calculus
linear algebra
number theory
real analysis
IMC
college contests

Does the IMC involve calculus based questions? Also, is it harder than the IMO?

Of course it includes calculus. It includes any university topics: calculus, linear algebra, abstract algebra, advanced number theory, ...

The difficulty is wider: the easier questions are easier than IMO, the harder questions are harder than IMO.

Would I be able to start preparing now? I'm a junior in high school. What's the selection process for the IMC team? And how would I start preparing? I am in AP Calculus BC right now and will have finished Real Analysis by my senior year in high school.

There is no selection. Everyone who wants to go can go.

I don't know what a "junior in high school" means, but you can take a peek at http://www.imc-math.org where you can find all questions of all past years.

Well I guess you can say that I'm in the 11th grade, or the 11th standard.


geometry
3D geometry
geometric transformation
rotation

The first answer is correct, however the second situation is actually significantly more complicated.  In general you can't just divide the number of ways to do it without counting rotations by the number of ways to rotate the cube.


There have been threads about favorite classic books and favorite modern books, but here's a slightly different question: what books do you recommend for students to read?  (Not necessarily math or science books, although those would certainly be interesting.)

Here are some that come to mind.  (My biases may show.)

1) Surely You're Joking, Mr. Feynman
2) What Do You Care What Other People Think?  (kind of a sequel to #1)
3) by Ayn Rand: The Fountainhead; Atlas Shrugged; Capitalism: The Unknown Ideal.
4) Economics in One Lesson
5) The Making of Modern Economics
6) 1984
7) The Singularity is Near, by Ray Kurzweil (but this will get less interesting in a couple years when the cutting-edge technology he describes becomes out of date.)
8) The History of Physics, by Asimov
9) Ender's Game and the "Shadow" sequels
10) the Dark Materials trilogy, by Philip Pullman
11) the Harry Potter books
12) A Brief History of Chemistry, by Asimov
13) The Demon Haunted World: Science as a Candle in the Dark, by Carl Sagan
14) Six Easy Pieces, a sample of easier lectures from the Feynman Lectures on Physics.

Since you've said the harry potter books, I would also recommend Lord of the Rings and Eragon........ :lol:

P.S:
 If a student is able to comprehend with every single character in LOR , you know there's a genius out there..  :D

I won't recommand Harry Potter. It is a book for entertainment but not a must-read one.


algebra
polynomial
linear algebra
matrix
system of equations

[hide="Easy"]
Let $ f(x) \equal{} Ax^3 \plus{} Bx^2 \plus{} Cx \plus{} D \minus{} (2x \minus{} 1)^2$. Then by the system, $ f(1) \equal{} f(2) \equal{} f(3) \equal{} f(4) \equal{} 0$. But since $ f(x)$ is a polynomial with degree at most 3 and has at least 4 roots, we can only conclude that $ f(x)\equal{}0$ for all $ x$. Hence, $ Ax^3 \plus{} Bx^2 \plus{} Cx \plus{} D \equal{} (2x \minus{} 1)^2 \equal{} 4x^2 \minus{} 4x \plus{} 1$, and matching coefficients, we get $ (A,B,C,D) \equal{} (0,4, \minus{} 4,1)$.[/hide]

Please do you know how to do this question without using "f() "   -> I'm very new, and I don't learn this yet.

[hide="solução(solution)"]essa é uma forma meio bruta de fazer, porém achei melhor do que escalonar o sistema.(this is a crude way of doing, but thought better than the staggered system).

também da pra fazer utilizando a matriz de Vandermonde, considerando que
(also to make use of the matrix of Vandermonde)

$ a\equal{}\dfrac{D_{a}}{D},b\equal{}\dfrac{D_{b}}{D},c\equal{}\dfrac{D_{c}}{D},d\equal{}\dfrac{D_{d}}{D}$

é muito chato!(is very annoying!)

vou fazer da forma bruta mesmo(I make the same crude way) :D

$ \begin{cases}a\plus{}b\plus{}c\plus{}d\equal{}8(line 1),\\ 8a\plus{}4b\plus{}2c\plus{}d\equal{}9(line 2),\\ 27a\plus{}9b\plus{}3c\plus{}d\equal{}25(line 3),\\ 64a\plus{}16b\plus{}4c\plus{}d\equal{}49(line 4).\end{cases}$

diminuindo a linha 2,3,4 da linha 1(decreasing the line 2,3,4 Line 1):

$ \begin{cases}7a\plus{}3b\plus{}c\equal{}8(line 5),\\ 26a\plus{}8b\plus{}2c\equal{}24(line 6),\\ 63a\plus{}15b\plus{}3c\equal{}48(line 7).\end{cases}$

diminuindo a linha 6 da linha 5(decreasing the line 6 Line 5):

$ \begin{cases}7a\plus{}3b\plus{}c\equal{}8(line 5),\\ 26a\plus{}8b\plus{}2c\equal{}24(line 6).\end{cases}$

$ 12a\plus{}2b\equal{}8$

$ 6a\plus{}b\equal{}4(line8)$

diminuindo a linha 7 da linha 5(decreasing the line 7 Line 5):

$ \begin{cases}7a\plus{}3b\plus{}c\equal{}8(line 5),\\ 63a\plus{}15b\plus{}3c\equal{}48(line 7).\end{cases}$

$ 42a\plus{}6b\equal{}24$

$ 7a\plus{}b\equal{}4(line9)$

finalmente(finally):

$ \begin{cases}6a\plus{}b\equal{}4(line 8),\\ 7a\plus{}b\equal{}4(line 9).\end{cases}$

$ a\equal{}0,b\equal{}4,c\equal{}\minus{}4,d\equal{}1$

__________________________

- kl -
[/hide]

Here's a reworded form of my solution without using $ f(x)$:

[hide]
Consider the polynomial $ Ax^3\plus{}Bx^2\plus{}Cx\plus{}D\minus{}(2x\minus{}1)^2$. According to the system of equations, this polynomial has 4 roots at $ x\equal{}1,2,3,4$. However, this polynomial has degree at most 3, and since it has 4 roots, it must be the zero polynomial. Hence, $ Ax^3\plus{}Bx^2\plus{}Cx\plus{}D\minus{}(2x\minus{}1)^2\equal{}0$ and so $ Ax^3\plus{}Bx^2\plus{}Cx\plus{}D\equal{}(2x\minus{}1)^2$, and expanding and matching coefficients gives the desired result.[/hide]


A man drives from his home at 30 miles per hour to the shopping center which is 20 miles from his home.  On the return trip he encounters heavy traffic and averages 12 miles per hour.  How much time does the man take in driving to and from the shopping center?

[hide]30 mph--2 mins per mile
2*20=40 mins to get there

12 mph--5 mins per mile
5*20=100 mins to get back

40+100=140 mins or [b]2 hours 20 mins[/b][/hide]

Hmmmmm...I'm trying to recall how to do this...yes! It may have been cross multiplying where the answer is[hide]10/9[/hide] m.p.h. But I'm not really sure. I'm sure that if you'd research it you'd find it. It's not particulary hard-I just hardly need it. :evilgrin:

going $30$ mph for $20$ m will take $40$ min
going $12$ mph for $20$ m will take $100$ min
$40 + 100 = 140$ $minutes$

Um...the answer is 140 minutes...$40+100\neq 140$  :P
Just remember $s=\frac{d}{t}$

[quote="akalra1"]$40 + 100 = 100$ $minutes$[/quote]

[quote="ch1n353ch3s54a1l"]Um...the answer is 140 minutes...$40+100\ne140$[/quote]
I think what you both are trying to say is that $40+100=140$ ;)

i got [hide]$2$ hours $20$ minutes[/hide]

140 minutes, in other words (standard)

hey, WHERE IS UR SOLUTION!!!!!!!!!


hi,
prove for every $ x \in (0 , \frac{\pi}{2})$ the inequality
\[ \sec{(x)}^{2n} \plus{} cosec{(x)}^{2n} \geq 2^{n\plus{}1},\]
where $ n \in \mathbb{N}$.

[quote="doicu"]hi,
prove for every $ x \in (0 , \frac {\pi}{2})$ the inequality
\[ \sec{(x)}^{2n} \plus{} cosec{(x)}^{2n} \geq 2^{n \plus{} 1},
\]
where $ n \in \mathbb{N}$.[/quote]
You would normally write $ \csc$ instead of $ cosec$ ;)

By Hölder:

$ \sec(x)^{2n} \plus{} \csc(x)^{2n} \equal{} (\sec(x)^{2n} \plus{} \csc(x)^{2n})(\sin(x)^2 \plus{} \cos(x)^2)^n \ge 2^{n \plus{} 1}$.


inequalities
geometry
3D geometry
tetrahedron
trigonometry
similar triangles
geometry unsolved

Let $ \triangle ABC$ have $ S_{ABC} \equal{} S$ and $ G$ is the centroid .
Theline to throught  $ G$  meets $ AB$ at  [b]M[/b] and $ AC$ at [b]N[/b].
Find the minimum of $ S_{AMN}$
( :maybe: )
[i]Louis Latin and Vicky[/i]

It is very well-known that: $ \frac{AB}{AM}\plus{}\frac{AC}{AN}\equal{}3$.
Then easy to solve your problem:
$ 3\geq 2\sqrt{\frac{AB.AC}{AM.AN}}\Rightarrow \frac{AM.AN}{AB.AC}\geq \frac{4}{9}\Rightarrow S_{AMN}\geq \frac{4}{9}S$

PS: Prove that: $ \frac{AB}{AM}\plus{}\frac{AC}{AN}\equal{}3$:
Call $ D$ be midpoint of $ BC$ we have:
$ \frac{S_{AMN}}{S_{ABC}}\equal{}\frac{S_{AMG}}{2S_{ABD}}\plus{}\frac{S_{AGN}}{2S_{ADC}}$
$ \Leftrightarrow \frac{AM.AN}{AB.AC}\equal{}\frac{1}{2}(\frac{AM.AG}{AB.AD}\plus{}\frac{AN.AG}{AC.AD})$
$ \Leftrightarrow \frac{AM.AN}{AB.AC}\equal{}\frac{1}{3}(\frac{AM}{AB}\plus{}\frac{AN}{AC})\Leftrightarrow \frac{AB}{AM}\plus{}\frac{AC}{AN}\equal{}3$

Similar problem appears in Bulgarian Math Olympiad 1972 or 1973 III round.

1).$ S_{AMN}\ge\frac {4S}{9}\Rightarrow \min S_{AMN} \equal{} \frac {4.S}{9}$.
2).$ S_{AMN} \equal{} \frac {4S}{9}\Leftrightarrow |MG| \equal{} |GN| \Leftrightarrow MN\| BC$.
View: Nicholas D. Kazarinoff: GEOMETRIC INEQUALITIES, 1961, the problem 43 solution.

[quote="mihai miculita"]1).$ S_{AMN}\ge\frac {S}{4}\Rightarrow \min S_{AMN} \equal{} \frac {S}{4}$.
2).$ S_{AMN} \equal{} \frac {S}{4}\Leftrightarrow |MG| \equal{} |GN| \Leftrightarrow MN\| BC$.
View: Nicholas D. Kazarinoff: GEOMETRIC INEQUALITIES, 1961, the problem 43 solution.[/quote]

What is about condition: $ G\in MN$??? :(
With this condition please check that min is $ \frac{4S}{9}$ not $ \frac{S}{4}$

OK! Sorry, my error!
$ G\in [MN] \land MN\| BC\Rightarrow \frac {|MN|}{|BC|} \equal{} \frac {2}{3} \Rightarrow \frac {S_{AMN}}{S} \equal{} {\left( \frac {2}{3} \right)}^2 \equal{} \frac {4}{9}$.

Take a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171914
[hide="Solution From The Link"]
Here, I prove that the area is smallest when $ MO \equal{} NO$ for any random O.  When O is G, then we have that $ MG \equal{} NG$.  With this in mind, we can show that $ MN\parallel BC$ giving us similar triangles $ \triangle ABC\sim \triangle AMN$.  We now extend $ AG$ to meet $ BC$ at $ X$.  Since $ \frac {AX}{AG} \equal{} \frac {2}{3}$, we have that the scale factor of $ \triangle AMN$ to $ \triangle ABC$ is $ \frac {2}{3}$, and thus, the $ [AMN] \equal{} \frac {4}{9}([ABC])$
[/hide]

Let tetrahedron $ ABCD$  and $ G$ is the centroid .
The plane to throught $ G$ meets the sides $ [AB], [AC]$ and $ [AD]$ at $ M$, $ N$ and $ P$.
The minimum of volume tetrahedron $ AMNP$ is $ \frac {9}{16}.V_{ABCD}.$

[quote][color=darkred][b][u]An easy extension.[/u][/b] Let $ P(\alpha ,\beta ,\gamma )$ be a [b]interior[/b] point of the triangle $ ABC$, where $ (\alpha ,\beta ,\gamma )$ is the ordered set of the barycentrical coordinates of the point $ P$ w.r.t. the triangle $ ABC$, i.e. $ \{\alpha ,\beta ,\gamma )\subset \mathcal R^*_ +$, $ \alpha + \beta + \gamma = 1$ and $ \frac {\alpha}{[BPC]} = \frac {\beta}{[CPA]} = \frac {\gamma}{[APB]}$. Suppose $ \boxed {\beta < \frac 12\ ,\ \gamma < \frac 12}$. Consider two mobile points $ M\in (AB)$, $ N\in (AC)$ so that $ P\in MN$. Prove that the area of the triangle $ MAN$ is minimum if and only if $ \frac {AM}{AB} = 2\beta$ and $ \frac {AN}{AC} = 2\gamma\ .$[/color][/quote]

[b][u]Particular cases :[/u][/b]

$ G(\frac 13,\frac 13,\frac 13)$ $ \implies$ $ MN\parallel BC$ ([b]the our problem[/b]).

$ I(\frac {a}{2p},\frac {b}{2p},\frac {c}{2p})$ $ \implies$ $ MN\perp AP$, i.e. $ AM = AN\ .$

$ O(k\sin 2A,k\sin 2B,k\sin C)$, where $ k = \frac {R^2}{2S}$ $ \implies$ $ \frac {AM}{\cos B} = \frac {AN}{\cos C}\ .$

$ H(\cot B\cot C, \cot C\cot A,\cot A\cot B)$ $ \implies$ $ \frac {AM}{\cos C} = \frac {AN}{\cos B}\ .$

a.s.o.

[quote][color=darkred][b][u]A large extension.[/u][/b] Given are an angle $ \widehat {XAY}$ and a fixed its interior point $ P$. Find
$ M\in (AX$, $ N\in (AY$ so that $ P\in MN$ and the area $ [MAN]$ is minimum.[/color][/quote]
[hide="Answer."] The point $ P$ is the middle of the segment $ [MN]$.[/hide]
[hide="Proof."]
Denote $ \|\begin{array}{c} B\in (AX\ ,\ PB\parallel AY \\
\ C\in (AY\ ,\ PC\parallel AX\end{array}$. Observe that $ \|\begin{array}{c} B\in (AM) \\
\ C\in (AN)\end{array}$

and $ ABPC$- parallelogram. Denote $ \|\begin{array}{c} [ABP] = [ACP] = a \\
\ [BMP] = x\ ,\ [CNP] = y\end{array}$.

Thus, $ \triangle BMP\sim\triangle CNP$ and $ \frac {a + x}{a + y} = \frac {[AMP]}{[ANP]} = \frac {PM}{PN} = \frac {\sqrt x}{\sqrt y}$. Hence

$ \frac {a + x}{a + y} = \frac {\sqrt x}{\sqrt y}$ $ \implies$ $ x = y$ or $ xy = a^2$. In conclusion, $ [MAN]$ is minimum

iff $ x + y$ is minimum. But $ xy$ is constant. Thus, $ x = y$, i.e. $ PM = PN$.[/hide]


Let $f: \mathbb{R}\to\mathbb{R}$ satisfy:
$f(2x)=f(\sin\frac{x+y}{2}\pi)+f(\sin\frac{x-y}{2}\pi)$
Find $f(2007+\sqrt{2007})$

The expression is supposed to be satisfied for any arbitrary inputs $x, y$.  So $y$ is any real.

Do we know if $f$ is continuous or not?
If yes, then we can prove that $f(x)=0, \ \forall x \in \mathbb{R}$

[hide]
$f(2x)=f(\sin\frac{x+y}{2}\pi)+f(\sin\frac{x-y}{2}\pi)$

Setting $y=x$ we get $f(2x)=f(\sin \pi x)+f(0) \ \ (1)$

and then for $x=0 \ \ (1)$ gives $f(0) = f(0)+f(0) \Rightarrow f(0)=0$

So we have $f(2x) = f(\sin \pi x)\Rightarrow \boxed{f(x) = f(\sin \frac{\pi x}{2})} \ \ (2)$

$f(\sin\frac{x+y}{2}\pi) = f(x+y)$

$f(\sin\frac{x-y}{2}\pi) = f(x-y)$

So the initial equation becomes $f(2x) = f(x+y)+f(x-y) \ \ (3)$

for $y=0$ we get $f(2x) = 2f(x) \ \ (4)$

$f$ is periodic with period $4$. Indeed:

$f(x+4) = f(sin \frac{\pi (x+4)}{2}) = f(sin (\frac{\pi x}{2}+2\pi)) = f(sin \frac{\pi x}{2}) = f(x)$

(and we can find more symmetries)

We will show that, if $f$ is continuous then $f(x) = 0, \ \ \forall x \in \mathbb{R}$

Suppose that there is $a \in [0,4]$ such that $f(a) \neq 0$
W.L.O.G suppose that $f(a)>0$
If $f$ is continuous, then $f$ has a maximum value in $[0,4]$

Say that $f(b)$ is this maximum. Of course $f(b)$ is the global maximum of $f$ (the max value $\forall x \in \mathbb{R}$)

$f(b)\geq f(a) \Rightarrow f(b)>0$

from $(4)$ we get $f(2b)=2f(b)>f(b)$, which yields a condradiction

So $f(x) = 0, \ \ \forall x \in \mathbb{R}$

I don't know how we get in contradiction if $f$ is not continuous (can someone prove it?)[/hide]

Did I do some mistake?  :huh:
Please, check my work


The MOP Handbook says:
"Bed sheets will be provided, but students will need to bring their own blanket and towels.  Sheets can be exchanged for clean ones at the front desk.  Additional items such as a favorite pillow can be brought to make your stay more comfortable."

I feel really stupid, but I can't figure out if that means that pillows are provided or that they aren't.  Does anyone who has been to MOP know?

Thanks!

It's a dorm. There will be pillows there, and you may or may not like them.

You don't need to bring a pillow.  The ones provided are fairly nice and fluffy.  :)

that's good to know because I wasn't sure how I was going to fit mine in my bag.
stupid restrictive airline restrictions...

pillows are provided, and they come in handy during lectures :D

and in general, if you're ever near Arnav.

They say you can bring your own blanket, but seriously, it's going to be hot all summer.


function
LaTeX
calculus
calculus computations

My prob is to solve x^2 + 3xy + y^2 = 6y by implicit differentiation.

Im having trouble understanding how to diff the function with respect to x & y. Let me just show you:

ok first I diff each term w/respect to x:

d/dx(x^2) + d/dx(3xy) + d/dx(y^2) = d/dx(6y),

so now, diff func of x w/respect to x and fcn of y w/respect to x so:

3x + d/dy(3xy) * dy/dx + d/dy(y^2) * dy/dx = d/dy(6y) * dy/dx

2x+3y+(3x)dy/dx+(2y)dy/dx-(6)dy/dx=0

(3x)dy/dx+(2y)dy/dx-(6)dy/dx=-2x-3y

(3x+2y-6)dy/dx=-2x-3y

dy/dx=(-2x-3y)/(3x+2y-6)

dy/dx=-(2x+3y)/(3x+2(y-3))

Does that look correct??? If not, can you point me in the right direction? Thanks in advance.
Chris

I worked it out on paper, and I got the same answer as you

Mine reads

$y' = \dfrac{-2x-3y}{3x+2y-6}$

I suggest, as a more useful way to display your questions, that you learn to write in LaTeX. There is a forum dedicated to learning it at the bottom of the main page, and it is really helpful.

Thanks again ankur, I will research LaTeX for later use!


With a only a little touch of suspicion Kyla watched as my Golden Lab- mix [img]http://www.artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg[/img] worked on the key pad  to “ring-up” her purchase. “Say!” she said at last, “Your dog is cute, and he seems to know how to use the key pad, BUT he multiplied the three amounts instead of adding them!”

“So he did, but that’s okay” I replied, “The total comes out to be $\$5.70$ either way.

Indeed that  was right (other wise I would not be posting the problem here). So what were the three amounts?

[hide="Solution"]
I'm assuming the multiplication is done in dollars.

$a + b + c = abc = 5.70$

Substitute $x = 100a, y = 100b, z = 100c$ (to get into cents):

$x + y + z = 570, xyz = 5700000 = 2^5 \times 3 \times 5^5 \times 19$

The sum of $x + y + z$ needs to be divisible by 2 and 5.  This means two of them are odd and one is even, or they are all even.  This also means that all of them must have a power of 5 (the other option, that only one of them has a power of 5, produces $x + y + z > 5^5 > 570$.

We have already established that all of the amounts must be divisible by 5, so substitute yet again:  $x = 5p, y = 5q, z = 5r$.  We have

$p + q + r = 114, pqr = 2^5 \times 3 \times 5^2 \times 19$

WLOG, $19 | p$.  Then we have $q + r \equiv 0 \bmod 19$.  We find that $5^2 + 2^5 = 25 + 32 = 57$ satisfies this requirement, so we have $p = 57, q = 25, r = 32$.

Backtracking, $p = 20a, q = 20b, r = 20c$, so we find that our solution is $\boxed{ \$2.85, \$1.25, \$1.60 }$.
[/hide]


function
trigonometry
limit
algebra unsolved
algebra

$ f(x) \equal{} 0$ for all integers $ x$.

$ f'(x) \equal{} \frac{1}{x^{2}}$ for all non-zero integers $ x$.

[quote="AndrewTom"]$ f(x) \equal{} 0$ for all integers $ x$.

$ f'(x) \equal{} \frac {1}{x^{2}}$ for all non-zero integers $ x$.[/quote]

Where did you find such a question ?

There are infinitely many solutions. One is :

Let $ a\in\mathbb R$

$ \forall x\in[ \minus{} 1,0)$ : $ f(x) \equal{} (\frac {1}{[x]^2} \plus{} a)\{x\}^3 \minus{} (\frac {2}{[x]^2} \plus{} a)\{x\}^2 \plus{} \frac {1}{[x]^2}\{x\}$
$ \forall x\in[0,1)$ : $ f(x) \equal{} (a \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^3 \minus{} (2a \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^2 \plus{} a\{x\}$
$ \forall x\in\mathbb R\backslash[ \minus{} 1,1)$ : $ f(x) \equal{} (\frac {1}{[x]^2} \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^3 \minus{} (\frac {2}{[x]^2} \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^2 \plus{} \frac {1}{[x]^2}\{x\}$

And you can find all the solution by adding to this one any function $ g(x)$ such that $ g(n) \equal{} g'(n) \equal{} 0$ $ \forall n\in\mathbb Z$, so infinitely many solutions :
$ g(x) \equal{} \{x\}^2(\{x\} \minus{} 1)^2h(x)$ for example
$ g(x) \equal{} \sin^2(\pi x)$ for example too
...

Someone sent it to me.

The actual question was to find a possible function f.

I presumed that the function was defined for integers $ x$.

(Perhaps there was a typo.)

[quote="AndrewTom"]Someone sent it to me.

The actual question was to find a possible function f.

[/quote]

So you got a lot of :)

[quote="AndrewTom"]
I presumed that the function was defined for integers $ x$.

(Perhaps there was a typo.)[/quote]

What would be the definition of $ f'(x)$ in such a case ?

Oops.  :blush:

However, consider $ f(x) \equal{} g(x)\tan(\pi x)$

$ f'(x) \equal{} g(x) \cdot \pi \cdot \sec^{2}(\pi x) \plus{} g'(x)\cdot \tan (\pi x)$.

For integers $ x$, $ \tan (\pi x) \equal{} 0, \sec (\pi x) \equal{} \pm 1$ and so $ f'(x) \equal{} \frac{1}{x^{2}} \implies \frac{1}{x^{2}} \equal{} \pi \cdot g(x)$

$ \implies g(x) \equal{} \frac{1}{\pi x^{2}}$.

Therefore $ f(x) \equal{} \frac{\tan (\pi x)}{\pi x^{2}}$.

[quote="AndrewTom"]Oops.  :blush:

However, consider $ f(x) \equal{} g(x)\tan(\pi x)$.[/quote]

I dont understand this "Oops.  :blush: However, ..."

I gave you infinitely many solutions. Are my proposal wrong ?

[quote="AndrewTom"]
$ f'(x) \equal{} g(x) \cdot \pi \cdot \sec^{2}(\pi x) \plus{} g'(x)\cdot \tan (\pi x)$.

For integers $ x$, $ \tan (\pi x) \equal{} 0, \sec (\pi x) \equal{} \pm 1$ and so $ f'(x) \equal{} \frac {1}{x^{2}} \implies \frac {1}{x^{2}} \equal{} \pi \cdot g(x)$

$ \implies g(x) \equal{} \frac {1}{\pi x^{2}}$.

Therefore $ f(x) \equal{} \frac {\tan (\pi x)}{\pi x^{2}}$.[/quote]
Your $ f(x)$ is not defined over $ \mathbb R$

If this is allowed (and without precision, the functions are generally supposed to be defined over $ \mathbb R$), just take $ f(x)$ defined over $ (n\minus{}\frac 13,n\plus{}\frac 13)$ as $ f(x)\equal{}\frac{x\minus{}n}{n^2}$ for any $ n\in\mathbb Z^*$ plus $ f(0)\equal{}0$

[quote="AndrewTom"]pco, first of all, thanks very much for your responses. I am impressed and grateful.

I wasn't suggesting that there is anything wrong with your solution; indeed, I don't feel qualified to judge it.
I was just offering something which I thought would work (and thanks for your correction).

Once, again, thanks a lot. I appreciate your help (as always).[/quote]

No problem :)

[quote="pco"][quote="AndrewTom"]$ f(x) \equal{} 0$ for all integers $ x$.

$ f'(x) \equal{} \frac {1}{x^{2}}$ for all non-zero integers $ x$.[/quote]

There are infinitely many solutions. One is :

Let $ a\in\mathbb R$

$ \forall x\in[ \minus{} 1,0)$ : $ f(x) \equal{} (\frac {1}{[x]^2} \plus{} a)\{x\}^3 \minus{} (\frac {2}{[x]^2} \plus{} a)\{x\}^2 \plus{} \frac {1}{[x]^2}\{x\}$
$ \forall x\in[0,1)$ : $ f(x) \equal{} (a \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^3 \minus{} (2a \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^2 \plus{} a\{x\}$
$ \forall x\in\mathbb R\backslash[ \minus{} 1,1)$ : $ f(x) \equal{} (\frac {1}{[x]^2} \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^3 \minus{} (\frac {2}{[x]^2} \plus{} \frac {1}{[x \plus{} 1]^2})\{x\}^2 \plus{} \frac {1}{[x]^2}\{x\}$

And you can find all the solution by adding to this one any function $ g(x)$ such that $ g(n) \equal{} g'(n) \equal{} 0$ $ \forall n\in\mathbb Z$, so infinitely many solutions :
$ g(x) \equal{} \{x\}^2(\{x\} \minus{} 1)^2h(x)$ for example
$ g(x) \equal{} \sin^2(\pi x)$ for example too
...[/quote]

It seems this need some explanations.

Let $ \{a_n\}_{n\in\mathbb Z}$ the sequence defined as $ a_0 \equal{} a$ and $ a_n \equal{} \frac 1{n^2}$ $ \forall n\ne 0$.

My proposal is $ f(x) \equal{} (a_{[x]} \plus{} a_{[x \plus{} 1]})\{x\}^3 \minus{} (2a_{[x]} \plus{} a_{[x \plus{} 1]}))\{x\}^2 \plus{} a_{[x]}\{x\}$.

Proof :
$ \forall x\in[n,n \plus{} 1)$ : $ f(x) \equal{} (a_{n} \plus{} a_{n \plus{} 1})(x \minus{} n)^3 \minus{} (2a_{n} \plus{} a_{n \plus{} 1})(x \minus{} n)^2 \plus{} a_{n}(x \minus{} n)$
And $ f'(x) \equal{} 3(a_{n} \plus{} a_{n \plus{} 1})(x \minus{} n)^2 \minus{} 2(2a_{n} \plus{} a_{n \plus{} 1})(x \minus{} n) \plus{} a_{n}$

$ f(n) \equal{} 0$ and $ \lim_{x\to (n \plus{} 1)^ \minus{} }f(x) \equal{} (a_{n} \plus{} a_{n \plus{} 1}) \minus{} (2a_{n} \plus{} a_{n \plus{} 1}) \plus{} a_{n} \equal{} 0$
$ f'(n^ \plus{} ) \equal{} a_n$ and $ \lim_{x\to (n \plus{} 1)^ \minus{} }f'(x) \equal{} 3(a_{n} \plus{} a_{n \plus{} 1}) \minus{} 2(2a_{n} \plus{} a_{n \plus{} 1}) \plus{} a_{n} \equal{} a_{n \plus{} 1}$

So $ f(x)$ is continuous and $ f(n) \equal{} 0$
And $ f'(x)$ is defined and continuous and $ f'(n) \equal{} a_n$

Hence the result.

And, when you got two solutions, it's easy to see that $ f_1(x) \minus{} f_2(x) \equal{} g(x)$ is such that $ g(n) \equal{} 0$ $ \forall n$ and $ g'(n) \equal{} 0$ $ \forall n\ne 0$

So, any solution may be written as $ f(x) \plus{} g(x)$ where $ f(x) \equal{}$ the above particular solution and $ g(x)$ is any function such that $ g(n) \equal{} g'(n) \equal{} 0$ $ \forall n$ (I included $ g'(0) \equal{} 0$ since we lose no generality by choosing $ a$ as we want).

Hope this is clearer.

Thanks pco. By the way, I think Oops is Oups in French or perhaps Zut! (I was thinking :oh no, I've made a silly mistake again.)


The following sum can be expressed as $ \binom{a}{b}$, where $ b<a\minus{}b$. Find $ a\plus{}b$:
\[ \binom{6}{6} \plus{} \binom{7}{6} \plus{} \binom{8}{6} \plus{} \ldots \plus{} \binom{40}{6}\]

This is a direct application of the Hockey Stick Identity, which states that \[ \binom{r}{r}\plus{}\binom{r\plus{}1}{r}\plus{}\binom{r\plus{}2}{r}\plus{}\cdots \plus{}\binom{n}{r}\equal{}\binom{n\plus{}1}{r\plus{}1}\]
Here, we have $ r\equal{}6$, $ n\equal{}40$, so the sum is equal to $ \binom{41}{7}$ and the answer is $ 41\plus{}7\equal{}\boxed{48}$

If you are familiar with summation notation, you can write the identity as $ \sum_{i\equal{}r}^{n}\binom{i}{r}\equal{}\binom{n\plus{}1}{r\plus{}1}$


Prove that 1+2n <= 3^n by using mathematical induction

Let it be true for $ n\equal{}k$. We have $ 2k\plus{}1<3^k\Rightarrow 2k\plus{}3<3(2k\plus{}1)<3^{k\plus{}1}$

Can you explain with words (or more steps) what you did?

Define the statement $ P(n)$ as the assertion that $ 2n\plus{}1 \leq 3^n$

We can clearly see that $ P(1) \iff 3\equal{}3$ is true.

Now suppose the inequality is true for $ n\equal{}k > 0$. That is, suppose $ P(k) \iff 2k\plus{}1 \leq 3^k$ is true.

It follows that:
$ 2(k\plus{}1)\plus{}1 \equal{} 2k\plus{}3 \leq 6k\plus{}3 \equal{} 3(2k\plus{}1) \leq 3(3^k) \equal{} 3^{k\plus{}1}$
Transitively, the above implies that $ P(k\plus{}1)$ is true.

By the principle of mathematical induction, since $ P(1)$ is true and $ P(k) \implies P(k\plus{}1)$ for all non-negative $ k$, it must be true that $ P(n)$ is true for all positive integers $ n$.


I'm homeschooled this year, so I need to find a way to take the AMC. The only way I know of is to take it with another school, since I don't think my ISP even has 10 people who have taken Algebra/Geometry and so would generally know the topics on the AMC, much less 10 people who would be interested in doing it, and I'm not even sure if an ISP can "host" the test. Is it possible for me to take it on my own, or will I have to find another school? If so, when should I check to see what schools in my area are registered, and what will I have to do to register myself with a school?

[b]EDIT:[/b] Also, is every round beyond the AMC automatically covered if I qualify for it?

You have several options:
1.  You can register as a home school, it is an option right on the registration form.  Several things to be aware of here:  Registering as a home school you will be purchasing a bundle of 10 contests, even if only you as the single student at the school taking the contest.  Another thing is that you must have someone other than a parent administer the contest, and the contest must be administered in a public place such as a library, a church, or a city activity center.  We send the contest package to a location other than the home, such as the library, church, etc.
2.  I am not sure what an ISP is, but some cities have Home School Association or similar group that registers for the contest as a group activity.  This eliminates several of the problems mentioned above.  You might also wish to contact the Math Department of a local college, University, or two-year college.  This also eliminates some of the problems above.
3.  You can register as a "Visitor" at a school in your area administering the contest.  Contact our office at amcinfo@unl.edu for details and schools in your area.  There is a fee for registering as a Visitor.

Yes, if you qualifiy for AIME and the USAMO, you will receive information from the AMC about arrangements which are basically the same as described above.

Steve Dunbar
AMC Director


for the 2nd look at the solution I gave for a simmilar problem! :D

http://www.mathlinks.ro/phpBB/viewtopic.php?t=151

cheers! :D :D


Sometimes, when the class crashes or when i accidently click out, and I try to log back in, it willl say my login name is still in use.  I can clear my cookies and logout and back in without any change.  The only solution I have found is to restart my computer, but that's really a pain!  Any better ideas?

What's happening there is that your 'ghost' is still in the classroom.  What you should do when you fall out of the classroom is immediately email us if you can't get back in, and we can kick your ghost out.  You don't have to restart your computer (what's happening while you are restarting your computer is that enough time is passing that your ghost finally falls out of the room).

When I was logged into both the usamts classroom and the geometry class, my classroom crashed as well. This happened right after the usamts class ended (before the geometry). Then i logged back into the geometry classroom, experiencing the same problem. When I logged in and out, it solved the problem, altought I guess this does not work for everybody.

[quote="Ravi B"]This happened to my daughter a couple of times because we had a wireless connection.  (After those first couple of times, we converted the connection to a "wired" connection, and it never happened again.)  We didn't mind logging in again, but waiting for the ghost to disappear was a nuisance.  When logging in a second time, wouldn't it make more sense for the classroom software to kick out the ghost, and allow the second login to succeed?[/quote]

I'm going to talk to the software vendor about how to do this - I think it is set by default this way because most chat applications have people choose their name as the member enters.  Kicking 'ghosts' in such an environment would be mayhem!

And, most of the people who have connection problems are a result of wireless connections.  Not all, though...

We've come up with an approach which will solve the ghosts problem and allow students to be in more than one classroom at once.

When a student receives the 'Login name already in use', the student will be able to click a link on the Classroom page which will reload the page and append a '_2' to the student's username so he/she can get into the Classroom.  This link will be just below the Classroom box on the Classroom page.

Students who wish to enter two classes at once (for example, Independent Study Office Hours and a Math Jam) can open the Classroom window from 2 different browsers.  By clicking this link in one of them, the student should then be able to access a second class.


[quote="mkkool64"]let me guess-graphing all of these on the same plane makes some pretty looking picture, right?  or maybe spells something out....[/quote]

Maybe you got it!  mkkool64.

Could you draw the picture?

kunny

I dont really wanna get out my graphing calc, what does it show?

This is what I got:

[img]http://img.photobucket.com/albums/v213/joml88/tama.jpg[/img]

You are right, thank you, joml88.

Tama means cat's name. :)

Thats fun. I've made somethings like that, they are just sines and lines at different thicknesses of nothing in particular.

I once made strong bad's face using the pen, i was going to send it in as weekly fanstuff. But someone did the cheat alot better...


Prove that the probability that event $A$ will happen before event $B$ is $\frac{P(A)}{P(A)+P(B)}$.

you know what, lingomaniac?
I don't care if I can prove it now that I know it.
Wow - that makes a lot of problems a whole lot easier

Either event A will happen before event B or event B will happen before event A.

In equation form, \[\frac{P(A)}{P(A)+P(B)}+\frac{P(B)}{P(B)+P(A)}=1\] which is obviously true.

You sort of have to qualify events $A$ and $B$... such as saying that they're independent events ;)

Yeah what if A and B are the same event? If they're the same you can't really have one before the other.  :P

Yes, A and B are independent events, and they are not the same event.

That A and B are independent events implies that they aren't the same event..  :wink:

BTW: How many of you here know the *definition* of independent event?

[hide]
$P(\mbox{A and B}) = P(A) \cdot P(B)$.

if they are independent.

Independent events are events in which one event doesn't affect the outcome of the other event in anyway.

Is that the definition?

I also know that $P(A|B) = P(A)$ if they are independent. I think...[/hide]


Could anyone help me identify this spider that I found?

It measures approximately 3.5cm in length. It looks to have a mostly yellow body, a white abdomen, a red head, and black feet.

I took some pictures and posted them.

Thanks!

I do not know but I have found this site (while searching for a mantis I found): http://www.spiders.com
It might help you

If you were to ask me off the top of my head, I would assume it to be either a hobo spider or grass spider.

It's a hobo spider, if you live in Washington, Oregon, Idaho, Montana, Wyoming, Utah or Colorado.
The unknown specimen falls under a grass spider, if you live in the Pacific Northwest states.

Of course, there is always room for error. (I am no reliable arachnologist)

If you wish for expert help in identification, [url]http://www.badspiderbites.com/spider-identification/[/url] accepts pictures and will detail the spider.


What is the product of $ 3 \times 10^6$ and $ 4 \times 10^3$? Express your answer in scientific notation.

if we multiply these two, we will get $ 12\times10^9$ (do you see why?)

but when we express things in scientific form-$ a\times10^x$, the $ a$ has to be at least one and less than 10
so our answer is not acceptable

we solve this problem by multiplying by another 10, to get $ 1.2\times10^10$

hmm....then there has to be something wrong with the ftw answer...it said 1.2*10^8.
WEIRD


Let $ x_1, \ldots, x_n$ be $ n>1$ real numbers satisfying $ A\equal{}\left |\sum^n_{i\equal{}1}x_i\right |\not \equal{}0$ and $ B\equal{}\max_{1\leq i<j\leq n}|x_j\minus{}x_i|\not \equal{}0$. Prove that for any $ n$ vectors $ \vec{\alpha_i}$ in the plane, there exists a permutation $ (k_1, \ldots, k_n)$ of the numbers $ (1, \ldots, n)$ such that \[ \left |\sum_{i\equal{}1}^nx_{k_i}\vec{\alpha_i}\right | \geq \dfrac{AB}{2A\plus{}B}\max_{1\leq i\leq n}|\alpha_i|.\]

We do a weighting trick. Let $\pi$ denote a permutation $(\pi(1), \pi(2), \dots, \pi(n))$, and let $\sum_{\pi}$ sum over all permutations of $\{1, \dots, n\}.$ Assume that $|\vec{\alpha_1}| \ge 1.$ Then it suffices to show the problem with the RHS replaced with $\frac{AB}{2A+B}.$ Consider the following equation. \[ \left(\sum_\pi |w_\pi| \right) \max_{\pi} \left| \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| \ge \sum_{\pi} |w_{\pi}| \left| \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| \ge \left| \sum_\pi w_\pi \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| = \left| \sum_{i = 1}^n \vec{\alpha_i} \sum_\pi w_\pi x_{\pi(i)} \right| \] for some real weights $w_\pi$ for all permutations $\pi.$

Now choose the weights $w_{\pi}$ such that \[ \sum_\pi w_\pi x_{\pi(i)} = \begin{cases} 1 \textnormal{ if } i = 1 \\ 0 \textnormal{ otherwise} \\ \end{cases}. \] Substituting this in, we have that \[ \max_{\pi} \left| \sum_{i = 1}^n x_{\pi(i)} \vec{\alpha_i} \right| \ge \left(\sum_\pi |w_\pi| \right)^{-1}, \] so our goal is now to minimize $\sum_\pi |w_\pi|$ given the previous system of equations.

My next claim is a duality claim. Let $c_1, \dots, c_n$ be real numbers. Then we know that \[ \sum_{i = 1}^n c_i \sum_\pi w_\pi x_{\pi(i)} = c_1 \implies \sum_\pi |w_\pi| \ge \frac{|c_1|}{\max_\pi \left| \sum_{i = 1}^n c_i x_{\pi(i)}\right|} . \] Essentially, by the duality of linear programming, one can show that equality holds, i.e. \[ \min \sum_\pi |w_\pi| = \max_{\substack{c_1, \dots, c_n \in \mathbb{R} \\ \sum |c_i| > 0}} \frac{|c_1|}{\max_{\pi} \left|\sum_{i = 1}^n c_i x_{\pi(i)}\right|} . \] So we have reduced it to an optimization problem on the right hand side of the equation. The denominator can be one of two things, by using the rearrangement inequality. Assume without loss of generality that $x_1 \le \dots \le x_n$, and $\sum x_i > 0.$ Also assume that $c_1 \le \dots \le c_n$, and instead we show that \[ \max_{\substack{c_1, \dots, c_n \in \mathbb{R} \\ \sum |c_i| > 0}} \frac{ \max_i |c_i|}{\max_{\pi} \left|\sum_{i = 1}^n c_i x_{\pi(i)}\right|} \le \frac{2A+B}{AB}. \] This statement [b]has to be true[/b], because it is the same as the original problem, except with the vectors $\alpha_i$ replaced by real numbers $c_i$! So we have strictly simplified the problem. Now, looking at the denominator, it equals \[ \max\left(\sum c_i x_i, -\sum c_i x_{n+1-i}\right), \] because $x_1 \le \dots \le x_n$ and $c_1 \le \dots \le c_n.$

[b]Lemma 1: [/b] If $c_n > 0$ and $c_1 \le 0$, we are done.

This is because \[ \sum c_i x_i-\sum c_i x_{n+1-i} = \sum_{i = 1}^{n/2} (c_{n+1-i} - c_i)(x_{n+1-i}-x_i) \ge (c_n - c_1)B \ge \max_i |c_i| B. \] Therefore, \[ \max\left(\sum c_i x_i, -\sum c_i x_{n+1-i} \right) \ge \frac{B}{2} \max_i |c_i| \ge \frac{AB}{2A+B} \max_i |c_i|, \] as desired.

[b]Lemma 2:[/b] If $0 < c_1 \le \dots \le c_n$, we are also done.

It suffices to consider the case where are $c_i$ are positive by symmetry. Also assume WLOG that $c_n = 1$, so $\max_i |c_i| = 1.$ We want to show that \[ \max\left(\sum c_i x_i, -\sum c_i x_{n+1-i} \right) \ge \frac{AB}{2A+B}. \] Consider a real $0 \le d \le 1.$ Consider the sum \[ d\sum c_i x_i - (1-d) \sum c_i x_{n+1-i} = \sum c_i (dx_i - (1-d)x_{n+1-i}). \] I will show that the last expression will be at least $\frac{AB}{2A+B},$ which will finish. Note that the sequence $\{dx_1 - (1-d)x_n, dx_2 - (1-d)x_{n-1}, \dots, dx_n - (1-d)x_1\}$ is an increasing sequence. Since we have no restrictions on $c_1, \dots, c_{n-1}$ other than they are between in the range $[0, 1]$, we can smooth them to all be the same value, since we want to minimize the previous quantity. After we smooth to $c_1 =  \dots = c_{n-1}$, we can then smooth all of them to equal $0$ or $1$, as the expression is linear in those variables. Therefore, \[ \sum c_i (dx_i - (1-d)x_{n+1-i}) \ge \min((2d-1)\sum x_i, dx_n - (1-d)x_1) = \min((2d-1)A, (x_1+x_n)d - x_1). \] ow we make our choice of $d$. A good choice is $2d-1 = \frac{B}{2A+B} \implies d = \frac{A+B}{2A+B}.$ Then clearly $0 \le d \le 1.$ To show that $\min((2d-1)A, (x_1+x_n)d - x_1) \ge \frac{AB}{2A+B}$, it suffices to show that  \[ (x_1+x_n)\frac{A+B}{2A+B} - x_1 \ge \frac{AB}{2A+B} \] \[ \frac{A+B}{2A+B}x_n - \frac{A}{2A+B}x_1 \ge \frac{AB}{2A+B} \] \[ (A+B)x_n - Ax_1 \ge AB = A(x_n-x_1) \Leftrightarrow Bx_n \ge 0, \] which is true as we assumed that $\sum x_i > 0 \implies x_n > 0$. So we're done.

[b]Remark:[/b] There must be an easier way... I think the beginning manipulation was all fairly roundabout, so I am hoping that there is a more direct solution.

[quote=mathocean97] Essentially, by the duality of linear programming, one can show that equality holds, i.e. \[ \min \sum_\pi |w_\pi| = \max_{c_1, \dots, c_n \in \mathbb{R}} \frac{|c_1|}{\max_{\pi} \left|\sum_{i = 1}^n c_i x_{\pi(i)}\right|} . \] So we have reduced it to an optimization problem on the right hand side of the equation. The denominator can be one of two things, by using the rearrangement inequality. (Finish to come)[/quote]

Can you show me the last step,plz  :maybe:  :blush:

WLOG assume that $ \left | x_n - x_1 \right| = \max_{1\leq i<j\leq n} \left| x_j-x_i \right| = B, $ $ \left| \vec{\alpha_n} - \vec{\alpha_1} \right| = \max_{1\leq i<j\leq n} \left|  \vec{\alpha_j} -  \vec{\alpha_i} \right| $ and let \begin{align*}
\vec{\beta} = x_1 \vec{\alpha_1} + x_2 \vec{\alpha_2} + \dots + x_{n-1} \vec{\alpha_{n-1}} + x_n \vec{\alpha_n} \\
\vec{\gamma} = x_n \vec{\alpha_1} + x_2 \vec{\alpha_2} + \dots + x_{n-1} \vec{\alpha_{n-1}} + x_1 \vec{\alpha_n} &.
\end{align*} Then $$ \mathbf{M} =  \max_{\substack{\{ k_1, \dots, k_n \} \\ = \{ 1, 2, \dots, n \}} } \left | \sum_{i=1}^{n}x_{k_i}\vec{\alpha_i} \right | \geq  \frac{1}{2} \left( \left| \vec{\beta} \right| + \left| \vec{\gamma} \right| \right) \geq  \frac{1}{2} \left| \vec{\gamma} - \vec{\beta} \right| = \frac{B}{2}  \left| \vec{\alpha_n} - \vec{\alpha_1} \right|. $$

Let $ \left| \vec{\alpha_k} \right| = \max _{1\leq i\leq n}\left|\vec{\alpha_i} \right| $ ($ k \in \mathbb{N}, 1 \leq k \leq n $) and $ \left| \vec{\alpha_n} - \vec{\alpha_1} \right| = t \left| \vec{\alpha_k} \right| $ ($ t \in \mathbb{R}, 0 \leq t \leq 2 $), then we get
$$  \mathbf{M} \geq \frac{Bt}{2} \left| \vec{\alpha_k} \right| \qquad \qquad \qquad (\dagger). $$

On the other hand, let $ \vec{\varsigma_i} = x_i \vec{\alpha_1} + x_{i+1} \vec{\alpha_2} + \dots + x_n \vec{\alpha_{n-i+1}} + x_1 \vec{\alpha_{n-i+2}}+ \dots + x_{i-1} \vec{\alpha_n}, $ then $$\mathbf{M} \geq \frac{1}{n} \sum_{i=1}^{n} \left| \vec{\varsigma_i} \right| \geq \frac{1}{n} \left| \sum_{i=1}^{n} \vec{\varsigma_i} \right| = \frac{A}{n} \left| n\vec{\alpha_k} - \sum_{j \neq k} \left( \vec{\alpha_k} - \vec{\alpha_j} \right)\right|, $$ so $$ \mathbf{M} \geq \frac{A}{n} \left( n \left| \vec{\alpha_k} \right| - \sum_{j \neq k} \left| \vec{\alpha_k} - \vec{\alpha_j} \right| \right) \geq \frac{A}{n} \left( n \left| \vec{\alpha_k} \right| - \left( n-1 \right) \left| \vec{\alpha_n} - \vec{\alpha_1} \right| \right) =  \left( A - \frac{\left(n-1\right) At }{n} \right) \left| \vec{\alpha_k} \right| \qquad (\ddagger). $$ Combining $ (\dagger) $ and $ (\ddagger) $ we conclude that $$ \mathbf{M} \geq \max \left\{ \frac{Bt}{2}, \left( A - \frac{\left(n-1\right) At }{n} \right) \right\} \left| \vec{\alpha_k} \right| \geq \left( \frac{\frac{Bt}{2} \cdot \frac{\left(n-1\right)A}{n} + \left( A - \frac{\left(n-1\right) At }{n} \right) \cdot \frac{B}{2}}{\frac{\left(n-1\right)A}{n} + \frac{B}{2}} \right) \left| \vec{\alpha_k} \right|. $$
i.e. there exists a permutation $ \left( k_1, \dots, k_n \right ) $ of the numbers $ \left(1, \dots, n \right ) $ such that $$ \left |\sum_{i=1}^{n} x_{k_i}\vec{\alpha_i}\right | \geq \frac{AB}{\frac{2\left(n-1 \right)}{n}A+B} \max_{1\leq i\leq n}\left|\vec{\alpha_i}\right|. $$


Boo. I am too lazy to draw a diagram, so here is a nice, crude diagram...

* = air outside system
! = mercury
| and _ = boundaries of tubes

         |*******|
         |*******|
         |*******|    <-- outer tube
         |**___**|
         |**|**|**|
         |**|**|**|     <-- inner tube
         |**|**|**|
         |**|__|**|    <-- mercury level inside inner tube (B)
         |**|!!!|**|
         |**|!!!|**|
         |**|!!!|**|
         |__|!!!|__|     <-- mercury level (A)
         |!!!|!!!|!!!|
         |!!!!!!!!!!!|
         |!!!!!!!!!!!|

So, B is the height of the mercury inside of the inner tube, and A is the height of the mercury inside of the outer tube and outside of the inner tube. The difference in height is h (h = B - A) (where the unit mmHg comes from).

Outside pressure: $ p_{\mathrm{air}}$.
Density: $ \rho \equal{} \frac{m}{V}$
$ p_{\mathrm{air}} \equal{} p_{\mathrm{Hg}}$.
$ p \equal{} \frac{F}{A}$.

$ p_{\mathrm{Hg}} \equal{} \frac{F_{\mathrm{Hg}}}{A_{\mathrm{inner}}}$
$ F_{\mathrm{Hg}} \equal{} mg \equal{} gV\rho$ and $ V \equal{} hA_{\mathrm{inner}}$
So, $ p_{\mathrm{Hg}} \equal{} \frac{ghA\rho}{A} \equal{} gh\rho$

But this irs really physics!
If we consider the toxicity of the mercury vapors, it could even be toxicology.[quote="aufha"]Show, with proof, that the pressure reading of a barometer doesn't depend on the radius length of the tube used.

Apparently, my brother says I'm supposed to use the fact that pressure= force/area but I don't get it...[/quote]
The pressure exerted by the air over the open mercury surface is the same as that produced by the column of mercury over that level in the inverted tube. The pressure of the column of mercury is
pressure=weight (of mercury column)/area (of tube cross section)
weight (of mercury column)=volume (of mercury column) X density (of mercury)
volume (of mercury column) = area (of tube cross section) X height (of mercury column)
If you put it all together, area (of tube cross section) simplifies out, and you're left with:
pressure=height (of mercury column) X density (of mercury)
Now, three more questions:
How high would be the liquid column, if you used water instead of mercury?
Why do you have to put the pump at the bottom of your water well, if your well is too deep?
How deep would be too deep (theoretically speaking)?


[quote="CCMath1"]For each positive integer n, let S(n) denote the sum of the digits of n . For how many values of  n is n+S(n)+S(S(n))=2007? [A].1;[B].2;[C].3;[D].4;[E]5.
SEE THIS PROBLEM AT:
http://www.mathlinks.ro/resources.php?c=182&cid=43&year=2007[/quote]

$ n<2007$, so $ S(n)\leq 28$, so $ S(S(n))\leq 10$, so $ S(n)\plus{}S(S(n))\leq 38$, so $ n\geq 1969$

From there, it's very easy to find that $ n\in\{1977,1980,1983,2001\}$ and that the answer is $ 4$.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Can somebody prove of disprove the following statement?:
Let $ G$ be a solvable group of order $ mn$ with $ (m,n)\equal{}1$ such that there exist a nilpotent normal subgroup $ H$ of order $ m$. Then there exist at most one subgroup of $ G$ of order $ n$.

The existence of such subgroup is a consequence of P.Hall's theorem: Let $ G$ be a finite solvable group of order $ mn$, with $ (m,n)\equal{}1$. Then $ G$ contains a subgroup of order $ n$ and any two subgroups of $ G$ of order $ n$ are conjugate.

So the statement is equivalent to: If $ A$ is a subgroup of $ G$ of order $ n$ then $ A$ is normal in $ G$.

I am interesting in the cases:
(a)$ n$ prime, or more generally
(b)$ G/H$ abelian.

It is easy to construct a counterexample, because what you state is that any semi-direct product $ G\equal{}VU$ with normal subgroup $ V$ and $ (|V|,|U|)$ is a direct product.

[b]Counterexample[/b]: Let $ p>2$ and $ \mathbb{F}_p^*\equal{}\,<t>\,$ and define $ G\equal{}\,<x,y,z>$ via relations $ 1\equal{}x^p\equal{}y^p\equal{}[x,y]\equal{}z^{p\minus{}1}\equal{}$ $ [x,z]x^{1\minus{}t}$ $ \equal{}[y,z]y^{1\minus{}t}$. Then $ N\equal{}\,<x,y>\,$ is elementary-abelian of order $ p^2$ and normal in $ G$ with factor group $ G/N\simeq\,<z>\simeq\mathbb{F}_p^*$, i.e. $ G\equal{}N<z>$. Clearly, $ G\not\equal{}N_G(<z>)\equal{}C_G(<z>)\equal{}<x^{\minus{}1}y>\times <z>$

I don't know if I understand your counterexample.
Supose that $ F_p^{*}\equal{}<2>$ (for example $ p\equal{}3,5$). Then the ralations $ [x,z]x^{1\minus{}t}\equal{}1$ and $ [y,z]y^{1\minus{}t}$ with $ t\equal{}2$ are $ xzx^{\minus{}1}z^{\minus{}1}x^{\minus{}1}\equal{}1$ and $ yzy^{\minus{}1}z^{\minus{}1}y^{\minus{}1}\equal{}1$ wich is the same that $ x^{\minus{}1}\equal{}y^{\minus{}1}\equal{}1$ or $ x\equal{}y\equal{}1$ and then $ N$ doesn't have order $ p^2$.

the commutator definition I used is $ [a,b]\equal{}a^{\minus{}1}b^{\minus{}1}ab$. In order to avoid confusion work with the relations $ z^{\minus{}1}xz\equal{}x^t$ and $ z^{\minus{}1}yz\equal{}y^{t}$


I was wondering if anyone knows where I can find the PDFs or such files of actual Mathlinks Contests from the past.

I heard there were seven editions that have occurred so far.


limit
inequalities proposed
inequalities

$ \{a_n\}$ is an infinite sequence of positive reals.
If $ \sum_{k \equal{} 1}^{\infty}\frac {1}{a_k}$is bounded
Can $ \sum_{k \equal{} 1}^{\infty}\frac {k}{a_1 \plus{} a_2 \plus{} ... \plus{} a_k}$ be unbounded?

This is what I thought:
Denote $ x_n \equal{} \frac {1}{a_n}$
use Hm-Gm
Ineq becomes $ \sum_{k \equal{} 1}^{\infty}x_k\le C$
Then$ x_1 \plus{} \sqrt {x_1x_2} \plus{} ... \plus{} \sqrt [n]{x_1x_2...x_n}$ is bounded
I remember $ RHS\le eC$
$ e \equal{} \lim_{k \rightarrow \infty}(1 \plus{} \frac {1}{k})^k$

What's your solution,my dear friend :)

[quote="shfdfzhjj"]Method 1:\[ \sum_{k\equal{}1}^n\sqrt[k]{\prod_{i\equal{}1}^k a_i}\plus{}n\sqrt[n]{\prod_{i\equal{}1}^n a_i}<e\sum_{i\equal{}1}^n a_i\](use induction)
Method 2:$ a_1$, $ a_2$,\ldots,$ a_n$ are positive reals,prove that
\[ \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} ... \plus{} \frac {n}{a_1 \plus{} a_2 \plus{}... \plus{}a_n} < \frac {2}{a_1} \plus{} \frac {2}{a_2} \plus{} ... \plus{} \frac{2}{a_n}\][/quote]


calculus
integration
trigonometry
symmetry
logarithms
calculus computations

Calculate

\[\int^{\pi}_0 \frac{x\sin^3 x}{4-\cos^2 x}dx\]

Let $y=\pi-x$, and the integral becomes
$
I = \int_0^\pi \frac{(\pi-y)\sin^3 y}{4-\cos^2 y}dy
$
Hence
$
2I = \int_0^\pi \frac{\sin^3 x}{4-\cos^2 x}dx = \int_0^\pi \frac{\sin^2 x}{4-\cos^2 x}d(-\cos x) = \int_{-1}^1 \frac{1-t^2}{4-t^2}dt = 2-\frac{3\ln 3}{2}$
Thus
$I=1-\frac{3\ln 3}{4}$.

You forgot to multiply by $\pi$, liyi. :?


AMC AIME USA(J)MO USAMO ARML Duke college

Post any important contest dates and/or websites here.  I will keep a complete list in the first post:
[hide="2006-2007 contest information"]
[hide="AMC:"]
[url=http://www.unl.edu/amc/]Website[/url]

AMC 8 is held on Tuesday, November 14, 2006.
AMC 10/12A is held on Tuesday, February 6, 2007
AMC 10/12B is held on Wednesday, February 21, 2007
AIME is held on Tuesday, March 13, 2007
Alternate AIME is held on Wednesday, March 28, 2007
USAMO is held on April 24 & 25, 2007
[/hide]

[hide="ARML"]
[url=http://www.arml.com/]Website[/url]

The contest is most likely the first Saturday in June.

The second practice will be on May 19.[/hide]

[hide="Duke Math Meet"][url=http://www.math.duke.edu/dumu/contests/2005/]Website[/url]

The contest is November 11, 2006[/hide]

[hide="MathCounts"]
[url=http://www.mathcounts.org/]Website[/url]

Chapter contests are any time during February, depending on the chapter.
State Contest is Friday, March 23, 2007 (first time on a weekday)
National is yet to be announced.[/hide]

[hide="NEMO"]
Email for support:  nemosupport@gmail.com

The contest will be held on Saturday, December 9th at Enloe High School (in Raleigh) from 8:00 AM to 3:15 PM.

There are 3 rounds:
Individual Round (30 problems/50 minutes)
Team Round (15 problems/50 minutes)
Application Round

Teams are made of up to 6 students, with a cost of 15 students per team.

Middle School Competition ONLY[/hide]

[hide="State Comprehensive Contest"]
The state comprehensive contest will be held on April 19.

[url=http://courses.ncssm.edu/goebel/statecon/state.htm]Website[/url][/hide][/hide]

I'll add more as information becomes available.

OH MAN WE ACTUALLY HAVE A POST HERE CRAZY

remind me again why we have this thing

if the NC forum is still fairly inactive, you guys are always welcome back to the SC forum :wink:

WE SHALL SURVIVE!  :D

Well, ARML is in early June, State Math Contest is in May, (regional sites in March), USAMTS round 1 due October 10.

More?

Apparently 11/11. Please add that and ARML stuff.

State MathCounts will be held Friday, March 23, 2007 at NCSSM in Durham.  First ever weekday administration of the competition.

Information about MathCounts, Duke Math Meet, and ARML has been added.

When I add more information, the topic will be edited so you can see it without viewing the thread (can't think of what that is called right now).

[hide="Duke Math Meet"][url=http://www.math.duke.edu/dumu/contests/2005/]Website[/url]

The contest is November 11, 2007[/hide]
[/quote]

Don't we only care about the '06 math meet right now? :maybe:

Also, MathCounts is one word  :wink:

Sorry, I'm kind of picky about grammar (about 3/4 of the time)

The 2007 thing was a typo, and I forgot about MathCounts being 1 word at the time.

b-flat,

      National Assessment & Testing administers five mail-in contests during the year, starting with the Fall Startup Event on September 28th.  We've got a forum under "Other Contests" with more information, and I'd be happy to answer any questions you might have.

Thank you,

we take the USAMO at our school right? I heard from someone that we go to MIT or something.

MIT? no, you are proctored by a teacher at your school.

The only tests you don't take at your school are MOP tests and IMO.

there was one year USAMO was held in a centralized location; it might have been MIT, I dunno

[quote="b-flat"]NEMO has been added to the first topic (I finally remembered to add it).[/quote]

you mean the fish?

edit: nvm

[quote="b-flat"]Teams are made of up to 6 students, with a cost of 15 students per team.[/quote]

Ooh. So that's where they went. I wonder what they did with them.

What?

The above post has been confusing in every way. I am left dazed and confused. (6=15??!?)

That would be a small typo on my part  :blush:

Will edit.

Add state math contest (April 19) and ARML practice (May 19).

Thanks, the front page has been updated again (I also included the website for the comprehensive contest).

Information for 2007-08.

[hide="AMC:"]
[url=http://www.unl.edu/amc/]Website[/url]
AMC 8 is held on Tuesday, November 13, 2007.
AMC 10/12A is held on Tuesday, February 12, 2008.
AMC 10/12B is held on Wednesday, February 27, 2008.
AIME is held on Tuesday, March 18, 2008.
Alternate AIME is held on Wednesday, April 2, 2008.
USAMO is held on April 29 & 30, 2008.[/hide]
[hide="ARML"][url=http://www.arml.com/]Website[/url]
May 31, 2008.[/hide]
[hide="Duke Math Meet"][url=http://www.math.duke.edu/dumu/contests/2005/]Website[/url]
November 10, 2007[/hide]
[hide="Princeton Math Meet"][url=https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition]Website[/url]
November 17, 2007[/hide]
[hide="MathCounts"][url=http://www.mathcounts.org/]Website[/url]
The regional contests are any time in February.
The state contest is TBA (as far as I know... I'm not really looking for this date).
The national contest is May 8-11, 2008.[/hide]
I'll add more as information becomes available.

[hide="State Math Contests"][url=http://courses.ncssm.edu/goebel/STATECON/state.htm]Website[/url]

The State Comprehensive Contest will be held on April 24, 2008.

The state subject contests (Algebra, Algebra II, Geometry) will be held on May 1, 2008.

Regional contests either have already been held or will be held sometime in the near future, depending on the site.[/hide]

[hide="AMC"]
In-school competition, probably the most commonly taken in America.

[url]http://www.unl.edu/amc[/url]

AMC 8: November 18, 2008
AMC 10/12 A: February 10, 2009
AMC 10/12 B: February 25, 2009
AIME I: No info yet
AIME II: No info yet
USAMO: No info yet
[/hide]

[hide="ARML"]
On-site competition in either Pennsylvania (hopefully) or Georgia. North Carolina sends 2 teams, A and B. Each team has 15 people. Some (2-3) alternates also go sometimes. There are 4 rounds - Power (team, proof-based, some number of problems in an hour); Team (team, numeric answers, 10 problems in 30 minutes, calculators (yuck) allowed [b]EDIT: OR NOT[/b]); Individual (individual :O, numeric answers, 4 rounds of 2 problems in 10 minutes); Relay (sub-teams of 3, every answer is based on what your relay teammates pass you... numerical answer, 2 of them, 6 minutes each.)

[url]http://www.arml.com/index.php[/url]
ARML: May 29-30, 2009

[url=http://courses.ncssm.edu/goebel/STATECON/]NC ARML website[/url] (Note that this is the state math contest website, just click ARML links for more info)
[/hide]

[hide="State Math Contest"]
The... state math contest. 30-40 problems in about 1.5-2 hours (depending on region). There are also Alg I, Geometry, Alg II division for those taking those classes.

[url=http://courses.ncssm.edu/goebel/STATECON/]State Math Contest[/url]
[/hide]

[hide="Duke Math Meet"]
An on-site meet held at Duke University (of all places!) Similar to ARML. You go with your school, or a local team of 6 people.

Un-updated website (as of post):
[url]http://www.math.duke.edu/dumu/mathmeet/[/url]
[/hide]

Princeton and HMMT, I am currently too lazy to post.
MathCounts, I don't know about and don't really wanna look it up.

For arml:

[quote="the website"]
[b]they will be prohibited in 2009 on the entire competition[/b]
[/quote]

I did not change anything (thats right, it even had the bold)

[url]http://arml.com/arml_registration/index.php[/url]


Ok I found a song and thought it was pretty funny. Im sure you will love this song if you think indian accents are funny :lol: . The link is-  http://www.andrew.cmu.edu/user/gkumar/WelcomeToIndia.htm

Speaking of Indian accents... From  MIT students this is really awsome:
Get videos directly
Small (10 MB): http://www.mit.edu/~kapoor/mile_sur_small.wmv
Big(50 MB): http://web.mit.edu/anoop/www/msur/milesur50.mpg

To the first: hahahaha that was funny.
To the second-???


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that a finite subgroup of a field is cyclic.


How do you download the USAMO trial question?

Close your eyes and tap your heels together three times. Then the "trial question" will appear on your computer screen.

There is no link... wonder why it is called "trial"??

[quote="paladin8"]I think they are just testing the link or something... its a 'trial' run for what they are actually gonna do on USAMO day.[/quote]

I don't think so, there should be a trial question,

[quote="Letter to USAMO qualifier"]A trial question will be posted one week before the USAMO for teachers & students to practice obtaining & downloading.[/quote]

The link to the HTML Format is up; but all it says is 'On April 19th this page will be replaced with questions 1, 2 & 3 of the 2005 USAMO.' when you go to it...


Let $ n$ be an integer greater than $ 3$, and let $ a_1, a_2, \cdots, a_n$ be non-negative real numbers with $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 2$. Determine the minimum value of \[ \frac{a_1}{a_2^2 \plus{} 1}\plus{} \frac{a_2}{a^2_3 \plus{} 1}\plus{} \cdots \plus{} \frac{a_n}{a^2_1 \plus{} 1}.\]

The minimum value is 3/2, attained when $a_1 = a_2 = 1$ and $a_3 = \cdots = a_n = 0$.

[quote="dgreenb801"]When does equality hold then? By your proof, it seems no matter how many variables there are, it will always be less than a fixed value, but if we make all the variables equal, then it depends on how many variables there are.[/quote]I think Altheman's proof is correct. Both lemmas have equality cases $\forall x,y\in\{0,1\}$ and $a_1 = a_2 = 1,a_3 = \cdots = a_n = 0$ respectively. Nothing to do with the case of all $a_i$ equal. ;)

The problem requires that $n>3$ however.

Any complete and more detailed solution?

How do you find such old threads and why do you revive them?

[quote=Altheman]$ \frac{x}{1\plus{}y^2}\ge x\minus{}\frac{xy}{2}$

Then we use the lemma $ (x_1\plus{}x_2\plus{}...\plus{}x_n)^2\ge 4(x_nx_1\plus{}x_1x_2\plus{}...\plus{}x_{n\minus{}1}x_n)$ and we're done.

(Prove by induction)[/quote]

Could also straight up use a two-step smoothing process to find maximum of $a_1a_2 + a_2a_3 + \dots a_na_1$:

Suppose $n \geq 5$. Else it is trivial.

i) Push an element to 0:
WLOG assume $a_1 + a_3 \geq a_3 + a_5$.
Then push $a_4$ to $a_2$.

ii) Apply chain reaction to push all irrelevant elements to 0:
Reorder elements to $a_1a_2 + a_2a_3 + \dots a_{k-1}a_{k}$
Push $a_1$ to $a_3$ and repeat (reordering to make indices nicer) until we have $a_1a_2 + a_2a_3$ left, at which point it becomes trivial.

Let $a_1,a_2,\cdots,a_n$ $(2\leq n\leq 4)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$ Prove that $$\frac{a_1}{a^2_2+1}+\frac{a_2}{a^2_3+1}+\cdots+\frac{a_{n-1}}{a^2_n+1}+
\frac{a_n}{a^2_1+1}\geq\frac{n}{2}.$$
Let $a_1,a_2,\cdots,a_n$ $( n\geq 5)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$ Prove that $$\frac{a_1}{a^2_2+1}+\frac{a_2}{a^2_3+1}+\cdots+\frac{a_{n-1}}{a^2_n+1}+
\frac{a_n}{a^2_1+1}>\max\{\frac{8n}{n^2+8} ,n-\frac{n^2}{8}\}.$$
[url=https://artofproblemsolving.com/community/u148231h1671756p10636795](Zhangyanzong)[/url]

Let $a_1,a_2,\cdots,a_n$ $( n\geq 2)$ be positive real numbers such that $a_1+a_2+\cdots+a_n=n.$   Find the minimum possible value of $$\frac{a_1}{a^3_2+4}+\frac{a_2}{a^3_3+4}+\cdots+\frac{a_{n-1}}{a^3_n+4}+
\frac{a_n}{a^3_1+4}.$$
[hide=2017 USAMO]Let  $a$, $b$, $c$, $d$ are nonnegative real numbers such that $a+b+c+d=4 .$ Then
$$\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\geq\frac{2}{3}.$$
https://artofproblemsolving.com/community/c5h1434574p8117097[/hide]

[quote=peregrinefalcon88]Note that the lemma being used is only true for $n \ge 4$ so for n = 3 you can use smoothing to show $\frac{4}{3}$ is the minimum.[/quote]
https://artofproblemsolving.com/community/c6h1820982p12167038:
Let $a, b, c$ be nonegative real numbers such that $a + b + c = 2$. Then$$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq\frac{4}{3}$$
[b]Proof of DievilOnlyM:[/b]$$2-\sum\frac{a}{1+b^2}=\sum\frac{ab^2}{1+b^2}\leq\sum\frac{ab^2}{2b}=\frac{1}{2}\sum ab\leq\frac{2}{3}$$

The maximum is $\frac 32$, achieved at $(1, 1, 0, 0, \dots, 0)$. The key is to use the tangent line at $x=1$, which gives us the estimate $$\frac 1{x^2+1} \geq 1-\frac x2 \iff \frac{x(x-1)^2}{2(x^2+1)} \geq 0.$$ Thus, $$\sum_{i=1}^n \frac{a_i}{a_{i+1}^2+1} \geq \sum_{i=1}^n a_i - \frac 12 a_ia_{i+1} \geq \frac 32.$$

We want to show that the minimum is $\frac{3}{2}$, which is attainable when all of the terms are $0$ except for two consecutive ones with value $1$.

First, define $f:=\dfrac{1}{x^2+1}$.We now have (tangent line trick) :

$$f(x) \geq \frac{-x}{2}+1 \iff \frac{(x-1)^2x}{2(x^2+1)} \geq 0 \quad \forall x \in \mathbb{R}$$

Hence:

$$\sum  \frac{a_i}{1+a_{i+1}} \geq \sum a_i - \frac{a_ia_{i+1}}{2} \geq \frac{3}{2}$$

Where the last inequality comes
$$\sum a_ia_{i+1} \leq \sum a_ia_j \leq \frac{1}{4} \left( \sum a_i \right)^2 $$

$$\mathbb{Q.E.D.}$$

We claim that the minimum possible value is $\boxed{\frac{3}{2}}$. We can see that this is given by $(1, 1, 0, 0, \ldots , 0)$. We claim that \[\frac{1}{a^2+1} \geq \frac{-a+2}{2}\] for $a \in [0, 2]$. Multiplying both sides by $2a^2+2$ we get \[a^3-2a^2+a \geq 0 \Rightarrow a(a+1)^2 \geq 0\] which is obvious for $a \in [0, 2]$.

Now we can see that since $\frac{a_n}{a_{n+1}^2+1} \geq a_n - \frac{a_na_{n+1}}{2}$ We simply need \[\sum_{\text{cyc}} a_1a_2 \leq 1\] We claim that \[4 \sum_{\text{cyc}} a_1a_2 \leq (a_1+a_2+\dots+a_n)^2\] Which obviously proves the problem. This last inequality is true because consider $(a_i, a_{i+2})$ we can see that we can increase the sum the most if we let one of these pair be $0$. So set this to either $(a_i+a_{i+2}, 0)$ or $(0, a_i+a_{i+2})$ (based on which of $a_{i-1}$ and $a_{i+3}$ is bigger). Thus, once we do this enough times for $0 \pmod 2$ and $1 \pmod2$ indices we will get only zero indices besides two indices. Thus we need that $4xy \leq (x+y)^2$ which is obvious. $\blacksquare$

For some reason many people are omitting the very untrivial proof of the last lemma


algebra
polynomial
number theory
number theory theorems

prove htat every positive integer can be written as the sum of not more than 53 fourth power of integers. :(

[b]Hint[/b]
using [url=http://mathworld.wolfram.com/LagrangesFour-SquareTheorem.html]Lagrange's four-square theorem[/url] and [url=http://mathworld.wolfram.com/LiouvillePolynomialIdentity.html]Liouville polynomial identity[/url].

PS This problem is case $ g(4)$ of [b]Waring's Problem[/b].

This was posted before. It may be hard to find, but I remember posting a solution.

More interestingly, any sufficiently large integer is the sum of just 16 4th powers. This is usually denoted as $ G(4)\equal{}16$, as opposed to $ g(4)\equal{}53$. Amazingly, $ G(3)$ is (as far as I know) still unknown.

I answer it in Japanese.
Please translate it somewhere.

$ n$ を自然数とする. $ n$ を $ 6$ で割ったときの商を $ k$, 余りを $ r$ とすると
\[ n=6k+r \ \ (k \in Z \ \text{and} \ k\ge 0, \ 0 \le r \le 5) \cdots \cdots \ooalign{\hss \footnotesize{1} \hss \crcr \bigcirc}\]
とかける.
このとき, [url=http://mathworld.wolfram.com/LagrangesFour-SquareTheorem.html]Lagrange's four-square theorem[/url] により $ k$ は $ 4$ つの平方数の和として表せる. それを
\[ k = a_1{}^2 + a_2{}^2 + a_3{}^2 + a_4{}^2\]
とすると
\[ 6k = 6a_1{}^2 + 6a_2{}^2 + 6a_3{}^2 + 6a_4{}^2\]
である. よって, $ \ooalign{\hss \footnotesize{1} \hss \crcr \bigcirc}$ より
\[ n = 6a_1{}^2 + 6a_2{}^2 + 6a_3{}^2 + 6a_4{}^2 + r \cdots \cdots \ooalign{\hss \footnotesize{2} \hss \crcr \bigcirc}\]
である. ここで, $ a_1$, $ a_2$, $ a_3$, $ a_4$ はすべて $ 0$ 以上の整数である. そのそれぞれがまた, $ 4$ つの平方数の和となっているから, 例えば
\[ a_1 = x_1{}^2 + x_2{}^2 + x_3{}^2 + x_4{}^2\]
とすれば
\[ 6a_1{}^2 = 6 \left(x_1{}^2 + x_2{}^2 + x_3{}^2 + x_4{}^2 \right)^2\]
であるから, [url=http://mathworld.wolfram.com/LiouvillePolynomialIdentity.html]Liouville polynomial identity[/url] を使うと, $ 6a_1{}^2$ が, $ 12$ 個の $ 4$ 乗数の和であることが分かる. 同様に $ 6a_2{}^2$, $ 6a_3{}^2$ および $ 6a_4{}^2$ それぞれも $ 12$ 個の $ 4$ 乗数の和である.
また, $ r$ は $ 5$ 個以下の $ 4$ 乗数の和であるから, 結局 $ n$ は $ \ooalign{\hss \footnotesize{2} \hss \crcr \bigcirc}$ より
\[ 12 + 12 + 12 + 12 + 5 = 53\]
以上より, すべての自然数は $ 53$ 個以下の $ 4$ 乗数の和で表せる.


geometry
trigonometry
integration
Euler
symmetry
calculus
calculus computations

The following curve:

\[ |x|^a + |y|^a = r^a \]
where $r$ is a constant, and $a > 0$, describes a family of closed geometric figures.

Can you find a formula for the area under this curve? This formula is a generalization of the formulae for the square, circle, diamond and astroid.

Express the curve in parametric form:
\[ x=r\cos^{\frac2a}\vartheta\qquad y=r\sin^{\frac2a}\vartheta\qquad 0\le\vartheta\le\frac{\pi }2\]
then from $ \frac yx=\tan^{\frac2a}\vartheta$ we get
\[ d\left(\frac yx\right)=\dfrac{2\sin^{\frac2a-1}\vartheta}{a\cos^{\frac2a+1}\vartheta}\]
The area is given by Green formula:
\[ S= \frac12\underset{\gamma}{\int}xdy-ydx\]
But \[\frac12(xdy-ydx)=\frac12x^2d\left(\dfrac yx\right)=\dfrac{r^2}a\left(\sin^{\frac2a-1}\vartheta\cos^{\frac2a-1}\vartheta\right)\]
Then
\[S=\dfrac{r^2}a\int_0^{\frac {\pi}2}\sin^{\frac2a-1}\vartheta\cos^{\frac2a-1}\vartheta\,d\vartheta=\frac12\dfrac{r^2}aB\left(\frac1a,\frac1a\right)\]
where B is the Euler beta function.

[quote="Nené"]
...
Then
\[S=\dfrac{r^2}a\int_0^{\frac {\pi}2}\sin^{\frac2a-1}\vartheta\cos^{\frac2a-1}\vartheta\,d\vartheta=\dfrac{r^2}aB\left(\frac2a,\frac2a\right)\]
where B is the Euler function beta. [/quote]
 :huh:
Counterexample : $a=2$ describes a circle centered at $0$ with radius $r$. In this case the area is $\pi r^2$. But
$4S$ (Finding $S$ is sufficient since the figure has four-fold symmetry)  is clearly not $\pi r^2$.

O.K Skanderbeg. Sorry for the typo. We must write:
\[ S= \frac12\dfrac{r^2}aB\left(\frac1a,\frac1a\right)\]

Thank you Nene. Your solution is correct. When $a \rightarrow \infty$, we have a square with side $2r$, So the area is
$4r^2$ (So a quarter of it is $r^2$). Can you use your result to verify this ?

It is possible to solve this problem without using Green's formula. You may be interested to read [url=http://www.geocities.com/mrtfkhang/TFKHANG2003.pdf]this[/url] , for a more elementary solution and some other discussions.  ;)

Thanks skanderbeg for your useful link suggestion.
For the other question, if $ a\to+\infty$ then we must  write:
\[B(\epsilon,\epsilon)=\dfrac{\Gamma(\epsilon)\Gamma(\epsilon)}{\Gamma(2\epsilon)}=\frac2{\epsilon}\dfrac{(\Gamma(1+\epsilon))^2}{\Gamma(1+2\epsilon)}\]
where $\frac1a=\epsilon$ and I used the known property between gamma and beta functions. Then I think that the desired result, $ S\to r^2\; \;if\;\;\epsilon \to 0^+$,  follow.


geometry
circumcircle
angle bisector
geometry unsolved

The problem is as follows:
Let $ABCD$ be a square and $E$, $F$ points on $BC$, $CD$ respectively so that $\angle FAE = 45^\circ$. Let $Q$ be the intersection of $BD$ and $AE$, and $R$ the intersection of $BD$ and $AF$. Let $l_1$ be a line through $R$ parallel  to $AB$ and $l_2$ be a second line through $Q$ parallel to $AD$. Let $M$ be the intersection of $l_1$ and $l_2$ . Prove that $F$, $M$ and $E$ are collinear.

Let a line through the point $F$ and forming a counter-clockwise angle of $45^o$ with the line $AF$ intersect the square diagonal $BD$ at a point $Q'$ Since the angles $\angle ADQ' = \angle AFQ' = 45^o$ are equal, the quadrilateral $AQ'FD$ is cyclic. Since the angle $ADF = 90^o$ is right, the circumcenter of this quadrilateral is the midpoint of its diagonal $AF$. Hence, the angle $\angle AQ'F = 90^o$ is also right. It follows that the remaining angle of the right angle triangle $\triangle AQ'F$ is equal to $\angle FAQ' = 180^o - (\angle AQ'F + \angle AFQ') = 45^o$. Consequently, the points $Q' \equiv Q$ are identical and the angle $\angle FQF = 180^o - \angle AQF = 90^o$ is right. Similarly, by considering a line through the point $E$, forming a clockwise angle of $45^o$ with the line $AE$ and intersecting the square diagonal $BD$ at a point $R'$, we can deduce that the points $R' \equiv R$ are identical and the angle $\angle FRE = 90^o$ is right. Thus the quadrilateral $EFRQ$ is cyclic and its circumcenter $I'$ is the midpoint of the diameter $EF$. The angles $AFQ = \angle AER = 45^o$ span the major arc $QR$ of the circle $(I')$, hence, the central angle of this arc is equal to $\angle QI'R = 90^o$. The triangle $\triangle QI'R$ is isosceles with $I'Q = I'R$ being radii of the circle $(I')$ and the angles $\angle QRI = \angle RQI' = 45^o$. Since the chord $QR \equiv BD$, one of the square diagonals, bisects the right angles $\angle ABC, \angle CDA$, the radii $I'Q, I'R$ are parallel to the opposite square sides, $I'Q \parallel BC \parallel DA$ and $I'R \parallel AB \parallel CD$. As a result, the points $I' \equiv I$ are identical and the points $E, I, F$ collinear, $EF$ being a diameter of the cuircle $(I') \equiv (I)$.

Obviously, $ \angle FAD \equal{} \angle EAC$ ( 1 ), from $ \angle FAE \equal{} \angle RAE \equal{} \angle DBC \equal{} \angle RBC \equal{} 45^\circ$ we get $ RABE$ cyclic and, as $ BR$ is an angle bisector, it follows $ AR \equal{} RE$; with $ \angle ARE \equal{} \angle ABE \equal{} 90^\circ$, we see that $ \triangle ARE$ is rightangled isosceles, consequently $ \frac {AR}{AE} \equal{} \frac {1}{\sqrt 2}$; similarly $ \frac {AQ}{AF} \equal{} \frac {1}{\sqrt 2}$ ( 2 ), hence $ FRQE$ is cyclic - $ EF$ being a diameter of its circumcircle - and $ \angle AEF \equal{} \angle ARQ \equal{} \angle DAR \plus{} \angle ADR \equal{} \angle DAR \plus{} 45^\circ$, so $ \angle AEF \equal{} \angle EAD$ which, in turn is equal to $ \angle AEB$, therefore $ AE$ is the bisector of the angle $ \angle BEF$, that is, $ A$ is the excenter of the triangle $ \triangle FEC$.
From (2) and following we see that $ RQ$ is the side of the square inscribed into the circle of diameter $ EF$, consequently, if $ O$ is the midpoint of $ EF$, the triangle $ \triangle RQO$ is rightangled and isosceles, hence $ QO \parallel{} BC$ and $ RO \parallel{} CD$, hence $ O \equiv M$, done.

Best regards,
sunken rock


I don't know if this is the right place to post this.. :roll:

anyway, I'm a 10th grade student (1st year of High school), I'am a beginner in olympiad but I love math,  this year My teacher chooses me to participate in a math contest, so I wonder if someone give me some training materials( for 10th grade), and show me some tricks

Thank you very much my dear Friends

hy anas,hy rachid,my name is Driss,i am really verry happy,i appreciate highly ur present here,and i encourage u to still here
i loved Olympiade problems too and i still do.contact me si vs avez besoin de nimporte quelle besoin.
may that is what ur looking for anas.

[url]http://www.imo.org.yu/[/url]

thank you my brother ( choukrane a khoya driss  :) )


hmm....

suppose it's possible to exist mutants. just like X-men.

who is the most possible X-men!?

I think is Professor Charles Francis Xavier and Magneto.

because if you agree with the eletromagnetic theory of mind, it's possible to manipulate those eletromagnetic, but it is too weak.

but if it could be strong enough we could make a Electron "levitates" in a distance of some centimeters of our brain

:)

i think Wolverine is possible to, in a near future maybe...

I have heard that if someone get some strange disease, it is possible for him to generate electricity and levitate metal objects :huh:

I also heard that people do something to their head, and they become a little bit telepathic.

I heard that if you eat broccoli you can become the Hulk.

so there was this kid I knew back in 6th grade

he was really stupid

so we all threw radioactive polonium at him

and then he died


Prove, that equation $ x^n\plus{}y^n\equal{}z^n$ doesn`t have positive integer solutions, if $ n>2$ and $ z<1\plus{}\frac{1}{2^{\frac{1}{n}}\minus{}1}$ .

Real It is a good problem if the Fermat last theorem wasn't be solved.
The first prove that $ z\geq n\plus{}1$
If $ z<n\plus{}1$ then $ x,y<n$
Let $ x\equal{}\max{y,z}$ then we have
$ z\geq x\plus{}1$
So we must have $ 2x^n>(x\plus{}1)^n,\Longleftrightarrow 2>(1\plus{}\frac{1}{x})^n\geq 1\plus{}\frac{n}{x}$  (tradition)!
So must have $ z>n\plus{}1$
But $ n\plus{}1>1\plus{}\frac{1}{\sqrt[n]{2}\minus{}1}$

[quote="TTsphn"]Real It is a good problem if the Fermat last theorem wasn't be solved.
The first prove that $ z\geq n \plus{} 1$
If $ z < n \plus{} 1$ then $ x,y < n$
Let $ x \equal{} \max{y,z}$ then we have
$ z\geq x \plus{} 1$
So we must have $ 2x^n > (x \plus{} 1)^n,\Longleftrightarrow 2 > (1 \plus{} \frac {1}{x})^n\geq 1 \plus{} \frac {n}{x}$  (tradition)!
So must have $ z > n \plus{} 1$
But $ n \plus{} 1 > 1 \plus{} \frac {1}{\sqrt [n]{2} \minus{} 1}$[/quote]

I'm sorry, but that's wrong. We always have  : $ n \plus{} 1 < 1 \plus{} \frac {1}{\sqrt [n]{2} \minus{} 1}$ $ \forall n\geq 2$

Sorry pco I have see my mistake . It is true if replace $ 2\to e$


calculus
integration
function
geometry
3D geometry
trigonometry
calculus computations

Consider a solid cone. The radius of its circular base is 2 and its height is 2. If the density function is ${\sigma}(x,y,z)=x^{2}z$,  find the mass.

Would the limits of integration be:

$\int_{0}^{2{\pi}}\int_{0}^{2}\int_{0}^{4-r^{2}}r^{3}cos^{2}({\theta})dzdrd{\theta}=\frac{16{\pi}}{3}$?.

I was thinking, since $x^{2}=r^{2}cos^{2}({\theta})$ and $z=r$ and we're using $rdzdrd{\theta}$, maybe we should use $r^{4}cos^{2}({\theta})$. Which would make the integral equal to $\frac{256{\pi}}{35}$

If I wanted to use, say, $drdzd{\theta}$ or $d{\theta}dzdr$ how would I determine the limits of integration then?.

$\int_{0}^{2{\pi}}\int_{0}^{2}\int_{0}^{4-r^{2}} (\cdot) dz\,dr\,d\theta$

represents an integral over a parabaloid, not a cone. $z=4-r^2$ as boundary makes it a parabaloid. For the cone, set $z=4-2r$ or

$\int_{0}^{2{\pi}}\int_{0}^{2}\int_{0}^{4-2r} (\cdot) dz\,dr\,d\theta .$

Thank you very much. That r^2 was a typo(I meant 4-r). Do I still use the $r^4cos^{2}({\theta})$ since the density function is $x^{2}z$?.

No. $x^2z=r^2z\cos^2\theta,$ and $dV=r\,dr\,d\theta\,dz.$

Don't set $z$ equal to anything - just let it be $z,$ where it is one of the variables  you're integrating over.


The sum of the six whole number factors of 32 is

easy! 1+2+4...32 equals 63 which is one less than 32*2

[quote="pianoforte"]The sum of the factors of $2^{n}$ is $2^{n+1}-1$, which in this case is 63.[/quote]
[hide="Here is why"]The factors are: $1,2,4,8,...,2^{n}$. Since the geometric series has first term $1$, ratio $2$, and $n+1$ terms, the answer is: $\frac{1-2^{n+1}}{1-2}= 2^{n+1}-1$.[/hide]

Just add one to the smallest and biggest number outside of the number and one (this case 2 and 16) so you get 3 and 17.  Now you add the bigger number by the second lowest number outside of the number and one (this case 4) and you get 21.  Multiply 21 and 3 and you get 63.  This is a special number in which this is one of the few in which this will work.  Four is similar but the same formula if you think about it. Six you just don't bother with the second step, as with 8.

Please don't revive old topics, especially ones that have been solved multiple times already.

No one's probably going to read mine though, but
[hide]the factors of 32 are 1, 2, 4, 8, 16, 32.
They add up to 63.
[b]1+2+4+8+16+32.[/b][/hide]

[b]Same answer as everyone else's.[/b]

To the people in the begginning, PLEASE HIDE YOUR ANSWERS!


ratio
geometry
similar triangles
angle bisector

Altitude $CH$ and median $BK$ are drawn in an acute triangle $ABC,$ and it is known that $BK=CH$and $\angle KBC= \angle HCB.$ Prove that triangle $ABC$ is equilateral.

[hide="My Solution"]Let the intersection of $CH$ and $BK$ be $P.$ Since $\angle KBC= \angle HCB,$ triangle $BPC$ is isosceles and $BP=CP.$ However, we're given that $BK=CH,$ so $PH=CH-CP=BK-BP=PK.$

Note that $\angle BPH=\angle CPK$ by vertical angles. Hence, $\triangle BPH \cong \triangle CPK \implies\angle CKP=90^\circ.$ It follows that $\triangle APH \cong \triangle APK\cong \triangle BPH \cong \triangle CPK.$ Hence, $AB=AC$ and $CH$ and $CK$ are both medians of triangle $ABC.$ Let the median from $A$ meet $BC$ at $I.$ Now, note that $\triangle APH \sim \triangle ABI$ with side ratio $1: 2.$ Hence, $BI=\frac12 AB.$ Similarly, $CI=\frac12 AC$ by similar triangles $APK\sim ACI,$ so triangle ABC is equilateral.
[/hide]

Any other solutions?

Mine is the same as yours up to a point.

[hide="Quicker conclusion"]$CH$ and $BK$ are both medians, but also heights, hence both are symmetrial axes, giving $AC=CB$ and $BC=BA$.[/hide]

Oh, right. I forgot you could do that.  :D

Different solution[hide]Let $KS$ be perpendicular to $AB$. Then as $KS=\frac{CH}2=\frac{BK}2$ from $\Delta BKS$ we see that $\angle SBK=30^{o}$. So $\angle HPB = 60^{o}$ ($P$ is the intersection point of $CH$ and $BK$) but $\angle HPB = \angle PCB+\angle PBC$. This means $\angle PBC=\angle PCB =30$. So $BK$ is also angle bisector and $AB=CB$ but as $\angle ABC=60^{o}$ $\Delta ABC$ is equilateral.[/hide]


trigonometry
induction
algebra unsolved
algebra

Find $ f: N\to N$ satisfy
$ f(f(f(n)))\plus{}f(f(n))\plus{}f(n)\equal{}3n$

Define the series $ a_{1} \equal{} f(m), a_{n \plus{} 1} \equal{} f(a_{n})$. We have the following linear recursion: $ a_{t \plus{} 3} \plus{} a_{t \plus{} 2} \plus{} a_{t \plus{} 1} \equal{} 3a_{t}$
We can find the general term by solving $ x^3 \plus{} x^2 \plus{} x \minus{} 3 \equal{} 0$. x=1 is a root, so it's easy to find the other roots: $ \minus{} 1\pm i\sqrt {2}$
The general term is $ a_{n} \equal{} A_{1} \plus{} A_{2}( \minus{} 1 \plus{} i\sqrt {2})^{n} \plus{} A_{3}( \minus{} 1 \minus{} i\sqrt {2})^{n}$
$ (1) a_{1} \equal{} A_{1} \plus{} A_{2}( \minus{} 1 \plus{} i\sqrt {2}) \plus{} A_{3}( \minus{} 1 \minus{} i\sqrt {2})$
$ (2) a_{2} \equal{} A_{1} \plus{} A_{2}( \minus{} 1 \minus{} 2i\sqrt {2}) \plus{} A_{3}( \minus{} 1 \plus{} 2i\sqrt {2})$
$ (3) a_{3} \equal{} A_{1} \plus{} A_{2}(5 \plus{} i\sqrt {2}) \plus{} A_{3}(5 \minus{} i\sqrt {2})$
From 1 and 3: $ 6(A_{2} \plus{} A_{3}) \in Z$. From 1 and 2: $ 3(A_{1} \minus{} A_{2} \minus{} A_{3}) \in Z$. Hence, $ 6A_{1} \in Z$. From 1 and 2: $ 3i\sqrt {2}(A_{2} \minus{} A_{3}) \in Z$.

If we put $ \minus{} 1 \plus{} i\sqrt {2} \equal{} \sqrt {3}cis\theta, \minus{} 1 \minus{} i\sqrt {2} \equal{} \sqrt {3}cis( \minus{} \theta),$ we have: $ a_{n} \equal{} A_{1} \plus{} 3^{\frac {n}{2}}((A_{2} \plus{} A_{3})\cos n\theta \plus{} (A_{2} \minus{} A_{3}) i\sin n\theta))$.
If $ A_{2} \equal{} A_{3}$ (not zero): $ A_{2}3^{\frac {n}{2}}\cos n\theta \in Q$ for every n, which implies that $ \frac {\cos 3\theta}{\cos\theta}$ is rational which is easy to disprove. Else, $ A_{2} \equal{} A_{3} \plus{} i\sqrt {2}a, a \in Q$
I'm stuck... I am sure that there are other ways of solving it but I wanted to use this recently accuired technique... The goal is to show that $ A_{2} \equal{} A_{3} \equal{} 0$ (by showing that otherwise, $ a_{n}$ can get negative\irrational values), and thus $ f(n) \equal{} n$.

$ (*)\quad f(f(f(m))) \plus{} f(f(m)) \plus{} f(m) \equal{} 3m$

Putting $ m \equal{} 1$ we get $ f(1) \equal{} f(f(1)) \equal{} f(f(f(1))) \equal{} 1$, because for all $ n\; f(n)\ge1$.

Assume that for all $ k < n\; f(k) \equal{} k$.
Suppose that $ f(n) < n$. Then $ f(f(n)) \equal{} f(n)$ (from the inductive hypothesis) and $ f(f(f(n))) \equal{} f(f(n)) \equal{} f(n)$.
Putting $ m \equal{} n$ in $ (*)$ we get $ 3f(n) \equal{} 3n$ which is contradiction.

If $ f(n) > n$ at least one of the numbers $ f(f(n))$ and $ f(f(f(n)))$ must be smaller then $ n$.
$ 1)$ If $ f(f(n)) < n$, putting $ m \equal{} f(n)$ in $ (*)$ and using $ f(f(f(n))) \equal{} f(f(n))$ we get $ 3f(f(n)) \equal{} 3f(n)$ which is contradiction, because $ f(f(n)) < n < f(n)$.
$ 2)$ If $ f(f(n)) > n$, then $ f(f(f(n))) < n$. We put $ m \equal{} f(f(n))$ and because $ f(f(f(f(n)))) \equal{} f(f(f(n)))$ we get $ 3f(f(f(n))) \equal{} 3f(f(n))$ which also can't be true.

So we conclude that $ f(n) \equal{} n$ for any $ n\in\mathbb{N}$.

[quote="Algadin"]Find $ f: N\to N$ satisfy
$ f(f(f(n))) \plus{} f(f(n)) \plus{} f(n) \equal{} 3n$[/quote]

Here is a slightly different twist on the induction method.

[hide=" Solution"]

After confirming that $ f(1) \equal{} 1$. Let's prove that $ f(n)$ is injective. Say there exist $ a,b \in \mathbb{N}$ s.t. $ f(a) \equal{} f(b)$ hence $ f(f(a)) \equal{} f(f(b))$ and $ f(f(f(a))) \equal{} f(f(f(b)))$ hence $ 3a\equal{}3b$ therefore $ a\equal{}b$.

Say $ f(k) \equal{} k$ for all $ k < n_0$ and since $ f(n)$ is injective therefore $ f(n_0) \neq f(k)$ hence $ f(n) \geq n_0$ for all $ n \geq n_0$.

So we have $ f(n_0) \geq n_0 \Rightarrow f(f(n_0)) \geq n_0 \Rightarrow f(f(f(n_0))) \geq n_0$ hence $ f(f(f(n_0))) \plus{} f(f(n_0)) \plus{} f(n_0) \geq 3n_0$ and since equality holds therefore $ f(n_0) \equal{} n_0$.

Therefore $ f(n) \equal{} n$.[/hide]


How can I become an administrator or moderator?

[quote="Arne"]You can't.[/quote]
but why i can't and which people can?

*Sigh*
Just read this:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=28268[/url]

It's not that you can't become a moderator, but we have hundreds of very active forum members and we choose the most active and responsible according to the needs of the community -- at least by our best judgment.

If you remain an active participant, it's possible that you would one day be asked to participate in moderation of the forums.


inequalities
search
algebra
polynomial
inequalities proposed

[color=red][b][i]Prove that inequality holds for every positive real numbers a,b,c,d [/i][/b][/color]

$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+bcd+bda+cda}{4}}.$

Davron

A better view:

[quote="Davron"]Prove that inequality holds for every positive real numbers $a,b,c,d$:
\[ \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq \sqrt[3]{\frac{abc+bcd+bda+cda}{4}}  \] [/quote]

to Davron: try \sqrt[n]{...} $\to\ \sqrt[n]{\cdots}$.

[quote="dule_00"]square both sides and use muirhead[/quote]
[b]square[/b]?

[quote="dule_00"]square both sides and use muirhead[/quote]

I did the same thing, but starting with the taking to the power of six... then expanding the terms i am using Muirhead and the second proof which i wrote in the Olympiad i used Am- Gm step by step many times took  my hour in the contest of total three hours ... and we had 12 problems... the second day 4 problems two hours duration time of problem solving.

Davron

This is not a strictly ineq, you also solve it, use the idea in Vietnam 1996. Try to search in this forum.

yes of cpurse not squaring. it's not soo hard to get that on third degre, you nedd at most 5min if you are treained, and with polinomial formula you can do it faster

You can find a solution to this problem in Engel's book (problem 3 of Chapter 7). As far I know, the problem is from the 1964 Hungarian Kürschák competition  :)

i think Manclaurin Inequality works. :harhar:

[quote="Davron"][color=red][b][i]Prove that inequality holds for every positive real numbers a,b,c,d [/i][/b][/color]

$ \sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2}{4}}\geq\sqrt [3]{\frac {abc \plus{} bcd \plus{} bda \plus{} cda}{4}}.$

Davron[/quote]
See here: http://canhang2007.wordpress.com/2009/11/06/inequality-52-maclaurin/, a proof by Cauchy-Schwarz and Chebyshev for the stronger one (a special case of MacLaurin's Inequality):
$ \sqrt {\frac {ab \plus{} ac \plus{} ad \plus{} bc \plus{} bd \plus{} cd}{6}} \ge \sqrt [3]{\frac {abc \plus{} bcd \plus{} cda \plus{} dab}{4}}.$


if the area of a hexagone is $ 486\sqrt{3}$, find the area of a circle circumscribed about the hexagon.

[hide="answer choices availble"] 81pi, 243pi, 324pi, 432pi[/hide]

[hide="Solution"]
The side of the hexagon can be found:

$ 486\sqrt {3} \equal{} \frac{3 \sqrt{3}}{2}t^2$
$ 972\sqrt {3} \equal{} 3 \sqrt{3}t^2$
$ 324 \equal{} t^2$

Now, since it's regular, the side length of the triangle is also the radius of the circumscribing circle, so

$ 324\pi \equal{} \boxed {\text{Choice 3}}$

[/hide]

[quote="DaReaper"]what does the $ t$ stand for?[/quote]
The $ t$ stands for a sidelength of the hexagon.

[quote]
Extension: Find the area of the equilateral triangle inscribed in the hexagon.
[/quote]
Assuming the hexagon is regular,

The triangles area is half that of the hexagons. Thus:

[hide]$ \frac {486\sqrt {3}}2$  $ \rightarrow$ $ \boxed{243\sqrt {3}}$[/hide]


We'll conside the new problem (but it's accidental a problem in a contest of Vietnam)

"O is the circumcentre of triangle ABC. Choose P, different from B and C on the line connecting BC.
The circumcircle of triangle ABC meets AP at N and the circle whose diameter is AP at E (N,E are both different from A).
BC and AE intersect at M. Prove that MN always passes through a fixed point!"

I think it's a nice problem and i need its proof very much!
Please help me to solve it. Anyone can help me by sending me a proof in a private message!
Thank you very much!
[b]ricardokaka![/b]

I knew that it's an old problem and there was a solution for it in Mathlinks.ro, but i can't search it! Please resolve it, or send me the link of the source problem! Thanks!

[quote="ricardokaka"]We'll conside the new problem (but it's accidental a problem in a contest of Vietnam)[/quote]

you should have been more precise: in a RUNNING contest in vietnam.

mods, please lock the topic (the problem has been posted some 4 times in other topics as well).

  darij


Does there exist a natural number $ n$ for which the number $ \left(\frac{2008}{n}\right)^3\plus{}\frac{2008}{n}$ is a square of a natural number?

The answer is no.
It's easy to check that $ a\equal{}\frac{2008}{n}$ is a natural number.
So $ a^3\plus{}a\equal{}a(a^2\plus{}1)$ is a square.$ a,a^2\plus{}1$ are squares as $ gcd(a,a^2\plus{}1)\equal{}1$
As $ 2008\equal{}2^3*251$, only possible values of $ a$ are $ 1$ and $ 4$, which are impossible too.

[quote="BG Yoda"]The real problem is to prove that $ \left(\frac {2008}{n}\right)^3 \plus{} \frac {2008}{n}$ can't be square of a rational number.
However, the proof is still elementary.[/quote]

For $ a \equal{} \frac {2008}{n} \equal{} \frac {q}{p}$ rational, where $ gcd(p,q)\equal{}1$
Then $ a^3 \plus{} a \equal{} \frac {q^3 \plus{} qp^2}{p^3}$, $ gcd(p^3,q^3 \plus{} qp^2) \equal{} 1$, so $ p \equal{} k^2$ is a square.$ q(q^2 \plus{} p^2)$ is a square too.But $ gcd(q,q^2 \plus{} p^2) \equal{} 1$, so $ q$ is a square.As $ q|2008$, $ q$ can only be $ 1$ or $ 4$

Case 1:When $ q \equal{} 1$, $ 1^2 \plus{} p^2$  is a square too, which is impossible.
Case 2:When $ q \equal{} 4$, $ 4^2 \plus{} p^2 \equal{} 16 \plus{} k^4 \equal{} x^2$  is a square ,Then $ (x \minus{} k^2)(x \plus{} k^2) \equal{} 16$ for some $ x,k$, which is impossible too.

[quote="BG Yoda"]The real problem is to prove that $ \left(\frac {2008}{n}\right)^3 \plus{} \frac {2008}{n}$ can't be square of a rational number.
However, the proof is still elementary.[/quote]
Thanks! I was also thinking that the problem should not be that straight forward.


Currently, I'm working through [u]C Programming: A Modern Approach[/u] by K. N. King.  I've gotten to Chapter 5, and am currently working on the Programming Projects.

Specifically, here is a paraphrased version of the problem that I need help on (you do [b]not[/b] need to own the book in order to help me; you just need to know C):
[quote]7. Write a program that finds the largest and smallest of four integers entered by the user and displays them.  Use only four 'if' statements.[/quote]

so this is pretty easy with, say, 6 if's, but we need to get it in just 4.

[hide="hint"]
assuming the four integers are a, b, c, and d...

your first 2 if statements are:

[code]
int w,x,y,z;

if(a < b)
{
    w = a;
    x = b;
} else {
    w = b;
    x = a;
}

if(c < d)
{
    y = c;
    z = d;
} else {
    y = d;
    z = c;
}

[/code]

[/hide]

Thank you A LOT for that hint, tjhance.  I've finally written the program now, and it works.

I was thinking what you were thinking, except that I was unable to form my thoughts into something concrete (i.e., to consider adding four new variables w, x, y, z to the program).

However, I also have a question.  Is there any way to write a shorter program?  Are the extra variables really necessary if we want the shortest program possible?  In other words, will more 'if' statements simplify the program?

[quote="MathAndKnowledge"]However, I also have a question.  Is there any way to write a shorter program?  Are the extra variables really necessary if we want the shortest program possible?  In other words, will more 'if' statements simplify the program?[/quote]

Well, we could just do

[code]
int s;

if(a > b)
{
    s = a;
    a = b;
    b = s;
}

if(c > d)
{
    s = c;
    c = d;
    d = s;
}
[/code]

Basically if a>b then you just switch a and b, and same with c and d. Of course, this still uses another variable s to help with switching. If you really want to eliminate the extra variable, you could use this fancier swap algorithm with the xor operator

[code]
a = a^b;
b = a^b;
a = a^b;
[/code]

I am usually content with the first swap algorithm, though.

Of course, maybe there is another solution that involves no swapping at all, and just uses clever if statements.


Suppose there are 18 lighthouses in the Mexican Gulf.  Each lighthouse lights an angle of 20 degrees.  Prove that we can choose the directions of the lighthouses such that the whole gulf is lit.

Anyone please solve this problem.It's surprisingly hard and strange for me.I saw it on MOSP 2006 but without any solution.:(


calculus
function
derivative
calculus computations

I have the following function and i need to find f '(x). I have the following workings. can anyone say if this is correct or not and be able to show some clear workings to solve the question.
[b]x=√(e^y)[/b]
x=(e^(y))^0.5
x=e^(0.5y)

[b]f '(y)=e^(0.5y) . 0.5[/b]

Your notation makes no sense as is.  What you've computed is $ \frac {dx}{dy}$, but the notation $ f'(x)$ means $ \frac {df}{dx}$, and you haven't specified whether $ f$ is supposed to be $ y$ or some other function entirely.  Alternately, you're trying to differentiate $ x$ with respect to itself, which is absurd.

Are you trying to compute the derivative of the inverse function of $ y \equal{} \sqrt {e^x}$?


function
trigonometry
advanced fields
advanced fields open

It has been asked quite a few times whether the equation $ x\sin x\equal{}y$ can be explicitly solved for $ x$. The answer is, of course, "No" but I don't have the slightest idea how to [b]prove[/b] that the inverse function is not elementary (i.e., cannot be obtained as some crazy combination of roots, exponents, and logarithms). Can anybody do it?

Your function reminds me of Lambert's W function, defined by $ W e^W \equal{} x$.  It is known that W is not elementary.  See for example
[url]http://www.apmaths.uwo.ca/~djeffrey/Offprints/ITSF2006.pdf[/url].

Perhaps a similar argument will show that your function is not elementary.

Thanks! This is a good idea (though the differential equation here is not that simple :( ).


[quote="MrM"]Many students considering MathPath also consider AwesomeMath, because it is one of the few other (inter)national all-math residential summer programs that takes middle-school-age students.  (MathPath is really the only such camp that takes [i]only[/i] middle-school-age students.) If you are trying to decide between these programs, I urge you to read the subforum of the Awesome Math forum titled "AwesomeMath vs MathPath".

Mr M
MathPath Academic Director[/quote]

Just wanted to add the link :) :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261914


OK...so as a second year college student I should know this but I'm drawing a total blank.  How do you simplify sqrt(20) into 2sqrt(5)?  What is the method for that?  Here's another, sqrt(40) = 2sqrt(10)!

Help! :idea:

[b]Definition:[/b]  Given a non-negative real $a$, we say that $\sqrt{a}= r$ iff $r \ge 0, r^{2}= a$.

[b]Lemma:[/b]  Given two non-negative reals $ab$, we have $\sqrt{ab}= \sqrt{a}\sqrt{b}$.

Proof:  Let $\sqrt{a}= p, \sqrt{b}= q$.  Then

$p^{2}= a, q^{2}= b \implies$
$p^{2}q^{2}= (pq)^{2}= ab \implies$
$\sqrt{ab}= pq = \sqrt{a}\sqrt{b}$

Hence $\sqrt{20}= \sqrt{4 \cdot 5}= \sqrt{4}\sqrt{5}= 2 \sqrt{5}$, and

$\sqrt{40}= \sqrt{4 \cdot 10}= \sqrt{4}\cdot \sqrt{10}= 2 \sqrt{10}$


search
number theory unsolved
number theory

¿Does there exist an infinite set S of natural numbers such that if a,b are in S  then $a^{2}-ab+b^{2}$ divides $(ab)^{2}$?

Search for IMO shorlist  :wink: If I am not wrong then it from 1998.


Is it possible to make an archive of all the MathCounts problems posted here?Some reasons:
1)People could easily look at problems and find new ways of solving them.
2)Problems wouldn't be posted more than once.

that would take like FOREVER.

if i have any free time tonight (not likely) i will

I did the first 2 pages (most recent)
It took me a while and its 7 pages long

[quote="easyas3.14159..."]Maybe (if anyone wants to contribute) give everyone a few pages to work on? There's only like 41 pages on here.[/quote]

Yea its [i]ONLY[/i] 41 pages. You  make it sound as if thats not much. OMG thats a lot. Look at how many pages BIFFANDDOC came up with in just the first 2

Thanks biffandoc! I'll work on some now too. Why don't you do a couple of pages dogseatcheese? :roll:

If we get 8 people to do 5 pages each, it shouldn't be that hard..
I'll help, once we have enough people willing XD

But all/most problems in this forum are from MATHCOUNTS, so the job would be to try to find those that aren't from the competitions.

I believe tz is trying to make a site with all the competitions.


I don't know, but I'd really to learn more about mods.  Seeing as though I know nothing about it, like a really basic one, that goes into the basics and maybe goes into some harder stuff like Fermat's Little Theorem (Um, IS that advanced or intermediate? :-P)

haha, I'm getting/doing both... but I can't for the fall class for the MATHCOUNTS/AMC Number Theory...:-D


Let $a,b,c$ be real numbers with $a > 0$ such that the quadratic polynomial

$T(x) = ax^2+bcx+b^3+c^3-4abc$

has nonreal zeros.

Prove that exactly one of the polynomials $T_1(x) = ax^2+bx+c$ and $T_2(x) = ax^2+cx+b$ has only positive values.

The discriminant is $d_0 = b^2 c^2 - 4a(b^3 + c^3 - 4abc) < 0$.  The discriminants of the other two quadratics are $d_1 = b^2 - 4ac$ and $d_2 = c^2 - 4ab$, respectively.

Now, $d_1 \cdot d_2 = b^2 c^2 - 4ac^3 - 4bc^3 + 16a^2 bc = d_0 < 0$, hence one or the other must be negative.  The quadratic that has a negative discriminant has no real roots, and since $a > 0$ this quadratic must only take positive values.  Q.E.D.

(Lot easier than I thought it would be.  I could've sworn the solution would've required using the symmetry in $T(a)$.)


Find all pairs $ (a,b)$ of positive real number such that:$ [a[bn]]\equal{}n\minus{}1$ for all positive integer n.

[quote="math10"]Find all pairs $ (a,b)$ of positive real number such that:$ [a[bn]] \equal{} n \minus{} 1$ for all positive integer n.[/quote]

1) $ ab\equal{}1$ and $ a,b>0$
We have $ bn\minus{}1<[bn]\leq bn$, and so $ abn\minus{}a<a[bn]\leq abn$ and so $ abn\minus{}a\minus{}1<[a[bn]]\leq abn$ and so $ abn\minus{}a\minus{}1<n\minus{}1\leq abn$
So $ \minus{}1\leq(ab\minus{}1)n<a$ And so $ ab\equal{}1$ and $ a>0$
Q.E.D.

2) $ a,b\notin\mathbb Q$
If $ b\equal{}\frac pq$, then $ a\equal{}\frac qp$, then $ [a[bq]]\equal{}q\neq q\minus{}1$
Q.E.D.

3) $ a<1$
If $ a>1$, then $ \frac 1a<1$ and, since $ b\notin\mathbb Q$, it is always possible to find $ n$ such that $ 1>\{bn\}>\frac 1a$
Then $ [bn]<bn\minus{}\frac 1a$ and so $ a[bn]<abn\minus{}1\equal{}n\minus{}1$ and so $ [a[bn]]<n\minus{}1$
So $ a<1$
Q.E.D.

4) these necessary conditions are enough
Let $ a,b\notin \mathbb Q$ such that $ 1>a>0$ and $ ab\equal{}1$
Since $ b\notin \mathbb Q$, $ [bn]<bn$ and so $ a[bn]<abn\equal{}n$ and so $ [a[bn]]<n$
We also have $ [bn]>bn\minus{}1$, so $ a[bn]>abn\minus{}a>n\minus{}1$ (since $ a<1$) and so $ [a[bn]]\geq n\minus{}1$
So $ n>[a[bn]]\geq n\minus{}1$
So $ [a[bn]]\equal{}n\minus{}1$
Q.E.D

5) synthesis of solutions:
Take any $ (a,b)$ such that
$ a,b\notin \mathbb Q$
$ 1>a>0$
$ ab\equal{}1$

[quote="pco"][quote="math10"]Find all pairs $ (a,b)$ of positive real number such that:$ [a[bn]] \equal{} n \minus{} 1$ for all positive integer n.[/quote]

1) $ ab \equal{} 1$ and $ a,b > 0$
We have $ bn \minus{} 1 < [bn]\leq bn$, and so $ abn \minus{} a < a[bn]\leq abn$ and so $ abn \minus{} a \minus{} 1 < [a[bn]]\leq abn$ and so $ abn \minus{} a \minus{} 1 < n \minus{} 1\leq abn$
So $ \minus{} 1\leq(ab \minus{} 1)n < a$ And so $ ab \equal{} 1$ and $ a > 0$
Q.E.D.

2) $ a,b\notin\mathbb Q$
If $ b \equal{} \frac pq$, then $ a \equal{} \frac qp$, then $ [a[bq]] \equal{} q\neq q \minus{} 1$
Q.E.D.

3) $ a < 1$
If $ a > 1$, then $ \frac 1a < 1$ and, since $ b\notin\mathbb Q$, it is always possible to find $ n$ such that $ 1 > \{bn\} > \frac 1a$
Then $ [bn] < bn \minus{} \frac 1a$ and so $ a[bn] < abn \minus{} 1 \equal{} n \minus{} 1$ and so $ [a[bn]] < n \minus{} 1$
So $ a < 1$
Q.E.D.

4) these necessary conditions are enough
Let $ a,b\notin \mathbb Q$ such that $ 1 > a > 0$ and $ ab \equal{} 1$
Since $ b\notin \mathbb Q$, $ [bn] < bn$ and so $ a[bn] < abn \equal{} n$ and so $ [a[bn]] < n$
We also have $ [bn] > bn \minus{} 1$, so $ a[bn] > abn \minus{} a > n \minus{} 1$ (since $ a < 1$) and so $ [a[bn]]\geq n \minus{} 1$
So $ n > [a[bn]]\geq n \minus{} 1$
So $ [a[bn]] \equal{} n \minus{} 1$
Q.E.D

5) synthesis of solutions:
Take any $ (a,b)$ such that
$ a,b\notin \mathbb Q$
$ 1 > a > 0$
$ ab \equal{} 1$[/quote]
Nice solution :D .
I think this is a natural solution.you can specify reasons why you have such a solution as :?: .


invariant
combinatorics proposed
combinatorics

Here is a nice problem:

 There is a truck driver with a truck that can transport 1000 apples at a time. This driver has the habit of eating 1 apple per each km while he is driving. His/her task is to transport 10000 (10^4) apples from one city to another. The distance between the cities is 500km.

 He/she may use any strategy of transporting apples. He/she may have less than full loads, may unload/reload at any place along the way from the first city to the second.

 (So he/she may take a full load for 20km and then unload the remaining 980 apples and then go back for another 40, which will be 20 by the time the driver gets back to the unload point; then he may load the 1000 apples back to the truck and continue.)

 The question is: what is the greatest number of apples that will survive the trip?

 One trivial solution is to transport full loads from one city to the other, in which case 5000 apples will survive.

 My task for you is to prove that the greatest possible number of apples that survive is 5868.
 (What strategy produces this?)

Sincerely,
Andrei Jorza

Anybody has idea for this nice problem??I like invariant and semi-invariant very much.But my English is not enough to understand this!!