AMC USA(J)MO USAMO geometry AIME MATHCOUNTS function

It looks like I've qualified for the USAMO this year, as a senior (AMC/AIME 114/11). My problem is that while my problem-solving and even proof-writing skills are quite decent, my background in theory is limited to that I've learned from AoPS1, some of AoPS2, and whatever else I've picked up randomly. I take it, though, that pure intuition isn't going to cut it for the USAMO, so I'll need to brush up on some more advanced concepts. I'm thinking of:
1) Going through AoPS2 and trying to get a grasp on the relevant chapters I haven't done yet, and reviewing the ones I've only partially done.
2) Going through the lectures/problems from the WOOT class I took three years ago and reviewing what I learned then, and learning the stuff that was over my head when I was taking the class.
3) Maybe getting a copy of Art and Craft of Problem Solving and working through some of that.

I definitely don't want to overdo it on the preparation, though, since burning myself out is what cost me USAMO last year.

So, what does everyone suggest?
Thanks for the advice.

You do have a decision to make, but it will depend on really how much you can handle math (I mean, sometimes, if you've spent a whole day doing it, you might go a bit crazy  :D )

I think, in preparation for the USAMO, its definitely important to work through AOPS2, since thats mostly basic concepts that everyone needs to understand for the USAMO. That WOOT class stuff looks really important, considering thats supposed to help you a lot with Olympiad stuff.

The Art and Craft of Problem Solving should be third priority in comparison to AOPS and WOOT.

OK, so basically, with the prep, you should really only do as much as you feel you can. In my case, it bores me to death reading AOPS and other stuff, so I have to take a break (meaning taking a contest :D)
And really, it also depends on how far you want to go with USAMO. You said you were a senior, so its not likely you will be accepted to MOP, so the other "ambitious" goal is to become one of 12 winners. And in that case, you've got a lot of work on your hands. Perhaps 3 hours+ a day?

Hope I helped.

Becoming a winner with a month of work is entirely unrealistic. Generally, 1 and 4 on USAMO don't require advanced knowledge. The most important thing is to do problems, especially in the areas you feel you are weak.

One thing people don't realize is how much of a gap there is between the AIME and the USAMO, significantly bigger between that of, for example, the AIME and the AMC, and the AMC and MathCounts. I've basically been dedicating a year of work (or at least I've tricked myself into thinking I'm dedicated) toward being able to solve one more problem on the USAMO. Of course I don't work 3 hours a day, but I try to spend as much time during class (Bio, History, and Calc, my three AP classes - coincidence?) working on problems, and I try to devote significant homework-free periods to math (although piano gets in the way too). It's a very slow process. That being said, I think it takes alot less time to work oneself up to the point where he/she can solve #1/4 than to work from that level to being able to solve one of #2 and #5.

I agree.  It's not just the size of the gap and tremendous increase in difficulty between the AIME and USAMO that makes it hard.  For example, from Mathcounts to AMC to AIME, there is a rather gradual process of continuous learning where one is able to slowly increase one's scores because of the large number (30, 25, 15) of relatively easy problems.  But then on the USAMO, there is a tiny amount of problems including some that one simply may not good enough to solve and cannot make any progress on.  It's kind of similar to comparing the graphs of $ y\equal{}x$ (MC-->AMC-->AIME) and the greatest integer function (USAMO).

CatalystOfNostalgia is completely right here - I've essentially been doing the same thing. Problem difficulty increases [b]much[/b] more quickly on USAMO than on other contests. For most people, even those who are really good, being able to solve 3 or 6 is pretty much unattainable.

Another thing is, the skill needed to quickly solve the easier ones to have enough time to work out and write up the harder ones on USAMO. The difference between being able to get 21 and being able to get 14 is huge. To be able to get 14 on a good day you can work on 1 and 4 for most of your time, and work them out, but you won't be able to do much more (sadly I think I fit in this category, due to not doing any math for a year). To get 21 you need to be able to get rid of the easier ones quickly, which takes a good deal more experience. The skill it takes to be able to consistently solve harder USAMO problems is also much much greater than the skill it takes to be able to consistently solve the easier ones. That is why although a very good portion of USAMO takers can get around 2 problems, getting 3 brings you to blue :o

[quote="sponge008"]I take it, though, that pure intuition isn't going to cut it for the USAMO, so I'll need to brush up on some more advanced concepts.[/quote]

Actually, that's far from the truth.  Most of the easier USAMO problems, the ones you'll probably be spending more time on anyway, don't require "advanced concepts."  This means that you shouldn't try to learn too many advanced concepts, because you probably won't need them, and if you learn them quickly you probably won't see how to apply them.  Also, this means that you should never be intimidated by a USAMO problem and say, "I can't solve this; I probably just haven't learned the technique that's needed," because this is hardly ever true.

Since you have only two months to prepare, I think a reasonable goal would be to fully solve one or two problems, ideally #1 and #4.  The best thing you can do for yourself is to try to solve some #1's and #4's, but if you can't, you could try working on easier Olympiad problems (Canada's for example).  AoPS Volume 2 probably won't help much for the USAMO unless you're still truly learning a lot of new stuff from it - if you are, that's great.  You should also be really comfortable with all the standard proof "tricks": induction, strong induction, contradiction, extremal principle, coloring, infinite descent, pigeonhole, and so on.  Engel's book, Problem Solving Strategies, is good for this.  But the best thing you can do, again, is to practice solving problems.

I want to point out that Art and Craft of Problem Solving probably is helpful for someone in your position - it has a good chapter (3, I think?) on induction, pigeonhole, etc.  - the techniques that you really do need to be familiar with.

A digression: wasn't #1 on last year also solvable by a quick factorization?

For the OP: though I'm not experienced as the other people here, I agree that #1 and #4 would be an ideal goal for you to shoot.

[quote="veezbo"] You said you were a senior, so its not likely you will be accepted to MOP, so the other "ambitious" goal is to become one of 12 winners.[/quote]

Aren't the two the same thing? I thought Black MOP consisted of the USAMO winners.

[quote="MellowMelon"]As a slight counterpoint to matt's post (or maybe we're just interpreting "advanced concepts" differently), anyone not familiar with the Chinese Remainder Theorem and not experienced enough with modular arithmetic to be able to rederive it would have stood no chance on last year's #1. There is a basic ground level of theory that you just have to know - it's just lower than most people seem to think.[/quote]
I beg to differ. I didn't make USAMO, but I tried day 1 as soon as the problems were out, and after 3 hours of very hard and very painful thinking I got the non-CRT solution for number 1, even though I had almost no knowledge of number theory, besides a basic idea of how mods work.
This is what I got, copied from last year's thread on that problem:
[hide]Problem 1

Firstly, we see that there are $ n \plus{} 1$ numbers $ k_i$. Since $ n\geq1$, there are at least 2 values of $ k_i$.
Define a relatively prime partition to be a set of relatively prime numbers such that their product is equal to some natural number.

Define $ P$ to be the greatest possible cardinality of a relatively prime partition for that number.

Lemma 1: All cardinalities of the relatively prime partitions of a number up to $ P$ can be attained.

Proof:
$ P \equal{} P \minus{} 0$ satisfies the properties. For any $ P \minus{} n$ which satisfies the properties, we can take any of the 2 $ P \minus{} n$ numbers and multiply them together. Because they are both relatively prime to all the other numbers, their product is relatively prime to all the other numbers as well, and that results in $ P \minus{} n \minus{} 1$ numbers which satisfy the conditions, unless $ P \minus{} n \equal{} 1$, because there is only one number left. Therefore, all numbers of numbers of relatively prime factors from $ 1$ to $ P \minus{} 1$ are attainable, if $ P$ is attainable as well.

End Lemma

Lemma 2: $ a^2 \plus{} a \plus{} 1$ can have arbitrarily many prime factors for some value of $ a$.

Proof:
Let $ f(x) \equal{} x^2 \plus{} x \plus{} 1$. Let $ a^2 \plus{} a \plus{} 1 \equal{} k$. Then $ f(a \plus{} k) \equal{} a^2 \plus{} 2ak \plus{} k^2 \plus{} a \plus{} k \plus{} 1 \equal{} 2ak \plus{} k^2 \plus{} 2k$.
After factoring, we get $ k(k \plus{} 2 \plus{} 2a) \equal{} (a^2 \plus{} a \plus{} 1)(a^2 \plus{} 3a \plus{} 3)$.
$ a^2 \plus{} 3a \plus{} 3 \geq a^2 \plus{} a \plus{} 1$, because $ a$ is strictly positive.
Let their GCD=g.
$ g|2a \plus{} 2$
$ g|2a^2 \plus{} 2a$
$ g|a^2 \minus{} 3a \minus{} 3 \plus{} a^2 \plus{} 3a \plus{} 3, g|2a^2$
$ g|2a$, and so $ g|2$, but $ g|a^2 \plus{} a \plus{} 3$ which is strictly odd, so g=1.
$ (a^2 \plus{} a \plus{} 1)(a^2 \plus{} 3a \plus{} 3)$ therefore must contain prime factors not in $ a^2 \plus{} a \plus{} 1$. Upon repeating this an arbitrary number of times, we get a number of the form $ a^2 \plus{} a \plus{} 1$ which has arbitrarily many distinct prime factors.

End Lemma

Then $ a(a \plus{} 1) \plus{} 1 \equal{} a^2 \plus{} a \plus{} 1$ would not have $ n \plus{} 1 \equal{} P$ elements in the relatively prime partition for some value of n, and any value of a.
It is possible to choose a value of $ a$ such that $ a^2 \plus{} a \plus{} 1$ has arbitrarily many unique prime factors, by our second Lemma, and so it is possible for $ P$ to be arbitrarily high. By our first Lemma, all numbers up to P are possible values for the cardinality of some relatively prime partition, and so there always exists some number that has an arbitrary number of elements in a relatively prime partition. Because $ a^2 \plus{} a \plus{} 1 \equal{} a(a \plus{} 1) \plus{} 1$, $ a(a \plus{} 1)$ is the product of 2 consecutive integers, we see that the given statement is true because if $ k_0k_1...k_n \equal{} a(a \plus{} 1) \plus{} 1$, then $ k_0k_1...k_n \minus{} 1 \equal{} a(a \plus{} 1)$.
[/hide]

With that said, knowing Chinese Remainder Theorem definitely kills the problem much much faster. I do like this solution more though because it is easier for someone who doesn't know math to understand (and because I came up with it :P). I suppose it does show that experience is pretty darn important if you want to have time to do more problems than just number 1.

Alternatively, look at [url=http://www.artofproblemsolving.com/Wiki/index.php/2008_USAMO_Problems/Problem_1]here[/url]. Solution 1 was the factorization method to which I referred, though it does use Euclidean Algorithm.

Well basically everything comes down to proving that those 2 factors are relatively prime, and you probably can figure Euclidean algorithm out from trying to get that their GCD is 1.

did you make it this year serialkllr? if i remember you were pretty close last year..

Yea, I made it this year (132/8, or 144/8 with amc10), last year I got 127.5/7 (bubbled 2 wrong on amc wth). I'm a sophmore currently.

Blah, for some reason I thought the factorization solution (which I was aware of) needed CRT. Maybe not. Throw CRT out of the list of "things you need to know", though I do still claim that list exists.


Each of the Fion brothers has the same number of sisters and brothers, but each of the Fion sisters has twice as many brothers as sisters. What is the total number of children in the family?

[hide="Solution"]
So we have the two equations (where $ b\equal{}boys$ and $ g\equal{}girls$):

$ b\minus{}1\equal{}g$

$ 2(g\minus{}1)\equal{}b$

Substituting:

$ 2(b\minus{}1\minus{}1)\equal{}b$
$ 2(b\minus{}2)\equal{}b$
$ 2b\minus{}4\equal{}b$
$ b\equal{}4$

We plug that back into the first equation:

$ 4\minus{}1\equal{}g$
$ g\equal{}3$

And then we add them to get:

$ 3\plus{}4\equal{}\boxed{7}$

[/hide]


integration
logarithms
real analysis
real analysis unsolved

$ {\text{Express }}\sum\limits_{n \equal{} 1}^\infty  {\sum\limits_{m \equal{} 1}^\infty  {\frac{1}
{{nm^2  \plus{} mn^2  \plus{} 2mn}}} } {\text{ as a rational number}}{\text{.}}$

Here is an elementary approach, the first thing that came to mind:

[hide]

Let $ f(m,n) \equal{} \frac {1}{m^2 n \plus{} m n^2 \plus{} 2mn} \equal{} \frac {1}{n(n \plus{} 2)}\left(\frac {1}{m} \minus{} \frac {1}{m \plus{} n \plus{} 2}\right)$.  Then

$ S \equal{} \sum_{n \equal{} 1}^\infty \sum_{m \equal{} 1}^\infty f(m,n) \equal{} \sum_{n \equal{} 1}^\infty \frac {1}{n(n \plus{} 2)} \sum_{n \equal{} 1}^\infty \left(\frac {1}{m} \minus{} \frac {1}{m \plus{} n \plus{} 2}\right)$,

for which the innermost sum is obviously a telescoping series.  This simplifies to

$ S \equal{} \sum_{n \equal{} 1}^\infty \frac {1}{n(n \plus{} 2)} \sum_{m \equal{} 1}^{n \plus{} 2} \frac {1}{m}$.

We proceed to interchange the order of summation:

$ S \equal{} \sum_{m \equal{} 1}^2 \frac {1}{m} \sum_{n \equal{} 1}^\infty \frac {1}{n(n \plus{} 2)} \plus{} \sum_{m \equal{} 3}^\infty \frac {1}{m} \sum_{n \equal{} m \minus{} 2}^\infty \frac {1}{n(n \plus{} 2)}$,

the splitting of the sum being necessary since the index of n is truncated below for $ m < 3$.  The first double sum is easy to evaluate:

$ \left(1 \plus{} \frac {1}{2}\right) \sum_{n \equal{} 1}^\infty \frac {1}{2}\left(\frac {1}{n} \minus{} \frac {1}{n \plus{} 2}\right) \equal{} \frac {3}{4} \left(1 \plus{} \frac {1}{2}\right) \equal{} \frac {9}{8}$.

The second sum is slightly trickier, but yields to the same telescoping technique after a shift of index on m:

$ \sum_{m \equal{} 3}^\infty \frac {1}{m} \sum_{n \equal{} m \minus{} 2}^\infty \frac {1}{n(n \plus{} 2)} \equal{} \sum_{m \equal{} 1}^\infty \frac {1}{m \plus{} 2} \sum_{n \equal{} m}^\infty \frac {1}{2}\left(\frac {1}{n} \minus{} \frac {1}{n \plus{} 2}\right)$,

which in turn yields

$ \frac {1}{2} \sum_{m \equal{} 1}^\infty \frac {1}{m \plus{} 2} \left(\frac {1}{m} \plus{} \frac {1}{m \plus{} 1}\right) \equal{} \frac {1}{2} \sum_{m \equal{} 1}^\infty \frac {1}{2}\left(\frac {1}{m} \minus{} \frac {1}{m \plus{} 2}\right) \plus{} \left(\frac {1}{m \plus{} 1} \minus{} \frac {1}{m \plus{} 2}\right)$,

again conveniently telescoping.  This gives the value of the second part of $ S$ as

$ \frac {1}{2} \left( \frac {3}{4} \plus{} \frac {1}{2} \right) \equal{} \frac {5}{8}$.

Therefore,

$ S \equal{} \frac {9}{8} \plus{} \frac {5}{8} \equal{} \frac {7}{4}$.

[/hide]

Of course, to be completely rigorous, one must show $ S$ is convergent, but I leave this as an exercise for the reader.

[quote="sos440"]\begin{eqnarray*} & & \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac {1}{nm^2 + mn^2 + 2mn} \\
& = & \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac {1}{mn}\frac {1}{m + n + 2} \\
& = & \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac {1}{mn}\int_{0}^{\infty} e^{ - (m + n + 2)x} \, dx \\
& = & \int_{0}^{\infty} \left( \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac {e^{ - mx}}{m}\frac {e^{ - nx}}{n} \right) e^{ - 2x} \, dx \\
& = & \int_{0}^{\infty} \left( \sum_{n = 1}^{\infty} \frac {e^{ - nx}}{n} \right)^2 e^{ - 2x} \, dx \\
& = & \int_{0}^{\infty} ( - \ln(1 - e^{ - x}))^2 e^{ - 2x} \, dx \\
& = & \int_{0}^{\infty} e^{ - 2x} \ln^{2}(1 - e^{ - x}) \, dx \\
& = & \left[ \frac {1}{2}(1 - e^{ - 2x}) \ln^{2}(1 - e^{ - x}) \right]_{0}^{\infty} - \int_{0}^{\infty} (1 + e^{ - x}) e^{ - x} \ln(1 - e^{ - x}) \, dx \\
& = & \int_{0}^{1} (t - 2) \ln t \, dt \quad \quad (\text{where } t = 1 - e^{ - x}) \\
& = & \frac {7}{4}. \end{eqnarray*}[/quote]

Very nice!  :clap:  :first:

[quote="atomicwedgie"]Here is an elementary approach, the first thing that came to mind:

[hide]

Let $ f(m,n) = \frac {1}{m^2 n + m n^2 + 2mn} = \frac {1}{n(n + 2)}\left(\frac {1}{m} - \frac {1}{m + n + 2}\right)$.  Then

$ S = \sum_{n = 1}^\infty \sum_{m = 1}^\infty f(m,n) = \sum_{n = 1} \frac {1}{n(n + 2)} \sum_{n = 1}^\infty \left(\frac {1}{m} - \frac {1}{m + n + 2}\right)$,

for which the innermost sum is obviously a telescoping series.  This simplifies to

$ S = \sum_{n = 1}^\infty \frac {1}{n(n + 2)} \sum_{m = 1}^{n + 2} \frac {1}{m}$.

We proceed to interchange the order of summation:

$ S = \sum_{m = 1}^2 \frac {1}{m} \sum_{n = 1}^\infty \frac {1}{n(n + 2)} + \sum_{m = 3}^\infty \frac {1}{m} \sum_{n = m - 2}^\infty \frac {1}{n(n + 2)}$,

the splitting of the sum being necessary since the index of n is truncated below for $ m < 3$.  The first double sum is easy to evaluate:

$ \left(1 + \frac {1}{2}\right) \sum_{n = 1}^\infty \frac {1}{2}\left(\frac {1}{n} - \frac {1}{n + 2}\right) = \frac {3}{4} \left(1 + \frac {1}{2}\right) = \frac {9}{8}$.

The second sum is slightly trickier, but yields to the same telescoping technique after a shift of index on m:

$ \sum_{m = 3}^\infty \frac {1}{m} \sum_{n = m - 2}^\infty \frac {1}{n(n + 2)} = \sum_{m = 1}^\infty \frac {1}{m + 2} \sum_{n = m}^\infty \frac {1}{2}\left(\frac {1}{n} - \frac {1}{n + 2}\right)$,

which in turn yields

$ \frac {1}{2} \sum_{m = 1}^\infty \frac {1}{m + 2} \left(\frac {1}{m} + \frac {1}{m + 1}\right) = \frac {1}{2} \sum_{m = 1}^\infty \frac {1}{2}\left(\frac {1}{m} - \frac {1}{m + 2}\right) + \left(\frac {1}{m + 1} - \frac {1}{m + 2}\right)$,

again conveniently telescoping.  This gives the value of the second part of $ S$ as

$ \frac {1}{2} \left( \frac {3}{4} + \frac {1}{2} \right) = \frac {5}{8}$.

Therefore,

$ S = \frac {9}{8} + \frac {5}{8} = \frac {7}{4}$.

[/hide]

Of course, to be completely rigorous, one must show $ S$ is convergent, but I leave this as an exercise for the reader.[/quote]

 :coolspeak:  :clap2:

Generally, $ \sum_{i\equal{}1}^{\infty}\sum_{j\equal{}1}^{\infty}\frac{1}{i^2 j\plus{}ij^2\plus{}nij} \equal{} \frac{2}{n} \sum_{k\equal{}1}^{n} \frac{H_k}{k} \equal{} \frac{1}{n} (H_{n,1}^2 \plus{} H_{n,2})$.

Where $ H_{n,m}$ is the generalized harmonic number, which is given for positive integer $ n$ by $ H_{n,m} \equal{} \sum_{k\equal{}1}^{n} \frac{1}{k^m}$


[quote="TTsphn"]Let n is an positive integer .
Find all function $ f: [0, \plus{} \infty)\to [0, \plus{} \infty)$ such that
$ f(\sqrt [n]{x^n \plus{} y^n}) \equal{} f(x) \plus{} f(y)$[/quote]

So $ f(\sqrt [n]{x \plus{} y}) \equal{} f(\sqrt[n]{x}) \plus{} f(\sqrt[n]{y})$

So $ f(\sqrt[n]{x})$ is solution of the additive Cauchy equation.

So $ f(x)\equal{}h(x^n)$ where $ h(x)$ is any solution of Cauchy equation $ h(x\plus{}y)\equal{}h(x)\plus{}h(y)$

For example, continuous solutions are $ f(x)\equal{}a\cdot x^n$


$A={a_{1},a_{2},...,a_{n}}$ ma rog ... nush de ce n-apare ca multime, dar e o multime din astea n elemente
f:A---->A ; f - bijectiva
am uitat sa zic ca : $a_{1}<a_{2}<...<a_{n}$
stiind ca $a_{1}+f(a_{1})<a_{2}+f(a_{2})<...<a_{n}+f(a_{n})$, sa se arate ca $f(a_{k})=a_{k}$, oricare ar fi k , de la 1 la n ...

stiu ca e simpla si clasica dar chiar nu imi dau seama

Presupunem ca $f^{-1}\left( a_{1}\right) = a_{j}$, pentru un $j \geq 2$. Pentru orice $k \in \overline{1,j-1}$ avem $a_{k}+f \left( a_{k}\right) < a_{j}+f \left( a_{j}\right) = a_{j}+a_{1}$. Din moment ce $a_{k}\geq a_{1}$, trebuie neaparat sa fie adevarata inegalitatea $f \left( a_{k}\right) < a_{j}$. Cum exista exact $j-1$ numere mai mici decat $a_{j}$, inseamna ca $f \left( \left\{ a_{1},\ldots,a_{j-1}\right\}\right) = \left\{ a_{1},a_{2},\ldots,a_{j-1}\right\}$. Acest lucru nu este posibil deoarece $a_{1}$ este deja imaginea unui alt numar ($a_{j}$). Contradictie. Rezulta ca $f^{-1}\left( a_{1}\right) = a_{1}$ si acum putem reduce numarul variabilelor la $n-1$. Dup-aia aplicam aceiasi pasi si concluzia ar trebui sa iasa.

Sper ca n-am gresit. :maybe:


trigonometry
limit
geometry
geometric transformation
rotation
function
real analysis

Consider the following differential equation:

$y''+\sin y=0;\,y(0)=0, \,y'(0)=A$ where $A>0.$

Find $\inf\{A: \lim_{t\to\infty}y(t)=\infty\}$

Hint: on another recent topic hin this forum, jmerry said:
[quote="jmerry"]Physical insights can be useful on strictly mathematical problems.[/quote]

I think
\[ \inf\{A: \lim_{t\to\infty}y(t)=\infty\}=2  \]

Actually, Nené's answer is correct. I would like to see both Nené explain how he got that and heartwork explain his objections to that.

Dear Kent I have followed this line of reasoning:
The equation $y''+\sin y=0$ is  the equation of a pendulum of lenght l equal to the modulus of gravity acceleration $\vec{ g}$. The pendulum must flip over the top, then the total energy must be greater then potential energy at the top. The pendulum motus is a no uniform rotation about a horizontal axe. Then
\[ \frac12ml^2A^2>2mgl\Rightarrow A>2  \]
and then the result I am looking for.

Here's my version of it. In essence, this is the same as the argument Nené used; the differences are matters of language.

Consider the quantity $E=\frac12(y')^2-\cos y+1$

(Yes, this is total energy. I used knowledge of physics to discover what I wanted this function to be, but with that in place, there will be no mention of physics for the rest of the problem.)

Then $\frac{dE}{dt}=y'y''+(\sin y)y'=y'(y''+\sin y)=0$ as long as $y(t)$ is a solution to the differential equation. Hence, $E(t)$ is constant.

$E(t)=E(0)=\frac12A^2-1+1=\frac12A^2.$

Now, can we ever have $y(t)=\pi$ ? If so, then $\frac12A^2=\frac12(y')^2+2$ at that point, or $(y')^2=A^2-4$

If $A<2,$ then this would be impossible; we can never have $y(t)=\pi,$ or for that matter, $y(t)=-\pi.$ In that case, the solution $y(t)$ remains in the interval $(-\pi,\pi)$ and is bounded.

However, if $A>2$ then for all $t,\,y'(t)\ge\sqrt{A^2-4}$ Hence $y(t)\ge t\sqrt{A^2-4}$ and tends to $\infty$ as $t\to\infty.$

That's enough to determine the infinum asked for; we don't have to decide what happens when $A=2.$ What does happen when $A=2$ is that $\lim_{t\to\infty}y(t)=\pi$ but $y(t)<\pi$ for all $t.$


Fie $A,B  \in M_2 (R)$ astfel incat $AB=BA$ si $det(A^2+B^2)=0$.
Aratati ca $detA=detB$.

A,B commute we have $det(A+iB)det(A-iB)=0$

$p(x)=det(A+xB)$ since the entries are reals $p(i)=p(-i)=0$
$p$ is polynomial degree  2
$p(x)=det(A)+qx+det(B)x^2$ where q is real

$p(i)=det(A)-det(B) +iq=0$ this gives $q=0$

and $det(A)=det(B)$


This is pretty easy actually, we just have to divide it in 2 parts: (a) 1/3>= xyz(x+y+z)/(x^2+y^2+z^2)(xy+yz+zx), which is easy to prove, just use x^2+y^2+z^2>=xy+yz+zx and the fact that ab+bc+ca<= ((a+b+c)^2)/3 for (a, b, c) = (xy, yz, zx).                                                                      (b)( 3^1/2)/9>= xyz((x^2+y^2+z^2)^1/2)/(x^2+y^2+z^2)(xy+yz+zx), where we just apply AM-GM.


inequalities integration calculus algebra polynomial function limit

Consider the following sequence:

$ a_k \equal{} \frac {k}{\sqrt [3]{(k \minus{} 1)^4} \plus{} \sqrt [3]{k^4} \plus{} \sqrt [3]{(k \plus{} 1)^4}}$

Prove that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} \cdots \plus{} a_{999} < 50$

I'm having trouble racionalizing the denominator, I tried working on another sequence:

$ \frac {k}{\sqrt [3]{(k \minus{} 1)^4} \plus{} \sqrt [3]{k^4} \plus{} \sqrt [3]{(k \plus{} 1)^4}}\; < \;\frac {k}{\sqrt [3]{(k \minus{} 1)^4} \plus{} \sqrt [3]{(k \plus{} 1)^4(k \minus{} 1)^4} \plus{} \sqrt [3]{(k \plus{} 1)^4}}$

But I'm having trouble with the denominator, how do you rationalize it? I tried using the factorization $ a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc \equal{} (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} ac \minus{} bc)$ but with no avail. Any help is appreciated.

Sorry, I haven't yet learned calculus, do you think you could solve this problem without the use of calculus?

Thank you for your solution.

How could I use telescopic cancelation on that? The denominator isn't a product..

Actually, there is no calculus involved, depending on your definition. For instance, you can directly prove that $ \Delta_k: \equal{}k^{\frac{1}{3}}\left[k^{\frac{2}{3}}\minus{}(k\minus{}1)^{\frac{2}{3}}\right]\geq\frac{2}{3}$ holds. If you have performed* this, then

$ a_1\plus{}\sum_{k\equal{}2}^{999}a_k\leq\frac{1}{1\plus{}2^{\frac{4}{3}}}\plus{}\sum_{k\equal{}2}^{999}\frac{1}{3k^{\frac{1}{3}}}\equal{}\frac{1}{1\plus{}2^{\frac{4}{3}}}\plus{}\sum_{k\equal{}2}^{999}\frac{k^{\frac{2}{3}}\minus{}(k\minus{}1)^{\frac{2}{3}}}{3\Delta_k}$ $ \leq\frac{1}{1\plus{}2^{\frac{4}{3}}}\plus{}\sum_{k\equal{}2}^{999}\frac{k^{\frac{2}{3}}\minus{}(k\minus{}1)^{\frac{2}{3}}}{2}\equal{}\frac{1}{1\plus{}2^{\frac{4}{3}}}\plus{}\frac{999^{\frac{2}{3}}\minus{}1}{2}$ $ <50$ as before.

*) For the proof of the $ \Delta_k$-inequality observe the identity $ (\Delta_k\minus{}k)^3\plus{}k^3\minus{}2k^2\plus{}k\equal{}0$ with corresponding polynomial function $ G(x)\equal{}(x\minus{}k)^3\plus{}k^3\minus{}2k^2\plus{}k$. Clearly, $ G$ is monotone increasing with limits $ \lim_{x\rightarrow\pm\infty}G(x)\equal{}\pm\infty$ and as a consequence the root $ \Delta_k$ is unique. Because of $ G(\frac{2}{3})\equal{}\frac{8\minus{}9k}{27}<0$ we have $ \Delta_k>\frac{2}{3}$

Thank you for the clarification! I have a question and a request. What is a monotone function, is it the same as a bijector function (does that word exist in english)?

Here is a solution that was posted today that I am trying to understand:

$ a_n<\frac{k}{\sqrt[3]{(k\minus{}1)^4}\plus{}\sqrt[3]{(k\plus{}1)^2(k\minus{}1)^2}\plus{}\sqrt[3]{(k\plus{}1)^3}}$

Rationalizing, we have:

$ a_n<\frac{k(\sqrt[3]{(k\plus{}1)^2}\minus{}\sqrt[3]{(k\minus{}1)^2})}{(k\plus{}1)^2\minus{}(k\minus{}1)^2}$

From which the rest is trivial. My question is, how did he rationalize that fraction??

[quote="triplebig"]What is a monotone function, is it the same as a bijector function (does that word exist in english)?[/quote] "Bijector" should be "bijective," meaning "invertible."  "Monotone" means "always increasing or always decreasing."

Substitute $ u_k\equal{}(k\plus{}1)^{\frac{2}{3}},v_k\equal{}(k\minus{}1)^{\frac{2}{3}}$, in order to check his rationalization $ \frac{k}{v_k^2\plus{}v_ku_k\plus{}u_k^2}\equal{}\frac{k(u_k\minus{}v_k)}{(v_k^2\plus{}v_ku_k\plus{}u_k^2)(u_k\minus{}v_k)}$ $ \equal{}\frac{k(u_k\minus{}v_k)}{u_k^3\minus{}v_k^3}$

How did I miss that..
Thank's a lot, appreciate it!


By the way,this was given at "Arhimede International Mathematics Competition"

I am struggeling here,HELP PLEASE!!!! :starwars:

[quote="Inequalities Master"]Let $ x,y$ be reals s.t. $ x^2\plus{}y^2\leq1$ and $ n$ a natural number.Prove that:
$ (x^n\plus{}y)^2\plus{}y^2\geq\dfrac{1}{n\plus{}2}(x^2\plus{}y^2)^n$[/quote]
Using the Cauchy-Schwarz inequality, we have \[(x^n+y)^2+\frac{2}{n+2}y^2 \ge \frac{(x^n+y-y)^2}{1+\frac{n+2}{2}}=\frac{2}{n+4}x^{2n}.\] Therefore, \[\begin{aligned} (x^n+y)^2+y^2&=\left[(x^n+y)^2+\frac{2}{n+2}y^2\right] +\frac{n}{n+2}y^2 \ge \frac{2}{n+4}x^{2n}+\frac{n}{n+2}y^2\\ &\ge \frac{2}{n+4}x^{2n}+\frac{n}{n+2} y^2(x^2+y^2)^{n-1}. \end{aligned}\] It suffices to proe that \[\frac{2}{n+4}x^{2n}+\frac{n}{n+2}y^2(x^2+y^2)^{n-1} \ge \frac{1}{n+2}(x^2+y^2)^2,\] or \[\frac{2n+4}{n+4}x^{2n} \ge (x^2+y^2)^{n-1} \big[x^2-(n-1)y^2\big].\] If $x^2 \le (n-1)y^2,$ then the inequality is obviously true. Otherwise, using the AM-GM inequality, we have \[\begin{aligned} (x^2+y^2)^{n-1}\big[x^2-(n-1)y^2\big] &\le \left[\frac{(n-1)(x^2+y^2)+x^2-(n-1)y^2}{n}\right]^n\\ &=x^{2n} \le \frac{2n+4}{n+4} x^{2n}.\end{aligned}\] This ends our proof.


group theory
abstract algebra
vector
binomial coefficients
superior algebra
superior algebra unsolved

Was going through a textbook and I found a question regarding the number of subgroups of a direct product.
Are there $ p^n\minus{}1$ subgroups of order $ p$ (prime) in $ G\equal{}\mathbb{Z}/p\mathbb{Z} \oplus \hdots \oplus \mathbb{Z}/p\mathbb{Z}$? There are $ n$ copies.

Every nonzero element generates such a group. Conversely, every group is archieved that way. But note that you counted every subgroup $ p\minus{}1$ times. So the total number of them is $ \frac{p^n\minus{}1}{p\minus{}1}$.

Equivalently, $ \mathbb{F}_p^n$ is a vector space of dimension $ n$ and the subgroups of order $ p$ are precisely the subspaces of dimension $ 1$.  The generalization is provided by the [url=http://mathworld.wolfram.com/q-BinomialCoefficient.html]q-Binomial coefficients[/url].


Hi, I was wondering if you guys can help me with this problem. I've been trying to do it for a while, but I've been getting no where? Is there some trick to doing this other than the long way? Well here it is. Thanks again

[img]http://img58.imageshack.us/img58/3118/42755334hk6.gif[/img]

Rationalize the denominator of each term of the sum... That'll help.

[hide]$ \frac {1}{\sqrt {1} \plus{} \sqrt {3}} \equal{} \frac {\sqrt {3} \minus{} \sqrt {1}}{(\sqrt {1} \plus{} \sqrt {3})(\sqrt {3} \minus{} \sqrt {1})} \equal{} \frac {\sqrt {3} \minus{} \sqrt {1}}{3 \minus{} 1}$[/hide]

Hmm, can you expand it to give me the solution?

After rationalizing we have

$ \frac{1}{\sqrt{1}\plus{}\sqrt{3}} \equal{} \frac{\sqrt{3}\minus{}\sqrt{1}}{2},\\ \frac{1}{\sqrt{3}\plus{}\sqrt{5}} \equal{} \frac{\sqrt{5}\minus{}\sqrt{3}}{2},\\ \frac{1}{\sqrt{5}\plus{}\sqrt{7}} \equal{} \frac{\sqrt{7}\minus{}\sqrt{5}}{2},\\ \frac{1}{\sqrt{7}\plus{}\sqrt{9}} \equal{} \frac{\sqrt{9}\minus{}\sqrt{7}}{2}.$

Hence,

$ \frac{1}{\sqrt{1}\plus{}\sqrt{3}} \plus{} \frac{1}{\sqrt{3}\plus{}\sqrt{5}} \plus{} \frac{1}{\sqrt{5}\plus{}\sqrt{7}} \plus{} \frac{1}{\sqrt{7}\plus{}\sqrt{9}} \equal{}\\ \left(\frac{\sqrt{3}\minus{}\sqrt{1}}{2}\right)\plus{}\left(\frac{\sqrt{5}\minus{}\sqrt{3}}{2}\right)\plus{}\left(\frac{\sqrt{7}\minus{}\sqrt{5}}{2}\right)\plus{}\left(\frac{\sqrt{9}\minus{}\sqrt{7}}{2}\right) \equal{}\\ \frac{\sqrt{9}\minus{}\sqrt{1}}{2}.$

Hope it helps...

Ah so basically the answer is 1. But what is rationalizing and how do you do it?

Take a look at the following site:

 [url]http://www.math.unt.edu/mathlab/emathlab/How%20to%20Rationalize%20the%20Denominator%20of%20a%20Fraction.htm[/url]

The process I mentioned is explained there...  :wink:

hmm.. I understand it a little, but can you show me how to rationalize the third one for example, 1/(radical 5 + radical 7), write out like the complete process please. thanks a lot

OK:

$ \frac {1}{\sqrt {7} \plus{} \sqrt {5}} \equal{} \left(\frac {1}{\sqrt {7} \plus{} \sqrt {5}}\right)\left(\frac {\sqrt {7} \minus{} \sqrt {5}}{\sqrt {7} \minus{} \sqrt {5}}\right) \equal{} \frac {(1)(\sqrt {7} \minus{} \sqrt {5})}{(\sqrt {7} \plus{} \sqrt {5})(\sqrt {7} \minus{} \sqrt {5})} \\
\equal{} \frac {\sqrt {7} \minus{} \sqrt {5}}{\sqrt {7}\sqrt {7} \minus{} \sqrt {7}\sqrt {5} \plus{} \sqrt {5}\sqrt {7} \minus{} \sqrt {5}\sqrt {5}} \\
\equal{} \frac {\sqrt {7} \minus{} \sqrt {5}}{7 \minus{} 5} \\
\equal{} \frac {\sqrt {7} \minus{} \sqrt {5}}{2}.$

Are things clearer now?

Understand it completely, thanks a bunch man.

Mhmm try this as well :). It is very easy to remember and most useful.

1) Rationalize  $ \frac{1}{\sqrt{a}\minus{}\sqrt{b}}$

2) Rationalize $ \frac{1}{\sqrt{a}\plus{}\sqrt{b}}$


calculus
integration
limit
trigonometry
calculus computations

Evaluate

\[\lim_{x\rightarrow a} \frac{1}{x-a} \int^{\sqrt{x}}_{\sqrt{a}} te^t\sin t dt\ (a>0)\]

It is like derivate

 $\frac{1}{2}e^{\sqrt a}\sin{\sqrt a}$


calculus
integration
LaTeX
function
logarithms
calculus computations

Hi I ran into some confusion when trying to integrate the following indefinte integral:

Int of  [(dx)/x(log10x)]

that is log base10 x ^

now Im not sure if this is correct but I simplified that to:

Int of ln10(xlnx)^-1 dx

At this point I am lost any hints are appreaciated.  Thanks

P.s. sorry I am still shaky w/ the LaTeX

Let u = log x
du = dx/(xln10)
du(ln10) = dx/x

so the function becomes

Integral of du(ln10)/u

ln10(ln u) + C

[quote="kunny"]To illcrowflu:Is this your problem?

$\int \frac{1}{x\log 10x}dx$[/quote]
I think it may be $\int \frac{1}{x\log_{10} x}dx$

[quote="liyi"]
I think it may be $\int \frac{1}{x\log_{10} x}dx$[/quote]

Thank you, liyi.

$\int \frac{1}{x\log_{10}x}dx=\int \frac{1}{x\cdot \frac{ln x}{ln {10}}}dx$

$=\ln {10}\cdot \int \frac{1}{x\ln x}dx=\ln {10}\cdot \int \frac{1}{\ln x}(\ln x)' dx=\ln {10}\cdot \ln |\ln x| +Const.$


Alicia earns $ \$20$ per hour, of which $ 1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?

$ \textbf{(A)}\ 0.0029\qquad
\textbf{(B)}\ 0.029\qquad
\textbf{(C)}\ 0.29\qquad
\textbf{(D)}\ 2.9\qquad
\textbf{(E)}\ 29$

[hide="no calculations"]
because all the answer choices differ by a power of 10, just use common sense....
by looking at the numbers, you can get E without any calculations....
[/hide]

[quote="#H34N1"]Alicia earns $\$20$ per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?

A. $0.0029$
B. $0.029$
C. $0.29$
D. $2.9$
E. $29$[/quote]

[hide="solution"]The answer choises are all $29\cdot 10^{n}$.  To find how many cents she pays, we divide 20 by about 100 to get .20 dollars or 20 cents.  Therefore it is 29.[/hide]

[hide="Rounding"]$1\%$ of $\$20$ is 20 cents so then the answer must be $E, 29 \text{cents}$[/hide]

[quote="rohit.is.the.lebron23"]THat was easy. Who did not ge that :rotfl:[/quote]

maybe you because you didn't post a solution?

$20 \times 0.0145 = 29$ or E

Yeah, considering you're the one who goes "that does not make sense" on 2001 AMC 10 #2.
As for why I'm not posting a solution, I suppose it's because it'd be sort of redundant... :oops:

[quote="#H34N1"]Alicia earns $\$20$ per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?

A. $0.0029$
B. $0.029$
C. $0.29$
D. $2.9$
E. $29$[/quote]
[hide="Answer"]It is obviously E. You don't even have to work it out. 1.45% would be around 1/100. That's about 20 cents. 29 cents is the closest one, so it is E.[/hide]

[quote="mathmaster46"][quote="#H34N1"]Alicia earns $\$20$ per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?

A. $0.0029$
B. $0.029$
C. $0.29$
D. $2.9$
E. $29$[/quote]
[hide="Answer"]It is obviously E. You don't even have to work it out.[/hide][/quote]And since it was obvious, why did you even bother posting? It just makes your post as an obvious spam :spam:

Yes.  Spam such as that (answer without solution) will be deleted in the future.  (this is a warning).


I typed the wrong answer to problem 20 and now when I hit try again the input box for entering the solution is gone.  Something needs to get reset - or the scripting for the 2nd try on problem 20 is wrong.

(I restarted my browser to be sure it was not on my own computer)

Thanks

Browser? Operating System? What do you mean by problem 20?


[b]Solution.[/b] The problem is equivalent to simplifying $S=\sqrt[3]{\frac{\sum_{i=1}^{n}i2i4i}{\sum_{i=1}^{n}i3i9i}}$.  By simplifying the numerator and denominator, we obtain
\[S=\sqrt[3]{\frac{8\sum_{i=1}^{n}i^{3}}{27\sum_{i=1}^{n}i^{3}}}=\frac{2}{3}.\]
And the conclusion follows.$\Box$


Let $ A$ be the set of all  $ 7$-lettered strings which has letters that are distinct. Find the probability that a randomly chosen string from $ A$ has a $ 3$-letter sub-string which has letters that are consecutive (wrapping included). Note: A string is a sequence of letters.

Examples:

$ ABCDEFG$ - this contains several sub-strings with consecutive letters, namely: $ ABC$, $ BCD$, $ CDE$, $ DEF$, $ EFG$
$ YZATALO$ - this contains a single substring with consecutive letters: $ YZA$ (This is what wrapping means. If you place $ A$ after $ Z$, it would be considered consecutive.)

[quote="TachyonPulse"]Let >>  $ A$ << be the set of all  $ 7$-lettered strings which has letters that are distinct. Find the probability that a randomly chosen string from $ A$ has a $ 3$-letter sub-string which has letters that are consecutive (wrapping included). Note: A string is a sequence of letters.

[/quote]

You need to use a character that is *different* from one of the
regular 26 capital English letters of the alphabet to name your set,
because you're already using the capital letters in your strings.

Or, use the capital letter to name the set, and then use lower
case letters exclusively in your seven-lettered strings, as a
possibility.

Does something backwards, like ASDFZYX (in this case ZYX), count?

I can't edit my post anymore, so here's the edited problem:

Let $ \lambda$ be the set of all  $ 7$-lettered strings which has letters that are distinct. Also, the letters in each sequence are all capital. Find the probability that a randomly chosen string from $ \lambda$ has a $ 3$-letter sub-string which has letters that are consecutive (wrapping included).

@dragon96: Yes that counts too, I forgot to make an example of that.

I would use complementary counting + PIE (which is probably the bad way), but I don't think this is hsb, so I'll move to hsi.

[hide="my try"]
There are $ 26\cdot 25\cdot 24\minus{}26\cdot 3!$  3-letter strings which don't consist of 3 consecutive letters so the probability of a random 3-letter string not to consist of consecutive letters is

$ \frac{26\cdot 25\cdot 24\minus{}26\cdot 3!}{26\cdot 25\cdot 24}\equal{}\frac{99}{100}$.
7-letter string has $ 5$ 3-letter sub-strings, so the probability that none of them has consecutive letters is $ \left (\frac{99}{100}\right )^5$. Using complementary probability gives us $ 1\minus{}\left (\frac{99}{100}\right )^5\approx 4.901\%$[/hide]


When I run the Latex and build a PDF with it, it builds the file, but it doesn't open with the error message:

[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]

Cannot execute the command.

on a popup.  I suspect that the problem is that I downloaded adobe reader 8 which seemed to replace my adobe reader 7.  I can still open the document, but I have to open it from adobe reader (so this isn't such a big problem, just for convience sake).

I get this a lot. Just close texnic center and adobe, then re-open them and everything should work fine. Otherwise, there is an error in your document, in which case you need to locate the source of the problem (should be error messages when you compile your document).

Ok, it worked.  But is Texnic center supposed to automatically open Adobe reader as I never had to have it running before?  I know my document had no errors because I tried last years USAMTS round 4 solutions I sent in.

In the build menu, select "define output profile."  In the pop-up that appears, select the "Viewer" tab.  The first field is labeled Path of Executable.  I'm guessing that this path still says adobe reader 7.  It needs to be changed to the path for adobe reader 8.  Then you should be able to build a pdf without trouble, even if adobe reader is closed.

Ok, now I have another questions, what is a bad file descriptor?  It is not letting me build a new PDF document.  So after I take the tex file that I made for USAMTS round 4 last year (so we can assume it is error free) and type in "Rohit is mighty" on the first line of the first solution, it doesn't show up and opens as it the pdf file that was saved.  It has the error message below:

-------------------------------Output Profile: LaTeX=>PDF-------------------------
This is pdfeTeX, Version 3.141592-1.30.6-2.2 (MikTex 2.5)
entering extended mode
! undefined control sequence.
<*>C:/Shared
                       Documents\for school\High School Math Competitions\USAMTS\06-...
? pdflatex.exe: Bad File Descriptor

LaTeX Result: 1 Error(s), 0 Warning(s), 0 Bad Box(es), 0 Page(s)

Edit: I found a way to solve the problem temporarily.  I can switch the outprofile to DVI which somehow builds it in adobe reader but doesn't open because there is no DVI file or something.  Then I switch back to pdf and it opens.  But if I want to see my changes again, I have to do that all over again.

[quote="ehehheehee"]! undefined control sequence.
<*>C:/Shared
                       Documents\for school\High School Math Competitions\USAMTS\06-...
? pdflatex.exe: Bad File Descriptor[/quote]
Is this a path in your document? LaTeX sees it as a command which it isn't. If you need to use \ for any reason then use \textbackslash instead.

I don't think I see that anywhere in my document.  That path is where my latex file is located (with all its folders it is in), but that is not written on my document.  Latex worked for me last school year and the only difference I thought happened since then was that I downloaded Adobe Reader 8.  I took roadnottaken's suggestion and changed the path of executable to adobe reader 8 and it started opening it automatically like it should.  But then this error, which wasn't there before I changed the path started popping up.  And it isn't just this file, every file I create has this problem.

File paths with spaces in them are a pain and should be avoided. Make a new folder like c:\latex and put your tex file there then try compiling it and see what happens.

I did and the same error showed up with:

<*> C:\Tex
                  \solutions2.tex-interaction=nonstopmode C:\Tex\solutions2.tex
?pdflatex.exe: Bad file descriptor

With C:\Tex\solutions2.tex as my path.

Can you post a screenshot of the build profile? Build->Define Output Profiles->LaTeX=>PDF, (La)TeX tab. Maybe there's an error there somewhere.

Aha! A stray %pm (which is the current file name*) has crept into the second box. That's why the filename has appeared when compiling.
Change[code]%pm-interaction=nonstopmode "%pm"[/code]to[code]
-interaction=nonstopmode "%pm"[/code]and all should be well.

* In TexnicCenter's help:
[quote]%pm
Will be replaced by the full path of the current project's main file.
["C:\My Documents\TxcTest\JustATest.tex"][/quote]

It works.  Thanks.  Maybe they should make a LaTeX for dummies for people like me.


trigonometry
calculus
integration
logarithms
calculus computations

$ \begin{array}{l}
 I \equal{} \int {\tan } x(1 \plus{} \sec x)\;dx \equal{} \int {\left( {\tan x \plus{} \sec x\tan x} \right)dx}  \equal{}  \\
  \equal{}  \minus{} \ln \left| {\cos x} \right| \plus{} \sec x \plus{} c \\
 \end{array}$


1)Αν $ a\plus{}b\plus{}c\equal{}3$ και $ a,b,c>0$ να δείξετε ότι

$ a^2b^2c^2\ge{(3\minus{}2a)(3\minus{}2b)(3\minus{}2c)}$

2) Αν a,b,c  είναι πλεύρες τριγώνου τόε να δείξετε ότι

$ a^3\plus{}b^3\plus{}c^3\plus{}3abc\minus{}2b^2a\minus{}2c^2b\minus{}2a^2c\ge{0}$

και

$ 3a^2b\plus{}3b^2c\plus{}3c^2a\minus{}3abc\minus{}2b^2a\minus{}2c^2b\minus{}2a^2\ge{0}$

Ίσως όχι κα τόσο δύσκολες ασκήσεις  ασκήσεις

[quote="kritikos"]1)Αν $ a \plus{} b \plus{} c \equal{} 3$ και $ a,b,c > 0$ να δείξετε ότι

$ a^2b^2c^2\ge{(3 \minus{} 2a)(3 \minus{} 2b)(3 \minus{} 2c)}$

[/quote]

Λύση στην πρώτη.

Ετσι όπως είναι δοσμένη η ανισότητα νομίζω δεν ισχύει...πάρε για παράδειγμα το $ (0.1, 0.1, 2.8)$ αν δεν έχω κάνει κανα λάθος στις πράξεις... Μάλλον η ανισότητα στην οποία αναφέρεσαι είναι η:
$ a^2b^2c^2\ge(3 \minus{} 2a)^2(3 \minus{} 2b)^2(3 \minus{} 2c)^2$

Θέτω $ x\equal{}3 \minus{} 2a$, $ y\equal{}3\minus{}2b$, $ z\equal{}3\minus{}2c$ και άρα παίρνω $ x\plus{}y\plus{}z\equal{}3$ και $ a\equal{}\frac{3\minus{}x}{2}$, $ b\equal{}\frac{3\minus{}y}{2}$, $ c\equal{}\frac{3\minus{}z}{2}$

Αρα η ανισότητα γίνεται:

$ (\frac{3\minus{}x}{2})^2(\frac{3\minus{}y}{2})^2(\frac{3\minus{}z}{2})^2\geq x^2y^2z^2\Longleftrightarrow$

$ (y\plus{}z)^2(z\plus{}x)^2(x\plus{}y)^2\geq 64x^2y^2z^2$

το οποίο ισχύει απ' τη γνωστή ανισότητα $ (y\plus{}z)^2\geq 4yz$ αν την πάρουμε κυκλικά για τα αντίστοιχα ζευγάρια.

Ισχύει για (0.1,0.1,2.8) ξαναδές το.Αυτό που γράφεις νομίζω ότι δεν ισχύει ειδικά για (0.1,0.,2.8)

sry έχεις δίκιο... δεν ξέρω τι μου ήρθε και έβαλα αυτό το παράδειγμα...είναι τελείως ατυχές :blush:
Απέδειξα μια διαφορετική ανισότητα lol :P [/list]

Λοιπόν η ανισότητα είναι ισοδύναμη με

$ (a \plus{} b \plus{} c)^3( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c) \leq 27a^2b^2c^2$

Θέτω  $ x \equal{} \minus{} a \plus{} b \plus{} c, y \equal{} a \minus{} b \plus{} c, z \equal{} a \plus{} b \minus{} c$ .

Η μόνη περίπτωση που έχει ενδιαφέρον είναι αυτή στην οποία x,y,z >0  (οι άλλες είναι τελείως προφανείς)
Τότε θέλουμε νδο
$ 64xyz(x \plus{} y \plus{} z)^3 \leq 27(x \plus{} y)^2(y \plus{} z)^2(z \plus{} x)^2$

Όμως  $ 9(x \plus{} y)(y \plus{} z)(z \plus{} x) \geq 8(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)$ και  $ (xy \plus{} yz \plus{} zx)^2 \geq 3xyz(x \plus{} y \plus{} z)$.

Συνδιάζοντας τις δύο αυτές παίρνουμε το ζητούμενο


calculus
integration
limit
trigonometry
function
real analysis
real analysis unsolved

How do you justify something like
\[ \lim_{M\rightarrow\infty}\int_{0}^{\pi}e^{\minus{}M\sin\theta}\, d\theta \equal{}\int_{0}^{\pi}\lim_{M}e^{\minus{}M\sin\theta}\, d\theta \equal{} 0\]  ?

The concept is [b]uniform convergence[/b].

[b]Definition:[/b] Let $ \{f_{n}(x)\}$ be a sequence of functions on set (non-empty) $ S$ we say $ f_{n}(x)$ converges uniformly to a function $ f(x)$ (called the "limit") iff for any $ \epsilon >0$ there exists a $ N>0$ we have that if $ n>N$ this would imply $ |f_{n}(x)\minus{}f(x)|<\epsilon$ for all $ x\in S$.

[b]Theorem:[/b] The uniform limit of continous functions is continous. Meaning if $ f_{n}(x)$ are continous on $ S$ and $ f_{n}(x)$ converges uniformly to a limit function $ f(x)$ then $ f(x)$ must be too continous.

[b]Theorem:[/b] Let $ \{f_{n}(x)\}$ be a sequence of continous functions on $ [a,b]$. If $ f_{n}(x)$ converges uniformly to $ f(x)$ then we have:
$ \lim\ \int_{a}^{b}f_{n}(x) dx \equal{}\int_{a}^{b}f(x) dx$.

[b]Proof:[/b] Notice that $ f(x)$ is continous by first theorem so it means it is integrable on $ [a,b]$.
For $ \epsilon >0$ there exists an $ N>0$ so that:
$ |f_{n}(x)\minus{}f(x)|<\frac{\epsilon}{b\minus{}a}$ for all $ n>N$ and $ x\in [a,b]$.
Okay this means,
$ \left|\int_{a}^{b}f_{n}(x)dx\minus{}\int_{a}^{b}f(x) dx\right|\leq\int_{a}^{b}|f_{n}(x)\minus{}f(x)| dx\leq\int_{a}^{b}\frac{\epsilon}{b\minus{}a}dx \equal{}\epsilon$
This shows that $ \int_{a}^{b}f(x) dx$ is $ \lim\ \int_{a}^{b}f_{n}(x) dx$.

Uniform convergence on a finite interval is a sufficient condition, and a useful condition, and it applies to this example. (I would consider the theorem sylow_theory quoted as being central to the Riemann theory of integration.)

However: it's a lot to ask for. Sometimes you're integrating over an infinite domain. Sometimes the convergence isn't uniform - and yet you have a strong interest in this interchange and think it might be justified anyway.

There is a more comprehensive theory, and it lies at the heart of the Lebesgue theory of integration. Indeed, for me justifying such interchanges of limit and integral (or sum and integral, or integral and integral) is the main motivation for developing the Lebesgue theory in the first place.

Within the Lebesgue theory, the two most useful theorems are called the Dominated Convergence Theorem and the Monotone Convergence Theorem. Stated briefly:

1. The Monotone Convergence Theorem

If the sequence of functions $ f_{n}$ is nonnegative and nondecreasing at each point (That is, $ 0\le f_{n}(x)\le f_{n\plus{}1}(x)$ for all $ x$), then

$ \lim_{n}\int_{E}f_{n}(x)\,dx\equal{}\int_{E}(\lim_{n}f_{n}(x))\,dx$

(Note: this theorem allows the possibility that both sides could be $ \plus{}\infty.$)

2. The Dominated Convergence Theorem.

Suppose $ \lim_{n}f_{n}(x)\equal{}f(x)$ for almost every point $ x\in E.$ (That means that the set of exceptions to this has measure zero.) Suppose also that there is a function $ g$ such that for all $ x\in E$ and all $ n,$ $ |f_{n}(x)|\le g(x).$ Suppose also that $ \int_{E}g(x)\,dx$ is finite. Then

$ \lim_{n}\int_{E}f_{n}(x)\,dx\equal{}\int_{E}f(x)\,dx$

(I left out some hypotheses about measurability, and there are some slightly more general ways of stating these - but this catches the main points.)

----

The Dominated Convergence theorem does apply to arp's example, but we didn't need it, since we also have uniform convergence. But consider this:

$ \lim_{n\to\infty}\int_{0}^{\infty}\frac{\sin\left(\frac{x}{n}\right)}{\sqrt{x}(1\plus{}x^{2})}\,dx$

That tends zero. It could be proved in several ways, but the D.C.T. makes it easy.

We don't actually have uniform convergence here- we only have it away from zero and $ \pi$. If we use that argument, we need to slice off the endpoints and estimate them separately- which ends up looking a lot like how you prove the DCT.

Thanks both for the detailed replies.

One question though, which is probably stupid: how is my function uniform; i.e how can we choose $ N$ given $ \epsilon$ so that $ |e^{\minus{}M x}| <\epsilon$  for $ M > N$ and $ x\in (0,1]$.

Edit: jmerry got to it first... this is actually the reason i didn't know how to show the interchange orginally (I already knew about the uniform convergence stuff)

jmerry is right; we don't have uniform convergence in arp's original example. Which reinforces my point about uniform convergence being a lot to ask for.

A quick way to see that the convergence is not uniform: the pointwise limit of these functions is $ 1$ at $ 0$ and $ \pi$ but $ 0$ everywhere else; that means that the pointwise limit is discontinuous. But sylow_theory stated another theorem in his post: the uniform limit of a sequence of continuous functions is uniform. Hence this couldn't have been uniform convergence.

wait, so as a riemann integral the limit is still $ 0$, but it's just hard to show?  I'm going to think about how to estimate the ends like jmerry said

The Dominated Convergence Theorem does apply to the original example - or a more restricted corollary called the Bounded Convergence Theorem. But let's skip that and find a specific argument.

Claim: $ \sin\theta\ge\frac{2}{\pi}\theta$ for $ 0\le\theta\le\frac{\pi}{2}.$ Then:

$ \int_{0}^{\pi}e^{-M\sin\theta}\,d\theta= 2\int_{0}^{\frac{\pi}{2}}e^{-M\sin\theta}\,d\theta$

$ \le 2\int_{0}^{\frac{\pi}{2}}e^{-\frac{2M\theta}{\pi}\,d\theta}$

$ <2\int_{0}^{\infty}e^{-\frac{2M\theta}{\pi}\,d\theta}$

$ =\frac{\pi}{M}.$

Clearly, that tends to zero as $ M$ tends to infinity.

Nice, that's pretty much what i ended up doing too.


If the operation $F$ is applied to an odd integer, the result is one plus three times that integer.  When $F$ is applied to an even integer, the result is that integer divided by two.  What is the result of applying the operation $F$ 99 times in a row starting with the number 5?

[hide]When F is repeatedly applied to 5, the results look like this:

16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1...

So after applying F twice, the results repeat in cycles of 3. If you apply F n times, the remainder in (n-2)/3 will tell the answer. If n-2 is a multiple of 3, the answer is 1. If n-2 is one more than a multiple of 3, the answer is 4. If n-2 is 2 more than a multiple of 3, the answer is 2. 97 is one more than a multiple of 3, so after 99 applications of F to 5, the result is 4.[/hide]


$x,y,z\ge 0 ,x + y + z = 1$,find the max of
$s=x^2 y + y^2 z + z^2 x$

Max=$\frac{4}{27}$ when $x=0 y=\frac{1}{3},z=\frac{2}{3}$ and its permutations.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=77686 .

  darij


algebra
polynomial
binomial theorem
complex numbers
sum of roots
product of roots

All the roots of the equation $x^6 - 6x^5 + 15x^4 - ax^3 + 15x^2 - bx + 1 = 0$ are positive.  Find $a$ and $b$.

just institution. The answer is $a=10,\ b=-6.$ :D

ok...The funda behind the question can be put as :
Since all the roots of the eqn are +ve then we can have a,b>0
(using Descarte's Rule of signs)

Now i started by plugging the values in the eqn:
find f(1) we get -a-b+26=f(1)>0 therrefore (a+b)<26
find f(2) we get 4a+b < 86.5
find f(3) we get  9a+b< 207

Using the above eqns m getting 0< b <3.3
and 0 < a< 22

Now plz tell us that are a and b integers,then if they are then i guess we can get a soln by using the above method.....

There certainly is an easier and MUCH better and correct method than that Varun. ;)

[hide="really big hint"]
[size=59]AM-GM[/size]
[/hide]

[hide="hint"]
binomial theorem
[/hide]

[hide="simple solution"]
Notice that $(x-1)^6 = x^6 - 6x^5 + 15x^4 - 20x^3 +15x^2 - 6x + 1 = 0$ is a polynomial with root 1 occuring 6 times, which matches the form of the polynomial in the question.  Therefore $a = 20$ and $b = 6$.

[/hide]

This  reminds me of an old story from Mahabharata.

"All  the roots of this equation are real and positive" Guru Dronacharya thundered, and this thundring woke our hero the great  Bhima (who liked to sleep in the class) up. He hurried to copy this 20 degree equation from the  board, but he got only up to the first two terms before Guru D. wiped the black board clean. He however remembered that the constant term was 1. Thus the equation looked like:
$x^{20}-20x^{19} \dots +1 = 0$

Can you help, our hero (lazy as he may be) solve the equation and find all the roots. :)

An obvious solution would be to factorise to $(x-1)^20 = 0$ .. but are there any others!?

[quote]. but are there any others!?[/quote]
[hide]The key thing is to prove that there are no "others" ... Hint: See scorpius119 "really big hint" . [/hide]

[quote="shadysaysurspammed"]If $a,b,c,d,e,f....$ are the roots,then by AM-GM

$\frac{a+b+c+d+e....}{20}\geq 1$

So $1\geq 1$

So all the roots are equal to 1[/quote]

I don't see how that proves anything. If $b$ were $2$ and $a$ were $\frac12$, the arithmetic mean would still be greater than 1.

no,since L.H.S=R.H.S,hat implies that the roots are equal......equality is holding

I still don't quite understand...

You're saying that:

$(x-a)(x-b)(x-c)(x-d)\cdots=0$

implies that

$\frac{a+b+c+d+\cdots}{20} = \sqrt[20]{abcd\cdots}$?

Can you explain a bit more?

Or are you saying that because AM=GM, all roots are equal? I think I understand. Is it becuase the coefficient of $x^{19}$ is the sum of the roots so $a+b+c+d+\cdots=20$? If so, it'd have been clearer if you had explained your application of AM-GM a bit more.

Yes, The key point is that you are given:   [quote]all the roots are [b]real and[color=red] positive [/color][/b][/quote]
and  their  arithmetic mean = 1 (look at the coefficient of $x^{19}$)
and the geometric mean = 1     (look at the coefficient of $x^0$
But .... Arithmetic mean is always greater than geometric mean unless all the numbers are equal,  (and in that case AM=GM)
So here all the roots have to be equal.

(Remove the condition of them (the roots) being positive (or real)  and then one can not say any such thing)

[quote="surge"]The coefficient of the last term of a polynomial is the geometric mean of the roots? (as the arithmetic mean is the $x^{n-1}$ coefficient where $n$ is the degree of the polynomial?)[/quote]

Given that the $x^n$ term is $1$ (where $n$ is the degree of the polynomial), then the last term is the (signed) product of the roots and the $x^{n-1}$ coefficient is the (negative) sum of the roots.

Does the fact that a number is positive imply that it is real? For instance, would $i$ be considered a positive number or just complex? If it'd be considered positive, what would $1-i$ be? I ask this because the first question didn't specify real, only positive.

@surge: if you look at how that result is proven, it will be clear that it holds in general, where the general form is that the sum of the roots is $-\frac b a$ and the product is $\pm \frac za$ where $z$ is the last (constant) coefficient and the sign depends on whether $n$ is even or odd.

@Hamster: The complex numbers have no canonical linear ordering, so there is no equivalent concept to "positive."  So, using the word positive implies that you are working with some subset of the reals.  (Often people say "positive real," to be totally unambiguous.)

[quote="surge"]Given
$ax^2 + bx + c = 0$

The sum of the roots is $-b/a$ and the product of the roots is $c/a$.

Can this be applied to any polynomial (of a higher degree) where

$ax^n+bx^{n-1}...+cx^0=0$? (Or are there certain exceptions for higher degree polynomials)[/quote]

Yes you can but most of the time the coefficient of $x^n$ is kept $1$,so that's why sum of roots would be coefficient of $x^{n-1}$ and product of roots is the constant term in the end.

Generally a polynomial would be written like(lets take degree 4)

$x^4-(a+b+c+d)x^3+(ab+ac+bc+ad+dc+bd)x^2-(abc+abd+acd+bcd)x+abcd=(x-a)(x-b)(x-c)(x-d)$

Remember the alternating signs


Hello. I am trying to do well on the exams listed above but I have yet to be able to compile a list of algebraic manipulations useful in problems and in proofs even. Sorry if the name I have attributed to them seems unclear, but what I know as algebraic manipulations are the following.

$ a^{2}\minus{}b^{2}\equal{}(a\plus{}b)(a\minus{}b)$
$ a^{3}\minus{}b^{3}\equal{}(a\minus{}b)(a^{2}\plus{}ab\plus{}b^{2})$
$ \frac{1}{n(n\plus{}1)}\equal{}\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$

There are many others that I know, however they all are pretty basic. Are there any advanced ones?

Generalization of the first 2:

$ a^n\minus{}b^n\equal{}(a\minus{}b)(a^{n\minus{}1}\plus{}a^{n\minus{}2}b\plus{}\ldots\plus{}b^{n\minus{}1})$

$ a^n\plus{}b^n\equal{}(a\plus{}b)(a^{n\minus{}1}\minus{}a^{n\minus{}2}b\plus{}\ldots\plus{}b^{n\minus{}1})$, for $ n$ odd.

Practice. Knowing some methods will help, but not much.

Like the one that the above poster just gave, that one is no use if you don't know how to use it; the insight to know how to use these "methods" will come from practices. As you go higher in the hierarchy of math competitions, you soon will realize that there are less problems you can rely on special algebraic manipulation to solve.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=455307#455307]Here[/url] is an example of problem that has a methodical way of manipulation, but is not really something that everyone knows from head (a lot of people know about it AFTER solving the problem though).

Note that all these are special cases of more general formulas.


geometry
AMC
AMC 12
algebra
polynomial
probability
AMC 10

message from mathfanatic: I will return your post tomorrow.  Until then, it will sit in a word document on my computer.  I intend to enforce the no-discussion-of-scores-until tomorrow rule as strictly as possible.

Just in case, I'm going to lock this thread until tomorrow.  You can discuss the difficulty later.

I don't think anything will happen, but it's probably best if we play it safe.  Besides, you can cite specific questions later (i.e. #xx was harder than usual).

I don't know, but I thought this year's was easier than usual.  Like 21, 22, 23, 24, and 25 all were much easier than usual.  20 was a small challenge, and 19 was the only one I really had trouble on (possibly because of misreading the question).  I eventually had to leave #19 blank.

interesting... i solved 19 and 20 with little trouble, and didn't get any of 21-25 right (which I thought were much harder than usual)...

I suspect that part of the reason so many students felt that this year's exam was more difficult relates partially to the fact that Geometry is the most poorly developed curriculum in American mathematics education.  The last few problems on both tests were heavily geometry and the hardest problem (in my opinion) on the AMC12-A was number 24 with the segment and the region traced by the circle around it.

314159265358979323846 - did you think 19 and 20 were easy? i didnt answer either and though they were quite hard. my opinion was that 21 and 23 were quite easy, especially 23, a typical polynomial question. i spent a lot of time on 24 but couldnt get it.

19 was a matter of finding the correct right triangle. It was hard if you couldn't see the two triangles needed to solve the problem. Took me a couple minutes to get them.

#20- The two numbers could be either between 0 and .5 and between .5 and 1. There was a 1/2 chance that both would be in the same category, and a 1/2 chance that the conditions of the problem would be satisfied if they were in the same category. 1/2*1/2 = 1/4. If one was between 0 and .5 and the other one was between .5 and 1 (1/2 probability), it would always work. 1/4 + 1/2 = 3/4. I'm not sure if my reasoning is correct, but I didn't find it too difficult.

On the other hand, for some reason I had no idea how to do anything after 20. Now that I've read the solutions, 21 and 23 weren't too bad.

for the AMC 10, i thought most of them were easy and the only "difficult" ones were 21, 23, and 25 but i got 21 and 25 anyways and 23 turns out to be not too hard at all. but overall, definitely harder than the last three years...

i thought 20, 21, and 23 were pretty easy, but of course i ran out of time screwing around with 25.

[quote="confuted"][quote="Tare"]Way hard, compared to last years this year was really hard...at least for me, and this assumes I didn't become stupider :lol:[/quote]
You said you've never taken one [url=http://artofproblemsolving.com/Forum/viewtopic.php?p=44981#44981]here[/url]. How are you qualified to say anything about the difficulty of the test?[/quote]
I thought this poll was for the AMC 10...isn't it?


Find all polynomials $ P(x) \in R[x]$ such that :
                             $ P(x^{2}+1) =(P(x))^{2}+1$

These problem posted before. It is equavalent to $ P(P_{2}(x))=P_{2}(P(x))$, were $ P_{2}(x)=x^{2}+1$.
Obviosly if $ P(x)$ is solution, then $ P(P_{2}(x))$ is solution. For example $ P_{2}^{(k)}(x)=P_{2}(P_{2}^{(k-1)}(x))$ is solution.

Sorry I did a stupid mistake before :blush: .
We discuss $ 2$ cases:when $ P(x)$ has a fix point, i.e $ \exists x_{0}\in\mathbb{R}$, $ P(x_{0})=x_{0}$, and when it hasn't got one.
If it hasn't, observe that putting $ x=0$ we get $ P(1)\ge1$.
So we can assume $ P(x)>x$ for all real $ x$ (else we would have a fix point).
Put $ x\rightarrow-x$, we get $ P^{2}(x)=P^{2}(-x)\rightarrow (P(x)+P(-x))(P(x)-P(-x))=0$ for all $ x$. If there exists $ a$ so that $ P(a)+P(-a)=0$, we get a contradiction with $ P(a)>a$ and $ P(-a)>-a$, so we have $ P(x)=P(-x)$ for all $ x$.
Using [b]Rust[/b]'s idea now we observe that $ P_{2}^{(n)}$ are solutions with $ grad=2n$ and these are unique for every $ n$ (we can show this by brute-force, identifying the coef. ,if we write $ P(x)=\sum_{k=0}^{2n}a_{k}x^{k}$, if $ k$ is odd $ a_{k}=0$...)
Ok, now if we have a fix point, $ x_{0}$, we have $ x_{0}^{2}+1>x_{0}$(just put $ x_{0}$ in the initial equation) is a fix point, so we can construct an infinite sequence of fix points, so $ P(x)-x$ is the null polynomial $ \rightarrow P(x)=x$. So the solutions:
$ P(x)=x$
$ P_{2}^{(k)}(x)$


algebra
binomial theorem
algebra proposed

Let a_n = n + [Xn], where X = 1 +  :sqrt: 2. Show that a_n is odd for all positive integers n.

Let $U_n = (1+ \sqrt 2 )^n + (1- \sqrt 2 )^n.$
Then $U_0 = U_1 = 2$ and for all $n$, we have $U_{n+2} = 2U_{n+1} + U_n.$
It follows easily that $U_n$ is even for all $n$.

Now, since $-1 < 1- \sqrt 2 < 0$, it follows that :
- If $n$ is even $U_n - 1 \leq U_n - (1- \sqrt 2 )^2 = (1+ \sqrt 2)^n < U_n$ so that $[X^n] = U_n - 1.$
In that case $a_n = U_n + n - 1$ is odd.

-Similarly, if $n$ is odd $U_n = [X^n]$ so that $a_n = U_n + n$ is odd, and we are done.

Pierre.

I'm assuming [X^n] means the floor function.
First let $Y = \sqrt{2}-1$, note that $0 < Y^n < 1$
There are two cases to consider:
When n is even, X^n + Y^n is an even integer.
To see why, expand using binomial theorem and cancelling terms you get:
\[2(\sqrt{2}^n + \sqrt{2}^n-2 + ... + 1)\]
Thus [X^n] = X^n + Y^n - 1 and is odd.  So $a_n = n + [X^n]$ is odd.
When n is odd, similarly X^n - Y^n is an even integer.
Thus [X^n] = X^n - Y^n and is even.  $a_n = n + [X^n]$ again is odd.
Q.E.D.   :)

Edit: oops i'm too slow at typing :P


Show that an isolated singularity of f can not be a pole of e^f.

Thx for ur help

If $a$ is an isolated singularity of $f$ then $f(z) \rightarrow z_{0} \in \mathbb {C}$ when $z \rightarrow a$. Now assuming $a$ is a pole for $e^{f(z)}$, you can write $e^{f(z)}=\frac{g(z)}{(z-a)^{m}}$ where $m>0$ is an integer, and $g$ is analytic in a neighborhood of $a$ (with $g(a) \neq 0$). But then $|e^{f(z)}| \rightarrow +\infty$ as $z \rightarrow a$, meaning in particular $Re(f(z)) \rightarrow +\infty$, and $|f(z)| \rightarrow +\infty$ as $z \rightarrow a$. This is of course absurd.

If $f$ has a pole of order $n$ at $a$ and $e^f$ has a pole of order $m$ at $a$, then we get that

$\frac{d}{dz} f(z)$ has a pole of order $n+1$ at $a$ and

$\frac{d}{dz} e^{f(z)}$ has a pole of order $m+n+1$ at $a$.

We conclude that $n=0$.  But a pole of order $0$ means that $f$ is analytic and hence $e^f$ does not have a pole.  So $f$ does not have a pole at $a$.

We can exclude the case where $f$ has a removable singularity, so it follows that $f$ must have an essential singularity.  By Caseorati-Weierstrass, there are points arbitrarily close to $a$ which are mapped into points arbitrarily close to the origin which are mapped into points arbitrarily close to $1$ (in the complex plane).  So $e^f$ cannot have a pole (since then we would have to have points near $a$ getting sent to $\infty$)


[quote="piano1321"]A circular table is pushed into the corner of a square room so that a point P onthe edge of the table is 8" from one wall and 9" fromt he other wall.  Find the radius of the table in inches.

p.s. there should have been a picture to go with it, but as you all know, my scanner broke. :([/quote]

[hide]This is a mathcounts problem? I thought i saw it somewhere else.. oh well.

r=radius.
the line from the intersection of the circle with the square to the place where the rectangle forms (the one 8x9) is of length r-8. Same with the other one, it is r-9. We draw parallel lines inside the circle with length r-8 and r-9 with hypotenuse r and solve the equation $(r-8)^2+(r-9)^2=r^2$. We get $r^2-34r+145=0, r=29,5$ but 5 is obviously not the answer,so the answer is $\boxed{r=29}$[/hide]


Just idea.

$P=(x+y+xy)\left(\frac{1}{x+y}+\frac{1}{x}+\frac{1}{y}\right)$

or

Let $x+y=s,\ xy=t\Longrightarrow P=\frac{1}{s}+\frac{s}{t}$ under condition that $s+t=1,\ s>0,\ t>0,\ s^{2}-4t\geq 0$.

I can show that $s \ge 2(\sqrt{2}-1)$.
Then?

[quote="Aryabhatta"]Looks like the following will work:

Using kunny's notations $s = x+y$ and $t = xy = 1-s$ we have

$P = \frac{1}{s}+\frac{s}{1-s}= \frac{1}{s}+\frac{1}{1-s}-1 = \frac{1}{s(1-s)}-1$

Now we can easily show that $s \ge 2(\sqrt{2}-1)$.

If $s < 2(\sqrt{2}-1)$, then $4xy = 4(1-s) > 4(3-2\sqrt{2}) = (2(\sqrt{2}-1))^{2}> s^{2}= (x+y)^{2}$ which is not possible.

Also, for any $1 > s \ge 2(\sqrt{2}-1)$ we can find corresponding positive real $x,y$ which satisfy $x+y+xy = 1$

Also $2(\sqrt{2}-1) > \frac{1}{2}$

Since $s(1-s)$ is decreasing in $(2(\sqrt{2}-1), \infty)$ $P$ is increasing. The minimum value thus occurs at $s = 2(\sqrt{2}-1)$[/quote]
This solution isn't nice!

[quote="Aryabhatta"]Looks like the following will work:

Using kunny's notations $s = x+y$ and $t = xy = 1-s$ we have

$P = \frac{1}{s}+\frac{s}{1-s}= \frac{1}{s}+\frac{1}{1-s}-1 = \frac{1}{s(1-s)}-1$

Now we can easily show that $s \ge 2(\sqrt{2}-1)$.

If $s < 2(\sqrt{2}-1)$, then $4xy = 4(1-s) > 4(3-2\sqrt{2}) = (2(\sqrt{2}-1))^{2}> s^{2}= (x+y)^{2}$ which is not possible.

Also, for any $1 > s \ge 2(\sqrt{2}-1)$ we can find corresponding positive real $x,y$ which satisfy $x+y+xy = 1$

Also $2(\sqrt{2}-1) > \frac{1}{2}$

Since $s(1-s)$ is decreasing in $(2(\sqrt{2}-1), \infty)$ $P$ is increasing. The minimum value thus occurs at $s = 2(\sqrt{2}-1)$[/quote]
I think this solution wrong
Equality occurs when $x+y=2(\sqrt{2}-1)$ and $xy=3-\sqrt{2}$
So,$x;y$ are root of equation :$z^{2}-2(\sqrt{2}-1)+3-\sqrt{2}=0$
But $\Delta =-4\sqrt{2}<0$

Do Jonhs Friend take out those article whence so? According to body known it show yourself on a Viet Nam's underpinning high school mathematical magazine sheets . For Me who don' t know you held up by design but it lies In this May edition  .management ought to behave this cause immediately

[hide="this is rather ugly but still solution"]
Condition implies $y=\frac{1-x}{1+x}$ so expression to minimize is equal to
\[\frac{1}{x}+\frac{1+x}{1-x}+\frac{1+x^{2}}{1+x}\]
Now $0<x<1$, so let $k$ be the positive $\frac{1-x}{x}$. Then $x=\frac{1}{1+k}$, and the expression in terms of $k$ is
\[(1+k)+\frac{2+k}{k}+\frac{(k+1)^{2}+(k+1)}{(k+1)^{2}+1}=3+k+\frac{2}{k}+\frac{k}{k^{2}+2k+2}\]
It's rather interesting that $k+\frac{2}{k}$ gets minimized at the same value that $\frac{k}{k^{2}+2k+2}$ gets maximized: $k=\sqrt{2}$. So it makes sense that the minimum occurs when $k=\sqrt{2}$ which would mean a minimum of $\boxed{\frac{5+5\sqrt{2}}{2}}$. To show that this works, the difference can be factored as
\[\left( k-\sqrt{2}\right)^{2}\left( \frac{1}{k}-\frac{1}{\left(2+\sqrt{2}\right) (k^{2}+2k+2)}\right) \geq 0\]
as the second factor is clearly positive.

Having the minimum occur for $k=\sqrt{2}$ means $x=y=\sqrt{2}-1$.
[/hide]

[quote="chien than"]
I think this solution wrong
Equality occurs when $x+y=2(\sqrt{2}-1)$ and $xy=3-\sqrt{2}$
So,$x;y$ are root of equation :$z^{2}-2(\sqrt{2}-1)+3-\sqrt{2}=0$
But $\Delta =-4\sqrt{2}<0$[/quote]

if $x+y=2(\sqrt{2}-1)$ then $xy = 1-(x+y) = 1-2(\sqrt{2}-1) = 3-2\sqrt{2}$ (not $3-\sqrt{2}$)

It is easy to see that $x=y=\sqrt{2}-1$ satisfy the quadratic.

[quote="Jonhs"][quote="Jonhs"]Let $x;y$ be positive real numbers satisfying $x+y+xy=1$.Find min of:
$P=\frac{1}{x+y}+\frac{1}{x}+\frac{1}{y}$[/quote]
This is hard problem.
Nobody can do it :rotfl:[/quote]

I believe this has been solved (I think my solution is correct), though not upto a standard you would like.

This problem is not hard!

I post just answer; $\frac{5+5\sqrt{2}}{2}\ (x=y=\sqrt{2}-1)$. :D

Good luck!

[hide="better solution"]
The condition implies $(1+x)(1+y)=2$, so by AM-GM on $1+x,1+y$, we get $x+y\geq 2\sqrt{2}-2$, and then $xy\leq 3-2\sqrt{2}$ follows. Now we have the inequality
\[\frac{1}{x}+\frac{1}{y}+\frac{4}{x+y}=\frac{x+y}{xy}+\frac{4}{x+y}\geq \frac{4}{\sqrt{xy}}\]
by AM-GM, so that
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{x+y}\geq \frac{3}{4}\cdot \frac{x+y}{xy}+\frac{1}{\sqrt{xy}}\]
Now use $x+y\geq 2\sqrt{2}-2$ and $xy\leq 3-2\sqrt{2}$ to finish.
[/hide]

Here is my solution.

Let $x+y=s,\ xy=t$, we have $s+t=1\Longleftrightarrow t=1-s$.

$\boxed{\left\{ \begin{array}{ll}x>0 &\quad \\ y>0 &\quad \end{array}\right. \Longleftrightarrow \left\{ \begin{array}{ll}x+y>0 &\quad \\ xy>0 &\quad \\ Discriminant\geq 0 &\quad \end{array}\right.}\Longleftrightarrow \left\{ \begin{array}{ll}s>0 &\quad \\ 1-s>0 &\quad \\ s^{2}-4(1-s)\geq 0 &\quad \end{array}\right.$

$\therefore 2(\sqrt{2}-1)\leq s <1\ \cdots [*]$.

Express $P$ in terms of $s$, we have $P=\frac{1}{s}+\frac{s}{t}=\frac{1}{s}+\frac{s}{1-s}: =-1+\frac{1}{f(s)}$,

where $f(s)=s(1-s)$. Thus from $[*]$, we have $f(1)<f(s)\leq f(2(\sqrt{2}-1))\Longleftrightarrow 0<f(s)\leq 2(\sqrt{2}-1)^{3}$,yielding

$P\geq-1+\frac{(\sqrt{2}+1)^{3}}{2}=\boxed{\frac{5+5\sqrt{2}}{2}}$. The equality holds $s=2(\sqrt{2}-1)$ and $s+t=1\Longleftrightarrow s=2(\sqrt{2}-1)$ and $t=(\sqrt{2}-1)^{2}$,

yielding $x=y=\sqrt{2}-1$.


Silouan it's my turn!...Einai apli kai sthrizetai se gnwsta. Apla einai mia endiaferousa kinisi kai gia auto tin postarw.

Estw kiklos $ (O)$, simeio $ A$ ektos autou kai simeio $ B$ panw se auton kai $ C$ i tomi tis $ AB$ me ton kiklo. Estw twta $ P$ [b]metablito [/b] simeio tou kiklou kai $ L\equal{}PA\cap (O)$, $ R\equal{}BP\cap LC$. An o perigeggrammenos kiklos $ \triangle PBA$ temnei tin $ AR$ sto $ S$ tote na brethei i megisti tou mikous tou tmimatos $ OS$.

[quote="vittasko"]Είπα μήπως σας προλάβω μία φορά και ελπίζω ότι δεν έχω λάθος. :)

[hide]Το σημείο $ R,$ κείται πάντοτε επί της πολικής του $ A$ ως προς τον κύκλο $ (O).$

Η ευθεία αυτή ως γνωστόν, ορίζεται από τα σημεία επαφής, έστω $ X,\ Y,$ των εφαπτομένων του κύκλου $ (O),$ από το σημείο $ A.$

Τα σημεία λοιπόν $ X,\ Y,$ όπου έχουμε $ P\equiv L\equiv R\equiv S,$ μας δίνουν το μέγιστο του $ OS,$ ίσο με την ακτίνα του δοσμένου κύκλου $ (O)$ και το πρόβλημα έχει λυθεί. [/hide]
Κώστας Βήττας[/quote]

Kurie Vitta geia sas...

Euxaristw gia to post sas, alla den katalava akribws giati i eidiki periptwsi ton $ X,Y$ na dinei to megisto sto $ OS$. To lew auto dioti kai epoptika an to $ A$ einai arketa markia apo ton kiklo tote to $ OS$ tha einai ''arketa'' megalo. Tha mporousate na anevasete ena sxima, opote mporeite kai exete xrono, gia na bebaiwthoume oti milame gia to idio pragma?

Sas euxaristw...

Κύριε Βήττα,
έχετε πάρει μόνο την περίπτωση όπoυ το S , είναι εσωτερικό του κύκλου και σαφώς σε αυτην την περίπτωση το μέγιστο είναι αυτό που αναφέρεται αλλά υπάρχει και η περίπτωση το S να είναι εξωτερικό του κύκλου .Ο γεωμετρικός τόπος του S αποδεικνύεται ότι είναι κύκλος διαμέτρου ΟΑ , οπότε $ {OS}\leq{OA}$ όμως  για να συμβεί το ίσον πρέπει BP//LC οπότε το R πάει στο άπειρο δηλαδή στην ουσία δεν υπάρχει μέγιστο.

Βάζω την απόδειξη σε hint .
[hide] Αν τα σημεία τομής του AR με τον κύκλο O τα ονομασουμέ Z kai N τότε η τετράδα ARZN είναι αρμονική .Από τη σχέση $ RS*RA\equal{}RZ*RN$ προκύπτει ότι το S είναι μέσο του OZ. Άρα, η γωνία OSA είναι ορθή. :lol: [/hide]

Συγχωρέστε με κύριοι, αλλά σαφέστατα το μέγιστο του $ OS$ ίσο με την ακτίνα του κύκλου $ (O),$ ισχύει μόνο όταν το $ L,$ βρίσκεται μεταξύ των $ A,\ P,$ δηλαδή όταν το $ P$ κινείται επί του τόξου $ XY,$ που περιλαμβάνει το $ C.$

Σας ευχαριστώ που μου ανόιξατε τα μάτια.  :blush:

Κώστας Βήττας

[quote="kritikos"]
 Αν τα σημεία τομής του AR με τον κύκλο O τα ονομασουμέ Z kai N τότε η τετράδα ARZN είναι αρμονική .Από τη σχέση $ RS*RA \equal{} RZ*RN$ προκύπτει ότι το S είναι μέσο του $ ZN$ (κι όχι $ OZ$). Άρα, η γωνία OSA είναι ορθή. :lol: [/quote]

Σε αυτήν την απόδειξη αν καταλαβαίνω σωστά ούτε εσύ παίρνεις την περίπτωση το $ R$ να είναι εξωτερικό. Βασικά αν το $ R$ είναι εξωτερικό δεν θα υπάρχουν τα "σημεία τομής" $ Z,N$ (κι οπότε η απόδειξη δεν μπορεί να είναι ίδια). Oπότε η ερώτηση μου είναι, αυτό το $ \angle{OSA}\equal{}90$ ισχύει πάντα? (πάντως δεν καλύπτονται όλες οι περιπτώσεις με αυτήν την απόδειξη!)

Oso afora tin apodeiksi tou [b]kritikos[/b] exei apolito dikio se oti leei mono pou simfwnw me ton Mimi sto oti i apodeiksi kaliptei mono tin periptwsi pou i $ AR$ temnei ton kiklo. Wstoso, otan den ton temnei tote i apodeiksi ginetai pio endiaferousa (opote iponow oti i kathetotita isxuei kai se autin tin periptwsi ). Episis, mias kai to $ P$ kineitai opoudipote sto kiklo auto simainei oti mporei na ginei kai to simmetriko tou $ B$ os pros tin $ AO$ (opote to $ R$ paei sto apeiro), alla estw kai se autin tin ekfilismeni periptwsi exoume isotita. Me autin tin logiki egrapsa megisto kai oxi sup.

Oso afora gia ton Kurio Vitta, fusika kai den exoume na tou sunxwresoume tipota. Anthrwpoi eimaste kai oloi mperdeuomaste. Eksallou, o kurios Vittas exei postarei sto parelthon analytiki sunthetiki lusi se kati pou sxetizetai poli amesa me to parapanw.

Όταν το $ P$ κινείται επί τόξου που περιλαμβάνει το $ B$ $ ($ $ P,$ μεταξύ των $ A,\ L$ $ ),$ από το εγγράψιμο τετράπλευρο $ BCLP,$ συμπεραίνουμε ότι το $ S$ είναι το σημείο [b][size=100]Miquel[/size][/b] για το πλήρες τετράπλευρο $ BCLPAR,$ το οποίο ως γνωστόν (αποδεικνύεται εύκολα) κείται επί της ευθείας που συνδέει τα σημεία τομής των απέναντι πλευρών του εγγραψίμου $ ($ εδώ τα $ A,\ R$ $ ),$ και επίσης είναι γνωστό ότι $ OS\perp AR.$

Δηλαδή το σημείο $ S,$ κινείται επί της περιφέρειας διαμέτρου $ AO,$ οπότε έχουμε μέγιστη τιμή για το $ OS,$ όταν $ S\equiv A,$ δηλαδή όταν $ P\equiv B',$ όπου $ B'$ το συμμετρικό του $ B,$ ως προς την ευθεία $ AO.$

Η τιμή του $ OS$ μηδενίζεται όταν $ S\equiv O,$ δηλαδή όταν $ P\equiv C',$ όπου $ C'$ το συμμετρικό του $ C,$ ως προς την ευθεία $ AO$ $ ($ στο ισοσκελές τραπέζιο $ BCC'B',$ αποδεικνύεται ότι ο περιγεγραμμένος κύκλος περί το τρίγωνο $ \bigtriangleup ABC',$ περνάει από το κέντρο του κύκλου $ (O)$ και προφανώς τότε, η ευθεία $ AR,$ ταυτίζεται με την $ AO$ $ ).$

Κώστας Βήττας.

Στεφανε σε ευχαριστώ πολύ για την ωραία αφιέρωση . Η απόδειξή μου για το όταν η AR τέμνει τον κύκλο είναι περίπου ίδια με του kritikou οπότε δεν την βάζω . Για την περίπτωση που δεν την τέμνει ακολουθώ το εξής λήμμα

Έστω $ ABCD$ εγγράψιμο τετράπλευρο και ότι οι $ AB,\ CD$ τέμνονται στο $ E$ και οι $ AC,\ BD$ στο $ F$ .
Αν οι περιγεγραμμένποι κύκλοι των $ AFD$ ,$ BFC$ τέμνονται στο $ H$ τότε $ \angle{EHF}\equal{}90$

(αν θέλετε πείτε μου να βάλλω τη λύση του)


modular arithmetic
number theory unsolved
number theory

I help a friend of me to post his problem on this field (since he cannot join the forum)

Find all integers $ x,y,z$ satisfying the following equation
$ (x\minus{}y)^3(x\plus{}3y)\equal{}z^2.$

Sorry, but my friend's problem is
$ (x\minus{}y)^3(x\plus{}3y)\equal{}z^2$
not
$ (x\minus{}y)(x\plus{}3y)\equal{}z^2.$


If $ a,b,c>0$ prove the following :
\[ \sqrt{3\plus{}\frac{b}{a}}\plus{}\sqrt{3\plus{}\frac{c}{b}}\plus{}\sqrt{3\plus{}\frac{a}{c}}\geq 6\]

Just use AM-GM several times:
Let $ x \equal{} \frac {b}{a},y \equal{} \frac {c}{b},z \equal{} \frac {a}{c}$ then $ xyz \equal{} 1$ and we need to prove:
$ \sum_{cyc}\sqrt {3 \plus{} x}\ge 6$.
Have:
$ \sum_{cyc}\sqrt {3 \plus{} x}\ge 3\sqrt[6] {27 \plus{} 9(x \plus{} y \plus{} z) \plus{} 3(xy \plus{} yz \plus{} xz) \plus{} xyz}\ge 3\sqrt [6]{27 \plus{} 9*3 \plus{} 3*3 \plus{} 1} \equal{} 6$ QED

[quote="rem"]Just use AM-GM several times:
Let $ x \equal{} \frac {b}{a},y \equal{} \frac {c}{b},z \equal{} \frac {a}{c}$ then $ xyz \equal{} 1$ and we need to prove:
$ \sum_{cyc}\sqrt {3 \plus{} x}\ge 6$.
Have:
$ \sum_{cyc}\sqrt {3 \plus{} x}\ge 3\sqrt {27 \plus{} 9(x \plus{} y \plus{} z) \plus{} 3(xy \plus{} yz \plus{} xz) \plus{} xyz}\ge 3\sqrt {27 \plus{} 9*3 \plus{} 3*3 \plus{} 1} \equal{} 6$ QED[/quote]

Maybe typing mistakes. It should be

$ \sum \sqrt {3\plus{}x}\ge 3\sqrt[3]{\sqrt{3\plus{}x}\cdot\sqrt{3\plus{}y}\cdot\sqrt{3\plus{}z}}\\
\equal{}3\sqrt[6]{27\plus{}9(x\plus{}y\plus{}z)\plus{}3(xy\plus{}yz\plus{}xz)\plus{}xyz}\ge 3\sqrt[4]{27\plus{}9\cdot3\plus{}3\cdot3\plus{}1}\equal{}6$

we can prove that

$ \sqrt[6]{\left(3\plus{}\dfrac{b}{a}\right)\left(3\plus{}\dfrac{c}{b}\right)\left(3\plus{}\dfrac{a}{c}\right)}\geq 2$.

in fact, from Holder's inequality we have that

$ \sqrt[6]{\prod\left(1\plus{}1\plus{}1\plus{}\dfrac{b}{a}\right)}\geq \sqrt{1\plus{}1\plus{}1\plus{}1}\equal{}2$

and the conclusion follows easily...


I wonder if it is possible to add a search function that takes a few user names and returns threads in which all of them posted. The reason is that I often remember fairly well who took part in the discussion I want to find but not the exact words they used.

No. While it is possible to search for posts by a specific user or for threads in which that user has posted, there's no way to search for several users at once.

oh ok, sorry !!
i think its a good idea, though

I will try to accommodate that option, but it will have to wait at while (probably after the IMO and IMC).


You forgot "No, because I've never heard of it."

Is there any multiplayer (not campaign), where players can play each other?

Yes. There is an online version where players compete. You need multiplayer tho which is not in demo

woo i play crysis on my computer !
runs rather well, imo.
really fun too, and i laugh at some of the cutscene conversations lol.
ive yet to try multiplayer. too busy on single player going everywhere messing around lawl


trigonometry
ratio
function
calculus
derivative
vector
Pythagorean Theorem

there is not really a good way to prove it without the pythagorean theorem without using circular reasoning, but i think that you could use the taylor series to show it

You could use the unit circle...oh wait, that would be using pythagorus anyway.

That's why they call it the Pythagorean Identity! :)

[quote="Jackya"]How do u prove: ((sin(x))^2)+((cos(x))^2)=1 without pythagorean's theorem?[/quote]

This is equivalent to the Pythagorean Theorem - what you ask for is axiomatically impossible.

Edit:  When I say axiomatically impossible, I mean that if you define $\sin, \cos$ by ratios of lengths in right triangles, then the above is a direct result of the Pythagorean Theorem.  In principle, if you defined $\sin x$ either as:

$\sin x = \frac{e^{ix} - e^{-ix}}{2}$

Or as

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} +-...$

Then it could be proved differently, but this relies on your original methods of defining $\sin, \cos$.

the sin and cos functions are not defined based on the pythagorean theorem, but they are based on similar triangles

Well, last year, I found one way to solve it without using Pythagora, but, it uses a bit of calculus, so...

[hide="if you still want to see it ;)"]
Let $f(x) = (\sin(x))^2 + (\cos(x))^2$.
Then : $f'(x) = 2\sin(x) \cdot \cos(x) - 2\sin(x) \cdot \cos(x) = 0$
Since the derivative equals 0, the function is constant, so it suffices to have one value.
Just plug "0" in it : $(\sin(0))^2 + (\cos(0))^2 = 1.\\ \mathbb{Q.E.D.}$[/hide]

I am nearly certain that the proofs of the derivatives of $\sin$ and $\cos$ rely on the Pythagorean Theorem, at least very implicitly... :(

In fact, we can generate [hide="nearly arbitrary proofs using calculus"]
Consider the vector function $\vec{R}(t) = <\cos t, \sin t>$.  Then the quantity we desire is given by $| \vec{R}(t) |$.  Note that $\vec{R'}(t) = <- \sin t, \cos t>$, and hence $\vec{R}(t) \cdot \vec{R'}(t) = 0$.  Integrating, we have $\frac{1}{2} \vec{R}(t) \cdot \vec{R}(t) = C_1 \implies| \vec{R}(t) |^2 = C_2 \implies | \vec{R}(t) | = C_3$ for some constant.  Q.E.D.
[/hide]

This does not constitute a proof, since my use of $| \vec{R} (t) |$ is, in fact, the use of the Pythagorean Theorem.

Yes, you're right... :(
But, at least, these two proofs do not use directly the Pythagorean theorem (unlike chess64's one for example), and I think that's what Jackya was looking for :)

Still hidden because it's not the good place to post this...
[hide]
Well, just keep in mind that : $(u^2)' = 2u \cdot u'$ and apply that to $u(x) = \sin{x}, \cos{x}$.[/hide]

 :)

i don't see how the derivations of the formulas rely on the pythagorean theorem, they rely on similar triangles, and the fact that when you draw the unit circle, their sum squared =1 is not crucial to their definition, yet it is true, but i think that some of the derivative things are based on that specific relationship so i don't think you really have a good proof other than using the pythagorean theorem

$\\\sin^2\theta+\cos^2\theta\\ =\left(\frac{e^{i\theta}-e^{-i\theta}}{i2}\right)^2+\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^2\\ =\frac{-e^{i2\theta}-e^{-i2\theta}+2}{4}+\frac{e^{i2\theta}+e^{-i2\theta}+2}{4}\\ =\frac{4}{4}=1\\ \mathbb{QED}$

Note: $e^{ik}$ is deduced from power series.

Masoud Zargar

i am pretty sure that you cannot show that
$D_x (\sin x) =\cos x$ without the trig sum and difference identities, which are derived using they pythagorean theorem

[quote="t0rajir0u"][quote="Jackya"]How do u prove: ((sin(x))^2)+((cos(x))^2)=1 without pythagorean's theorem?[/quote]
$\sin x = \frac{e^{ix} - e^{-ix}}{2}$
[/quote]

You forgot the $i$ in the denominator:
$\sin x = \frac{e^{ix} - e^{-ix}}{i2}$

Masoud Zargar


Let $ a,b,c,x,y,z\geq0$ s.t. $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z$. Prove that:
$ ax(a\plus{}x)\plus{}by(b\plus{}y)\plus{}cz(c\plus{}z)\geq 3(abc\plus{}xyz)$

Aply CBS $ (a^2x\plus{}b^2y\plus{}c^2z)(yz\plus{}zx\plus{}xy)\geq xyz(a\plus{}b\plus{}c)^2$

  Then , following $ (a\plus{}b\plus{}c)^2\equal{}(x\equal{}y\plus{}z)^2\geq3(xy\plus{}yz\plus{}zx)$ we have $ a^2x\plus{}b^2y\plus{}c^2z\geq3xyz$

 and similarly   $ ax^2\plus{}by^2\plus{}cz^2\geq3abc$

  Adding up these inequalities yields the desired result.


modular arithmetic
number theory
number theory unsolved


I think this two identities will do just fine: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
 and $(a+b+c)^3- (a^3+b^3+c^3)=3(a+b)(b+c(c+a)$ :D


Show that every positive integer n can be written in a [b]unique[/b] way as : n = 2^p (2q + 1) where p and q are integers. (Say btw how this implies  IN² ~ IN !)
Say then why : Hn = 1 + 1/2 + ...  + 1/n is not integer once n is > 1

Consider $ m = 2^{[log_2(n)]}$. If $ n > 1$ and $ H_n$ integer, then $ H_n*N, \ N = m*N_1, \ N_1 = \prod_{p\ge 3}p^{[log_p(n)]}$ must be even, but $ H_n*N = \sum_{k\not = m}\frac {N}{k} + N_1$ is odd - contradition.

thanks much Rust . Excuse me but firstly ,  when i was given this exercise for the 1rst time , it was in Logic lesson and we hadn't studied x --> log_a(x) yet neither the neperian logarithm itself , additionnally , i think your proof is rather hard and doesn't comply with the term of the problem .


[quote]Find all functions from positive reals to positive reals satisfying \[ f(xf(y))\equal{}f(xy)\plus{}x\]
[/quote]

For $ f(x)$ instead of $ x$, you get
$ f(f(x)f(y)) \equal{} f(f(x)y)\plus{}f(x) \equal{} f(xy)\plus{}y\plus{}f(x)$.
Now, using symmetry between the variables, you also have $ f(f(x)f(y)) \equal{} f(xy)\plus{}x\plus{}f(y)$.
It follows that $ g(x) \equal{} f(x)\minus{}x$ is a constant, so that $ f(x) \equal{} x \plus{}c$ for some nonnegative constant $ c$.
Back to the initial equation leads to $ f(x)\equal{} x\plus{}1$, which is a solution.

Pierre.

mathema*, $ f$ is only defined on positive reals, you cannot let $ y\equal{}0$.

[quote="xxp2000"]mathema*, $ f$ is only defined on positive reals, you cannot let $ y \equal{} 0$.[/quote]

sorry i didn't saw it !!

Is this solution OK?:
Let $ x\equal{}1$. Then we have $ f(f(y))\equal{}f(y)\plus{}1$. Letting $ f(y)\equal{}a$, we have $ f(a)\equal{}a\plus{}1$.
It seems too short...

[quote="Brut3Forc3"]Is this solution OK?:
Let $ x \equal{} 1$. Then we have $ f(f(y)) \equal{} f(y) \plus{} 1$. Letting $ f(y) \equal{} a$, we have $ f(a) \equal{} a \plus{} 1$.
It seems too short...[/quote]

you'd need $ f$ is onto $ \mathbb R^\plus{}$...


A math field day competition is held in a room with many tables, and there are 6 stools at each table. Each stool has 3 legs, and each table has 4 legs. If there is a total of 484 legs on all the tables and stools in the room, how many tables are in the room?

Call the number of tables $ t$.  There are a total of $ 6t$ stools, and $ 3\cdot6t\equal{}18t$ legs that the stools account for.  The number of legs the tables account for is just $ 4t$.  So we can set up and equation and solve:
\[ 4t\plus{}18t\equal{}484\implies22t\equal{}484\implies t\equal{}\boxed{22}\]


If $ p$ is an odd prime, let $ q \equal{} (p\minus{}1)/2$.  Prove that

$ (q!)^2 \plus{} (\minus{}1)^q \equiv$ (mod $ p$)

$ (p\minus{}1)!$
$ \equal{}1\times 2\times \cdots \times (q\minus{}1)\times q\times (q\plus{}1) \times \cdots \times (p\minus{}2)\times (p\minus{}1)$
$ \equiv 1\times 2\times \cdots \times (q\minus{}1)\times q\times (\minus{}q) \times \cdots \times (\minus{}2)\times (\minus{}1)\pmod{p}$
$ \equal{}(\minus{}1)^q \cdot (q!)^2$
and now by Wilson's theorem $ (p\minus{}1)!\equiv \minus{}1\pmod{p}$

hence we have

$ (q!)^2\plus{}(\minus{}1)^q\equiv (\minus{}1)^q\cdot (p\minus{}1)!\plus{}(\minus{}1)^q\equiv (\minus{}1)^{q\plus{}1}\plus{}(\minus{}1)^q\equal{}0\pmod{p}$,and we are done.


Here is a problem from one of Titu Andreescu and Zuming Feng's books, 101 Problems in Algebra:

Evaluate

\[ \frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2001}{1999!+2000!+2001!}. \]

$\frac{3}{1!+2!+3!} = 1/3$
$\frac{4}{2!+3!+4!} = 1/8$
$\frac{5}{3!+4!+5!} = 1/30$

$\frac{1}{3} + \frac{1}{8} = \frac{11}{24}$
$\frac{11}{24} + \frac{1}{30} = \frac{59}{120}$

From this, the sum of the first $n-2$ terms of the series appears to be
\[\frac{\frac{n!}{2} - 1}{n!}\]

or, after dividing the numerator and denominator by $n!$:
\[\frac{1}{2} - \frac{1}{n!}\]

and we know it works for the sums of the first two and first three terms.

Using my newly acquired skill of mathematical induction, we can prove that this will work for any number of terms.

\[\frac{1}{2} - \frac{1}{n!} + \frac{n+1}{(n-1)! + n! + (n+1)!} = \frac{1}{2} - \frac{1}{(n+1)!}\]

The $\frac{1}{2}$ on each side cancels out, and we move the $\frac{1}{n!}$ to the right side.

\[\frac{n+1}{(n-1)! + n! + (n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}\]

Put $\frac{1}{n!}$ and $\frac{1}{(n+1)!}$ into one fraction, and we can set the means equal to the extremes to get rid of the fractions.

\[\frac{n+1}{(n-1)! + n! + (n+1)!} = \frac{(n+1)! - n!}{(n+1)!n!}\]

\[(n+1)(n+1)!n! = ((n+1)! - n!)((n-1)! + n! + (n+1)!)\]

Proving the sides as equal involves a lot of manipulating the factorials and factoring the products. Starting with the left side, $(n+1)!$ can be written as $(n+1)(n!)$, so we have $(n+1)^2(n!)^2$.

On the right side, $(n+1)! - n! = (n+1)n! - n! = n!(n+1-1) = n!n$. Now we multiply $n!n$ and $(n-1)! + n! + (n+1)!$. For the first term, $n!n(n-1)! =  n!n! = (n!)^2$. Now the second term: $n!(n)(n!) = (n!)^2 n$, and the third: $n!n(n+1)! = n!n(n+1)n! = (n!)^2n(n+1)$. We factor these three terms to get $(n!)^2(1 + n + n(n+1)) = (n!)^2(n+1)(1+n) = (n!)^2(n+1)^2$. So we finally have equality for both sides

\[(n+1)^2 (n!)^2 = (n+1)^2 (n!)^2\]

proving that our formula for the sum will work for any number of terms. Now we just plug in 2001 for $n$ for the answer.

\[\frac{1}{2} - \frac{1}{2001!}\]

Nice work.  You do a lot of things right here including playing around with the first few terms and making a conjecture.  At some point you will be able to see through the algebra without even doing that much work.  You might do this by jumping straight to the general form of each term.

You don't really need induction here.  After you show the way each general terms turns into a difference of the reciprocals of factorials you could simply write this as a telescoping series, but it's nice to have tools like induction that allow you to solve problems even when you don't stumble upon the most efficient solution.  I can't count the number of contest problems I solved using a less-efficient method that I knew really well only to think of the more efficient solution later.

Ah, I see..

Each term is $\frac{1}{(n-1)!} - \frac{1}{n!}$ when $n$ starts at 3. So all but $\frac{1}{2} - \frac{1}{2001!}$ cancel. Bah, that does look really easy now.

Thanks for the feedback.


[size=150]На плоскости, тангенс угла наклона которой равен коэффициенту трения, лежит монета. В горизонтальном направлении вдоль плоскости монете сообщили скорость 0,2м/с. Найти установившуюся скорость монеты.

Bản dịch: Trên mặt phẳng nghiêng, tgα=k (α là góc nghiêng của mặt phẳng nghiêng, k là hệ số ma sát), có để một đồng tiền. Người ta truyền cho nó một vận tốc ban đầu v=0,2 m/s theo phường nằm ngang dọc theo mặt phẳng nghiêng. Sau một thời gian nào đó đồng tiền sẽ chuyển động với vận tốc không đổi v’. Tìm v’.
2. В цирковом аттракционе мотоциклист движется по внутренней поверхности сферы радиуса 10м. Разогнавшись, он начинает описывать горизонтальную окружность в верхней полусфере. Определить минимальную скорость мотоциклиста, если коэффициент трения шин о поверхность сферы равен 0,5, а угол между вертикалью и направлением к мотоциклисту из центра сферы равен 60°.
Bản dịch: Trong một tiết mục xiếc, một người đạp xe (diễn viên) để nó chuyển động theo bề mặt trong của một quả cầu bán kính 10m. Sau khi lấy đà diễn viên bắt đầu vẽ lên một đường tròn ở nửa trên của quả cầu. Xác định vận tốc nhỏ nhất của người diễn viên này, nếu như hệ số ma sát giữa vành xe đạp với bề mặt của quả cầu là k=0,5 còn góc giữa phương thẳng đứng với phương nối tới người diễn viên tính từ tâm quả cầu α=60 độ.
3. В K-системе отсчета частица, движущаяся со скоростью 0,99c пролетела от места своего рождения до точки распада расстояние 2км. Определить собственное время жизни этой частицы.

Bản dịch: Trong hệ quy chiếu K, một hạt chuyển động với vận tốc v=0,99c (c là vận tốc ánh sáng trong chân không) từ vị trí nó bắt đầu được sinh ra cho tới điểm nó bị phân rã, đã đi được quãng đường 2km. Hãy tính thời gian “sống” của hạt này gắn với hệ quy chiếu K chuyển động cùng với hạt.
[/size]

Никто  интересует тем ????? :D
Помогите  мне !!!!  Спасибо

Зачем просить решить задачку из физики на математическом форуме?Для этого есть по моему более подходящие места,к примеру вот это:
http://lib.mexmat.ru/forum/viewforum.php?f=29

Или:

[url]http://www.mathlinks.ro/index.php?f=405[/url]
[url]http://www.mathlinks.ro/index.php?f=332[/url]

Ну просто дело в том что условие задач опубликованы на русском,и я предположил что и решение он хочет получить на том же языке, поэтому и дал ссылку на российский форум,кстати там также присутствует и математический форум и многие его участники также активны на ML.

Да, я понимаю ;)


[quote="thegod277"]Let a, b, c be positive numbers such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 1$. Prove that
$ \sum \frac {a^3}{2b^2 \plus{} c^2}\geq 1$
 :D ![/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt[4]3}.$ :wink:

[quote="thegod277"]Let a, b, c be positive numbers such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 1$. Prove that
$ \sum \frac {a^3}{\sqrt {2b^2 \plus{} c^2}}\geq 1$
 :D!
Sorry, I had edited the inequalities.[/quote]
By Cauchy-Schwarz inequality, we have:
$ \sum \frac {a^3}{\sqrt {2b^2 \plus{} c^2}}\ge \frac{(a^4\plus{}b^4\plus{}c^4)^2}{\sum a^5\sqrt{2b^2\plus{}c^2}}\equal{}\frac{1}{\sum a^5\sqrt{2b^2\plus{}c^2}}$
We need to prove:
$ \sum a^5\sqrt{2b^2\plus{}c^2} \le 1$
By Cauchy-Schwarz inequality:
$ (\sum a^5\sqrt{2b^2\plus{}c^2})^2 \le (a^4\plus{}b^4\plus{}c^4)(\sum a^6(2b^2\plus{}c^2))\equal{}\sum a^6(2b^2\plus{}c^2)$
Let $ a^2\equal{}x,b^2\equal{}y,c^2\equal{}z, x^2\plus{}y^2\plus{}z^2\equal{}1$
Inequality becomes:
$ x^2(2y\plus{}z)\plus{}y^2(2z\plus{}x)\plus{}z^2(2x\plus{}y) \le1\equal{}(x^2\plus{}y^2\plus{}z^2)^2$
Which is clearly true. (because by Vasc's inequality:
$ (x^2\plus{}y^2\plus{}z^2)^2 \ge 3(x^3y\plus{}y^3z\plus{}z^x)$ and: $ (x^2\plus{}y^2\plus{}z^2)^2 \ge 3(xy^3\plus{}yz^3\plus{}zx^3)$)


Let $O$ be circumcenter of triangle $ABC$. Let $BO$ meet $AC$ at $E$ and let $CO$ meet $AB$ at $F$. If $\angle CFE= \angle AFE = \angle BAC$ find the value of that angle.

by angle chasing,i got it $60$,

but if the angle is $60$ , there not wxist triangle $ABC$,

because $\angle ACB=0$


Use this factoring to prove that every number has a unique base-2 representation.  ;)


Three Players play 4 rounds of a game and end with a tie score of 80.  A different player wins each of the final 3 rounds.  When a player wins, s/he doubles his or her points, and the two losers each must subtract the amount the winner gains from their scores.  What was the highest score at the end of round 1?

[hide]Without loss of generality, we can assume that player 1 wins game 2, 2 wins 3, and 3 wins 4. Since each has 80 to start with, player 3 had 40 at the end of round 3, and players 1 and 2 had 120. Therefore, player 2 had 60 at the end of round 2, player 3 had 100, and player 1 had 180. At the end of round 1, player 1 had 90, player 2 had 150, and player 3 had 190. Therefore, the highest score was 190.[/hide]

[hide]I drew a table, then worked backwards:
[color=green]190[/color]  [color=orange]100[/color]  [color=red]40[/color]    [color=darkred]80[/color]
[color=green]150[/color]  [color=orange]60[/color]    [color=red]120[/color]  [color=darkred]80[/color]
[color=green]90[/color]    [color=orange]180[/color]  [color=red]120[/color]  [color=darkred]80[/color]

[color=darkred][][/color]=1st
[color=red][][/color]=2nd
[color=orange][][/color]=3rd
[color=green][][/color]=4th
The answer is 190.[/hide]

Nothing more to say other than that I do have the 1992 answer key and 190 is correct. I solved it the same way as mathgeek.


[hide="solution"]
(45/50)(44/49)(43/48)
[i]calculator-calculator-calculator[/i]
0.72
[/hide]

[quote="MCrawford"]Five of the 50 computers in a lab have a virus.  Three of the computers are selected at random.  What is the probability that none of the selected computers are infected?  Express your answer as a decimal to the nearest hundredth.[/quote]

[hide]Okay the question is really very interesting ...

Now the proablity is viertually zero if:
A)  .. These computeres are not connected to internet. (Else won't all computers will be in danger - specially if the use IE?)
B   ..  They use firefox.
C       They do not run  Miccosoft Windows..

What do the experts think? [/hide]

[quote="Gyan"]Okay the question is really very interesting ...

Now the proablity is viertually zero if:
A)  .. These computeres are not connected to internet. (Else won't all computers will be in danger - specially if the use IE?)
B   ..  They use firefox.
C       They do not run  Miccosoft Windows..

What do the experts think? [/quote]

Sorry Gyan. Didn't mean to be rude like that...

Farazy -  That comment was rude.. better edit it before moderators  may  notice it.  And Don't  "unhide" other comments unless  you have a  question.

MCrawford:

[hide]I was hoping that someone here would notice that "five machine has viruses"  does not imply that other 45 don't  (I would have worded it  some thing like "exactly five" or "only five"  .. oh well :) [/hide]

y do we do this?

[quote=Treething and dwx314 wrote](5/50)(4/49)(3/48) [/quote]

ok so 5 of the fifty have viruses so 45 don't:
no its a simple porbability question:
45/50 * 44/49 *43/48
then we use a calculator...


find the point of intersection algebraically with the given equations

y=x^3+2x-1
y=-2x+15

I set them equal to each other and set them equal to zero but i can't do anything else because it is a cubic and not a quadratic (can't factor) :(

x^3+2x-1=-2x+15
x^3+4x-16=0

Graphically the point if (2,11)
How can i solve this algebraically :?:

You can use the rational root test.  Cardano's method works, but it's sort of messy.

I also vote for the RRT because 16's not too bad of a number to do it with.

I think if you could do this problem by calculator, you knew the root of $ x^3\plus{}4x\minus{}16\equal{}0$ is $ x\equal{}2$ like you said. we could factor it and we get $ x^3\plus{}4x\minus{}16\equal{}(x\minus{}2)(x^2\plus{}2x\plus{}8)$ after that expend the $ (x\minus{}2)(x^2\plus{}2x\plus{}8)$ to show your work algebraically.

You don't need a calculator to find 2.  Just use RRT.


The operation & is defined as $ a \& b \equal{} (a\plus{}b)(a\minus{}b)$. What is the value of $ 6 \& 3$?