Please, give me a solution for this problem

This seems to be a counterexample: $x=21$, $y=85$ and $z=-7$.

Sorry, sorry.
This problem is incorrect.

This is a correct problem.
Please, give me a solution.  :blush:

This problem is Hard  :(

pleease, give me a solution  :blush:

Hi Lopes, where did you get this problem? (and do I know you?)

I see you are a teacher, where do you teach?

I got this one:

using the cuadratic formula for $z$, we find $z=(2x+2y)/5+-(4x+y)^{1/2}(4y+x)^{1/2}/5$ (1). Now clearly $(4x+y)(4y+x)=m^2$ since $(4x+y,4y+x)=(4x+y,3x-3y)$ wich is $1$, $3$ or $5$. if it is $5$, it means $5$ divides $x-y$, but an easy check at the beggining modulo $5$, shows that $5$ divides $x$ or $y$ so it divides both of them, contradiction. So we have 2 cases: $4x+y=3a^2$ and $4y+x=3b^2$. and from (1) it follows that $10z+x+y=5x+5y+-2.3ab=3a^2+3b^2+-2.3ab=3(a+-b)^2$, and result follows in this case. The secound case is $4x+y=a^2$ and $4y+x=b^2$ so from (1) $10z+x+y=5x+5y+-2ab=a^2+b^2+-2ab=(a+-b)^2$ as we also wanted. so result follows.


All right... this one's been bugging me for a while now.  Find:

$ \displaystyle\prod_{n\equal{}1}^{\infty}\frac{2^n\minus{}1}{2^n}$

I brute forced the thing on Mathematica, and I got 0.288788..., but I want to know the significance of this number.  Anyone have any ideas?

[quote="sunnyboy780"]All right... this one's been bugging me for a while now.  Find:

$ \displaystyle\prod_{n \equal{} 1}^{\infty}\frac {2^n \minus{} 1}{2^n}$

I brute forced the thing on Mathematica, and I got 0.288788..., but I want to know the significance of this number.  Anyone have any ideas?[/quote]

I do not know.
I am just rewriting it as this for others to think about:

$ \displaystyle\prod_{n \equal{} 1}^{\infty}(1 \minus{} \frac {1}{2^n})$ = $ (1 \minus{}\frac {1}{2})(1 \minus{} \frac {1}{4})(1 \minus{} \frac {1}{8})(1 \minus{} \frac {1}{16}) \cdot \cdot  \cdot$

I thought of logarithm function perhaps. Like this:

$ \displaystyle\prod_{n \equal{} 1}^{\infty} \left(1 \minus{} \frac{1}{2^n}\right) \equal{} e^{\displaystyle\sum_{n\equal{}1}^{\infty}\ln\left(1\minus{}\frac{1}{2^n}\right)}$

but I'm not sure how it should be proceeded from here.

I don't know any reason to suppose that it should turn out to be a nice number.  (I also feel like it's been posted on the forum before, but it's not clear what to search for to find it.  In any event, if there's going to be any hope of solving it then it should be moved to the calculus forum.)

Well, while I'm here, here's one that should be solvable.  Compute the product
\[ \prod_{n \equal{} 0}^\infty \frac{2^{2^n} \plus{} 1}{2^{2^n}}.
\]

By the [url=http://en.wikipedia.org/wiki/Pentagonal_number_theorem]pentagonal number theorem[/url], this product is equal to

$ \sum_{k \equal{} \minus{} \infty}^{\infty} ( \minus{} 1)^k \frac {1}{2^{\frac {k(3k \minus{} 1)}{2} } }$.

At least infinite sums are better understood than infinite products, and you've got the additional advantage here that this sum converges very quickly.  But if you want anything like a closed form, it'll have to be in terms of (or otherwise involve) [url=http://en.wikipedia.org/wiki/Theta_function]theta functions[/url].

Thanks for the help, guys.  I didn't think it was too possible, either.  Mathematica only gave me a number after I made it approximate the answer, not find the exact value.  It's just that it converges to something, and I thought that it couldn't be just some random number that it converged to.


induction
logarithms
calculus
derivative
real analysis
calculus computations

prove:$ \frac {k}{n \plus{} k} < ln(1 \plus{} \frac {k}{n})$
in fact, i have a solution which i think is quite concise(or maybe it is wrong),so i hope you can give me a better solution.

[hide="my solution:"]
$ \Longleftrightarrow (1 \plus{} \frac {k}{n})^{n \plus{} k} > e^k$
$ \Longleftrightarrow (1 \plus{} \frac {k\plus{}1}{n})^{n \plus{} k\plus{}1} > e(1\plus{}\frac{k}{n})^{n\plus{}k}$
$ \Longleftrightarrow (1 \plus{} \frac {k \plus{} 1}{n})^{n \plus{} k }\cdot(1 \plus{} \frac {k \plus{} 1}{n}) > e$
AND..
$ RHS \equal{} (1 \plus{} \frac {1}{n \plus{} k})^{n \plus{} k}(1 \plus{} \frac {k \plus{} 1}{n}) > (1 \plus{} \frac {1}{n \plus{} k})^{n \plus{} k}(1 \plus{} \frac {1}{n \plus{} k}) \equal{} (1 \plus{} \frac {1}{n \plus{} k})^{n \plus{} k \plus{} 1} > e$
[/hide]

$ \log(1\plus{}k/n)\equal{}\minus{}\log((k\plus{}n)/k)\equal{}\minus{}\log(1\minus{}n/(k\plus{}n))> \minus{}(\minus{}n/(k\plus{}n))\equal{}n/(k\plus{}n)$.
I've used that $ \log(1\plus{}x)< x$ for $ x\neq 0$.

[quote="zhoraster"]$ \log(1 \plus{} k/n) \equal{} \minus{} \log((k \plus{} n)/k) \equal{} \minus{} \log(1 \minus{} n/(k \plus{} n)) > \minus{} ( \minus{} n/(k \plus{} n)) \equal{} n/(k \plus{} n)$.
I've used that $ \log(1 \plus{} x) < x$ for $ x\neq 0$.[/quote]

i am afraid that i can't prove the ineq:
$ ln(1\plus{}x)<x$
when x is a irration or even a negative without using derivatives.so how to use elementary ineq (like AM-GM ineq )or some basic method to solve it?

Moreover,i am sorry for having too many strict request.
but i think while learning the real analysis,it is good for us to use the things we know how to prove to solve the problem~~

At last, thank you for replying.

[quote="wxydx00"]
i am afraid that i can't prove the ineq:
$ ln(1 \plus{} x) < x$
[/quote]
Can you prove that $ e^x>1\plus{}x$?  :wink: (If you can't,[hide="hint"] $ e^x\minus{}1 \equal{} e^\theta x$, where $ \theta$ is strictly between $ x$ and $ 0$[/hide].)

[quote="Jolie"][quote="wxydx00"]prove:$ \frac {k}{n \plus{} k} < ln(1 \plus{} \frac {k}{n})$
in fact, i have a solution which i think is quite concise(or maybe it is wrong),so i hope you can give me a better solution.

$ \Longleftrightarrow (1 \plus{} \frac {k}{n})^{n \plus{} k} > e^k$
$ \Longleftrightarrow (1 \plus{} \frac {k \plus{} 1}{n})^{n \plus{} k \plus{} 1} > e(1 \plus{} \frac {k}{n})^{n \plus{} k}$
[/quote]

sorry, could you tell me why the two are equivalent？Thank you[/quote]

mathematical induction

[quote="zhoraster"][quote="wxydx00"]
i am afraid that i can't prove the ineq:
$ ln(1 \plus{} x) < x$
[/quote]
Can you prove that $ e^x > 1 \plus{} x$?  :wink: (If you can't,[hide="hint"] $ e^x \minus{} 1 \equal{} e^\theta x$, where $ \theta$ is strictly between $ x$ and $ 0$[/hide].)[/quote]

 :blush: i'm sorry i can't find the reason why the equality $ e^x \minus{} 1 \equal{} e^\theta x$ holds.
but anyway,it must be a good way to solve the problem.
but i think in the book the author may ask us to use the other ways because in front of it, the things they taught us is only the stolz and some difinitions.so i try to find the way the author ask to find...

anyway,thank you for replying such a strange and annoyable question![/quote]

That's the mean value theorem or Taylor's theorem.

Fix $ n > 0$. Consider $ f(x) \equal{} \ln \left(1 \plus{} \frac {x}{n}\right) \minus{} \frac {x}{n \plus{} x}$ on [0, k].

By MVT, there exists $ \theta \in (0, k)$ such that $ f(k) \equal{} f(0) \plus{} kf'(\theta) \equal{} kf'(\theta) > 0 \iff f'(\theta) > 0$.

But $ f'(\theta) \equal{} \frac {\theta}{(n \plus{} \theta)^2}$.

Edited after mav's post;-

With all the typos and miscalculations in most of the posts, the result is correct  :)

[quote="akech"]Fix $ n > 0$. Consider $ f(x) \equal{} \ln \left(1 \plus{} \frac {x}{n}\right) \minus{} \frac {x}{n \plus{} x}$ on [0, k].

By MVT, there exists $ \theta \in (0, k)$ such that $ f(k) \equal{} f(0) \plus{} kf'(\theta) \equal{} kf'(\theta) > 0 \iff f'(\theta) > 0$.

But $ f'(\theta) \equal{} \frac {\theta}{(n \plus{} \theta)^2}$.

[/quote]

well done!but if it is allowed to use derivatives, isn't it too simple and easy?
the purpose i post the passage is find some other ways which can go by using elementary method.
anyway,thank you for replying

[quote="jmerry"]Every [i]definition[/i] of the natural logarithm I know uses derivatives. If you exclude the standard tools of calculus (and the MVT is the most important and broadly useful of those), you don't even have a question.[/quote]

i'm sorry but if my solution is right,it works without using derivatives.(for we use $ e\equal{}lim_{n \rightarrow \infty}(1\plus{}\frac{1}{n})^n$ to define e without the concept of derivatives, so it is possible to solve it without using derivatives.)
by the way, what is Mean value theorem.

thx for replying

[quote="zhoraster"][quote="wxydx00"]
by the way, what is Mean value theorem.
[/quote]
[url=http://www.google.com/search?q=mean+value+theorem]JFGFI[/url][/quote]

oh,i know what is MVT now,sorry for my poor vocabulary.
but i think it should also about derivation which i want to keep away from.


articles
MATHCOUNTS
percent
inequalities
summer program
MathPath
AMC 10

What strategies are there when you are actually TAKING the test? Not just preparation, but during the 40 minute sprint round and 6 minute target rounds? How would you avoid stupid mistakes, keep calm, check for mistakes, and pace yourself?

Well, obviously, if there were one, people who made Nationals wouldn't tell you, because they want to win :P

To avoid stupid mistakes, try reading this article: http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php

By pacing, try to work through the easy problem fast and spend don't spend too much time on hard ones unless you have sufficient time to check it. try to leave the last couple minutes to check answers.

For Sprint, you should try not to spend too much time on one question. If you can't solve it, skip and and move on to other questions. Then go back to the questions you skipped and try them again later. Also, be sure to check your answers as thoroughly as possible when there's time remaining. Also be sure that you completed as many questions as you possibly could.

For Target, use your calculator only when necessary. Sometimes there can be special tricks that help you solve a problem faster without punching in so many numbers on your calculator. In addition, when you tackle one pair of problems, do the easier one first. That'll save you more time for the harder one.

that comes from the guy who didn't make nationals  :)

Anyways, I basically agree with all that he said. For target, be sure to read every single word carefully when you are checking. There's been many problems that I've gotten wrong because I didn't read them right.  :wink:

Anyways, how did you do at states? What place and state did you get/are you in?

My strategy is...
[hide]Nope, I won't tell you it. :) I want to win just like you do, or at least get in cd. And also get top 3 team.[/hide]

That's setting the goal too low for yourself.

We want to get the top 1/2 team!

wow...i didnt even make STATES :(

i failed AMC 10 this year (i did worse than last year by 24 points hehe....:] i know...thats a lot of points :(

and i failed it because i was panicing (its spelled wrong...i dont know how to spell it)

well i usually skip the problems i dont know.....maybe thats not the best way for some people though but it works for me

and you shouldnt panic I LEARNED THAT THE HARD WAY >.<

I failed chapter.

I failed AIME.

*Cries in a corner*

But in MATHCOUNTS, there is usually a nice fast way of doing things. So try to find it. This goes for both Sprint and Target.

[quote="gauss1181"][quote="tinytim"]ONLY do this if your REALLY good: On the sprint, start with the last 10 questions. Then do the first 20.

Target: Read both questions first, then pick the one that you think will take you shorter time to do first.[/quote]
HOW DOES THAT REALLY HELP FOR SPRINT??? :o[/quote]

That way you get the harder questions right. But, like I said, you have to be REALLY good.

That also helps in tiebreakers.

Right now, I'm trying to get my target scores higher, since I usually do decent on sprint, but I always get owned in target.

Unlike me, I do horrible in sprint and do good in target.

I usually do good at target and suck at sprint. (last year at state I got 7 on target and like 16 on sprint). But this year it was exact opposite. (5 on target and 23 on sprint)

We play calculator games with 16 minutes, and then notice that we were being idiots on #1, 3, and 7 in the last minute.  We then hurriedly correct them, only to miss #2 because we forgot to divide by 4. xD

[quote="frost13"]Knowing both coaches well, I think their strategies are to have all 4 kids do all 10 problems each, and all get the same answers.  Then discuss what to do with the remaining 17 minutes :lol:[/quote]

lol i still like the florida strategy......better....

Yeah but if TX and IN's strategy works, then shouldn't you use it?

[quote="mathking123"]what do you mean it works?

i mean, it only got texas 1st place and indiana second place last year?

eitherway, i like florida's strategy better!!!!!![/quote]

You haven't even told us florida's strategy. This is super spam (meaning worse than my posts). I htink one strategy was to eat sugar before test so that you get all worked up.

ehh, no thats not the florida strategy...if i told you all the rest of the florida team (only one besides me has an account here lol) would mob me  :rotfl:

but someone would copy our pwnage strategy, so im keeping it secret :ninja:

[quote="Ihatepie"]yeah, we are so good, we get around 30th place ever year! (we might actually make top 10 this year)[/quote]

Keep hoping. Our strategy is to have one man teams. =P.

Run a mile everyday! It gets your body used to low air pressure.

Tinytim, we are GEEKS! We can't run an inch (2.54 cm) let alone a MILE! :rotfl:

[quote="leoxnlin"]Make sure you get some of the easier tiebreaker questions toward the end of the test. (A strategy that maybe has already been mentioned and only works for people who actually have any skill at all, unlike me  :( )

And mewto, I believe you are mistaken. There are 12 inches in a foot, and 5280 feet in a mile, so a mile is 12*5280*2.54 = 2.25 cm + or - a millimeter. And that is a substantial 11.4 percent less than an inch. By the Ssuag-Tamref inequality, that IS a runnable distance.[/quote]

Hmmm... I've never heard of the Ssuag-Tamref inequality, but that is an obvious case of made-up theorems.
Also, I wish a mile were shorter than an inch... :|

We're getting off topic. Guess it was my fault though...

Stategy: Don't miss the first or second problem in the sprint round! I missed #1 last year and #2 this year!

[quote="tinytim"]Common stereotype: Geeks suck at athletics[/quote]

It always amazed me how many kids at MathPath were good at soccer, although unlike P.E., you are welcome to play even if you can't run an inch!

Ergh...this is way off topic.  Let's get back to the strategies.


A tunnel in the Smoky Mountains is semicircular. At a distance of 12 feet from the center of the tunnel, the tunnel has a height of 16 feet. How many feet tall is the tunnel at its center?

[asy]draw((0,10)..(-10,0)--(10,0)..cycle);
fill((-10,0.1)--(10,0.1)--(10,-0.1)--(-10,-0.1)--cycle,white);
draw((-10,0)--(10,0),dashed);
dot((0,0));
draw((0,0)--(6,0)--(6,8));
label("12'",(3,0),S);
label("16'",(6,4),E);[/asy]

pythagorean theorem: 12-16-20 triangle, radius is [b]20 feet[/b]


Here are the problems of the IMC, hope they are bug-free.

i tried to load the file it 's not work
could you repair it ?
please

here his file works... but it's only questions, no solutions.


geometry
3D geometry
real analysis
real analysis unsolved

Prove that the (double) cone $ \{(x,y,z) \in \mathbb{R}^{3} : x^{2} \plus{} y ^{2} \equal{} z^{2}\}$ is path connected.

After a looking at the shape of two cones connected at their "topmost" point, I can clearly see that it's possible to draw a line from one point on one of the cones to another on the other cone so long as that line crosses through the point that connects the two cones. The intuition is here ( I hope), but how would I formally prove this?

Take two points $ P,Q$ on the cone. The segment from $ P$ to the origin is on the cone. The segment from the origin to $ Q$ is on the cone. We can parametrize that: $ f(t)\equal{}\begin{cases}(1\minus{}2t)P&0\le t\le\frac12\\(2t\minus{}1)Q&\frac12<t\le 1\end{cases}$ is an explicit path from $ P$ to $ Q$, entirely contained in the cone.

So I don't have to use $ x^{2} \plus{} y^{2} \equal{} z^{2}$ in this proof?

I didn't say that. You do need to show that all of the points on the curve I gave are on the cone; that uses the equation.

Why is it necessary to show that $ f[0,1]$ maps to our double cone in this way?

Duh.

Seriously, this is just a random curve through $ \mathbb{R}^3$ that has nothing to do with anything, until proven otherwise.

So would I have to propose $ P, Q \in \{(x,y,z) \in\mathbb{R}^{3} | x^{2} \plus{} y^{2} \equal{} z ^{2}\}$ where $ P \equal{} (a,b,c)$ and $ Q \equal{} (d,e,f)$. Then, say for $ P$, show that $ (a(1 \minus{} 2t))^{2} \plus{} (b(1 \minus{} 2t))^{2} \equal{} (c(1 \minus{} 2t))^{2} \subset \{(x,y,z)\in\mathbb{R}^{3} | x^{2} \plus{} y^{2} \equal{} z ^{2}\}$?


[color=#FF0000][moderator edit: do not quote the whole post immediate before you.][/color]

$ 2 > 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}$ Then $ 2(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ...\frac {1}{a_n}) > (1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n})(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ...\frac {1}{a_n}) > \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} \frac {3}{a_1 \plus{} a_2 \plus{} a_3} \plus{} ... \plus{} \frac {n}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}$

[quote="quangpbc"][quote="super_math"]$ 2 > 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n} 2(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ...\frac {1}{a_n})$[/quote]

What do you mean ?  :(

To prove this inequality you can use Cauchy-Schwarz  :wink:[/quote]
I mean $ 2 > 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}$

[quote="super_math"]
$ 2 > 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}$
[/quote]
But this is not true. :rotfl:

[quote="super_math"]$ (1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n})(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ...\frac {1}{a_n}) > \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} \frac {3}{a_1 \plus{} a_2 \plus{} a_3} \plus{} ... \plus{} \frac {n}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n}$[/quote]
could you please explain this step in details?

If that were true then $ \sum_{n\equal{}1}^{\infty}\frac{1}{n}$ would converge, which however is known to diverge.

From Secrets in Inequalities :

$ \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} ... \plus{} \frac {n}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n} < 4\left(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n}\right)$

Prove by Cauchy-Schwarz , too !

Just a comment:
In the inequality
$ \frac{1}{a_1}\plus{}\frac{2}{a_1\plus{}a_2}\plus{}\dots\plus{}\frac{n}{a_1\plus{}a_2\plus{}\dots\plus{}a_n}<c(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}\dots\plus{}\frac{1}{a_n})$,
$ c\equal{}2$ is the best constant.If someone want i could post a proof for it.

it follows that 2 is best from
$ a_i\equal{}2i$ :roll:


This is an interesting inequality!
Let $ x,y,z$ be positive real numbers such that:$ xyz\equal{}x\plus{}y\plus{}z\plus{}2$.
Prove that: $ \sqrt{\frac{xy\plus{}yz\plus{}zx}{3}}\geq1\plus{}\frac{2}{3}(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})$.

[quote="ntkhang-vn"]
Let $ x,y,z$ be positive real numbers such that:$ xyz \equal{} x \plus{} y \plus{} z \plus{} 2$.
Prove that: $ \sqrt {\frac {xy \plus{} yz \plus{} zx}{3}}\geq1 \plus{} \frac {2}{3}(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})$.[/quote]
Let $ xy \plus{} xz \plus{} yz \equal{} 12t^2,$ where $ t > 0.$ Hence,
$ 8t^3 \equal{} \left(\frac {xy \plus{} xz \plus{} yz}{3}\right)^{\frac {3}{2}}\geq xyz \equal{} x \plus{} y \plus{} z \plus{} 2\geq\sqrt {3(xy \plus{} xz \plus{} yz)} \plus{} 2 \equal{} 6t \plus{} 2,$
which gives $ t\geq1.$
$ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {xy \plus{} xz \plus{} yz}{x \plus{} y \plus{} z \plus{} 2}\leq\frac {xy \plus{} xz \plus{} yz}{\sqrt {3(xy \plus{} xz \plus{} yz)} \plus{} 2} \equal{} \frac {6t^2}{3t \plus{} 1}.$
Id est, it remains to prove that $ 2t\geq1 \plus{} \frac {4t^2}{3t \plus{} 1},$ which is indeed true.

Thank you!
But I have an another solution :D !
Let's continue to solve this problem!


inequalities
function
logarithms
calculus
derivative
calculus computations

Sorry for all the questions. None of these are homework per se, but they are old exam problems. I hope you don't mind

Show that the function
\[ \ln\left(1 + \frac{1}{x}\right) - \left(x + \frac{1}{2}\right)^{-1}  \]
is decreasing for x > 0 and that
\[ e < \left(1 + \frac{1}{x}\right)^{1 + \frac{1}{2}}  \]
when x > 0.

I need some kind of hint for the second part. What exactly does the first function have to do with the second? The derivative of the second function is close to that of the first, but not the same. I keep trying to show that the second function is decreasing, but I'm not doing well with its derivative. So I figure there has to be some kind of trick connecting the first function and  the second. What is it?

Thanks!

edit: Thought I'd include the derivative of the second function so you don't have to differentiate it yourselves:
\[{ \left(1 + \frac{1}{x}\right)^{1 + \frac{1}{2}}\left(\ln(1 + 1/x) - \frac{2x + 1}{2x(x + 1)}\right)}  \]

I figured this out. You just take logarithms of both sides in  the original equation and then you're pretty much done. I feel very dumb.


Let n be a positive integer. And let p be a prime number such that n divides p-1, and p divides n^3 - 1. Show that 4p-3 is a square.

I can show that p divides n^2+n+1, but then what??

Posted many times before
http://www.mathlinks.ro/Forum/viewtopic.php?t=5863

Pierre.


Marshall and Alex are playing a game. Marshall starts with 8 chips and Alex with 6 chips. Alex rolls his fair die, and the number on top is the number of chips he takes from Marshall. Marshall then rolls his fair die, and the number on top is the number of chips he takes from Alex. Whoever has the most number of chips wins. If there is a tie, they roll another round(but then they both start with 7 chips). This process continues until a winner is determined. What is the probability Alex wins?

[hide="Solution"]Let $ a$ be the number of Alex's die, and $ m$ be the number on Marshall's die.

[color=darkred][b]Case 1:[/b][/color] $ a \equal{} m \plus{} 1$ (probability $ \frac {5}{36}$). After the first round (which will be a tie), the probability of winning will be $ \frac {1}{2}$ for both players.

[color=darkred][b]Case 2:[/b][/color] $ a > m \plus{} 1$ (probability $ \frac {10}{36}$). In this case, Alex wins.

[color=darkred][b]Case 3:[/b][/color] $ a < m \plus{} 1$ (probability $ \frac {21}{36}$). In this case, Marshall wins.

Considering the $ 3$ cases together, the total probability that Alex will win is $ \frac {5}{36}\cdot\frac {1}{2} \plus{} \frac {10}{36} \equal{} \boxed{\frac {25}{72}}$.[/hide]


Given positive integers $ a$ and $ b$
Prove that if $ a^2 \plus{} b^2\over ab \minus{} 1$  is an integer then it is 5

A similar problem appeared in IMO 88 which was solved using descent. I tried descent without much success  :(
Please help.

[hide]
Suppose there is $ c\neq 5$ such that the exist $ a,b$ with $ \frac {a^2 \plus{} b^2}{ab \minus{} 1} \equal{} c$. WLOG let $ a\geq b$, and consider a solution with least value of $ a \plus{} b$. Then $ a^2 \minus{} bca \plus{} b^2 \plus{} c \equal{} 0$, let $ a_1 \equal{} bc \minus{} a \equal{} \frac {b^2 \plus{} c}{a}$. Then $ a_1$ is a positive integer with $ \frac {a_1^2 \plus{} b^2}{a_1b \minus{} 1} \equal{} c$, so we must have $ a_1\geq a$. This means
\[ bc\geq 2a\Rightarrow b(a^2 \plus{} b^2)\geq 2a(ab \minus{} 1)\Rightarrow b^3 \plus{} 2a\geq a^2b\Rightarrow 2a\geq b(a\plus{}b)(a\minus{}b)
\]
If $ a>b$ and $ b\geq 2$, this is false. So either $ b\equal{}1$ or $ a\equal{}b$. In the first case,
\[ 2a\geq (a\plus{}1)(a\minus{}1),a<a\plus{}1\Rightarrow 2>a\minus{}1\]
so $ a\leq 2$. This means $ a\equal{}2$ and $ c\equal{}5$. In the second case,
\[ c\equal{}\frac{2a^2}{a^2\minus{}1}\equal{}2\plus{}\frac{2}{a^2\minus{}1}\]
But for $ a>1$, $ a^2\minus{}1>2$ so $ c$ can't be an integer. Either way, there's a contradiction.
[/hide]

[quote="scorpius119"]
\[ bc\geq 2a\Rightarrow b(a^2 \plus{} b^2)\geq 2a(ab \minus{} 1)\Rightarrow b^3 \plus{} 2a\geq 2a^2b
\]
[/quote]

Should it be
\[ b(a^2\plus{}b^2)\geq 2a(ab\minus{}1)\Rightarrow b^3\plus{}2a\geq a^2b
\]
and so the ensuing arguments fail ?

This is rather old problem of journal Kvant (year about 1991?). The author of the problem as I remember Petrosyan.

This problem can be translated to the problem: Show that there exist infinetely many integers $ a,b$ such that $ \frac{a^2\plus{}5}b,\frac{b^2\plus{}5}a$ are integers.

[quote="baysa"]This problem can be translated to the problem: Show that there exist infinetely many integers $ a,b$ such that $ \frac {a^2 \plus{} 5}b,\frac {b^2 \plus{} 5}a$ are integers.[/quote]
I have post this .
Prove that for each $ m\in Z$ there exist infinite $ (x,y)\in N$ satisfy
$ x|y^2\plus{}m$ and $ y|x^2\plus{}m$
Prove by use the Pell's equation.

I mean from $ \frac{a^2\plus{}b^2}{ab\minus{}1}\equal{}5$ implies $ \frac{a^2\plus{}5}b\equal{}5a\minus{}b$ and $ \frac{b^2\plus{}5}a\equal{}5b\minus{}a$.


A family has septuplets (the birth of seven children):  four boys and three girls.  For the boys they have chosen the names:  Jerry, Jim, Jack , and John.  For the girls, they have chosen the names:  Janice, Joyce, and Jane.  In how many different ways could they name their septuplets?

A.3 0	B.105	C.144	D. 5040   E.35

[hide]Obviously the boys and the girls are to be named independently. The boys can be named in $ 4! \equal{} 24$ ways and the girls in $ 3! \equal{} 6$ ways, so $ 24 \times 6 \equal{} 144$. C.[/hide]


AMC
AMC 12
trigonometry
Gauss
geometry
3D geometry
octahedron

I'm just gonna copy what I wrote in the AMC forum (note this is for AMC12):

144 for me.

I got the last one wrong because I said $ \tan\pi \equal{} \minus{} 1$.  :ewpu:

I got a 126....I obviously can't do quadrilateral problems and made a few stupid mistakes.

I epicly failed the AMC 10A with a 129.  Made stupid mistakes on the tangent circles problem and #25...and skipped #22 and #23 for no reason even though they were so easy...

My head totally blanked out that morning since I only got 6.5 hours of sleep the night before(  ) so I kinda like feel asleep/absentminded during the exam...

And what sucks worse is that I feel sick right now due to sleep deprivation  it feels strange because everything seems dizzy aroud me

but normally when i don't have lack of sleep, I can usually get 140s to a perfect... :|

And Bedford beats Acton for top score in the AMC 10! :winner_first:

Lol, but not team score :ewpu:

MUST... TRAIN... OTHER... PEOPLE...

First of all, seeing gauss' comment, can people not say they "failed" unless they got a score of 30 or lower? Thanks. It's disrespectful.

On the other hand, I got a 136.5 on the 12A. That octahedron problem...scared me.

30 or lower would be truly fail. Even leaving everything blank would grant you a 37.5.

Ack, 126 on 12. Made a lot of idiot mistakes like spending 20 minutes on 23, finally getting it and then doing $ 50/2 \equal{} 50$ as part of the last step.

Btw, Acton 8th grader got 150 on AMC10, I'm not sure if that counts as ABRHS or junior high tho.

[quote="perfect628"]Haha 144 on the 10...I really need to stop taking 10s...(which I am don't worry...) Missed 17, stupid inability to read...Anyway Andrew got 144 too. He missed 22.

Anyway what was up with the no combo on the 12? Does this mean it will be filled with it on the 12B? (Hope so...)

PS The only ones I couldn't get on the 12 were 22 and 23...Evil problems. 24 took me like 10 minutes...[/quote]

Again, 22 scared me. Hopefully there isn't any of this 3D bull on the 12B. 24 was definitely doable, I just didn't work on it during the test (spent the last 10 minutes on 23).

SamL have you not heard the phrase "Quit while you're ahead?"

Indeed I have, but I'm so sporadic I felt I needed a guarantee for USAMO. So I guess I'll have to quit while I'm behind, as I'm done with 10s...

Also is it just me or was 19 on the 12 [i]really[/i] mathcountsy? (the heptagon/pentagon one)

I "did not not fail" with a 144 on amc10A...

And, right now, I have the lowest score of anyone posting: 115.5 on the 12. Heh I did better last year.

 :wallbash_red: [b][i]...must...do...better...on...b...[/i][/b] :wallbash_red:


MATHCOUNTS
FTW
analytic geometry
function
geometry
3D geometry
ratio

I will be hosting a Mock MathCounts this summer.

It will have four rounds, similar in difficulty to the school, chapter, state, and national rounds at the real MathCounts.  Some problems will be taken from previous MathCounts.

The first 50 people who sign up will be allowed to enter.  Anyone else is free to use the problems and submit answers, they just will not advance.

30 people will advance after the school round.  There will be a Sprint round and a Target round, each with the same rules and scoring.  PM me answers, and do not take longer than the allotted time for the two rounds please!  We will be going by the honor system, so PLEASE follow the rules!

The same rules will apply to the Chapter round.  Problems will be slightly more difficult than on the school round.  The top 12 people will advance to the State round.

3 people will advance to the mock National round.  There will be a Sprint round and a Target round, a little bit harder than the ones in the Chapter round.

The National round will be the same.  Additionally, there will be a Masters round.  I will decide on scoring later.  I will not score the normal way because we definitely cannot have people tape presentations and such.  You will need to submit a full solutions, not just answers.  One person will be eliminated from these three tests, and the top two spots will be decided by an FTW Countdown.

If there are ties for the last qualifying spot in any of the rounds, they will be settled the same way as in MathCounts.

This mock MathCounts cannot exist without your cooperation and integrity.  There is no way for me to tell if you used the alloted 40 minutes for the sprint round, or a few hours.  You may only use tools and aids that you are allowed to use in the real MathCounts competition.

Please join this mock MathCounts!

[size=59]AIME15 is not responsible for anything posted on this forum.  If something is found offensive, please contact the moderators.[/size]

I'll join. About how hard is the difficulty?

i'll sign up
hopefully, i won't fail dreadfully =)

[quote="isabella2296"]MC difficulty.

I'll sign up too.

@AIME15: Since you copied my avatar, I made a new panda one for you and pmed it to you - use it, please! Everyone thinks I copied from you!![/quote]

Whoa you made that panda? That's skillful.

People signed up:
moogra
isabella2296
vallon22
27r
Yongyi781
BlackSwordsman
Urc
pinkpiglet

If you want to sign up, please POST on this forum, do not PM me until we start.

[quote="Yongyi781"]If nobody else who actually made Nationals signs up, I'm afraid I might spoil this tournament! :(

But... I DON'T CARE! :D I'm joining![/quote]

I think some people who made Nats will sign up.  And I'll try to make the level of math not too difficult until the mock state and nats, so many people will have a fighting chance.

[quote="moogra"]

Whoa you made that panda? That's skillful.[/quote]

Sorry I wasn't clear. I used an image from a website and I tweaked it and added some stuff. Almost all credit goes to the person who made the website.

OK, back to signups.

Join this tournament for learning experience with MathCounts!

I will PM everybody before the tournament starts.

Just wondering - where are you getting the questions?

I'm not telling--then some people will look there for answers.

Some will be past MathCounts problems, some maybe from AMC 8's and VERY easy AMC 10's.  I will change around the numbers so you can't look up the answers.  Maybe from some math books that i have here.

[quote="AIME15"][quote="vallon22"]here are the rankings again because nobody seems to have posted them in 2 pages...

1. RisingFlame: 45
2. Moogra: 44
3. johnnybeard: 43
4. Yongyi781: 42
4. 27r: 42
4. shentang: 42
7. vallon22: 41
8. Ajain: 40

____________________
9. ernie: 39
10. isabella2296: 38
11. kookamunga: 33
12. Urc: 31[/quote]

Kookamunga is out.[/quote]

Looks like you may have a tie between the 42's.  What will yo do?  Ugh... I'm 9th... That sucks.

the way they break ties in real mathcounts is:
look at the sprint round, starting from #30.
go backwards, first person who got a question wrong is out

Sorry gonna have to drop.
I'll still Latexify the q's though.

wow
if that's the way they do it in mathcounts, then if i'm in a tie, i'll be out...
i know i got the last question wrong....

Actually, the real way that MATHCOUNTS breaks ties is to look at the sprint round score first. The lowest SCORE is out. Then if the scores are the same, they look at the individual problems. :)

All right, I'm back.  Finally.

State will be posted on...whenever I can get on.  If you are eliminated, and have a few original state-level problems, please PM me the questions and solutions.

:( we got 9th at track and field nationals :( top 8 get all american :((

If nobody is posting here, should I cancel this?

No... we are waiting for the state competition... (obviously!) :!:

AIME,
You need to PM the questions to be Latexified.
~dojo

BTW, is this still the standings?

1. RisingFlame: 45
2. Moogra: 44
3. johnnybeard: 43
4. Yongyi781: 42
4. 27r: 42
4. shentang: 42
7. vallon22: 41
8. Ajain: 40

____________________
9. ernie: 39
10. isabella2296: 38
11. Urc: 31

or did someone else get in?

I'm not sure.  I could make a state for you guys, though.

I don't think I can do the Nats, though. (haven't made it :()

this was dead for a long time

let's just take the standings right now and continue it
who wants to take over?

@ernie
i'll just pull some problems from this hard math book i have for nats if i don't make it

people moving on
1. RisingFlame: 45
2. Moogra: 44
3. johnnybeard: 43
4. Yongyi781: 42
4. 27r: 42
4. shentang: 42
7. vallon22: 41
8. Ajain: 40

since im not in this I"LL MAKE THE PROBLEMS! :D


With combinations; lame
[hide]For $ p$, there are $ \frac{10}{_40}C{_4}$
For $ q$, we can again choose $ 4$ slips in $ 40 \choose 4$ ways.  There are $ 40$ possibilities for the first slip, and since order doesn't matter, we can just say that there is $ 3$ ways for the second slip.  We then have $ 36$ possibilities for the third slip, and $ 3$ ways for the last slip.  Of course, we're counting each combination $ 2^{3}$ ways, since the first two and the last two are being counted twice for every single choice, and overall we're counting every combination twice.  So, there are $ \frac{\frac{40 \bullet 36 \bullet 3 \bullet 3}{2}}{8}\equal{}1620$ ways for the numerator.  Our answer is:
$ \frac{1620}{10}\equal{}\boxed{162}$ [/hide]
WITHOUT COMBINATIONS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
[hide]Assume that order matters and all slips are distinct.  For $ p$, there are $ (40)(39)(38)(37)$ ways to get $ 4$, of which $ 40*3*2*1$ are all of the same number.  Thus, $ p\equal{}\frac{6}{(39)(38)(37)}$.  Now for the fun part! There are, once again $ (40)(39)(38)(37)$ ways to get $ 4$ slips.  There are $ 40 \bullet 3 \bullet 36 \bullet 3 \bullet 3$ ways to get $ 2$ of one kind and $ 2$ of another (the last $ 3$ coming from the fact that you can arrange the slips as $ aabb, abab$ and $ abba$.  Thus, our answer is $ \frac{3 \bullet 36 \bullet 3 \bullet 3}{6}\equal{}\boxed{162}$.[/hide]

[quote="math154"][hide="Solution"]
We can find the ratio of the number of ways instead of the ratio of the probabilities (assume that order does not matter).

The number of ways for them to be the same is $ 10$, and there are $ \dbinom{10}2\dbinom42\dbinom42 \equal{} 1620$ ways for there to be two distinct pairs of the same number, so our answer is $ \frac {1620}{10} \equal{} \boxed{\textbf{(A)}\ 162}$.
[/hide][/quote]
How do we get $\dbinom{10}2\dbinom42\dbinom42 \equal{} 1620$?

There are $\binom{10}{2}$ ways to choose the values of $a$ and $b$, and $\binom{4}{2}$ ways to choose the slips of each of those numbers. Thus there are $\binom{10}{2}\binom{4}{2}\binom{4}{2}=1620$ ways that two of the slips bear a number $a$ and the other two bear a number $b\not = a$.

[quote="TachyonPulse"]Forty slips are placed into a hat, each bearing a number $ 1$, $ 2$, $ 3$, $ 4$, $ 5$, $ 6$, $ 7$, $ 8$, $ 9$, or $ 10$, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $ p$ be the probability that all four slips bear the same number. Let $ q$ be the probability that two of the slips bear a number $ a$ and the other two bear a number $ b\not\equal{} a$. What is the value of $ q/p$?

$ \textbf{(A)}\ 162\qquad
\textbf{(B)}\ 180\qquad
\textbf{(C)}\ 324\qquad
\textbf{(D)}\ 360\qquad
\textbf{(E)}\ 720$[/quote]

The wording to me is a bit ambigous, why does the question say 40 slips are placed into a hat , then later each number is entered on four slips?

Each number from 1 to 10 is written 4 times.

Edit: Typing fail.

No, each number from 1 to 10 is written 4 times.


1) Didn't it say that if you made a program, it would be acceptable as long as you explain and prove every single little step to it?

2) Wouldn't that be asking someone else to help, i.e. cheating?

3) Can't you do it yourself if you are going to write a program?

4) Wouldn't that be destroying the whole point of competing in a MATH proof competition?

5) Don't you doubt that you would get a 5 even if you wrote a program because programs are inelegant especially in a MATH proof competition?

6) Didn't it also say that programs shouldn't make up the whole proof, but instead be a insignificant part of the proof?

7) Don't you wonder how long I can just ask questions?

[quote="BOGTRO"]1. If I were to write a program to simply count every possibility, would that be acceptable?

2. If so, can I ask someone to write the program (not participating in USAMTS?)[/quote]
"Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, if a key step of a participant's solution uses these tools, the participant must demonstrate that the tool completed the step correctly."

So, yes to 1, and you must be able to explain it if you are to do that.

"Participants may use resources such as books or the internet to do mathematical research to try to solve the problems. Participants may not use 'live' online help, e.g. they may not ask for help with the problem on online forums or tutorial services."
For 2, I think the programmer would count as "live" help.

[quote="BOGTRO"]1. If I were to write a program to simply count every possibility, would that be acceptable?

2. If so, can I ask someone to write the program (not participating in USAMTS?)[/quote]
1. This is acceptable, but you must prove that your program solves the problem.
2. No -- you can never ask any other individual for assistance on a USAMTS problem.  All the work must be your own.

FYI - I have gotten several 5's for programs - but never a commendation.  And I think that once you can understand 1) how to count (combinatorics) and 2) how to do a case study, there is little reason to use a program at all.  I find that it is quicker to do the problems with no programming most of the time, than to program, because with programming I usually have issues with them not enumerating the possibilities correctly.

Having said that, it can be useful occasionally.  For instance, I used a program to find my answer to round 4 problem 5 last year.  But then I worked backwards from my answer to what I knew $ s$ to equal, and then reversed that.  So in the end my proof relied only on geometry and algebra ,was completely direct, and stood alone without any program.  That proof got the commendation.


[hide="1."]
$ \frac {7!9!}{7! \plus{} 8! \plus{} 9!} \equal{} \frac {N!}{9}$
Denom of LHS = $ 7!(1 \plus{} 8 \plus{} 8*9) \equal{} 7!81$.
Mult. both sides by 9:
$ \frac {7!*9!}{7!*9} \equal{} N!$
$ 8! \equal{} N!$
$ N \equal{} \boxed{8}$.
[/hide]

[hide="2."]
$ n \equal{} 2\sqrt {\frac {n(n \minus{} 3)}{2}}$.
$ n^2 \equal{} 2n(n \minus{} 3)$
$ n^2 \equal{} 2n^2 \minus{} 6n$
$ n^2 \minus{} 6n \equal{} 0$
$ n(n \minus{} 6) \equal{} 0$
$ n \equal{} \boxed{6}$.[/hide]

[hide="3."]
The blocks with exactly 2 painted sides are the ones on the edges but not on the corners.
So there are $ 4(28 \plus{} 14 \plus{} 10) \equal{} 208$ of these cubes, out of a total of $ 30*16*12 \equal{} 5760$.
$ \frac {208}{5760} \approx \boxed{3.6\%}$[/hide]

[hide="4."]
find the small angles...
$ 180 \minus{} < A/2 \minus{} < B \minus{} ( < A \plus{} < C)/2 \equal{} 50$
$ < C/2 \equal{} 50$
$ < C \equal{} \boxed{100}$
[/hide]

[hide="5."]
3 congruent rectangles side by side; let each have length y and width x.
$ 6x \plus{} 4y \equal{} 60$
$ y \equal{} \minus{} \frac {3}{2}x \plus{} 15$.
$ A \equal{} 3x( \minus{} \frac {3}{2}x \plus{} 15)$
$ A \equal{} \minus{} \frac {9}{2}x^2 \plus{} 45x$
Vertex at $ x \equal{} \minus{} \frac {b}{2a} \equal{} 5$.
Plug back in for $ A \equal{} \boxed{112.5}$.[/hide]

meanwhile to hide stuff type [code][hide]*message*[/hide] or [hide=*message name*]*message*[/hide][/code]
the preceding would translate into
[hide]*message*[/hide] or [hide="*message name*"]*message*[/hide]

Or you can use the bar of commands right under "Subject Description".
Click the "Hide" button to hide something, then when you're at the end of the section you want to hide, click it again for the closing tag.


[color=green][size=200]            In the name of God the Merciful[/size][/color]

If every vertex of a regular pentagon is connected to every other vertex,

how many triangles are formed?

regards
salahcool

Hi salahcool. There are $ 35$: see http://www.cs.uwaterloo.ca/journals/JIS/sommars/newtriangle.html

This is true,

thank you very much AndrewTom ,

God bless you

REGARDS
SALAHCOOL


I'm sorry if this is the wrong forum, but I didn't know where else to post this. And also, apologies for the lengthy post. :blush:

The thing is, I've never participated in a real (i.e. not online) math olympiad or competition because there are none where I live. I do, however, have some problem-solving experience. I think I fit best in the intermediate forum, but I've managed to solve a question or two from the pre-olympiad forum.

Anyway, I believe I'll be heading to the University of Waterloo in the fall, and I'd really like to participate in the Putnam and actually do well. I don't expect to kick ass on my first or second attempt, but I want to begin building up my skills so I could hopefully kick ass on my following attempts. To do that, I've decided to buy a few olympiad books and go through them this summer. What I'm looking for is a book that teaches the necessary topics and a couple of problem books, or just a few problem books instead.

Here are some books I'm considering:
- [url=http://www.amazon.com/exec/obidos/ASIN/0198501056/qid%3D1114162523/sr%3D11-1/ref%3Dsr%5F11%5F1/102-8371051-2688163]The Mathematical Olympiad Handbook: An Introduction to Problem Solving Based on the First 32 British Mathematical Olympiads 1965-1996[/url]
- [url=http://www.amazon.com/exec/obidos/ASIN/0486277097/qid%3D1114162556/sr%3D11-1/ref%3Dsr%5F11%5F1/102-8371051-2688163]The USSR Olympiad Problem Book : Selected Problems and Theorems of Elementary Mathematics[/url]
- [url=http://www.amazon.com/exec/obidos/ASIN/0387947434/qid%3D1114364659/sr%3D11-1/ref%3Dsr%5F11%5F1/102-8371051-2688163]Winning Solutions[/url]
- [url=http://www.amazon.com/exec/obidos/ASIN/0817641556/qid%3D1114354263/sr%3D11-1/ref%3Dsr%5F11%5F1/102-8371051-2688163]Mathematical Olympiad Challenges[/url]
- [url=http://www.amazon.com/exec/obidos/ASIN/052154677X/qid=1114373541/sr=2-3/ref=pd_bbs_b_2_3/102-8371051-2688163]The Cauchy-Schwarz Master Class : An Introduction to the Art of Mathematical Inequalities[/url]

Unless it's totally crap, I think I'm definitely going with the BMO book since I've done British A-levels in math and further math, so it probably "bridges the gap" between the math I'm used to and olympiad math.

The USSR book seems like it's worth the $11... Right? Or should I forget it and get a Dover linear algebra book instead (I need one)?

Winning Solutions and Mathematical Olympiad Treasures both seem like excellent books, but unfortunately I'm not made of money so I can only pick one of them. Has anyone used either one? If so: any comments? From which one will I benefit more? Or should I just get both and forget about the BMO book?

Finally, The Cauchy-Schwarz Master Class. I found this book accidentally, but it seems very interesting. I really like inequalities (and am capable of solving low-level olympiad ones), so I thought this book might be an interesting and valuable read. Does anyone know anything about it? I've read the table of contents and the first few pages on Amazon, and it seems pretty good; but what do I know? :)

I'll appreciate any comments or even new book recommendations.

Thanks.

So, speaking as someone who was never the real tops at high school olympiads but who has done quite well on the Putnam, I would say that it's definitely worth while to include a book of actual Putnam questions on your list, or do those which can be found in a variety of places on the web (but usually with not-such-great solutions).  A large part of any mathematical contest is being comfortable with the format.  For the Putnam, because the time constraint is fairly tight, it's very important that you have enough experience to spot the questions you'll be able to do which are sometimes the numbers 3 or 4 instead of numbers 1 or 2.  (I believe A4 was one of the easier questions on this past year's exam.)

Well, it's nice to know what the exam you're prepping for is going to look like, both in terms of knowing what kind/difficulty of problems to expect so you aren't psyched out and for purposes of your own test strategizing.  The most recent one (by Kedlaya and some others) is a really nice problem book.  Whether it's "worth it" depends how valuable ou think having a question book is and how much it costs.

[quote="fk"]You mean like one of those three books published by the MAA? Is one really worth it?[/quote]I would suggest you work [b]all[/b] previous putnam questions and all the questions in the [url=http://www.mathlinks.ro/Forum/index.php?f=218]College Playground[/url] section of this site ;)

I do intend to work through all the Putnam questions eventually, but should I start like that? I don't think I have the necessary skills yet, which is why I'm looking for easier problem-solving books.

My goal is to finish these books over the summer, and then start doing Putnam problems when I begin college. I know I'm not gonna do well the first time, so I want to maximize my chances of doing well the next times.

Thanks for the replies btw. :)

[quote="Valentin Vornicu"]Start with the easier (older) Putnam questions and build your way up from there ;)[/quote]
All in good time... :D

Anyone else want to add anything?

Btw, I decided not to get the BMO book after going through some of its pages on Amazon.

There are a lot of great resources online to supplement books.  Try browsing this forum and you should find some of them.


Prove without Bertrand's Postulate that there is a prime between N and N^2

I think it says something a little stronger than that  :wink:
http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate


geometry
geometric transformation
rotation
calculus
integration
function
conics

I have another question about volume and integral also:

Say I have a region bounded by y = x^2, y = 1, and the y-axis. Find the volume of the solid obtained by rotating the region around the y-axis. How do I solve this

I'm a little rusty with this so somebody please correct me if I'm wrong:
[hide]
What you can do is express $y$ as a function of $x$ and then integrate with respect to $y$:
If $y = x^{2}$ and we are looking at the 1st quadrant, then $x=-\sqrt{y}$ and so now we find the volume of the solid:

$\int^{1}_{0}[(-\sqrt{y})^{2}\pi \, ]dy$

[/hide]

EDIT: flipped around the 1 and 0...

In the first quadrant, everything is positive. I'd have written that integral as

$\int_{0}^{1}\pi\left(\sqrt{y}\right)^{2}\,dy.$

The alternative computation would set it up as

$\int_{0}^{1}2\pi x(1-x^{2})\,dx$

Both integrals are straightforward, and give the same answer.

I understand that you'd have to take the absolute value of my integral to find the volume, but the reason why I put in $-\sqrt{y}$ was because (and this is the least mathematical argument ever) if you draw out the half of the parabola that lies in the first quadrant and flip it over on its side (that is, you look at y as the horizontal axis and x as the vertical axis) the function lies under the y axis. Am I visualizing this wrong?

Why flip? Just let the whole thing stand in its original position. You're slicing into horizontal slices and rotating those slices about the vertical axis to form disks, and then you're integrating $dy.$ Nothing wrong with an integral $dy,$ and you don't have to reorient anything.

[This is what a typical calculus book these days would call the "disk" or "washer" method; my second integral represents the "shell" method.]


How many digits does Pi^250(to the power of 250) have?

please explain to me ...thank you

I think you look for the integral value, since the fractional part has infinit length

You only have to do some logarithm calculations, since the number of digits of an integer is the integral value of it's decadical logarithm plus one:
$digits(\pi ^{250})=[log_{10} (\pi^{250})]+1=[250 \cdot log_{10}(\pi)] + 1$ where $[.]$ means integral value

Note that all integers from 10^1 to 10^2-1 have 2 digits
10^2 to 10^3-1 have 3 digits,
10^(n-1) ti 10^n-1 have n digits

So logically,if x is a base 10 number and [j] denotes the integer part of j, then the number of digits in x is [log x] + 1 (where the logarithm is base 10)

Now read ZetaX's post and see if it makes sense.


Let $ n\in \mathbb{Z}_\plus{}$, $ t_1,t_2\in \{0,1,\ldots,9\}$, $ P(n)\equal{}\{p\in\mathbb{P}: p\leq n\}$ and define $ f(n,t_1)$ to be the number of $ t_1$'s in the decimal representation of elements of the set $ P(n)$. Is always $ \limsup_{n\to\infty} \frac{f(n,t_1)}{f(n,t_2)}\equal{}1$ as $ t_1$ and $ t_2$ are fixed so are the digits of primes equally distributed?

Yes, but for a rather trivial reason: there are very few numbers (much fewer than the number of primes) for which the distribution of their decimal digits is noticeably skewed. So, the vast majority of primes have about equal numbers of all digits in their decimal representations.


I had some trouble on some of these.  How do you unfold the prism in #30 to get 2 :sqrt: 113?

And #27 with the thing with the side view.  Anyone want to draw some pictures of them?

I'd help but I don't have the problems. Where can you get them?

My coach made copies of the problems, answers, graphs etc. and sent them out to me and the teammates.

I can help you with 30 though i don't have the question
Try unfolding the cube both ways and one of them comes out to be 4root29 and the other way comes out to be 2root113 which is clearly smaller than the first

But I don't see the other way to unfold it

Try rotating the entire prism 90 degrees on its flat base before unfolding it

[quote="nat mc"]OH![/quote]

Yep, that's right, cheer for good ol' Ohio...

Choose one rectangle on the left for the unfolded left side wall.  Ignore the other rectangles in the left column.  (Gecko's position is shown as a red plus.)  Choose one rectangle on the right for the right side wall.  Ignore the other rectangles in the right column.  (Fly's position is shown as a red plus.)

The 12 foot wide vertical section in the center consists of floor (shaded), which the gecko may not cross, front wall, ceiling, back wall, and floor (shaded).

The shortest distance between two points is a straight line.  So, we find the shortest paths from geckos on the left to flies on the right.  Five possible paths (with 4 distinct lengths) are shown.  All other paths are longer than those shown.

Which of the five paths is shortest?

I wasn't fast enough to even get to this problem...

But when I tried the other way, I got an answer larger than 4sqrt29...

I got 2root452 (14sq +16sq=452) 21.26feet and seems the same whether you use floor or ceiling as it is symmetrical??


[quote="SUPERMAN2"]Solve the system of equations:
$ x^3 \minus{} 3x \equal{} y$
$ y^3 \minus{} 3y \equal{} z$
$ z^3 \minus{} 3z \equal{} x$[/quote]
Posted many times before,You can use Search function.
By the way, the main idea is proving that $ x , y , z \in [ \minus{} 2;2]$.
And use the substitution  $ x \equal{} 2cos \alpha$
we have the equation $ cos(27\alpha) \equal{} cos(\alpha)$
Solving it is not a hard work


Any body knows who is qualified for CMO? I got 71/80... but i'm not sure how do we get to know when to write the test?

Yes.  Private Message me with your credit card number and I will acess the appropriate CEMC files.

  Regards,

  G-UNIT

More seriously. If you are an official contestant, then I'm pretty sure that you've qualified. Within a few weeks, you'll get a letter from your math teacher (or whoever proctored your COMC) with the details for CMO as well as a book prize for qualification.

Dude G-Unit, what are you saying? lol... Who are you really? 50 cent? Tony yayo? Olivia? XD (joking)

What score is considered qualified to write the CMO?

Yeah, does anyone know what score is required to qualify?
You can qualify from COMC or AMC right?

I would say you're qualified for the CMO if you scored a 69 on the COMC.  The CMS might qualify some kids who scored a 68 (maybe those with cool names...I ono), or even some who scored less than a 68 (frequent OLYMON contributors, past CMO qualifiers, people with a good record I would think).

You can't qualify for the CMO through the AMC.


A frog is placed at the origin on a number line, and moves according to the following rule: in a given move, the frog advanced to either the closest integer point with a greater integer coordinate that is a multiple of 3, or to the closest integer point with a greater integer coordinate that is a multiple of 13.  A [i]move sequence[/i] is a sequence of coordinates which correspond to valid moves, beginning with 0, and ending with 39.  For example, 0, 3, 6, 13, 15, 26, 39 is a move sequence.  How many move sequences are possible for the frog?

[hide]Define $ F(n)$ as the number of ways to get to $ n$. Obviously, $ F(n)$ is equal to the sum of $ F$ applied to every number from which you can reach $ n$.
$ F(0) \equal{} 1$
$ F(3) \equal{} 1$
$ F(6) \equal{} 1$
$ F(9) \equal{} 1$
$ F(12) \equal{} 1$
$ F(13) \equal{} F(0) \plus{} F(3) \plus{} F(6) \plus{} F(9) \plus{} F(12) \equal{} 5$
$ F(15) \equal{} F(12) \plus{} F(13) \equal{} 6$
$ F(18) \equal{} 6$
$ F(21) \equal{} 6$
$ F(24) \equal{} 6$
$ F(26) \equal{} 29$
$ F(27) \equal{} 35$
$ F(30) \equal{} 35$
$ F(33) \equal{} 35$
$ F(36) \equal{} 35$
$ F(39) \equal{} 169$[/hide]

Here's a brute force method:
[hide]
Using the jump of 13 a total of 3 times:
5*5*4 = 100 ways to reach 39

Using the 13 jump only 2 times:
3 possibilities-
13 and 26: 5*5= 25 ways to reach 39
26 and 39: 4*4 = 16 ways
13 and 39: 5*3 = 15 ways

Using the 13 jump only once:
13: 5 ways
26: 4 ways
39: 3 ways

Not using the 13 jump at all:
1 way

100+25+16+15+5+4+3+1=169[/hide]

Another way
[hide]Going through 13 and 26: 5 to 13, 5 to 26, 5 to 39.  5x5x5=125

Going through 13 or 26: one side is 5 the other is 4. 2x5x4=40

Going through neither: 4 ways to jump to 39. 4=4

125+40+4=169[/hide]

(I had 5x5x5+4(5+5+1) on my paper but i forgot about the 4 and got the answer 136)

ya, i did it the way above, but i kept bubbling it wrong and erasing. i hope the scantron doesnt mistaken my answer :maybe:

Is it a coincidence that the answer is $13^{2}$ or is there some deep connection?

[quote="Ignite168"]Try it for 3 and 17 instead and see.[/quote]
and make it go to 51 and see what happens then

i think it will be 6x6x6+2x6x5+5=279 close though

[quote=mysmartmouth]A frog is placed at the origin on a number line, and moves according to the following rule: in a given move, the frog advanced to either the closest integer point with a greater integer coordinate that is a multiple of 3, or to the closest integer point with a [b]greater integer [/b][b]coordinate[/b] that is a multiple of 13.  A [i]move sequence[/i] is a sequence of coordinates which correspond to valid moves, beginning with 0, and ending with 39.  For example, 0, 3, 6, 13, 15, 26, 39 is a move sequence.  How many move sequences are possible for the frog?[/quote]

I don't understand what is meant by the bolded parts.

I think the greater integer coordinate means that when you go to the nearest integer, you have to go to the larger one.

For example going from 6 to 13 is allowed, but 13 to 6 is not allowed.

I believe RamanConjecture was just making a point that it has to go up; I think that 6 to 9 does work.

Ah I see , thanks for the clarification both of you!

O oops so it would be 1+2^3 but this still gives 9^3 = 729 which is not right??

[quote=eagles2018]@above, to get from 26 to 39, the multiples of 3 are 27, 30, 36, and [b]39[/b], so you overcounted for that case.

Does anyone else hate frogs, or is it just me?[/quote]

I also despise frogs #frogblocked

[quote=OlympusHero]We will use recursion. Let $p_x$ be the amount of ways to get a move sequence ending with $x$. Then $p_{x+13} = p_x(1 + 2^4)$ because we can either directly go $13$ or we can go in four 3's, with two choices for the number each time, giving $2^4$ ways. But $p_0 = 1$, so then we'd get $p_{39} = 17^3$, which isn't right :maybe: What did I do wrong?[/quote]
I don't think your recursion works (even the updated one), since you don't neccesarily have to hit $26$ or $13$ along the way.

[b]Solution:[/b] I tackled this with casework. You can view this problem as hopping through multiples of $3$ and pitstopping at a $13$ if desired. Case 1: 0, smth, 13, smth, 26, smth, 39. Let's consider the possibilities for 0 -> ?? -> 13. We can skip directly to it in one move, or go to 3 then 13, or go to 3 to 6 then 13, or to 3, 6, 9, 13, or 3, 6, 9, 12, 13. This gives us 5 ways. However notice that 13 -> ?? -> 26 and 26 -> ?? -> 39 are basically the same setups. So for case 1 there are $5^3 = 125$ possibilities. Now we have 0 -> ?? -> 26 -> ?? -> 39 (without ever hitting 13). So, for this 0, 3, 6, 9, 12, 15 is required (for obvious reasons). And then from 15 we can directly skip to 26, or go to 18 then to 26, or 15, 18, 21, 26, or lastly 15, 18, 21, 24, 26. That gives us 4 ways. Notice that 26 -> ?? -> 39 is 5 possibilities as we already found. So, that gives us $4 \cdot 5 = 20$ ways. For 0 -> ?? -> 13 -> ?? -> 39, this is basically the same as the case we just covered, so $20$ ways for that as well. Lastly, 0 -> ?? -> 39 has 4 ways (27, 39; 27, 30, 39; 27, 30, 33, 39; 27, 30, 33, 36, 39). So, in total there are $5^3 + 2 \cdot 4 \cdot 5 + 4 = \boxed{169}$.

Now, while reading through other's solutions I realized that recursion is also a viable way to tackle this problem. The thought comes from analyzing these potential sequences and seeing they depend a lot on previous ones (0 to $n$). In this fashion, let $F(n)$ denote the number of ways that you can get from 0 to $n$ (as described in the problem statement). $F(0), 1, F(3) = 1, F(6) = 1, F(9) = 1, F(12) = 1, F(13) = F(0) + F(3) + F(6) + F(9) + F(12) = 5$. (hopefully this should be clear as to why this is the case.) Carrying on, $F(15) = F(12) + F(13) = 6$ (directly from problem statement). Continuing, $F(18) = 6, F(21) = 6, F(24) = 6, F(26) = F(13) + F(15) + F(18) + F(21) + F(24) = 29, F(27) = F(24) + F(26) = 35, F(30) = 35, F(33) = 35, F(36) = 35, F(39) = F(26) + F(27) + F(30) + F(33) + F(36) = \boxed{169}$. (This type of recursion should make sense as the only way to jump to a number $n$ is from $n-13$ or from a number $3k > n-13$.)


Hey,I have just found a hard game, http://godtower.com/
And I think all member here like it
I only passed level 10,lv 11 is too hard
Sorry if it is old for you

What? When I click on rules, the text isn't in English...
How do you play?

Indeed, this is quite old. Level 11 doesn't require such a big leap. Have you found the movie director?

@bleutofu: the goal is to find the password to get to the next level. There are several hints on the page, so you need to work it out.


geometry
incenter
geometric transformation
reflection
rotation
homothety
ratio

The incircle of a right triangle ABC with A = 90  touches BC at P. If AD is an altitude of triangle ABC and  K is  the incenter of triangle ABD , prove that PK is perpedicular to  AB.

( I ' ll post some more problems - unsolved in this book. Please post your nice solutions. I'm sure that many friends of geometry would like to see them or send theirs)

Let $I$ be the incenter of $ABC$ and suppose $IP$ intersects $AB$ at $Q$. Reflect the triangle $ABC$ across $BC$ (see attached figure). A rotation of angle $B$ followed by a homothety of ratio $\cos {B}$, both centered at $B$ sends triangle $A'CB$ to triangle $DAB$. This transformation sends the points $P,I',Q'$ to the points $L,K,P$, respectively. It follows that $L,K,P$ are collinear and $PK$ is perpendicular to $AB$.

In fact, it is very easy - in a right-angled triangle, everything is similar...

[color=blue][b]Problem.[/b] Let ABC be a right-angled triangle with A = 90°. The incircle of triangle ABC touches the side BC at P. Let D be the foot of the altitude of triangle ABC from the vertex A, and let K be the incenter of triangle DBA. Prove that the line PK is perpendicular to AB.[/color]

[i]Solution.[/i] Let L be the point where the incircle of triangle DBA touches the side AB. Then, $KL\perp AB$, since K is the center of this incircle. Now, if we succeed to show that $PL\perp AB$, then it will follow that PL || KL, thus the points P, L and K lie on one line, and since $KL\perp AB$, this line is perpendicular to AB. Consequently, the line PK is perpendicular to AB. This solves the problem.

Hence, in order to complete the solution, it remains to show that $PL\perp AB$.

Since < CAB = A = 90° and < ADB = 90°, we have < CAB = < ADB. Also, trivially, < CBA = < ABD. Thus, the right-angled triangles ABC and DBA are similar. The points P and L are corresponding points in these two similar triangles (in fact, P and L are the points where the incircles of these triangles ABC and DBA touch their hypotenuses BC and AB). Since corresponding points in similar triangles form corresponding segments and these make equal ratios, we conclude that $\frac{BL}{BP}=\frac{BA}{BC}$. Thus, by Thales, PL || CA. Now, since < CAB = 90°, we have $CA\perp AB$, and thus $PL\perp AB$. And the problem is solved.

[[b]EDIT:[/b] Sorry, Enescu, didn't see your post when I wrote mine.]

  darij

I note $S\in (AB),\ KS\perp AB$. The points $S,P$ are homologously in the similitude
$\triangle DBA\sim \triangle ABC\Longrightarrow \frac{SA}{SB}=\frac{PC}{PB}\Longrightarrow PS\parallel AC\Longrightarrow PS\perp AB\Longrightarrow$ $PK\perp AB.$


linear algebra
matrix
linear algebra unsolved

Hey Guys,

I have a linear algebra question for you. I'm just learning how to use Maple, so this question will probably be really stupid. The project isn't supposed to be hard, it's just supposed to get us used to Maple.

I'm supposed to (a) generate 50 5x5 matrices with entries in R if 1<R<50. Then, I have to find the determinant of each matrix and print all of this.

Then, (b) I have to do the same but now "with entries in Z/mod 5 {0,1,2,3,4}"

The code that our instructor gave us (which I can't decipher) is:

with(linalg):
for i in 1 to 50 m=RandomMatrix(5,5,Real)
do m Print(m,detm); od

Does anybody have any insight into this problem for Maple 12? Using his code definitely isn't required, though. I would appreciate any help!

Thanks so much!
Craig

This does the first part.

with(LinearAlgebra):
for i from 1 to 50 do m:=RandomMatrix(5,5,generator=1.0..50.0);
print(m,Determinant(m)); od:

  The first line loads the linear algebra package, giving you access to commands such as RandomMatrix. The second one assigns m to be a random matrix, and then prints it together with determinant. Notice that the colon after od (instead of semicolon) suppresses the output of m:=... command, to avoid m displayed twice. The argument "generator=..." indicates what kind of numbers should be used to fill the matrix. I put 1.0 and 50.0 with decimal dots to make the entries real rather than integers. If you use 0 and 4 instead, you'll get matrices filled with numbers {0,1,2,3,4}, as required for (b). I will leave it to you to compute determinant in (b) correctly.


Prove that if $ n>1$ then $ n^4 \plus{} 4^n$ is composite.

[quote="barcelona"]Prove that if $ n > 14$ then $ n^4 \plus{} 4^n$ is composite.[/quote]
[hide]
If $ n$ is even then the question is obviously true.

If $ n$ is odd then $ n^4 \plus{} 4^n \equal{} n^4 \plus{} 4 \cdot 4^{n \minus{} 1}$,$ \boxed{1}$

Setting $ n \equal{} 2k \plus{} 1$,then $ \boxed{1}$ can be rewritten on the form $ a^4 \plus{} 4b^4$ which is composite.....

Dimitris
[/hide]

If n is even it is trivial. If n is odd, write n=2k+1.  Then $ n^4\plus{}4^n\equal{}n^4\plus{}4\cdot{(2^k)}^4\equal{}(n^2\minus{}2^{k\plus{}1}n\plus{}2^{2k\plus{}1})(n^2\plus{}2^{k\plus{}1}n\plus{}2^{2k\plus{}1})$, by Sophie Germain. Both of these terms are clearly more than 1 for n>1, so it is composite.


.1. Let m and n be positive integers with n >12m(m+1). Show that the number of partitions of n into mdistinct parts equals the number of partitions of n −12m(m + 1) into at most m parts. (Korea, 1995)
Can you help me solve those problems? Thanks you very much !

[quote="dtdong91"]
9. For each positive integer n, let f(n) denote the number of partitions of n into powers of 2. Prove that,for any n ≥ 3, 2n2/4< f(2n) < 2n2/2. (IMO, 1997)4
[/quote]

see , pls  :wink: http://www.kalva.demon.co.uk/imo/isoln/isoln976.html

some these problems have this this forum, I think.


AMC AIME probability geometry inequalities ratio

1. For each non-empty subset of {1, 2, 3, 4, 5, 6, 7} arrange the members in decreasing order with alternating signs and take the sum. For example, for the subset {5} we get 5. For {6, 3, 1} we get 6 - 3 + 1 = 4. Find the sum of all the resulting numbers.

2. n people vote for one of 27 candidates. Each candidate's percentage of the vote is at least 1 less than his number of votes. What is the smallest possible value of n?

3. Let S = {1, 2, ... , 10}. Find the number of unordered pairs A, B, where A and B are disjoint non-empty subsets of S.

4. S is a set of positive integers containing 1 and 2002. No elements are larger than 2002. For every n in S, the arithmetic mean of the other elements of S is an integer. What is the largest possible number of elements of S?

5. P lies on 8y = 15x, Q lies on 10y = 3x and the midpoint of PQ is (8,6). Find the distance PQ.

6. 5N points form an array of 5 rows and N columns. The points are numbered left to right, top to bottom (so the first row is 1, 2, ... , N, the second row N+1, ... , 2N, and so on). Five points, P_1, P_2, ... , P_5 are chosen, P_1 in the first row, P_2 in the second row and so on. P_i has number xi. The points are now renumbered top to bottom, left to right (so the first column is 1, 2, 3, 4, 5 the second column 6, 7, 8, 9, 10 and so on). Pi now has number yi. We find that x_1 = y_2, x_2 = y_1, x_3 = y_4, x_4 = y_5, x_5 = y_3. Find the smallest possible value of N.

7. You have 8 different rings. Let n be the number of possible arrangements of 5 rings on the four fingers of one hand (each finger has zero or more rings, and the order matters). Find the three leftmost non-zero digits of n.

8. A bag contains 12 tiles marked 1, 1, 2, 2, ... , 6, 6. A player draws tiles one at a time at random and holds them. If he draws a tile matching a tile he already holds, then he discards both. The game ends if he holds three unmatched tiles or if the bag is emptied. Find the probability that the bag is emptied.

9. ABC is a paper triangle with AB = 36, AC = 72 and angle B = 90 degrees. Find the area of the set of points P inside the triangle such that if creases are made by folding (and then unfolding) each of A, B, C to P, then the creases do not overlap.

Have fun!

Question:

for number 1, there are two possible interpretations.  Do we take the sum over all subsets, or do we take the sum of all resulting numbers?  That is, {6, 3, 1} gives us 4.  {4} gives us 4.  Do we add 4 in our sum once or twice?

And, if we're supposed to add it twice (which I suspect is what the question wanted), is the answer [hide]448[/hide]?

By the way, if it's not too inconvenient, could you edit the questions to include what number they were on the AIME?

For #1...

We did this in our class

[hide]

There is a obvious pattern if you list out the first few,

For 1, there is one set - {1}=1

For 2, there are 3 sets - {2}, {1}, {1,2}= 2+2

For 3, there are 7 sets - {1}, {2}, {3}, {3,1}, {2,1}, {3,2}, and {3,2,1} = 3+3+3+3

The pattern would seem to be 2^(n-1)*n where n is the greatest integer in the set.  For n=7, it would be 2^6*7=448.

Right?

[/hide]

#5  Not bad with some easy algebra

[hide]

Although I probably did this the stupid way, I used my brute algebra

Line P, y=15/8a and line Q, y=3/10b

Midpoint: ((a+b)/2,(15/8a+3/10b)/2)=(8,6)

With much algebra, I got the points (32/7,60/7) and (80/7, 24/7)

With some more algebra, the distance comes to a nice 60/7.  Right?

[/hide]

Answer to 4 depends on whether the integers in S have to be distinct or not - the two cases are quite different. This might depend on how much you read into the word "set" i.e. is {1,3,3,3,2002} a set because the 3s are "different" 3s, or is it not a set because it contains repeated elements.

I [i]think[/i] the AIME requires answers in the range 000 to 999, in which case the implication is that the integers in S are distinct. But for extra credit you could try to solve it in both cases.

Different way to do #1: [hide]for 1 to be positive, there are 1+6C2+6C4 + 6C6, and there are  6C1+6C3+6C5 ways for 1 to be negative. (6C0-6C1+6C2-6C3+6C4-6C5+6C6)1=0. Same for 2,3,4,5,6, which will all be 0.For 7, it's always positive, with 2^6 ways containing 7. so 7*2^6=448.[/hide] QED

That's how I did number 1, too, beta.  Gandalf -- the word set almost always is taken to mean "with no repeating elements."  If they want repeating elements, they usually have to say it.

For number 4, what i did was:

 [hide]Call the sum of other numbers S,and we have n+1 member,  we have (S+2002)mod n :equiv: 0. (S+1)mod n :equiv: 0. S mod n :equiv: (n-1). 2002 mod n :equiv: n+1, which means 2001 mod n :equiv: 0. The largest n will be 2001. The answer should be 2002. This means there are 2001 1s, and one 2002, this works.[/hide]

If they don't allow repeat elements, the answer should be [hide]30[/hide]

tetrahedr0n wrote:2. n people vote for one of 27 candidates. Each candidate's percentage of the vote is at least 1 less than his number of votes. What is the smallest possible value of n?

[hide]Set up an inequality, we have p_1 :le: p_1(n)/100-1, where p_1 is the percentage of candidate 1. Same thing for candidate 2,3...27. Add all the inequality and then simplify it we have n :ge: 127. 127 doesn't work since splitting it evenly, we have 4s, 4/127=3.14%. n has to be larger. When everything is 5 we have n=135 which works because 5/135=3.7% so 127 :le: n :le: 135. So n=134 which is the answer.[/hide]

I need some help on number 9. I think you need Ceva's theorem to solve it, which I totally forgot other than that it somehow relates to a point inside of a triangle and some ratios add up to 1.
Thanks!

Huh, number 9 is interesting -- I'm not sure if I've seen it before.

You might want to reprove Ceva's theorem for yourself, just to make sure you know how to do it.  The theorem states that given any point P and triangle ABC, such that AP meets BC at A', BP meets CA at B', and CP meets AB at C', we have
BA'/A'C * CB'/B'A * AC'/C'B = 1.  The nicest proof I know involves areas.

Thanks. For number 8, I think I solved it, although I almost never get a probability problem right on my first try. [hide]The first one you pick can be anything, the second one you pick must be either same or different(1/11 or 10/11). If the first two are same, then the probability is anything again, if the first two are different, we must have the third one to match with one of them, or 2/10... the same kind of thinking should get you some thing like 10/11*2/10*8/9*2/8....1/2 or something like 1/11*1*1/9*1*6/7*2/6..........note that when the numerator is 2 and denominator is even(except 2 because you can pick any one the 2 tiles, but for 3, it works just like other evens), then the denominator cancels out with the numerator infront of it. When the denominator is even and numerator is not 2, the whole fraction has to be 1, the fraction in front of it must be 1/odd(same for 3). so the denominator of the final probability is 11*9*7*5*3. The numerator

equals 1+5C1(2)+5C2(4)+5C3(8)+5C4(16)+5C5(32)=1+10+40+80+80+32=243. 243/11*9*7*5*3=9/385, which is the answer. [/hide].


probability
expected value
probability and stats

E[g(X)|X E B] =  E[g(X) I (X E B)]  /  P[ X E B]

I am trying to figure out the meaning, in words of part of this formula. Specifically  E[g(x) I (X E B)] . In my book the "I" sort of leans to the right. I don't know what this means and there's no explanation.
Help?

That "I that sort of leans to the right" must be the letter $ I$ being used as an indicator function. Let me change to notation slightly to use $ \mathbf{1}$ for that rather than $ I.$

In particular, define $ \mathbf{1}_B\equal{}\mathbf{1}_{X\in B}$ to be the random variable that takes the value $ 1$ on the set $ B$ and the value $ 0$ on the complement of $ B.$ The expression you're trying to figure out should then be:

$ E[g(X)\,|\,X\in B]\equal{}\frac{E[g(X)\cdot \mathbf{1}_B]}{P(B)}.$

Here's an illustration. Let $ X$ take on the values $ 0,1,$ and $ 2$ with probabilities $ \frac12,\frac13,$ and $ \frac16.$

Now, compute $ E[X\plus{}2\,|\,X>0].$ By the expression above, this should be

$ \frac{E[(X\plus{}2)\cdot \mathbf{1}_{\{X>0\}}]}{P(X>0)}\equal{}\frac{2\cdot \frac12\cdot 0\plus{}3\cdot \frac13\cdot 1\plus{}4\cdot\frac16\cdot 1}{\frac13\plus{}\frac16}\equal{}\frac{10}3.$

To return to what had you puzzled: $ E[(g(X)I(X\!\in B)]$, or as I wrote it, $ E[g(X)\cdot I_B],$ that is the expected value of the random variable that equals $ g(X)$ when $ X\in B$ and equals zero otherwise.


Let ABC be an acute angled triangle; let D,F be midpoints of BC AB respectively.Let the perpendicular from F to AC and perpendicular at B to BC meet in N.Prove that ND is equal to the circumradius of Triangle ABC

here's the answer.... simply genius  (AKASHNIL).
http://www.mathlinks.ro/viewtopic.php?t=238343


What is the maximum number of points of intersecton of the graphs of two different fourth-degree polunomial functions y=P(x) and y=Q(x), each with leading coefficient 1?

[hide]Well, at the intersection points, $P(x)-Q(x) = 0$.
The LHS is a polynomial whose maximum degree $3$, and the maximum number of real roots it can have is $3$, so the answer is $\boxed{3}$.[/hide]

EDIT: Example:
$P(x) = x^{4}+2(x-1)(x-2)(x-3)$
$Q(x) = x^{4}+(x-1)(x-2)(x-3)$

Extensions:
What is the maximum number of points of intersecton of the graphs of two different $n^{\textrm{th}}$-degree polynomial functions $y=P(x)$ and $y=Q(x)$, each with leading coefficient $1$?

What is the maximum number of points of intersecton of the graphs of two different $n^{\textrm{th}}$-degree polynomial functions $y=P(x)$ and $y=Q(x)$?

n-1, by the same logic

n, by the same logic...


Find all the sequences like $x_1,x_2,x_3,...$ so that every $x_k$ is natural and:

$gcd(x_m,x_n)=x_{gcd(m,n)}$

Did you get the idea for this problem looking at Fermat numbers? :)

These sequences are called Mersenne sequences.
When I remember correctly, they can be written as some product of a special kind (I've to think about which it was) and only those.

Well, every sequence defined by $x_n : = \alpha^n- \beta^n$ with $\gcd(\alpha, \beta)=1$ and some similar ones should work, giving e.g. Fibonacci numbers or also the Mersenne numbers itself (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=54800 ).


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}4(ab\plus{}ac\plus{}bc).$ Prove that
\[ a^3\plus{}b^3\plus{}c^3\geq12abc\]

\begin{align*}
a^3 + b^3 + c^3 - 3abc &= (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \\
&= 3(a + b + c)(ab + ac + bc) \\
&\ge 3 \cdot 3 \sqrt[3]{abc} \cdot 3 \sqrt[3]{a^2 b^2 c^2} \\
&= 27abc,
\end{align*}
so $ a^3 + b^3 + c^3 \ge 30abc$.  And since $ a$, $ b$, and $ c$ cannot all be equal, even 30 is not the best constant.

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 + b^2 + c^2 = 4(ab + ac + bc).$ Prove that
\[ a^3 + b^3 + c^3\geq12abc
\]
[/quote]

Certanly, nsato! I think, the "best constant" $ <40.$ :wink:

hmm... the constant I get is much bigger than 40...
[hide]
Normalize by letting $ x\equal{}\frac{6a}{a\plus{}b\plus{}c}$ etc. Because the condition and inequality are homogenous, the inequality in $ a,b,c$ reduces to one in $ x,y,z$ where $ x,y,z$ nonnegative and $ x\plus{}y\plus{}z\equal{}xy\plus{}yz\plus{}zx\equal{}6$. This also means $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}108$.

Now the polynomial $ P(t)\equal{}(t\minus{}x)(t\minus{}y)(t\minus{}z)\equal{}t^3\minus{}6t^2\plus{}6t\minus{}p$ (for $ p\equal{}xyz$) has $ x,y,z$ as roots. But also,
\[ t^3\minus{}6t^2\plus{}6t\minus{}4\left( \sqrt{2}\minus{}1\right) \equal{}\left( t\minus{}\left( 2\minus{}\sqrt{2}\right) \right)^2\left( t \minus{}\left( 2\plus{}2\sqrt{2}\right) \right)\]
So we must have $ p\leq 4\left( \sqrt{2}\minus{}1\right)$. Otherwise, the above expression is positive for $ x,y,z$, which would mean $ x,y,z>2\plus{}2\sqrt{2}$ and then $ x\plus{}y\plus{}z$ would be much greater than 6.

Now $ xyz\leq 4\left( \sqrt{2}\minus{}1\right)$ and $ x^3\plus{}y^3\plus{}z^3\minus{}3xyz\equal{}108$ implies
\[ \frac{x^3\plus{}y^3\plus{}z^3}{xyz}\equal{}3\plus{}\frac{108}{xyz}\geq 30\plus{}27\sqrt{2}\]
So the best constant is $ 30\plus{}27\sqrt{2}$, with $ x\equal{}2\plus{}2\sqrt{2},y\equal{}z\equal{}2\minus{}\sqrt{2}$ producing equality.
[/hide]

Very nice reasoning, scorpius119! My mistake. Thank you!


I got this right by my instinct, but I want to know the proper way of solving it.

12. For what value of $ x$ does $ \frac{8}{9}\equal{}\frac{x}{x\plus{}\frac{x}{x\plus{}x}}?$

[quote="DonkeyKong"]I got this right by my instinct, but I want to know the proper way of solving it.

12. For what value of $ x$ does $ \frac {8}{9} \equal{} \frac {x}{x \plus{} \frac {x}{x \plus{} x}}?$[/quote]

[hide="My Solution"]
You solve from the bottom up.
$ \frac{x}{x\plus{}\frac{x}{x\plus{}x}}$
Goes to
$ \frac{x}{x\plus{}\frac{x}{2x}}$
Goes to
$ \frac{x}{x\plus{}\frac{1}{2}}$
Multiply the top and bottom by 2
$ \frac{2x}{2x\plus{}1}$
So then, x must equal 4.
Because
$ \frac{2(4)}{2(4)\plus{}1} \equal{} \frac{8}{9}$
[/hide]

[hide="alternate solution"]
By repeatedly inverting fractions you can do it this way too

$ \dfrac{8}{9} \equal{} \dfrac{x}{x \plus{} \dfrac{x}{x \plus{} x}}$

$ \dfrac{8}{9} \equal{} \dfrac{x}{x \plus{} \dfrac{x}{2x}}$

$ \dfrac{9}{8} \equal{} \dfrac{x \plus{} \dfrac{x}{2x}}{x}$

$ \dfrac{9}{8} \equal{} \dfrac{x}{x} \plus{} \dfrac{\dfrac{x}{2x}}{x}$

$ 1 \plus{} \dfrac{1}{8} \equal{} 1\plus{} {\dfrac{1}{2x}}$

$ \dfrac{1}{8} \equal{} {\dfrac{1}{2x}}$

$ \dfrac{8}{1} \equal{} {\dfrac{2x}{1}}$

$ 8 \equal{} 2x$

$ 4\equal{}x$

I always really liked this technique for solving these problems.

Hope that helps

[/hide]

the first thing I did was replace x/x+x with 1/2 then cross multiplied and got 9x=8x+4, subtracting 8x yeilds x=4

ZOLOJETTO

[hide="Click for my solution"]Cross multiply and solve.
$ 8x \plus{} \frac {8x}{2x} \equal{} 9x$
$ 8x \plus{} 4 \equal{} 9x$
$ \boxed{x \equal{} 4}$[/hide]
Try not to revive un-necessary posting topics.
BTW, Welcome to AoPS/Mathlinks!

I agree that the mods should lock all topics older than 1 month.

i disagree because then someone that has a question about that problem/topic has to make another post about the same thing. senseless

Yeah, but most of the time dead threads are revived by this person who just posts another solution for it, which has already been posted or something like that...


can anyone help me write a plain language program and eventually an actually program on a graphing calculator? i need to display an animation of an air balloon (or something closely resembling it) floating across the screen, there also needs to be a person in the balloon that at some point falls out and descends on the screen. if anyone has any suggestions or anything at all please let me know!!

well this is ASM programming in which you need to knwo a variety of other  comptuer programming (i think like C) in order to do it well...

Well you could use sprites but it's kind of tricky. Calculator programming is only really fun if you know assembly (ASM) but I would definitely not attempt learning it as your first language. You should have lots of experience with C or Java or something before you attempt to learn it.

[quote="funcia"]wait a minute :blush:
if you know just java can you program a calculator?[/quote]

umm

[url]http://nwps.ws/~dragonfire/Asmin28/lesson/day01.html[/url]

[quote="mathchick33"]umm yeah im not really sure what all that stuff is, but i just need a program with like line commands, loops, circle commands, etc. to draw the animation. anyone??[/quote]

what, do you want us to actually write it for you? :P

nah i just need some help putting commands in to make the balloon shift positions and move up to the right and across the screen

well it's generally a good idea to learn how to program before you start programming

at least i always thought so


logarithms
calculus
integration
calculus computations

for which $ p$ does $ \sum_{4}^{\infty}\frac{1}{n(\log{n})(\log{\log{n}})^p}$ converge?

The integral test is really well suited. Make change of variables to reduce to easy case

The condensation test is also well suited. (This is not an accidental pairing. The integral test and the condensation test share a key hypothesis - that the terms are monotone decreasing.)

For decreasing $ a_n,$ the condensation test says that $ \sum a_n$ is comparable to $ \sum 2^ka_{2^k}.$

In this case,

$ \sum_n\frac1{n(\log n)(\log\log n)^p}$ compares to $ \sum_k\frac{2^k}{2^k(ak)(\log k\plus{}b)^p}$ for some constants $ a$ and $ b$ that I don't care about.

Cleaning it up this is comparable to $ \sum_k\frac{1}{k(\log k)^p}$

We run the condensation test again, to find that this is comparable to

$ \sum_j\frac{2^j}{2^j(cj)^p}$ which is a constant times $ \sum_j\frac{1}{j^p}.$

This is the familiar "$ p$-series," so it converges for $ p>1$ and diverges for $ p\le 1.$ (We could even go on to settle the $ p$-series issue with one more use of the condensation test.)


[quote="mathkid"]This is purely theoretical, but I am wondering if at any point in time, a temperature function of points on earth would be a continuous function.

Seems to me like it would, but nature does some mysterious things.[/quote]

I'm not completely convinced, since if temperature is the movement of discrete objects, I would expect it to be discrete in some way. (Although I think everything gets complicated at that level (of smallness))

[quote="mathkid"]I'm asking because of the proof that at any time, there are two polar opposite points on earth with the same temperature.

It uses the intermediate value theorem and assumes the continuity of the temperature function, so I was curious.[/quote]

Yeah, I've heard that before, yet I am not convinced of the truth of the continuity of temperature (although it may appear that way with all theoretical equipment (since they must be constructed of atoms)

On a quantum level, nothing is continuous.  The "proof" that two polar opposite points on Earth have the "same" temperature is a proof about a [i]model[/i] of the temperature on the Earth at any point in time.


If $a,b,c$ are real positive numbers. Prove:

$|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}|\leq 1$

Easy enough. The left hand expression is
\[|\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}| \le 1. \]


superior algebra
superior algebra theorems

Hello!

I am studying error correcting codes. I would like to solve the following problem:

[b]1.[/b] Show that there is no code with 4 infomation bits and 2 extra bits (hence a total of 6 bits) that can correct 1 error.

Similarly:

[b]2.[/b] Show that there is no code with 7 information bits and 6 extra bits that can correcting two errors.

How does one go about doing this? I am asking the question in the context of studying Hamming codes and BCH codes. Therefore I assume that one needs to show that there are no such things as BCH or Hamming codes that satisfy the conditions.  Is this correct, or is there some more general and algebraic way of showing that these code words can't exist?

Thanks in advance!

Los

Let's look at 1. There are 16 possible 4-bit messages; I would assume that code with 4 information bits must be able to transmit all 16 of these messages.

There are 4 possible 1-bit errors in a 4-bit message; if you include the 2 extra bits, there are actually 6 possible errors. That means that there are 7 possible received messages that have to be resolved to this one message. They can't "belong" to any other decoded message. So each decoded message has a set of 7 received messages that belong to it, and these sets cannot overlap. Multiply that by the 16 possible decoded messages, and we need 112 received messages to account for all of these sets. But there are only 64 6-bit messages. We don't have room to do that.

Note that with 3 extra bits, and allowing for the extra bits to be in error, then each message must have 8 received messages that belong to it. Multiply that by 16 decoded messages and we get 128, which is precisely the total number of received 7-bit messages. Hence, this simple count reveals the possibility of a single error correcting code with 4 information bits and 3 extra bits - but the code would have to be "perfect", with no wasted space. (In fact, there is such a code, but this argument does not construct it.)

Let's look at the second problem in the same way. A 13-bit message could have two errors in it in $ \binom{13}{2}\equal{}78$ ways and one error in $ 13$ ways, so (note that I am assuming the extra bits can have errors), we need $ 78\plus{}13\plus{}1\equal{}92$ possible received messages for each decoded message, and there are $ 2^7\equal{}128$ possible decoded messages. To do that, the required space is $ 128\cdot 92\equal{}11776.$ The amount of room we have to fit this is $ 2^{13}\equal{}8192.$ Since $ 11776>8192,$ there isn't enough space.

In other words, your impossibility results are simple counting. You should be able to see how to generalize and automate such arguments.


Fie a,b,c trei numere complexe astfel incat |a-b-c|=|a|=|b|=|c| si (b-c)(a+b)(a+c) diferit de 0 .Sa sa arate ca (b+c)(a-b)(a-c)=0 .

Ps:am rezolvat problema,dar am presupun ca toate numerele sunt de modul 1 si merge usor.Este corecta aceasta presupunere? Va multumesc

pai ideea e sa vezi ce se intampla daca inlocuiesti pe $a,b,c$ cu $a',b',c'$, astfel incat $a'=\lambda a$, etc, si $|a'|=1$.
in cazul de fata, o sa iti iasa $\lambda$ in fata si o sa se reduca, deci problema ramane ok.


Where can you make use of: if $ c|a$ and $ c|b$ then $ c|(a,b)$?

This does not follow from Bezout's identity, this is the [b]definition[/b] of the greatest common divisor.


does anyone know a good site for solving quintics? one that gives an exact answer.

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic

i know that site but it gives a decimal answer,

Quintics can't be solved exactly.  Haven't we told you this already?

[quote="t0rajir0u"]Quintics can't be solved exactly.[/quote]  Rather, quintics with rational coefficients cannot in general be solved exactly in terms of a combination of addition, multiplication, radicals and rational numbers.  I could define a function $T(a, b, c, d, e)$ which gives one real root of the equation $x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e = 0$ -- this would be an exact solution (by definition!) but it wouldn't have any nice properites (like being able to be expressed in terms of radicals, etc.), and in particular all the information you are likely to want would be a result of writing down a decimal approximation for it.

i thought so,
imagine a quintic has one real root  that we know, and other  four roots are like $\sqrt{2}$, so we divide the quintic by that rational root, it becomes quartic, then of course we can use the quartic formula to find an exact answer.so that means even the quintic has exact roots.

[quote="binomial_4eva"]imagine a quintic has one real root  that we know[/quote]

Yes, but this is, in the general case, not possible.  And there is no practical reason to ever use the cubic or quartic formulas - approximate methods work much faster and give all the accuracy you'll need.

There are formulas solving quintics in terms of Thetafunctions for example. And a lot of other ways, all of them using nontrivial functions.
Note that there is not only a lack of a formula always giving the solutions, but also there are very specific equations whose roots cannot expressed as stated above (note that this is really a stronger condition): one example is $x^{5}-4x+2$.

Another method for solving quintic equations is by "Bring Radicals".
Search for it online.

Which is essentially equivalent to what I wrote: if you invent a function which solves quintics exactly, you'll have a function which solves qunitics exactly.  (Really, the interesting content of your statement is that if we can solve qunitic equations of the form $0 = x^{5}+x+a$ then we can solve all quintic equations.)


calculus
integration
logarithms
real analysis
real analysis unsolved

Say I have the integral from 2 to infinity of dx/(x((ln x)^p)). How do I solve for p so that this integral:
1. converges
2. diverges

I am totally stuck with this problem

$I=\int_{2}^{+\infty}\frac{dx}{x\log^{p}x}$
If $I$ -converges, then series $\sum_{n=2}^{\infty}\frac{1}{n\log^{p}n}$ - is converges
If $I$ -diverges, then series $\sum_{n=2}^{\infty}\frac{1}{n\log^{p}n}$ - is diverges.

In series $\sum_{n=2}^{\infty}\frac{1}{n\log^{p}n}$, $0< a_{n+1}<a_{n}$ and $a_{n}\to 0$.

So $\sum_{i=1}^{\infty}a_{i}<\infty$ if and only if $\sum_{i=1}^{\infty}2^{i}a_{2^{i}}< \infty$.

$\sum_{i=1}^{\infty}2^{i}a_{2^{i}}= \sum_{i=1}^{\infty}2^{i}\frac{1}{2^{i}\log^{p}2^{i}}= \sum_{i=1}^{\infty}\frac{1}{i^{p}\log^{p}2}= \frac{1}{ \log^{p}2}\sum_{i=1}^{\infty}\frac{1}{i^{p}}< \infty$ iff $p>1$.

The exact $I$'s calculation is very easy!

The integral analogue of the condensation test: substitute $u=\ln x$.

to tell you the truth I don't really understand the 2nd post here. Can someone explain it more clearly??

It's an unnecessarily complicated method, using a test for series convergence known as the condensation test. Everything that series does can be done for integrals by a logarithm substitution.


Be [tex]ABC[/tex] triangle with circumcircle  [tex](O,R)[/tex] and external tangent circle [tex](O_1)[/tex] to [tex](O)[/tex] and the lines[tex]AB[/tex], [tex]AC[/tex] at points [tex]D,E[/tex] and  [tex]F[/tex] respectively. If $P$ is a point on the line $BC$ such that [tex]DP[/tex] is orthogonal to [tex]BC[/tex], and [tex]M[/tex] is midpoint of [tex]EF[/tex], prove that: [tex]DM^2 = 2R.DP[/tex]

First of all, let me just say that $M$ is, in fact, $I_a$, the excenter of $ABC$ corresponding to $A$. I won't prove this, it's a well-known problem, and the only proof I know is computational and rather messy.

Here's a sketch of what I'm trying to do:

It's clear that the conclusion is equivalent to the fact that the image of $(O)$ through the inversion of pole $D$ and power $DM^2$ must be a circle with the same radius. In order to show this, it would suffice to show that $DM^2=DB\cdot DC$ (I've noticed that this holds). After that, I've also notied that something even stronger holds: the triangles $DCM,DMB$ are similar, and this is, basicly what I want to show.

The result would follow easily if I could show that $DCFM$ is cyclic. By an angle-chase this would show that $DCM,DFE$ are similar, and in the same manner we would have $DMB,DFE$ similar.

Here's something else about the configuration: all of the above is equivalent to the fact that $D$ is on the symmedian of $BMC$, or that $DM$ cuts $(O)$ in a point equidistant from $B,C$. I'll be back later.

I've managed to prove my last assertion, but if you sum it all up you get a pretty nasty proof. Here's what I did: it's fairly easy to see that $EF$ is tangent to the circumcircle of $BMC$, so $EF$ is the exsymmedian of $BMC$, which means that we should show that $MB,MD,MC,MF$ form a harmonic quadruple. You cut these four lines with $FD$ and do the computations on that line, so they're easier. The rest consists of an application of Menelaus in $D EF$ with transversal line $MC$.


pigeonhole principle
inequalities
trigonometry
modular arithmetic

(1) Given 7 real numbers. Prove that there exist two denoted by $ x, y$ satisfy $ 0<\frac{x-y}{1+xy}<\frac1{\sqrt3}$.

(2) Prove that there exists an element in $ \{2^i-1|2\leqq\leq n\}$ is divided by $ n$, where $ n\geq1$ is an odd number.

[quote="ifai"](1) Given 7 real numbers. Prove that there exist two denoted by $ x, y$ satisfy $ 0 < \frac {x - y}{1 + xy} < \frac1{\sqrt3}$.[/quote]
Are there any more conditiong on $ x,y$? Maybe say $ x,y>0$ because otherwise the denominator can be zero  :o

[quote]
(2) Prove that there exists an element in $ \{2^i - 1|2\leqq\leq n\}$ is divided by $ n$, where $ n\geq1$ is an odd number.[/quote]
Maybe you mean that there exists two elements who difference when divided by $ n$ is an odd number?
Because just let $ n=2k$ then among $ 2^1-1,...,2^n-1$ none of these numbers are divisible by $ n$ because it is even.

[hide="(1)"]I assume that the real numbers are distinct. Because otherwise we can choose $ x \equal{} y$ and the inequality doesn't hold.

Assume that the numbers are $ x_1 < x_2 < \dots < x_7$. Let $ x_i \equal{} \tan X_i,X_i\in[0,\pi]$. Then we need to show that $ \exists i,j|0 < \tan(X_i \minus{} X_j) < \frac {1}{\sqrt 3}$ or $ 0 < X_i \minus{} X_j < \frac {\pi}{6}$. Consider the differences $ X_i \minus{} X_j$ where $ 1\le j < i\le 7$. All of them are $ > 0$. So we'll be done if we can show that one of them is $ < \frac {\pi}{6}$. Assume the contrary that $ X_i \minus{} X_j > \frac {\pi}{6}$ for all $ i > j$. Then we have $ X_2 > X_1 \plus{} \frac {\pi}{6}\implies X_7 > X_1 \plus{} 6\cdot\frac {\pi}{6} \equal{} X_1 \plus{} \pi$, which is a contradiction.[/hide]

EDIT: [hide="(2)"]Consider the numbers $ 2,2^2,\dots,2^{n \plus{} 1}$. All of them are prime to $ n$, since $ n$ is odd. There are $ n$, namely $ 0,1,2,\dots,n \minus{} 1$, least residues $ \mod n$, but we have $ n \plus{} 1$ integers to deal with. Hence two of them must yield the same remainder upon division by $ n$. Let $ 2^i\equiv 2^j\pmod n$. This implies that $ 2^j(2^{i \minus{} j} \minus{} 1)\equiv 0\pmod n$. But $ n\not|\,\ 2^j$. Thus $ n\mid 2^{i \minus{} j} \minus{} 1$, which is what we wanted to prove.[/hide]

[quote="nayel"]
 So we'll be done if we can show that one of them is $ < \frac {\pi}{6}$. Assume the contrary that $ X_i \minus{} X_j > \frac {\pi}{6}$ for all $ i > j$.
[/quote]
I really like you solution :).

But the second part about proving $ X_i\minus{}X_j < \frac{\pi}{6}$ can be accomplished more elegantly. Consider the interval $ [0,\pi]$ break it up into $ [0,\pi/6),[\pi/6,2\pi/6),[2\pi/6,3\pi/6),[3\pi/6,4\pi/6),[4\pi/6,5\pi/6),[5\pi/6,6\pi/6]$. There are six subintervals and seven numbers $ X_k$ numbers placed into them. So two of them end up in the same subinterval and so thier max difference is $ < \frac{\pi}{6}$.


[quote="C'est lavie"]Comparison between $sin^{\alpha}A+sin^{\alpha}B+sin^{\alpha}C$ and $3(\frac{\sqrt{3}}{2})^{\alpha}$ with $\alpha>2$[/quote]
we can easily see that $sin^{\alpha}A+sin^{\alpha}B+sin^{\alpha}C\leq 3(\frac{\sqrt{3}}{2})^{\alpha}$ with 0<=alpha<=2

[quote="imathsvn"][quote="C'est lavie"]Comparison between $sin^{\alpha}A+sin^{\alpha}B+sin^{\alpha}C$ and $3(\frac{\sqrt{3}}{2})^{\alpha}$ with $\alpha>2$[/quote]
we can assert that if  \alpha >2 then n $sin^{\alpha}A+sin^{\alpha}B+sin^{\alpha}C \geq 3(\frac{\sqrt{3}}{2})^{\alpha}$ not true[/quote]did you use the Power Mean Inequality? or Jensens? :maybe:

If LHS>=RHS then
$A\to \pi-$
$B,C\to 0+$
$LHS\to 0$->False

who moved my topic ????This problem is observed by high school level,pls transfer it!!!!

nobody can solve it ????i am waitting 4 a effective way  :D  :D  .Are u superman ??? i think you are,come on !!!  :)

ok,i think this problem's very interesting,why dont you like it???So,it's really really hard and it need to be discovered .Ok,i need your replies  :)


How do you prove

Sum (cyclic): a^2/((a+2b)^2+(a+2c)^2)<=1/2 ?

Could it be the it's a strict inequality? ($<$ instead of $\leq$)

\[\sum_{cyc} \frac{a^2}{(a+2b)^2+(a+2c)^2} \leq \sum_{cyc} \frac{a^2}{2(a+b+c)^2} = \frac12 \cdot \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac12\]

Nice question :)

Numerator is the one above the fraction line right?

Anyway I rather like the problem as it was... now it's too trivial

\[\sum_{cyc} \frac{a^2+2bc}{(a+2b)^2+(a+2c)^2} \leq \sum_{cyc} \frac{a^2+2bc}{2(a+b+c)^2} = \frac12 \]