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In class, I am doing a project on transitive forces and collisions.  We're going to have a ball traveling down a ramp to impact another ball at the bottom of the ramp and take measurements and that kind of stuff.

Does anyone know of instances where this project has been done in the past?  If you have information on the design and/or results of the project, please post/tell me them.

We did it in school. Just like you said - build a ramp, have a ball travel down the ramp and hit another ball. You should measure the height from which you had the first ball travel and the speed of the second ball after the collision. To do it properly, you need the first bal to hit the second ball exactly in the center so you'd better build your ramp with some kind of groove in it and put both balls in that groove.
What should happen is that the first ball accelerates to speed $v=\sqrt{2gh}$ and then comes to complete stop when it hits the second ball. The second ball takes all of the first ball's energy and should travel at the same speed $v=\sqrt{2gh}$ as the first ball did.

We did that too!!!! What's the big deal???

Do it then :) . As to air resistance, you're talking about laminar and turbulent air flow. You can forget about turbulent - that's only at higher velocities and there's almost no way to calculate it. So I'd try to measure resistance of the ball to laminar air flow - you'll probably notice that it's very low as long as you don't use any light or fluffy material for the ball :) . Same thing with friction - you have rolling friction between the ball and the groove / ramp / table. If you use steel or iron ball on some hard surface, this friction is also quite low and generally doesn't have to be considered. (Rubber tends to have greater friction coefficients. Don't use soft surface.) Finally, collision or rubber / steel balls at reasonable velocities are considered perfectly elastic (= no energy is lost). The collision wouldn't be perfectly elastic only if the balls would deform, but deforming steel balls is not really that easy and rubber balls tend to get back to their former shape after deformation (which means perfectly elastic collision as well). Just search the net for some values or do your own experiments to measure then to show your teacher how useless it is to consider any of these factors.

Okay, here's a few more things.  The teacher mentioned the following factors as well, although some of them might have been for our other project (although it was extremely similar), we're not quite sure:

Air resistance - turbulent & laminar flow, Reynold's Number, magnus force
Heating due to impact (??? - talk about negligible)
Universal gravitation
Stress and strain
Elastic materials theory
Elastic limits
Rotational motion and energy (torque)
Friction (static and kinetic)

Yeah, a whole load of stuff.  I believe we're supposed to do the experiment as well as research these topics, and since we're not allowed human/measurement error, our results had better be exact.  Though I don't know why we aren't allowed measurement error.  After all, we're not using top-of-the-line measurement devices...  Feedback?

I don't exactly understand, whether you are to explore and investigate in detail those phenomena you stated through the ball experiment, or just assess their influence on the whole experiment (and deviations from the simple model). I would guess the latter is the case.

Secondly, I don't really get the "we're not allowed measurement error" part. You should do your experiment as carefully as possible (=not doing large errors), maybe even as precisely as possible (choosing better materials, techniques, more precise tools), but there ALWAYS is an error in the measurement. Not only it is always there, but it is very important to know its magnitude. Being a teacher, if my student stated that they had zero measurement error, I would look at their work (and data analysis) with suspicion;-) So I am not sure what you mean by that.

I've been told the focus of this experiment is actually not experimental, but research.  We're in the *research phase*, and he says this is one of the most important phases.  And it is, after all, a research project.  The teacher is only looking for the trend, actually, and he'll accept some variation in the data.  I think he meant that we can't get the wrong trend or no trend at all when we're supposed to be getting one and blame it on human error or measurement error.  I guess we're supposed to do the research and then make our hypothesis graph based on it.  Then we run the experiment and see how everything matches up.

I think our teacher wants us to test a range of values for which factors like air resistance and other small forces might have a noticeable effect.  What heights, slopes, etc. have you used for which this might be true?  Or at least what heights have you used for which air resistance and other small factors have negligible effect?

I think that anything you can fit into a room is of dimensions small enough for these effects to be negligible. Well, perhaps if you made your ramp really long - say 6 ft. or more - than you could probably measure some air resistance / friction. Yes, I think that making the ramp long enough will do. Then you can also use different balls - not metallic - which have worse friction / aerodynamic / elastic properties. Anyway, you'd probably do best if you just measure it all and find the right dimensions. The bigger the thing, the better you'll see the things you're supposed to see.

Do you really think 6 feet will be enough?  I don't think we'll use different balls, since then they wouldn't be uniform and that could cause all sorts of problems.  We'd have to do more research, too.  :o  We were planning on using ping-pong balls...

Ping-pong ball is light enough, that could work. The problem is that it's really hard to calculate the air resistance because you need to know some coefficients which are pretty specific for every given setup... anyway, give it a try  ;) .

[b]What are transitive forces???[/b] (your header!!!)

[quote="CLee"]Well, it's not quite that simple.  The teacher wants us to research air resistance (laminar and some other kind of flow), elasticity, friction, etc... because we are doing it in an applied environment.  And he absolutely will not accept measurement/human error as an excuse for anything.  Quite annoying.[/quote]Nor should he. Human error can be corrected. And where its causes can be anticipated (which I would guess is almost everywhere in this kind of experiment) human error can be avoided altogether. The business of physics is measurement. To the extent that one's measurements contain avoidable errors one is doing junk science!

To answer the question of your original post, the experiment in question is one of the most basic in high school physics. The University of Washington, for example, uses a variant of it -offering little explanation to the student- in its investigation of cognitive models for learning physics.


[hide="partial solution"]
Let $ x^2\equal{}e^a y^2\equal{}e^b z^2\equal{}e^c$ With a+b+c=ln64

$ f(m)\equal{}\frac{1}{\sqrt{e^m\plus{}1}}$ is convex when $ e^m\ge2$ Which is the same as $ x,y,z\ge\sqrt{2}$ And we have by jensens the whole thing greater than or equal to $ \frac{3}{\sqrt{e^\frac{a\plus{}b\plus{}c}{3}\plus{}1}}\equal{}\frac{3}{\sqrt{5}}$ When each of $ x,y,z\ge\sqrt{2}$

If 2 are less than $ \sqrt{2}$ Then The L.H.S. is immediately greater than 1 as well.

Therefore, if the inequality is to be false, we have to have one of them less than $ \sqrt{2}$ and the other 2 greater than or equal to $ \sqrt{2}$

Then WLOG let x be the smallest one. Then we can do jensens on the larger 2 terms.

Have to go, someone else can complete this proof.
[/hide]


A sequence $ (a_n)$ is defined by $ a_1>0, a_{n\plus{}1}\equal{}\frac{a_n}{1\plus{}a_n^2} (n\ge 1)$
Prove that there exists a natural number $ k$ such that $ a_k>\frac{7}{10\sqrt{k}}$

$ \frac{1}{\sqrt{2n\plus{}1}}\not\equal{}\frac{2n\sqrt{2n}}{2n\plus{}1}$
and, no initial conditions given


function
inequalities
calculus
derivative
inequalities proposed

[quote="nguoivn"]Given $ a, b, c \geq\ 0$; $ k \geq\ 3$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:

$ \sqrt [k]{\frac {a^{k \minus{} 2}}{(b \plus{} c)^3}} \plus{} \sqrt [k]{\frac {b^{k \minus{} 2}}{(c \plus{} a)^3}} \plus{} \sqrt [k]{\frac {c^{k \minus{} 2}}{(a \plus{} b)^3}} \geq\ \frac {3}{\sqrt [k]{8}}$
 :)[/quote]
We assume the $ f(x)\equal{}\sqrt[k]{\frac{x^{k\minus{}2}}{(3\minus{}x)^3}}$
If $ f''(x) \ge 0$,for $ a,b,c \in(0,3)$(i'm sort of time so i cant check it :mad: ),we have:
$ f(a)\plus{}f(b)\plus{}f(c) \ge 3f(\frac{a\plus{}b\plus{}c}{3})\equal{}3f(1)$
So $ LHS \ge 3\cdot \sqrt[k]{\frac{1}{2^3}}\equal{}\frac{3}{\sqrt[k]{8}}$

what the fun, "i cant check it",  :rotfl:

woa, i have checked it, f''(x) not greater than 0 for all x   :lol:

[quote="Dimitris X"][quote="nguoivn"]Given $ a, b, c \geq\ 0$; $ k \geq\ 3$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:

$ \sqrt [k]{\frac {a^{k \minus{} 2}}{(b \plus{} c)^3}} \plus{} \sqrt [k]{\frac {b^{k \minus{} 2}}{(c \plus{} a)^3}} \plus{} \sqrt [k]{\frac {c^{k \minus{} 2}}{(a \plus{} b)^3}} \geq\ \frac {3}{\sqrt [k]{8}}$
 :)[/quote]
We assume the $ f(x) \equal{} \sqrt [k]{\frac {x^{k \minus{} 2}}{(3 \minus{} x)^3}}$
If $ f''(x) \ge 0$,for $ a,b,c \in(0,3)$(i'm sort of time so i cant check it :mad: ),we have:
$ f(a) \plus{} f(b) \plus{} f(c) \ge 3f(\frac {a \plus{} b \plus{} c}{3}) \equal{} 3f(1)$
So $ LHS \ge 3\cdot \sqrt [k]{\frac {1}{2^3}} \equal{} \frac {3}{\sqrt [k]{8}}$[/quote]

I am not sure that the function as you define it has non-negative second derivative.  You may however write the inequality as $ af(a) \plus{} bf(b) \plus{} cf(c)\geq(a \plus{} b \plus{} c)f\left(\frac {a \plus{} b \plus{} c}{3}\right)$ for $ f(x) \equal{} \frac {1}{\sqrt [k]{x^2(3 \minus{} x)^3}}$, and this one, if I am not mistaken, DOES satisfy $ f''(x)\geq0$ for $ 0<x<3$ - if my calculation is correct you do get something of the sort
\[ f''(x) \equal{} \frac {k \plus{} 1}{k} (f(x))^{k \plus{} 1}(3 \minus{} x)\left(\frac {(g(x))^2}{k} \plus{} h(x)\right) ,
\]
where $ g(x)$ is a binomial in $ x$ and $ h(x)$ is a trinomial, positive at $ x \equal{} 0$ and with two complex roots.  The solution was so ugly that I posted it, changed my mind and deleted it two minutes later...  I am only reposting it in a loose way in order to show that Jensen is in fact a possible path towards the solution, but I want to think that a more elegant solution is feasible...

EDIT: I rechecked my calculations, you do get
\[ f''(x) \equal{} \frac {(f(x))^{k \plus{} 1}(3 \minus{} x)}{k}\left(\frac {(6 \minus{} 5x)^2}{k} \plus{} 5x^2 \minus{} 12x \plus{} 18\right),
\]
and the trinomial has no real roots since $ 12^2 \equal{} 144 < 360 \equal{} 4\cdot5\cdot18$...

EDIT again: well, although this solution is actually quite ugly for my taste, it looks like it does prove the inequality for any positive $ k$...  Am I wrong?

I think your differentiation is correct, but how did you apply Jensen in this case? Is it weighted Jensen?

I thought it needs to be:
$ af(a) \plus{} bf(b) \plus{}cf(c) \geq 3f(\frac{a^2\plus{}b^2\plus{}c^2}{3})$

[quote="hendrata01"]I think your differentiation is correct, but how did you apply Jensen in this case? Is it weighted Jensen?

I thought it needs to be:
$ af(a) \plus{} bf(b) \plus{} cf(c) \geq 3f(\frac {a^2 \plus{} b^2 \plus{} c^2}{3})$[/quote]
Hmm.. I think that:
$ w_1f(x_1) \plus{} w_2f(x_2) \plus{} ... \plus{} w_nf(x_n) \geq$ $ (w_1\plus{}w_2\plus{}...\plus{}w_n)f \left (\frac{w_1x_1 \plus{} w_2x_2\plus{}...\plus{}w_nx_n}{w_1\plus{}w_2\plus{}...\plus{}w_n} \right )$
So it should be:
$ af(a) \plus{} bf(b) \plus{} cf(c) \ge (a\plus{}b\plus{}c) f \left ( \frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c} \right )$

So it would only holds if $ f(x)$ is increasing, which it isn't unfortunately..

[quote="Mathias_DK"][quote="hendrata01"]I think your differentiation is correct, but how did you apply Jensen in this case? Is it weighted Jensen?

I thought it needs to be:
$ af(a) \plus{} bf(b) \plus{} cf(c) \geq 3f(\frac {a^2 \plus{} b^2 \plus{} c^2}{3})$[/quote]
Hmm.. I think that:
$ w_1f(x_1) \plus{} w_2f(x_2) \plus{} ... \plus{} w_nf(x_n) \geq$ $ (w_1 \plus{} w_2 \plus{} ... \plus{} w_n)f \left (\frac {w_1x_1 \plus{} w_2x_2 \plus{} ... \plus{} w_nx_n}{w_1 \plus{} w_2 \plus{} ... \plus{} w_n} \right )$
So it should be:
$ af(a) \plus{} bf(b) \plus{} cf(c) \ge (a \plus{} b \plus{} c) f \left ( \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c} \right )$

So it would only holds if $ f(x)$ is increasing, which it isn't unfortunately..[/quote]
Yes because $ a\plus{}b\plus{}c\equal{}3$ so that's why I wrote it that way. And because $ f$ is not monotone increasing, that approach wouldn't work as well.

[quote="hendrata01"][quote="Mathias_DK"][quote="hendrata01"]I think your differentiation is correct, but how did you apply Jensen in this case? Is it weighted Jensen?

I thought it needs to be:
$ af(a) \plus{} bf(b) \plus{} cf(c) \geq 3f(\frac {a^2 \plus{} b^2 \plus{} c^2}{3})$[/quote]
Hmm.. I think that:
$ w_1f(x_1) \plus{} w_2f(x_2) \plus{} ... \plus{} w_nf(x_n) \geq$ $ (w_1 \plus{} w_2 \plus{} ... \plus{} w_n)f \left (\frac {w_1x_1 \plus{} w_2x_2 \plus{} ... \plus{} w_nx_n}{w_1 \plus{} w_2 \plus{} ... \plus{} w_n} \right )$
So it should be:
$ af(a) \plus{} bf(b) \plus{} cf(c) \ge (a \plus{} b \plus{} c) f \left ( \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c} \right )$

So it would only holds if $ f(x)$ is increasing, which it isn't unfortunately..[/quote]
Yes because $ a \plus{} b \plus{} c \equal{} 3$ so that's why I wrote it that way. And because $ f$ is not monotone increasing, that approach wouldn't work as well.[/quote]
Ohh. Didn't pay attention to the constraint :oops: So we agree :)

[quote="nguoivn"]Indeed, with $ 0 \leq\ k \leq\ 2$, the ineq is obviously trues.
With $ k \geq\ 3$, I had an easy proof by Am-Gm. Example, with $ k \equal{} 3$, we obtain a very nice result:
$ \frac {\sqrt [3]{a}}{b \plus{} c} \plus{} \frac {\sqrt [3]{b}}{c \plus{} a} \plus{} \frac {\sqrt [3]{c}}{a \plus{} b} \geq\ \frac {3}{2}$   ($ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$)  :)
But with $ 2 < k < 3$, I have no solution  :maybe:
Example, with $ k \equal{} \frac {5}{2}$, we obtain a hard ineq (I think so):
$ \frac {1}{b \plus{} c}\sqrt [5]{\frac {a}{b \plus{} c}} \plus{} \frac {1}{c \plus{} a}\sqrt [5]{\frac {b}{c \plus{} a}} \plus{}$ $ \frac {1}{a \plus{} b}\sqrt [5]{\frac {c}{a \plus{} b}} \geq\ \frac {9}{2\sqrt [5]{2}(a \plus{} b \plus{} c)}$[/quote]
This problem only prove for k=3.Because
$ \frac{\frac{k\minus{}2}{k}}{\frac{3}{k}} \ge \frac{1}{3}$.By power mean inequality,this problem only need prove for k=3.


A set of disks is being placed in a stack. The thickness of each sucessive disc added to the stack is 1.5 times the thickness of the disk immediately below it. What is the number of disks that must be placed in a stack in order for the thickness to be 25 times the thickness of the original disc?

I am thinking of how I could use my information (and my formula :D)  to find the answer, any ideas? :?

[hide="hint"]
Solve for $n$
$\displaystyle\sum_{x=0}^{n}1.5^x=25$

[hide="answer"]
6
[/hide][/hide]

[hide="hint"] You can set up a geometric sequence. Then use the formula for the sum to solve it. I'm gonna try it now.[/hide]

I don't know how to use or solve with the sum sign... :mad:

[hide="solution"]The sequence is $x, x*\frac{3}{2}, x*(\frac{3}{2})^2,...$
with x=the thickness of the first disk.
The formula for sum of a geometric sequence is
$\frac{t_{1}(1-r^n)}{1-r}$ where $t_{1}$ is the first term and $r$ is the common ratio ($\frac{3}{2}$ in this case)
$\frac{x(1-(\frac{3}{2})^n)}{1-\frac{3}{2}}=25x$
$\frac{(1-(\frac{3}{2})^n)}{-\frac{1}{2}}=25$
$-(1-(\frac{3}{2})^n)=\frac{25}{2}$
$-1+(\frac{3}{2})^n=\frac{25}{2}$
$(\frac{3}{2})^n=\frac{27}{2}$
$\frac{3^n}{2^n}=\frac{3^3}{2}$
$\frac{3^{n-3}}{2^{n-1}}=1$
$3^{n-3}=2^{n-1}$
This may be too hard but... using logs:
$(n-3)\log{3}=(n-1)\log{2}$
Then just use a calculator to solve for n. :D [/hide]

[quote="hoshattack"][hide] After examining the problem for a few minutes, I decided to come up for a formula for this to the [b][i]nth[/i][/b]term.  For $\displaystyle{x_n}$ it is $\displaystyle{(\frac{1}{2}n)+(\frac{1}{2}n)(\frac{1}{2})}$.   I checked this on a few small problems to make sure it worked.
Then I put it into the equation $\displaystyle{(\frac{1}{2}n)+(\frac{1}{2}n)(\frac{1}{2})=25}$  Solve for [b][i]n[/i][/b], the number of disks.
$\displaystyle{(\frac{1}{2}n)+(\frac{1}{4}n)=25}$  Simplify.
$\displaystyle{(2n)+(n)=100}$  Multiply by 4 to get rid of all the fractions. :lol:
$\displaystyle{3n=100}$  Combine  Like-Terms.
$\displaystyle{n=33\frac{1}{3}}$  Isolate n.
Round $\displaystyle{33\frac{1}{3}}$ up to 34.  So the answer is $\displaystyle{\boxed{34}}$ disks.
Edit:   34 disks until the disk is 25 times the orginal, I didn't solve for the problem. :blush:[/hide][/quote]

I don't think this is a correct formula. Your formula is equivalent to $\frac{3}{4}n$ when the equation should actually be $(\frac{3}{2})^n$. My solution doesn't work, I don't think. I get a solution of about $6.42$, which means I would need $7$ disks, which isn't correct. surge, did you just use guess and check for your summation? I think that is the best way. My solution is too long for a solution like 6.  :oops:

I'm still confused... can you explain the steps that either of you took to solve it?  :huh:

Argh... for some reason it posted it twice and it won't let me delete one.

Pay no attention to this  :spam:

My formula worked fine for finding how many disks it would take before that DISK was 25 times the original... and not the SUM of all the disks.

But still I know what you are saying, it is somewhat irrelevant to this problem.

Ok, thanks :) .
I just hope someone will show me how to solve those sum things...maybe my brother. :mad:

[quote="lotrgreengrapes7926"][quote="surge"][quote="lotrgreengrapes7926"]Wait, I am right.  :D Surge, you forgot to account for the original disk of thickness 1, according to your solution.

hoshattack, I think I should explain surge's answer as it's easier. If the first disk is thickness $1$, the second must be thickness $1.5$, and the third one must be $1.5*1.5$. I don't know if I can explain this well but, the summation means that you replace all the integers between 1 and n into the expression $1.5^n$ and add them. We're trying to find the lowest possible $n$ where this summation is greater than 25. ***Note: to make it easier, we could just have the summation from 0 to n, to include the initial disk of thickness 1. I don't know how to solve this summation except for guess and check. Just keep entering in numbers.  :D Hope that helped.[/quote]

How so? $1.5^0=1$ and my summation starts at zero. Use finite geometric series.[/quote]

There are 7 integers between 0 and 6, so there are 7 disks.[/quote]

oh, got it.


I just got done with the chapter and i cant tell what i got but lets just say that i could have done a little better on the sprint, i dont know if it was my laszyness or my lack of mathematical skill or a comboniation of both, but i somehow got into state because of my great school team. Can anyone tell me some tips for success in the state cometition, btw my state is Kansas. I have a month till state comp so let your ideas be practical in my time constraints. (Please dont tell me to do more tests because i have already done state tests from 00-07).

Thx In Advance  :D

Well, you can review formulas, but you have to understand them unless the proof is too complicated. Another way is just do practice problems or problems you had trouble on. I am sure that there are 1990-1999 States too, but I wouldn't know how to get them.
Hey, nice job making it to states. I should have made it, but I wasn't on the official team. :(

Hao, anyone on our team would've made it to States.
Seeing as we had the top three scorers.
Meaning that basically even if Johnathan had gotten a 20 I think we still would have made it to States.

to get those tests from 1990-1999, ask your coach.  Don't do anything before 1990 seriously as those had a different format.  Do them for fun if you wish

the all-time greatest mathcounts problems book can be pretty helpful

Psst, alan, how do I scan things?
Two guys asked me for the tests we got but I can't scan them...

dude, considering that you aren't a coach, I really don't think you should...

dude, what's with all the clarke people???

search on the forum for state problems if you want, but don't reply if it's too old.

Also, old problems are sort of confusing...
They sometimes ask questions that aren't hard, but have a ridiculous wording, making it confusing.


let $ f(z)\equal{}cosz\minus{}\frac{sinz}{z}$ for non zero z and $ f(0)\equal{}0$ also let $ g(z)\equal{}sinhz$ for z be a complex number
then what will be the order of zero of $ f(z)$ at $ z\equal{}0$
and what will be the order of pole of $ \frac{g(z)}{zf(z)}$
i have solved
by expansion
$ f(z)\equal{}(1\minus{}\frac{z^2}{2!}\plus{}\frac{z^4}{4!}\minus{}...)\minus{}\frac{1}{z}(z\minus{}\frac{z^3}{3!}\plus{}\frac{z^5}{5!}\minus{}...)$
$ f(z)\equal{}(1\minus{}\frac{z^2}{2!}\plus{}\frac{z^4}{4!}\minus{}...)\minus{}(1\minus{}\frac{z^2}{3!}\plus{}...)$
$ f(z)\equal{}\minus{}\frac{z^2}{3}\plus{}...$
so in taylor expansion we have
$ a_0\equal{}0$ and $ a_1\equal{}0$
so $ f(z)$ has zero of order 2
is it right?

That's correct.  So if I have a function with a zero of order n and I divide it by a function with a zero of order m, then the resulting expression will have either a pole or zero of order $ |n \minus{} m|$.  If $ n < m$, then it's a pole of order $ m \minus{} n$.  And if $ n > m$, then it's a zero of order $ n \minus{} m$.  If $ n \equal{} m$, then it's a removable singular point like $ \sin(z)/z$.  So you have $ \sinh(z)$ which you can check has a zero of order 1, and I'm dividing it by $ zf(z)$ which now has a zero of order 3, then the quotient has an order of 1-3 or a pole of order 2.


arithmetic sequence
recurrence relation
algebra
function
IMO Shortlist

For positive integers $ n,$ the numbers $ f(n)$ are defined inductively as follows: $ f(1) \equal{} 1,$ and for every positive integer $ n,$ $ f(n\plus{}1)$ is the greatest integer $ m$ such that there is an arithmetic progression of positive integers $ a_1 < a_2 < \ldots < a_m \equal{} n$ for which

\[ f(a_1) \equal{} f(a_2) \equal{} \ldots \equal{} f(a_m).\]

Prove that there are positive integers $ a$ and $ b$ such that $ f(an\plus{}b) \equal{} n\plus{}2$ for every positive integer $ n.$

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
we claim that:

$ f(4k \plus{} 3) \equal{} 2$.

$ f(4k) \equal{} k$ , $ (k\neq 2)$,and $ f(8) \equal{} 3$.

$ f(4k \plus{} 1) \equal{} 1$ , $ (k\neq 1,3)$,and $ f(5) \equal{} f(13) \equal{} 2$.

$ f(4k \plus{} 2) \equal{} k \minus{} 3$ , $ (k\neq 0,1,2,3,4,6)$,and $ f(2) \equal{} 1,f(6) \equal{} f(10) \equal{} 2,f(14) \equal{} f(18) \equal{} 3,f(26) \equal{} 4$.

we will prove our claims using induction on $ n$.it's easy to compute the values of $ f$ for $ n\leq 31$,so let $ n>31$:

if $ n \equal{} 4k$ ($ k\geq 8$) then $ f(3) \equal{} f(7) \equal{} \ldots \equal{} f(4k \minus{} 1) \equal{} 2$ hence $ f(4k)\geq k$,on the other hand we have $ f(n)\leq\max\{f(m): m < n\} \plus{} 1$,thus $ f(4k) \equal{} k$.

now if $ n \equal{} 4k \plus{} 2$ ($ k > 7$),we have $ f(17) \equal{} f(21) \equal{} \ldots \equal{} f(4k \plus{} 1) \equal{} 1$ hence $ f(4k \plus{} 2)\geq k \minus{} 3$.on the other hand if we denote $ d$ as the common difference of the progression and $ d\neq 4$ then obviously $ d\geq 8$,hence the length of the progression would be less than or equal to $ \frac {4k \plus{} 1}8$ which with the assumption $ k > 7$ yields us $ k \minus{} 3 > \frac {4k \plus{} 1}8$ hence $ f(4k \plus{} 2) \equal{} k \minus{} 3$.

also for $ n \equal{} 4k \plus{} 1$,because of $ k\geq 7$,for every $ m < 4k$ we have $ f(m) < k$,hence $ f(4k \plus{} 1) \equal{} 1$.

last but not least,for $ n \equal{} 4k \plus{} 3$,note that because of $ k\geq 8$ and $ f(4k \plus{} 2) \equal{} k \minus{} 3$,also for every $ k\geq 8$,there exists one and only one $ m < 4k \plus{} 2$ for which $ f(m) \equal{} k \minus{} 3$,hence $ f(4k \plus{} 3) \equal{} 2$.

now according to the proved claims we can see that $ f(4n \plus{} 8) \equal{} n \plus{} 2$,thus $ a \equal{} 4,b \equal{} 8$.

QED

[/quote]


we define S = {1,2,3...,4018}
Let $ A$ be a 2009 member-subset of $ S$ such that for any $ x,y\in S$  $ x$ can't be divided by $ y$
find the smallest member in $ A$

Maybe it's ...

[hide="maybe"]2010
because the set is $ 2010\le x\le 4018$
[/hide]

[quote="JBL"]But $ 2009 \leq x \leq 4017$ works fine, too ;)  And actually we can push it quite a lot smaller than that.

(I'm assuming that the problem was meant to ask, "Find the smallest possible value for the smallest element of $ A$.")[/quote]
yes you're right JBL ...sorry for my bad English  :blush:


quadratics
number theory proposed
number theory

Determine whether $ 200000.......003000000....0004$ is a perfect square where there are 100 zeroes between 2 and 3 as well as between 3 and 4.

I dont think this number is a perfect square coz it is 6 mod 7. But -1 is not a quadratic residue mod 7.


use the same method i did here: http://www.mathlinks.ro/viewtopic.php?p=1702603#1702603

the solution is almost exactly the same and the answer is exactly the same.


geometry
rectangle
rotation
geometric transformation
AoPSwiki
AMC

Rectangle PQRS lies in a plane with PQ=RS=2 and QR=SP=6. The rectangle is rotated 90 degrees clockwise about R, then rotated 90 degrees clockwise about the point that S moved to after the first rotation. What is the length of the path traveled by point P?

A) (2 sqrt 3 + sqrt 5)pi
B) 6 pi
C) (3 + sqrt 10) pi
D) (sqrt 3 + 2 sqrt 5) pi
E) 2 sqrt 10 pi

Thanks in advance!!

[hide] Situate the rectangle such that S is the origin and P is on the x-axis. Now, $ PR \equal{} 2\sqrt {10}$. The distance traveled by P during the first rotation is a fourth of the circumference of a circle with radius $ 2\sqrt {10}$, which is $ \sqrt {10}\pi$. Now, S is the origin, and PS=6, so the next rotation causes P to travel a fourth of the circumference of a circle with radius 6, which is $ 3\pi$. Our answer is $ (3 \plus{} \sqrt {10})\pi$.[/hide]
That seems so easy now...it took me so long to do on the real test.

...Why aren't my hide tags working?

None of the hide tags are working right now...not even in my classes.

When I took this on the real test...I thought we needed to find the distance from where P originated to where P ended...instead of the arcs.  Then I saw the answer choices...

Many problems can be found on the AoPSWiki - [[2008 AMC 10A Problems/Problem 19]].


logarithms
trigonometry
limit
calculus
calculus computations

Show that the series $ \sum^{\infty}_{n\equal{}2}\ln{\left(1\minus{}\frac{1}{n^2}\right)}$ converges and hence find the sum .

To prove convergent,i compare it with $ \frac{1}{n^2}$ and hence it is convergent . But how can we evaluate the sum?

A sum of logarithms is the logarithm of a product. Try writing out the first few terms of the product. Is there any chance that it telescopes (or something like that)?

Why I do not think of telescoping  :o ,I get the answer now. Thanks   :D

sorry people I forget to post link where you can chek your answer :
http://www.mathlinks.ro/Forum/viewtopic.php?t=170829
 :blush:

[quote="Kent Merryfield"]That's a different problem, Extremal. The one in this topic is easier (and shyong says he has the answer now.)[/quote]
Yes I agree with you, I only wanted to say that in this link
http://www.mathlinks.ro/Forum/viewtopic.php?t=170829
 product $ \prod (1\minus{}\frac{1}{n^{2}}$ I have already calculated and answer is $ \frac{1}{2}$

[quote="shyong"]Show that the series $ \sum^{\infty}_{n \equal{} 2}\ln{\left(1 \minus{} \frac {1}{n^2}\right)}$ converges and hence find the sum .

To prove convergent,i compare it with $ \frac {1}{n^2}$ and hence it is convergent . But how can we evaluate the sum?[/quote]

Under assumption that
\[ \sum^{\infty}_{n \equal{} 2}{\ln{\left(1 \minus{} \frac {1}{n^2}\right)}}
\]
converges, we can make transformation like this one:
\[ S \equal{} \sum^{\infty}_{n \equal{} 2}\ln{\left(1 \minus{} \frac {1}{n^2}\right)} \quad \Longrightarrow \quad e^{S} \equal{} \prod^{\infty}_{n \equal{} 2}{\left(1 \minus{} \frac {1}{n^{2}}\right)} \text{.}
\]
Now, as we know that
\[ \sin{x} \equal{} x \left(1 \minus{} \frac {x}{\pi} \right) \left(1 \plus{} \frac {x}{\pi} \right) \left(1 \minus{} \frac {x}{2\pi} \right) \left(1 \plus{} \frac {x}{2\pi} \right) \dots\text{,}
\]
putting $ x \leftarrow t\pi$ we get
\[ \sin{t\pi} \equal{} t\pi \left(1 \minus{} t^{2}\right) \prod_{n \equal{} 2}^{\infty}{\left(1 \minus{} \frac {t^{2}}{n^{2}}\right)} \text{.}
\]
So, it follows that
\[ e^{S} \equal{} \lim_{t \rightarrow 1}{\frac {\sin{t\pi}}{t\pi \left(1 \minus{} t^{2}\right)}} \text{.}
\]
Because of $ \lim_{t \rightarrow 1}{\sin{t\pi}} \equal{} 0$ and $ \lim_{t \rightarrow 1}{\left(1 \minus{} t^{2}\right)} \equal{} 0$, we can use L'Hospital's rule which brings out:
\[ e^{S} \equal{} \lim_{t \rightarrow 1}\frac {\pi \cos{t\pi}}{\pi \left(1 \minus{} 3t^{2} \right)} \equal{} \frac {1}{2} \quad \Longrightarrow \quad S \equal{} \minus{} \ln{2} \text{.} \quad \checkmark
\]

That's really overkill: we have
\begin{align*} \prod_{i = 2}^n 1 - \frac {1}{i^2} & = \prod_{i = 2}^n \frac {i^2 - 1}{i^2} \\
& = \frac {(1 \cdot 3)(2 \cdot 4)(3 \cdot 5) \cdots ((n - 1)\cdot(n + 1))}{(2 \cdot 2)(3 \cdot 3) \cdots (n \cdot n)} \\
& = \frac {n + 1}{2n} \\
& \to \frac {1}{2}. \end{align*}

(Edited typo caught by Fermat's Little Turtle.)

Just to note the index of the second product symbol, it should be 2?

$ \ln\left(1 \minus{} \frac{1}{n^2}\right) \equal{} \ln\left(\frac{n^2\minus{}1}{n^2}\right) \equal{} \ln \left[\frac{(n\minus{}1)(n\plus{}1)}{n \cdot n}\right]$

$ \equal{} \ln\left(\frac{n\minus{}1}{n}\right) \minus{} \ln\left(\frac{n}{n\plus{}1}\right) \equal{} a_n \minus{} a_{n\plus{}1}$


calculus
integration
LaTeX
calculus computations

[quote="shanniepooh2"][code]\int_-infty^\-2\,\frac{\textbf dx}{3\left(x+1\right)^3\right)}[/code]

a. Does not converge
b. -1/2
c. 0
d. 1[/quote]

Before we actually evaluate it lets make sure it converges

$ \int_{ \minus{} \infty}^{ \minus{} 2}\frac {dx}{(x \plus{} 1)^3}\text{ }dx$

We see that

$ \bigg|\frac {1}{(x \plus{} 1)^3}\bigg|\leq\bigg|\frac {1}{x^3}\bigg|$

Which converges by the p-seriest test for integrals.

Ok now on to evaluating it

$ \int_{ \minus{} \infty}^{ \minus{} 2}\frac {dx}{(x \plus{} 1)^3}\text{ }dx$

Let $ x \plus{} 1 \equal{} \varphi\Rightarrow\varphi \minus{} 1 \equal{} x$

So $ dx \equal{} d\varphi$

And as $ x\to \minus{} \infty\Rightarrow\varphi\to \minus{} \infty$

And when $ x \equal{} \minus{} 2\Rightarrow{\varphi \equal{} \minus{} 1}$

So we have

$ \int_{ \minus{} \infty}^{ \minus{} 1}\frac {d\varphi}{\varphi^3} \equal{} \frac { \minus{} 1}{2\varphi^2}\bigg|_{ \minus{} \infty}^{ \minus{} 1} \equal{} \frac { \minus{} 1}{2}$
So we have

Note: the "code" tags are for spacing text, as in indented programming code. For $ \text{\LaTeX}$, enclose your code between dollar signs.


geometry
incenter
power of a point
radical axis
geometry proposed

[b][u]Generalization[/u].[/b] Let $ ABC$ be a triangle and consider the points $ D\in BC$ , $ E\in CA$ , $ F\in AB$ so that the perpendicular

lines from $ A$ , $ B$ , $ C$ on $ EF$ , $ FD$ , $ DE$ are concurrently in the point $ P$ (the triangles $ ABC$ , $ DEF$ are [u]orthologically[/u] !).

Define points $ \left\|\begin{array}{c} \{A_1,A_2\}\subset BC\ ,\ DA_1 \equal{} DA_2 \equal{} DP \\
 \\
\{B_1,B_2\}\subset CA\ ,\ EA_1 \equal{} EA_2 \equal{} EP \\
 \\
\{C_1,C_2\}\subset AB\ ,\ FA_1 \equal{} FA_2 \equal{} FP\end{array}\right\|$ . Prove that the points $ \{A_1, A_2,B_1,B_2,C_1,C_2\}$ are concyclic.

[b][u]Remark[/u].[/b] See [url=http://www.mathlinks.ro/viewtopic.php?p=1190553#1190553][color=red][b]here[/b][/color][/url] a particular case - [u]question 1 from day 1 at IMO 2008[/u]. Another particular case is

when $ P$ is the incenter and $ D$ , $ E$ , $ F$ are the tangent points of incircle with sides of the triangle $ ABC$ .

[b][u]Proof[/u].[/b] $ AP\perp EF$ $ \Longleftrightarrow$ $ FA^2 \minus{} FP^2 \equal{} EA^2 \minus{} EP^2$ $ \Longleftrightarrow$ the point $ A$ has same power w.r.t. the circles $ C(F,FP)$ and $ C(E,EP)$ $ \Longleftrightarrow$

$ \overline {AC_1}\cdot\overline{AC_2} \equal{} \overline{AB_1}\cdot\overline{AB_2}$ $ \Longleftrightarrow$ the quadrilateral $ B_1B_2C_1C_2$ is cyclically. Pove analogously the quadrilaterals $ C_1C_2A_1A_2$ and $ A_1A_2B_1B_2$

are cyclically. Since the radical axis $ A_1A_2$ , $ B_1B_2$ , $ C_1C_2$ aren't concurrently obtain that the points $ A_1,A_2,B_1,B_2,C_1,C_2$ are concyclically.


Λοιπόν έστω $ a,b,c$ πλευρές τριγώνου που ικανοποιούν την εξής συνθήκη:

[size=150][color=black]mean(a,b,c)>=median(a,b,c)[/color][/size]

Να δείξετε την ανισότητα:

$ \sum_{cyc}\frac {3a \minus{} 2\sqrt {a^2 \plus{} (b \minus{} c)^2}}{b \plus{} c \minus{} a} \geq 3$

mean είναι ο αριθμητικός μέσος, δηλαδή το $ \frac{a\plus{}b\plus{}c}{3}$ ενώ median είναι ο μεσαίος όρος απ' τους 3 , δηλαδή στη συγκεκριμένη περίπτωση αυτός που δεν είναι ούτε ό μικρότερος ο μεγαλύτερος.

π.χ. απ' τους αριθμούς $ 5,9,10$ το mean είναι το 8 ενώ το median το 9

ευχαριστω  :)

Λοιπόν ας ξεφτιλίσω τον Schur λέγοντάς σας ότι ισχύει:

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \geq ab\sqrt {2a^2 \plus{} 2b^2} \plus{} bc\sqrt {2b^2 \plus{} 2c^2} \plus{} ca\sqrt {2c^2 \plus{} 2a^2}$ :wink:

Η πρώτη είναι αρκετά μυστήρια αν και δεν έχω ασχοληθεί μαζί της . Αυτή που πρότεινες τώρα είναι μια απλή εφαρμογή της SOS
Λοιπόν η ανισότητα μπορεί να γραφεί ως

$ \sum a^3\plus{}3abc\minus{}\sum bc(b\plus{}c)\geq \sum bc(\sqrt{2b^2\plus{}2c^2}\minus{}b\minus{}c)$

ή ισοδύναμα $ \frac{1}{2}\sum (b\minus{}c)^2(b\plus{}c\minus{}a)\geq\sum bc\frac{(b\minus{}c)^2}{\sqrt{2b^2\plus{}2c^2}\plus{}b\plus{}c}$

και αφού $ 2(b^2\plus{}c^2)\geq (b\plus{}c)^2$ αρκεί νδο

$ \sum (b\minus{}c)^2S_a\geq 0$ όπου $ S_a\equal{}c\minus{}a\plus{}\frac{b^2}{b\plus{}c}$

Ε τώρα θεωρούμε $ a\geq b\geq c$ και $ S_b,\ S_c\ ,S_b\plus{}S_c\geq 0$ so qed  :D

Δεν την πρότεινα τυχαία... Χρειάζεται στο πρώτο. :wink:

Ό,τι ακολουθεί είναι αρκετά βοηθητικό, προειδοποιώ:

[hide]Λοιπόν μια βοήθεια:

Πρώτα λίγο algebraic manipulation (να ξεμπερδέψουμε από τη συνθήκη τρίγωνο... ξέρετε εσείς... Προσοχή κάτι αλλάζει και στην πρώτη συνθήκη με τα νέα δεδομένα. Αυτό είναι το ωραίο της άσκησης.)

Μετά, χρήση της παραπάνω stronger-than-Schur αφού διαιρέσετε και τα δυο μέλη της με $ abc$.

Τέλος, θυμηθείτε κάτι ανάλογο που είχα ποστάρει παλιά στο sticky ανισότητες...[/hide]

Καλό κυνήγι.

Αν δεν υπάρξει λύση μέχρι αύριο το βράδυ θα τη βάλω εγώ.

Καλή επιτυχία σε όσους δίνουν Μεσογειάδα. Γενικά θα είμαστε πολύ λίγοι απ' ό,τι φαίνεται.

Θέτω $ b \plus{} c \minus{} a \equal{} x, c \plus{} a \minus{} b \equal{} y, a \plus{} b \minus{} c \equal{} z$ τότε η ανισότητα γίνεται $ \frac {3}{2}\sum\frac {y \plus{} z}{x} \minus{} \sum\frac {\sqrt {2y^2 \plus{} 2z^2}}{x}\ge 3$. Όμως η ανισότητα είναι συμμετρική άρα μπορόυμε να υποθέσουμε $ x\ge y\ge z$ τότε η συνθήκη γίνεται $ 2y\ge x \plus{} z$. H ανισότητα γράφεται μετά από πράξεις ως

$ \frac {(2y \minus{} x \minus{} z)[2(z \minus{} x)^2 \plus{} (y \minus{} x)(z \minus{} y)]}{2xyz} \plus{} \frac {1}{xyz}[\frac {z(\sqrt {2y^2 \plus{} 2z^2} \plus{} z \minus{} y)}{\sqrt {2y^2 \plus{} 2z^2} \plus{} y \plus{} z}(y \minus{} z)^2 \plus{} \frac {z(\sqrt {2x^2 \plus{} 2z^2} \plus{} z \minus{} x)}{\sqrt {2x^2 \plus{} 2z^2} \plus{} x \plus{} z}(x \minus{} z)^2 \plus{} (2x \plus{} y \minus{} 3z \plus{} \frac {y(\sqrt {2x^2 \plus{} 2y^2} \plus{} y \minus{} x)}{\sqrt {2x^2 \plus{} 2y^2} \plus{} x \plus{} y})(x \minus{} y)^2]\ge 0$ που ισχύει.


ok im confused about how to solve this problem,, any help or step by step solution would be great so i can see how to solve. Thanks

Carbon-10 is a radioactive isotope with a half life of 19.255 seconds. Suppose you start with 1 gram of carbon-10. Then the amount f(t) of carbon-10 left after t seconds is
f(t) =_______________

the hint i was given is:
     Use a formula of the form [img]http://webwork2.math.utah.edu/webwork_system_html/tmp/equations/ce/95c4786bbd870e9abd8df8a5d4c5fb1.png[/img] with suitable values for A

even with that im lost.

[hide="Hope it helps!"]I did a bunch of this problem long time ago, dealing with half times

The first thing to know is what is half life.

Half life is the time required for half the a given amount of material decays.

If I remember well, the formula for the mass remaining after some time $ t$ is $ m(t)\equal{}(m_0) 2^{t/t_h}$. Where $ m_0$ is the initial mass and $ t_h$ is the half life.

So for this scenario, the formula you need is $ f(t)\equal{}2^{t/19.255}$[/hide]

[quote="superwizard188"]...um, I think it should be $ 1*2^{ \minus{} t/19.255}$? The amount of carbon-10 should not increase as t gets larger. (and if you think about it, it keeps getting cut in half, so it makes sense.)[/quote]

Yeah, you are absolutely right. Somehow I had it right in my mind, but I miss putting the neg sign.

Thks

The radioactive decay of a nucleous A is an "unimolecular" process. Therefore,

$ \minus{} \frac{d[A]}{dt} \equal{} k[A] \equal{}> \frac{d[A]}{[A]} \equal{} \minus{} k\,dt \equal{}> \ln \frac{[A]}{[A]_o} \equal{} \minus{}kt \equal{}> [A] \equal{} [A]_oe^{\minus{}kt}$.

The half-life $ \tau$ is defined as the amount of time necessary for $ [A] \equal{} \frac{[A]_o}{2}$, and so

$ \frac{[A]_o}{2} \equal{} [A]_oe^{\minus{} k \tau} \equal{}> e^{k \tau} \equal{} 2 \equal{}> k \equal{} \frac{\ln 2}{\tau}$.

Therefore, the law of decay will be $ [A] \equal{} [A]_oe^{\minus{} \ln 2 \cdot t/\tau} \equal{}> [A] \equal{} [A]_o2^{\minus{}t/ \tau}$.


inequalities
trigonometry
geometry proposed
geometry

[color=purple] In space let two triangle $ ABC$ and $ XYZ$ (with $ BC \equal{} a,CA \equal{} b,AB \equal{} c$ and $ YZ \equal{} x,ZX \equal{} y,XY \equal{} z$) and a point $ P$. Prove that :
  $ \frac {a}{x}.\frac {PX}{\sin\frac {A}{2}} \plus{} \frac {b}{y}.\frac {PY}{\sin\frac {B}{2}} \plus{} \frac {c}{z}.\frac {PZ}{\sin\frac {C}{2}}\ge\frac {a \plus{} b \plus{} c}{\sqrt {3}}$
  Actually we can make it stronger  :D  :) [/color]

Could you please make question more clear. In statement you wrote
 In space let two triangle  and  (with  and ) and a point M. Prove that :
$ \frac {a}{x}.\frac {PX}{\sin\frac {A}{2}} \plus{} \frac {b}{y}.\frac {PY}{\sin\frac {B}{2}} \plus{} \frac {c}{z}.\frac {PZ}{\sin\frac {C}{2}}\ge\frac {a \plus{} b \plus{} c}{\sqrt {3}}$. So what is P and what is M and how there are related to ABC and XYZ

[color=purple] Sorry all mathlinkers ! My problem must be
          $ \frac {a}{x}.\frac {PX}{\sin\frac {A}{2}} \plus{} \frac {b}{y}.\frac {PY}{\sin\frac {B}{2}} \plus{} \frac {c}{z}.\frac {PZ}{\sin\frac {C}{2}}\ge\frac {2(a \plus{} b \plus{} c)}{\sqrt {3}}$
 :D Sorry everyone ! [/color]


Solve for $ x$:

$ 32\cos^6x\minus{}\cos6x\equal{}1$

[hide="Bashy"]Note that $ \cos(2x) \equal{} 2 \cos^2 x \minus{} 1$, so $ 32\cos^6 x \equal{} 4(\cos 2x \plus{} 1)^3$. Letting $ u \equal{} 2x$, we see that $ 4(\cos u \plus{} 1)^3 \minus{} \cos 3u \equal{} 1$. Since $ \cos 3u \equal{} 4 \cos^3 u \minus{} 3 \cos u$, expanding yields $ 12\cos^2 u \plus{} 15 \cos u \plus{} 3 \equal{} 0$, that is, $ 4\cos^2 u \plus{} 5\cos u \plus{} 1 \equal{} 0$, so $ \cos u \equal{} \minus{} 1/4$ or $ \cos u \equal{} \minus{} 1$. In the latter case, we get that $ x \equal{} \pm 3\pi/4 \pm k\pi$. In the former, we get that $ x \equal{} \pm \frac {\cos^{ \minus{} 1}( \minus{} 1/4)}{2} \pm k\pi$. [/hide]

hello, rewriting your equation in the form $ \minus{}1\plus{}32\cos(x)^6\minus{}\cos(6x)\equal{}0$ it can be factorized in the form
$ 6\cos(x)^2(1\plus{}4\cos(2x))\equal{}0$ and from here we get the solutions $ x\equal{}\minus{}\frac{\pi}{2}\plus{}2k\pi$ or $ x\equal{}\frac{\pi}{2}\plus{}2k\pi$ or $ x\equal{}\frac{1}{2}(\minus{}\arccos(\minus{}\frac{1}{4})\plus{}2k\pi)$ or $ x\equal{}\frac{1}{2}(\arccos(\minus{}\frac{1}{4})\plus{}2k\pi)$.
Sonnhard.

[quote="triplebig"] [color=darkred]Solve the trigonometrical equation $ 32\cos^6x \minus{} \cos6x \equal{} 1$ .[/color] [/quote]

[color=darkblue][b][u]Method 1[/u].[/b] $ 32\cos^6x\minus{}\cos 6x\equal{}1\Longleftrightarrow 16\cos ^6x\equal{}\cos^23x\Longleftrightarrow$ $ 4\cos^3x\equal{}\cos 3x\ \vee\ 4\cos^3x\plus{}\cos 3x\equal{}0$ $ \Longleftrightarrow$

$ \cos x\equal{}0\ \vee\ 8\cos^2x\equal{}3$ $ \Longleftrightarrow$ $ \cos x\equal{}0\ \vee\ \cos 2x\equal{}\minus{}\frac 14$ a.s.o. I used the relations $ \left|\begin{array}{c}
2\cos^2\frac {\phi}{2}\equal{}1\plus{}\cos\phi\\\\
\cos 3\phi \equal{}\cos\phi\cdot \left(4\cos^2\phi \minus{}3\right)\end{array}\right|$ .

[b][u]Method 2[/u].[/b] Denote $ z\equal{}\cos x\plus{}i\cdot\sin x$ . Prove easily that for any $ n\in\mathbb N$ we have $ \cos nx\equal{}\frac {z^{2n}\plus{}1}{2z^n}$ . Therefore,

$ 32\cos^6x\minus{}\cos 6x\equal{}1\Longleftrightarrow$ $ \left(z^2\plus{}1\right)^6\equal{}\left(z^6\plus{}1\right)^2$ $ \Longleftrightarrow$ $ \left(z^2\plus{}1\right)^6\equal{}\left(z^2\plus{}1\right)^2\left(z^4\minus{}z^2\plus{}1\right)^2$ $ \Longleftrightarrow$

$ \left(z^2\plus{}1\right)^2\left[\left(z^2\plus{}1\right)^4\minus{}\left(z^4\minus{}z^2\plus{}1\right)^2\right]\equal{}0$ $ \Longleftrightarrow$ $ \left(z^2\plus{}1\right)\left[\left(z^2\plus{}1\right)^2\plus{}\left(z^4\minus{}z^2\plus{}1\right)\right]\left[\left(z^2\plus{}1\right)^2\minus{}\left(z^4\minus{}z^2\plus{}1\right)\right]\equal{}0$ $ \Longleftrightarrow$

$ z^2\left(z^2\plus{}1\right)\left(2z^4\plus{}z^2\plus{}2\right)\equal{}0$ $ \Longleftrightarrow$ $ z\left(z^2\plus{}1\right)\left[2\left(z^2\plus{}1\right)^2\minus{}3z^2\right]\equal{}0$ $ \Longleftrightarrow$

 $ z\left(z^2\plus{}1\right)\left(z^2\sqrt 2\minus{}z\sqrt 3\plus{}\sqrt 2\right)\left(z^2\sqrt 2\plus{}z\sqrt 3\plus{}\sqrt 2\right)\equal{}0$ $ \Longleftrightarrow$ $ z\in\left\{\ 0\ ;\ \pm i\ ;\ \frac {\pm\sqrt 3\pm i\sqrt 5}{2\sqrt 2}\ \right\}$ a.s.o.[/color]


induction
combinatorics proposed
combinatorics

There are $ 3$ schools. There are $ n$ students in every school. Every student knows at least $ n\plus{}1$ students out of the other $ 2n$ students from the other $ 2$ schools. Let a [i]trio[/i] mean $ 3$ students from $ 3$ different schools such that every one of them knows the other two. The relation "knowing" is symmetric.
Prove that there are at least $ n$ trios.

this is a fairly well-known problem and you can find it in various sources. it has also been discussed here before.

I know that it is a known problem. And i'm posting it to "Proposed & Own Problems" !!
(I also give the source, from which i got it.)
But i would prefer to see a solution or link to some solution rather than that!!
Isn't it a section that we may discuss different solutions to problems ? I have an easy one, and i'm waiting others. (or at least other links.)
Thanks

Here's a link: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=schools++students&t=40982 .

But the conclusion there is a bit different...

what about induction and graph !! since this problem is familiar to me !!  :lol:

[quote="Anto"]There are $ 3$ schools. There are $ n$ students in every school. Every student knows at least $ n \plus{} 1$ students out of the other $ 2n$ students from the other $ 2$ schools. Let a [i]trio[/i] mean $ 3$ students from $ 3$ different schools such that every one of them knows the other two. The relation "knowing" is symmetric.
Prove that there are at least $ n$ trios.[/quote]
stupid question:
hmm...why was the question stated like "there are at least $ n$ trios", whereas there are can not be more than $ n$ trios so you should have stated it like "prove that there are exactly $ n$ trios" or like "prove that all students can be partitioned into non-intersecting trios".

[quote="darij grinberg"]I think he wants $ n$ trios, not $ n$ disjoint trios.

  darij[/quote]

hmm...that's kinda obvious then: if we take any school and each of its students belongs to at least one trio, take any of them, and do the same for each student from the taken school.

on the other hand, for case of disjoint trios problem seems to be incorrect, counterexample is obvious.

Am I missing something ?

[quote="darij grinberg"][quote="Erken"]hmm...that's kinda obvious then: if we take any school and each of its students belongs to at least one trio, take any of them, and do the same for each student from the taken school.[/quote]

Why that?

  darij[/quote]

my stupidity, sorry...

[quote="ocha"][b]5)[/b] Each edge between $X_2$ and $X_2\cup X_1$ corresponds to a unique triangle with a vertex in $abc$[/quote]

What if the $X_2$-vertex is connected to $a$ and $b$ while the $X_1\cup X_2$-vertex is connected to $c$ only?

[b]EDIT:[/b] Since ocha deleted the post I quoted, let me quote the part of it where he paraphrases and corrects grobber's solution of the original problem (this part is not affected by the mistake):

[quote="ocha"]Let the students represent vertices of a triparite graph $G$ with vertex set $V=(A,B,C)$ representing the schools and edges between acquaintances.

[b]Lemma:[/b] There exists at least one triangle in $G$.

[i]Proof:[/i] (Grobber) Take the vertex $a$ that is adjacent to the minimum number, $m$, of vertices in a single class $A$, $B$ or $C$ (where its own class is not being counted). We know $m\ge 1$ because every vertex has degree at least $n+1$. Say $a\in A$ has $m$ neighbours in $B$. Then since $d_G(a) \ge n+1$, $a$ has at least $n+1-m$ neighbour in $C$. Let $b\in B$ be a neighbour of $a$. But by minimality $b$ also has at least $m$ neigbours in $C$. By PHP $a$ and $b$ must share a neighbour in $C$ because $n+1-m+m=n+1>|C|$. Thus there is a triangle $\square$

So we can find a triangle $abc$ in $G$.[/quote]

(I have taken the liberty to fix a few typos...)

We will construct a counter-example for arbitrarily large $n$ that only has $4$ triangles!

[b]Counterexample[/b]
Consider the tripartite graph $G$ with students represented as vertices in the set $V=(A,B,C)$. Each component contains $n$ vertices for some large $n\in\mathbb{N}$. Edges in the graph are between acquaintances. We will direct the edges from $A$ to $B$, from $B$ to $C$ and from $C$ to $A$. Then any triangles will be a directed cycle. For now let the edge set be empty and we will add then as we go. [i]Notation:[/i] the statement "draw edges $X \to Y$" will mean draw a directed edge from each vertex in set $X$ to each vertex in set $Y$.

[b]Construction[/b]
Let $G_0$ be a subgraph of $G$ with vertex set $V=(A_0,B_0,C_0)$ such that $3 \le |A_0| < |B_0| < |C_0| < n-3$. Let $G_1 = G\backslash G_0$ have vertex set $V=(A_1,B_1,C_1)$. The only edges between $G_1$ and $G_0$ will be directed edges from $G_1$ to $G_0$. Draw edges $A_1 \to B_0$, $B_1 \to C_0$, $C_1\to A_0$. This means that there can be no triangle in $G$ with vertices in both $G_0$ and $G_1$ (because if a directed path leaves $G_1$ it cannot go back).

Now consider $G_0$. We need each vertex $d_G(v)\ge n+1$. Therefore, taking in to account the edges between $G_1$ and $G_0$, we require vertices in $A_0,B_0$ and $C_0$ to have minimum degree $d_{G_0}(a) \ge |C_0|+1$, $d_{G_0}(b) \ge |A_0|+1$ and $d_{G_0}(c) \ge |B_0|+1$ respectively.

Similarly in $G_1$, we need vertices in $A_1,B_1$ and $C_1$ to have minimum degree $d_{G_1}(a) \ge |B_1|+1$, $d_{G_1}(b) \ge |C_1|+1$ and $d_{G_1}(c) \ge |A_1|+1$ respectively. Notice that if we reverse the direction of the edges in $G_1$ then $G_0$ and $G_1$ are similar (up to a relabelling of the vertex sets).

So it will suffice to show that we can limit $G_0$ to no more than $2$ triangles. Then if we do the same for $G_1$ we will be done. So let $a \in A_0, \{b,b'\}\subset B_0$ and $c\in C_0$. Draw edges:

$B_0\,\backslash \{b,b'\} \to C_0$

$C_0\,\backslash \{c\} \to A_0$

$A_0 \to \{b,b'\}$

$\{b,b'\} \to c$

$c\to a$

Then one can check that each vertex has the desired minimum degree and the only two triangles are $abc$ and $ab'c$, because any direct path eventually ends up in this triangles.

So we have constructed a counter example with only $4$ triangles and arbitrarily many vertices.

Hopefully my counterexample above is correct. Here is the problem again with an added restraint. I have a proof of this claim:

Let $G$ be a directed tripartite graph with vertex set $V=(A,B,C)$ and $|A|=|B|=|C|=n$. Edges are directed from $A$ to $B$, from $B$ to $C$ and from $C$ to $A$. Show that if each vertex has total degree at least $n+1$ and $G$ is strongly connected then there exists at least $n$ triangles.

Posting this just to confirm ocha's counterexample. Welldone, ocha!

I don't have a proof of your strongly connected version (yet?).


function
topology
real analysis
real analysis unsolved

Let \[T=\{(tq,1-t) \in\mathbb R^{2}| t \in [0,1],q\in\mathbb Q\}\]Prove that each continuous function $f: T\longrightarrow T$ has a fixed point.

The set is a bunch of segments which intersect only at $ (0,1)$. Look at where that point goes.

IMS is a competition for undergraduate students held by Iranian Mathematical Society.

[quote="Omid Hatami"]Here is  the [url=http://ims.ir]link[/url] for the official site, and I think you can find the Persian version of Problems and Solutions there.[/quote]

Oh, how can I understand....?  :blush:

I think there is no way, unless you learn Persian :D
But I will try to translate some of the problems in English, as I did some of the problems of recent years.

Very intersting problem. I really like these shapes with fixed point property.
Few fact that if anybody knows would solve this problem quickly.

First define a new topology on this set $ T$ generated with a basis consisting of open intervals on each of those segments [ these open intervals on the
segments should not include point $ (0,1)$ ] and the usual open basis around the point $ (0,1)$ induced on $ T$ by the usual topology of $ \mathbb{R}^2$.
It is obvious that this topology is finer that the usual topology induced on $ T$ by the usual topology of $ \mathbb{R}^2$.

Let $ (T,A)$ be the topology on $ T$ induced by the usual topology of $ \mathbb{R}^2$ and $ (T,B)$ be the new topology I introduced in the above paragraph.

Second fact is that if $ f: (T,A)\to(T,A)$ is continous function then $ f: (T,B)\to(T,B)$ is continous too. This might not be quite obvious and can be a good exercise.

Third and the most important fact is if a continous function $ f: (T,B)\to(T,B)$ sends $ (0,1)$ to $ (0,1)$ then it has a fixed point and if not, then $ f((0,1))\in
I\setminus \{(0,1)\}$ where $ I\equal{}\{(qt,(1\minus{}t)\times 1)|t\in [0,1]\}$ for some $ q\in\mathbb{Q}$ then I claim that $ f_{|I}$ has a fixed point in $ I$ and thus can be
proved by the old trick that if you have a topological space which is the union of two subsets $ U$ and $ V$ such that $ \bar{U}\cap V\equal{}U\cap\bar{V}
\equal{}U\cap V\equal{}\{ \mbox{ a point }\}$ such that these two sets with the induced topology have fixed point property then their union which is the original topological space has fixed point property.

I know that I didn't explain everthing in detail but I believe readers of mathlinks can fill in the gaps easily. I wanted to void my words to be boring.

$ (T,B)\equal{}\{\mbox{ a point }\}* (\mathbb{Q},\mbox{ usual topology})$

I just found that my approach can be generalized to my theorem:

[b]Theorem.[/b] Let $ A\in\mathbb{S}^n$ such that each connected component of $ A$ is homeomorphic to $ \mathbb{B}^i$ where $ i\in\{0,\ldots,n\}$. Then the set $ SG\equal{}\{ta|t\in[0,1] , a\in A\}$ has fixed point property.

You can see by stereographic projection that the IMS problem is a special case of my theorem. And in this case $ n\equal{}1$ and all connected components of A are just $ \mathbb{B}^0$ or points. There must be other sort of generalizations [ Banach spaces , ... ] .

Again you can prove the theorem by using [url=http://www.maa.org/reviews/mystic1.html]brouwer[/url]-fixed-point theorem and my approach of looking at a finer topology.

Other interesting questions:

[b]Question 1.[/b] Does the product of two topological space [ preferably embedded in some $ \mathbb{R}^m$ ] that they both have fixed point property, have fixed point property?

[b]Question 2.[/b] How about if in question 1 instead of product we use join of those spaces [ $ *$ ] ?

Here $ \mathbb{B}^i\equal{}\{x\in\mathbb{R}^i : \|x\|\leq 1\}$ and $ \mathbb{S}^i\equal{}\{x\in\mathbb{R}^{i\plus{}1} : \|x\|\equal{}1\}$ .


analytic geometry
geometry
3D geometry
sphere
function
topology
vector

I'm a student taking an intro to modern algebra course, so forgive me if my question does not make complete sense.

I was considering a three-dimensional sphere, defined in the customary way with rectangular coordinates. What kind of operation * can you apply to points on the sphere that will make the sphere closed under *.

Also, what kind of binary operation * can you apply to points in the cartesian plane.

I'm asking these questions because I'm looking at a function that maps points on a sphere to a plane, and I'm trying to see if the function can be an isomorphism if I define some binary operations on the sphere and the plane.

Thanks for any help

PS: Can polar coordinates help? I know that in the complex plane the set of all complex numbers with magnitude 1 is closed under complex multiplication. How could I apply similar ideas to R^3?

Is the function you speak of defined globally?

Usually, when we are speaking of the 3 dimensional sphere, we mean the sphere in $\mathbb R^{4}$, because this sphere has dimension 3 as a manifold. With this convention it is known tha the 1-dimensional sphere has a group operation (the multiplication of complex numbers of norm 1), the 3-dimensional sphere has a group operation (the multiplication of hamilton numbers of norm 1; I mean those numbers of the form $a+bi+cj+dk$ with $a,b,c,d\in\mathbb R$, $i^{2}=j^{2}=k^{2}=-1$ and $ij=k=-ji$, $jk=i=-kj$ and $ki=j=-ik$.)

If your sphere in $\mathbb R^{3}$, then I don't think that the map you speak of is defined globally, so you can't talk about isomorphisms.

Assuming you mean the sphere $S^{2}$ embedded in $\mathbb{R}^{3}$, and you want the sphere to be a group under *, there's no way to do it smoothly.
This really belongs to the theory of manifolds and Lie groups. You haven't seen those yet, so I'll try to describe what's going on in relatively elementary terms.

A smooth $n$-manifold is a topological space on which we have local coordinates in $\mathbb{R}^{n}$; for each point in the space, there is a bijection between an open subset of $\mathbb{R}^{n}$ and a neighborhood of that point. In addition, these coordinate charts are related by smooth functions from $\mathbb{R}^{n}$ to itself whenever they intersect. (This isn't the whole definition, but it's the most important part)
We say that a map between manifolds is smooth if it is smooth in the coordinates.
The Euclidean sphere is of course a 2-manifold. Spherical coordinates give a chart if we cut out the poles and a line of longitude; projection onto a plane gives a chart for any hemisphere.

A Lie group is a smooth manifold which is also a group, with the additional property that the group operation is smooth: $x\to x^{-1}$ and $(x,y)\to xy$ are smooth maps.

Now, we can define the tangent space to a manifold at a point. For the sphere, this is the space of vectors tangent to the sphere in the usual sense (and perpendicular to the radius). Any smooth map that sends point $A$ to point $B$ induces a linear map from the tangent space at $A$ to the tangent space at $B$. As an example, think of geographic directions; we can define north/south/east/west at any point that isn't a pole, and then move these definitions elsewhere by a rotation.

If the manifold is a Lie group, the multiplication maps $x\mapsto ax$ take us from the identity to $a$. These maps are invertible, so the induced linear maps are also invertible. If we pick a nonzero vector at the identity and look at what all of these maps do to it, we get a vector field tangent to the manifold. This vector field is smooth with the right definitions, and it's never zero.
Now, we have the famous "Hairy Ball Theorem": every smooth vector field tangent to $S^{2}$ is zero somewhere. Combined with the previous fact, it is impossible to make $S^{2}$ into a Lie group.

Probably the most important thing I realized long after my first abstract algebra course was over was that I shouldn't focus quite so much attention on the idea that "groups are" and much more attention on the idea that "groups act."

Many groups are realized as a set of functions on some particular set (perhaps all function, perhaps with some restrictions.) In a few cases, we can usefully identify the group with the set being acted on. We can think that way about the permutation group $S_{n}$ or about the group of permitted motions of a Rubik's cube. But the permutation group acts on the set $\{1,2,\dots,n\}.$ The group is not the same as that set. And while we we can identify the Rubik's group with the set of possible positions of the cube, that's a set of possible positions of the cube, not the cube itself.

We can't make the 2-sphere into a group, at least not smoothly. That's what jmerry just said. So "groups are" goes nowhere with the sphere.

But can we find a group that acts on the 2-sphere in a "nice" way? Certainly so, in several different ways. The most pleasing would be the actions of the group SO(3) of proper rotations of 3-space.

But note: The sphere is a locally 2-dimensional object (a 2-manifold). The set SO(3) is locally 3-dimensional. (That is in effect a count of how many "degrees of freedom" there are for rotations.)


function
quadratics
real analysis
real analysis unsolved

Let be given $a,x_0$ are positive real numbers.
$x_{k+1}=\frac{1}{3}(2x_k+\frac{a}{x_k^2})(k=0,1,2,...)$.
i)Find $L=limx_n$
ii)$z_k=x_k-L$,such that $z_1$ unequal $0$.Prove that
$z_k>0,z_{k+1}<\frac{2}{3}z_k,z_{k+1}<\frac{1}{\sqrt[3]{a}}z_k$.

This is Newton's method for solving $f(x) = 0$, applied to a very simple function $f$.
Thus, the limit is the solution of $f(L)=0$ and $z_{k+1} = O(z_k^2)$.

[quote="fleeting_guest"]... to a very simple function f.[/quote]
I suppose it doesn't really need to be said, but in case anyone didn't get that, fleeting_guest was referring to $f(x)=x^3-a$ so that $L=\sqrt[3]{a}.$

The quadratic convergence estimate is asypmtotically true, but if you make a spectacularly bad initial guess, so that $x_1$ is much larger than $a$, then the convegence is "linear" for a long time, with $x_{k+1}\approx \frac{2x_k}3$ as long as $x_k$ is still very large.

(Note: I used "linear" in the sense that the numerical analysts use the word, new error proportional to previous error. It actually results is an approximately geometric sequence.)

[quote="Moubinool"]$a,x_0$ are positive real numbers.
$x_{k+1}=\frac{1}{q}((q-1)x_k+\frac{a}{x_k^{q-1}})$
(k=0,1,2,...)

q>3

What is the limit of $x_{k}$ ?[/quote]

$L=\sqrt[q]{a}$


group theory
abstract algebra
function
superior algebra
superior algebra solved

Let $G$ be an infinite group. Prove that:
(i) If $H$ is a normal finite subgroup of $G$, then $C_G(H)$ is a normal subgroup of $G$ and $G/C_G(H)$ is finite. By definition, $C_G(H)=\{x\in G| xh=hx\ \forall h\in H\}$.
(ii) If $G$ has no nontrivial finite quotients then any normal finite subgroup of $G$ is central (in particular it is abelian).

For (i) ((ii) follows immediately):

If $xh=hx,\ \forall h\in H$, and $a\in G$, then $xa^{-1}ha=a^{-1}hax,\ \forall h\in H$, so $axa^{-1}h=haxa^{-1},\ \forall h\in H$, so $x\in C_G(H)\Rightarrow axa^{-1}\in C_G(H)$. This proves the normality of $H$.

If $xC_G(H)=yC_G(H)$, then $x^{-1}y\in C_G(H)$, which means that $x^{-1}yh=hx^{-1}y,\ \forall h\in H\Rightarrow xhx^{-1}=yhy^{-1},\ \forall h\in H$, so $x,y$ induce the same permutation on the elements of $H$ by conjugation. This means that there's an injection from $G/C_G(H)$ to the set of permutations of $H$, and, since $H$ is finite, $G/C_G(H)$ must be finite as well. Actually, the argument above shows that there is a function from $G/C_G(H)$ to the set of permutations of $H$, but it's clear, by employing the same type of argument, that the function is injective.

We should not however disregard (ii) since it states a pretty consistent fact which would not be easy to prove without (i).


The year 2002 is a palindrome (a number that reads the same
from left to right as it does from right to left).  What is the product of
the digits of the next year after 2002 that is a palindrome?

The next year that is a palindrome is obviously 2112.
The product of the digits is:
2*1*1*2=4.


Find the maximum value of
$\displaystyle S=\frac {x+x^2+...+x^{2n-1}}{(1+x^n)^2}$
for positive real $x$ and the values of $x$ at which the maximum is achieved.

Sorry for bumping this :) Notice that \[ \sum_{k\equal{}1}^{2n\minus{}1}x^{k}\equal{}x^{n}\plus{}\left(1\plus{}x^{n}\right)\sum_{k\equal{}1}^{n\minus{}1}x^{k}.\] Using AM-GM twice, we get that \[ \sum_{k\equal{}1}^{2n\minus{}1}x^{k}\leq x^{n}\plus{}\left(1\plus{}x^{n}\right)\sum_{k\equal{}1}^{n}\frac{kx^{n}\plus{}(n\minus{}k)}{n}\equal{}x^{n}\plus{}\frac{n\minus{}1}{2}\left(1\plus{}x^{n}\right)^{2}\leq\frac{2n\minus{}1}{4}\left(1\plus{}x^{n}\right)^{2}.\] Now it is clear that the maximum value, $ \frac{1}{4}(2n\minus{}1)$, is attained for $ x \equal{} 1$ and for no other values of $ x$.


Four positive integers are given. Select any three of these integers, find
their arithmetic average, and add this result to the fourth integer. Thus the
numbers 29, 23, 21 and 17 are obtained. One of the original integers is:

(A) 19    (B) 21    (C) 23    (D) 29    (E) 17

I don't really know why they ask for only 1 of the original integers. I thought 3 of those integers from the set of 4 numbers given must be in the original set, and all 4 of those numbers are in the answers. Maybe I'm just misunderstanding the question. Help please.

[hide="My try"]

Let $ a,b,c,d$ are the numbers
WLOG, $ a\ge b\ge c\ge d$

Then, we have

$ a\plus{}b\plus{}c\plus{}3d\equal{}51$
$ a\plus{}b\plus{}3c\plus{}d\equal{}63$
$ a\plus{}3b\plus{}c\plus{}d\equal{}69$
$ 3a\plus{}b\plus{}c\plus{}d\equal{}87$

Summing four equations above, and divide by 6, we get $ a\plus{}b\plus{}c\plus{}d\equal{}45$
Then substract each equation with $ a\plus{}b\plus{}c\plus{}d$ then divide them by 6, we get
$ a\equal{}21,b\equal{}12,c\equal{}9,d\equal{}3$

Then only $ a\equal{}21$ satisfy the answer

$ (B) 21$
 :)

[/hide]


inequalities
induction
inequalities proposed

Let $a, b, c$ be positive real numbers such that $abc=1$. Show that \[\frac {a^{n}}{b+c}+\frac {b^n}{c+a}+ \frac{c^n}{a+b}\geq\frac32\]

This one is rather trivial. It's known for n=1. Suppose n>1. Then
  $\frac{a^n}{b+c}+...=\frac{ a^{n+1}}{ab+ac}+...\geq\frac{(a^{\frac{n+1}{2}}+...)^2}{2(ab+bc+ca)} $. But it's trivial to prove using the power mean inequality that $ a^{\frac{n+1}{2}}+...\geq a+b+c$ and from here it is trivial.

use induction + Cebasev +AM-GM inequality. that is
\[ \sum \frac {a^{n+1}}{b+c } \geq \frac 13 \sum a \sum \frac {a^n}{b+c} \geq 1 \cdot \frac 32 \] are you are done :)

Of course for $n=1$ we have again AM-GM :P

What do you mean with Cebasev?? Is it Chebyshev?  :?  :)

Forget about it, I think you meant Cauchy, becuase aI already solved it with it, but I dind't really understand your last post  :?

I ment Chebyshev :) (I spelled it like in Romanian).

Why  we  use  induction??  Couldnt  we  use  chebichev  from  the  beggining  like  we  did  at  the  induction's   final  step??
Thanks

Sure we can.By Chebysev inequality:
 [tex] \sum \frac{a^{n+1}}{b+c}\ \geq \frac 13 \sum a^n \sum \frac{a}{b+c} \ \geq 1.\frac32\ [/tex]


Two distinct numbers are selected simultaneously and at
random from the set $ \{1, 2, 3, 4, 5\}$. What is the probability that
their product is an even number? Express your answer as a
common fraction.

The product is odd if the two numbers multiplied are odd.  So let's use complementary probability.

[i]Probability that both numbers are odd: $ \frac35 \times \frac12 \equal{} \frac3{10}$[/i]

$ 1 \minus{} \frac3{10} \equal{} \boxed{\frac7{10}}$[/i]

It actually is $ \frac{3}{5}\times \frac{2}{4}\equal{}\frac{6}{20}\equal{}\frac{3}{10}$, he just simplified the fraction first

[hide=Solution]Selecting 2 numbers: 5 choose 2 $\Rightarrow$ 10 ways
Here they are:
[center]
2 X
3
4 X
5
6 X
8 X
10 X
12 X
15
20 X
[/center]

All the ones with an "X" are even, so $\boxed{\frac{7}{10}}$.[/hide]

[hide=Casework Solution]
Note that odd x even = even (even x odd = even by commutative property) and even x even = even.
Case 1: We have odd x even (or even x odd)
Since we have odd x even = even and even x odd = even, we multiply the probability that we choose one of each type by 2. The number of ways to choose 2 numbers from the set is $5 \cdot 4 = 20$. The number of ways to choose an even and a odd is $3 \cdot 2 = 6$.
Hence, the probability that we have an odd x even to get an even product is $2 \cdot \frac{6}{20} = \frac{6}{10}$. We don't simplify here because it will be easier to work with our addition later.

Case 2: even x even
Same as before, the number of ways to choose the numbers from this set is 20. The number number of ways to choose 2 evens is $2 \cdot 1  = 2$.
Thus the probability of choosing 2 evens is $\frac{1}{10}$.

Adding the two cases together gives us the answer, $\boxed{\frac{7}{10}}$. [/hide]

[quote=bwu]kay.  i don't quite understand why it's 3 ways, not 6 (1,3; 1,5; 3,1; 3;5, 5;1, 5;3).[/quote]

(1, 3) and (3, 1) are the same thing


geometry
algebra
function
domain
Ring Theory
number theory
prime factorization

Hi I hope this is the correct forum to post this question.  I am especially weak in this area of mathematics.  I am just learning the fundamentals of elementary number theory.  I hope somebody can help me with this question.  Thank you.

Question: Prove that if $ n$ is composite then there are integers $ a$ and $ b$ such that $ n$ divides $ ab$ but $ n$ does not divide either $ a$ or $ b$.

[hide="Hint"] What does the word "composite" imply?[/hide]

The first thing to do before you even try to do a problem is to convince yourself it's true.  In this case, find examples of such $ n, a, b$.  Then pick prime values of $ n$ to convince yourself that the condition that $ n$ is composite is necessary.

[hide="For example"] $ 12 | 6 \cdot 6$ but $ 12$ divides neither $ 6$ nor $ 6$. [/hide]
Besides convincing yourself that a problem is true, analyzing the nature of examples can help you come up with a technique.

[hide="An example that might help you come up with a technique"] $ 35 | 70 \equal{} 7 \cdot 10$ but $ 35$ divides neither $ 7$ nor $ 10$.

[hide="Do you see it now?"] $ n | 2n$ but... [/hide] [/hide]

[quote="t0rajir0u"][hide="An example that might help you come up with a technique"] $ 35 | 70 \equal{} 7 \cdot 10$ but $ 35$ divides neither $ 7$ nor $ 10$.

[hide="Do you see it now?"] $ n | 2n$ but... [/hide] [/hide][/quote]

Or even...

[hide]$ 35|35\equal{}5\cdot 7$ but 35 divides neither 5 nor 7.  :wink: [/hide]

:oops:

I knew there was a simpler technique I'd forgotten.

Um... isn't this kind of trivial if you allow that?

It's not supposed to be a difficult problem.  The point is the reverse implication, which together with the problem tells us that $ p$ is prime if and only if $ p | ab \implies p | a \text{ or } p | b$.  The interesting question from here is to study whether this continues to be true over other rings of integers.

Perhaps the Gaussian Integers might be more interesting?

See [url=http://en.wikipedia.org/wiki/Integral_domain#Divisibility.2C_prime_and_irreducible_elements]here[/url].  The two keywords are "prime" and "irreducible," which mean the same thing over the integers but do not mean the same thing in general.

Edit:  [hide="The relevant statements you want to try proving"] $ \text{prime} \implies \text{irreducible}$ in an arbitrary ring but not the reverse.  (In other words, can you find a ring with an irreducible element that is not prime?)

$ \text{prime} \Leftrightarrow \text{irreducible}$ in a [url=http://en.wikipedia.org/wiki/Unique_factorization_domain]UFD[/url]. [/hide]
In any case, I suppose this is a little more than what you want to get into.  This alternate characterization of the primes is useful in the integers alone (for example, it makes the proof of prime factorization easier).

Nothing to do with the problem, but if there were a Nobel Prize for math literature, torajirou would win it!

T0r, doesn't a UFD require generalizing the fundamental Thm. of arithmetic?
I'm not sure how I'd go about doing it.

We prove that the Gaussian integers are a UFD by proving that they are a [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url] with the norm $ N(a \plus{} bi) \equal{} a^2 \plus{} b^2$.  You can find all this information in a standard text on algebraic number theory.

Edit:  Do you mean a proof of the second statement I gave?  Think about what prime factorization implies for prime and irreducible elements.


probability
expected value
probability and stats

Let $ (X_n)$ be independent and uniformly distribuited on $ (0,1)$. Let $ 0<x<1$ and define $ N\equal{}min(n \ge1 : X_1\plus{}X_2\plus{}...\plus{}X_n>x)$.
 Find $ Var(N)$.

Let's compute the law of $ N$ for $ 0 \leq x \leq 1$:
$ P(N\equal{}n)\equal{}P((x_1\plus{}x_2\plus{} \ldots \plus{} x_{n\minus{}1} \leq x) \cap (x_1\plus{}x_2\plus{} \ldots \plus{} x_{n} > x))$
$ P(N\equal{}n) \equal{} P(x_1\plus{}x_2\plus{} \ldots \plus{} x_{n\minus{}1} \leq x) \minus{} P(x_1\plus{}x_2\plus{} \ldots \plus{} x_{n\minus{}1} \plus{} x_n \leq x)$
 so $ P(N\equal{}n) \equal{} \frac{x^{n\minus{}1}}{(n\minus{}1)!} \minus{} \frac{x^n}{n!}$
Hence the expected value of $ N$ is $ E(N)\equal{}\sum_{n\equal{}1}^{\infty} P(N\equal{}n).n \equal{} \exp(x)$ and its variance is $ \sigma^2 \equal{} E(N^2) \minus{} E(N)^2 \equal{} \exp(x)(1\plus{}2x) \minus{} \exp(2x)$. (modulo computation mistakes ..)

and how to solve the general case ? for example, what if $ (X_n)_{n \geq 0}$ is a sequence of i.i.d gaussian variables ?

For Gaussian it is not hard - you know that the sum $ S_n$ is also a Gaussian random variable, so you can compute everything in terms of its distribution function.

A previous [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=20848]discussion[/url] of this and similar problems.


function
topology
real analysis
real analysis solved

Call a smooth closed curve that intersects itself [i]exactly[/i] once an [i]8-curve[/i]. Does there exist an uncountable family of pairwise non-intersecting 8-curves in the plane?

Hmm.. Is the answer "no"? Because I think it's been shown before on the forum that there can only be countably many topological $Y$'s in the plane, and since each such curve has a subset which is a topological $Y$..

However, I'm not entirely sure about this. Maybe I'm only imagining seeing that problem with the $Y$'s.

Edit: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=topological&t=6468]here[/url]'s the link to that problem

Oh, I didn't know that this had previously been discussed on the forum. But it's such a cute little problem, isn't it?  :P

[quote="grobber"]Just out of curiosity: what was your solution? I mean, if you used the fact that they have the shape of $8$, then maybe it's something completely different. And yes, it is cute :).[/quote]

The solution for the topological 8 uses the same idea as for the Y, but much, much simpler...

Indeed, yesterday I said to grobber my idea on this problem: we identify each '8' with two rational points - one in each 'o'. Then each pair of rational points is associated with at most one '8'.

There are some problems based on the same idea in Rudin's book (Principles of Analysis)
such as proving that a real function has at most a countable number of discontinuities of various kinds.

For the 8 problem there should also be another kind of solution where the 8's are organized into something like a partially ordered set.  Any set of mutually external 8's is countable, so the "width" is countable, but also any set of 8's that is linearly-ordered-by-inclusion is countable, so the "height" is countable.  Something along these lines (or maybe this is a complete solution by invoking Dilworth theorem on posets).

Actually, I think the second type of solution works for a topological
"circle with one hair",  where the rational point solution does not immediately work.
i.e. a homeomorphic copy of a circle with a line segment attached to it (but not crossing
the circle).


inequalities
algebra
polynomial
calculus
derivative
absolute value
inequalities proposed

While trying to prove Markov inequality for polynomials, I found an interesting fact. If the norm of a polynomial of degree 3 over [-1,1] ( considered in the sense that flip 2004 introduced) is at most 1, then for all x whose modulus is at least 1, the absolute value of the derivative of the polynomial calculated in x is at most $ 12 x^2-3$.

It can easily be deduced from the original Markov inequality and alternative property of the Chebyshev polynomials. Indeed, consider a polynomial \[g_n(x)=T_n(x)-P_n(x).\]
Since $\|P_n\|=1$ and $T_n$ alternates in between $-1$ and $1$ precisely $n$ times on $[-1,1],$ the MVT implies that $g_n$ has all its zeros belonging to $[-1,1].$ By Rolles theorem so does $g'_n,$ and hence out of $[-1,1]$ it does not change a sign. So, for instance, since $T_n(1)-P_n(1)=n^2-P_n(1)>0$ (if $P_n\ne T_n$), then  $T_n'(x)\ge P_n'(x)$ for all $x\ge 1.$


geometry
parallelogram
rhombus
geometry unsolved

$E$ is a point inside a square $ABCD$ such that $AE=BE$. If $\angle AEB=150^\circ$, find $\angle CED$.

I wanna see a solution without trigonometry.

answer is 60 degrees.
MEDHOD 1: Let's draw equilateral triangle CDE' inside the square.We will prove E=E' .since BC=DC=CE', triangle E'CB is isosceles.m(CE'B)=m(CBE')=75 .similarly, triangle DAE' is isosceles and m(DAE')=m(DE'A)=75.so, AE'B is isosceles and ,m(E'AB)=m(E'BA)=15,m(AE'B)=150.Clearly, E=E'.

MEDHOD 2: Let's draw triangle BCF such that BCF and BAE are congruent triangles.so, EA=EB=FB=FC, m(EAB)=m(EBA)=m(FBC)=m(FCB)=15 and m(AEB)=m(BFC)=150. in the square, ABC is right.m(EBF)=90-(15+15)=60. hence, EBF is equilateral.m(EFB)=60 and EF=FB=FC. EFC is isosceles and m(EFC)=150, m(ECF)=15.thus m(ECB)=15+15=30.
similarly, m(ADE)=30.so, DEF is a equilateral triangle.m(DEC)=60.
sorry, I can't draw figure, yet.

MEDHOD 3: Let's draw a eqilateral triangle ABF, outside the square.AFBE is kite.FA=FB=FE.since m(EBF)=m(EBC)=75 and FB=FC, parallelogram FBCE must be a rhombus! thus EC=BC.similarly, AD=DE.that is, DEC is equilateral.m(CED)=60.


Let $ a,b,c $ be three sides of a triangle and $m_a,m_b,m_c$ be its three medians. Prove the inequality :
$\sum\frac{a}{m_a}\frac{(b+c)^2}{4bc} \leq\frac{R\sqrt{3}}{r} $

I think this one comes directly from the following inequality $2Rm_a \geq\frac{b^2+c^2}$ and the equality $abc = 4RS$.  Ooops, and with AM - GM.

Dear treegoner,

Nice to see you again. Would you mind explaining more about your solution ? I had tried with your hint but it was still complicated for me.

Your friend

Hoang

PS : the inequality you used should be corrected to
$m_a \geq\frac{b^2+c^2}{4R} $

[quote="nickolas"]Well  yes , tranminhhoang is  right....following  the  instuctions  we get
$ LHS\leq\frac{a^2+b^2+c^2}{2A} $  but  $ \frac{a^2+b^2+c^2}{2A}\geq\frac{R\sqrt3}{r}  $   :?[/quote]

Excuse me, nickolas. May you post your solution for the last inequality you've proposed ? I don't know how to prove it.
Thank you very much.

In fact, the inequality of Nickolas is not right for all triangles.
  The original inequality, in my view, is a really challenging one. I 'll try it again in the near future. :?
    PS What have you been up to, Hoang ?

[quote="tranminhhoang"]Let $a,b,c$ be three sides of a triangle, $m_{a},m_{b},m_{c}$ be its three medians, $R$ be the radius of circumcircle and $r$ be the radius of incircle. Prove the inequality :
\[\frac{a(b+c)^{2}}{4bcm_{a}}+\frac{b(c+a)^{2}}{4cam_{b}}+\frac{c(a+b)^{2}}{4abm_{c}}\leq\frac{R\sqrt{3}}{r}. \]
[/quote]
In fact, we have the following inequality chain :
\[\sum\frac{a(b+c)^{2}}{4bcm_{a}}\leq\sum\frac{a}{w_{a}}\leq\frac{2\sqrt{3}}{9r}\sum{m_{a}}\leq\frac{2\sqrt{3}}{9r}\sum{r_{a}}\leq\frac{R\sqrt{3}}{r}, \]
where $w_{a},w_{b},w_{c}$ be the angle-bisectors and $r_{a},r_{b},r_{c}$ be the radii of excircles.

[color=purple]Here is the similar problem  [/color]
       http://www.mathlinks.ro/Forum/viewtopic.php?t=144807


geometry
circumcircle
geometric transformation
homothety

Σε τρίγωνο $ ABC$ έστω $ AB\plus{}BC\equal{}3AC$. Ο εγγεγραμμένος κύκλος (με κέντρο $ I$) τέμνει τις $ AB, BC$ στα σημεία $ D,E$ αντίστοιχα. Εστω $ K, L$ τα συμμετρικά των $ D, E$ (αντίστοιχα) ως προς το το $ I$. Να αποδείξετε ότι το $ ACKL$ είναι εγγράψιμο.

[quote="bilot4"]Σε τρίγωνο $ ABC$ έστω $ AB \plus{} BC \equal{} 3AC$. Ο εγγεγραμμένος κύκλος (με κέντρο $ I$) τέμνει τις $ AB, BC$ στα σημεία $ D,E$ αντίστοιχα. Εστω $ K, L$ τα συμμετρικά των $ D, E$ (αντίστοιχα) ως προς το το $ I$. Να αποδείξετε ότι το $ ACKL$ είναι εγγράψιμο.[/quote]

[u][b]ΛΥΣΗ[/b][/u]

Κατ' αρχήν ο εγγεγραμμένος κύκλος [u]εφάπτεται[/u] στις πλευρές στα σημεία $ D,E$.

[i]I shall continue my solution in english.[/i]

Draw the circumcircle of the triangle $ ABC$.
Denote by M the midpoint of the small arc $ AC$ and by $ M'$ the midpoint of $ BI$.

Applying [i]Ptolemeus Theorem[/i] to the inscribed quadrilateral $ AMCB$, we get $ BM \equal{} 3AM \equal{} 3MI\Longrightarrow \boxed{M'I \equal{} MI}$ . Now we use the geometric transformation of homothety $ H(I, \minus{} 1)$ and according to this trasformation $ M',D,E$ are mapped to the points $ M,K,L$ respectively. It is easy to prove that points $ D,I,E$ lie on a circle centered at $ M'$ with radius $ M'I$, and therefore we have shown that points $ L,I,K$ lie on a circle centered at $ M$ with radius $ MI$. It is also well known that $ A,I,C$ lie on the same circle and this finishes our proof.

Nick Rapanos

Εκπληκτική λύση Νίκο και καλά έκανες που την έβαλες στα Αγγλικά :)

Η δική μου λύση βασίζεται στο ότι αν $ X$ είναι η τομή των $ AL, BC$ τότε το $ X$ είναι και το σημείο τομής του αντιστοιχου παραγγεγραμμένου με τη $ BC$. Ομοια γαι το $ Y$ ως σημείο τομής των $ CK, AB$. Ομως είναι γνωστό ότι $ CX\equal{}AY\equal{}p\minus{}b\equal{}b$ (το $ p\minus{}b\equal{}b$ προκύπτει εύκολα απ' την υπόθεση. Μετά από αυτό με λίγο angle-chasing προκύπτει $ \angle{XLK}\equal{}KCA$ κι έπεται το ζητούμενο.

[quote="rokas"]Την άσκηση αυτή την πρότεινε στην IMO 2005 ο Κοντογιάννης. (τότε υπαρχηγός)
Τον θυμάμαι αμέσως μετά τον διαγωνισμό που ρώταγε με αγωνία να μάθει αν έπεσε η άσκηση που πρότεινε.
Πολύ τσαντίστηκε όταν έμαθε ότι δεν έπεσε.

Η άσκηση άρεσε πολύ στο jury, άλλα κάποιος αρχηγός την θεώρησε γνωστή και γι αυτό δεν μπήκε.[/quote]

Αγαπητές rokas,

 μου ζήτητσαν από την Αμερική τους δημιουργούς των ασκήσεων IMO short list. Ο Κος Θ.Μπόλης μου έγραψε :

 Μπάμπη,

Τα προβλήματα 1986-6, 1988-12, 1988-13 και 2006-G5 είναι του Δημήτρη
Κοντογιάννη. Το ίδιο ίσως ισχύει και για το 2005 - G1.
Τα 1989-10 και 1990-7 είναι δικά μου.
Το 1996-Α3 είναι μάλλον του Μιχάλη Λάμπρου.(Τελικά , σίγουρα. Το επιβεβαίωσε ο Μιχάλης).
Για το 2001 και 2003 ρώτα τον Φελλούρη.
Για το 1975 και το 1979 δεν έχω ιδέα.

  Μήπως ξέρεις τα υπόλοιπα θέματα που δεν ήξερε. Νομίζω μου είπε για κάποιο ο Αν. Φελλούρης , αλλά δεν ήταν σίγουρος. Μπορείς σε παρακαλώ να μάθεις ; Δεν έχω το τηλέφωνο του Κοντογιάννη.

 Σε ευχαριστώ - Μπάμπης Στεργίου

[quote="bilot4"]Εκπληκτική λύση Νίκο και καλά έκανες που την έβαλες στα Αγγλικά :)

Η δική μου λύση βασίζεται στο ότι αν $ X$ είναι η τομή των $ AL, BC$ τότε το $ X$ είναι και το σημείο τομής του αντιστοιχου παραγγεγραμμένου με τη $ BC$. Ομοια γαι το $ Y$ ως σημείο τομής των $ CK, AB$. Ομως είναι γνωστό ότι $ CX \equal{} AY \equal{} p \minus{} b \equal{} b$ (το $ p \minus{} b \equal{} b$ προκύπτει εύκολα απ' την υπόθεση. Μετά από αυτό με λίγο angle-chasing προκύπτει $ \angle{XLK} \equal{} KCA$ κι έπεται το ζητούμενο.[/quote]

Mim,
 βάλε σε παρακαλώ και την υπόλοιπη λύση με τη μετακίνηση των γωνιών , γιατί δεν προλαβαίνω να την κοιταξω αναλυτικά.Θέλω να μαζέψω όλες τις λύσεις σε αυτό το θέμα και να το βάλω στο αρχείο.

 Μπάμπης

Ευχαρίστως να βάλω το υπόλοιπο της λύσης μου με το angle-chasing κ. Στεργίου (παρεπιπτόντως όχι ότι έχω πρόβλημα με το mim αλλά προτιμώ το Δημήτρης :wink: ).

Bέβαια πιθανότατα δεν θυμάμαι αν αρχικά είχα σκεφτεί αρκιβώς αυτό αλλά σίγουρα είχα σκεφτεί κάτι παρεμφερές (βέβαια θυμάμαι ότι η αρχική λύση μου σε αυτό το σημείο ήταν σίγουρα πιο σύντομη...). Το "δύσκολο" κομμάτι που υπάρχει είναι απλά να αποδείξουμε ότι το σημείο τομής των ευθειών $ AX, CY$ βρίσκεται στον εγγεγραμμένο
Κατ' αρχάς $ CX\equal{}b\equal{}AC\iff \angle{XAC}\equal{}CXA$. Αρα έχουμε:
Αν $ M$ το σημείο τομής του εγγεγραμμένου με την $ AC$ και η $ MI$ τέμνει την $ AX$ στο $ P$, $ \angle{ILP}\equal{}90\minus{}AXC\equal{}90\minus{}XAC\equal{}IPL$. Αρα το $ P$ βρίσκεται στον κύκλο.
Ομοια αν $ S$ είναι η τομή $ MI, CY$ και το $ S$ βρίσκεται στον κύκλο αρα τα σημεία $ MPS$ είναι συνευθειακά και βρίσκονται σε ένα κυκλο, άρα $ P\equal{}S$ ίσον το σημείο τομής $ AX,CY$.

Μετά από αυτό έχουμε:
$ \angle{YCA}\equal{}CYA\equal{}90\minus{}DKY\equal{}90\minus{}DLX\equal{}XLK$ κι άρα το ζητούμενο.

[quote="bilot4"].... (παρεπιπτόντως όχι ότι έχω πρόβλημα με το mim αλλά προτιμώ το Δημήτρης :wink: ).

[/quote]

Δημήτρη ,

 το κόλησα από το φίλο μας το Σιλουανό ! Από την άλλη είναι και πιο σύντομα και .. είπα να δερδίσω χρόνο :)  !

 Και γω προτιμώ το Δημήτρης , που έχει και ...σημασία!

 Ευχαριστώ για την απάντηση .Μόλις πάω σπίτι θα μελετήσω και τις επίσημες λύσεις. ΕΠίσης στο ΙΜΟ compendium νομίζω ότι έχουν και κάτι άλλο.
 Καλό Σαββατοκύριακο !!!

 Μπάμπης

[quote="rokas"]Καλησπέρα Μπάμπη,

Το 2005 G-1 είναι σίγουρα του Δ Κοντογιάννη.
Αν υπάρχουν άλλα στο short list του Κοντογιάννη εκτός από αυτά που ανέφερες δεν το ξέρω, μπορείς να ρωτήσεις τον ίδιο. (Το τηλέφωνο του είναι στις πρώτες σελίδες του βιβλίου Γεωμετρία που έχει γράψει και νομίζω ισχύει ακόμα. Μπορώ να στο στείλω σε private μήνυμα)

Πάντως αν υπάρχει ελληνική γεωμετρία στο shortlist θα είναι ή του Δ. Κοντογιάννη ή του Σ. Λουρίδα - δεν είμαι σίγουρος αν ο Λουρίδας έχει άσκηση στο shortlist ομως.

- Κωνσταντίνος, USA[/quote]

NAI !

 Ο [b]Σωτήρης Λουρίδας [/b]έχει προτείνει το [b]G5 [/b]το 2001 στην Αμερική . Μου το είπε ο  σημερινός αρχηγός της ομάδας , αλλά μου το επιβεβαίωσε και ο ίδιος ο κ. Σ. Λουρίδας.
 Να τα λέμε αυτά , διότι σε όλο τον κόσμο συνηθίζεται να καταγράφονται τα ονόματα των δημιουργών των προβλημάτων στις Short list.Είναι μια υποχρέωση προς όλους αυτούς τους ανθρώπους που με τόσο κάπο μας χαρίζουν αυτά τα πολίτιμα στολίδια των μαθηματικών.

 [u]Μπάμπης Στεργίου[/u]


The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length. If the width is 2, how many units are in the length of the diagonal?

[hide="Answer"]$ 1\plus{}\sqrt{5}$[/hide]

[quote="DonkeyKong"]The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length. If the width is 2, how many units are in the length of the diagonal?

[hide="Answer"]$ 1 \plus{} \sqrt {5}$[/hide][/quote]
$ \frac {l}{2} \equal{} \frac {d}{l}$
$ d \equal{} \sqrt {l^2 \plus{} 4}$
$ d^2 \equal{} l^2 \plus{} 4$
$ 2d \equal{} l^2$
$ d^2 \minus{} 2d \equal{} 4$
$ d^2 \minus{} 2d \minus{} 4 \equal{} 0$
$ \frac {2\pm\sqrt {4 \minus{} (4\cdot \minus{} 4\cdot1)}}{2\cdot1}$
$ \frac {2\pm\sqrt {20}}{2}$
$ \mbox{Positive Root: } \frac {2 \plus{} \sqrt {20}}{2} \equal{} 1 \plus{} \frac {\sqrt {20}}{\sqrt {4}} \equal{} 1 \plus{} \sqrt {5}$

add 1/2 to the answer and it's phi!!! yea i'm bored today


Slove system of equation: $ \begin {cases}x\plus{}y\plus{}z\equal{}7\\x^2\plus{}y^2\plus{}z^2\equal{}21\\xz\equal{}y^2\end{cases}$

[hide]x=1
y=2
z=4

or

x=4
y=2
z=1[/hide]

I am sorry I am unable to provide any explanation. I am afraid that this answer just sort of jumped out at me  :!:  so I didn't do any sorts of working- other than checking that the solutions I found worked.  :maybe:

[quote="GioMott"][quote="thanhnam2902"]Slove system of equation: $ \begin {cases}x \plus{} y \plus{} z \equal{} 7 \\
x^2 \plus{} y^2 \plus{} z^2 \equal{} 21 \\
xz \equal{} y^2\end{cases}$[/quote]

So we have
$ \displaystyle 7^2 \equal{} (x \plus{} y \plus{} z)^2 \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2xz \equal{} 21 \plus{} 2xy \plus{} 2yz \plus{} 2y^2$
and simplifying
$ xy \plus{} yz \plus{} y^2 \equal{} y(x \plus{} y \plus{} z) \equal{} 7y \equal{} 14$
so $ y \equal{} 2$.
Subsitute the value of y in the original system and you get a very simple system in the two unknowns x and z. The two solutions are $ (x,z) \equal{} (1,4)$ and $ (x,z) \equal{} (4,1)$. That's all folks...

Bye
GioMott[/quote]
Thank you! :lol:


$ x_{n\plus{}3}\equal{}2x_{n\plus{}2}\plus{}2x_{n\plus{}1}\minus{}x_{n}$ and $ x_{1}\equal{}x_{2}\equal{}1 , x_{3}\equal{}4$.
Prove that all $ x_{n}$ are perfect squares

Let $ F_n$ be the fibonacci sequence. Then
$ 2F_{n\plus{}2}^2 \plus{} 2F_{n\plus{}1}^2 \minus{} F_{n}^2$
$ \equal{}2F_{n\plus{}2}^2 \plus{} 2F_{n\plus{}1}^2 \minus{} (F_{n\plus{}2} \minus{} F_{n\plus{}1})^2$
$ \equal{}(F_{n\plus{}2} \plus{} F_{n\plus{}1})^2$
$ \equal{}F_{n\plus{}3}^2$

Thus by induction, $ x_n \equal{} F_n^2$


One commercially available ten-button lock may be opened by depressing -- in any order -- the correct five buttons.  The sample shown below has $\{1, 2, 3, 6, 9\}$ as its combination.  Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations.  How many additional combinations would this allow?

[asy]
path p=origin--(18,0)--(18,5)--(0,5)--cycle;
draw(p^^shift(0,8)*p^^shift(22,0)*p^^shift(22,16)*p^^shift(22,24)*p);
filldraw(shift(0,16)*p^^shift(22,8)*p^^shift(22,32)*p^^shift(0,32)*p^^shift(0,24)*p, black, black);
draw((-1,-1)--(41,-1)--(41,38)--(-1,38)--cycle, linewidth(2));
int i;
for(i=1; i<6; i=i+1) {
label(string(6-i), (-3,8*i-5.5), W);
label(string(11-i), (43,8*i-5.5), E);
}[/asy]

What about pushing $0$ buttons? One more thing... not sure if I have this correct:

Let's say we have $x$ is as few as 2 and $x$ is as great as 10. Does this mean that $2 \le x \le 10$?

[quote="lingomaniac88"]Here's what I would do, based on how I interpreted the question:
[hide]$\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}$
It's a lot of arithmetic, but you still get 770.[/hide][/quote]

A plausibly easier method?

[hide]It'd be a bit easier if you noticed that:

$\binom{10}{0} + \binom{10}{1} + \ldots + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} = 2^{10}$

and then subtracted out $\binom{10}{0}$ and $\binom{10}{10}$. That would give you:

$2^{10} - 1 - 1$. But then you have to subtract out the $\binom{10}{5}$ because it asks you how many [i]more[/i]. So we have for our grand total, $2^{10} - 1 - 1 - \binom{10}{5}$. I think I just didn't understand what as great as meant  :D[/hide]

[hide="Complete Solution"]To form a combination, we can either choose to press a button, or leave it unpressed. Since there are $10$ buttons, this can be done in $2^{10}$ ways. Subtracting the number of $0$, $5,$ and $10$ button combinations, the answer is $ 2^{10}-1-\binom{10}{5}-1=770.$[/hide]

Combo AIME grind: After redesigning, there are $2^{10}-1-1=1022$ combinations. Before, there were $\binom{10}{5}=252$ combinations. The answer is $1022-252=\boxed{770}$.

After redisigning, we have $2^{10}-2=1022$, combinations. In the original, there are $\binom{10}{5}=252$, ways, hence, $1022-252=\boxed{770}.$


algebra
polynomial
number theory proposed
number theory

Not really hard, but at least not posted I think :D :

Let integer polynomials $p_{1}(X), p_{2}(X), ..., p_{m}(X)$ be given, where $X = (x_{1}, x_{2}, ..., x_{n})$ is short for the tupel of variables.

a) Show that there is an integer polynomial $p(X)$ such that \[\forall i: \ p_{i}(X)=0 \ \iff \ p(X)=0.\] b) Show that for a given prime $q$, there is an integer polynomial $p(X)$ such that \[\forall i: \ q|p_{i}(X) \ \iff \ q|p(X).\] c) Find all integers $q$ such that, independent how the $p_{i}(X)$ look like, we can always find an integer polynomial $p(X)$ with \[\forall i: \ q|p_{i}(X) \ \iff \ q|p(X).\]
For all these problems, $X$ runs through all tupels of integers.

I have a strange solution for the part b)
We have the polynomials $p_{i}(x)$
define $r_{1}(x)=\sum_{i=1}^{q-1}p_{i}(x)^{q-1}$ and so on
then $k_{i}(x)=\sum_{i=1}^{q-1}r_{i}(x)^{q-1}$and so on ...
in the end we will be left with less then q polynomials $f_{i}(x)$
and we take $P(x)=\sum f_{i}(x)^{q-1}$ . the proof is obvious from:
$\forall i<q ,q|a_{i}\longleftrightarrow q|\sum_{i=1}^{m<q}a_{i}^{q-1}$

Well, I think you missed some definitions for general $r_{i}$, but I see what you mean.
In fact, my first approach was the same.
But note that $q=2$ makes problems.

yeah I noticed the q=2 case before and I don't have a solution for that now. I admit that I didn't write very clearly above but its understandable :)
[edit]
no I actually found a very simple solution for q=2:
$P(x)=1+\prod (p_{i}(x)+1)$ :D

now only the part c is left :maybe:


algebra
polynomial
superior algebra
superior algebra unsolved

1.Z[i] = {a+bi | a,b E Z}

What is the quotient field of Z[i]

2. Prove there are infinitely many irreducible polynomials in Z[x] of any fixed degree d.

1.  Plainly $ \mathbb{Q}[i]$.  Note that $ \frac{1}{a \plus{} bi} \equal{} \frac{a \minus{} bi}{a^2 \plus{} b^2}$.

2.  A cheap solution is to construct infinite families of irreducible polynomials using Eisenstein's criterion or Perron's criterion.  There's probably a simpler way.


[quote="luiseduardo"]"The Boss" always wants the numbers 1 to 11 of the wheel are arranged so that each time they are three numbers in a straight line, their sum is 18.

Can you put the numbers like that?

[img]http://img2.imageshack.us/img2/9803/cals.jpg[/img][/quote]

[hide]

There is more information given in the problem than need be.  It doesn't have to tell
that the sum is 18.

The middle number is 6.

1 + 6 + 11
2 + 6 + 10
3 + 6 + 9
4 + 6 + 8
5 + 6 + 7

EDIT:

It is similar to forming a Magic Square having nine cells and state that the
numbers are to go into the cells, be they consecutive integers or that they
have a different common difference, for instance.
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if $ 4^n \plus{}2^n \plus{}1$ is a prime then prove that n is a power of 3

Can you please tell what you are trying to convey .
If you wanted to give a counterexample .$ 73$ is prime when $ n\equal{}3$
So I  dont think this is a correct counterexample

[quote="srinath.r"]I am not sure if my solution is correct ,anyway here it is
[hide="Solution"]
If we prove the contrapositive of the statement we are done
That is if $ n$ is not a power of $ 3$ then $ 4^{n} \plus{} 2^{n} \plus{} 1$ is not prime .
Let $ n \equal{} 3^{\alpha} \plus{} k$ where $ k \equiv 0,1\bmod{3}$
We got to prove $ 4^{3^{\alpha} \plus{} k} \plus{} 2^{3^{\alpha} \plus{} k} \plus{} 1$ is not prime .
We observe that $ 4^{3m \plus{} 1} \equiv 4\bmod{7}$
Or $ 4^{3m \plus{} l} \equiv 4\bmod{7}$ if $ l \equiv 1\bmod{3}$ (1)
Case 1
$ k \equiv 1 \mod 3$
Then from $ 1$ we have $ 4^{3^{\alpha} \plus{} k} \equiv 4\bmod{7}$  $ \boxed{1}$
Now considering mod 7 for powers of $ 2$
We observe that $ 2^{3m \plus{} l} \equiv 2\bmod{7}$ if $ l \equiv 1\mod 3$ (2)
So from (2) we have $ 2^{3^{\alpha} \plus{} k} \equiv 2\bmod{7}$ $ \boxed{2}$
From $ \boxed{1}$ and $ \boxed{2}$ we have
$ 4^{3^{\alpha} \plus{} k} \plus{} 2^{3^{\alpha} \plus{} k} \plus{} 1 \equiv 7\equiv 0\mod 7$
Similarly for $ k \equiv 2\bmod{3}$ we can prove that
$ 4^{3^{\alpha} \plus{} k} \plus{} 2^{3^{\alpha} \plus{} k} \plus{} 1 \equiv 7\equiv 0\mod 7$
So the given expression is not prime if $ n$ is not a power of $ 3$
[/hide][/quote]I quoted because you might change your post later. Your proof only proved that $ n$ must be a multiple of 3. Maybe if you cover the case when $ k\equiv0\mod 3$. For example, if $ \alpha\equal{}1$, you still got to prove that $ k\equal{}3$ yields a composite. However, when $ k\equal{}2\cdot3^\alpha$, something interesting happens. Just trying to patch up your solution.

It is easily proved that $ x^2\plus{}x\plus{}1|x^{2m} \plus{} x^m \plus{} 1$ iff $ m \not \equiv 0 \pmod 3$

So, if $ n \equal{} 3^k q$ where q is prime to 3, then let $ x \equal{} 2^{3^k}$ and from the above result we see that $ 4^{3^k} \plus{} 2^{3^k} \plus{} 1 | 4^n\plus{}2^n\plus{}1$

Hence if the expression is a prime it is only when n is a power of 3.

10000th user you are right , I realised the case  I left $ k \equiv 0\mod3$ when I switched off my computer  :oops: .
@hsbhatt, I have one doubt
You said that $ x^{2} \plus{} x \plus{} 1 \mid x^{2m} \plus{} x^{m} \plus{} 1$ iff $ m \not\equiv 0\mod 3$
Now you said let $ x \equal{} 2^{3^{k}}$ so $ (2^{3^{k}})^{2} \plus{} 2^{3^{k}} \plus{} 1 \mid (2^{3^{k}})^{2n} \plus{} (2^{3^{k}})^{n} \plus{} 1$  .As per your post this happens only when $ n \not\equiv 0\mod 3$ ,but you have set $ n \equal{} 3^{k}q$

I should have further said, let m = q

$ 4^{3^k} \plus{} 2^{3^k} \plus{} 1 | \left(2^{3^k} \right)^{2q} \plus{} \left(2^{3^k} \right)^q \plus{} 1 \equal{} 4^n \plus{} 2^n \plus{} 1$


Let A={a1=0,a2,a3,...} be a set of no negative integers s.t.
every integer is the sum of k elements (not necessarily distincts) of A.
Let A(n) be the number of elements of A less than or equal to n.
Prove that there is r>0 s.t. A(n)>r[n^(1/k)].

If I am not mistaken... there are at most $ A(n)^k$ ways of picking $ k$ numbers to add together, and you have to have a way to get all of $ 0$ through $ n$, so $ A(n)$ clearly must be of order at least $ n^{1/k}$...


Is it OK if I don't mail/fax my entry form until tomorrow?


Let n be a fixed positive integer.  How many ways are there to write n as the sum of a sequence n = a[1] + a[2] + . . . + a[k] for an arbitrary integer k :ge: 1 such that a[1] :le: a[2] :le: ... :le: a[k] :le: a[1] + 1?
For example, with n = 4, there are 4 ways: 4 = 4 = 2 + 2 = 1 + 1 + 2 = 1 + 1 + 1 + 1

tom_clymer wrote:I believe that there are [hide]      n      [/hide] ways.  I haven't rigorously proven it, but this is definitely a maximum because [hide]for each possible number of terms in the sequence, 1-n, there cannot be two or more ways to make a sum of n[/hide].  I believe that there will always be this many ways because [hide]for any number of terms 1 :le: b :le: n, you should be able to distribute terms around the average, n/b, to produce a sum of n[/hide].

[hide]Yes, you are right. And you can prove this by induction.[/hide]

[hide]I did it roughly the same way Tom did -- it really works out just to the division algorithm -- a[1] = quotient, # of extra 1's = remainder.[/hide]

I didn't even think of doing it this way on the contest. I just found [hide]a one-to-one correspondence between a way to get n and a way to get n+1 that wasn't all 1s.[/hide]


hello, do you mean
$ 9a\plus{}16b\plus{}25c\plus{}36d\plus{}49e\plus{}64f\plus{}81g\equal{}123$
or
$ 9a\plus{}16b\plus{}25c\plus{}36d\plus{}49e\plus{}64f\plus{}81f\equal{}123$
Sonnhard.

[color=red]Now fixed in original post by mod.[/color]

This problem (from the 2004 Indonesia MO) is a copy of Problem 8 from the 1989 AIME.

[quote=wangsacl]\[ a\plus{}4b\plus{}9c\plus{}16d\plus{}25e\plus{}36f\plus{}49g\equal{}1\]
\[ 4a\plus{}9b\plus{}16c\plus{}25d\plus{}36e\plus{}49f\plus{}64g\equal{}12\]
\[ 9a\plus{}16b\plus{}25c\plus{}36d\plus{}49e\plus{}64f\plus{}81g\equal{}123\]
Determine the value of $ 16a\plus{}25b\plus{}36c\plus{}49d\plus{}64e\plus{}81f\plus{}100g$.[/quote]

This is a problem for human, or for a computer?  :roll:


[quote="cincodemayo5590"]Solve for $x$:

$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5$

[hide="hint"]
It's the same thing (in my mind) as $\sqrt{x-\sqrt{x-\ldots \sqrt{x-5}=5}}$ (With any number of radical signs).[/hide][/quote]

[hide]
first square both sides, we get:
$25=x-\sqrt{x-\sqrt{x-\sqrt{x-\cdot\cdot\cdot\sqrt{x-5}}}}$.
the thing in the radical is equal to $5$, and $x=30$.[/hide]

[hide]
$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5 \Leftrightarrow \sqrt{x-5}=5 \Leftrightarrow x=30$
[/hide]

[quote="vvr1590"]midorya: that's a cool method, but how do you get the second statement from the first?[/quote]
$a_{1}=\sqrt{x-5}$
$a_{n+1}=\sqrt{x-a_{n}}$
$a_{n}=5 \Rightarrow a_{n}=a_{1}=\sqrt{x-5}=5$


i know its an easy problem, but i can't remember how to solve it in algebric way.

[img]http://img171.imageshack.us/img171/3563/1av9.gif[/img]

any help will be appreciated

thanks.

:mad:
you do not owe poster in this section


Let $\{3,4,12\}$ be a set.  In each step of conversion, you may choose two numbers $a,b\in\{3,4,12\}$ and convert them into $0.6a-0.8b$ and $0.8a+0.6b$.  Is it possible to acquire the set $\{4,6,12\}$ through a finite number of conversions? Is it possible to reach $\{x,y,z\}$ such that $|x-4|, |y-6|, |z-12|\in\left[ 0,\frac{1}{\sqrt{3}}\right]$?

Masoud Zargar

[hide="Solution"]
It is easy to see $a^2 + b^2 + c^2$ is an invariant.  $\{ 4, 6, 12 \}$ has a different value of the invariant, hence it is not reachable.

The actual value of the invariant with our given problem condition is $13^2$.  Because $\{ 4, 6, 12 \}$ has too large a value, we wish to determine whether $\{ 4-k, 6-k, 12-k \}$ satisfies the condition where $k \in \left[ 0, \frac{1}{ \sqrt{3} } \right]$.

$(4 - k)^2 + (6 - k)^2 + (12 - k)^2 = 14^2 - 44k + 3k^2$

Clearly a decreasing function in $k$ for the relevant domain.  For maximal $k$, it gives the value $14^2 - \frac{44}{ \sqrt{3} } + 1 > 13^2$ (I can't think of a neat way to show this but it's true), so no such value is possible.
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Edit:  This is from Engel.  :)

[quote="t0rajir0u"][hide="Solution"]
It is easy to see $a^2 + b^2 + c^2$ is an invariant.  $\{ 4, 6, 12 \}$ has a different value of the invariant, hence it is not reachable.

The actual value of the invariant with our given problem condition is $13^2$.  Because $\{ 4, 6, 12 \}$ has too large a value, we wish to determine whether $\{ 4-k, 6-k, 12-k \}$ satisfies the condition where $k \in \left[ 0, \frac{1}{ \sqrt{3} } \right]$.

$(4 - k)^2 + (6 - k)^2 + (12 - k)^2 = 14^2 - 44k + 3k^2$

Clearly a decreasing function in $k$ for the relevant domain.  For maximal $k$, it gives the value $14^2 - \frac{44}{ \sqrt{3} } + 1 > 13^2$ [b](I can't think of a neat way to show this but it's true), so no such value is possible.  [/b][/hide]

Edit:  This is from Engel.  :)[/quote]

Haha :lol: very rigorous.

:P [hide="Fine"]We wish to show $- \frac{44}{ \sqrt{3} } + 1 > -25 \Leftrightarrow 26 > \frac{ 44 } \sqrt{3} } \Leftrightarrow \sqrt{3} > \frac{22}{13}$.

Squaring gives

$3 > \frac{484}{169}$
$507 > 484$,

Which is true.
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