A team averages 7 losses for every 13 games it wins. If ties are not permitted, what is the probability that the team will win its next game? Express your answer as a percent.

[quote="GameBot"]A team averages 7 losses for every 13 games it wins. If ties are not permitted, what is the probability that the team will win its next game? [b]Express your answer as a percent.[/b][/quote]
Finishing off, $ \frac{13}{20}\equal{}\boxed{65}\%$

[quote=\$LaTeX\$]The chance that a team wins is $ \frac{13}{13\plus{}7}\equal{}\frac{13}{20}$. So the probability that the team will win its next game is $ \boxed{\frac{13}{20}}$.[/quote]

That is what I did. Bu we both forgot the question is asking for a percent. 13/20 as a percent is 65/100 =65%


modular arithmetic
Pascal\u0027s Triangle

After doing a little reading on "fun facts" about Mersenne primes, I came across one that stated that the rows in Pascal's Triangle that correspond to Mersenne primes (i.e. row 3, 7, 31, etc...) have only odd entries.

Can anyone formalize this or put their finger on why this is?

Lucas' Theorem: if $ p$ is a prime and $ n \equal{} n_dn_{d \minus{} 1}\dots n_1n_0$ and $ k \equal{} k_dk_{d \minus{} 1}\dots k_1k_0$ are the base $ p$ representations of $ n$ and $ k$, then
\[ \binom{n}{k}\equiv \binom{n_d}{k_d}\binom{n_{d \minus{} 1}}{k_{d \minus{} 1}}\cdots\binom{n_{0}}{k_0}\pmod{p}.
\]
applying this result with $ p \equal{} 2$ shows that the only completely odd rows in pascal's triangle are the ones numbered $ 2^n\minus{}1$.

No they are not, every $ 2^n\minus{}1$-th row is an odd row, independent from that number being prime.
By the way, it can be done more directly as using Lucas' theorem.

Please note that the original claim did not say that [i]only[/i] those rows corresponding to Mersenne primes have this property. It's true that this hold for all numbers of the form $ 2^n\minus{}1$, but these include the Mersenne primes, which is the original claim.

I was only responding to pleurestique who originally explicitely said "only the Mersenne-primes".

Hamster, thank you for the enlightening comment ;)... the obvious error came from writing fast, and doing this at 5 in the AM, when the obvious often escapes the best of us. anyway, i edited my post and there's no problem anymore :)


Given a stack of n different cards we wish to reverse the order of the cards by a sequence of moves. Each move is to take a group of adjacent cards from anywhere in the stack and put it somewhere else in the stack without separating any cards or turning over any cards. Show that this can be done in 5 moves for n = 9. Show that it can be done in 27 moves, but not in 26 moves for n = 52.

You know, a long time ago, during a rather long train ride, I found ways to do it in $n+1$ moves for $2n+1$ cards, when $2n+1=5,7,9$, but never actually gave any thought to the general case, although it seems very interesting. Anyway, here's a possible answer for $9$ cards (the first question of the problem):

[b]1)[/b] Put $3,4,5,6$ between $8,9$;

[b]2)[/b] Put $6,9$ between $7,8$;

[b]3)[/b] Put $1,2,7,6$ between $4,5$;

[b]4)[/b] Put $2,7,6,5$ between $3,4$;

[b]5)[/b] Finally, put $3,2$ between $4,1$.

[quote="Myth"]This problem was invented by my supervisor!!![/quote]

So who is the author of the problem ? Do you know of any other problem he invented for TOT ?  :)

As far as I know, it was unique creation for ToT.


function
real analysis
real analysis unsolved

Suppose $ f$ is a non-decreasing function on $ [a, b]$ such that
$ f' \equal{} 0$ a.e.
Prove that, for every positive, $ \epsilon$, $ \delta$, there exists a finite collection of non-overlapping intervals [$ {x_{k} , y_{k} }$] such that $ \sum{|y_{k} \minus{} x_{k} |} < \delta$ and $ \sum {f(y_{k} ) \minus{} f (x_{k} )} > f (b) \minus{} f (a)\minus{} \epsilon$.

I guess you are reading Royden's book. :lol:
This problems is already proved by using a lema. Just try to read the lemma carefully you can prove it

It was on my final exam;- Yes, the lemma your are referring to is one way to do it.


Euler
real analysis
real analysis unsolved

... probably one of the few ones of it's kind that i haven't been able to sum yet...   :oops:

show

$ \sum_{n\equal{}1}^\infty\;\dfrac{\left(\sum_{k\equal{}1}^n\;\dfrac{1}{k^4}\right)}{n^2}\;\equal{}\;\boxed{\dfrac{37}{12}\zeta(6)\minus{}\left[\zeta(3)\right]^2}$

As $ n\rightarrow\infty$, doesn't the top sum go to $ \frac{\pi^{4}}{90}$ and the bottom to $ \frac{\pi^{2}}{6}$?

[quote="JRav"]As $ n\rightarrow\infty$, doesn't the top sum go to $ \frac {\pi^{4}}{90}$ and the bottom to $ \frac {\pi^{2}}{6}$?[/quote]

well...... ummm..... errrr....... pssstttt....... sighhhh.... [b]no.[/b]  :oops:   the sum is not going to be $ \zeta(4) \times \zeta(2)$ if that is what you're thinking.   :D

notice that the terms of the series are

$ 1,\quad \dfrac{1\plus{}\dfrac{1}{2^4}}{2^2},\quad  \dfrac{1\plus{}\dfrac{1}{2^4}\plus{}\dfrac{1}{3^4}}{3^2},\quad \dfrac{1\plus{}\dfrac{1}{2^4}\plus{}\dfrac{1}{3^4}\plus{}\dfrac{1}{4^4}}{4^2},\quad \dfrac{1\plus{}\dfrac{1}{2^4}\plus{}\dfrac{1}{3^4}\plus{}\dfrac{1}{4^4}\plus{}\dfrac{1}{5^4}}{5^2},\quad \ldots$

Oh I definitely know that it wouldn't go to the product of the two.  I was just thinking out loud.  Do you have any idea how to solve this?

Maybe looking at the missing terms after considering $ \zeta(2) \times \zeta(4)$ might help... I'll try my hand at that later and post here if I have anything!  :)

Apparently there are some very general results regarding these multi-zeta values, they are introduced as Euler originally did [url=http://www.math.ou.edu/~kujawa/CalcIIIpmF07/EulerSeries.pdf]here[/url] and the result you want is given [url=http://mathworld.wolfram.com/EulerSum.html]here[/url] on this Mathworld article on Euler sums (the modern definition of these is slightly different from what Euler considered so the expressions have to be modified slightly.


so how much pay does one get in a carrer in math??
wat r some examples???
 :blush:
yup, i kinda am not that involved in disscussions and am too lazy to read them, so yeah. :oops:
[quote]All 2nd Violins are awesome NINJAS!!! :ninja: [/quote]
[url]http://www.thenamewhichescapesme.blogspot.com[/url]
[quote]Dont die. Ur not worth saving, anyway.[/quote]

Uh, what career?  Obviously, a career in academia will pay less than a high-flying investment banker, and even in academia, large research institutions are likely to pay more than smaller schools.  People who go into a science/applied math/engineering field will make a different salary than someone who goes into business.

And math is not the money, it's about the joy of math.

[b]@ shanmug1000 [/b]: You haven't explicitly mentioned which kind of 'career' in Math you intend to pursue.

A career in academia gets lesser pay than, say, a career in Investment Banking (There are various designations in I-banking, say, like -- Quant,Trader,etc). If you want to know how much pay one receives in the latter, refer to the post below by AoPS founder Richard Rusczyk in "Careers in Mathematics Announcements section."

[quote="rrusczyk"][quote="ffdbzathf"]do traders and/or quants often or even
ever have to travel?  i know that investment bankers are constantly being sent
by plain or train or however to meet with clients.  also, if it's appropriate for
this forum, whats the approximate beginning salary for a trader or quant, and
what sort of bonus and salary increases are there?[/quote]

Travel: Depends on the job.  I went to Russia, Argentina, and Brazil as part of
my trading job.  Others in my group went to Japan and London.

Salary: I'll defer to Sandor on starting salary, but I'll note that if expected
value of year five-ten salary is what's most important to you, you will not find
a better profession anywhere than going into trading, provided you have the
requisite skills and desire.  If you want to be a multi-billionaire, you'll have to
start your own Google or Dell, but if you'll be happy with at least a
mid-6-digit salary after being in a career for several years, then finance/trading
is a good bet.  It's no guarantee this will happen, to be sure, but if you get with
a good company and do good work, you won't be in a position to complain
about the pay.[/quote]

I think as a result of the "Global Economic Downturn" salaries will no longer be as what they used to be in the past. (at least in Investment Banking)

what about like an anesthesiologist,for example? And how much of that job relates to math?

That's a very stressful job, and it requires a lot of very hard work.  My friend's father is an anesthesiologist, and he's always working odd hours, and always on call.  Plus, you are responsible for the patient who's under-you mess up and you might have a very big lawsuit on your hands.

lol. How much does a radiologist get paid? And how much of a radiologists job involves math?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group, and suppose that $ H$ is a proper subgroup of $ G$.
Show that the number of elements of $ G$ that do not lie in any conjugate of $ H$ is at least $ \left|H\right|$.

I think you already have a solution and are just looking for alternative ways of thinking about it?

I'll describe the way I thought about it in case it is interesting.  The quick summary is that we can imagine that G is a permutation group acting on the cosets of H, and then it is just the famous Burnside lemma on the average number of fixed points of a permutation group.  However, there is a nice research article from 1992 talking about it and gives a strengthening of the bound and a converse.

First I said this is easy, everyone knows that the conjugates of H intersect in at least the identity, so one only has [G:H](|H|-1)+1 = |G| - ([G:H]-1) elements.  Oops, [G:H] could be small while |H| is large.

Ok, so lets make sure it was not a typo and try to construct a counterexample.  The nonabelian group of order 6 is a nice example, and the union of its Sylow 2-subgroups has 4 elements, and 6-4 = 2 = |H| is smaller than [G:H].  Ok, check a few more pq groups and yes this keeps working out.  But wait, if we replaced H cyclic of prime order with H cyclic of prime power order, it would radically increase |H| without affecting the structure of the union at all!

So take 4 |x 3, with normal subgroups of order 2 and 3.  The union of the 4s has size 4+2+2=8, and 12-8=4=|H|, another miracle.  Ok, that can't always work out.  Let G be a finite group, H a subgroup, and K the intersection of the conjugates of H.  Let X be the union of the conjugates of H.  Suppose x in X and k in K.  There is some g in G with x in H^g.  Since K ≤ H^g, one has that x, k, and xk are all in H^g so all in X.  In other words, X is a union of cosets of K.  Assuming the result for K=1, we get |G-X| = |K|*|(G/K) - (X/K)| = |K|*|H/K| = |H|.

Wow, so we can assume H contains no normal subgroups.  This means that the action of G on the cosets of H is faithful, so we can assume that G is a permutation group and H is the stabilizer of a point.  H^g is the stabilizer of some other point.  X is the set of permutations fixing at least one point.  G-X is the set of fixed point free permutations.  Oh, and now this is just a well-known part of orbit-stabilizer shenanigans familiar from say character theory.

At any rate, trying to find a clean reference for this version of the result (that I had learned as just part of Burnside's lemma), I found an entire paper dedicated to the problem.  It even talks about how surprised they were not to find an earlier reference.

Cameron, Peter J. and Cohen, Arjeh M. "On the number of fixed point free elements in a permutation group." Discrete Math. 106/107 (1992), 135-138.
http://www.ams.org/mathscinet-getitem?mr=1181907
http://dx.doi.org/10.1016/0012-365X(92)90540-V

It has the small advantage of tightening the bound a little and noting exactly when the bound is sharp.

At any rate, I thought you might find the journey interesting.  I suspect the travel itself was more interesting than the story, but at least the paper can serve as a nice postcard.


Just browsing thru the classroom math forum, I've noticed that many of the posts are probably just people asking for help on homework that is due the next day and they are too lazy to do it.

Many of the posts ask, "Please answer quickly", or the original person takes no part in the discussion following. I agree that this forum is a good one for [i]discussing[/i] classroom problems, but maybe there should be a new forum for people who just want homework help? 'Cause people in the classroom math forum will post full solutions for discussion. If a new forum is made, people will know not to do the whole problem for the student. In classroom math, people can post whole solutions, in a new homework help forum, people can just give, like, hints or something...

If someone wants to be given a hint, rather than a solution, they can say so. And those who do want a full solution will still be posting in a "discussion" forum instead of "homework" forum.

A better solution is to encourage people to [i]avoid[/i] answering posts that say "please answer quickly" or "help!!!!" or "due tomorrow!" (or posts that are just obviously homework problems) for 24 hours or so.  This maintains the value of the discussion while discouraging the flooding of the forum by homework cheaters.  One way you can do this is to make posts of the form,

"This forum is not a homework-help forum.  Please show what steps you've attempted so far.  Other users, please restrict yourselves to hints for the time being."

Make posts like that every so often and eventually other members of the forum will catch on.

Ok thanks, but i just didn't want to spend all those posts trying to get poeple to do just that. i didn't know what the mods would do.

idk...maybe some people could still use a homework help forum :?:

A homework help forum would dramatically rise in post count. People who are lazy, just don't know, or are sort of confused will gain and learn nothing if even given the slight hint. I use resources such as my mathbook (which is online, also), dictionary (for definition of math vocabulary), and my brother (when all else fails). If you are struggling with something, get off the computer and forum and then work it out with your math book or even a friend who does not also understand it. Plus, many people would be strained to add to their post:

-"Please reply quickly!"
-"Help plzzzzzzzzzzzzzzz!"
-"I NEED HELP!!! AHH!"

I have seen this occur in the Classroom Math forum.


I wish to know whether there is any procedure to recover my Password.I had registered last year but since i didn't log on for a week or so, I forgot my password.luckily , I had clicked  ([b]LOG ME ON AUTOMATICALLY ON NEXT VISIT[/b]) after my first visit. But now i wish to change my password and email id which i am not able to do unless i enter my password.

Please help me.

I'm against recovering of passwords if there is not very much evidence that it is the same person. Just imagine someone who forgot to log out and some guy wanting to hijack the account...

What happended to your E-mail¿

No need for it anymore I have finallybeen able to recall my password from my lost memory......

Some people use fake emails, so maybe verification links may work?


logarithms
analytic geometry
limit
trigonometry
graphing lines
slope
calculus

(1) Prove that $ \sqrt{2x}>\ln x$ for $ x>0.$

(2) Let $ A,\ B,\ P$ be the points on $ y\equal{}\ln x$ and the $ x$ coordinates be $ 1,\ 2,\ t$ respectively.Find $ \lim_{t\to\infty}\cos \angle{BAP}.$

($ 1$): We define $ f(x)\equal{}\sqrt{2x}\minus{}\ln{x}$.

$ f^{\prime}(x)\equal{}\frac{\sqrt{x}\minus{}\sqrt{2}}{\sqrt{2}x}$

$ \Rightarrow f^{\prime}(x)\equal{}0 \Leftrightarrow x\equal{}2$.

So there is just one local extreme. If we show, that $ f(2)>0$ and that there is a minimum of $ f$ at $ x\equal{}2$, we are done.

$ f(2)\equal{}2\minus{}\ln{2}>2\minus{}\ln{e^2}\equal{}0$ and

$ \lim_{x \to 0}{f(x)}\equal{}\infty$ (Since $ 0\minus{}(\minus{}\infty)\equal{}\infty$.)
$ \lim_{x \to \infty}{f(x)}\equal{}\infty$ (Since $ \lim_{x \to \infty}(\frac{\ln{x}}{\sqrt{x}})\equal{}0 \Rightarrow \ln{x}$ grows much slower than $ \sqrt{2x}$.)

($ 2:$) The slope of $ AP$ is $ \lim_{t \to \infty}{\frac{\ln{t}\minus{}\ln{1}}{t\minus{}1}}\equal{}0$. Hence $ \angle{BAP}\equal{}\angle{BAx\minus{}\mathrm{axis}}\equal{}{\frac{\ln{2}\minus{}\ln{1}}{2\minus{}1}}\equal{}\ln{2}$.

So $ \lim_{t\to\infty}\cos\angle{BAP} \equal{} \cos{\ln{2}}.$

Your answer to (1) is correct, but to (2), is incorrect.

[url=http://img11.imageshack.us/my.php?image=imageyrg.png][img]http://img11.imageshack.us/img11/6379/imageyrg.th.png[/img][/url]

From the image:

$ \tan\alpha(t) \equal{} \frac {\ln t \minus{} \ln 2}{t \minus{} 2}$

So:

$ \lim_{t\to\infty}\tan \alpha(t) \equal{} \lim_{t\to\infty}\frac {\ln t \minus{} \ln 2}{t \minus{} 2} \equal{} 0 \Rightarrow \alpha(\infty) \equal{} 0$

And since $ \tan \beta \equal{} \frac {1}{\ln2}$ we have that $ \angle{BAP} \equal{} \frac {\pi}{2} \plus{} \tan^{ \minus{} 1} \frac {1}{\ln2}$. We are looking for $ \cos \angle{BAP}$, so we have:

$ \cos\left(\frac {\pi}{2} \plus{} \tan^{ \minus{} 1} \frac {1}{\ln2}\right) \equal{} \minus{} \sin \left(\tan^{ \minus{} 1} \frac {1}{\ln2}\right)$

Since on the interval $ \left[ \minus{} \frac {\pi}{2} , \frac {\pi}{2}\right]$ we have $ \sin x \equal{} \frac {\tan x}{\sqrt {1 \plus{} \tan^2 x}}$ we can calculate the above as:

$ \minus{} \sin \left(\tan^{ \minus{} 1} \frac {1}{\ln2}\right) \equal{} \minus{} \frac {1}{\ln2}\frac {1}{\sqrt {1 \plus{} \left(\frac {1}{\ln2}\right)^2}} \equal{} \minus{} \frac {1}{\sqrt {1 \plus{} \ln^2 2}}$

$ \lim_{t\rightarrow \infty }\cos BAP\equal{}\lim_{t\rightarrow \infty }\frac{(1\minus{}t)^{2}\plus{}(\ln t)^{2}\plus{}1\plus{}(\ln 2)^{2}\minus{}(2\minus{}t)^{2}\minus{}(\ln \frac{t}{2})^{2}}{2\sqrt{\left( (1\minus{}t)^{2}\plus{}(\ln t)^{2}\right) \left( 1\plus{}(\ln 2)^{2}\right) }}\equal{}\allowbreak \frac{1}{\sqrt{1\plus{}(\ln 2)^{2}}}$

IndoChina, I believe you are missing a minus sign in front of your solution, since $ \angle{BAP} > \frac{\pi}{2}$, so $ \cos \angle{BAP} < 0$.

[quote="IndoChina"]$ \lim_{t\rightarrow \infty }\cos BAP \equal{} \lim_{t\rightarrow \infty }\frac {(1 \minus{} t)^{2} \plus{} (\ln t)^{2} \plus{} 1 \plus{} (\ln 2)^{2} \minus{} (2 \minus{} t)^{2} \minus{} (\ln \frac {t}{2})^{2}}{2\sqrt {\left( (1 \minus{} t)^{2} \plus{} (\ln t)^{2}\right) \left( 1 \plus{} (\ln 2)^{2}\right) }} \equal{} \allowbreak \frac {1}{\sqrt {1 \plus{} (\ln 2)^{2}}}$[/quote]

How did you get the result?

Simply by cosine rule...

$ \cos BAP \equal{} \frac {AB^{2} \plus{} BP^{2} \minus{} AP^{2}}{2(AB)(BP)}$

I see as I expected, after that how did you find the limit value?

Usual technique.. multiply numerator and denominator by 1/t^2 .. then bring t to inf :)


Έστω τρίγωνο $ ABC$ και $ M$ το μέσον της $ BC$. Εαν $ D$ ένα σημείο της $ AB$ για το οποίο η $ CD$ τέμνει την $ AM$ στο $ E$ και ισχύει $ DE\equal{}AD$, να αποδείξετε ότι $ CE\equal{}AB$.

[hide]η ΑΕΜ ειναι διατεμνουσα του τριγωνου ΒΓΔ,
αρα απο θεωρημα του Μενελαου,
$ \frac{AD}{AB}\frac{MB}{MC}\frac{EC}{ED}\equal{}1 \leftrightarrow EC\equal{}AB$...[/hide]


geometry
angle bisector
perpendicular bisector

I was having trouble with this problem. So far I have realized that if the 25 degrees gets bigger, then so does x degrees. If the 75 degrees gets bigger, x degrees gets smaller.

I was thinking that it might have something to do with the triangle in a circle idea where the middle angle is twice the angle of the angle measured at the circumference.

I don't really know much more than that.

Thanks a lot to people who help!

See the attached diagram.

$\angle C=2\cdot 25^\circ=50^\circ$

$\angle A=180^\circ-\angle B-\angle C=55^\circ$

$\angle CAF={1\over 2}\angle A=27.5^\circ$

Now $x$ is an external angle in $\triangle CAF$, which gives $x=\angle CAF+\angle ACF=52.5^\circ$

[quote="Treething"]Actually, angle x is only dependant on what you set that angle that's currently 75 degrees to. I believe that $x = \frac{90-\theta}{2}$.[/quote]
Actually, I think it's $90-\frac{\theta}{2}$.

I don't understand how that second angle is 25 degrees. Could you elaborate some more please?

Here's my thinking on it shown in the picture.

Do you see what's written in the bottom left corner of your first diagram? "Two lines in the middle of triangle are bisectors." Now, try to recall (or learn) what does "angle bisector" mean.

Bisector

A line segment, line, or plane that divides a geometric figure into two congruent halves.

wow lol the problem is easy now. My dad and I were slaving away at this problem for hours trying to figure it out. I didn't realize that that was the definition of a bisector.

see my picture to see what i thought a bisector was.

No, but you see, each segment is half of the original segment!

.5 of AB is equal to .5 of AB.

Cutting something (anything) in half makes two parts, yes, but it's important to know that those 2 parts are equal.

if we have a number X
and X/2 = Y
Then Y essentially "cuts" X in half. We can get X/2 * 2 = Y * 2, which means X = Y*2. So "both" Y's are equal here.

An angle bisector divides an angle in half.

A perpendicular bisector divides a segment in half and is perpendicular to the segment.

A median is a segment within a triangle which connects a vertex and the midpoint of the opposite side.

cool i just did a little testing on my own and it turns out that even in a weird triangle like 89 degrees 89 degrees 2 degrees, the bisector still cuts it into equal parts. I never knew that. Well cool!!


This is something that my friend showed me and knowing me of course, asked me about it, and i didn't have an anwer. To start with, if you punch in $\sqrt{2}$ in your calculator, the answer'll take the whole screen, it being irrational. Now, punch in $32\sqrt{2}$ or $\sqrt{2048}$and the answer doesn't fill the whole screen and you get 45.254834, meaning it's rational, right? Does that make sense or am i forgetting something vital that i should have told my friend?

No, this can't be unless the rational numbers are 0. In another case you will have irrational = rational/rational = rational.

Calculators can show only finite precision approximation of irrational numbers. You should always remember that you can't always trust your calculator since there might e some rounding errors. It is always better if you can calculate problems with pencil and paper.

$32 \sqrt{2}$ won't fill the screen because the calculator will only store the first $n$ digits of $\sqrt{2}$ (usually $10 \le n \le 14$, depends on the calculator) and then multiply it by $32$, so some precision will be lost.

Irrational * Rational = Rational is impossible.  It would imply

Irrational = Rational / Rational = Rational

well, here is the thing

we all know that pi is irrational and takes up the entire screen in calculators right?

mutiply pi by about 100000 and subtract the whole numbers in front of the decimal

you'll get a limited remainder

same thing with that

Also,

We all know that $\log_2{16}=\frac{\log 16}{\log 2}=4$
And if you type that in your calculator you get the same thing.

Now take the greatest integer less than or equal to that, and you get 3.

Just a happy reminder that calculators aren't always very accurate... :lol:

[quote="t0rajir0u"]Irrational * Rational = Rational is impossible.[/quote]
It is possible, $\pi \cdot 0=0$.

If $x \in \mathbb{Q}$ and $y \not\in \mathbb{Q}$ then unless $x = 0$, $xy \not\in \mathbb{Q}$.

But calculators can get fooled by accuracy error, which happens if any irrational number has a lot of $0$ digits in a row. For example try $e^{\pi} \times 462$. The answer seems like an integer but in fact it is only an apporximation to the integer.

[quote="236factorial"][quote="Centy"]But calculators can get fooled by accuracy error, which happens if any irrational number has a lot of $0$ digits in a row. For example try $e^{\pi} \times 462$. The answer seems like an integer but in fact it is only an apporximation to the integer.[/quote]

Or on a 8-digit 4 function calculator, try $\sqrt{2499}$.

On the computer calculator, it's
49.989998999799949985995798679571

But a simple 4-function one doesn't have that many digits, so it rounds and looks rational  :D[/quote]
Or $\pi\times\frac{3501408748}{11}$. :D


search
function
number theory unsolved
number theory

Find all $ (x,y,z)\in\mathbb{N}^{3}$ such that $ 4^{x}\plus{}4^{y}\plus{}4^{z}$ is a perfect square.

If $ x\le y\le y\le z$, then x+z odd and $ x\plus{}z\plus{}1\equal{}2y$.

I'm wondering why people don't go to the bother of using the internal search function, before deciding to post any problem... Maybe too lazy? :maybe: Click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160202]here[/url], please.

I used the same idea I used to solve #4 of IMO 2006.

Suppose that $ x\le y\le z$. Then we have $ n \equal{} 4^{x}\plus{}4^{y}\plus{}4^{z}\equal{} 4^{x}(1\plus{}4^{y\minus{}x}\plus{}4^{z\minus{}x})$. Now the divisors of $ 4^{x}$ are all powers of two, while those, except $ 1$, of $ 4^{y\minus{}x}\plus{}4^{z\minus{}x}\plus{}1$ are not. Hence $ (4^{x},1\plus{}4^{y\minus{}x}\plus{}4^{z\minus{}x}) \equal{} 1$. So if $ n$ is a perfect square, then so is $ 4^{x}$ and $ 1\plus{}4^{y\minus{}x}\plus{}4^{z\minus{}x}$. Let $ 1\plus{}4^{a}\plus{}4^{b}\equal{} k^{2}$, where $ a \equal{} y\minus{}x$ and $ b \equal{} z\minus{}x$. Then verify that $ a$ and $ b$ must be positive, and so $ k$ must be odd. Letting $ k \equal{} 2q\minus{}1$, our equation becomes, $ 4^{a}(1\plus{}4^{b\minus{}a}) \equal{} 4q(q\minus{}1)$. Which is, after some work, equivalent to $ q(q\minus{}1) \equal{} 4^{a\minus{}2}(4^{b\minus{}a}\plus{}1)$. Now it's not difficult to check that if $ 0\le a\le 2$, then $ 1\plus{}4^{a}\plus{}4^{b}$ is a perfect square only for $ (a,b) \equal{} (1,1),(2,3)$ that is, $ (x,y,z) \equal{} (t,t\plus{}1,t\plus{}1),(t,t\plus{}2,t\plus{}3)$. [hide="Proof"]i. $ a \equal{} 0\Rightarrow (2^{b})^{2}\plus{}2 \equal{} k^{2}$, impossible.
ii. $ a \equal{} 1\Rightarrow (2^{b})^{2}\plus{}5 \equal{} k^{2}$. Now $ k^{2}\minus{}(2^{b})^{2}\ge k^{2}\minus{}(k\minus{}1)^{2}\equal{} 2k\minus{}1\quad(*)\Rightarrow 2k\minus{}1\le 5,k\le 3$. Checking for $ k$ it's easy to see that $ k \equal{} 3$ and thus $ (a,b) \equal{} (1,1)$.
iii. $ a \equal{} 2\Rightarrow (2^{b})^{2}\plus{}17 \equal{} k^{2}$. Using $ (*)$, we get $ k\le 9$, also $ k$ is odd and $ k\ge 5$. Checking for $ k \equal{} 5,7,9$ gives $ k \equal{} 9,b \equal{} 3$. Thus $ (a,b) \equal{} (2,3)$. [/hide]
So now we can assume that $ a\ge 2$ and consider the solutions of $ q(q\minus{}1) \equal{} 4^{a\minus{}2}(4^{b\minus{}a}\plus{}1)$. It's not difficult to see that either $ 4^{a\minus{}2}\le q\minus{}1\le q\le 4^{b\minus{}a}\plus{}1$ or $ 4^{b\minus{}a}\plus{}1\le q\minus{}1\le q\le 4^{a\minus{}2}$.

Case 1: $ 4^{a\minus{}2}\le q\minus{}1\le q\le 4^{b\minus{}a}\plus{}1$. As only one of $ q$ and $ q\minus{}1$ is even, $ 4^{a\minus{}2}\mid q$ or $ q\minus{}1$.

Sub-case 1: $ q \equal{} 4^{a\minus{}2}m$. Then our equation becomes $ m(4^{a\minus{}2}m\minus{}1) \equal{} 4^{b\minus{}a}\plus{}1$.

I will continue later as I have exams tomorrow, sorry. :blush:


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

$ h'(x) \equal{} \frac{d}{dx}(f(x))^6) \equal{} 6 (f(x))^5 \cdot f'(x).$

So now you can get $ f(3)$ and $ f'(3)$ from the picture given.

[quote="ben_b"]I don't understand how you get $ f(3)$ and $ f'(3)$ from the picture above?[/quote]
f'(x)=rate of change of f=slope of tangent atx=3 to f(x).
btw Ben_b,which class are you in/ :huh:


I get six, with Gene getting a 2, 3, 2, 3, 3, and a 2 before getting a 4.  As for a general solution, i have not found one yet.

i also got 6 and i dont think there's a general formula... i tried varying the number of people in the circle and the total number of times gene said the numbers was pretty weird (not always increasing)... maybe someone better can find something  :idea:

Can someone tell me best way to solve and how you arrive at 6 instead of 5? remember there are 9 kids in circle - so count off should be 1 2 (gene) 3 4 1 2 3 4 and ninth kid is 1 before going back to John...


function
combinatorics unsolved
combinatorics

Find the number of non-decreasing functions $f:\{1..n\} \longrightarrow \{1..n\}$ such that for every $k=1,2,...,n$ we have:
$f(k)\le k$
I know that this number is equal to $n^{th}$ [i]Catalan Number [/i], but i don't have a beautiful, or a natural proof.

Yes! I have invented it.

Consider a sequence with $2n$ terms of '0' and '1'. Place '1' on the places $f(k)+k-1$ and '0' on the other places. Then for each $m\in\{1,2,...,2n\}$ the number of units within first $m$ terms is not less than the number of zeros (also you can interpret '1' as opening bracket and '0' as closing bracket).
And vice-versa. From any proper sequence you can construct a corresponding function $f$.

[quote="pleurestique"]yep; exactly what i was gonna say.  and then to find the number of such 0-1 sequences... a method was outlined in
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33009[/quote]I don't think it was a source for me.


Assuming we have a southern site again next year, I think we should try and organize something specific to our site (inspired from the Penn State song contest and the other things at other sites which I don't what they are =p).  Any ideas?

Also, it'd be nice to find a better field to play ultimate on =p

Amazing, 4+ hours and Billy hasn't suggested a song contest yet!  :D

I think we should have a song contest. I'll volunteer to judge, if so desired :).

If judging doesn't involve singing, then you judging would be desirable.  :wink:


The circular region of the sign (below, left) has an area of 154 square inches. Vanessa would like to place a tiny ribbon (shaded) around the circle's edge. To be sure she has enough ribbon, she decides to buy 2 inches more of the ribbon than the original circle's circumference. How many inches of ribbon will Vanessa need to buy if she estimates $ \pi \equal{} \frac{22}{7}$?

[asy]import graph;
size(125,72.5);
picture p;
draw(p,unitsquare);
filldraw(p,Circle((.5,.5),.3),white);
label(p,"Enter",(.5,.5),ZapfChancery("m","n"));
add(p);
filldraw(Circle((2,.5),.4),gray(.6));
add(shift(1.5*right)*p);
draw((1.1,.5)--(1.4,.5),EndArrow(5,25));[/asy]

$ \pi r^2$=154.

$ \pi \equal{}\frac{22}{7}$.

$ \frac{22}{7}*r^2\equal{}154.$

$ r^2\equal{}49, r\equal{}7.$

original circle's circumference : $ 7*2*\frac{22}{7}\equal{}44$

2 inches more : 46

answer : 46


algebra
polynomial
number theory proposed
number theory

Find all integers $ m>1$ satisfying the following condition: There is a polynomial $ F(x)$ with integer coefficients such that for any integer $ a$ either $ F(a)\equiv 0(modm)$ or $ F(a)\equiv 1(modm)$ and each of these two cases is satisfied for at least one integer $ a$.

if $ m$ prime we take $ F(X) \equal{} \minus{} \prod_{k \equal{} 1}^{m \minus{} 1}(k \minus{} X)$


If [i]p[/i] and [i]q[/i] are positive integers, then

(5^-p)(5^q+1)^P is equivalent to:

A) 5
B) 5^pq+p
C) 5^pq
D) 5^pq-p
E) 5^q+1

Okay. First you simplify the (5^-p).
That would simplify to (1/5^p).
Then, we see that the (5^q+1) can't be simplified further.
So we multiply the (1/5^p) and the (5^q+1) to get 5^q+1/5^p

But how do I do the final "^P" step?

Please help. I take the test in January! :help:

Just remember that $(5^{a})(5^{b})=5^{a+b}$ and $(5^{a})^{b}= 5^{ab}$.

[hide="Hint"]
Leave $-p$ as it is.
[/hide]

(5^-p)(5^q+1)^P is equivalent to:

A) 5
B) 5^pq+p
C) 5^pq
D) 5^pq-p
E) 5^q+1

the SAT does not give problems that take a lot of stepts to solve.
i scored all perfect scores on all 3 tests remembering this
(it can't be complicated as the AMC)

just get the exponents

$5^{-p}(5^{q+1})^{p}$
$=5^{pq+p}* 5^{-p}$
$=5^{pq+p+(-p}$
$=5^{pq}$

to check just subsitute the numbers to p and q


Find all pairs $ (x,y)$ of integers such that:
\[ xy+\frac{x^{3}+y^{3}}{3}=2007\]


HI for all! I'M A NEW USER AND NEED SOME HELPS! CAN YOU HELP ME?[/u][/hide][/hide][/list] :)


wht do u mean total radians..like a circle has $360^\circ\implies 2\pi$ (in radian measure)

so the conversion basically is

$\left(\frac{A}{180}\right)\cdot\pi$, where $A$ is an angle in degrees..

[quote="rickyc1"]If I have a circle with 12'' diameter what is the total radian?  What is the formula?[/quote]
Not to be dumb or anything but if you have a whole circle than the answer is just 2pi no matter what the diameter...at least i was taught that :maybe:

ya thats why i asked what he meant by "total radian"


trigonometry AMC AIME USA(J)MO USAMO ARML Duke

Hi. I recently moved here from NC last summer, and I'm attending a public high school in Fairfax County. I just heard about the "Math League" (and VAML), and I'm thinking about participating in them. How hard are they? They seem to be very easy (even compared to AMC 10.) Are they harder, and what fields of math are covered in the high school level? (combinatorics, number theory, trigonometry, etc)

i did vaml last year. If i remember correctly it's split into 7th grade and 8th grade tests.

the problems are probably around amc 8 level, for both 7th and 8th grade.

topics i think were basic algebra and geometry.

im sure if u have qualified for AIME that this test will be a breeze for u

i did take the first test, and apparently it seemed extremely easy (while surprisingly, almost everyone else had to struggle with the problems.) i remember there was just one problem that tests the understanding of an identity of pascal's triangle, and the rest were school algebra 1+2 level.

i guess i'll just stick to the prevalent national competitions like amc, usamo, and arml. :wink:

Hey remy1140, how are you doing in Virginia?  What high school competitions do you have there besides Math League, AMC-AIME-USAMO, and ARML?  There's a few back in NC (Duke Math Meet at least).

Also, are you taking 10B or 12B?  (I know, your A got cancelled  :P )


I've reflected on my last post, and I see why the moderator deleted it.  I'm not mad or anything at the moderator; but I do have a question.
Why was unimpossible's post not deleted?
I don't really see how his post is considered "relevant" to the "Math Awards" topic if all it contains is a provocation for me to fight with him (composed of insults, even though I didn't insult him before that).
Just for those of you who don't know what my post contained, it responded to unimpossible's sarcasm (he said he got a 19 on AMC 8 and thinks he can make it to IMO, supposedly) and hurled back some insults, but told him not to take offense (or if he did take insult, then he'll understand how I felt when he started the argument by insulting me).
His post just remains on that topic, mocking me, and being irrelevant.

And hopefully, this post won't be considered irrelevant to the topic because this is the topic.
And please, if anyone else besides the Indiana mod wants to comment, don't try to pick a fight with me about the argument at the "Math Awards" topic; that is locked for a reason.  All I have is a question about it; I do not wish to continue the argument.

Please answer, mod.

I think the mod's undefined now cause he told us to get back on topic... unimpossible's post could've been taken seriously, but we know that he didn't get a 19 on the amc 8, nor a 100 on the amc 10. If his is deleted, probably half of the other posts should be deleted also. and i think sooey is biased cause he likes yuqing  :love: lol. if the mod isn't undefined, uh oh. time for me to go hide xD

I saw no provocations of any kind nor any insults in unimpossible's post. I think what he was trying to say was that anything is "achievable," while actually turning a certain goal into reality is a challenge. Not necessarily insulting, I'd say. :|

mathcrazed, I want to murder that smiley =P

Well, anyways, wherever unimpossible is, if he read my deleted post, I hope he doesn't feel animosity, because now, looking back at the argument, it was pointless anyways.

But, um, I don't think unimpossible was being particularly encouraging in his post, and even if you think it's somehow not provoking, it's still unrelated.
But as mathcrazed said, well, so many other posts on the Indiana forum are unrelated.  I guess mine was the unfortunate one to be deleted.

Yeah, I'll just forget about the Math Awards topic and not post in it anymore.  I wouldn't have posted my last post in it if my PM system [i]actually worked[/i], because then I could just talk to unimpossible with a PM rather than spamming up that topic.

But, yes, I talked about my messed up PM system in the Site's Announcements/Questions/Etc. Thread, so hopefully people can tell me how to fix it.

To undefined117: I received your PM, but obviously I can't PM you back since it won't work.  My original PM to you was a question asking if you are the Indiana Moderator, but apparently you never received it.  But I'm pretty sure that you are the mod by now.

[quote="MathAndKnowledge"][quote="logic812"]So the question is, is undefined a mod (even though he's not in IN)?[/quote]
Well, that wasn't the initial question.  The initial question was why unimpossible's post wasn't deleted, but I don't really care about that anymore.

So this topic should now die (unless unimpossible has something to say).

But as an aside, since I lack the ability to send PMs, I am forced to "PM" undefined117 on this forum in order to ask him if he's the mod.  Of course, he can PM a reply (since I am able to receive his PMs, but not yours, logic812).[/quote]
I fail to see your situation.
First of all, why can you not PM people?
Second of all, if there were to be a problem with your PM box in general, it would probably disable all functions that has to do with private messaging, not just the composing of PMs.
Third, the fact that you know I sent you a PM maked me believe that you "segregate" you PMs, which doesn't make a 100% sense to me.
Yes, this thread should die, but just for the sake of it, let's hope that undefined tells us whether he's the mod.

er.. I thought it was pretty obvious by now that I am the mod, especially with my previous reply above.

MathAnd, you shouldn't have taken my statement that I had removed all "irrelevant" posts so literally ;) If I really did that, it'd be like, authoritarian or something. We don't want that - the point here is for everybody to be mature enough to moderate their own thoughts to provide quality posts, and everybody's been doing a very decent job of it already anyway. :) It's just that one topic that was beginning to get out of hand.

Also MathAnd, if you want to PM me, you can just post in my blog shoutbox for the time being until your PM function gets fixed.

Now if anyone has anything math-related to discuss, feel free.. *cough* in another topic.


At a party recently Chunlin and some of his friends sat in a circle and passed around a bag containing 19 hard pretzels. Each took one pretzel as the bag passed. Chunlin got the first pretzel and the next to last pretzel. How many people were in the circle, including Chunlin?

If Chunlin got the first pretzel and the next to last one, he got the 1st and 18th ones.

The pretzel immediately before the 18th was given to the last person. Thus, there were $ \boxed{17}$ people.


Hello, I was wondering if a player could run out of problems to do on Alcumus. Since there's a limited amount, does Alcumus just stop giving you problems? I fso, can you still eatch videos?

It will feed you old problems that you have missed.  And you'll still be able to access the videos.

I have $ 8$ problems missed still, and it says "no more problems available"...

The 8 problems you still have marked as wrong are no longer active within Alcumus.

so.... what does happen when you've run out of problems and you don't have any old ones you've missed
(not that that's happened to me yet!!! Just wondering!!!)  :wink:

then it'll probably say that there are no more problems when you click the get problem tag


How can I make a multiple column?

For example

$ 1) a+b\quad\qquad 2)x+y\qquad3)x-y\\[3mm]4) a+b\quad\qquad 5)x+y\qquad6)x+z$

[quote="Healker"]How can I make a multiple column?

For example

$ 1) a \plus{} b\quad\qquad 2)x \plus{} y\qquad3)x \minus{} y \\
[3mm]4) a \plus{} b\quad\qquad 5)x \plus{} y\qquad6)x \plus{} z$[/quote]

The multicols package will help you here.

[code]
\usepackage{multicols}

\begin{document}
\begin{multicols}{2}
Hello!
\columnbreak

Column 2
\columnbreak

Columnn 3
\end{multicols}
\end{document}
[/code]

Unfortunately the numbering will be advancing vertically rather than horizonally. For enumerated lists that advance horizontally, either the multenum package or use minipages like this:

[code]
\begin{enumerate}
\begin{minipage}[t]{0.45\linewidth}
\item Stuff
\end{minipage} \hspace{1em}
\begin{minipage}[t]{0.45\linewidth}
\item More Stuff
\end{minipage}
\end{enumerate}
[/code]

changing 0.45\linewidth to 0.3\linewidth when using 3 columns etc.

Okay thanks but, something gone wrong 'cause it gives the following error:

TeX capacity exceeded, sorry [save size=5000].

(Ahhh, I see you edited, I'll see what happens now.)

[quote="Healker"]How can I make a multiple column?

For example

$ 1) a + b\quad\qquad 2)x + y\qquad3)x - y \\
[3mm]4) a + b\quad\qquad 5)x + y\qquad6)x + z$[/quote]This is best done using the align* environment.
[code]\begin{align*}
1)\ a+b && 2)\ x+y && 3)\ x-y \\
4)\ a+b && 5)\ x+y && 6)\ x+z
\end{align*}[/code] which gives
\begin{align*} 1)\ a + b & & 2)\ x + y & & 3)\ x - y \\
4)\ a + b & & 5)\ x + y & & 6)\ x + z \end{align*}

You can also automate the question numbering with
[code]\newcounter{question}
\newcommand{\qu}{\stepcounter{question}\thequestion )\ }
\begin{align*}
\qu a+b && \qu x+y && \qu x-y \\
\qu a+b && \qu x+y && \qu x+z
\end{align*}[/code]so every time you want a new question number you just type \qu.

Thanks steve, but, if I want an image instead \qu? I'm currently using

\psshadowbox[linecolor={},framearc=0.5,linewidth=0.7pt,fillstyle=gradient,
gradbegin={}, gradend={},framesep=5pt,
shadowcolor={}, gradmidpoint=1]{\mathcal P_1}

which generates the image I want, then, instead by putting a number, how can I insert such image?

This document should give what you want.
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{pstricks}
\usepackage{pst-grad}
\begin{document}
\newcounter{question}
\newcommand{\qu}{%
\psshadowbox[framearc=0.5,linewidth=0.7pt,fillstyle=gradient,framesep=5pt,gradmidpoint=1]{\stepcounter{question}\mathcal{P}_{\thequestion}}\ }
\begin{align*}
\qu a+b && \qu x+y && \qu x-y \\
\qu a+b && \qu x+y && \qu x+z
\end{align*}
\end{document}[/code]
I have removed some of the blank options like linecolor={} as they are errors.

Okay, one more thing: I'll be putting in first page the problems that will be solved below. (New page.) For example, if I click on P1 (which is the image I put in last post), it should direct me into that problem.

I've seen that this can be created with \hyperlink and \hypertarget, but I dunno how to make it.

You need the hyperref package which has a huge number of options. I strongly recommend that you read the [url=http://www.tug.org/applications/hyperref/manual.html]manual[/url].

Meanwhile here is a basic version for you to study. I have shortened \psshadowbox to \psbox. and have changed \qu.
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{pstricks}
\usepackage{pst-grad}
\usepackage{hyperref}
\begin{document}

\newcommand{\psbox}[1]
{\psshadowbox[framearc=0.5,linewidth=0.7pt,fillstyle=gradient,framesep=5pt,gradmidpoint=1]{\ensuremath{#1}}}

\newcounter{question}
\newcommand{\qu}{\psbox{\stepcounter{question}\mathcal{P}_{\thequestion}}\ }
%links for first page
\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}
\newpage

%reset counter for list of questions
\setcounter{question}{0}
\newcommand{\rqu}{\hypertarget{P\thequestion}{\qu}}

\begin{align*}
\rqu a+b && \rqu x+y && \rqu x-y \\
\rqu a+b && \rqu x+y && \rqu x+z
\end{align*}
\end{document}[/code]

About blank link, it's not working, I have:

[code]\newcounter{question}
\newcommand{\qu}{\psbox{\stepcounter{question}\mathcal{P}_{\thequestion}}\ }

\setcounter{question}{0}
\newcommand{\rqu}{\hyperlink{P\thequestion}{\qu}}
\renewcommand{\minalignsep}{50pt}
\newcommand{\target}{\hypertarget{P\thequestion}{}\stepcounter{question}}
\newcommand{\repeattarget}{\target \target \target \addtocounter{question}{-3}}
\repeattarget
\begin{align*}
&\rqu \int\frac{dx}{x^2-1}.  &&\rqu \int\frac{x+\cos x}{1+\os x}\,dx.  &&\rqu \int \sec^3 x\,dx.\repeattarget\\[4mm]
&\rqu \int\os^2 x\,dx.  &&\rqu \int e^{\alpha x}\os(\beta x)\,dx.  &&\rqu \int\frac{dx}{1+x^2}.\repeattarget\\[4mm]
&\rqu \int\frac{x}{1+x^2}\,dx.  &&\rqu \int\frac{x^2}{1+x^2}\,dx. &&\rqu \int\frac{x^3}{1+x^2}\,dx.
\end{align*}[/code]

and it's not giving the expected result. :(

[quote="Healker"]Okay well, the second you posted is more or less what I want.

What I want is:

[img]http://img140.imageshack.us/img140/1843/xdghkxgss1.png[/img]

Those circles are replaced by another image, so, that doesn't matter.[/quote]

Use the multicols package as described? i.e. \begin{multicols}{3} \end{multicols}

[quote="Healker"]About blank link, it's not working, I have:
...
and it's not giving the expected result. :([/quote]
What do you get instead? It works fine for me in this document:
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{pstricks}
\usepackage{pst-grad}
\usepackage[bookmarks=false,pdfstartview={FitH}]{hyperref}
\newcommand{\os}{\operatorname{sen}}
\begin{document}

\newcommand{\psbox}[1]
{\psshadowbox[framearc=0.5,linewidth=0.7pt,fillstyle=gradient,framesep=5pt,gradmidpoint=1]{\ensuremath{#1}}}

\newcounter{question}
\newcommand{\qu}{\psbox{\stepcounter{question}\mathcal{P}_{\thequestion}}\ }
%links for first page
\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}

\hyperlink{P\thequestion}{\qu}
%\newpage

%reset counter for list of questions
\setcounter{question}{0}
\newcommand{\rqu}{\hyperlink{P\thequestion}{\qu}}
\renewcommand{\minalignsep}{50pt}
\newcommand{\target}{\hypertarget{P\thequestion}{}\stepcounter{question}}
\newcommand{\repeattarget}{\target \target \target \addtocounter{question}{-3}}
\repeattarget
\begin{align*}
&\rqu \int\frac{dx}{x^2-1}.  &&\rqu \int\frac{x+\cos x}{1+\os x}\,dx.  &&\rqu \int \sec^3 x\,dx.\repeattarget\\[4mm]
&\rqu \int\os^2 x\,dx.  &&\rqu \int e^{\alpha x}\os(\beta x)\,dx.  &&\rqu \int\frac{dx}{1+x^2}.\repeattarget\\[4mm]
&\rqu \int\frac{x}{1+x^2}\,dx.  &&\rqu \int\frac{x^2}{1+x^2}\,dx. &&\rqu \int\frac{x^3}{1+x^2}\,dx.
\end{align*}
\vspace{5in}\\
More text
\end{document}[/code]
Perhaps you need to post/PM more of your document if something else is preventing it from working.


This is an easy problem!
In a tourism area, there are 4 people that 2 of them tell the truth, and the others tell lie. They said that:
Person 1st: "At least, 1 of us tell lie"
Person 2nd: "At least, 2 of us tell lie"
Person 3rd: "At least, 3 of us tell lie"
Person 4th: "Nobody in us tell lie"
Find out who tell the truth, and who tell lie.
Good luck!

Isn't it obvious that person $1$ and $2$ are telling the truth, and the other two lying?

This problem is a special case of a Canadian problem with $n=4$ and without the given number of liars, JBL is right.

[quote="10000th User"]This problem is a special case of a Canadian problem with $n=4$[/quote]

So, given persons $1, 2, 3, ... n$, persons $i = 1, 2, 3, ... n-1$ say that at least $i$ people are lying and person $n$ says that nobody is lying.  And we are to deduce which of them are telling the truth and which are lying?

[quote="JBL"]The more interesting fact to note is that you don't need to be given the number of liars in order to answer this question.[/quote]
JBL's right.  You can start with Person 4's statement, and the rest is clear.


geometry
analytic geometry
quadratics
Pythagorean Theorem
linear algebra

[quote="dgreenb801"]I found this really cool website with 75 proofs of the Pythagoren theorem:  [url]http://www.cut-the-knot.org/pythagoras/index.shtml[/url] . I was wondering, is there any other theorem that has any many proof of it like that?[/quote]

 There is a book which contains more that 300 proofs of Pyth. theorem (by Elisha Loomis, I think).About your question , I think that the answer is negative!

I think that's because the Pythagorean theorem is so well-known

I think it's because it's a geometry problem with just the right complexity. If it was too simple, then there would be an obvious proof, but if it was too complex, there would only be one proof. Also, geometry problems have multiple solutions more often than other areas of math.

[quote="MellowMelon"][quote="dgreenb801"]Also, geometry problems have multiple solutions more often than other areas of math.[/quote]
I have a feeling that this is because if you reduce each of those 75+ proofs to equations, many would be the same proofs with very minor variations, and in other topics they are generally not considered distinct proofs.[/quote]

Even if that is the case, the elegance of Euclidean Geometry lies in the fact that it is not so algebraically oriented (like coordinate or other higher geometry).

P.S. http://physicsworld.com/cws/article/print/23892

This is probably disputed, but a lot of sources call Quadratic Reciprocity the most proved theorem ever. And unlike the Pythagorean Theorem, this really is nontrivial.

By the way, Euclidean geometry is very much algebraic. It seems you have a wrong expression of what "algebraic geometry" really is, it's not that dumb coordinate calculating stuff from linear algebra and can be rather interesting.


Find all functions $ f : R\rightarrow R$ that satisfy $ f ((x\minus{}y)^2) \equal{} f (x)^2 \minus{}2x f (y)\plus{}y^2$.

[quote="aviateurpilot"]$ (f(x) \minus{} x)^2 \equal{} f(x)^2 \minus{} 2xf(x) \plus{} x^2 \equal{} f(0)$ and $ f(0)^2 \equal{} f(0)$ gives $ f(0) \equal{} 0\ or 1$
if $ f(0) \equal{} 0$ then $ \forall x\in R: \ f(x) \equal{} x$
if $ f(0) \equal{} 1$ then $ \forall x\in R: \ f(x) \equal{} x \plus{} 1\ or x \minus{} 1$
and $ y \equal{}$ gives $ \forall x\in R \minus{} \{ \minus{} 1/2,0\}: \ f(x^2) \equal{} f(x)^2 \minus{} 2x$
if $ f(x) \equal{} x \minus{} 1$ we take $ f(x^2) \equal{} x^2 \minus{} 4x \plus{} 1\not\in\{x^2\pm 1\}$
so $ \forall x\in R \minus{} \{ \minus{} 1/2,0\}: \ f(x) \equal{} x \plus{} 1\Rightarrow \forall x\in R \minus{} \{ \minus{} 1/2\}: \ f(x) \equal{} x \plus{} 1$
and $ 5/4 \equal{} f(1/4) \equal{} f( \minus{} 1/2)^2 \plus{} 1$ gives $ f( \minus{} 1/2)^2 \equal{} 1/4$ then $ f( \minus{} 1/2)\neq ( \minus{} 1/2) \minus{} 1$

$ \forall x\in R: \ f(x) \equal{} x \plus{} 1$[/quote]
VERY GOOD
THNAKE YOU VERY MUCH


Let $ F(x)\equal{}x^2000\minus{}x^1000\plus{}1$.Does there exist $ a_1,a_2,...,a_8002 \in Z_\plus{}$ which are pairwise distinct such that
$ F(a_i)F(a_j)F(a_k)$ is divisible by $ a_ia_ja_k$ for all $ 0<i<j<k<8003$

easy.
$ a_1\equal{}1$
$ a_n\equal{}F(a_{n\minus{}1})$


Problema pe care v-o propun spre rezolvare este urmatoarea:
"Fie ABC un triunghi oarecare, iar A1, B1 şi C1 simetricele vârfurilor sale A, B si C, faţă de dreptele support ale laturilor sale opuse. Notăm cu A2, B2 şi cu C2 punctele de intersecţie ale dreptelor BC1 şi CB1, AC1 şi CA1, respective AB1 şi BA1. Să se arate că dreptele A1A2, B1B2 şi C1C2 sunt paralele." (A. Zaslavski)

Iata si analoaga ei spatiala gasita de catre mine:
"Fie ABCD un tetraedru ortocentric(!). Notam cu A1, B1, C1 si cu D1 simetricele varfurilor A, B, C si respectiv D fata de planele fetelor opuse acestor varfuri; iar cu A2 punctul de intersectie al planelor (B1CD), (C1BD) si (D1BC); in mod analog se definesc punctele B2, C2 si D2. Sa se arate ca: A1A2||B1B2||C1C2||D1D2."

Intentia mea initiala, era sa va postez pe FORUM si 2 solutii oficiale ale problemei + o solutie a mea mai simpla decat fiecare dintre cele 2 solutii oficiale, solutie a carei analoaga, pas cu pas, este solutia analoagei spatiale. Nu am putut duce intentia mea initiala la bun sfarsit, fiind blocat, intrucat ATTACHMENTELE mele de pana acum+ noul attachment ar fi depasit astfel 10MB. Aceasta solutie (pe care n-o pot posta acum), contine o noua solutie sintetica a problemei 1121 din Kvant, mai simpla decat cea solutia mea sintetica de pe FORUM!

ASA CA,  DEOCAMDATA, NU-MI RAMANE DECAT SA VA UREZ SUCCES IN REZOLVAREA ACESTOR 2 NOI PROBLEME!

Pentru problema in plan, iata rezolvarea pe care am gasit-o:
Din faptul ca pct. $ A, A_{1}$ respectiv $ B, B_{1}$, resp. $ C,C_{1}$ sunt simetrice fata de laturi, daca notam laturile cu $ a, b, c$
avem ca $ BC_{1}\equal{}BC\equal{}CB_{1}\equal{}a$, $ AC_{1}\equal{}AC\equal{}CA_{1}\equal{}b$, $ AB_{1}\equal{}AB\equal{}BA_{1}\equal{}c$. Avem deci o congruenta de triunghiuri $ A_{1}BC, ACB_{1}, ABC_{1}, ABC$. De aici, $ \angle{C_{1}AB}\equal{}\angle{BAC}\equal{}\angle{ACB_{1}}$ si analog...
Daca ne uitam la unghiurile triunghiurilor $ BAC_{2}$, $ ACB_{2}$, $ BCA_{2}$ avem de exemplu $ \angle{BC_{2}A}\equal{}180\minus{}2\angle{ACB}$ s.a.
De aici triunghiurile de mai sus sunt asemenea, avand, unghiuri egale.
Ele au si laturile proportionale... Din asemanarea primelor doua obtinem ca: $ \frac{AC_{2}}{AC}\equal{}\frac{AB}{AB_{2}}$
$ \rightarrow$ $ AC_{2}\cdot AB_{2}\equal{} bc$. Cum am vazut ca $ AC_{1}\equal{}b, AB_{1}\equal{}c$ $ \rightarrow$
$ \frac{AC_{2}}{AB_{1}}\equal{}\frac{AC_{1}}{AB_{2}}$ $ \rightarrow$ $ B_{1}B_{2}\parallel C_{1}C_{2}$ si analoagele...

Ma voi uita mai tarziu si peste var. in spatiu...


I'm looking for an introductory number theory book.  Can any of you make any reccomendations?  "Number Theory" By George Andrews looks pretty good but I here it's kinda focused on partitions.  I've heard LeVeque's books are sort of unreadable.  Thanks

Buy Andrews.  I have it, and don't particularly like it, but it's $10 and has a lot of exercises, partial solutions, and his nonstandard combinatorial proofs will be quickly forgotten when you develop a sufficient amount of mathematical knowledge.

If you were at a higher level, Niven's book would be a good choice.


trigonometry
calculus
calculus computations

Solve $ \frac{d^2}{dt^2}\left(
\begin{array}{cc}
x_1 &  \\
x_2 &
\end{array}
\right)=\left(
\begin{array}{cc}
-a^2-b & b  \\
b & -a^2-b
\end{array}
\right)\left(
\begin{array}{c}
x_1 &  \\
x_2
\end{array}
\right)$

hello, we have
$ x_1(t)\equal{}\minus{}C_3\sin(\sqrt{a^2\plus{}2b}t)\minus{}C_4\cos(\sqrt{a^2\plus{}2b}t)\plus{}C_1\sin(at)\plus{}C_2\cos(at)$
$ x_2(t)\equal{}C_1\sin(at)\plus{}C_2\cos(at)\plus{}C_3\sin(\sqrt{a^2\plus{}2b}t)\plus{}C_4\cos(\sqrt{a^2\plus{}2b}t)$
Sonnhard.

It would probably be helpful to people to post solutions, not just answers.


Given an acute triangle $ ABC$.Let $ P$ be the interior point of $ ABC$ such that $ \widehat{APB}\equal{}\widehat{BPC}\equal{}\widehat{CPA}\equal{}120$. Prove that the Euler lines of triangles $ ABP,BCD$ and $ CAP$ are concurrent.

Maybe there is a typo and should read BCP instead BCD? Thanks

it's an old problem:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6188[/url]

HINT:
1) If T=Torricelli point of the triangle ABC($ \widehat{ATB} \equal{} \widehat{BTC} \equal{} \widehat{ATC}$), and $ G_a, G_b, G_c$ the bayricentres of the triangles TBC, TAC and TAB, respectively $ \Longrightarrow$ the lines $ AG_a, BG_b, CG_c$ concurrent.
2). If $ e_a, e_b, e_c$ Euler lines of the triangles TBC, TAC and TAB, respectively, them $ e_a\parallel TA, e_b\parallel TB, e_c\parallel TC$.


Okay, I do not actually know the answer, but I would like to. I made it up myself, but I am sure it has already been asked before;

If x and y are positive integers greater than 1, and x is greater than y, then prove that y^x is bigger than y^x , or give a counter example.

In that case, counterexample: $ 1^5 > 5^1$ That is false.

@isabella: $ x \neq 1$, and $ y \neq 1$.

Restatement of problem: Given that $ x \in (1, \infty)$ and $ y \in (1, \infty)$, with $ x > y$, show that $ y^x > x^y$.

Okay, well, here's a counterexample:

$ 2^4 \equal{} 4^2$

[quote="isabella2296"]By the reflexive property, $ a \equal{} a$, $ y^x \equal{} y^x$. Thus that is false.

Counterexample: $ 2^3 \equal{} 2^3$[/quote]

I said that x is greater than y.

[quote="isabella2296"]In that case, counterexample: 1^5 > 5^1  That is false.[/quote]

I said that x and y are greater than 1.
Please read the problem.

Okay I am changing the problem to this;

Okay, I do not actually know the answer, but I would like to. I made it up myself, but I am sure it has already been asked before;

If x and y are positive integers greater than 1, and x is greater than y, then prove that y^x is bigger than [b]or equal to [/b]y^x , or give a counter example.

$ x^y < y^x$ when $ x > y> e$, where $ e \equal{} 2.71828...$
If not, then take $ x \equal{} 3$ and $ y \equal{} 2$, then $ x^y > y^x$

[quote="Bugi"]$ x^y < y^x$ when $ x > y > e$, where $ e \equal{} 2.71828...$
If not, then take $ x \equal{} 3$ and $ y \equal{} 2$, then $ x^y > y^x$[/quote]

I said that they had to be greater than 1. You could use 2.

$ (3, 2)$ is actually the only counterexample (the way you've stated it).  The problem statement is true for $ y \equal{} 2$ and $ x > 3$, and it's true for $ y \ge 3$ because of the claim Bugi made, which is essentially based on restating the problem as

$ x \ln y \ge y \ln x \Leftrightarrow$
$ \frac{x}{\ln x} \ge \frac{y}{\ln y}$.

The plot of $ x^y\equal{}y^x$ divides the 1st quadrant into 4 sections. See if you can figure out for which two sections $ x^y>y^x$ and for which two sections $ x^y<y^x$.

[url=http://mathforum.org/library/drmath/view/53229.html]Here is a page[/url] explaining the solution set of $ x^y\equal{}y^x$. See if you can figure out what they look like, and thus determine, roughly, what those four sections of the first quadrant look like!

You may want to find an implicit grapher if you find this difficult; or, someone could plot it and upload it if you can't find one.

[quote="Wickedestjr"]
If x and y are positive integers greater than 1, and x is greater than y, then prove that [b]y^x is bigger than or equal to y^x[/b], or give a counter example.[/quote]

The reason isabella2296 used the reflexive property is that you said $ y^x \ge y^x$ which is true for any numbers $ x, y$.

But I AM sure you meant $ y^x > x^y$.


inequalities
AwesomeMath
summer program
inequalities theorems

http://reflections.awesomemath.org/2007_4/on_the_am_gm.pdf

My best thank is conveying to Mr Lascu and the Editor of Awesomemath magazine who helped me alot to revise this article, especially the English. This is an article on AM-GM inequality,

I don't know if this is particularly helpful, but I found that the S.O.S form:

$ \sum_{cyc}\frac{a}{b}\equal{}\frac{1}{6abc}*\sum_{cyc}(3a\plus{}b\minus{}c)(b\minus{}c)^{2}$

It is very helpful actually, but the following form is better

\[ \sum\frac{a}{b}\minus{}3\equal{}\frac{(a\minus{}b)^{2}}{ab}\plus{}\frac{(c\minus{}a)(c\minus{}b)}{ac}.\]

Hello, I don't understand 'fully mixing variables method.' You used this in problem 5.


integration
trigonometry
real analysis
real analysis unsolved

$ I\equal{}2\int_0^{\infty} \frac{1}{1\plus{}x^2}\ dx\equal{}2\left[\tan ^ {\minus{}1}x\right]_0^{\infty}\equal{}\pi$.


I really need a good cube that is meant for speedcubing. One with tiles would be preferred because the stickers keep coming off.

What's wrong with a traditional Rubik's brand? I think those are fine, and easy to repair.

That's a joke right? You don't want to be mean... that's got to be the WORST lube. Most speedcubers use silicone.

:( fine must stop being sarcastic...

I used wd-40 and had to get a new cube...it also screwd up my stickers.

I got mine at Target :blush:

I got my 2x2, 4x4 and 5x5 online though.  Save on shipping if you buy from the same source.

I'm way to lazy to leanr how to solve 4x4, at least until I get sub30 on 3x3. ANy recommnedations on how? On good trials i get about a minute.

Get silicone spray online. (Do NOT use WD-40) For a good cube BUY ONLINE. use puzzleproz or cube4you, those are really good sites. Always use cubesmith for stickers. DO NOT buy from target <_<. Rubik's is sometimes okay, but frequently not. You could try though.

[quote="Poincare"]One with tiles would be preferred because the stickers keep coming off.[/quote]
sounds like you have paper stickers. you can get PVC stickers on most sites that sell cubes (rubiks.com, cubesmith)

[quote="AIME15"]I got mine at Target :blush:[/quote]
preassembled cubes are usually too stiff to speedcube with. DIY's are really nice.

[quote="Poincare"]k, so i WD-40'd mine. :([/quote]

What even after I said not to?

no, i had done a so a looooooooooooooooooooooooooooooooooong time ago (almost over 5 months ago) but it should have evaporated by now. I did it to my tiled cube over a year ago and it works just fine (*touches wood*) right now.

Lube doesn't "evaporate", unelss I'm mistaken.  My old cube with WD-40 still turns ok...

I meant wear off, not really evaporate sorry

Oh, and another thing, I don't recommend Vaseline (some people actually use it) as lube it RUINED my rubix. Yeah, silicone FTW! :D

[quote="rhodium"]
[quote="AIME15"]I got mine at Target  :oops:  [/quote]

preassembled cubes are usually too stiff to speedcube with. DIY's are really nice.[/quote]

Hmm. When My dad turns my preassembled cube it like wh.i..z...z....e.....s......

Does anyone here, by any chance, know how to turn the cube well and quickly? I always misalign the faces and the cube keeps getting stuck. Also, my fingers are quite clumsy. Any thoughts?

Please don't revive 6 year old threads as it takes up extra forum space.


A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac{7}{17}$. What number was erased?

$\textbf{(A) } 6\qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) } 9\qquad \textbf{(E) }\text{cannot be determined}$

[quote="4everwise"]A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac 17$. What number was erased?

$\text{(A)} 6\qquad \text{(B)}7 \qquad \text{(C)}8 \qquad \text{(D)} 9\qquad \text{(E)}\text{cannot be determined}$[/quote]
[hide]
I suppose it's not E. It usually isn't. :| but since the number of terms is not stated, and the common difference isn't stated, I'm going to say (E) :dry:[/hide]

[quote="ch1n353ch3s54a1l"]
I suppose it's not E. It usually isn't. :| but since the number of terms is not stated, and the common difference isn't stated, I'm going to say (E) :dry:[/quote]
Ah, but the common difference is stated :) The problem says they are consecutive integers.

That's basically what happened to me too... :(

4everwise, are you sure the numbers are right? :D

Then possibly the right answer is (E)?  ;)

according to eryaman:

[quote="eryaman"][hide="I think I messed up somewhere..."]So we get the equation $\frac{\frac{n(n+1)}{2}-x}{n-1}=\frac{246}{7}$ where $x$ is the removed number.  Since $7$ is prime, $n-1\equiv 0 \pmod 7$, so $n\equiv 1\pmod 7$.  Since the arithmetic mean of the numbers is around $35$, the $n$ must be somewhere from $50$ to $78$.  Testing out $50$, $57$, $64$, $71$, and $78$ as $n$ gives values of $x$ that are either: negative, or greater than $n$, both of which are impossible.

So I probably screwed up somewhere.[/hide][/quote]

The original sum is $S=\frac{n(n+1)}{2}$

and the original arithmetic mean is $a=\frac{S}{n}=\frac{n+1}{2}$

I will use the expression $na$ for the sum instead of $S$

The removed number is the number $a+x$.
So $x$ could be negative

$1 \leq  a+x \leq  n \Rightarrow$

$1-a \leq x \leq n-a \ \ (1)$

the new sum is $S-(a+x)=na-(a+x)$ and we divide it by $n-1$

$\frac{na-(a+x)}{n-1}=\frac{na-a-x}{n-1}=\frac{(n-1)a-x}{n-1}=a-\frac{x}{n-1}=35\frac{1}{7}$

$\Rightarrow a=35\frac{1}{7}+\frac{x}{n-1}$

If $n$ is odd, $a$ is an integer
If $n$ is even, then $a$ has $.5$ as decimal part

from $(1)$ we get
$1-a \leq x \leq n-a \Rightarrow$

$\frac{1-a}{n-1} \leq \frac{x}{n-1} \leq \frac{n-a}{n-1}$

the lowest limit is $\frac{1-a}{n-1}=\frac{1-\frac{n+1}{2}}{n-1}=\frac{1-n}{2(n-1)}=-\frac{1}{2}$

the upper limit is $\frac{n-a}{n-1}=\frac{n-\frac{n+1}{2}}{n-1}=\frac{n-1}{2(n-1)}=\frac{1}{2}$

so we have:

$-\frac{1}{2} \leq \frac{x}{n-1} \leq \frac{1}{2}$

and from $a=35\frac{1}{7}+\frac{x}{n-1}$ we get that

$35\frac{1}{7}-\frac{1}{2} \leq a \leq 35\frac{1}{7}+\frac{1}{2}$

$34\frac{9}{14} \leq a \leq 35 \frac{9}{14}$

since $a$ can have either zero or $0.5$ decimal part we find out that $a=35$ or $a=35.5$

$a=\frac{n+1}{2}\Rightarrow n=2a-1$

$a=35 \Rightarrow n=69$
$a=35.5 \Rightarrow n=70$

these values are not acceptable because we want $n-1\equiv 0 \pmod 7$

hence I think that the answer is not $undetermined$ but $\boxed{impossible}$

for example, take the numbers $1,\cdots,71$

if you remove number $1$, you get an average $36,5$
if you remove number $71$, you get an average $35,5$

for $n=64$ you'll get worst results.
The original average is $32,5$ and the maximum you'll get (if you remove number $1$) is $33$

Am I missing something??  :|

[hide="Interesting problem."]
$\frac{ \frac{n(n+1)}{2} - k }{n-1} = 35 \frac{1}{7}$ for some $k \le n$.  Clearly $n \equiv 1 \bmod 7$ so write $n = 7p + 1$.

Our expression becomes $(7p+1)(7p+2) - 2k = 492p$.  Expanding, $49p^2 + 21p + 2 - 2k = 492p \implies 49p^2 - 471p + 2 = 2k$ for $k \le 7p + 1$.  This is impossible - the smallest positive value comes when $p = 10, k = 96 > 7p + 1$ hence no solution.
[/hide]

Edit:  Honestly, I don't know why this is #28  :huh:

[quote="nsato"]That number should be 35 7/17, not 35 1/7.  Is there a way to correct that in our contest database?[/quote]All you have to do is edit the post, and the problem in the database is automatically edited.  :)

For some reason, my source for this test gave me the wrong number.  :mad:

[quote="nsato"]That number should be 35 7/17, not 35 1/7.  Is there a way to correct that in our contest database?[/quote]

Well let's rewrite that solution :)

[hide="New solution"]
$\frac{ \frac{n(n+1)}{2} - k}{n-1} = 35 \frac{7}{17} \implies n = 17p + 1$
$(17p+1)(17p+2) - 2k = 1204p$
$289p^2 + 51p + 2 - 2k = 1204p \implies 289p^2 -1153 p + 2 = 2k$.  Observing that $289 \approx \frac{1153}{4}$ we guess that the first positive value occurs at $p = 4 \implies k = 7$, so, at last, our answer is $\boxed{B}$.

(I have a faint suspicion that this problem can be done with mods.)
[/hide]

[hide]$\frac{\frac{n(n+1)}{2}-x}{n-1}=\frac{602}{17}$

Notice that the left side is approximately $\frac{\frac{n^{2}}{2}}{n}=\frac{n}{2}$, so $n$ is about $35\times 2=70$.  Also, we know that $n-1$ is a multiple of $17$.  Therefore, $n=69$.  Thus,

$\frac{\frac{69(70)}{2}-x}{68}=\frac{602}{17}$

which we solve to find $x=7$.[/hide]

Hint:

[/hide]since the arithmetic mean is 35 1/17, it is kind of obvious that the number of consec. integers is divisble by 17.  Also, if only one number is erased, this will not affect the mean all that much.  We see that the mean is 35 1/17, telling us that that is near the middle of the number.  17*4=68, and 68/2 is close to the mean given.  From here, just sum up all of the consec. integers to 68.  Then multiply the given mean by 68.  Find the difference between the two numbers.  I haven't run the numbers, but 7 seems to be a popular answer...

Hint:

[hide]since the arithmetic mean is 35 1/17, it is kind of obvious that the number of consec. integers is divisble by 17.  Also, if only one number is erased, this will not affect the mean all that much.  We see that the mean is 35 1/17, telling us that that is near the middle of the number.  17*4=68, and 68/2 is close to the mean given.  From here, just sum up all of the consec. integers to 68.  Then multiply the given mean by 68.  Find the difference between the two numbers.  I haven't run the numbers, but 7 seems to be a popular answer..[/hide]

hm.. seems i made an error in there somewhere...  the concept is correct though i believe...

Let me fix my answer too

[quote]$35\frac{1}{7}-\frac{1}{2} \leq a \leq 35\frac{1}{7}+\frac{1}{2}$

$34\frac{9}{14} \leq a \leq 35 \frac{9}{14}$[/quote]becomes

$35\frac{7}{17}-\frac{1}{2} \leq a \leq 35\frac{7}{17}+\frac{1}{2}$

$34\frac{31}{34} \leq a \leq 35 \frac{31}{14}$

since $a$ can have either zero or $0.5$ decimal part we find out again $a=35$ or $a=35.5$

$a=35 \Rightarrow n=69 \equiv 1 \pmod {17}$
$a=35.5 \Rightarrow n=70 \equiv 2 \pmod {17}$

Since $n\equiv 1 \pmod {17}$ we can only accept $n=69$

So if $w$ is the removed number and $S$ the original sum we have $S-w=68 \cdot 35\frac{7}{17} = 2408$

But $\sum_{n=1}^{69}{n}=69=2415\Rightarrow$

$w=2415-2408=7$

heh

[hide]

AMC is with calculator right? well then we can easily reach like

$\frac{\frac{(n+1)(n)}{2} - x}{n-1} = \frac{602}{17}$

and then couldnt we just guess and check almost? of course for the purpose of learning it would be best to actually do the whole thing and understand the problem and everything... but on a timed test i think i wouldve just plugged in 6,7,8, and then 9 for $x$, and then find the value that gives me an integer value for $n$

doing this, we wouldve arrived at $7$ pretty fast, so yea just some insight i guess

[/hide]

I have an interesting insight no one mentioned yet.  :)

[hide="Solution"]

As people mentioned, the number of terms after the deletion is divisible by $17$, we'll call that $17k$. Now consider the average of $1+2+\cdots+(n-1) \Rightarrow \frac{n(n-1)}{2(n-1)}=\frac{n}{2}$ and the average of $1+2+\cdots+n \Rightarrow \frac{n(n+1)}{2n}$.

What's interesting is that their difference turns out to be $\frac{1}{2}$. So no matter what list of consecutive integers we're given, deleting one of the integers will never lower the average by $\frac{1}{2}$ and adding an integer will never raise the average by more than $\frac{1}{2}$.

Since the fractional part is $35 \frac{7}{17}$, this means that the original average was $35$. Now we can set up the equation $$\frac{35 \frac{7}{17} \cdot 17k +j}{17k+1}=35$$ where $17k+1$ is the number of terms in the original list and $j$ is the deleted number. Simplifying yields $$7k=35-j$$, and since $j$ can't be $0$, a quick check with $j=7, k=4$ gives us the answer.

Wasn't this a mathcounts question? Or is it a similar one on a different test...

Looks like 2015 MathCOUNTS Nationals target #5


Two eight-sided dice each have faces numbered $ 1$ through $ 8$. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?

$ \textbf{(A)}\ \frac{1}{2}\qquad
\textbf{(B)}\ \frac{47}{64}\qquad
\textbf{(C)}\ \frac{3}{4}\qquad
\textbf{(D)}\ \frac{55}{64}\qquad
\textbf{(E)}\ \frac{7}{8}$

[hide="Click for solution"]
There are $ 8 \times 8\equal{}64$ ordered pairs that can represent the top numbers on the two dice.  Let them be $ x$ and $ y$.  Thus $ xy>x\plus{}y \implies xy\minus{}x\minus{}y>0$, so $ 1<xy\minus{}x\minus{}y\plus{}1\equal{}(x\minus{}1)(y\minus{}1)$.  This is satisfied except when $ x\equal{}1$, $ y\equal{}1$, or $ x\equal{}y\equal{}2$.  This totals $ 16$ ordered pairs, so the desired probability is $ \frac{64\minus{}16}{64}\equal{}\frac{3}{4}$, or $ \boxed{\textbf{(C)}}$.
[/hide]

No, I said it DOESN'T work, so we subtract it out.

We no that it cannot work when it there is a one so there are 15 combinations. 1,...8 and back but we only count 1,1 once. and the other one is 2,2 so that is 16. 64-16/64=3/4.

[hide]Well it'll work for everything except for when a number is paired with a 1.  So we have $ \dbinom{8}{2}$ pairs and out of those only $ \dbinom{8}{2} \minus{} 7$ work.  (It's 7 because we have 8 * 1, 7 * 1,..., 2 * 1).  Therefore our answer is:

$ \frac{28 \minus{} 7}{28} \equal{} \frac{21}{28} \equal{} \boxed{\frac{3}{4}}$[/hide]


inequalities
algebra
polynomial
inequalities proposed

Let $ a,b,c,d$ be non-negative such that $ (a+b+c+d)^{2}=3(a^{2}+b^{2}+c^{2}+d^{2})$. Prove that
\[ (a+b+c+d)^{3}\le 27(abc+bcd+cda+dab).\]

[quote="hungkhtn"]Let $ a,b,c,d$ be non-negative such that $ (a+b+c+d)^{2}=3(a^{2}+b^{2}+c^{2}+d^{2})$. Prove that
\[ (a+b+c+d)^{3}\le 27(abc+bcd+cda+dab). \]
[/quote]
I think, it's very nice, but easy for IMO.

It is unlikely that questions of this kind might appear at IMO as a contest problem due to its artifical restriction. I bet that inequality of this year will be multivariable one with special inequality case. People who propose problems for IMO also look at problem solving forum like this and they know which ones are posted and known. Some guess that anything like IMO 2001 inequality or IMO 2000 inequality will not appear in future IMOs anymore.

[quote="silouan"][quote="arqady"] I think, it's very nice, but easy for IMO.[/quote]

Arqady what is your solution ?[/quote]
$ (a+b+c+d)^{2}=3(a^{2}+b^{2}+c^{2}+d^{2})\Leftrightarrow a^{2}+b^{2}+c^{2}+d^{2}=ab+ac+ac+bc+bd+cd.$
Hence, $ (a+b+c+d)^{3}\le 27(abc+bcd+cda+dab)\Leftrightarrow$
$ \Leftrightarrow(a+b+c+d)(a+b+c+d)^{2}\le 27(abc+bcd+cda+dab)\Leftrightarrow$
$ \Leftrightarrow(a+b+c+d)(ab+ac+ad+bc+bd+cd)\leq9(abc+bcd+cda+dab)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{2}b+a^{2}c+a^{2}d)\leq6(abc+bcd+cda+dab).$
But $ (a+b+c+d)(ab+ac+ad+bc+bd+cd)=(a+b+c+d)(a^{2}+b^{2}+c^{2}+d^{2})\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{3}-3abc)=0.$
Hence,$ \sum_{cyc}(6abc-a^{2}b-a^{2}c-a^{2}d)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(6abc-a^{2}b-a^{2}c-a^{2}d)+\frac{3}{2}\cdot\sum_{cyc}(a^{3}-3abc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3a^{3}-2a^{2}b-2a^{2}c-2a^{2}d+3abc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{3}+b^{3}+c^{3}-a^{2}b-a^{2}c-b^{2}a-b^{2}c-c^{2}a-c^{2}b+3abc)\geq0,$ which true by Schur.

I solved it using the standard method of decreasing the number of variables by Roll's theorem.

So from it we have that there exist positive real numbers $ x,y,z$ so that \[ \frac{a+b+c+d}{4}=\frac{x+y+z}{3}\] \[ \frac{ab+ac+ad+bc+bd+cd}{6}=\frac{xy+xz+yz}{3}\] \[ \frac{abc+abd+bcd+acd}{4}=xyz\]. Now the problem is to show that \[ (x+y+z)^{3}\leq \frac{27^{2}}{16}xyz\] when $ 8(x+y+z)^{2}=27(xy+xz+yz)$. As our inequality and the condition are homogeneous we can assume w.l.o.g  that $ x+y+z=3$. Hence we need to prove that $ xyz\geq \frac{16}{27}$  when $ x+y+z=3$ and $ xy+xz+yz=\frac{8}{3}$. The greatest of the variables should be greater than $ 1$,let it to be $ x$. Then $ yz=\frac{8}{3}-x(y+z)=\frac{8}{3}-x(3-x)$ hence \[ xyz=\frac{8}{3}x-3x^{2}+x^{3}=(x-\frac{1}{3})(x-\frac{4}{3})^{2}+\frac{16}{27}\geq\frac{16}{27}\].

P.S. So what will happen? I will solve it :D .

[quote="Svejk"][quote="vardanch"]
 WLOG $ d=1$
[/quote]

I don't see why.I think it's wrong[/quote]

If $ d=min(a,b,c,d)$ and $ d<>0$ => $ a,b,c,d>0$.
Since this equality and inequality are homogenous we can $ WLOG$ assume that $ d=1$.

[hide]
Let  $ a1=\frac{a}{d}$, $ b1=\frac{b}{d}$, $ c1=\frac{c}{d}$ (since d<>0)
the equality <=> $ d^{2}(a1+b1+c1+1)^{2}=d^{2}*3(a1^{2}+b1^{2}+c1^{2}+1)$  <=> $ (a1+b1+c1+1)^{2}=3(a1^{2}+b1^{2}+c1^{2}+1)$
and we must prove that $ d^{3}(a1+b1+c1+1)^{3}\leq 27d^{3}(a1b1c1+\sum{a1b1})$ <=> $ (a1+b1+c1+1)^{3}\leq 27(a1b1c1+\sum{a1b1})$
=> we can WLOG assume that $ d=1$
[/hide]

Now do you understand why we can WLOG $ d=1$, Svejk ?

[quote="vardanch"]

Now do you understand why we can WLOG $ d=1$, Svejk ?[/quote]
got it

Yes, very nice solution, Arqady and Tiks (especially Tiks, with special solution). However, I think even my solution is shorter (hi, also nice, too). I use Cauchy-Schwarz inequality, in 4 line-solution.

Let $ a,b,c,d$ be non-negative real numbers such that $ (a+b+c+d)^{2}=3(a^{2}+b^{2}+c^{2}+d^{2})$. Prove that
$ (a+b+c+d)^{3}+3(min[a,b,c,d])^{3}\leq 27(abc+bcd+cda+dab)$.

OK. And here is my solution (I feel it is abit similar to Arqady's one)
\[ (a+b+c+d)^{2}=3(a^{2}+b^{2}+c^{2}+d^{2}) \Rightarrow a^{2}+b^{2}+c^{2}+d^{2}=\sum ab \]

\[ \Rightarrow \left(\sum a^{2}\right)\left(\sum a\right) = \left(\sum ab\right)\left(\sum a\right) \Rightarrow \sum a^{3}=3\sum abc \]
By Cauchy-Schwarz inequality
\[ 3\left(\sum abc\right)\left(\sum a\right) = \left(\sum a^{3}\right)\left(\sum a\right) \ge \left(\sum a^{2}\right)^{2}=\frac{1}{3}\left(\sum a\right)^{4}\]
So we have done.

[quote="Tiks"]I solved it using the standard method of decreasing the number of variables by Roll's theorem. .[/quote]
Nice solution.

Please, can you tell me about $ Roll's$ theorem ?

As this is a repetition of this question (Silouan also asked about this in a PM), I am posting here the main idea of the mentioned method.

[color=blue]Roll's Theorem: If $ P(x)$ is a polynomial given on $ R[x]$ and has two real zeros($ P(a)=P(b)=0$ and $ a\geq b$) then there exist a real number $ c$ such that $ P'(c)=0$ and $ a\geq b \geq c$.[/color]

Now let's consider the fourth degree polynomial $ P(t)=(t-a)(t-b)(t-c)(t-d)$ with $ a\geq b\geq c\geq d$. From the fact mentioned above we can conclude that there exist three real numbers $ x\geq y\geq z$ such that $ a\geq x\geq b\geq y\geq c\geq z\geq d$ and \[ P'(t)=p(t-x)(t-y)(t-z)=pt^{3}-pt^{2}(x+y+z)+pt(xy+xz+yz)-pxyz\]. On the other hand,

\[ P'(t)=4t^{3}-3t^{2}(a+b+c+d)+2t(ab+ac+ad+bc+bd+cd)-(abc+abd+acd+bcd)\]

,thus $ p=4$ and

\[ P'(t)=4t^{3}-3t^{2}(a+b+c+d)+2t(ab+ac+ad+bc+bd+cd)-(abc+abd+acd+bcd)=4t^{3}-4t^{2}(x+y+z)+4t(xy+xz+yz)-4xyz\]
equalizing the respective coefficients we find that

\[ \frac{a+b+c+d}{4}=\frac{x+y+z}{3}\]

\[ \frac{ab+ac+ad+bc+cd+bd}{6}=\frac{xy+xz+yz}{3}\]

\[ \frac{abc+abd+acd+bcd}{4}=xyz\] :wink:

Yes, by Tick's method, we can prove the following general result

For $ 0<r<1$. If $ a,b,c,d$ are non-negative and
\[ (a+b+c+d)^{2}=\frac{4(r+2)^{2}}{16r^{2}-8r+28}(a^{2}+b^{2}+c^{2}+d^{2})\]
then
\[ (a+b+c+d)^{3}\le \frac{16(r+2)^{3}}{27r}(abc+bcd+cda+dab).\]

a nice problem, Hungkhtn! :D I think the method is very useful. We using Roll's theorem to reduce variable - From 4-variable to 3 variable and after that we can solve the inequaities easily. Many problems was solved by the method. For example:
Problem [hungkhtn]
Let $ a,b,c,d$ be non-nagative real numbers such that $ a+b+c+d=1$. Prove that:
\[ a^{3}+b^{3}+c^{3}+d^{3}+10(ab+bc+cd+da+ac+bd)\le 64 \]
Problem [pvthuan's book]
Let $ a,b,c,d$ be non-nagative real numbers such that $ a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that:
\[ a+b+c+d\ge a^{3}+b^{3}+c^{3}+d^{3}+ab+bc+cd+db+ac+bd \]
I am not sure pvthuan's problem is correct because I don't remember it, I solved it some day ago :P And remain examples are interesting ...

[quote="hungkhtn"]Let $ a,b,c,d$ be non-negative such that $ (a \plus{} b \plus{} c \plus{} d)^{2} \equal{} 3(a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2})$. Prove that
\[ (a \plus{} b \plus{} c \plus{} d)^{3}\le 27(abc \plus{} bcd \plus{} cda \plus{} dab).
\]
[/quote]
By the way, with the same conditions we obtain:
\[ (a \plus{} b \plus{} c \plus{} d)^{3}\geq 21.6(abc \plus{} bcd \plus{} cda \plus{} dab)
\]
The equality for example when $ a\equal{}b\equal{}c\equal{}1$ and $ d\equal{}3.$

By the way, the following inequality, appeared as problem 96, page 239, "secrets in Inequalities", is very very nice

Let $ a,b,c,d$ be non-negative real numbers such that
\[ (a\plus{}b\plus{}c\plus{}d)^2\equal{}3(a^2\plus{}b^2\plus{}c^2\plus{}d^2).\]
Prove the following inequality
\[ a^4\plus{}b^4\plus{}c^4\plus{}d^4 \ge 28abcd.\]

I am waiting for a solution.

[quote="Tiks"]
[color=blue]Roll's Theorem: If $ P(x)$ is a polynomial given on $ R[x]$ and has two real zeros($ P(a) \equal{} P(b) \equal{} 0$ and $ a\geq b$) then there exist a real number $ c$ such that $ P'(c) \equal{} 0$ and $ a\geq b \geq c$.[/color]
[/quote]

Shouldn't it be $ a \geq c \geq b$?

P.S. Nice solution.

So we reduce a 4-variable inequality to a 3-variable one. What stops us from taking that 3-variable inequality and making it a 2-variable inequality, or a 1-variable inequality for that matter?

All you can do is to reduce the number of variables to $ 3$. You can not change it to a 2-variables inequality by this way. But 3-variables is much easier than 4-variables one.

Yes, Roll's theorem is really interesting. Using  Roll's theorem, we can solve my own inequality very easily  :wink:

http://www.mathlinks.ro/viewtopic.php?t=178404

All of them are not hard to prove by Mixing Variable method but I am also interest in this technieque :)

[quote="hungkhtn"][quote="zaizai-hoang"]Of course, If the inequality have $ abcd$ or $ abcde$ etc, this technique will fail. When it happen, we should use some substitution  :D[/quote]

Let's guess what is the case of the equality?[/quote]
I can guess one case of equality is $ a\equal{}b\equal{}c\equal{}d\equal{}0$. Remain case, I will check later  :lol:

[quote="hungkhtn"][quote="hungkhtn"]
Let $ a,b,c,d$ be non-negative real numbers such that
\[ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2).
\]
Prove the following inequality
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \ge 28abcd.
\]
[/quote]

By the way, you can not use Roll theorem to reduce number of variables to solve this problem. It is highly complicated and harder than the original one I posted.[/quote]

I am really curious to know what you used to prove above inequality... I think we can transform above inequality to the following inequality:
Let $ a,b,c,d$ be non-negative real numbers such that
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\equal{}28abcd.
\]
Prove the following inequality
\[ (a \plus{} b \plus{} c \plus{} d)^2 \le 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2).
\]

I am really interested in the conditon of problem. It is rather like an well-known inequality:
Let $ a,b,c,d$ be non-negative real numbers. Prove the following inequality

\[ (a \plus{} b \plus{} c \plus{} d)^2 \le 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)\plus{}4\sqrt{abcd}\]

[quote="zaizai-hoang"][quote="hungkhtn"][quote="zaizai-hoang"]Of course, If the inequality have $ abcd$ or $ abcde$ etc, this technique will fail. When it happen, we should use some substitution  :D[/quote]

Let's guess what is the case of the equality?[/quote]
I can guess one case of equality is $ a \equal{} b \equal{} c \equal{} d \equal{} 0$. Remain case, I will check later  :lol:[/quote]

Let's check $ a\equal{}3,b\equal{}c\equal{}d\equal{}1$ first, Hoang, before think anything else,

[quote="Altheman"]Okay I see, so we basically have symmetric sums, like say we have $ n$ variables, and we have a sum choosing $ k$ at a time.

So we can reduce the n C k to n-1 C k

So we can reduce n C k into k C k or k variables

Hence our theorem has the strongest effect on a problem like

$ (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2\ge K\cdot \sum_{1\le i < j\le n}a_ja_j$

In particular, we can reduce this to a two variable inequality?!

This weakens one of the symmetric sums to a product[/quote]

This is not true, Altheman.

@ Hoang. Anh có giúp gì được em đâu. Hi vọng em sẽ sớm vượt qua mọi chuyện. That is all my wish!

[quote="Altheman"]Okay so lets see lets prove

$ (a \plus{} b \plus{} c \plus{} d)^2\ge \frac {8}{3}(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)$

Suppose that $ a\ge b\ge c\ge d$

Then say $ P(t) \equal{} (t \minus{} a)(t \minus{} b)(t \minus{} c)(t \minus{} d)$

Then $ P'(t) \equal{} 4t^3 \minus{} 3(a \plus{} b \plus{} c \plus{} d)t^2 \plus{} 2(ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd)t \minus{} (abc \plus{} bcd \plus{} cda \plus{} dac)$

But by rolle's theorem, we have that there are $ a\ge x\ge b\ge y\ge c\ge z\ge d$ for which

$ P'(t) \equal{} 4(t \minus{} x)(t \minus{} y)(t \minus{} z)$...

Then we have $ \frac {a \plus{} b \plus{} c \plus{} d}{4} \equal{} \frac {x \plus{} y \plus{} z}{3}$, $ \frac {ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd}{6} \equal{} \frac {xy \plus{} yz \plus{} zx}{3}$.

Hence our inequality is equivalent to $ \frac {16}{9}*(x \plus{} y \plus{} z)^2\ge \frac {8}{3}*2*(xy \plus{} yz \plus{} zx)$  or

$ (x \plus{} y \plus{} z)^2\ge 3(xy \plus{} yz \plus{} zx)$

Let $ Q(t) \equal{} (t \minus{} x)(t \minus{} y)(t \minus{} z)$. Then $ Q'(t) \equal{} 3t^2 \minus{} 2(x \plus{} y \plus{} z)t \plus{} (xy \plus{} yz \plus{} zx)$

But also by rolle's theorem, $ Q'(t) \equal{} 3(t \minus{} r)(t \minus{} s)$ where $ x\ge r\ge y\ge s\ge z$.

Equating coefficients, we have $ \frac {x \plus{} y \plus{} z}{3} \equal{} \frac {r \plus{} s}{2}$ and $ \frac {xy \plus{} yz \plus{} zx}{3} \equal{} rs$.

Hence our inequality is equivalent to $ \frac {9}{4}*(r \plus{} s)^2\ge 3*3rs$ or $ (r \plus{} s)^2\ge 4rs$...

??? I am pretty sure that works. Is there a problem?[/quote]

So, let's prove the very simple one

\[ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\ge 9abc.\]

I think we both understand. I agree that your inequality cannot be 'rolle reduced' because there is a 3 C 3 term.

However, if we were to take

$ \left(\sum_{1\le i\le n} a_i\right)\left(\sum_{1\le i<j \le n} a_ia_j\right)\ge \frac{\binom{n}{1}\binom{n}{2}}{\binom{n}{3}}*\left(\sum_{1\le i<j<k\le n}a_ia_ja_k\right)$

where $ n>3$

then we could reduce that to the inequality that you gave because we have a
n C 1, n C 2, n C 3 terms

and since 3 is the highest, we can make it a 3 variable inequality

Yes, sorry. I misunderstand you ;) That explanation is true,

[quote="zaizai-hoang"]Yes, they are rather like argady's equality. Anyway, it seems to be new inequality's form with a complicated condition. I think Mixing variables method can help us although maybe we can not get a nice solution.

@ my brother: For a long time, I have felt tired of living... My friend sadden me. I have felt a kind of remorse what I did. I want to find out new happiness in the life so I will try prove above inequality in weekend when I passed school's examinition . Thanks you for helping when I was feeling sad. You are really kind :wink:[/quote]

For a hint, notice that the inequality holds for $ (a,b,c,d)\sim (3,1,1,1)$ or $ (a,b,c,d)\sim (2 \plus{} \sqrt {3},2 \plus{} \sqrt {3},2 \minus{} \sqrt {3},2 \minus{} \sqrt {3})$ up to permutation. So we have 8 cases of equality. Hope that it is useful to find a solution.

See you and Mathlinkers later.  :) My Spring Break is going to... end.

[quote="hungkhtn"][quote="hungkhtn"]
Let $ a,b,c,d$ be non-negative real numbers such that
\[ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2).
\]
Prove the following inequality
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \ge 28abcd.
\]
[/quote]

By the way, you can not use Roll theorem to reduce number of variables to solve this problem. It is highly complicated and harder than the original one I posted.[/quote]

But if you set $ f(a,b,c,d) \equal{} a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \minus{} 28abcd$ then we have $ f(ta,tb,tc,td) \equal{} t^4f(a,b,c,d)$. Indeed the inequality is homogenous and can't we suppose $ abcd \equal{} 1$ ?

[quote="zweig"]
But if you set $ f(a,b,c,d) \equal{} a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \minus{} 28abcd$ then we have $ f(ta,tb,tc,td) \equal{} t^4f(a,b,c,d)$. Indeed the inequality is homogenous and can't we suppose $ abcd \equal{} 1$ ?[/quote]

The goal is to take three variables, x,y,z and find an equivalent problem. Even if abcd=1, you still don't have a way of writing that piece of information using x,y,z

[quote="King Wang JJang"]I have a question!!

So, if some inequality is symmetric and polynomial,

we can always reduce the number of variables???

4->2, 3->2 ???

I think rolle reducing(??) doesn't have any error.

If it is, it can be very powerful, I think.[/quote]

Of course, it is not always true, sorry. The last problem I posted in this topic is one example.

[quote="hungkhtn"][quote="hungkhtn"]
Let $ a,b,c,d$ be non-negative real numbers such that
\[ (a \plus{} b \plus{} c \plus{} d)^2 \equal{} 3(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2).
\]
Prove the following inequality
\[ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \ge 28abcd.
\]
[/quote]

By the way, you can not use Roll theorem to reduce number of variables to solve this problem. It is highly complicated and harder than the original one I posted.[/quote]
No one post solution, because it is so hard  :(  I have read your solution in Secrets in Inequality.

[quote="hungkhtn"]By the way, the following inequality, appeared as problem 96, page 239, "secrets in Inequalities", is very very nice

Let $ a,b,c,d$ be non-negative real numbers such that
\[ (a\plus{}b\plus{}c\plus{}d)^2\equal{}3(a^2\plus{}b^2\plus{}c^2\plus{}d^2).\]
Prove the following inequality
\[ a^4\plus{}b^4\plus{}c^4\plus{}d^4 \ge 28abcd.\]
[/quote]
I think it was problem 96, page 246. :wink:

[quote="hungkhtn"]
Let $ a,b,c,d$ be non-negative real numbers such that
\[ (a\plus{}b\plus{}c\plus{}d)^2\equal{}3(a^2\plus{}b^2\plus{}c^2\plus{}d^2).\]
Prove the following inequality
\[ a^4\plus{}b^4\plus{}c^4\plus{}d^4 \ge 28abcd.\]
[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition gives $d^2-3ud+9u^2-9v^2=0$.
Id est, $d$ does not depend  on $w^3$.
Hence, our inequality is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove our inequality for $b=c=1$ (if $w^3=0$ our inequality is obviously true).
Hence, we need to prove that $a^4+d^4+2\geq28ad$ for all non-negatives $a$ and $d$ such that $a^2-ad+d^2+1=2a+2d$.
Let $a+d=2p$ and $ab=q$.
Hence, $a^2-ad+d^2+1=2a+2d\Leftrightarrow4p^2-4p+1=3q^2$, which gives $q^2=\frac{(2p-1)^2}{3}$ and $p^2-4p+1=3(q^2-p^2)\leq0$.
Finally, $a^4+d^4+2\geq28ad\Leftrightarrow8p^4-8p^2q^2+q^4+1\geq14q^2\Leftrightarrow(p^2-4p+1)(p-2)^2\leq0$. Done!


Prove that there exist a dense open subset $S$ of $M(4,C)$ such that

i) $A$ is nonsingular for all $A\in S$
ii) $A$ has distinct eigenvalues for all $A\in S$
iii) for each $A\in S$ the element $aI_{4}+bA+cA^{*} \in M(4,C)$ has rank $\geq 3$
for all $(a,b,c)\neq (0,0,0)$ in $C^{3}$

Here $A^{*}=det A\cdot A^{-1}$ or $A^{*}_{i,j}$ is the conjugate of $a_{ji}$?

here $A^{*}$ is the transconjugate $A^{*}=\overline {^{t}A}$


:lol: How can I make 659 using all of these numbers: 1 50 7 3 25 2

Are we allowed to repeat numbers or must they be used once?

Which operations?
And can we use numbers like this:

Put 1 & 7 together to get 17 or 71?

...659 is a prime.

Which obviously means it can't be created with the given numbers by multiplication..

Nor addition, because that would give you infinite possibilities.


analytic geometry
graphing lines
slope
ratio

Point $ A$ is at $ (0, 0)$ and point $ B$ is on the line $ y \equal{} 4$. The slope of segment $ AB$ is $ \frac{2}{3}$. What is the sum of the $ x$- and $ y$-coordinates of point $ B$?

Let the $ B=(x,4)$. Then $ \frac{4-0}{x-0}=\frac23\implies x=6\implies x+4=\boxed d{10}$.

Also if it is (x,y) the ratio of x to y is 3 to 2 which sums to 5. Thus it is 5/2 y=10


Among 25 apparently equal coins 3 are fakes. All true coins have equal masses, and so do all fake coins, though these are lighter than the true ones. How to find six true coins by two measurings on a balance without weights?

Use the balance with 11 coins on each plate.
If it is balance then on each group there is a fake coin and the left coin is fake.
    Pick one group of 11 (10 true ones and 1 fake)
    Use the balance with 4 coins on each plate.
    If it is balance then the fake coin is on the left group of 3 coins, pick 6 coins out of the two 4-coins group.
    If it is not balance then the fake coin is on the lower group of 4 coins, pick 6 coins out of the left group and the upper group.

If it is not balance then on the upper group there is either no fake coin or one fake coin on it and the left coin is true.
    Pick the upper group of 11 (0 or 1 fake coins)
    Use the balance with 4 coins on each plate.
    If it is balance then pick 6 coins out of the two 4-coins group.
    If it is not balance then the fake coin is on the lower group of 4 coins, pick 6 coins out of left group and upper group.

[quote="mszew"]Use the balance with 11 coins on each plate.
If it is balance then on each group there is a fake coin and the left coin is fake.[/quote] $ 2 \times 11 \plus{} 1 \equal{} 23 \neq 25$.

The idea was sound, even if the arithmetic was not.  Balance 12 against 12.  If one side is heavier, it has only 0 or 1 fake coin; if the two sides balance, both have 1 fake coin.  In either case, you get 12 coins with at most 1 fake coin.  Balance 6 against 6 -- the heavier side (or both sides, if they balance) contains all true coins.


geometry
trapezoid
parallelogram
geometry unsolved

prove that the trapezoid ABCD(AB is paralel to CD) in which AB+CD=CD+AD then this trapezoid is paralelogram .. i couldn't find any solution .

[color=darkblue][quote]which AB+CD=CD+AD[/quote]
I think you ment $ AB\plus{}BC\equal{}CD\plus{}AD$
Very easy problem:
[img]http://i011.radikal.ru/0805/ff/2925c143d10a.jpg[/img]
Let $ D \in PC, B \in AQ$, such that $ PD\equal{}AD, BC\equal{}BQ$.
Then $ AQ\equal{}AB\plus{}BC\equal{}AD\plus{}CD\equal{}PC$, so $ PAQC$ is paralelogram and $ \angle{DPA}\equal{}\angle{BQC}$. Then $ \angle{ADP}\equal{}\angle{QBC}\equal{}\angle{BCD}$, so $ AD \parallel BC$ and $ ABCD$ is a parallelogram.[/color]

you are a very good student as i can see and also helpful for your friends and non friends


15=0.33(1+R)^25

how do i solve for R?
R=the rate.

[quote="anyone"][hide]$15=0.33(1+R)^{25}$
Divide both sides by .33
$45\frac{5}{11}=(1+R)^{25}$
Take the 25th root of both sides (calculator)
$1.1649=1+R$
$R=.1649$[/hide][/quote]

could you do the ln(500/11) / ln(25)?

[hide]
$15 = 0.33(1+R)^{25}$
$\frac{500}{11}= (1+R)^{25}$
$R = \sqrt[25]{\frac{500}{11}}-1 \approx 0.1649$
[/hide]

I was trying to do this with logs but you end up with something like

$\log500-\log11 = 25\log(1+R)$

and as far as I know theres no way to break up $\log(1+R)$


can anyone explain to me why this is true?

if, for a polynomial with integer coefficients with degree :ge: 2, f(1), the leading coefficient, and the constant term are all odd, then the polynomial has no rational roots.

for html, use <>, not [].  Nicely done, though.

So, how did my post just end up before yours, even though it was responding to it?

ahaha, sorry about that, I knew there was something wrong with the braces, but I couldn't remember what, ended up deleting it and posting again :)

hmm i dont understand a couiple things here:

"Suppose an(x/y)n + ... + a1(x/y) + a0 = 0.
Multiply all by yn and we get
anxn + ... + a1xyn-1 + a0yn = 0.

Assuming x/y is in lowest terms. If x is even, then from above the last term must be even, and since a0 is odd, y must be even, contradiction. Same applies to y being even. So they both must be odd, and that big expression is congruent to the sum of the coefficients which is odd and can't equal 0."

(That is wat Triple said.)
If x/y is in lowest terms and x is even, then y must be odd. But why does this make the last term (im assuming u mean teh constant) even? we already assumed a_0 = odd and we know that y^n = odd, so a_0y^n = odd. I dont see where we get our contra.

All but the last term are divisible by x, and thus even. 0 is even. Thus the last term must be even.


We were learning about electronic configurations, and I stumbled on the word "characteristic electron". What does it mean?

Thanks.

I also have never heard that expression before. Perhaps it has the same meaning has valence electrons?

No, it doesn't because we had to fill 2 separate slots for practice problems: valence electrons and characteristic electrons. Apparently, they are electrons on the highest energy level of atom.

My understanding of "energy level" refers only the to principal quantum number $ n$, meaning that all the electrons in the highest energy level would exactly be the valence electrons.  Did "energy level" mean something different in this context?

Just a guess. Highest energy level as in maybe the electrons occupying the highest atomic/molecular orbitals?

it is perhaps the unpaired valence electron(s) :maybe:


TV?  maybe, say 3-5 hours per week, mostly depending on what football games are on  :D

Internet like 15-20 hours.

I barely watch any TV. So 1 hour per week.

20-30 hour of internet / week.

Taking this survey online biases the results, don't you think?

In my current situation, I don't even have a TV. Internet time is hard to estimate; no final routine yet.

0 hours of TV weekly, ~30-40 hours on Internet weekly.

Wow..some people dont watch TV at all

As for me- I watch about 2 hours of TV every weekday. And 4 hours of TV on Saturday and Sunday each...so that totals to 18 hours of TV each week.

Computer..usually if i dont need it for school projects and all, i spend about an hour each weekday, and unless i have a project or something, i dont usually use the computer much on weekends...so that totals to 5 hours of computer.

TV: 90 minutes per week, usually less

internet: 4 hrs per week now.

0 hours for TV
4 Hrs for Comp  :?  :?

Go Naruto!

Ever since I've been going on AoPS, my tv time has dramatically decreased, and my internet time increased. And I agree that this is kind of biased. 1 hr tv, 30 hr. online.

Maybe 2-4 hours weekly of TV.... and about 15 on the internet- well, actually , 15 on the 'computer'... maybe 14 of those hours are on the internet.

Wow... I'm not different...

1) -100 to -10 hours of tv weekly

2) 20 to 30 hours of internet weekly

TV: usually never on weekdays. maybe 6-8 on the weekend.
internet: probably around 17 hours a week.

Television: Slim to none.

Computer: around 15 hours.

TV: three and a half minutes
Internet: 30-40

0<tv time<0.000000001 hrs

12hrs<computer time<20hrs

tv: hmm... about 5-6 hours a week
internet: 20+ hours a week

tv... 5 hours on saturday...and an hour on sunday...so 6 weekly

internet... about 12 hours a day...so something like 72 weekly

[quote="blivy"]hey, i'm not sure where this goes, but i'm taking a survey for my stat hw on how many weekly hours people spend on tv versus the internet. so if you could just answer these two questions, i'll be forever indebted to you!
(a rough estimate is fine.)

1) how many hours of tv do you watch a week?
2) how much time do you spend online a week?

 thanks![/quote]

1) 0
2) 168

Kidding! More like...30, if you include FPS.

1=2hr

2=3hr

i like your avatar blivy  :D

1) 1hr
2)10-15 hrs

this is kind of biased, cuz ppl on AoPS tend to go on the comp way more.