \frac{7}{15} \equal{} \boxed{46\frac{2}{3}\%}$


[quote="The soul of rock"]Prove that for all $a,b,c > 0$ we have
$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le \frac{1}{3}$[/quote]
Let $a=\min\{a,b,c\}.$ Then $\sum_{cyc}\frac{a}{4a+4b+c}\leq\frac{1}{3}\Leftrightarrow$
$\Leftrightarrow9a(a^{2}+b^{2}+c^{2}-ab-ac-bc)+(4b+c-5a)(a+b-2c)^{2}\geq0.$ :)

[quote="The soul of rock"]Prove that for all $ a,b,c > 0$ we have
$ \frac {a}{4a \plus{} 4b \plus{} c} \plus{} \frac {b}{4b \plus{} 4c \plus{} a} \plus{} \frac {c}{4c \plus{} 4a \plus{} b}\le \frac {1}{3}$[/quote]
This inequality is [b]Problem 2.18 - page 56[/b] in new book (where it is presented with a nice solution by Cauchy-Schwarz):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://can-hang2007.blogspot.com/2009/12/new-book-inequalities-with-beautiful_20.html][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.


algebra
polynomial
calculus
derivative
calculus computations

Here is a problem that I came up with after looking at the graphs of different polynomial functions.  Is it a well-known statement or something?

Show that if $ x_0$ is a root with multiplicity two or greater of a polynomial $ P(x)$, then $ y \equal{} P(x)$ has a critical point at the point $ (x_0,P(x_0))$.

[hide="What I Did"]
Let $ P(x) \equal{} a (x \minus{} x_0)^n Q(x)$, where $ n \geq 2$ is the multiplicity of the root $ x_0$, $ Q(x)$ is a polynomial in $ x$ (it basically contains any other factors that $ P(x)$ may contain when it is in factorized form), and $ a$ is the leading coefficient of $ P(x)$.

Then,

$ P'(x) \equal{} a[n(x \minus{} x_0)^{n \minus{} 1}Q(x) \plus{} (x \minus{} x_0)^n Q'(x)]$
$ P'(x) \equal{} a(x \minus{} x_0)^{n \minus{} 1}[nQ(x) \plus{} (x \minus{} x_0)Q'(x)]$

$ P'(x_0) \equal{} 0$ (for $ n \geq 2$, which means that $ n \minus{} 1 \geq 1$), so $ P(x)$ has a critical point at $ x \equal{} x_0$.[/hide]

Can you show under what conditions that that critical point is a relative maximum, a relative minimum, or neither?

It is well-known that if $ Q$ has a root of multiplicity $ k$ at $ x_0$, then $ Q'$ has a root of multiplicity $ k\minus{}1$ at $ x_0$. (assuming $ k\ge 2$).
Concerning your second question: consider the value of $ Q''$, and possibly of higher derivatives, at $ x_0$.

So, like I did above:

$ P(x) \equal{} a (x \minus{} x_0)^n Q(x)$

$ P'(x) \equal{} a[n(x \minus{} x_0)^{n \minus{} 1}Q(x) \plus{} (x \minus{} x_0)^n Q'(x)]$
$ P'(x) \equal{} a(x \minus{} x_0)^{n \minus{} 1}[nQ(x) \plus{} (x \minus{} x_0)Q'(x)]$

Now looking at the second derivative (for $ n \geq 2$):

$ P''(x) \equal{} a[ (x\minus{}x_0)^{n\minus{}2} [n Q(x) \plus{} (x\minus{}x_0)Q'(x)] \plus{} (x\minus{}x_0)^{n\minus{}1}[nQ'(x) \plus{} Q'(x) \plus{} (x\minus{}x_0)Q''(x)] ]$
$ P''(x) \equal{} a(x\minus{}x_0)^{n\minus{}2} [ n Q(x) \plus{} (x \minus{} x_0) Q'(x) \plus{} (x \minus{} x_0)((n\plus{}1) Q'(x) \plus{} (x \minus{} x_0) Q''(x)) ]$
$ P''(x) \equal{} a(x\minus{}x_0)^{n\minus{}2} [ n Q(x) \plus{} (x \minus{} x_0) ((n\plus{}2)Q'(x) \plus{} (x \minus{} x_0) Q''(x)) ]$

If $ n \equal{} 2$, $ P''(x_0) \equal{} n Q(x_0)$.
If $ n > 2$, $ P''(x_0) \equal{} 0$.

Hmmmm....


How do you hide stuff, like highlight in blue so that when you click it, it expands.

-rts2007

Honestly, there's a "Hide" button right there when you're posting.  :D

[code][hide=Click to show stuff]Stuff that is hidden.[/hide][/code]

In the future, you can avoid having to make a new topic about these kinds of things by hitting "Quote" on someone else's post (who is using what you want to be using) and look at the "source code" of their post to see how they do it.


[hide]Subtract exponents, keep the base number.

Exponents: $2x+3-(3-2x) = 4x$

Answer: $3^{4x}$[/hide]

[quote="Xantos C. Guin"]$\frac{3^{2x+3}}{3^{3-2x}} = 3^{(2x+3)-(3-2x)} = 3^{4x} = 81^{x}$[/quote]

I wonder why none of us above thought of that final step yet.


Any positive integer can be written as the sum of two or more distinct Fibonacci numbers ($ F_0 \equal{} F_1 \equal{} 1$ are not distinct), but generally not uniquely.

How many ways can $ F_{2008} \plus{} F_{2006}$ be written as the sum of distinct Fibonacci numbers?

[hide="You probably want to..."] Write everything in [url=http://www.google.com/search?hl=en&q=fibonacci+base&btnG=Google+Search]Fibonacci base[/url].  :) [/hide]


If you would like to sit at the MO table at nats, please sign up here (4 spots per meal):

Thursday Dinner

Friday Breakfast

Friday Lunch

Friday Dinner

Saturday Breakfast

Saturday Lunch

Sunday Brekfast?

but missouri is so beastly that every1 would want to sit with them

[quote="dnkywin"]Dude. This topic is spam even by Missouri standards. You all know that the following  teams are much beastlier:
[list]California
Maryland
Massachusetts
North Carolina
Texas
Washington
[/list]

There is no dispute. No one likes Missouri. They flock here only to spam, or to correct it.[/quote]

i love how north carolina is arbitrarily entered in there? or i could be mistaken, they might be well known on aops

[quote="Mewto55555"][quote="CubeX"]Uhh.. dude, i think we would rather sit at our own team tables, wouldn't we? :maybe:[/quote]

I forget you are a nats newbie. At most of the meals, there are like tables of 8, and there are no team tables persay, more of a general mingling.

Plus if I remember correctly MO totally whooped NC last year at nats?[/quote]

hmm.... by .75 of a point. and I did better than all of the people on the MO team ('cept 1)

[quote="xpmath"]NC<MO, unless Stella fails.

However, Calvin>anybody from MO.

Dude I will just come and sit with you guys at random times.[/quote]

Part 1: I think we had better worry about me. I get either cd, or >57th place =O

2. Not true, Harriet Truman was from MO.

3. At first I thought that with said "on", and I was like, im glad you arent ubey!

Dude everyone knows that CA team is pwnage this year. We have 2 potential moppers  :o

[quote="xpmath"]

Dude I will just come and sit with you guys at random times.[/quote]

Me too....

[quote="xpmath"]
EDIT: In response to that "'cept 1" I will comment that quite a lot of people beat everybody on TX 'cept 1.  [/quote]

Hey, same here!  :wink:

[quote="xpmath"]No Runpeng came 16th, you came 21st.

EDIT: In response to that "'cept 1" I will comment that quite a lot of people beat everybody on TX 'cept 1.  :wink:

EDIT2: If you read this, you should get on FTW.[/quote]

darn, i wish i was part of this exclusive "'cept 1"

lol mewto.  People are really paying attention to your sign up sh__t...I'm pretty sure there will be [b]at max[/b] four people sitting at the "Missouri Table," out of the eight seats.  Not that the tables are labeled or anything.

No, seriously, I will be sitting with Missouri quite a lot. Probably as far away from Max as possible, but still.


calculus
integration
limit
logarithms
search
calculus computations

No, that doesn't work. Don't go looking for an elementary antiderivative- there isn't one.

The limit (the improper integral) is finite, and the source is a hint.

Yes jmerry , the source is a hint and the answer is :$- \frac{{\pi ^2 }}{6}$ but how to show it using integral calculus?!

What is the power series of $\ln(1-t)?$

You should be looking for a way to turn the integral into a series.

$ln(1-t)=-\sum_{k \geq 1} \frac{t^k}{k}$
it's a classical inversion integral-sum

[quote="maskman"]Find $\mathop {\lim }\limits_{x \to 0} \int\limits_x^{1 - x} {\frac{{\ln (1 - t)}} {t}dt}$[/quote]

simple - integrate Taylor series, apply limit and calculate Fourier series to y = x^2n to calculate Zeta(2).

[quote]... calculate Fourier series to y = x^2n to calculate Zeta(2).[/quote]
... or any of a dozen or so other methods, all of which work. Search the site and you'll find more.


trigonometry
analytic geometry
graphing lines
slope

Im trying to write a proof for a problem and I end up using $\tan{(90-\theta)}$ and im trying to solve in terms of another value and just $\tan{\theta}$
I know that
$\tan{a \pm b}= \frac{\tan{a}\pm \tan{b}}{1 \mp \tan{a}\tan{b}}$

thus
$\tan{90-\theta}= \frac{\tan{90}-\tan{\theta}}{1+\tan{90}\tan{\theta}}$

but $\tan{90}$ cannot be calculated.. so what would I end up doing here?

I was going to use the fact that $\tan{\theta}= \frac{\sin{\theta}}{\cos{\theta}}$ but was looking for a shorter method and if the above can be done without converting to cosine and sine, then thatd be awesome :o

What I am trying to simplify is:

$\frac{1}{AB}= \frac{1}{AB(1+\tan{\theta})}+\frac{1}{AB(1+\tan{(90-\theta)})}$

Thanks :o

sin(90-theta)=cos(theta)
cos(90-theta)=sin(theta)

tan(90-theta)=cos(theta)/sin(theta)=cot(theta)

I think that should be the right answer :)

well, if you know your proof for the tangent identity, you know that the problem arises from dividing by cos (90) cos -theta=0...the whole (sin a cos b+sin b cos a) / (cos a cos b- sin a sin b) reduces to what rofler said

I always think of tangent as the slope between the origin and the point on the unit circle

that gives me a graphical way for my mind to derive identities for tangent


function
real analysis
real analysis theorems

Please someone give a hint to the following.
If $ f$ is a function of bounded variation on $ [a,b]$ and if $ f$ is absolutely continuous on $ [a,c]$ for some $ c$ s.t. $ a<c<b$ then show that $ f$ is absolutely continuous on the entire interval $ [a,b]$

That makes no sense.

Let $ f(x)\equal{}\begin{cases}0&0\le x\le 1\\1&1<x\le 2\end{cases}$

Then $ f$ is of bounded variation on $ [0,2]$ and absolutely continuous on $ [0,1],$ but certainly not absolutely continuous on $ [0,2].$

I think you didn't say what you meant to say. In particular, I question the words "for some."

What did you really mean?

Yes it is "for all $ c$". I m extremely sorry.

Then it's still false unless you add in the condition that $ f$ is continuous on $ [a,b].$

If $ f(x)\equal{}0$ on $ [a,b)$ and $ f(b)\equal{}1,$ then $ f$ is absolutely continuous on $ [a,c]$ for all $ c\in(a,b).$

I think you need continuity at $ b$ to have anything.

[quote="Kent Merryfield"]
I think you need continuity at $ b$ to have anything.[/quote]
Sorry. Yes the function is continuous  :(


The product of 2 consecutive two-digit numbers is 8010. What is the sum of the two numbers?

hello, let $ n$ and $ n \plus{} 1$ two consecutive natural numbers, then we have the equation $ n(n \plus{} 1) \equal{} 8010$, solving the quadratic equation we get $ n_1 \equal{} 89$ or $ n_2 \equal{} \minus{} 90$, but $ n_2$ is not a natural number and our sum is $ 89 \plus{} 90 \equal{} 189$.
Sonnhard.

hello, yes that's what i meant, sorry.
Sonnhard.


Two swimmers positioned in opposite sides of a rectangular swimming pool and in adjacent lines, begin to swim in the same instant, with constant speeds.
In the first two times in which both were side by side, they was swimming in opposite directions: in the first time, to $ 15$ meters of an edge and, in the second time, to $ 12$ meters of another edge.
Find the length of this swimming pool.

Letting $ x$ denote the length, in meters, of the pool, we have:

$ \frac {2x \minus{} 15 \minus{} 12}{15} \equal{} \frac {27}{x \minus{} 15} \Rightarrow\frac {2x \minus{} 27}{15} \equal{} \frac {27}{x \minus{} 15}$

Solving, and discarding the extraneous solution, we get: $ x \equal{} \frac {57}{2} \equal{} 28.5$ meters.

I have a feeling that I made this problem way too complicated.

Suppose the length of the swimming pool is $ L$ metres. Let the swimmers’ speeds be $ v_A$ and $ v_B$ metres per second. Suppose they meet first time round after $ t_1$ seconds, 15 metres from swimmer A’s side. Then $ v_At_1 \equal{} 15$ and $ v_Bt_1 \equal{} L \minus{} 15$ and so $ \frac {v_B}{v_A} \equal{} \frac {L \minus{} 15}{15}$.

Suppose they next meet again $ t_2$ seconds after their first meeting. During this time, swimmer A will have swum a distance of $ (L \minus{} 15) \plus{} 12$ metres and swimmer B $ 15 \plus{} (L \minus{} 12)$ metres. Hence $ v_At_2 \equal{} L \minus{} 3$ and $ v_Bt_2 \equal{} L \plus{} 3$ and so $ \frac {v_B}{v_A} \equal{} \frac {L \plus{} 3}{L \minus{} 3}$.

$ \therefore\ \frac {L \minus{} 15}{15}\ \equal{} \ \frac {L \plus{} 3}{L \minus{} 3}$

Solving that should give $ L \equal{} 33$.

Oh wow. I forgot that the distance from a wall the second they meet is measured from the other wall...stupid me.


calculus
integration
function
real analysis
calculus computations

What integral are we talking about ? Clearly the function $x \rightarrow [x]$ is not Lebesgue-integrable (not Riemann-sommable), and even the limit $\int_{-a}^{a}{[x]dx}$ as $a \rightarrow +\infty$ does not exist.

Probably, socrates wants the indefinite integral (the antiderivative)  ;)


induction
modular arithmetic
number theory proposed
number theory

Given any positive integer n, there exists a power of 2 such that it last n digits are either 1 or 2.

The proof follows from the following two facts:

i) There exists (unique) $ n$-digit number $ m$, consisting of digits 1 or 2 only, that is divisible by $ 2^n$. It is easy to prove by induction on $ n$.

ii) 2 is a primitive root modulo $ 5^n$ for any positive integer $ n$. This is a classic fact, exposed in every textbook on number theory.

To get an answer to the original question - take $ m$ defined in i) and find such $ k\geq n$ (existing by ii) that:
\[ 2^k\equiv m\pmod{5^n}
\]
from where it also follows that
\[ 2^k\equiv m\pmod{10^n}
\]
as required.

P.S. the digit 1 in the this problem can be replaced by any other odd digit.


Anyone have the state tests for previous years?  I need them for extra practice on the state competitions.  If so, e-mail them to smarteryaman@yahoo.com

Try this address if you didn't try:
mathcounts.org/Problems/2002Competitions.html
It's tests for 2002

i went there, but when I clicked on the link it said page not found...

I forgot to tell... You are able to see questions but I don't know why the solution doesnt' come up...
2003 works definitely fine, though...


summer program
Mathcamp
logarithms
MATHCOUNTS
algorithm

now that its way past deadline to reply and prolly most of the waitlisters have been contacted... who is for sure going to be in portland?  Introduce urself.

Hi, I'm Susan

i play piano and violin, am currently taking er hu lessons in china, and i like chess (though i'm very bad at it.)

I'll listen to any music cept country.

looking forward to seeing you all in portland!

I am. I got taken off the waitlist!  :lol:  :lol:  :lol:

So yeah. I like math. A sophomore in Iowa. Yeah. Okay I'm out of ideas.

I'm going! I'll DEFINITELY be the least smart person there... but hey, that's why I'm going.
If a girl gets all excited when Mexico or Latin America is mentioned, that's me. :)

Counting down the days...

[quote="bubala"]I am :)

Hi, I'm bubala.  I don't give personal info via the internet.  Lo siento mucho.

But I'll tell you my first name is.......[hide][hide][hide]I told you I don't give out personal info on the internet, silly[/hide][/hide][/hide][/hide][/quote]
you have an extra [/hide]  :) [/pickiness]
hi bubala, I hope to see you at mathcamp! (although I'm not 100% sure, I would really like to visit mathcamp for one weekend - preferably during the puzzle hunt)

sure, but there will be like $120 jet blue fee.

by the way, it took me three years to get rid of that "-" between gee and hoon, but now you're writing it again  :)

I am  :D

And btw, Bubala's comment brings up an exellent point. You are bound to dicusover someones identity on a given forum that your both are active while at mathcamp/some other place. You need to be very sure you dont start assoiated that persons username with their real name, as a large section of people dont like their real name getting released on the web   :) .

Anyways, see you all in 23 days 12 hours +/- 12 hours  :D

I'm coming!

If you find that the day of the week has suddenly changed to Wednesday and there is some watermelon in front of you, along with some music and tea, I'm probably around.

I go by Yasha. After a nonexistent day, I will wear a blue nonexistent item over a green similarly nonexistent item, and perhaps a third one. Until then you can recognize me by...um people calling me Yasha I guess. Or that I look like my picture on the faceboard.

I look forward to seeing everyone coming there!

Me!

Er...can someone teach me how to speedcube? :-D Hehe...I can uh...teach them frisbee! or bridge! or uh...well, not math...but nevm.

Anyways, I'm Daniel. And yeah...if you see Susan owning someone at pool, it'll probly be me. :-)

I'm going too.  As I'm a rather... umm... unnoticeable and unintelligent person I doubt saying anything will make myself identifiable.  Unless maybe a little blue light appears, but it probably won't.

...but yes, I will see you all on July 3.

[quote="Yasha"]After a nonexistent day, I will wear a blue nonexistent item over a green similarly nonexistent item, and perhaps a third one. [/quote]

Hahahaha...  :lol:

[quote="probability1.01"]I'm going!  If you see a tall guy there's a $1-\ln (\sinh 1 + \cosh 1)$ chance it's me :P[/quote] :D  :lol: That's funny. Any other clues as to your looks? :)

I'll be coming also to visit from 20th July to 26th of July (or something like that) :) See y'all there.

Well, I'm Chinese and have a lot of MathCounts shirts.  I'm afraid that doesn't help distinguish much.

[quote="MithsApprentice"]Er...can someone teach me how to speedcube? :-D Hehe...I can uh...teach them frisbee! or bridge! or uh...well, not math...but nevm.[/quote]

I would think you can teach yourself to speedcube easier than someone else.  Just go here:

http://www.ws.binghamton.edu/fridrich/cube.html

Well, my friend made me show him algorithms, I guess because he didn't want to bother with notations.  But, you are mathematically inclined, so notation should be no problem.  I think the two biggest parts to speedcubing are memorizing algorithms and developing finger shortcuts.
****

I'm there.  My name's Aaron.  A hairy white nerd, I guess.  I just graduated high school.  Thank God.  If idiots could fly that place would be a freakin' airport.

I love chess.  Well, I used to.  I also like billiards, juggling, basketball, math, minesweeper, joggling, pretty landscapes, games, science, sports, cold climates, puzzles, reading, biking, and I'm allergic to excersice after eating.  Maybe I can break out my hives for you at camp.

I await with great anticipation.

Read the first post...

[quote]
who is for sure going to be in [b]portland[/b]?

[...]

looking forward to seeing you all in [b]portland[/b]![/quote]

The location changes from year to year however.

oh, i was just wondering if there was other locations.


Four set equation below, each distinct letter represents a different non-zero digit(AB mean A represent ten digit, B represend one digit)

AB*F=JG
AD^E=AHJ
C^E=EC
B^D=HB

what is the value of G^E when expressed as in integer?

1. First, We note that [b]E must be 2[/b] because AD^E is a 3 digit integer. Quickly we find that [b]C is 5[/b] (since 25 is the only square that begins with 2). Next, we look at AD^E = AHJ. They both start with A, so [b]A is obviously 1. [/b]Therefore, D must be either 3 or 4 (if D was 5, 15^2 = 225). However, if D was 4, then B would equal 3, and this does not satisfy B^D = HB. Therefore, [b]D = 3[/b]. Now we also know that [b]H = 6 and J = 9[/b]. We also have: B^3 = 6B (sixty-b). Therefore, [b]B is 4[/b]. Finally, we look at the first equation: 14*F = 9G. 14*7 = 98, so [b]F = 7 and G = 8[/b]

A = 1
B = 4
C = 5
D = 3
E = 2
F = 7
G = 8
H = 6
J = 9

Therefore, G^E = 8^2 = [b]64.[/b]

Even though it seems like a lot, it would only take a minute to complete if you worked quickly.

[hide]
E must be two, because AD is a two digit number and if E was 3 or higher it would be too much for a three digit integer.

We quickly see that A is 1, because there is no other possibility of the tens digit being equal to the square's hundreds digit.

Let's list what we've got:

E=2
A=1
Then, 1B*F=JG
1D^2=1HJ
We want H to be either 2 or 6 or 8, because others are not possible.  13^2=169
C^2=2C
B^D=HB.
B^3=6B
B=4
14*F=JG
14*F=9G
F is most definitely 7 and G is 8.
Obviously, C is just a distraction.
G^2 is what we're aiminig for.
64 is the answer[/hide]

[quote="Big M"]Four set equation below, each distinct letter represents a different non-zero digit(AB mean A represent ten digit, B represents one digit)

AB*F=JG
AD^E=AHJ
C^E=EC
B^D=HB

what is the value of G^E when expressed as in integer?[/quote]
[hide]E must be 2 since the only integer power that allows two digit numbers to become a 3 digit number is 2.
from this we can ascertain A as 1, because 2 has already been used and 3 doesn't work.
Using the last equation we know that since D is at the very least 3, B is relatively easy to get.
Therefore B is 4.
Also, we now know the value of H, 6.
so therefore J is 9 because the square with 6 in the middle is 169. and D is 3.
so 14*F=9G
F must be 7(only one working) and G is therefore 8.
Therefore 8^2 = 64 and 64 is the answer.
[/hide][/hide]

[quote="usaha"]hmm.. talk about luck.. i jsut assumed ad was 11 (since 11^2 produces a hundreds value of 1)-once you get that you work from there by guess and checking.. dont use this way though..[/quote]

AD isn't 11. For one A and D cannot both be the same integer

sry i meant 13..too late.. i cant write.. once you get that.. something to the second power equals 2 something.  that something is 5.. since there distinct now we have limited numbers to work with for b and d.. and only 4^3=64, making distinct values of B and H.  so we have a has 1, b as 6, so 14x=9y where y is a digit.. only multiple of 14 in the 90s is 98 so G is 8, checking we have different values for the numbers.. so 8^2=64

[hide="solution"]We test stuff for C^E=EC, ut we find only 5^2=25 works.

For AD^E=AHJ, A obviously is 1.

For D=2, H and J would be equal.

For D=4, B^4=9B, which is impossible.

D≥5 means A is 2, but everyone is therefore sad.

So D=3, H=6, and J=9.

B^3=6B

The only solution is B=4

14*F=9G

F can only be 7 ad G can only be 8.

G^E=8^2=64[/hide]


Let $ PQRS$ be a square piece of paper. $ P$ is folded onto $ R$ and
then $ Q$ is folded onto $ S$. The area of the resulting figure is 9
square inches. Find the perimeter of square $ PQRS$.
[asy]/* AMC8 1998 #20P */
size(1inch,1inch);
label("$P$", (0, 72), SE);
label("$Q$", (72, 72), SW);
label("$S$", (0, 0), NE);
label("$R$", (72, 0), NW);
draw((0,0)--(72,0)--(72,72)--(0,72)--cycle);[/asy]

When a paper is folded like this, the result is an isoceles right triangle, with a leg being half the length of the diagonal.

So you have the equation: $ s \plus{} s \plus{} s\sqrt{2} \equal{} 9$

$ s (1 \plus{} 1 \plus{} \sqrt{2}) \equal{} 9$

$ s \equal{} \frac{9}{\sqrt{2} \plus{} 2}$

Now double that to find the diagonal.

$ d \equal{} \frac{18}{\sqrt{2} \plus{} 2}$

$ d \equal{} 18 \minus{} 9\sqrt{2}$

Find the length of a side and multiply by 4.

$ \text{side} \equal{} \frac{18 \minus{} 9\sqrt{2}}{\sqrt{2}}$

$ \text{side} \equal{} 9\sqrt{2} \minus{} 9$

$ \text{perimeter} \equal{} 36\sqrt{2} \minus{} 36$

Hmmm, it seems wrong...

I know for a fact the answer is 24, but I don't know how to solve it.

@ernie - the area is 9, not the perimeter :wink:

[hide="solution"]
The first fold cuts the area in half, and the second fold cuts the area in half, so a fourth of the original area is 9.  Thus, the area of the square is 36, and the side length is 6.  The perimeter is 6(4)=24
[/hide]

The result should be a 45-45-90 triangle with area 9, right?  And if the short side is denoted s, the area would be $ \frac {s^2}{2} \equal{} 9$ which after solving gives us $ x \equal{} 3\sqrt{2}$.  Then the long side of that triangle is $ 3\sqrt{2}\cdot\sqrt{2} \equal{} 6$.  The long side of that triangle is the edge of the big square, so the perimiter of the big square is 4 times that:  24.


Determine all integers $n > 1$ such that \[\frac{2^{n}+1}{n^{2}}\] is an integer.

Obviosly n is odd. Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n\plus{}1$ and $ n\equal{}mp$, then $ p|2^m\plus{}1$.
If $ n\equal{}p^km, v_p(m)\equal{}0, k\ge 1$, then must be $ v_p(2^m\plus{}1)\ge k$.
Therefore from all solution n we get $ p|2^1\plus{}1$ and find minimal solution (n>1) $ n\equal{}3$, Because $ 2^3\plus{}1$ had not another prime factors. And all solutions are $ n\equal{}1$ and $ n\equal{}3$.

[quote="Rust"]Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n \plus{} 1$ and $ n \equal{} mp$, then $ p|2^m \plus{} 1$.[/quote]Who says there is such a prime divisor?

[quote="Rust"]If $ n \equal{} p^km, v_p(m) \equal{} 0, k\ge 1$, then must be $ v_p(2^m \plus{} 1)\ge k$.[/quote]Why?

For notations, always take a look at http://www.mathlinks.ro/viewtopic.php?t=76610 .

[quote="kimnimalar"][quote="Rust"]Obviosly n is odd. Let $ p$ is maximal prime divisor n, suth that $ n^2|2^n \plus{} 1$ and $ n \equal{} mp$, then $ p|2^m \plus{} 1$.
If $ n \equal{} p^km, v_p(m) \equal{} 0, k\ge 1$, then must be $ v_p(2^m \plus{} 1)\ge k$.
Therefore from all solution n we get $ p|2^1 \plus{} 1$ and find minimal solution (n>1) $ n \equal{} 3$, Because $ 2^3 \plus{} 1$ had not another prime factors. And all solutions are $ n \equal{} 1$ and $ n \equal{} 3$.[/quote]

 please write full, I can't understand  [i]vp(m) [/i]and others... what's mean this function?[/quote]
Full solution:
Because for $ n>1$ $ 2^n\plus{}1$ is odd, n is odd. Let p is maximal prime divisor of n and $ n\equal{}mp^k,p\not |m$. Then $ 2^{mp^k}\equal{}\minus{}1\mod p^{2k}m^2$ and $ (\phi(m^2),p)\equal{}1$. If $ m\equal{}1$ obviosly $ m^2|2^m\plus{}1$. Let $ m>1$.Therefore exist odd u ($ \phi(m^2)$ - even if m>1 and odd), suth that $ p^ku\equal{}1\mod \phi(m^2)$.
It give $ 2^{mp^ku}\equal{}(\minus{}1)^u\equal{}\minus{}1 \mod m^2$. Because $ p^ku\equal{}1\mod \phi(m^2)$, we get $ m^2|2^m\plus{}1$. It give new solution m.
If $ p\not |2^k\plus{}1$, then $ 2^{kp}\plus{}1\equal{}(2^{p\minus{}1})^k*2^k\plus{}1\equal{}2^k\plus{}1\mod p$, therefore $ p\not |2^{kp}\plus{}1,...p\not |2^{kp^l}$.
Because $ p|2^{mp^k}\plus{}1$, we get $ p|2^m\plus{}1$. Exactly $ p^k|2^m\plus{}1$.
From solution $ n_0\equal{}n\equal{}p^km$ we get solution $ n_1\equal{}m^2|2^m\plus{}1$, suth that $ p|2^m\plus{}1$ and $ n_1$ had less prime divisors, then $ n_0$.
From $ n_1$ we get $ n_2$,...$ n_l\equal{}p^k$ and $ n_{l\plus{}1}\equal{}1$. Therefore $ p^k|2^1\plus{}1\equal{}3\to p\equal{}3,k\equal{}1$.
It give solution $ n\equal{}3$. If $ n\equal{}p^k*3$, then $ p^k|2^3\plus{}1\equal{}9$. Therefore we had not another solutions.

it's very easy to prove it by the LTE
By the LTE we have $v_3(2^n+1)=v_3(2+1)+v_3(n)$
so $v_3(2^n+1)=v_3(3n)$
and as we know $2^n+1$ is divided by $n^2$ so $v_3(2^n+1) \ge v_3(n^2)$
so we replace the LHS by : $v_3(3n) \ge v_3(n^2)$ so we deduce that $n^2$ divides 3n so n divides 3 but n is different to 1 then n must be 3

[quote="soundouss"]so we replace the LHS by : $v_3(3n) \ge v_3(n^2)$ so we deduce that $n^2$ divides 3n [/quote]

totally wrong!!!

By LTE
Vn(2^n+1)=Vn(3)
Following  n|3
n=1,3
N=3


geometry
parallelogram
analytic geometry
graphing lines
slope
linear algebra
matrix

If ad-bc=+-1, prove that (a+b)/(c+d) is an irreducable fraction.

Suppose that $ p|(a+b)$ and $ p|(c+d)$, $ p\in\mathbf{N^{*}}$.
Then $ p|d(a+b), \ p|b(c+d)\Rightarrow p|(ad+bd-bc-bd)\Rightarrow p|(ad-bc)\Rightarrow p|\pm 1\Rightarrow p=1$

Nice solution.

[hide="Another in a more geometric spirit"] The parallelogram $ (0, 0), (a, c), (b, d), (a+b, c+d)$ has area $ |ad-bc| = 1$.  By Pick's Theorem, it has no internal lattice points.  There are therefore no lattice points on the point between $ (0, 0), (a+b, c+d)$, so the slope is irreducible. [/hide]

[hide="Something equivalent..."]We have that \[ \pm 1 = \begin{vmatrix}a & b \\ c & d \end{vmatrix}= \begin{vmatrix}a+b & b \\ c+d & d \end{vmatrix}= \gcd \left( a+b, c+d \right) \cdot \begin{vmatrix}\frac{a+b}{\gcd \left( a+b, c+d \right)}& b \\ \frac{c+d}{\gcd \left( a+b, c+d \right)}& d \end{vmatrix},\] which obviously implies $ \gcd \left( a+b, c+d \right) = 1$.[/hide]


Let be given $ ABC$ with the lengths of the sides $ a;b;c$
Prove that:
$ \sum_{cyc} \sqrt[3]{(\frac{3a}{3b\plus{}3c\minus{}a})^2} \geq \frac{3\sqrt[3]{45}}{5}$

[hide]
Substitute $ a\equal{}y\plus{}z$ etc. so that
\[ \left( \frac{3a}{3b\plus{}3c\minus{}a}\right) ^\frac{2}{3} \equal{}\left( \frac{3(y\plus{}z)}{6x\plus{}2(y\plus{}z)}\right) ^\frac{2}{3}\]
Now by AM-GM, $ (6x\plus{}2y\plus{}2z)^5(5y\plus{}5z)^4\leq \frac{1}{9}\left( \frac{10}{3}(x\plus{}y\plus{}z)\right)^9$. Raise both sides to the -2/15 power to get
\[ \frac{1}{(6x\plus{}2y\plus{}2z)^\frac{2}{3}(5y\plus{}5z)^\frac{8}{15}}\geq 9^\frac{2}{15}\left( \frac{3}{10(x\plus{}y\plus{}z)}\right)^\frac{6}{5}\]
\[ \left( \frac{3(y\plus{}z)}{6x\plus{}2(y\plus{}z)}\right) ^\frac{2}{3} \geq \frac{\sqrt[3]{45}}{5}\left( \frac{3(y\plus{}z)}{2(x\plus{}y\plus{}z)}\right)^\frac{6}{5}\]
Summing cyclically and applying power mean gives the desired inequality.
[/hide]


A full bottle of milk weighs 32 ounces and a half-full bottle of milk weighs 19 ounces. What is the weight, in ounces, of an empty bottle?

------------------------------------------------------

I let the weight in ounces of the empty bottle be $ x$ and did the following.

$ x \plus{} 32 \equal{} 2x \plus{} 38$

$ \minus{}x \equal{} 6$

$ x \equal{} \minus{}6$

What did I do wrong?

Let the bottle be x and the weight the milk by itself be y.

y+x=32....(1)

y/2+x=19....(2)

2*(2)
y+2x=38....(3)

(3)-(1)
x=6

[quote="isabella2296"]A full bottle of milk weighs 32 ounces and a half-full bottle of milk weighs 19 ounces. What is the weight, in ounces, of an empty bottle?

------------------------------------------------------

I let the weight in ounces of the empty bottle be $ x$ and did the following.

$ x \plus{} 32 \equal{} 2x \plus{} 38$

$ \minus{} x \equal{} 6$

$ x \equal{} \minus{} 6$

What did I do wrong?[/quote]

Or you know the weight of the bottle plus milk is 32, and half full plus the weight is 19.  $ 32\minus{}19\equal{}13$, so half of the milk weighs 13 ounces.  $ 32\minus{}2(13)\equal{}\boxed{6}$

[quote="isabella2296"]A full bottle of milk weighs 32 ounces and a half-full bottle of milk weighs 19 ounces. What is the weight, in ounces, of an empty bottle?

------------------------------------------------------

I let the weight in ounces of the empty bottle be $ x$ and did the following.

$ x \plus{} 32 \equal{} 2x \plus{} 38$

$ \minus{} x \equal{} 6$

$ x \equal{} \minus{} 6$

What did I do wrong?[/quote]

I think you set it up wrong. It should be set up as a system. x + 2y = 32, and x + y = 19. Then solve from there.

Solving Wickedestjr's system:

\begin{align*}x+2y&=32\\x+y&=19\end{align*}

Since we want the value of $ x$, we eliminate $ y$ by multiplying the second equation by $ 2$.  We subtract the first equation from this new one:

\begin{align*}2x+2y&=38\\x+2y&=32\\x&=\boxed{6}\end{align*}

Thus, the empty milk bottle weighs $ 6$ ounces.

[quote="isabella2296"]A full bottle of milk weighs 32 ounces and a half-full bottle of milk weighs 19 ounces. What is the weight, in ounces, of an empty bottle?

------------------------------------------------------

I let the weight in ounces of the empty bottle be $ x$ and did the following.

$ x \plus{} 32 \equal{} 2x \plus{} 38$

$ \minus{} x \equal{} 6$

$ x \equal{} \minus{} 6$

What did I do wrong?[/quote]

It should be $ 2x\plus{}32\equal{}x\plus{}38$

[quote="johnnybeard"][quote="isabella2296"]A full bottle of milk weighs 32 ounces and a half-full bottle of milk weighs 19 ounces. What is the weight, in ounces, of an empty bottle?

------------------------------------------------------

I let the weight in ounces of the empty bottle be $ x$ and did the following.

$ x \plus{} 32 \equal{} 2x \plus{} 38$

$ \minus{} x \equal{} 6$

$ x \equal{} \minus{} 6$

What did I do wrong?[/quote]

It should be $ 2x \plus{} 32 \equal{} x \plus{} 38$[/quote]

How did you even get to that equation as a result?

Izzy messed up because she assumed that the 32 oz didnt include the bottle weight.

johnny was wrong. My equation was right. I'd know, because I took this sprint round, and I got the right answer using the exact same equation that is shown above.

Wicked, I'm sure that there are many ways to solve this equation...

If you solve johnny's equation, you also get $ x\equal{}6$...so he's wrong how?

[quote="AIME15"]Wicked, I'm sure that there are many ways to solve this equation...

If you solve johnny's equation, you also get $ x = 6$...so he's wrong how?[/quote]

AIME is right...I solved it using:

\begin{eqnarray*}
32&=&38-x\\
-6&=&-x\\
6&=&x\\
\end{eqnarray*}

According to you,Wicked, I would be wrong, but I'm not.  :D  :lol:  :)

[quote="johnnybeard"][quote="isabella2296"]A full bottle of milk weighs 32 ounces and a half-full bottle of milk weighs 19 ounces. What is the weight, in ounces, of an empty bottle?

------------------------------------------------------

I let the weight in ounces of the empty bottle be $ x$ and did the following.

$ x \plus{} 32 \equal{} 2x \plus{} 38$

$ \minus{} x \equal{} 6$

$ x \equal{} \minus{} 6$

What did I do wrong?[/quote]

It should be $ 2x \plus{} 32 \equal{} x \plus{} 38$[/quote]

And solving my brother's (johnny's):

$ 2x\plus{}32\equal{}x\plus{}38$
$ x\equal{}6$

Subtract x+32 from both sides as the first and last step.

Well, how did you get that equation in the first place?

[color=darkred][size=75]Broken quote removed.
-Neal (nebula42)[/size][/color]

[quote="AIME15"]I don't really get it either, I'll ask johnny...

He said that he just took the constants in izzys equation and put them on the other side.

I doubt that should work though.[/quote]

Read the posts, Wicked...

Our point is that there are many ways to arrive at this solution, using different equations.  There isn't just one way to solve the problem.

[quote="james4l"]How's this:

A full bottle (32 oz), a half-full bottle (19 oz), and an empty bottle (? oz) form an arithmetic sequence. Therefore the question mark should be replaced with a 6. Easy.[/quote]

Ooh, I never thought about that. That was a good solution. Do you think that there is a way to apply knowledge of Arithmetic sequences to every system?

Yeah, the "empty - halffull - full" thing indicates that an arithmetic sequence may prove to be an elegant solution.

[quote="AIME15"]Noo....

8x+3y=675
761x+98y=32

I don't think arithmetic sequences will help there...[/quote]

Well, that is also more than one variable.

You said SYSTEMS...systems have more than one variable, usually.

I said the wrong word. Sorry. I meant equations with one variable, on both sides, and the variables being the same.

8x+12=3x-86

I don't think you'd use arithmetic sequences for that either...

Alright, I was wrong.  :oops:

I think that it would work if there was no coeffecient, and there were no negative numbers.

If there's no coefficient, we subtract $ x$ from both sides...and get constant=constant...so the equation would have to be x+y=x+y...or something of the sort.

Hmm, yeah, I don't know what I am saying.

???

I think this is dead.


geometry
cyclic quadrilateral
geometry unsolved

A point P in the interior of a cyclic quadrilateral ABCD satisfies ∠BPC = ∠BAP + ∠PDC. Denote by E, F and G the feet of the perpendiculars from P to the lines AB, AD and DC, respectively. Show that the triangles FEG and PBC are similar.

A point $P$ in the interior of a cyclic quadrilateral $ABCD$ satisfies $\angle BPC = \angle BAP + \angle PDC$. Denote by $E, F$ and $G$ the feet of the perpendiculars from $P$ to the lines $AB, AD$ and $DC$, respectively. Show that the triangles $FEG$ and $PBC$ are similar.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=49710&hilit=Polish][u][b]here[/b][/u][/url].

Best regards,
sunken rock