find all function $ f:\mathbb Q\rightarrow\mathbb R$such that:
$ (i)$ $ f(x+y)-yf(x)-xf(y) = f(x)f(y)-x-y+xy$ for all $ x,y\in\mathbb Q$
$ (ii)$ $ f(x) = 2f(x+1)+2+x$, for every $ x\in\mathbb Q$
$ (iii)$ $ f(1) >-1$

motmaenni soorate soalo doros neveshti?  :huh:

age gharar bedim $ x \equal{} y \equal{} 0$ natije mishe ke:

$ f(0) \equal{} f(0)^{2}$

pas $ f(0) \equal{} 0$ ya $ f(0) \equal{} 1$

khob hala be moadele aslimoon bargardim,bejaye $ y$ bezarid $ 0$ mishe:

$ f(x)\minus{}xf(0) \equal{} f(x)f(0)\minus{}x$

$ \Rightarrow (f(x)\plus{}x)(1\minus{}f(0)) \equal{} 0$

khob hala age $ f(0) \equal{} 0$ bashe natije mishe ke $ f(x) \equal{}\minus{}x$ be ezae har $ x$e gooya,ke khob age too soorate masale jagozari konim javabe yeni ham too sharte $ 1$ va ham too sharte $ 2$ sedgh mikone,vali in ba sharte $ 3$ dar tanaghoze,man inja shak daram ke asan sharte $ 3$vase chi oomade?  :maybe:

khob hala too halati ke $ f(0) \equal{} 1$ bashe ba tavajoh be rabete dovom natije mishe:

$ f(0) \equal{} 2f(1)\plus{}2$

pas $ f(1) \equal{}\minus{}\frac{1}{2}$ badesh natije mishe ke $ f(2) \equal{}\minus{}\frac{7}{4}$ va badesh $ f(3) \equal{}\minus{}\frac{23}{8}$ va dar halate kolli:

$ \forall n\in\mathbb{N}: f(n) \equal{}\minus{}n\plus{}\frac{1}{2^{n}}$

khob ta injasho raftam vali hanoozam too halati ke $ f(x) \equal{}\minus{}x$ bashe yekhorde sharte $ 3$ maskharas,hala nemidoonam,age motmaenni ke soorato doros neveshti begoo ta say konim roo baghye soal fek konim...  :maybe:

Soorate masale doroste . soal male olympiade riazie Balkan 2003 hastesh .
javabet rooye adade tabiyi doroste hala bayad bebarish roo gooya :)

dar har soorat ye khorde maskharas soalesh...  :huh:

idash too poste ghabli taghriban gofte shode...

biaid soorate masalaro yekhorde moratabesh konim,mese kari ke too poste ghabli kardimo be $ f(x) \equal{}\minus{}x$ residim:

biaid mosbataro negah darid,manfiaro befrestid oontaraf ta beshe:

$ f(x\plus{}y)\plus{}x\plus{}y \equal{} f(x)f(y)\plus{}xy\plus{}xf(y)\plus{}yf(x)$

amma darim

$ (f(x)\plus{}x)(f(y)\plus{}y) \equal{} f(x)f(y)\plus{}xy\plus{}xf(y)\plus{}yf(x)$

pas khahimd asht:

$ f(x\plus{}y)\plus{}(x\plus{}y) \equal{} (f(x)\plus{}x)(f(y)\plus{}y)$

khob hala bebinid ye seri jomle darim hame be shekle $ f(z)\plus{}z$ pas inike biaim tabe jadidi be esme $ g(z) \equal{} f(z)\plus{}z$ tarif konim khoobe,pas tasavie akharemoon mishe:

$ \boxed{g(x\plus{}y) \equal{} g(x)g(y) (I)}$

khob hala biaim tasavie $ (I)$ ro yekhorde barresi konim:
avvalan deghat konid ke $ f:\mathbb{Q}\to\mathbb{R}$ pas $ g:\mathbb{Q}\to\mathbb{R}$
hala deghat konid ke darvaghe moadele $ (I)$ hamoon moadele cauchye,ke ye moadele shenakhte shodaso javabesh be shekle $ g(x) \equal{} 0$ ya $ g(x) \equal{} 10^{cx}$ hastesh,vali baraye oonai ke hallesho nemidoonan,yebar ba ham hallesh mikonim...

[hide]pas mikhaim moadele $ g(x\plus{}y) \equal{} g(x)g(y)$ ro hal konim...

gharar bedid $ h(z) \equal{}\log g(z)$

pas $ g(z) \equal{} 10^{h(z)}$ hala biaid ino too moadele asli gharar bedid mishe:

$ 10^{h(x\plus{}y)}\equal{} 10^{h(x)}.10^{h(y)}\equal{} 10^{h(x)\plus{}h(y)}$

pas darim:

$ h(x\plus{}y) \equal{} h(x)\plus{}h(y)$

khob pas baraye peida kardane $ g$ kafie $ h$ ro az moadele bala peida konim...
-----------------------------------------------
amma deghat konid ke chon $ g:\mathbb{Q}\to\mathbb{R}$ pas dar vaghe $ x,y\in\mathbb{Q}$ pas $ h:\mathbb{Q}\to\mathbb{R}$...

khob hala aval az hame gharar bedid $ x \equal{} y \equal{} 0$ natije mishe ke $ h(0) \equal{} 0$

hala gharar bedid $ y \equal{}\minus{}x$ natije mishe $ h(\minus{}x) \equal{}\minus{}h(x)$

hala ba esteghra rooye $ n$ mishe sabet kard ke:

$ h(x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}) \equal{} h(x_{1})\plus{}h(x_{2})\plus{}...\plus{}h(x_{n})$

pas darim:

$ h(1) \equal{} h(\frac{1}{n}\plus{}\frac{1}{n}\plus{}...\plus{}\frac{1}{n}) \equal{} h(\frac{1}{n})\plus{}...\plus{}h(\frac{1}{n}) \equal{} n.h(\frac{1}{n})$

pas $ h(1) \equal{} n.h(\frac{1}{n})$

khob hala chon $ x\in\mathbb{Q}$ pas mishe nevesht $ x \equal{}\frac{p}{q}$ pas darim:

$ h(x) \equal{} h(\frac{p}{q}) \equal{} h(\frac{1}{q}\plus{}...\plus{}\frac{1}{q}) \equal{} h(\frac{1}{q})\plus{}...\plus{}h(\frac{1}{q}) \equal{} p.h(\frac{1}{q})$

pas darim:

$ h(x) \equal{} h(\frac{p}{q}) \equal{} p.h(\frac{1}{q}) \equal{} p.\frac{h(1)}{q}\equal{} h(1).\frac{p}{q}\equal{} h(1).x$

pas baraye har $ x$e gooya darim $ h(x) \equal{} h(1).x$

hala gharar bedid $ h(1) \equal{} c$ pas darim $ h(x) \equal{} cx$

amma dashtim $ g(x) \equal{} 10^{h(x)}$ pas khahim dasht:

$ g(x) \equal{} 10^{cx}$
--------------------------------------------------------
[/hide]

pas be das oomad $ g(x) \equal{} 10^{cx}$

pas $ f(x)\plus{}x \equal{} 10^{cx}$ pas $ f(x) \equal{} 10^{cx}\minus{}x$

hala biaid ino too sharte $ ii)$ jagozari konid mishe:

$ 10^{cx}\minus{}x \equal{} 2(10^{c(x\plus{}1)}\minus{}x\minus{}1)\plus{}2\plus{}x$

ke age sadash konim be in khatm mishe:

$ 10^{cx}\equal{} 2.10^{cx\plus{}c}$

$ \Rightarrow 1 \equal{} 2.10^{c}$

pas

$ 10^{c}\equal{}\frac{1}{2}$

yeni darim:

$ f(x) \equal{}\frac{1}{2^{x}}\minus{}x$

baraye har $ x$e gooya...


[quote="bryantacademy"]If you have a number and a variable in parentheses like (3x) with a zero exponent on the outside like (3x)^0, then is that expression evaluated as 3^0 * x^0 which equals 1, or  3x^0 which equals 3???

Just reassurring myself.

 :D[/quote]

$(3x)^{0}= 1$

[quote="bryantacademy"]If you have a number and a variable in parentheses like (3x) with a zero exponent on the outside like (3x)^0, then is that expression evaluated as 3^0 * x^0 which equals 1, or  3x^0 which equals 3???

Just reassurring myself.

 :D[/quote]
Whenever you have an expression like $(3x)^{0}$ with parentheses around the whole thing, you have to do the operation for everything, so it would be $3^{0}\times x^{0}=1$ (unless $x=0$ of course). If instead you had $3x^{0}$ without parentheses, it would just be $3\times x^{0}=3$ (unless $x=0$ of course).

[quote="math92"]I get confused sometimes with that also.

But again, $3x^{0}=1$[/quote]
No, that's not correct. $3x^{0}=3\times x^{0}=3$ (for $x\neq 0$). However, $(3x)^{0}$ [i]is[/i] $1$ (as long as $x\neq 0$). You have to make sure whether there are parentheses or not in your expression, because that could possibly change its value.  :wink:

If memory serves, and if anyone cares, I believe the property Neal is using (3x)^0 = (3^0)(x^0) = 1 is called The Distributive Property of Exponentiation over Multiplication.  I just liked the word "exponentiation" :lol:

In general:

$(ab)^{c}=a^{c}\times b^{c}$ but $ab^{c}=a\times b^{c}$

Please excuse my dear Aunt Sally.

When you have an exponent around parentheses you distribute that exponent--$(3x)^{0}=3^{0}\cdot x^{0}=1$


Does pairwise relatively prime mean:

1. Each pair of two numbers in a list are relatively prime?
2. Consecutive terms are relatively prime?

The first one.

Every pair of numbers is relatively prime.


function
calculus
integration
real analysis
real analysis solved

Let $f_n:R_+-->R$ defined by $f_n(x)=\frac{(x+n).ln(x+n)}{x^3 + n^3}$
Study the simply and uniform convergence of $\sum_{n>0}f_n(x)$ on $R_+$

Lemma: For large .

In spoiler. Sorry about not using TeX, but that just doesn't interact well with spoiler.

[hide]The series converges pointwise and uniformly for all x in [0, :inf:). The limit is a continuous function (call it g(x)), but, although each of the terms of the series is integrable on [0, :inf:), the sum g is not integrable. More specifically, for large x, g(x) looks like C (ln x)/x for some appropriate positive constant C.

For a given x, suppose m is the integer nearest to 2x (the 2 is arbitrary). Then for n <= m, x^3 + n^3 > x^3 > C m^3. (The letter C will stand for some convenient constant which may be different each time it's used. That's what analysts often do: a constant is a constant, don't sweat the details that don't matter.)  The numerator of the term is no bigger than (3m)ln(3m) < C m ln m. Hence each term of the sum, for n <= m, is no larger than C (ln m)/m^2. Adding up those m terms, we get C (ln m)/m, which is comparable to C(ln x)/x. For n > m we note that x + n < 2m + n < 3n, so the numerator of each term is < C n ln n. The denominator is > n^3. Hence each term is < C (ln n)/n^2. To sum those up from m+1 to infinity, use the integral test and the lemma above. The sum is < C (ln m)/m. With both parts of the sum satisfying the same estimate, we have the whole sum, g(x), less than C (ln x)/x. We note that this is not integrable from 0 to infinity.

What remains is showing uniform convergence on [0, :inf:). Given epsilon > 0, chose an N so large that C (ln N)/N < epsilon. (What C here? Whatever you need to make the rest of it work). Then for x > N, the difference between the whole sum and any partial sum is smaller than the whole sum, g(x), which is < g(N) < epsilon. For x between 0 and N, look at the sum from N+1 to :inf:. The nth term is (as above) < C (ln n)/n^2. Adding these up gives (as above) a sum comparable to C (ln N)/N, which is < epsilon. Hence the convergence is uniform, which in turn implies that g(x) is continuous.

There's a generalization here: if f(x) is a continuous, eventually decreasing function on [0, :inf:) that is integrable on [0, :inf:), then the sum from n = 1 to :inf: of f(x+n) converges uniformly on [0, :inf:). This doesn't quite match the exact problem, but the distinction between (x+n)^3 and x^3 + n^3 was unimportant to the analysis.[/hide]


[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=17041[/url]

Are you a member of a high IQ society?  Just participate in the poll.  :)

Maybe we even have one of the [url=http://giga.iqsociety.org/]Giga[/url] gus on the forum. For a history on [url=http://www.eskimo.com/~miyaguch/history.html]high IQ societies read the material here.[/url]

And here are two links to challenging IQ tests:

- [url=http://www.eskimo.com/~miyaguch/hoeflin.html]Uncommonly Difficult IQ Tests[/url]

- [url=http://httpd.chello.nl/~p.cooijmans/gliaweb/tests/]High-Range IQ Tests[/url]

Unfortunately, the poll let me just specify at most ten alternatives. If I did not mention your society please feel free to tell us. ;)

Comments are welcome.  :)

[i]Edited by admin because the width of your source field was screwing up the forum topic listing[/i]

I've scored very high on one of the Uncommonly Difficult IQ Tests site's tests, but I must point out right at the beginning of this thread that all of those tests have a HUGE problem, namely complete lack of proper validation. To learn about what proper IQ test validation is all about, read the books mentioned in the online bibliography "[url=http://learninfreedom.org/iqbooks.html]Books on IQ and Human Intelligence[/url]."

I should also point out that at least one member of a high-IQ society is a well known [url=http://www.crank.net/harris.html]Internet crank[/url], a major idiot about mathematics, so I have good reason to doubt the high-IQ society that admitted him, and to be skeptical about high-IQ societies in general.

What's really important about intelligence is intelligent accomplishments in the real world apart from taking IQ tests.

After edit: The poll should include my choice, namely that I am not a member and I know better than to apply.

It is well knowed that IQ test are'nt made for discrimination at high level (beyond the last percentiles).

[quote="beta"]My IQ is probably in the 40s because I'm so bad at math[/quote]

Actually, this is one of the key points attacking the theory of IQ measurements. There are theories that talk about seven or more different kinds of intelligence, and every person has different levels of intelligence on each.

The fact that you are so bad in math just say that your logical mathematical intelligence is not well developed, but probably at least two or three are very well developed on you.

My wife's degree is in education, and I remember when she was doing all that stuff. She made me take a couple of those test (I was here lab rat!), and I remember that I got

Very High:
* Logical-Mathematical
* Spatial

High:
* Bodily-Kinesthetic
* Linguistic
* Interpersonal

Normal:
* Intrapersonal
* Musical

Of course, the test was kind of lying, because I am VERY BAD in music. I can listen to it, I can appreciate it, but if I try to sing or play an instrument, probably would be sued by my neigboors, colleagues and even my wife for attempt to distroy their hearing organs! In general, I am baaad in any kind of art! Also, I really don't consider my abilities in linguistic something above the normal.

Best,

For better or worse, IQ tests overestimate the intelligence of people whose hobbies include solving large numbers of mathematical problems.

fleeting_guest makes a good point: IQ tests are "SAMPLES of LEARNED behavior" (statement by Alan S. Kaufman, emphasis mine) and therefore are as subject to biased sampling as any other sample when taken to be an estimate of "general intelligence." I've seen some high-IQ people do some amazingly stupid things (some of them done by me  ;)  ), so I don't assume that high IQ is a guarantee of high level performance in real-world tasks.

On another issue: I don't think being in a low-level IQ society is a great achievement. First of all at some top university almost anybody will get admitted, at least to the low-level IQ societies, if they want to and prepare for the tests. And secondly the regular meetings of low-level IQ society such as Mensa are just ridiculous and for the higher level societies meetings are hard to organize because people tend to be busy, are widely spread around the globe and especially in the case of Giga: a quick estimations yields that there can be just 6-7 people in the world having this IQ.  :)

Actually,

I would like to know how many of the famus leaders of the world have really a high IQ. In general I don't trust too much the IQ test freely available on the web. They are just very unconsistent. I have taken a few and they have given me from 133 to 196!, the truth is that I haven't really investigate if they have the same scale, but if so, I don't think that you should have that range variation.

On the other hand, I consider that many leaders should be considered very smart people, including some gang leader. I am not saying that what they do is OK, I think not, but the truth is that a 16 or 18 years old boy that have been the leader of a gang for two or three years, that have never been cought, and have never been killed by enemy gangs, is a smart guy. Probably many people (I believe I sholud include myself) wouldn't survive this experience. AKA, we should have a better definition of intelligence!

Best,


calculus
integration
trigonometry
symmetry
calculus computations

A couple homework problems that I need help with:

1)      Why can't you grow wheat in Z mod 6?

         ---I know the answer is "its not a field"   but i need to know why

2)      For what value(s) of x in this equation does y attain its minimum value if a, b, and c are all positive constants

3)(couldnt find a way to put the integral sign on here but..) integral sign with (pi)/2 on the top and 0 on the bottom.. to the right of it:
sq. root of(sin(x)+3) all divided by sq. root of(sin(x)+3) + sq. root of(cos(x)+3)
                  - if that makes sense..

1) It's not a field because $(2)(3) \mod 6 \equiv 0 \mod 6$ where $2, 3 \neq 0 \mod 6$ (i.e. everything must be a unit) another reason is that 2 has no inverse (because $\gcd(2,6)\neq1$).
2) What equation?

3: $\int_{0}^{\frac{\pi}2}\frac{\sqrt{\sin x+3}}{\sqrt{\sin x+3}+\sqrt{\cos x+3}}\,dx$

Ahh - the old symmetry trick. Try making the substitution $u=\frac{\pi}2-x,$ then after you make the substitution, turn the $u$s back into $x$s. Can you see what you'd have then?

[quote="Tummbler"]oh.. sorry..

the equation for #2:

y=sq root of(a^2+(b-x)^2)+sq. root of(c^2+X^2)[/quote]

$y = \sqrt{a^{2}+(b-x)^{2}}+\sqrt{c^{2}+x^{2}}$.

Consider the distances between the points $(a,b); (0,x); (c,0)$.

perhaps i shouldve mentioned that these are extra credit questions in my pre-calc class.. and i have no previous experience w/ these types of problems..

so.. I got what you guys said about #1 but I didnt really get what you guys said about #2 aat all and are the distances between those points the answer to #3 or..
help plz! :oops:

okay.. i dont really know what modulo means.. does Z mod 6 = Z(sub)6

={0,1,2,3,4,5}

and does a,b= 2,3???

im in precalc and this is an E.C. question so these are just bits and pieces that ive looked up..


On an $n$x$n$ chessboard, numbers $1$ to $n^{2}$ are put in randomly, each being used once. A token is on square $n^{2}$. In each move, it can only move to an adjacent square (two squares are adjacent iff they share a side). It will first move to square $1$, using the shortest path possible. Then it will move to $2,3,...,$ until it returns to $n^{2}$. What is the maximum of the length of its path?

[quote="JBL"]The total motions in the vertical and horizontal directions are independent, so the answer is just twice that of the 1-D case.[/quote]

Eh, what do you mean?

Also, what's the 1-D case?


A sequence of numbers $ a_1, a_2, a_3, ...$ satisfies

(i) $ a_1 \equal{} \frac{1}{2}$
(ii) $ a_1\plus{}a_2 \plus{} \cdots \plus{} a_n \equal{} n^2 a_n \ (n \geq 1)$

Determine the value of $ a_n \ (n \geq 1)$.

This is a rather straightforward proof so I would be really glad to see some ingenious solution to this problem!

[hide="Using Induction"]

$ a_1 \equal{} \frac{1}{2} \equal{} \frac{1}{1 \cdot 2}$

$ a_2 \equal{} \frac{1}{6} \equal{} \frac{1}{2 \cdot 3}$

$ a_3 \equal{} \frac{1}{12} \equal{} \frac{1}{3 \cdot 4}$

$ a_4 \equal{} \frac{1}{20} \equal{} \frac{1}{4 \cdot 5}$

....

So, let's establish a lemma.

[i]Lemma: For all $ n$, $ a_n \equal{} \frac{1}{n(n\plus{}1)}$[/i]

For $ n \equal{} 1,$ it obviously holds.

Now, let's assume that it holds for $ n \equal{} k$. Then:

$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k \equal{} k^2 a_k$ by the problem and applying $ a_k \equal{} \frac{1}{k(k\plus{}1)}$ (this is what we assumed), then we get

$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k \equal{} \frac{k}{k\plus{}1}$

So, we'll just have to prove it for $ n \equal{} k\plus{}1$. By the problem:

$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_k \plus{} a_{k\plus{}1} \equal{} (k\plus{}1)^2 (a_{k\plus{}1})$

$ \frac{k}{k\plus{}1} \equal{} [(k\plus{}1)^2 \minus{} 1] a_{k\plus{}1}$

$ \frac{k}{k\plus{}1} \cdot \frac{1}{(k\plus{}1)^2\minus{}1} \equal{} \frac{1}{(k\plus{}1)(k\plus{}2)}$

which is what we would get if $ n \equal{} k\plus{}1$ was true.

$ \therefore a_n \equal{} \frac{1}{n(n\plus{}1)} \ \blacksquare$. [/hide]

[quote="Poincare"]That i think is a really good solution aside from where you just "assume" that $ n \equal{} k$. That means i don't really get the induction part  :rotfl:  :D   :blush:[/quote]

In induction, you prove $ n \equal{} 1$ works then assume $ n \equal{} k$ to prove $ n \equal{} k\plus{}1$. This works because you can change $ k$ to any number to apply this. It's like a domino logic.

$ (a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_{n \minus{} 1}) \plus{} a_n \equal{} (n \minus{} 1)^2 a_{n \minus{} 1} \plus{} a_n \equal{} n^2 a_n$

$ \Rightarrow a_n \equal{} \frac {n \minus{} 1}{n \plus{} 1} \ a_{n \minus{} 1} \equal{} \frac {n \minus{} 1}{n \plus{} 1} \ \frac {n \minus{} 2}{n} \ a_{n \minus{} 2}$

$ \equal{} \frac {n \minus{} 1}{n \plus{} 1} \times \frac {n \minus{} 2}{n} \times ....\times \frac {2}{4} \times \frac {1}{3} \times a_1$

$ \equal{} \frac {1}{n(n \plus{} 1)}$

edited: typo observed by Zuton Force in the post below corrected

I do believe that should be:

$ (a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_{n \minus{} 1}) \plus{} a_n \equal{}$ $ \color{red}{(n \minus{} 1)}^2$ $ a_{n \minus{} 1} \plus{} a_n \equal{} n^2 a_n$

Thanks guys. I get the full solution now haha.  :D

I get the induction, but what if you did the following.
Lemma: $ a_{n} \equal{} \frac{1}{n^2 \plus{} n}$
Then
$ \displaystyle\sum_{i\equal{}1}^{n}a_i \equal{} \displaystyle\sum_{i\equal{}1}^{n}\frac{1}{i^2 \plus{}1}$
$ \equal{}\displaystyle\sum_{i\equal{}1}^{n}\left(\frac{1}{i} \minus{} \frac{1}{i\plus{}1}\right)$
$ \equal{} 1\minus{} \frac{1}{n\plus{}1}$
$ \equal{} \frac{n}{n\plus{}1}$
Now by our Lemma again $ a_n \equal{} \frac{n}{n\plus{}1}$.
And done? Will that be okay? Or do we have to induct?

Using the second condition, we have$$a_{n+1} = (n+1)^2a_{n+1} - n^2 a_n \implies \frac{a_{n+1}}{a_n} = \frac n{n+2}.$$Thus inductively $a_n = \frac 1{(n+1)(n+2)}$.


Putnam function limit integration algebra domain geometry

Let $ f: (1,\infty)\to\mathbb{R}$ be a differentiable function such that
\[ f'(x)\equal{}\frac{x^2\minus{}\left(f(x)\right)^2}{x^2\left(\left(f(x)\right)^2\plus{}1\right)}\quad\text{for all }x>1.\]
Prove that $ \displaystyle\lim_{x\to\infty}f(x)\equal{}\infty.$

Let me know if the following makes sense:

Subtracting 1 from both sides we get:

    $ f'(x) \minus{} 1 \equal{} \minus{} \frac {(f(x))^2(x^2 \plus{} 1)}{x^2((f(x))^2 \plus{} 1}) < 0$, therefore $ f'(x) < 1$.

In a similar fashion,adding 1 on both sides and also using the fact that f is defined for $ x > 1$ we prove that $ f'(x) > \minus{} 1$.

            Hence $ \minus{} 1 < f'(x) < 1$. (1)

Solve in terms of $ f(x)$ in the original equation and you will find it in terms of f'(x).
Let me know your thoughts.

We know that $ f'(f(x)) \equal{} 0$ for all $ x > 1$ and $ f'( \minus{} f(x)) \equal{} 0$ for all $ x > 1$. E.g. $ f'(x) \equal{} \frac {(x \minus{} f(x))(x \plus{} f(x))}{x^{2}(f(x)^{2} \plus{} 1}$.

I don't see how this can help to solve this problem but still i think there is some flaw in your argument because the domain of the function is $ (1,\infty)$ hence the substitution $ x\minus{}>f(x)$ is not always applicable because $ f(x)$ can take all the real values,i think.

That doesn't make sense, ragingbullrocky. This is a differential equation; a solution curve may happen to pass through a point where $ f(x)\equal{}x$, but it doesn't stay there.

[quote="new_member"]Let me know if the following makes sense:

Subtracting 1 from both sides we get:

    $ f'(x) \minus{} 1 \equal{} \minus{} \frac {(f(x))^2(x^2 \plus{} 1)}{x^2((f(x))^2 \plus{} 1}) < 0$, therefore $ f'(x) < 1$.

In a similar fashion,adding 1 on both sides and also using the fact that f is defined for $ x > 1$ we prove that $ f'(x) > \minus{} 1$.

            Hence $ \minus{} 1 < f'(x) < 1$. (1)

Solve in terms of $ f(x)$ in the original equation and you will find it in terms of f'(x).
Using (1) then you can see that the limit goes to infinity,i think.
Let me know your thoughts.[/quote]

Can you show me your f(x) in terms of f'(x)?

I got it but it did not seem to go to infinity.

I found :

    $ f(x) = \sqrt {\frac {1 - f'(x)}{f'(x) + \frac {1}{x^2}}}$.
Question: Can we use the fact that $ - 1 < f'(x) < 1$ on that relation to find the $ lim{x - > \infty}f(x)$?

That's not actually true, since $ f(x)$ can be negative. More importantly, I know of no techniques that work with expressing a differential equation that way. It just doesn't help us do anything.

Does the following make any sense. Clearly $ f(x)\to c$ for some constant $ c$ fails (since then $ f'(x)\to d > 0$ for some constant $ d$). If $ f(x)\to \minus{} \infty$ then $ f'(x)\to 0$ as $ \frac {1}{x^2}$, which implies $ f(x)\to c$, a contradiction. Thus $ f(x)\to\infty$.

What if it doesn't have a limit? Your three cases do not exhaust all functions.

you can prove that for every $ m$ there is a $ M$ so that $ x > M \implies f(x)\neq m$. limit has to be $ \pm \infty$.
then show it can not be $ \minus{} \infty$ as it would have to lie under(outside) the cone $ x > |f(x)|$ contradicting the diff eq we have given... This was a pain to write out, with all the cases and every thing (I also didn't write it in the most efficient form)

I don't know if this would be right but

Assume $ \displaystyle\lim_{x\to\infty}f(x) \equal{} / \equal{} \infty$

then $ \displaystyle\lim_{x\to\infty}f'(x) \equal{} \displaystyle\lim_{x\to\infty}1/((f(x))^2 \plus{} 1)$ which is greater than 0

so $ \displaystyle\lim_{t\to\infty} \int_1^t f'(x)dx \equal{} \infty$ which is a contradiction

what if that limit doesn't exist? you can't conclude it's positive...

I'm not sure about not existing but it should always be positive when it does exist (because it's it's f(x)^2)

After seeing all this fumbling on how to express that argument, I'm liking my explicit quantitative method more and more.

Writing out this sort of thing for a differential equation is hard.

Is it possible to receive points with an answer showing that the limit of the derivative is positive and has a greater rate of growth than the natural logarithm function? (aka withoutout actually doing any integrals or derivatives?)

WLOG, we may assume that there exists some $ x_0>1$ such that $ f(x_0)<x_0$, or else $ f(x)$ would clearly go to infinity. Also, note that $ f'(x)\le1$ for all $ x>1$. It follows that $ f(x)<x$ for all $ x\ge x_0$. This implies that $ f'(x)>0$ for all $ x\ge x_0$, so $ f(x)$ is eventually decreasing. Thus, by the Monotone Convergence Theorem, $ f(x)$ either diverges to infinity or converges to a finite limit. In the former case, we are done. If $ f(x)$ approaches a finite limit $ L$, then $ f'(x)$ approaches $ \frac{1}{L^2\plus{}1}>0$. Thus, in particular, $ f'(x)$ is bounded below by some $ b>0$ for $ x$ sufficiently large. But this implies that $ f(x)$ goes to infinity, so we are done.

I showed first that $ f$ is unbounded from above (if not, then $ f'$ is bounded from below for large $ x$ and integrating you can make $ f$ as large as you want).
Then assume for some $ M$ and for any $ k$ we can find $ l > k$ for which $ f(l) < M$. (If such $ M$ does not exist result follows immediately; I guess it is also possible to have $ f(l)\leq M$ but then we just take $ M \plus{} 1$ instead of $ M$.). But $ f$ is unbounded so we can find points as large as we want for which $ f$ is greater than $ M$.
So we have a sequence of points $ x_1 < x_2 < ...$ which approaches infinity, and for which $ f(x_i) < M$ for odd $ i$, $ f(x_i) > M$ for even $ i$. But then by continuity of $ f$ there is a sequence of points $ z_1 < z_2 < ...$ which also goes to infinity, and $ f(z_1) \equal{} f(z_2) \equal{} ... \equal{} M$, and $ f(x) > M$ for $ x\in (z_i, z_{i \plus{} 1})$ if $ i$ is odd (unfortunately this part I was not able to show very explicitly during the contest; one argument that I think works is that you cannot have the $ z_i$s forming an interval or else $ f'(x) \equal{} 0$ on that interval so $ f(x) \equal{} x$ or $ f(x) \equal{} \minus{} x$; impossible as $ f(x) \equal{} M$ on that whole interval). By Rolle's Theorem there is a sequence of points $ y_1 < y_2 < ...$ approaching infinity so that $ y_i \in (z_i, z_{i \plus{} 1})$ and $ f'(y_i) \equal{} 0$. Then $ f(y_i) \equal{} y_i$ or $ f(y_i) \equal{} \minus{} y_i$. In particular for sufficiently large odd $ i$ we then have $ f(y_i) > M > \minus{} y_i$ so we must have $ f(y_i) \equal{} y_i$. But after that $ f$ must get back to the line $ y \equal{} M$ which is impossible since $ f'(x) > 0$ for $ M < f(x) < x$, so $ f$ is increasing in that area and will not be able to get back down to the line $ y \equal{} M$. This gives a contradiction, and we are done.

[quote="jmerry"]My solution:

Write $ y'=\frac1{y^2+1}-\frac1{x^2}\cdot\frac{y^2}{y^2+1}.$ The latter term can take us down only by a total of 1 at most, while the first term inexorably pushes y up.
Suppose $ 1-A\le f(0),$ and choose $ M>A>0,$ $ B\ge (M+A)((M+1)^2+1)+1.$* Split into two cases: either (1) $ f(s)> M+1$ for some $ s\in [1,B],$ or (2) $ f(x)\le M+1$ on the whole interval.
Case 1:
$ f(B)=f(s)+\int_s^B\frac1{f^2(x)+1}\,dx-\int_s^B \frac1{x^2}\cdot\frac{f^2(x)}{f^2(x)+1}\,dx$
$ f(B)>f(s)+\int_s^B\frac1{f^2(x)+1}\,dx-\int_s^B \frac1{x^2}\,dx$
$ f(B)>f(s)-\frac1s+\frac1B>f(s)-1>M.$
Case 2:
$ f(B)=f(1)+\int_1^B\frac1{f^2(x)+1}\,dx-\int_1^B \frac1{x^2}\cdot\frac{f^2(x)}{f^2(x)+1}\,dx$
$ f(B)>f(1)+\int_1^B\frac1{f^2(x)+1}\,dx-\int_1^B \frac1{x^2}\,dx$
$ f(B)>f(1)-1+\int_1^B\frac1{f^2(x)+1}\,dx$
$ f(B)>-A+(B-1)\frac1{(M+1)^2+1}=M$
Either way, $ f(B)>M$ for all large enough $ B.$ Repeat for all $ M,$ and $ f(x)$ goes to $ \infty$ as $ x\to\infty.$

*I chose that number after writing the rest of the argument, of course. Standard practice in analysis estimates.[/quote]

A post from long time ago, but I think this solution is false.  The case 2 argument fails without demonstrating $f(x)>0$. (We could have $f(x) < -(M+1)$, i which case the inequality cannot hold). This isn't that difficult to prove, but it seems to me that if we were to go for an analytical solution for this differential equation, this step (proving $f(x)>0$ for all $x>N$ for some $N$) is the trickiest  step(or at least as tricky as other steps).

By my definitions, we have $f(0)\ge 1-A>1-M$. By the simple estimate of my first line, $f(x)>-M$ for all $x$. Case 2 actually becomes $-M<f(x)\le M+1$ on the whole interval, and the inequality holds.

I found this:

Case 1)
 If f(x) is greater than sqrt(x-2) for arbitrarily large values of x, then obviously f approaches infinity as x approaches infinity.
Case 2)
There is M such that if x is greater than M then f(x) is less than sqrt(x-2). In this case, one can easily calculate that f'(x)>1/x for x>M. This means that there is c such that f(x) +c > ln(x) and we're done.

First, we have $f'(x) > -\frac{1}{x^2}$, so $\int_2^x f'(t) dt > \int_2^x -\frac{1}{t^2} dt$ or $f(x) > \frac{1}{x} + O(1)$, so $f$ is bounded below.

Next, we have $f'(x) < \frac{1}{f(x)^2 + 1}$, so $\int_{2}^x [f(t)^2+1]f'(t) dt < \int_2^x dt$, or $\frac{f(x)^3}{3} + f(x) < x + O(1)$. Since $f$ is bounded below, we get $f(x)^3 < 3x+O(1)$, or $f(x) < \sqrt[3]{3x+O(1)}$.

Both of these imply that $|f(x)| = o(x)$, and in particular for large enough $x>C$ we have $|f(x)| < \frac{x}{2}$. Then for such $x$, $f'(x) > \frac{3/4}{f(x)^2+1}$, so $\int_C^x [f(t)^2+1] f'(t)dt > \int_C^x \frac{3}{4}dt$, or $\frac{f(x)^3}{3} + f(x) > \frac{3}{4}x + O(1)$. Now it is clear that $\lim_{x\rightarrow\infty} f(x) = \infty$ (in particular, $f=O(x^{1/3})$).

Since,$f'(x)>-\frac{1}{x^2}$ so by integrating from 2 to $x$ we can say $f$ has a lower bound for large $x$. Assume for all $x\ge M, f(x)\ge C$.

Now consider the function $g(x)= f(x)+\frac{1}{3}f(x)^3$.
We have ,$$g'(x)=f'(x)(1+f(x)^2)=1-\frac{f(x)^2}{x^2}\le 1\quad(\star)$$
This means that, $g(x)-x\le R$ for some constant $R$.
So
\begin{align*}
g(x)-x&= f(x)+\frac{f(x)^3}{3}-x\le R \\
&\implies \frac{f(x)^3}{3}-x\le R-f(x)\le R-C\quad \forall x\ge M\\
f(x)&\le [x+(R-C)]^\frac{1}{3}
\end{align*}
This means that $f(x)=o(x)$.From $(\star)$ we can say $\lim _{x\to \infty}g'(x)=1$.
But then,using L'Hospital rule,
$$\lim_{x\to \infty} g(x)=\lim_{x\to \infty} \frac{g(x)}{x}\lim_{x\to \infty}x=\lim_{x\to \infty}g'(x)\times \infty=\infty$$
But this implies $f(x)\to \infty,\text{as}\ \ x\to \infty$. $\blacksquare$

[quote=rogue](by Yevgen Bilokopytov, Kyiv Taras Shevchenko National University)

Let $ g(x) \equal{} f(x) \minus{} \frac {1}{x}.$ Then $ g'(x) \equal{} \frac {x^2 \plus{} 1}{x^2(f(x)^2 \plus{} 1)} > 0.$ Thus $ g$ is increasing. Consider 2 cases:

(1) $ \lim\limits_{x\to\infty}g(x) \equal{} c\in\mathbb{R}.$ Hence $ \lim\limits_{x\to\infty}f(x) \equal{} c,$ $ f'(x)\to \frac {1}{c^2 \plus{} 1} > 0$ and we have $ f'(x)\ge\frac {1}{2(c^2 \plus{} 1)} > 0,$ $ x\ge x_0,$ for some $ x_0.$ Then for $ x\ge x_0$ we have $ f(x)\ge f(x_0) \plus{} \frac {x \minus{} x_0}{c^2 \plus{}1}\to\infty,$ $ x\to\infty,$ a contradiction.

(2) $ \lim\limits_{x\to\infty}g(x) \equal{} \infty.$ Then $ \lim\limits_{x\to\infty}f(x) \equal{} \infty.$[/quote]

Can anyone motivate the substitution $g(x) = f(x) - \frac{1}{x}$. It's so ingenious
Also can anyone explain why if we have $x_0$ such that $f'(x) \ge \frac{1}{2(c^2 + 1)}$ for all $x \ge x_0$, then we must have
\[ f(x) \ge f(x_0) + (x - x_0) \cdot \frac{1}{c^2 + 1} \]
I thought it's by Tangent Line, but $f$ is not convex.

[quote=oty]It's just the limit definition.  if $g(x) \to L$ and $a\leq L$ you can always find $x_{0}$  such that $g(x) \geq a$.[/quote]

I get that this implies $f'(x) \ge \frac{1}{2(c^2 + 1)}$ for some $x \ge x_0$ since $f'(x) \to \frac{1}{c^2 + 1}$, but I was asking how the inequality below it (the one I said was by "Tangent line" --  but it's not even a convex function) to be true

It's just integration the inequality over $[x_{0},x]$.


Let $ABC$ be a triangle.Prove that:
$\displaystyle SinA^{SinB}+SinB^{SinC}+SinC^{SinA} > \frac {1}{\sqrt[e]{e}}+1$

if it comes from a book, the book will probably give you a solution for this... :?


Find all positive integer solutions $ x,y,z,p$, with $ p$ a prime, to the equation $ x^p\plus{}y^p\equal{}p^z$.

I've got a solution, but it's pretty horrendous and is long as well.

Let p be an odd prime, and let z be the smallest positive integer which admits a solution (x,y) for this equation with this fixed odd prime p.

Since p is an odd prime, $ x + y$ is a factor of the L.H.S., hence it is a factor of the right hand side.

Therefore, either $ x + y = 1$ or $ x + y = p^k$ for some $ 1\lek\lez$. The former does not admit a solution in the positive integers, so we have $ x^p + (p^k - x)^p = p^z$. k cannot equal z, because then the exponents force us to have p=1.

If $ z\ge k + 2$, consider the equation mod $ p^{k + 2}$.

$ x^p + (p^k - x)^p\equiv 0$
By the binomial theorem, every term will be $ \equiv 0$ except $ x^p - x^p + p(p^k)(x^{p - 1}) \equiv0$

$ p^{k + 1}x^{p - 1} \equiv 0$ so p is a factor of x, so it's also a factor of y, so dividing both sides of the original equation by $ p^p$ we have a smaller z, e.g. z-p, which we can use, contradicting the minimality of z.

Hence, $ k < z < k + 2$ so $ k = z - 1$.

$ x^p + (p^{z - 1} - x)^p = p^z$

But now, one of those two terms is $ \ge p^z/2$ since the parts under the exponent sum to p^z, and I claim that $ (p^z/2)^p > p^z$, thereby deriving that the L.H.S. > R.H.S.

This gives $ p^{pz - z} > 2^p$ Which is true since $ p^{pz - z} \ge p^{p - 1} > 2^p$ which isn't that hard to prove (you  could use induction).

Hence, the L.H.S. > R.H.S. contradiction!

So this doesn't work for odd primes. But maybe it works for p=2.

This boils down to $ x^2 + y^2 = 2^z$ Now, z=1 gives x=y=1, and z=2 gives no solution. So assume now that
$ z\ge 3$. Then simple mod 4 checking will convince you that x,y are both even. Then dividing both sides by 4, we get a smaller z. Continuing this way, we boil it down to z=1, or z=2. z=2 doesn't work, but z=1 gives x=y=1. Working backwards, we see that x,y must have been equal throughout this whole procedure to get x=y=1, so x=y, which gives $ 2*x^2 = 2^z$ so $ x = 2^{\frac {z - 1}{2}}$ so z is odd, let z=2k+1, putting it al together, we have:

(k,k,2k+1,2) as the complete set of ordered quadruples (x,y,z,p). Q.E.D.


geometry
trapezoid
trigonometry
analytic geometry
graphing lines
slope
symmetry

Nonagon ABCDEFGHI has side length 1. Equilateral triangle ADG is drawn on it. If you fold flaps down along the lines of ADG, a smaller equilateral triangle is formed. What is the side length of the smaller triangle?

cool problem
[hide]
Looking at trapezoid ABCD, $CDE=140, ADG=60, CDA=GDE \implies CDA = 40$.  Then $AD = 2\cos 40 + 1$ and the height $h = \sin 40$.  We can ignore the weird folds of the flaps and just concentrate on the intersections by thinking of it as an equilateral triangle with side $s = AD$, and drawing 3 lines parallel to each side length $h$ from the parallel side.
Construct such a triangle in the first quadrant with one vertex at $(0,0)$ and a second at $(s, 0)$.  If we can find the intersection between the two lines parallel to the sides meeting at $(0,0)$, we can find the side length.  The first line is $y = \sin 40$.  Let point $P$ be where the second line intersects the x-axis, and $Q$ the altitude from $P$ to the line it's parallel to.  Since $QOP = 60$ ($O$ being the origin) and $QP = \sin 40$, we have $OP = \frac{2 \sin 40}{\sqrt 3}$.  The slope is $\sqrt 3$, so the equation of this line is $y = \sqrt 3 (x - OP) = \sqrt 3 x - 2 \sin 40$.  Solving the system, $x = \sqrt 3 \sin 40$.  By symmetry, the length of the small triangle is twice this distance subtracted from the length of the whole triangle, or $2 \cos 40 + 1 - 2\sqrt 3 \sin 40$.  This simplifies to $4 \left( \frac 12 \cos 40 - \frac{ \sqrt 3}2 \sin 40 \right) + 1 = 4 \cos 100 + 1 = \boxed{1 - 4 \cos 80^\circ}$
[/hide]


quadratics
algebra
quadratic formula
special factorizations

[quote="t0rajir0u"]Any quadratic is factorable using the quadratic formula.  I'm not sure where those coefficients are coming from, though.  The resulting factorization does not have a nice form in terms of $ a$ and $ b$.[/quote]
Wait, Sorry,.. I made a typo. Can you look it over again?

I'm pretty sure that's on the step of how to get the quadratic formula from $ ax^2\plus{}bx\plus{}c$

[quote="DonkeyKong"]I was looking at something involved with completing the square and the process of getting quadratic formula, a question came to my mind.

Can you factor $ x^2 \plus{} \frac ba x \plus{} (\frac b{2a})^2$?

I think it may be possible...but I need you guys to help me.[/quote]

It is simply $ \left(x\plus{} \dfrac b{2a}\right)^2$.

You just have to treat $ (\frac{b}{2a})^2$ as an integer and plug in to the quadratic formual.

[quote="Brut3Forc3"][quote="DonkeyKong"]I was looking at something involved with completing the square and the process of getting quadratic formula, a question came to my mind.

Can you factor $ x^2 \plus{} \frac ba x \plus{} (\frac b{2a})^2$?

I think it may be possible...but I need you guys to help me.[/quote]

It is simply $ \left(x \plus{} \dfrac b{2a}\right)^2$.[/quote]
Oh.. yeah. That works.

Thanks

[quote="darkmatter47"]You just have to treat $ (\frac {b}{2a})^2$ as an integer[/quote]

Being an integer has nothing to do with it.  If we set $ k \equal{} \frac{b}{2a}$, the given expression is of the form $ x^2 \plus{} 2kx \plus{} k^2 \equal{} (x \plus{} k)^2$, which is a standard expansion.

[quote="t0rajir0u"][quote="darkmatter47"]You just have to treat $ (\frac {b}{2a})^2$ as an integer[/quote]

Being an integer has nothing to do with it.  If we set $ k \equal{} \frac {b}{2a}$, the given expression is of the form $ x^2 \plus{} 2kx \plus{} k^2 \equal{} (x \plus{} k)^2$, which is a standard expansion.[/quote]

But at first, it might seem to complicated to apply the quadratic formula to (if it does not factor well), so I suggested treating it as one woould do to an integer, or something else simpler looking (like $ k$)

Applying the quadratic formula to this is basically circular reasoning.  Factoring this expression is used in deriving the quadratic formula.

But, as we have clearly shown, the expression is factorable.

[quote="darkmatter47"]But at first, it might seem to complicated to apply the quadratic formula to[/quote]

It's perfect for applying the "quadratic formula" (by which I actually mean completing the square).

[quote="DonkeyKong"]I was looking at something involved with completing the square and the process of getting quadratic formula, a question came to my mind.

Can you factor $ x^2 \plus{} \frac ba x \plus{} (\frac b{2a})^2$?

I think it may be possible...but I need you guys to help me.[/quote]

well according to ax^2 + bx + c, since x has no coefficient, a must equal 1

so it is now x^2 + bx + b^2 which is pretty easy to factor

Ihatepie, your logic is kinda flawed. Nowhere does the question say $ a$ is the coefficient of $ x^2$.  Therefore, you are using $ a$ for two different values.


What would your math instructor (teacher/professor) get on the AIME?

My teacher is really cool and really good.  I said he would get 12-14.

Wow, I said my would get 0-1.  But then again, what would 1 expect from a bad school......... :(

My teacher's pretty experienced in math, but isn't always able to find an elegant solution. I'd say 8-9.  :)

This is just wrong! Teachers can't be that good! That would be better than me......! :mad:

I vote 2-3.  My teacher's not that smart, though I don't think she'd ever admit so.

2-3. I know mainly because he's tried to solve some of the Mandelbrot Regional test problems and it takes him like 5 minutes to get the first one...

[quote="t0rajir0u"]My teacher has freely admitted that she probably couldn't get many.  There are a couple reasons for this:

1)  She's out of practice,

2)  She's seven months pregnant.

Example:  Yesterday she claimed that $\frac{22}{7}$ was irrational.  :)[/quote]

Ditto that. We're in the same math class.

Anyone from Exeter gets to answer 15 :D. My math teacher is much better than average but can still probably only solve 2-3. My physics teacher can probably solve 10ish though.

Let's just say that being able to find out how those age things work (those ones where you start with your age, and multiply and subtract and a bunch of stuff to get a number that everyone else will get no matter what their age is) is a lot easier than most of the problems on the AIME.

Well...My math teacher actually TAKE the aime test unofficially with us when we took the test....
It happens that he got a 7

Our math instructor likes to say "Good luck, because I can't even do those problems, and I have the [i]answers[/i]."

they don't have the answers to the AIME ;)

i prefer not to comment on how well my teacher would do, but i did vote :)

Are we all forgetting our instructors right here, MCrawford, RRusczyk, DPatrick, NSato, and Valentin Vornicu?  They would all surely get 15's.  =)

I believe that there's a big difference between the knowledge involved in teaching a Geometry class using a teacher's edition textbook and actually doing AIME-style geometry problems.  After all, The AIME wouldn't be all that difficult if the contest people outlined some hints after each problem.

None of the math teachers at my school are into problem-solving  :( Not really sure how well they would do.

On the 2006 AIME I, advanced class teachers at my school would probably definitely get #1,3, and 5.  Then maybe a couple more depending on what they specialize in, so I'll say 4-6.  Maybe more, but I don't know, AIMEs tough.  They're pretty good at problem solving tho.  4-6 sounds about right.

Do you think that the average math teacher would do better than the average AIME math student?  ( I think for students, the average is aroudn a 2-3 or so)

My math coach tries to work on the AIME problenms every year as we take it. I think he got 7 or 8 this year. But he's an exception at our school by far. If the other math teachers had the AIME put in front of them, they'd run away screaming "IMPOSSIBLE!!!!" Of course, that might have to do with the fact that they can't use their vital TI-83's.

[quote="paladin8"][quote="Xantos C. Guin"]AlGI teacher: 5 max.
She did not know how to graph y=(3/5)x on a TI-82
She also said that the squareroot of seven is rational.

Geometry teacher: 9-11
She knows how to do geometry problems well and is somewhat proficient in trig.

ALGII teacher: 9-13
She can do advanced trig and makes less careless mistakes.[/quote]

I would be very impressed if a teacher could do 9-13 problems on the AIME... there are only a handful of students who can, and these are pretty much the best in the country. Knowing some trig doesn't quite equate to a 9-13 on the AIME... and the test doesn't only require math knowledge, but very strong problem solving ability (something I have yet to see in many of my teachers at school).[/quote]

Now that I think about that, Those are probably over estimates. Subtract around 3-4 from what I thought.

I just want to say that this is an awesome poll.

As for my school's math teachers, I'm not exactly sure where to place their problem solving skills, though I do know that careless mistakes would have a large effect.  ;)


If the surface area of a cube is 54, evaluate its volume.

[hide] The volume would 27. Since all faces of cubes are equal, each face would have an area of 9 (54/6). All sides of squares are equal so we know each side is 3. 3 cubed would give us the volume, 27.[/hide]
[color=darkred][size=200]Hide'd[/size][/color]

[hide]Let a be the length of the edge.
 6*a^2=54 then a=3. The Volume of the cube is equal to a^3=27  :D[/hide]

[hide]Let a be the length of the edge.
 6*a^2=54 then a=3. The Volume of the cube is equal to a^3=27  :D [/hide]

[hide="answer"]27..[/hide]

I have to leave so fast!!!!! :P

[hide="solution"]if the surface area is 54, each face will be 9 so a side legth is 3

then the volume is 3^3=27[/hide]

[hide]Each face has area 54/6 = 9.  So the side length is 3, and the volume is 3 x 3 x 3 = 27[/hide]

54 is the surface area. To find the area of one side, divide it by six. 54/6 is 9. Square root nine to find the length of one side. sqrt of 9 is 3. 3 cubed, or 3*3*3, is 27. The volume of the cube is 27.


Find the value of

$\frac{\left(\sin{56^{\circ}}\right)\left(\sin{34^{\circ}}\right)\left(\tan{22^{\circ}\right)}}{\cos{68^{\circ}}}$.


How many real solutions are there to the equation $ |x \minus{} |2x \plus{} 1\parallel{} \equal{} 3.$ (Here, $ |x|$ denotes the absolute value of $ x$: i.e., if $ x \geq 0,$ then $ |x| \equal{} |\minus{}x| \equal{} x.$)

A. 0
B. 1
C. 2
D. 3
E. 4

[hide]
\[ |x \minus{} |2x \plus{} 1|\ | \equal{} 3 \Longrightarrow \left\{ \begin{array}{lr} x \minus{} |2x \plus{} 1| \equal{} 3 \Longrightarrow \left\{ \begin{array}{lr} x \minus{} 3 \equal{} 2x \plus{} 1 \Longrightarrow x \equal{} \minus{} 4 \\
3 \minus{} x \equal{} 2x \plus{} 1 \Longrightarrow x \equal{} \frac 23 \end{array} \right. \\
x \minus{} |2x \plus{} 1| \equal{} \minus{} 3 \Longrightarrow \left\{ \begin{array}{lr} x \plus{} 3 \equal{} 2x \plus{} 1 \Longrightarrow x \equal{} 2 \\
\minus{} x \minus{} 3 \equal{} 2x \plus{} 1 \Longrightarrow x \equal{} \frac { \minus{} 4}{3} \end{array} \right. \end{array} \right.
\]
Of these, only $ x \equal{} \boxed{2, \minus{} 4/3}$ actually work, so the answer is $ \mathrm{(C)}$. [/hide]

[size=59]Solution hid by Silverfalcon[/size]


Let $ABCD$ be a convex quadrilateral, the middlepoints $M,N,P,R$
of the sides $[AB], [BC], [CD], [DA]$ respectively and the intersections
$I\in AC\cap BD,\ E\in AB\cap CD,\ F\in AD\cap BC.$
I denote the point $G$ for which $GE\parallel MP,\ GF\parallel NR$.
Prove that $G\in BD\Longleftrightarrow IA=IC.$
See and http://www.mathlinks.ro/Forum/viewtopic.php?t=31921

Let $I$ be the origin of the vectorial system, i.e. for any $X$ we have $X=\overrightarrow {IX}$ and $I=\overrightarrow {0}$.

$(\exists )m,n>0$ such that $IC=mIA,\ ID=nIB$, i.e. $C=-mA,\ D=-nB$. Thus,

$2M=A+B,\ 2N=-mA+B,\ 2P=-mA-nB,\ 2R=A-nB$.
**********************************************************
$E\in AB\cap CD\Longleftrightarrow (\exists )\alpha ,\beta \in R-\{ 0,1\}$ such that

 $E=\alpha +(1-\alpha )B=\beta D+(1-\beta )C$

$\Longleftrightarrow \alpha =\frac{m(1+n)}{m-n},\ \beta =\frac{1+m}{m-n},\ E=\frac{m(1+n)}{m-n}\cdot A-\frac{n(1+m)}{m-n}\cdot B.$
***********************************************************
$F\in AD\cap BC\Longleftrightarrow (\exists ) u,v\in R-\{0,1\},\ F=uC+(1-u)B=vD+(1-v)A\Longleftrightarrow$

$F=-umA+(1-u)B=-vnB+(1-v)A\Longleftrightarrow$

$u=\frac{1+n}{1-mn},\ v=\frac{1+m}{1-mn},\ F=\frac{m(1+n)}{mn-1}\cdot A+\frac{n(1+m)}{mn-1}\cdot B$.
************************************************************
$G\in BD\Longleftrightarrow (\exists )x,y,z\in R^*,\ G=zB=E+x(M-P)=F+y(N-R)\Longleftrightarrow$

$\frac{2m(1+n)}{mn-1}-y(1+m)=\frac{2m(1+n)}{m-n}+x(1+m)=0,$

$\frac{2n(1+m)}{mn-1}+y(1+n)=\frac{2n(1+m)}{n-m}+x(1+n)\Longleftrightarrow$

$x=\frac{2m(1+n)}{(n-m)(1+m)},\ y=\frac{2m(1+n)}{(mn-1)(1+m)},$

$\frac{n(1+m)}{mn-1}+\frac{m(1+n)^2}{(mn-1)(1+m)}=\frac{n(1+m)}{n-m}+\frac{m(1+n)^2}{(n-m)(1+m)}\Longleftrightarrow$
***********************************************************
$\left[ \frac{1}{mn-1}-\frac{1}{n-m}\right] \cdot \left[ \frac{m(1+n)^2}{1+m}+n(1+m)\right]=0\Longleftrightarrow$

$mn-1=n-m\Longleftrightarrow m=1\Longleftrightarrow IA=IC.$

[b]Remark. [/b]What are the relations between the four points $M\in AB,\ N\in BC,\ P\in CD,\ R\in DA$ such that $G\in BD$ ?.
.


AwesomeMath
summer program
email
Mathcamp

so um guys you know the old people at AwesomeMath? where by old people I mean the mentor guys or whatever you call them

like I think zach abel might have been there last year

er, how do they get picked to be mentor guys?

It seems that [url=http://www.awesomemath.org/staff.shtml#assistants]Abel[/url] is still a mentor.

I think your best path would be to email (titu@awesomemath.org) Dr. Titu Andreescu, the program director, and ask him how you can be hired.

[quote="MysticTerminator"]it is sort of incumbent upon me to point out that I might not be allowed to mentor you in how to get all the girls[/quote]
aww dang man

i've got no reason to go to AMP now

Personal Essay Question 1 Response:

My extracurricular activities include picking up girls... and doing math, duh...

[quote="Treething"]yufei's a decent enough mentor

i mean i wrote him a note saying i'd be back in my dorm by 7 and ardito changed the 7 to a 2

and when i came back at 6:30 he didn't flip out

btw this is 7 AM[/quote]
{0,0}
(__(|
-"-"-
NO WAI!!

what other stuff happened last year?

we were going to go have a party in the common house thingie.

but zuming was sitting there, watching TV

there were like 15 of us. we saw him sitting there and just turned back in silence. hahaha

i don't know man.  chinese people scare me.  i go to enloe, though, so i'm like paranoid.  (arnav gets this.)

although like every sophomore in the top 50 at enloe got into NCSSM and is going.  except for me and like 5 other kids.  so like i should just be like ok zuming i come to exeter like you want b/c all my friends leave me anyway.  screw them.

what friends?

no but i mean, david vincent (my AMSP mentor) didn't even care at all.  we just came back and if he was awake then he knew we were back - if he wasn't, he didn't, until the next morning.

[quote="Phelpedo"]I'm pretty sure Yufei was the only mentor who did anything. I saw zabel, but I'm not really sure why he was there (he wasn't a mentor).[/quote]

zach was probably just doing nothing, wasting time until he went off to college

Zach (we are talking about zabel right?) wasn't a mentor. I did see him a couple times, but I don't know what he was doing.

[quote="calc rulz"][quote="MysticTerminator"]eh he wouldn't have minded

HA JUST KIDDING

he got pretty pissed at us at MOP for staying up late I think

so that night alex and I stayed up until around 3?

yeah?[/quote]

Mathcamp pwns MOP because we can stay up late having problem parties.[/quote]

Definitely.  Last year 3 was about the earliest I got to bed the entire last week of camp (which was capped by two consecutive all-nighters).


Let $n$ be a positive integer. Find all monotone functions $f$ from $\mathbb R$ to $\mathbb R$ such that
\[f(x+f(y))=f(x)+y^{n},\]
for all $x,y \in \mathbb R$.

[quote="N.T.TUAN"]From hypothesis we have $f$ is injective.
Choose $y=0$ then $f(x+f(0))=f(x)$, so $f(0)=0$.
Choose $x=0$ we have $f(f(y))=y^{n}$, so $n$ is odd. Therefore $f$ is surjective and $f$ is additive, so $f(x)=ax$, check! We will find $a$.[/quote]

I'm sorry, could you explain how $f(f(y))=y^{n}$ implies $f$ is surjective and $f$ is additive?

1)If $n$ is odd then $\{y^{n}|y\in\mathbb{R}\}=\mathbb{R}$, therefore $f$ is surjective.
2)From above we have $f(x+f(y))=f(x)+f(f(y))$ and so $f$ is additive.

why $n$ must be odd?

and why if we have $f(x)$ bijective and additive,so we have $f(x)=ax$ ?
thx..

[quote="Yosh..."]why $n$ must be odd?[/quote]
Because $f(f(y))=y^{n}$ and $f_{o}f$ is increasing!

[quote="silouan"]Let $n\in N$ . Find all monotone functions f from R to R such that
$f(x+f(y))=f(x)+y^{n}$[/quote]

I don't agree with N.T.TUAN when he said that hypothesis imply $f(x)$ is an injective function. It's true but not immediate (a monotonous function may be non-injective). BTW, we don't need to show that $f(x)$ is injective.

I suggest a little modification to N.T.TUAn demo (which is rather good indeed) :

$E1(x,y)$ : $f(x+f(y))=f(x)+y^{n}$
$E1(x,0)$ : $f(x+f(0))=f(x)$ and if $f(0)\neq 0$, we have a monotonous periodic function, so a constant and no constant satisfy the problem, so $f(0)=0$
Then $E1(0,x)$ gives $f(f(x))=x^{n}$ and $n$ is odd since $f(x)$ monotonous implies $f(f(x))$ increasing and $x^{n}$ increasing implies $n$ odd.
Then $n$ odd and $f(f(x))=x^{n}$ implies $f(x)$ is a surjective function.
$E1(0,y)$ : $f(f(y))=y^{n}$ and so $E1(x,y)$ becomes $f(x+f(y))=f(x)+f(f(y))$ and since $f(x)$ is surjective $f(x+z)=f(x)+f(z)$ and $f(x)$ is additive.

If $f(x)$ is additive, it is well known that $f(x)=f(1)x$ for any $x\in\mathbb{Q}$ and then $f(x)=f(1)x$ for any $x\in\mathbb{R}$ since $f(x)$ is monotonous.

So, if we write $f(1)=a$, $ax+a^{2}y=ax+y^{n}$ and $n=1$ and $a\in\{-1,+1\}$

[quote="N.T.TUAN"]I mean that ''monotone functions''= increasing or decreasing. Therefore $f$ is injective![/quote]

OK, but I think that "monotonic" function is a "non-decreasing" or "non increasing" one.

See [url]http://en.wikipedia.org/wiki/Monotonic_function[/url], for example.

Let $P(x,y)$ be the assertion $f(x+f(y))=f(x)+y^n$.

[b]Case 1:[/b] $n$ is odd
$P(0,x)\Rightarrow f(f(x))=f(0)+x^n$, so $f$ is bijective
$P(0,0)\Rightarrow f(f(0))=f(0)\Rightarrow f(0)=0$ by injectivity.
$P(0,x)\Rightarrow f(f(x))=x^n$, and we have $f(f(f(y)))=f(y)^n=f\left(y^n\right)$
$P(x,f(y))\Rightarrow f\left(x+y^n\right)=f(x)+f(y)^n=f(x)+f\left(y^n\right)$, so $f$ is additive and thus linear since it's monotone. We get the solutions $\boxed{n=1,f(x)=x}$ and $\boxed{n=1,f(x)=-x}$, which both work.

[b]Case 2:[/b] $n$ is even
$P(0,x)\Rightarrow f(f(x))=f(0)+x^n$, so if $f(a)=f(b)$ then $a^2=b^2$.
$P(0,0)\Rightarrow f(f(0))=f(0)\Rightarrow f(0)^2=0\Rightarrow f(0)=0$
$P(0,x)\Rightarrow f(f(x))=x^n$, so $f(f(f(x)))=f(x)^n=f\left(x^n\right)$
$P(x,f(y))\Rightarrow f\left(x+y^n\right)=f(x)+f(y)^n=f(x)+f\left(y^n\right)$, thus $f\left(-y^n\right)=-f\left(y^n\right)$, which we obtain by setting $x=-y^n$. By oddness, $f$ is additive, thus linear, and we find no solutions in this case.


There are 14 red, 14 blue and 14 yellow marbles in a jar. What is the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 6 to 13 and the ratio of yellow to non-yellow marbles is 6 to 13?

[hide="ANS."]4[/hide]

[hide]
The number of red marbles must be 6 or 12, as does the number of yellow marbles.
Let's try to make 12 red and 12 yellow to minimize the number removed.
You find that blue = 14.
So you must remove 2 red marbles and 2 yellow marbles.[/hide]


IMO 2004
polyomino
rectangle
combinatorics
tilings
IMO
IMO Shortlist

Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.

[asy]
unitsize(0.5 cm);

draw((0,0)--(1,0));
draw((0,1)--(1,1));
draw((2,1)--(3,1));
draw((0,2)--(3,2));
draw((0,3)--(3,3));
draw((0,0)--(0,3));
draw((1,0)--(1,3));
draw((2,1)--(2,3));
draw((3,1)--(3,3));
[/asy]
Determine all $ m\times n$ rectangles that can  be covered without gaps and without overlaps with hooks such that

- the rectangle is covered without gaps and without overlaps
- no part of a hook covers area outside the rectangle.

This problem looks like a traditional Saint-Petersburg or 239MO problem, respectively. It is kind of common usage to make use of such problems over and over again.  :)

solution from Christian Sattler(Germany): first we define an involution on the hooks: send every hook to the one which includes the square where the hook goes around. it is clear that this is an involution, so the number of hooks is even. it immediately follows that 12 | mn. there are three cases(wlog 3 divides m): 3 divides m and 4 divides n(trivial since we can do 3x4-rectangles), 12 divides m, then n can be arbritarily, but not equal to 1,2 or 5(it is clear that n is not 1,2, and 5 is easy to exclude. in all other cases we may write n as linear combination of 3 and 4 with positive coefficients and we can use 3x4-rectangles again taking rows of 3x4-rectangles or 4x3-rectangles). so we are left with the case 6 | m  and 2 | n. so look again at our involution: it shows that our hooks appear in pairs of the possible forms: either 3x4-rectangles or something similar to an S:
.===
.===
===.
===.
now mark every second row. it is easy to see that there are always 6 out of the 12 squares of the S are marked, the same for horizontal 3x4-rectangles, but vertical 3x4-rectangles have either 4 or 8 squares marked, that shows that there's an even number of them. doing the same with columns, we get that the total number of 3x4-rectangles is even. now, look at the coloring of
.==.
=..=
.==.
=..=
and periodic. it is easy to see that 6 out of 12 squares are marked at 3x4-rectangles, the S upright, the S turned by 90 degree and even distance to the left boundary of the rectangle, but 4 or 8 for an S turned by 90 degree with odd distance to the left boundary. this shows that there's an even number of them and by translating the coloring by 1 to the right and turning it all around, we get that the number of S is even as well. this shows that 24 is a divisor of mn, so we get one of the two other cases.

i only found all solutions but couldn't prove that it are all.

Peter

Hi all,

I am the author of this problem (or the one who proposed it actually, the solution is a teamwork). I thought that it might be interesting for the readers of this forum to know the story behind its birth. So here it comes.

In autumn 2003 I gave a problem creation seminar for our young olympiad organizers, since they complained that they were not able to come up with new problems. As the first topic I chose combinatorics and said: "Problem creation in combinatorics is trivial -- take an arbitrary shape (I drew the hook above) and ask whether it is possible to fill some rectangle with these". I never imagined that this would ever become a real olympiad problem, yet to talk about IMO!

Jan

Dear Jan,

it would be great if you inform us where do you live, and to post the "official" solution to your problem.

Thanks in advance.

Kantor

I agree with Kantor: the more solutions the merrier.

So please, Jan, post the solution(s) you had in mind.

~Mike

[url]http://www.math.ucf.edu/~reid/Polyomino/j6_rect.html[/url] gives one proof.
This is indeed an easy sort of problem to create if you don't mind the solution being already known and online.

"There is a small hole in the proof of Part 1, pertaining to Figure 3C.  The astute reader should be able to find and close the hole."

Edit:  Actually, the hole was fatal.  The proof has been removed.  Sorry.

Here's a cute way of showing that if a rectangle $6a\times 2b$ can be tiled with $3\times 4$ rectangles then at least one of $a,b$ must be even: we start from a corner of the board and place a $1$ there. Then we continue to the right, placing $i\cdot k$ to the right of $k$ (so the numbers in the squares will be $1,i,-1,-i$). The $n+1$'th row is the $n$'th row multiplied by $i$ (so we write $i\cdot k$ below $k$). Each $3\times 4$ rectangle has the same number of occurences for $1,i,-1,-i$, but it's easy to see that a rectangle of the form $4m+2\times 4n+2$ doesn't satisfy this.

Well, the problem in itself is nice, but all these tiling-type problems have been studied a lot and many papers appeared on this kind of problems. Thus, the reference given by jsm28 is not so surprising and this probably appear somewhere else.
This should be sufficient for the jury to eliminate the problem, because some contestants may have known it before.

Pierre.

By some simple observations (but not trivial, in fact i think that the most important part is that of the invultion, there is the magic) the problem reduces to tile a rectangle of shape m*n (with m=6m and n=2nmand nodd) with 3*4 bricks and some S (which shape you probably know).
 What we want is to show that the number of S is even, and that the number of 3*4 bricks is even two. From here we deduce that 24 divide mn which contradict the shape of the rectangle.
 For the bricks, procede in this way, put the number i-j on the intersection of column i and row j. You can easy check that mod 12 the sum of every S is Zero and that of every brick is 6. But if you look every 2*6 rectangle have sum Zero mod 12, from where the whole rectangle have sum zero, so the number of bricks is even.
 For the S, we do the following, call a j-column a column numebr k with j=k mod 4. Put in each 1-column a 1 and in each 3-column a -1. It is easy to check that the sum of all number is 2 mod 4.But bricks have sum Zero mod 4, and the 3333 S too.Now, the S with 2442 square per column have sum +2 or -2 mod 4. Hence Suming over all S we get that there is an odd number of this type of S. For the other type of S (the one with 3333 squares per column) aply the same idea but with rows and we get that of this new type there are an odd number. Then the total number of S is even and we are done.

 Ok, when i wrote it i havent read yet that link solution, it make me feel like a fool such nice solution, anyway...here is what i did.

Hi all,

Sorry for not specifying my country in the first post -- I (and hence also the problem) come from Estonia.

I was afraid that something like the link posted by jsm28 might come up, but there is really no way of telling it before it actually does. Our own solution went pretty much along the lines of this link, so there is not much point in retyping it here.  There are several details (e.g. there is also a third way of covering the "center" of this hexamino, but that does not work), but they can be sorted out.

Of course there is a philosophical question "some students might have seen this link". But please consider that the amount of material on Internet is huge and the chance that some student has found Michael Reid's page and memorized the facts about J6 hexamino, is pretty small. Anyway, Michael Reid's page has a lot of information about other tile shapes as well, so why not remember all of them?

Jan

I didn't say that Jan. Your problem is nice, but I thought that an IMO problem had to be not 'well-known' (even if I didn't know it before myself).
In my opinion, tiling problems as this one have given a lot of papers, with quite often elementary solutions, the problem is quite always to find the good invariant/coloration (which indeed can be really hard) and then it probably appeared in some math magazine somewhere, and thus may be known to some contestants. This risk is sufficiently high to think about it.

Pierre.

I totally agree with Pierre, however the problem is very nice. Anyway, congratulations Jan for your second problem to be considered for an IMO, the first was problem 1 IMO 1999.

Kantor

I used combination of similar approach and coloring in my solution.

The only thing I do not get about this question is: why did they make it number 3?? I mean I've already tried day 1 at home, and I guess I'd have gotten a 7 for this one (which I'd certainly not get for the others).

It's simply
1. showing they can only come as rectangles or semi-tetros (the other form), this means 12 | mn (pretty easy)
2. showing it's ok if at least 1 side is a multiple of 4 (even easier)
(unless you got a 12x1 or 12x2 or 12x5 case -- the other number must be writable as 4m+3n)
3. finding any contradiction on the way that no side divides 4.
(I did the the same way as grobber: 1,i,-1,-i,1,... with sum and product contradicting.)

Is it really an easy question to IMO standards, or am I just lucky having found the right coloring immediately? I'm not used to solve problems being number 3! :P

[quote="Valentin Vornicu"]Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.
[img]http://www.mathlinks.ro/Forum/files/hook.png[/img]
Determine all $m\times n$ rectangles that can  be covered without gaps and without overlaps with hooks such that

- the rectangle is covered without gaps and without overlaps
- no part of a hook covers area outside the rectagle.[/quote]

Can the squares move randomly or they hook is regular
Abdurashid

[quote="Valentin Vornicu"]Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.
[img]http://www.mathlinks.ro/Forum/files/hook.png[/img]
Determine all $m\times n$ rectangles that can  be covered without gaps and without overlaps with hooks such that

- the rectangle is covered without gaps and without overlaps
- no part of a hook covers area outside the rectagle.[/quote]

Can the squares move randomly or the hook is regular
Abdurashid

You have that hook, given, a lot of exact copies of it. So 6 blocks glued together.

And by using only such hooks, you should overlap the m x n rectangle.

Can this problem be done using some idea of generating functions? That is, for the $ m\times n$ board, you consider each cell of the form $ x^i y^j$ where row wise you name the cell $ x^i$ and column wise you name the cells $ y^j$  and name the cells of the board correspondingly as $ x^iy^j$.
For the hook, there are lots of cases leading to lots of construction of polynomial functions??? Can someone suggest an easier way.

Sorry, I don't understand what Grobber means by Msub4. Could someone please explain?

[quote="grobber"]I like this problem, and I have the feeling a lot of solutions will come up :).

Here's another proof that a board $4m+2\times 4n+2$ can't be tiled with J-hexomino's:

First put $1,i,-1,-1$ in the squares of the board like in my first post. In each $3\times 4$ rectangle the elements add up to $0$, so they have $M_4+M_4i$, while in each shape formed by two intertwined hexomino's such that they don't form a $3\times 4$ rectangle (call this shape $S$) the sum is $2,2i,-2$ or $-2i$. The sum of the numbers on the entire board is $2,2i,-2$ or $-2i$ (depending on the number we put in the top left corner), so there must be an odd number of $S$ shapes $(*)$.

If the two $2\times 1$ bits of an $S$ shape are vertical we call that a vertical $S$ shape, and if they're horizontal we call it horizontal.

We now put $i$ on the first line of the board, and continue by putting $i^n$ on the $n$'th line. It's easy to see that the sum of the numbers in the board has form $M_4+2+(M_4+2)i$, and the sum of the numbers in a $3\times 4$ rectangle and in a vertical $S$ is $M_4+(M_4)i$, while for a horizontal $S$ the sum is $M_4+2+(M_4+2)i$, so we must have an odd number of horizontal $S$. By doing the same, but with columns instead of lines, we find an odd number of vertical $S$, so the total number is odd+odd=even, contradicting $(*)$.[/quote]

what do you mean by M sub4?

riemann, I'm pretty sure that this famous Msub4 means just "multiple of $4$", with classical notation $\mathcal{M}4$ (grobber talks about "... form $M_4$ ...). But do you noticed you are asking a question on a 2004 or 2005 post? Never too late ... !

The answer is: $4a \times 3b$ rectangles (with $a,b \ge 1$) and $12a \times b$ rectangles (with $b \neq 1,2,5$ and $a \ge 1$).

Proof:
Note that we can unite two hooks to form a $3\times 4$ rectangle. Using these rectangles it's easy to see $4a \times 3b$ rectangles (with $a,b \ge 1$) satisfy. Now, if we have a $12a \times b$ rectangle (with $b \neq 1,2,5$ and $a \ge 1$), then by stacking either horizontally or vertically the rectangles, we see that the rectangle can be tiled if a $12a \times (b-3)$ or $12a \times (b-4)$ rectangle can be tiled. So if $b=4x+3y$ ($x,y \ge 0$), then the rectangle can be tiled. It's easy to see that all numbers $b$ are of that form, except for $b=1,2,5$. Therefore, the rectangles I described above can be tiled.

Now I prove no other rectangle can be tiled.

Assume $n \times m$ is not one of the above rectangles, and that it can be tiled (we will prove a contradiction). Note that the hole of each hook must be covered by some other hook, and it's easy to see we can pair up the hooks like this. This forms $3\times 4$ rectangles and strange "snape" shapes, which look like two 2x3 rectangles stacked unevenly. Both of these shapes have 12 squares, and so $12 | nm$. Since $n,m$ is not of the above rectangles, then we have that

$n \times m$ = $12k \times 1$, $12k \times 2$, $12k \times 5$ or $6a \times 2b$, with $a,b$ odd and $k$ a positive integer.

In the last case ($6a \times 2b$), color the rectangle like a chess board, with the top left corner black. We have eight types of snapes:

$S_1$ snapes:

oXxx
oxxx
xxxo
xxXo

where the capital $X$'s are black.

$S_2$ snapes:

xxxo
xxXo
oXxx
oxxx

where the capital $X$'s are black.

$S_3$ snapes:

ooxx
Xxxx
xxxX
xxoo

where the capital $X$'s are black.

$S_4$ snapes:

xxoo
xxXx
xXxx
ooxx

where the capital $X$'s are black.

$S_5$ snapes:

oXxx
oxxx
xxxo
xxXo

where the capital $X$'s are white.

$S_6$ snapes:

xxxo
xxXo
oXxx
oxxx

where the capital $X$'s are white.

$S_7$ snapes:

ooxx
Xxxx
xxxX
xxoo

where the capital $X$'s are white.

$S_8$ snapes:

xxoo
xxXx
xXxx
ooxx

where the capital $X$'s are white.

Now, color the rectangle with $4$ colors, using southwest-northeast diagonals. The first diagonal (which in reality is only the square in the top left corner) is color $1$, the second diagonal is color $2$, third is color $3$, fourth is color $4$, fifth is color $1$, and so on. We see that there will be $Y$ squares $1$, $Y+1$ squares $2$, $Y$ squares $3$ and $Y-1$ squares $4$, for some $Y$. Each $3 \times 4$ rectangle will cover 3 squares of each color.

For all $k \ge 5$, each $S_k$ snape will cover 3 squares $1$ and 3 squares $3$.

For all $k \le 4$, each $S_k$ snape will cover 2 squares $1$ and 4 squares $3$, or 4 squares $1$ and 2 squares $3$.

Since the number of squares $1$ is equal to the number of squares $3$, we have $S_1+S_2+S_3+S_4$ is even.

Now, color again, but this time using southeast-northwest diagonals, so that the top left corner is color $1$. This time, the number of squares $2$ is equal to the number of squares $4$. With analogous reasoning, we see that $S_5+S_6+S_7+S_8$ is even.

Therefore, the number of snapes is even.

Finally, color with 4 colors, according to the parity of each coordinate of the squares. Each snape will cover 3 squares of each color, while each $3 \times 4$ rectangle will cover an even number of squares of each color. Since in total there are an odd number of squares of each color, there will be an odd number of snapes!

Contradiction.

Therefore we have

$n \times m$ = $12k \times 1$, $12k \times 2$ or $12k \times 5$

Since each hook takes up space from 3 rows and 3 columns, the first two options are not possible.

Now if $n=12k$, $m=5$, with $k$ a positive integer, look at the first column. (Columns have 5 squares). We consider each $3 \times 4$ rectangle and each snape as the union of two $2 \times 3$ rectangles. The top left and bottom left corners can't both be covered by vertical rectangles. If they are both covered by horizontal rectangles, then the 3rd square of the left-most column can't be covered. So, without loss of generality the top left corner is covered by a horizontal rectangle, and the bottom left is covered by a vertical one. Then it's easy to see that that horizontal rectangle can't be united with any other rectangle. Contradiction!

So we are done.

The answer is: $4a \times 3b$ rectangles (with $a,b \ge 1$) and $12a \times b$ rectangles (with $b \neq 1,2,5$ and $a \ge 1$).

How about coloring the board in 12 colors
ABCDABCD...
EFGHEFGH...
IJKLIJKL... on below and sideways.  I think the rest is clear of as each square covers only 1 of the colors.

Edit: this is a fakesolve

[hide=Fakesolve]Notice that if another hook fits inside the "hole" of a hook, they must create a $3 \times 4$ rectangle or two $3 \times 2$ rectangles, with one shifted one square. Every hook must pair with some other hook in this way, so the number of hooks is even. Thus, the number of squares is divisible by $12$. We have three cases:

[b]Case 1:[/b] $12 \mid m$. Then, we can create $3 \times 12$ rectangles and $4 \times 12$ rectangles with $3 \times 4$ rectangles, so by Chicken McNugget, we find that $n=3,4,6,7,\ldots$ work. It's easy to see that $n=1,2,5$ don't work.

[b]Case 2:[/b] $4 \mid m$ and $3 \mid n$. Then, we can tile with a grid of $3 \times 4$ rectangles.

[b]Case 3:[/b] $6 \mid m$, $2 \mid n$, and $m \equiv n \equiv 2 \pmod{4}$. Then, we can color the rectangle with the $4 \times 4$ pattern \begin{align*}
0 \quad 0 \quad 1 \quad 1\\
0 \quad 0 \quad 1 \quad 1\\
1 \quad 1 \quad 0 \quad 0\\
1 \quad 1 \quad 0 \quad 0
\end{align*} Notice that both of the $12$-square patterns cover the same number of zeroes and ones, but there are more zeroes than ones when $m \equiv n \equiv 2 \pmod{4}$. Thus, it cannot be tiled.[/hide]

For the $6a \times 2b$ case, I have a short coloring proof which I haven't seen anywhere above. I provide a complete solution:

By the manner in which the "hole" can be filled, there are two possible shapes: an S-shape and a $3 \times 4$ rectangle. These are the only two ways to cover the hole, and the holes of both participating hooks are filled, so hooks occur in pairs and are even in number, so $12 | mn$.

Clearly $3|m$ and $4|n$ works, since this can be tiled with $3 \times 4$ rectangles.

$12|m$ and $n \neq 1,2,5$ also works since $12 \times 3$ and $12 \times 4$ are possible. These can be placed one above the other to give $n$ which is a linear combination of $3$ and $4$, clearly giving $n \neq 1,2,5$, and all others work. The $12 \times 5$ rectangle can be shown to be un-tilable using some small casework for corner pieces.

There remains one case: $6a \times 2b$ for odd $a, b$. Note that $4$ doesn't divide either of $m, n$. Assume it is possible to fill the grid with hooks. Colour the grid with colours $1$ through $4$ with colour congruent to sum of row and column number modulo $4$. In colouring a $6a \times 2b$ rectangle for odd $a, b$, there is $1$ more square with colour $1$, $1$ more square with colour $3$, and $2$ more squares with colour $2$ than squares of colour $4$. However, in every S-shape and $3 \times 4$ rectangle, the parity of number of $2$-coloured and $1$-coloured squares is always the same, hence their difference cannot be $1$ when summed over several S-shapes and $3 \times 4$ rectangles, a contradiction.

Clearly, if there exists a hook, then the square in the middle must be part of another hook. The only valid options are the following two:
[center]
[asy]
size(6cm,0);
draw((0,0)--(3,0)--(3,3)--(2,3)--(2,1)--(1,1)--(1,2)--(0,2)--(0,0));
draw((1,2)--(1,4)--(4,4)--(4,2)--(3,2));
draw((6,0)--(6,3)--(10,3)--(10,0)--(6,0));
draw((7,0)--(7,2)--(8,2)--(8,1)--(9,1)--(9,3));
[/asy]
[/center]
In particular, there must be an even number of hooks, so the number of grid is a multiple of $12$. From now on, we will ban rotations and instead add two more figures that we can tile with. Reflections are still allowed. Denote the first one by $A$ and its ninety degree rotation by $B$. Denote the second one by $C$ and its ninety degree rotation by $D$.

If we color every other row red, each type of figure will cover six red squares, except figure $B$, which will cover four or eight. Similarly, if we color every other column red, each type of figure will cover six red squres, except figure $D$, which will cover four or eight. If we group two squares horizontally and apply chessboard to those groups, each type of figure will cover six red squares, except figure $C$, which will cover four or eight. If we do the same vertically we get similar result for $A$.

Thus, if both $m$ and $n$ are even, then we have proved that an even number of each figure is needed, so $24\mid mn$. Then, one of them must be divisible by $4$. If not both $m$ and $n$ are even, $12\mid mn$ forces that anyway. WLOG, assume $4\mid m$. Then, either we have $12\mid m$ or $4\mid m$, $3\mid n$. The second one is obviously possible and the first is possible for any $n=3a+4b$ for some nonnegative integers $a,b$. That is, $n$ can equal anything that isn't $1$, $2$, or $5$ and it's each to check that those don't work.

:wallbash_red:

The answer is all $3a \times 4b$ rectangles, as well as all $12a\times b$ rectangles, where $b \not \in \{1,2,5\}$. To see that these work, observe that we can use $3 \times 4$ rectangles only to tile the first class, and for the second we can use $3 \times 4$ rectangles to form both $3 \times 12$ and $4 \times 12$ rectangles, so by Chicken McNugget we can form any $12a \times b$ rectangle for $b \geq 6$. It is also easy to see that we can form $12a \times 3$ and $12a \times 4$, but clearly $b=1,2,5$ are not possible. We now show that these are the only answers.

It is clear that in any valid tiling, by considering the "divot" in each hook, we can either pair up the shapes to form a $3 \times 4$ rectangle or an "S-shape", which is a $4 \times 4$ square with the highest two cells on the left column and the lowest two cells on the right column removed, as well as rotations and reflections. This immediately implies that $12 \mid mn$. The rest of this solution will show that we cannot have $\nu_2(m)=\nu_2(n)=1$; suppose that this was true for the sake of contradiction.

Ban rotations (but allow reflections) and consider the four following shapes:
[list]
[*] Rectangles which are $3$ wide and $4$ tall (type A)
[*] Rectangles which are $4$ wide and $3$ tall (type B)
[*] S-shapes where we remove from the left/right columns (type C)
[*] S-shapes where we remove from the top/bottom rows (type D)
[/list]
Number the columns $0$ through $n-1$ in order, and for $0 \leq i \leq 3$ let $a_i$ denote the number of type A rectangles whose leftmost cells lie in a column whose label is congruent to $i \pmod{4}$. Define $c_i$ similarly. Now,
[list]
[*] By coloring the columns in the pattern $1212\ldots$, we get an equal number of $1$s and $2$s. Type B, C, and D shapes cover an equal number of $1$s and $2$s, so we get the equation $a_0+a_2=a_1+a_3$.
[*] By coloring the columns in the pattern $1221\ldots$, we get an equal number of $1$s and $2$s, since $\nu_2(n)=1$. Type B and D shapes cover an equal number of $1$s and $2$s, but type A shapes either have four extra $1$s or four extra $2$s, and certain C shapes can as well, so we get the equation $a_0+a_1+c_0=a_2+a_3+c_2$.
[*] By coloring the columns in the pattern $1122\ldots$, we get $2m$ extra $1$s; note that $\nu_2(2m)=2$. Type B and D shapes still cover an equal number of $1$s and $2$s, but type A shapes either have four extra $1$s or four extra $2$s, and certain C shapes can as well, so we get the equation $a_0+a_3+c_3=a_1+a_2+c_1+\tfrac{m}{2}$.
[/list]
Taking things modulo $2$, it follows that $c_0 \equiv c_2 \pmod{2}$ and $c_3 \equiv c_1+1 \pmod{2}$, since $\tfrac{m}{2}$ is odd. Therefore there are an odd number of type C shapes. Also, since $a_0+a_1+a_2+a_3 \equiv 0 \pmod{2}$, there are an even number of type A shapes.

By repeating this argument with rows and type B/D shapes, we also get that there are an odd number of type D shapes, and an even number of type B shapes. Therefore, there are an even total number of shapes, so the number of cells in the $m \times n$ rectangle must be divisible by $24$, but this contradicts $\nu_2(m)=\nu_2(n)=1$; done. $\blacksquare$

[b]Remark:[/b] As a wise man once said,
[quote=Inconsistent]This problem is similar to modern art, in the sense that I have no clue what the point of this problem even is.[/quote]


Re current USAMO selection rules:
(1)  top 160 AMC12 + 10xAIME
(2)  anyone who took AMC 10 and got AIME score at least as high as the lowest AIME score in (1)
(3) highest in any state with no qualifiers from (1) or (2)

a. How many years have these rules been in effect?

b.  What was the cutoff point for (1)  for '99-'03 (I've seen a     posting of 226 for '03 and 210 for '02) ?

c.  What was the floor for (2) for '99-03 I've seen '03 8, '02 6, '01 6, '00 7) ?

In the past they used a different system entirely. They took the top 100-and-something regardless of grade. Then the next 20 (or so) in grades 11 and below, then 20 in grades 10 and below, then 20 in grades 9 and below, then the state people.

There is a history of tweaking the rules after they have the test results.

And before that I believe it used to simply be anyone with an AHSME + 10*AIME index of 200 or above. Which is still basically what it turns out to be now, with the exception of some younger students and state representatives.

Yes, I now see the selection criteria for each year from the USAMO archive page on the AMC site.  The current selection criteria has been used since '02.

 Look at the 160th
 best score to help
 compare the years
 ___ AMC_12_AIME(a)_ (b)__(c)
'02__136_____9.43__210?__6
'03__140.5____10___226?__8
'04__133.5

(a) eg 10 means exactly 160 AIME scores were 10 or higher.
       9.5 would mean half the 9s fell in the top 160.
       9.1 would mean 90% of the 9s fell in the top 160.
(b) the cutoff for the 1st 160 selected from AMC12+AIMEx10
(c) the lowest (floor) AIME score of the 160 in (b)

The stated goal is to select ~250 for USAMO, but  In '02 327 qualified as apparantly one (or more) of the 37 perfect 150 AMC 12s got a 6 on AIME which opened the door for 94 10th graders and below who also had a 6.

For 6 to be the floor this year:
- top 160 cutoff is 210 or lower and 1 of the 17 AMC 12 150s has 6
- .......................206.5............................23...........>=146.5
- .......................204...............................34............>=144
-...etc

For 7 to be the floor this year:
- top 160 cutoff is 220 or lower and 1 of the 17 AMC 12 150s has 7
- .......................216.5............................23...........>=146.5
- .......................214...............................34............>=144
-...etc

My index is a 225.5--so if the cutoff this year is 226 again, I will break things.

From the above, the AMC 12 component of the index is lower this year.  And posts on topics such as "AIME cutoff", "AIME 1 scores", "Wording on AIME #1", "AIME scores for TJ, AAST..." suggest that the AIME component is lower as well this year.

210 may be a reasonable estimate for the cutoff.


calculus
integration
algebra
polynomial
derivative
calculus computations

Your signature is incomplete:

"For with God nothing shall be impossible." -Luke 1:37

I think that this integral can be calculated as follows:
let
$\int\sum_{k=0}^{n}a_{k}x^{k}e^{-x}dx=P_{n}e^{-x}$, where $P_{n}$ is a polynomial of degree $n$ which is to be calculated.

Taking the derivative of this equality we get that

$\sum_{k=0}^{n}a_{k}x^{k}=P'_{n}-P_{n}$, which is a linear differential equation which can be solved by standard techniques.
Didi


Moves are only allowed one segment to the right or one segment up. How many paths from $A$ and $B$ are possible?

[img]7592[/img]

[quote="jhredsox"]I got a 26 on this national sprint.

I get [hide]18[/hide] for this one.

can't explain it too mathematically though

jorian[/quote]

Off by a lot.

you just use the "block-to-block" method

[hide]i agree that the answer is 91[/hide]

I know this is hard to explain without a picture.  But, the average response here is about 5 words.  Doesn't anybody who gave a response want to give a little more detail and/or draw a picture?


For $ \sum_{k\equal{}3}^{\infty} k z^k$, determine the disc of convergence and determine at which points on the boundary of the disc of convergence the series converges.

The quotient criterium gives that the radius is $ 1$. And it doesn't converge at any point on the boundary (the summands are not bounded).


calculus
derivative
integration
advanced fields
advanced fields unsolved

Let $([0,1],\mathcal{M},m)$ be a Lebesque measure space on $[0,1]$. Let \[g(x,y)=\begin{cases}1 &\text{for }0\leq y\leq x\leq 1\\ 0&\text{for }0\leq x<y\leq 1\end{cases}\] and let \[\tau(A): =\int_{A\times[0,1]}g\,d(m\times m)\] (a) Show that $\tau<< m$ ( I got this)
(b) Find $d\tau/dm$  ( I am not sure about this)

Hint: The double integral defining $\tau(A)$ reduces to a single integral over $A$ with respect to $x$. The form of the integrand will give you the Radon-Nikodym derivative.


Solve the equation $(x^2-3x+3)^2-3(x^2-3x+3)+3=x$.

I found two solutions: one from my own and the other from the book. I think mine is easier! :P

[hide]
If $r$ is a root of $x^2-4x+3$ (which factors into $(x-3)(x-1)$), then $(r^2-3r+3)^2-3(r^2-3r+3)+3=r^2-3r+3=r$, so $r$ will be a root of the polynomial. Therefore 1 and 3 are roots. To find the other two roots, factor out $(x-3)(x-1)$. Since the $x^4$ coefficient of $(x^2-3x+3)^2-3(x^2-3x+3)+3-x$ is 1, the $x^3$ coefficient is -6, and the constant term is 3, we have
\[ (x^2-3x+3)^2-3(x^2-3x+3)+3-x=(x-3)(x-1)(x^2-2x+1)  \]
and this factors further into $(x-3)(x-1)^3$, so all of the roots are $\boxed{1,3}$.
[/hide]

[quote="scorpius119"][hide]
If $r$ is a root of $x^2-4x+3$ (which factors into $(x-3)(x-1)$), then $(r^2-3r+3)^2-3(r^2-3r+3)+3=r^2-3r+3=r$, so $r$ will be a root of the polynomial. Therefore 1 and 3 are roots. To find the other two roots, factor out $(x-3)(x-1)$. Since the $x^4$ coefficient of $(x^2-3x+3)^2-3(x^2-3x+3)+3-x$ is 1, the $x^3$ coefficient is $-6$, and the constant term is 3, we have
\[ (x^2-3x+3)^2-3(x^2-3x+3)+3=(x-3)(x-1)(x^2-6x+1)  \]
Using the quadratic formula, the other roots are $3\pm 2\sqrt{2}$, so all of the roots are $\boxed{1,3,3\pm 2\sqrt{2}}$.
[/hide][/quote]

I think you didn't see the $x$ in th RHS.

oops, i didn't deal with the $x^3$ coefficient correctly. i'll fix this.

If x are ony integers, i got $x = 1, 3$
I think there should be more solutions, not sure

[quote="Sir.Aaron"]Here is what I got. :?:  :huh:
$(x^2-3x+3)^2-3(x^2-3x+3)+3=x$

$x^4-9x^2+9-3x^2-3x+3=x$

$x^3-15x^5+9+3+3=x$

$16x^9+15=x$

$16x^9=16x$

$x=16x^8$[/quote]

You expanded it wrongly, if i'm not mistaken. try it again. I don't really understand some of your steps too

[quote="Sir.Aaron"]
squred everything in the ()
$x^3-15x^5+9+3+3=x$
Combined like terms
$16x^9+15=x$[/quote]

Since when is  $x^3 - 15x^5 = 16x^9$?   :|

Edit:  The previous line and the line after are also both completely wrong... I don't understand what you did at all.

[quote="Sir.Aaron"][quote="Sir.Aaron"]
You expanded it wrongly, if i'm not mistaken. try it again. I don't really understand some of your steps too[/quote]

I will explain what I did.
$(x^2-3x+3)^2-3(x^2-3x+3)+3=x$

$x^4-9x^2+9-3x^2-3x+3=x$
squred everything in the ()
$x^3-15x^5+9+3+3=x$
Combined like terms
$16x^9+15=x$
Combined like terms
$16x^9=16x$
Combined like terms, the devided by 16
$x=16x^8$

Do you[b] have to [/b]  factor for this problem?[/quote]

since it's polynomial of power four, of course it's easier if you factor. I still don't get you, i think it's wrong. and for the first 2 lines, are you sure it's correct? i got something different if i expand

[quote="Sir.Aaron"][quote="Sir.Aaron"]
You expanded it wrongly, if i'm not mistaken. try it again. I don't really understand some of your steps too[/quote]

I will explain what I did.
$(x^2-3x+3)^2-3(x^2-3x+3)+3=x$

$x^4-9x^2+9-3x^2-3x+3=x$
squred everything in the ()
$x^3-15x^5+9+3+3=x$
Combined like terms
$16x^9+15=x$
Combined like terms
$16x^9=16x$
Combined like terms, the devided by 16
$x=16x^8$

Do you[b] have to [/b]  factor for this problem?[/quote]

I think you have a few algebra mistakes in this... in general,

$(x^2-3x+3)^2 \neq x^4-9x^2+9$

$-3(x^2-3x+3) \neq -3x^2+3x+3$

$-9x^2-3x^2-3x \neq -15x^5$

$x^3-15x^5 \neq 16x^9$

$x - 15 \neq 16x$

$16x^9-16x \neq 16x^8$

The answers are 1 and 3
The given equation factored into $(x-1)^2(x-3)=0$
Also note that if $f(x)=x^2-3x+3$, then our equation becomes $f(f(x))=x$.


algebra
polynomial
quadratics
quadratic formula

[quote="maokid7"][quote="A M Y"]Not another one of those big polynomials  :o

fine... I'll do it...
[hide="ok"]
$x^{5}+5x^{4}-20x^{2}-2x+14=(x+1)(x^{4}+4x^{3}-4x^{2}-16x+14)$

So $x=-1$ solves.

Then, $x^{4}+4x^{3}-4x^{2}-16x+14=x^{4}+4x^{3}-4x^{2}-16x+16-2$

$=(x^{2}+2x-4)^{2}-(\sqrt{2})^{2}=(x^{2}+2x-4+\sqrt{2})(x^{2}+2x-4-\sqrt{2})$

And then you do the quadratic formula for a solution set of:

$\boxed{x=-1\;\text{or}\;x=-1\pm\sqrt{5\pm\sqrt{2}}}$
[/hide][/quote]

could you please explain your thought process for this solution.[/quote]

Well, basically, you use synthetic substitution at the first step (guess some simple values for x using your p/q values), then you get a 1st degree polynomial times a 4th degree.

Then you look at the 4th degree polynomial part.

You want to somehow manipulate it so that it factors into two quadratics (so you can use the quadratic formula to solve for each of the quadratics). The easiest way to do this is to add a value that will make the quartic a quadratic, then subtract it, thus creating a difference of two squares situation in which you can factor and apply the quadratic formula to both factors.


Express in the form $ z\equal{}u(x,y)\plus{}iv(x,y)$ where $ z\equal{}x\plus{}iy$.

$ f(z)\equal{}\frac{z\plus{}i}{z^{2}\plus{}1}$

Does anybody have an idea?  I know the answer should be
$ \frac{x}{x^{2}\plus{}(y\minus{}1)^{2}}\plus{}\frac{(1\minus{}y)}{x^{2}\plus{}(y\minus{}1)^{2}}i$

$ \frac {z \plus{} i}{z^2 \plus{} 1} \equal{} \frac {1}{z \minus{} i} \equal{} \frac {1}{z \minus{} i}\frac {\overline{z} \plus{} i}{\overline{z} \plus{} i} \equal{} \frac {\overline{z} \plus{} i}{|\overline{z} \plus{} i|^2} \equal{} \frac {x \minus{} i(y \minus{} 1)}{x^2 \plus{} (y \minus{} 1)^2}$
The idea, which works for simplifying any complex ratio, is to make the denominator real by multiplying the numerator and denominator by the denominator's complex conjugate.  I don't think problems of this level belong in the Olympiad section.


inequalities
triangle inequality
algebra proposed
algebra

for each n from N* and let $S_n$ the least value of the sum
\[ \sum^n_{k=1} \sqrt{(2k-1)+a_k}\]
such that $a_i$ are all positive reals and \[\sum^n_{k=1}\ a_k=17\]

prove that there exist one integer n such that $S_n$  is also an integer

Indeed, using triangle inequality we obtain
\[\sum_{k=1}^{n}\sqrt{(2k-1)^2+a_k^2}\geq \sqrt{(1+3+...+(2k-1))^2+(a_1+a_2+...+a_n)^2}=\sqrt{n^4+17^2},\]
and equality holds iff $\frac{a_k}{2k-1}=\textrm{const}$. Therefore, $S_n=\sqrt{n^4+17^2}$ and we are to find such $n$ that $n^4+17^2$ is a perfect square.
From Pythagorean triples we know that $17$ must be $a^2-b^2$ and $n^2$ must be $2ab$, but equation $17=a^2-b^2$ has unique solution $a=9$ and $b=8$, so $n^2=2\cdot 9\cdot 8=144$ and $n=12$.


trigonometry
calculus
derivative
function
Pythagorean Theorem
geometry

A weather balloon is released 60 feet from an observer with a camera. If the balloon is rising at 10 ft. per sec.
a) How fast is the angle of the camera changing when the balloon is 100 ft. high?
b) Is the angle of the camera changing more rapidly or more slowly as the balloon gets higher. (Prove it)

If you could help me with this problem that would be most appreciated :]

What I believe you can do is set up two related rates problems.  First, use the Pythagorean Theorem and differentiate, so we have $ 2x\frac{dx}{dt}\plus{}2y\frac{dy}{dt}\equal{}2z\frac{dz}{dt}$ but $ \frac{dx}{dt}\equal{}0$ so it goes away.  Then, using the information we already have ($ \frac{dy}{dt}$, $ y$, and $ x$), we can solve for $ z$ using the Pythagorean Theorem, and then solve for $ \frac{dz}{dt}$.  Now, once we know $ \frac{dz}{dt}$, we can say $ \cos{\theta}\equal{}\frac{x}{z}\equal{}\frac{60}{z}$.  From here, we can get $ \theta$.  Then, differentiating both sides, we get $ \minus{}\sin{\theta}\frac{d\theta}{dt}\equal{}\minus{}\frac{60}{z^{2}}\frac{dz}{dt}$.  We know everything here execpt $ \frac{d\theta}{dt}$, but we can then easily solve for it.  Hope that helps.

I still dont really understand it.  How did using the pythagorean theorem and then differentiating it get that?

Are you familiar with the concept of implicit differentiation?  Here what we are interested in are rates (that is, rates of changes with respect to time), so the relevant variable to differentiate with respect to is $ t$ (time).  In this interpretation, $ x, y, z$ are functions $ x(t), y(t), z(t)$ of time, and what we are doing is using chain rule on the expression $ x^2 \plus{} y^2 \equal{} z^2$.

And are you all familiar with the purpose of this subforum? [url=http://www.mathlinks.ro/viewtopic.php?t=26438]NO CALCULUS[/url]

Sorry, but he asked for a solution, and I gave him the first solution that came to mind.  I didn't notice what forum this was in.

Well, it is a calculus problem, so it would be pretty hard to solve without calculus ;)

@topic: The angle is given by $ \theta \equal{} \arctan\left(\frac{h}{60\mathrm{ft}}\right)$ and the chain rule gives
$ \frac{\mathrm d\theta}{\mathrm dt} \equal{} \frac{h'}{1\plus{}\left(\frac{h}{60\mathrm{ft}}\right)^2} \equal{} \frac{600\mathrm{ft}^2}{3600\mathrm{ft}^2\plus{}h^2}\frac{1}{s}.$ Plugging in 100 ft gives 0,044 radians per second and derivating the equation again gives b (the angle is increasing less and less).


vector
topology
abstract algebra
advanced fields
advanced fields unsolved

What is a bundle of Lie algebras associated to the adjoint representation?
It seems like a construction from a vector bundle, but I don't really know how to define the Lie algebra at each point and the transition function.

Aren't you leaving smth out? Smth like a Lie group? Bundle on/over what?

Sure, it is a bundle over a smth manifold, just a manifold, not Lie groups.
First we have a vector bundle, then by some ways we get a bundle of Lie Algebras associated to the ajoint representation...  And I don't know what the "some way" is...

How can you have an adjoint representation without a (Lie) group? What is being represented?

Yeah, you can construct a principal bundle from a vector bundle, I think it should be the way how Lie groups come in, but what about the Lie algebras?


Can u use the weighted am-gm inequality?