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Here's #1:

$\frac{n^{k}}{\binom{n}{k}}=\frac{n^{k}\cdot k!}{n(n-1)(n-2)\cdots(n-k+1)}$

$=\frac{k!}{1\cdot\left(1-\frac1n\right)\left(1-\frac2n\right) \cdots\left(1-\frac{k-1}n\right)}$

For fixed $k,$ let $n\to\infty$ in that expression. The denominator tends to $1,$ and hence the overall limit is $k!.$

I suppose on #4, if I say that the answer is obviously $1,$ you'll want more explanation?

Yes please, i don't see how to get it. I have tried $\sin 2x=2\sin x \cos x$ and hope that will turn something related to $\lim_{x \to 0}\frac{e^{x}-1}{x}=1$ but didn't got anything. Of course, thank you, i was really excited to see the solution of #1. I got the first step, but factoring $n$ in the denominator was out of my sight. Sorry for the english.

For small $u,\ \sin u=u+o(u)$ and for small  $u,\ e^{u}=1+u+o(u).$

Hence $\frac{e^{\sin2x}-e^{\sin x}}x=\frac{e^{2x+o(x)}+e^{x+o(x)}}x$

$=\frac{(1+2x+o(x))-(1+x+o(x))}x=\frac{x+o(x)}x=1+o(1).$

The limit is $1.$

Once you get used to that - to seeing $\sin 2x$ as $x\to 0$ and thinking "that might as well be $2x$" - you can actually do problems like this in your head.

#2 is more delicate than #4; there's too much going on there for me to do it in my head.

$12^{x}=12^{12}\cdot 12^{x-12}$ and

$x^{12}=12^{12}\cdot\left(\frac x{12}\right)^{12}$

Thus the numerator can be written as

$12^{12}\left[12^{x-12}-\left(\frac x{12}\right)^{12}\right].$

Continuing ...

$12^{x-12}=e^{(x-12)\ln(12)}=1+(x-12)\ln(12)+o(x-12).$

and

$\left(\frac x{12}\right)^{12}=\left(1+\frac{x-12}{12}\right)^{12}=1+12\cdot\frac{x-12}{12}+o(x-12)$

$=1+(x-12)+o(x-12)$

That leaves the function we're trying to take the limt of as

$\frac{12^{12}[(1+(x-12)\ln(12)+o(x-12))-(1+(x-12)+o(x-12)]}{x-12}$

$=\frac{12^{12}[(x-12)(\ln(12)-1)+o(x-12)]}{x-12}= 12^{12}[(\ln(12)-1)+o(1)]$

The limit is $12^{12}(\ln(12)-1).$

---

OK, I've probably overdone the "avoid L'Hôpital's rule" thing with this one; in this particular case, L'Hôpital is easier.

When trying to compute $\lim_{x \to x_{0}}\frac{f(x)-g(x)}{h(x)}$, where $\lim_{x \to x_{0}}f(x) = \lim_{x \to x_{0}}g(x) = a \in \mathbb R, \, \lim_{x \to x_{0}}h(x) = 0$, you can try to use the following trick: write the limit as $\lim_{x \to x_{0}}\left( \frac{f(x)-a}{h(x)}-\frac{g(x)-a}{h(x)}\right)$ and then hope that both fractions have a limit.

[quote="math_music"]
2)$\lim_{x \to 12}\frac{12^{x}-x^{12}}{x-12}$[/quote]
Make a change of variable $y = x-12$, so that the limit becomes
\begin{eqnarray*}\lim_{y \to 0}\frac{12^{y+12}-(y+12)^{12}}{y}&=& \lim_{y \to 0}\left[ 12^{12}\cdot \frac{12^{y}-\left( \frac{y}{12}+1 \right)^{12}}{y}\right] \\ \ &=& \lim_{y \to 0}\left[ 12^{12}\cdot \left( \frac{12^{y}-1}{y}-\frac{\left( \frac{y}{12}+1 \right)^{12}-1}{y}\right) \right] \\ \ &=& \lim_{y \to 0}\left[ 12^{12}\cdot \left( \frac{12^{y}-1}{y}-\frac1{12}\cdot \frac{\left( \frac{y}{12}+1 \right)^{12}-1}{\frac{y}{12}}\right) \right] .\end{eqnarray*}
Knowing that $\lim_{t \to 0}\frac{a^{t}-1}{t}= \ln a, \, \lim_{t \to 0}\frac{(1+t)^{a}-1}{t}= a$, the sought limit is $ 12^{12}\cdot \left( \ln 12-\frac1{12}\cdot 12 \right) = 12^{12}\cdot \left( \ln 12-1 \right)$.

[quote="math_music"]
4)$\lim_{x \to 0}\frac{\text{e}^{\sin 2x}-\text{e}^{\sin x}}{x}$[/quote]
We write the limit as $\lim_{x \to 0}\frac{\left( e^{\sin 2x}-1 \right)+\left( 1-e^{\sin x}\right)}{x}= \lim_{x \to 0}\left( \frac{e^{\sin 2x}-1}{\sin 2x}\cdot \frac{\sin 2x}{2x}\cdot 2+\frac{1-e^{\sin x}}{\sin x}\cdot \frac{\sin x}{x}\right)$, which, by using the "basic" limits, equals $2-1 = 1$.

PS: For (2), L'Hospital is indeed the fastest method.

Thanks perfect_radio,  very good trick. I never had saw it in that fashion. It had open my mind to solve other limits where i had no idea to start. By the way, te teacher doesn't allowed L'hospital rule in these limits (we don't had derivatives yet)

$\boxed{\ 0<a\ne 1\Longrightarrow L=\lim_{x \to a}\frac{a^{x}-x^{a}}{x-a}=a^{a}(\ln a-1)\ }\ .$

[b]Proof.[/b] $L=\lim_{x\to a}\frac{a^{x}-x^{a}}{x-a}=$ $\lim_{x\to a}x^{a}\cdot \frac{\frac{a^{x}}{x^{a}}-1}{x-a}=$ $a^{a}\cdot\lim_{x\to a}\frac{e^{(x\ln a-a\ln x)}-1}{(x\ln a-a\ln x)}\cdot \frac{x\ln a-a\ln x}{x-a}=$

$a^{a}\cdot 1\cdot\lim_{x\to a}\left(\ln a-a\cdot\frac{\ln x-\ln a}{x-a}\right)=a^{a}\cdot\lim_{x\to a}\frac{\ln\left(\frac{x}{a}\right)}{\left(\frac{x}{a}\right)-1}=a^{a}\cdot (\ln a-1)\ .$

[b]Remark.[/b] You can prove similarly that $\lim_{n\to\infty}\left(\sqrt [n+1]{(n+1)!}-\sqrt [n]{n1}\right)=\frac{1}{e}$ ([b][u]Lalescu's sequence[/u][/b]).

Here's a nice PDF about the so-called "Lalescu Sequence":
http://pefmath2.etf.bg.ac.yu/files/118/868.pdf

[quote="Virgil Nicula"][color=darkred][b]Remark.[/b] We can prove similarly that $L\equiv\lim_{n\to\infty}L_{n}=\frac{1}{e}\ ,$ where $L_{n}=(\sqrt [n+1]{(n+1)!}-\sqrt [n]{n1})$ ([b][u]Lalescu's sequence[/u][/b]).[/color][/quote]
$\{\begin{array}{c}a_{n}=\sqrt [n]{n!}\rightarrow\infty\\\\ b_{n}=\frac{a_{n}}{n}=\frac{\sqrt [n]{n!}}{n}\rightarrow \frac{1}{e}\\\\ c_{n}=\frac{a_{n+1}}{a_{n}}=\frac{\sqrt [n+1]{(n+1)!}}{\sqrt [n]{n!}}\rightarrow 1\\\\ \lim_{x\to 0}\frac{e^{x}-1}{x}=1\end{array}$ $\Longrightarrow$ \begin{eqnarray*}L_{n}& = & a_{n+1}-a_{n}\\\\ & = & a_{n}(c_{n}-1)\\\\ & = & nb_{n}(c_{n}-1)\\\\ & = & b_{n}\cdot\frac{e^{\ln c_{n}}-1}{\ln c_{n}}\cdot n\ln c_{n}\end{eqnarray*} $\Longrightarrow$ \begin{eqnarray*}L & = & \lim_{n\to\infty }b_{n}\cdot\frac{e^{\ln c_{n}}-1}{\ln c_{n}}\cdot n\ln c_{n}\\\\ & = & \frac{1}{e}\cdot 1\cdot\lim_{n\to\infty}n\ln c_{n}\\\ & = & \frac{1}{e}\cdot\lim_{n\to\infty}n\cdot[\frac{\ln (n+1)!}{n+1}-\frac{\ln n!}{n}]\\\ & = & \frac{1}{e}\cdot\lim_{n\to\infty}\frac{n}{n+1}\cdot\frac{n\ln (n+1)!-(n+1)\ln n!}{n}\\\ & = & \frac{1}{e}\cdot\lim_{n\to\infty}\frac{n\ln (n+1)-\ln n!}{n}\end{eqnarray*}

[b]Stolz's lemma for[/b] $\{\begin{array}{c}u_{n}=n\ln (n+1)-\ln n!\\\ v_{n}=n\nearrow \infty\end{array}$ $\Longrightarrow$ $\lim_{n\to\infty}\frac{u_{n+1}-u_{n}}{v_{n+1}-v_{n}}=\lim_{n\to\infty}[(n+1)\ln (n+2)-\ln (n+1)!-n\ln (n+1)+\ln n!\ ]=$

$\lim_{n\to\infty}(n+1)\cdot\ln\frac{n+2}{n+1}=\lim_{n\to\infty}\ln(1+\frac{1}{n+1})^{n+1}=1$ $\implies$ $\lim_{n\to\infty}\frac{u_{n}}{v_{n}}=1$ $\implies$ $\lim_{n\to\infty}\frac{n\ln (n+1)-\ln n!}{n}=1$ $\implies$ $\boxed{\ L=\frac{1}{e}\ }\ .$

Good generalisation Virgil Nicula, never saw that trick with $e^{x}$.  Lalescu's sequence was new to me too, thanks. The paper is really nice perfect_radio


do exists [i]n[/i] in set[i]N[/i] >1 such that:
[i]2007^n[/i]-[i]2006^n[/i] are divisble [i]n[/i]

I think this problem is n|(2007^n-2006^n)

2007^n-2006^n=(2006+1)^n-2006^n=1+n.A ,with A is in set N>1

So the result is only n=1

[quote="haitran1989"]I think this problem is n|(2007^n-2006^n)

2007^n-2006^n=(2006+1)^n-2006^n=1+n.A ,[/quote]
It is not true if n is not prime, for example $2007^{10}-2006^{10}=9\not=1(mod 10)$.
If p is prime divisor of n and (n,2006*2007)=1, then define $a(p)=2007/2006(mod p)\not =0(mod(p)$ and c(p): as minimal k,suth that $a(p)^{k}=1(mod \ p)$. Then we can define a(n) and c(n) for all (n,2006*2007)=1. It is easy to chek, that if $(n,2006*2007)\not =1$, then is not solution. It give that your condition, equavalent to $c(n)|n$.
Let p is minimal prime divisor of n then c(p)|n give c(p)=1, give contradition. Therefore solution is only n=1.


A regular chocolate bar weighs 7 ounces. A jumbo chocolate bar weighs 5 pounds.
How many regular chocolate bars would you have to eat in order to consume the same
total weight as a jumbo chocolate bar? Express your answer as a decimal to the
nearest tenth.

5 pounds is equal to 80 ounces

you would have to eat a total of 80/7 regular bars to eat the same amount of chocolate as a jumbo bar

we divide and we get our answer at 11.4
(i didn't use a calc or paper, so i'm not sure if i divided correctly)

Yes, you are correct.

Also keep in mind that there are 16 ounces in a pound, which is how vallon got 5 pounds is equal to 80 ounces. ($ 5 \times 16$)


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[quote="t0rajir0u"]Have you looked at the solution posted on AoPS for it?  The intended (that is, short) solution requires considerably more creativity than the solution to this problem  :huh:[/quote]

And the AoPS solution is too difficult for MathCounts. The obvious solution isn't, however.

Now please post a solution to this problem if you want to keep discussing this issue.

[quote="junggi"][quote="Ignite168"]Orihime Inoue spins her welsh onion once in 4 frames and the accompaning song last for 25 seconds. A video of this is played at 30 frames per second until the song cycles 8 times. The number of times the welsh onion is spunn the video is denoted $x$. What is $2006-x$?

I believe this is as difficult as the #11 on the 2006 AIME II.[/quote]
Orihime Inoue...rings a bell :ninja: ....Bleach???

anyways, how is this at the same difficulty as a #11 AIME(didn't look at the 2006 AIME II #11 tho)?
[hide]$25\cdot8=200$seconds for the thing to end.
$200\cdot30=6000$frames in the thing
$6000/4=1500=x$spins total.
$2006-1500=506$  :huh:  :huh:[/hide]

edit:I just looked at 2006 AIME II #11
1. It's nothing at all like this problem
2. The solution is not too difficult for mathcounts(meaning easy to understand)
3. As t0rajir0u said, that solution(for the AIME problem) does require a lot more creativity than this one...[/quote]

Correct!

Not talking about AoPS solution, there is a simple solution for #11 through subtraction... a LOT of it

This is considerably easier than any AIME problem, except for perhaps #1 on AIME I this year, but even that may have been more difficult than this.
Unless of course I'm wrong...
[hide]$(30)(25)=750\\ \\ (750)(8)=6000\\ \\ \frac{6000}{4}=1500\\ \\ 2006-1500=506\\ \\ \boxed{506}$[/hide]

This problem is much easier than any AIME problem I've ever seen...
All it is, is multiplication and subtraction...

[quote="SplashD"][quote]Not talking about AoPS solution, there is a simple solution for #11 through subtraction... a LOT of it
[/quote]

would you care to share with us that solution?[/quote]

If you can't figure out the solution I have no idea how you could make it to taking the AIME. The solution is obvious.

[quote="SplashD"]why don't you post your solution and i'll decide if it's obvious or not[/quote]

Because 1. It's extremely long, and 2. It's off topic.

I don't even think you read the problem if you can't see the obvious solution. The number in a sequence is the sum of the 3 numbers before it. You are given 3 numbers, what do you think you need to do to figure out the number before in a sequence? Maybe.. SUBTRACT?

uuhhhhh
you're saying that you would actually do subtraction with those FREAKISHLY huge numbers to find all $a_{i}$'s until i=28?
without a calculator?

[quote="Ignite168"]It's extremely long[/quote]

It's not a solution you should spend time on on the real thing.  You can brute-force a lot of problems that way (see AIME I 2006 #14, which I [b]almost[/b] solved with Pythagorean Theorem alone), but that doesn't make it a good solution at all.  Besides taking more time, long solutions are also easier to mess up.

You could brute-force a lot of AIME problems, but the fact that the brute-force solutions are extremely long tells you you shouldn't be solving that way ;)

[quote="Ignite168"]Yes. It's just subtraction.

I just did it for the last 3 numbers however as that was all that was necessary. I added 1000 whenever the number got into the negatives.[/quote]
the last 3 numbers are still freakishly huge, imo
and I would NOT want to waste my time doing those calculations on an AIME...
And what if you screw up doing the subtractions?A bunch of 3digit number subtractions usually ends in a screw-up..

This was hard because I didn't get the frame thing so I was forced to look at someone else's solution. Well at least I now know what to expect in Mathcounts when they ask Leekspin questions...

[quote="Ignite168"][quote="Treething"]err i brute forced #11 on the aime ii and got it correct, it only took around 10 minutes...[/quote]

Me too, expect it took be like 20 minutes.[/quote]
uuhhhh
did you guys get the right answer? :ninja:

[quote="t0rajir0u"]The neat solution takes less than $30$ seconds if you see it quickly enough ;)[/quote]

That's above MathCounts level. We're discussing MathCounts level solutions here and you have 3 hours on the AIME, no need to try a fancy solution and only use 30 seconds.

junggi: Yes, I did get it correct.

[quote="t0rajir0u"]The neat solution takes less than $30$ seconds if you see it quickly enough ;)[/quote]

yeah, i saw it after taking the test, it has something to do with adding only a couple terms but i forgot it =/

Why would it be beyond MathCounts? All it involves is some clever grouping. My solution was a very MathCounts solution: list out some small numbers and notice the pattern: $S_{n}+S_{n+1}=a_{n+3}$ (very easily verified, even though MathCounts people don't like to prove things).

Well some people who do MathCounts are just starting to right proofs or the term is entirely new to them.

Yes but I imagine that many MathCounts-ers know the strategy of "list out terms and recognize pattern".

[quote="Ignite168"][quote="t0rajir0u"]Is this a joke?  AIME II 2006 #11 is considerably harder unless you're already familiar with the corresponding lemma for the Fibonacci sequence.[/quote]

No it isn't, AIME II 2006 is lots of addition and subtraction. This problem requires multiplication (it could be done with repeated addition though)[/quote]

I challenge you to the following EXTREME PROBLEM!!! Find $\frac{1+2*3-4}{5}$!!!. Requires ADDITION, SUBTRACTION, MULTIPLICATION, aaannndd.....DIVISION!!!.

OH HEY I KNOW THAT ONE

actually... yeah... that's pretty much how i solved #11 the quick way... list out terms and recognize pattern...

Yes that's the first thing I got taught when I started these contests..


A circle of radius 10 is centred on the x-axis, on the positive side of it, and is tangent to $y = \frac{4}{3}x$. What's the x-coordinate of the center of that circle?

[hide]
Since the line has slope $\frac 4 3$, a triangle with vertices the origin, the point of tangency, and the center of the circle is a right triangle similar to a 3-4-5 right triangle.  Then
$\frac 4 5 = \frac {10} x$
$x = \frac{25} 2$
[/hide]


What is the greatest possible quotient using two distinct numbers
from the set $ \{ 0.2, 0.5, 0.8, 3.2 \}$?

To maximize the quotient, we want the first number to be as large as possible, and the divisor to be as small as possible. The largest number is $ 3.2$ and the smallest is $ 0.2$, so our answer is $ 3.2 \div 0.2 \equal{} \boxed{16}$. Note that if we wanted to do the opposite - that is, have the smallest quotient possible - we would have as small a number as possible divided by the largest number possible.

ill answer this in a series of steps

[hide="this is the answer"]1.we know the following facts

I.when we want the largest quatioent we want the 1st # to be as large as possible  and the 2nd # to be as small as possible
EX:9/5=1.8

II.we want to find the largest and smallest # in the set.

2.when we find the largest and smallest # we find the #'s 3.2,.2

3.dividing these #'s we get 3.2/.2=16

4.[hide="ANSWER"]16[/hide][/hide]


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We define $ S(n)$ as the sum of the decimal digits of the positive integer $ n$. We say $ n$ is a [i]green[/i] number if $ S(n)$ divides $ n$. For example, $ 370$ is a green number because $ 3\plus{}7\plus{}0\equal{}10$ divides $ 370$.
Does there exist a positive integer $ k$ such that $ k, 2k, 3k, ..., 2008k$ are all green numbers?

I'm not sure I completely understood that...

Anyway I've got 2 questions:

1) Does your solution work for any $ m$ we put instead of $ 2008$?
2) For the particular $ 2008$ case, can anyone find a smaller $ k$? Because there is  :wink:

Solution for any m (m=2008 in first condition).
Let $ k\equal{}10^c\frac{10^{ab}\minus{}1}{10^b\minus{}1}$, were $ b>log_{10}m$ . Then $ S(ik)\equal{}S(i)<a9(1\plus{}[log_{10}m])\equal{}d.$
We can take $ c\equal{}[log_2m]$. Then $ v_p(ik)\ge v_p(S(ik))$ for any $ i\le m, p\equal{}2,3,5$.
If $ p\ge 7$, then $ v_p(S(ik))\le v_p(a)\plus{}[log_p(m)].$ Therefore sufficienlly choose a, suth that $ p^{v_p(a)\plus{}[log_p(m)]}|10^a\minus{}1.$
Let $ a_1\equal{}lcm(\phi (p^{[log_p(m)]}),p\equal{}7,11,...,) p\le m$. Then work $ a\equal{}a_1*\prod_p p^{v_p(a_1)}$.

Why $ S(ik)\equal{}S(k)$? I think $ k$ is not a power of 10.

Anyway, I insist you should look for a much smaller $ k$...

In fact $ 999$ doesn't work because $ 999 \times 1011 \equal{} 1009989$ is not divisible by $ 1 \plus{} 0 \plus{} 0 \plus{} 9 \plus{} 9 \plus{} 8 \plus{} 9 \equal{} 36$.

But you're getting the idea  :wink:

[quote="uglysolutions"]Why $ S(ik) \equal{} S(k)$? I think $ k$ is not a power of 10.

Anyway, I insist you should look for a much smaller $ k$...[/quote]
$ S(ik)\equal{}aS(k)$ (I forget a), because $ k\equal{}10^c(10^{(a\minus{}1)b}\plus{}...\plus{}1)$ and $ S(k)\equal{}a$.


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Determine all $n$ for which the system with of equations can be solved in $\mathbb{R}$:

\[\sum^{n}_{k=1} x_k = 27\]

and

\[\prod^{n}_{k=1} x_k = \left( \frac{3}{2} \right)^{24}.\]

There is a solution for all $n\geq 3$. Proof by induction on n:
 $n=3$
$x_{1}=-1$ and there exist $x_{2},x_{3}$ such that
\[x_{2}+x_{3}=28\]
\[x_{2}x_{3}=-(\frac{3}{2})^{24}\]
(Indeed, the equation $X^{2}-28X-(\frac{3}{2})^{24}$ has two real solutions)
Suppose that a solution $x_{1},\dots,x_{n}$ exists, and that $x_{1}<0$. We can find two reals $x,y$ such that
\[x+y=x_{1}\]
\[xy=x_{1}\]
(again considering the equation of the second order, the discriminant being $>0$)
As the product $xy$ is strictly negative, $x$ or $y$ is strictly negative. Name $x$ the solution $<0$ and $y$ the solution $>0$.
Then the $n+1$-uple $x,y,x_{2},\dots,x_{n}$ is a solution, and the first coordinate is $<0$.
for $n=2$, there is no solution, because the equation $X^{2}-27X+(\frac{3}{2})^{24}$ has no real solution.


$ a^2(b\plus{}c\minus{}a)\plus{}b^2(a\plus{}c\minus{}b)\plus{}c^2(a\plus{}b\minus{}c)\le{3abc}$

a,b,c - triangle side lengths.

$ \Leftrightarrow a(a\minus{}b)(a\minus{}c)\plus{}b(b\minus{}a)(b\minus{}c)\plus{}c(c\minus{}a)(c\minus{}b)\ge 0$

which is Schur

I hope it's not a typo...  :maybe:

hello, your inequality is equivalent with
$ 3abc\minus{}a^2(b\plus{}c\minus{}a)\plus{}b^2(a\plus{}c\minus{}b)\plus{}c^2(a\plus{}b\minus{}c)\geq 0$
setting $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ we get
$ x^2y\plus{}xy^2\plus{}x^2z\plus{}z^2x\plus{}y^2z\plus{}yz^2\geq 6xyz$ which is true by AM-GM.
Sonnhard.

Factor the equation and it becomes

$ (b\plus{}c\minus{}a)(a\plus{}c\minus{}b)(a\plus{}b\minus{}c)\le abc$

Make the sub $ a\equal{}x\plus{}y...$

$ 8xyz \le (x\plus{}y)(y\plus{}z)(z\plus{}x)$

This is true by AM-GM  :)

[quote="enndb0x"][quote="ocha"]

$ 8xyz \le (x \plus{} y)(y \plus{} z)(z \plus{} x)$

This is true by AM-GM  :)[/quote]

How this is true by AM-GM ?[/quote]
I think he means Am-Gm for two variables
$ \frac {a \plus{} b}{2}\geq \sqrt{ab}$  :wink:

well,
applying AM-GM for positive reals,
$ x,y$
$ x \plus{} y \ge 2(x^\frac{1}{2})(y^\frac{1}{2})$
$ y,z$
$ y \plus{} z \ge 2(y^\frac{1}{2})(z^\frac{1}{2})$
$ y,z$
$ z \plus{} x \ge 2(z^\frac{1}{2})(x^\frac{1}{2})$
mulitilying these inequalities we get the required

[quote="Potla"][quote="ocha"]Factor the equation and it becomes

$ (b \plus{} c \minus{} a)(a \plus{} c \minus{} b)(a \plus{} b \minus{} c)\le abc$

Make the sub $ a \equal{} x \plus{} y...$

$ 8xyz \le (x \plus{} y)(y \plus{} z)(z \plus{} x)$

This is true by AM-GM  :)[/quote]

I do not think we need to do the substitution necessarily :roll: , because,
for all real $ a, b, c$ we have:
$ (b \plus{} (c \minus{} a))(b \minus{} (c \minus{} a)) \equal{} b^2 \minus{} (c \minus{} a)^2\le b^2$.
Getting the other two relations and taking the product yields the result Easily.... :rotfl:  :P
PS:[quote="Dimitris X"][quote="Bachukas"]$ a^2(b \plus{} c \minus{} a) \plus{} b^2(a \plus{} c \minus{} b) \plus{} c^2(a \plus{} b \minus{} c)\le{3abc}$

a,b,c - triangle side lengths.[/quote]
Do we need $ a,b,c$ to be sides of a triangle....
I don't think so... :huh:[/quote]
 The "sides of a triangle" is necessary to say the LHS to be POSITIVE, I think..

Your proof is nice, Ocha!! :lol:[/quote]
Take for instance $ a \equal{} 5,b \equal{} 2,c \equal{} 1$.Then the inequality is also true.But $ a,b,c$ are not sides of a triangle.....
Nevertheless the inequality is equivalent to the well known(Schur)
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc\geq a^2(b \plus{} c) \plus{} b^2(a \plus{} c) \plus{} c^2(a \plus{} b)$ ,so i think we don't need to assume that $ a,b,c$, are sides of a triangle :wink:

I think statement "a,b,c - sides of triangle" makes the solution of the problem become easier than usual.

If that is why the condition was given then why did we not use it ?
Perhaps it is unnecessary after all, but I am feeling creepy about it,......

I did not notice that
$ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)\le abc$ holds even if $ a\plus{}b<c$, so I told it in my post before to Dimitris X....
Thank you for your counter example, Dimitris X! :lol:

[b]A stronger inegality[/b] : $ \sum a^2(b\plus{}c\minus{}a)\le \frac 13\cdot\left[8abc\plus{}\prod (b\plus{}c\minus{}a)\right]\le 3abc$ .

[quote="Virgil Nicula"][b]A stronger inegality[/b] : $ \sum a^2(b \plus{} c \minus{} a)\le \frac 13\cdot\left[8abc \plus{} \prod (b \plus{} c \minus{} a)\right]\le 3abc$ .[/quote]
The second pair is easy:
If we put
$ a \plus{} b \minus{} c \equal{} x,a \minus{} b \plus{} c \equal{} y, \minus{} a \plus{} b \plus{} c \equal{} z$ So
$ a \equal{} \frac {x \plus{} y}{2},b \equal{} \frac {x \plus{} z}{2},c \equal{} \frac {y \plus{} z}{2}$
So our inequality can be written
$ (x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} xyz\leq\frac {9}{8}(x \plus{} y)(y \plus{} z)(z \plus{} x)$ which is equivalent to the
$ (x \plus{} y)(y \plus{} z)(z \plus{} x)\geq 8xyz$,which is true....(because $ x \plus{} y\geq2\sqrt {xy}$ etc..)
And the first pair
Using the same substitution we have to prove that
$ \sum (\frac {x \plus{} y}{2})^2 z\leq \frac {1}{3}[(x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} xyz] \rightarrow 4(x^2y \plus{} x^2z \plus{} y^2x \plus{} y^2z \plus{} z^2y \plus{} z^2x \plus{} 3xyz) \geq 3(x^2y \plus{} x^2z \plus{} y^2x \plus{} y^2z \plus{} z^2y \plus{} z^2x \plus{} 6xyz)$ which is
$ x^2y \plus{} x^2z \plus{} y^2x \plus{} y^2z \plus{} z^2y \plus{} z^2x\geq 6xyz$ ,which is true by AM-GM  :)


I am currently writing Mock AMC F, with (pledged) help from Joe (joml88).  I would like some feedback on what you guys would prefer.  I am going to call it a Mock AMC, and not distinguish between 10 and 12.  I think it is a bit harder than a regular AMC 12, but we'll see how it turns out, and then smooth it out.

Anyway, sign-up here if you plan to participate.

[b]UPDATE[/b]: TEST POSTED [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128689]HERE[/url]

Hopefully it will be good.  We're writing more than 25 problems, then refining it: eliminating anything where a calculator would trivialize the problem (a lot of trig), determining if problems are AMC-type, etc.

[u][b]List of Participants[/b][/u]
1) davidlizeng
2) adalton $\Omega$
3) bpms
4) Xaero
5) pianoforte
6) now a ranger
7) knexpert
8) Calculus
9) anirudh
10) robinhe
11) stupidityismygam
12) azndemon43
13) aahmadi
14) DanK
15) zptlimyw
16) white_horse_king88
17) Iversonfan2007
18) nutz_for2.718281828 $\Omega$
19) TheDFK
20) xxreddevilzxx
21) Xantos C. Guin
22) 10000th User
23) rnwang2
24) 1234567890
25) 1337h4x
26) EFuzzy
27) kstan013
28) CornOnTheCobbe
29) arkmastermind
30) life=tennis+math
31) bigboy82892
32) lotrgreengrapes7926

Don't expect this to stay totally updated, but I'll do my best.  As soon as the test is posted, I will put a symbol by your name once I have received your answers.

OK, I have finished writing it.  I will release it as soon as I get some really smart people to look over it.

That or write nutsfor e, that's what most people do.

The Mock AMC F has been posted:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=729827#729827

If you want to sign up, post in THIS thread.

Otherwise, please continue further discussion in the NEW thread.

I'm in
but just saying that isn't long enough for me to be allowed to post it...

[quote="goldendomer"]I'll do it.[/quote]THe results have been posted and every thing...


inequalities
function
search
calculus
inequalities proposed

Let $ a,b,c$ be positive numbers such that; $ a\plus{}b\plus{}c\equal{}1$ and
$ k\equal{}\frac{118098}{(299\plus{}5\sqrt{349})(\sqrt{374\minus{}10\sqrt{349}}\plus{}15)\plus{}4374}\equal{}7.556406208$
Prove that:
$ \frac{a^2\plus{}kbc^2}{b\plus{}c}\plus{}\frac{b^2\plus{}kca^2}{c\plus{}a}\plus{}\frac{c^2\plus{}kab^2}{a\plus{}b} \ge \frac{1}{2}\plus{}\frac{k}{6}$

[quote="nguoivn"][quote="quykhtn-qa1"]Let $ a,b,c$ be positive numbers such that; $ a \plus{} b \plus{} c \equal{} 1$ and
$ k \equal{} \frac {118098}{(299 \plus{} 5\sqrt {349})(\sqrt {374 \minus{} 10\sqrt {349}} \plus{} 15) \plus{} 4374} \equal{} 7.556406208$
Prove that:
$ \frac {a^2 \plus{} kbc^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} kca^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} kab^2}{a \plus{} b} \ge \frac {1}{2} \plus{} \frac {k}{6}$[/quote]
How did you get it ?  :maybe:
@quykhtn-qa1: [hide]Bdt nay` co dung' voi $ k \equal{} 8$ ko nhi?, cai nay` thi` anh chua thu?. Anh moi' chi? co' loi` giai cho $ k \equal{} 6$ thoi  :D  [/hide][/quote]
$ k\equal{}7.556406208$ isn't best,but this problem very difficult if you find the best k :) .I find $ k$ by $ p,q,r$ method.
Thank you very much ,my brother :lol:

$ \frac{a^{2}(a\plus{}b\plus{}c)\plus{}kbc^{2}}{b\plus{}c}\plus{}\frac{b^{2}(a\plus{}b\plus{}c)\plus{}kca^{2}}{c\plus{}a}\plus{}\frac{c^{2}(a\plus{}b\plus{}c)\plus{}kab^{2}}{a\plus{}b}$ $ \minus{} (a\plus{}b\plus{}c)^{2}\left(\frac{1}{2}\plus{}\frac{k}{6}\right)$ is equal to $ f(a,b,c)\plus{}kg(a,b,c)$ for some rational functions $ f,g$.
In order to find the maximal value of $ k$, we only need to find the minimal value of $ \frac{f}{g}$.
First, we search the value of $ \frac{f(0,b,1)}{g(0,b,1)},\frac{f(1,b,1)}{g(1,b,1)},\frac{f(a,0,1)}{g(a,0,1)},\frac{f(a,1,1)}{g(a,1,1)}$.
Also, if $ \frac{f(a,b,1)}{g(a,b,1)}$ has a maximal value where $ 0<a,b<1$, then $ \frac{\partial}{\partial a}\left(\frac{f}{g}\right)\equal{}\frac{\partial}{\partial b}\left(\frac{f}{g}\right)$. Then we have two equations. By solving them, we get the several values of $ a,b$. Then one of the correspond values of $ k$ is the maximum.

[quote="Parkdoosung"]$ \frac {a^{2}(a \plus{} b \plus{} c) \plus{} kbc^{2}}{b \plus{} c} \plus{} \frac {b^{2}(a \plus{} b \plus{} c) \plus{} kca^{2}}{c \plus{} a} \plus{} \frac {c^{2}(a \plus{} b \plus{} c) \plus{} kab^{2}}{a \plus{} b}$ $ \minus{} (a \plus{} b \plus{} c)^{2}\left(\frac {1}{2} \plus{} \frac {k}{6}\right)$ is equal to $ f(a,b,c) \plus{} kg(a,b,c)$ for some rational functions $ f,g$.
In order to find the maximal value of $ k$, we only need to find the minimal value of $ \frac {f}{g}$.
First, we search the value of $ \frac {f(0,b,1)}{g(0,b,1)},\frac {f(1,b,1)}{g(1,b,1)},\frac {f(a,0,1)}{g(a,0,1)},\frac {f(a,1,1)}{g(a,1,1)}$.
Also, if $ \frac {f(a,b,1)}{g(a,b,1)}$ has a maximal value where $ 0 < a,b < 1$, then $ \frac {\partial}{\partial a}\left(\frac {f}{g}\right) \equal{} \frac {\partial}{\partial b}\left(\frac {f}{g}\right)$. Then we have two equations. By solving them, we get the several values of $ a,b$. Then one of the correspond values of $ k$ is the maximum.[/quote]
Thank you very much :).Ofcourse,$ p,q,r$ method can solve it . :lol:

[quote="quykhtn-qa1"]Ofcourse,$ p,q,r$ method can solve it . :lol:[/quote]
The proof of Parkdoosong  is not elementary ( at least for me). It seems to be used some knowledges not for high school students ! :maybe: .
So dear quykhtn-qa1, can you show your proof. I give up with this problem!
Edit: Yes , I know. I don't think you can prove Parkdoosung's constant . But I want to mention yours; I am intersected in using $ pqr$ method to  prove it!

[quote="Allnames"][quote="quykhtn-qa1"]Ofcourse,$ p,q,r$ method can solve it . :lol:[/quote]
The proof of Parkdoosong  is not elementary ( at least for me). It seems to be used some knowledges not for high school students ! :maybe: .
So dear quykhtn-qa1, can you show your proof. I give up with this problem![/quote]
I think you don't understand. I speak my problem can solve by $ p,q,r$ me thod.
With $ k$ of Parkdoosung,I think he used calculator :wink:

Maybe you haven't read my edited post above!  :maybe: . Please give me your proof using $ pqr$ method for your constant.

[quote="nguoivn"]Can you post your proof, Quy ? Thank you  :)[/quote]
Dear nguoivn,now I'am very busy,so I can't post my solution. :(
I will post my solution then some days.Thank you :)

If use a little calculus, we can prove this ineq with $ k\equal{}9$, Quy  :)

[quote="nguoivn"]If use a little calculus, we can prove this ineq with $ k \equal{} 9$, Quy  :)[/quote]
dear nguoivn,can you post full your solutions for it? :)

[quote="nguoivn"]It's here, Quy  :lol:
http://www.mathlinks.ro/viewtopic.php?t=295955[/quote]
Thank you very much,nguoivn.I will try  :)


Do you use the MathLinks Forum or the AoPS skin?

I'm surprised so many use the AoPS one... personally, the MathLinks forum has a cleaner feel to me. I like the color scheme and everything.

[quote="Valentin Vornicu"][quote="math92"]Do you use the Mathlinks Forum or the AoPS forum?[/quote]Improper question. It is the same forum. You may be reffering to the skin. Poll changed.[/quote]

Well, in fact just by changing the skin I feel on a totally different forum. :)

Anyway I use the mathlinks one. No offence, but the AoPS looks like a kid who just got some new paint and wants to give everything a different color. :?
Good thing on the AoPS is the many mathematical animations throughout the site - but which then again slow down the site.

Well the AoPS one is designed to be more "fun" for kids and teens, and to present more important options for those students, while the MathLinks one is designed to be more practical, and give quicker access to important olympiad stuff.

Usually the AoPS skin format is easier, but the colors are annoying.  Like paladin8 said, the clean feel is why I use the MathLinks skin.

I'm too used to the AoPS skin to change.  Plus, I think the MATHLINKS skin is more designed to be practical for Olympiad students while the AoPS skin is best for intermediates and pre-olympiaders.

I use MathLinks skin. It seems to be more organized and neat.

i use mathlinks because its more futuristic

it surprises me everyone replies "mathlinks" while the poll says 50-50... :?

[and since the last 4 people replied mathlinks, the ML counter has only risen by one? oh well I must be sleeping... happens when you're doing algebra for 36 hours in a row.]

I prefer the AoPS skin--I think it's nicer looking. And the blue/yellow contrast is easier on the eyes.

[quote="xxreddevilzxx"]I like to use MathLinks, but it is annoying that not every page on the AoPS-MathLinks page have Mathlinks skin.... e.g. Resources Page, Online School PAge, LaTex Page, etc.[/quote]Damn, I never thought about that before. Well, at some point there will be for those too, I am just to occupied with other stuff (like :w00t: WooT :D) to deal with that right now, but it's been written on the (long) task list :P

woot is a popular leet-like exclamation, usually of happines

and they used that for a letterword for Worldwide Online Olympiad Training, more info on the announcement :)

AOPS all the way. It is much more professional looking I think, and it's way easier on the eyes. I feel like I'm floating on a cloud or something with the other skin, just gets annoying. Grey is like 1000% better for your eyes than is white, I value my 20-20 vision!

I use the MathLinks for the already-mentioned reasons :D

I wish we could make our own skins.  We have some people here interested in graphic design, so it'd be nice if we could write skins and upload them for other people to see.

[quote="Peter VDD"]woot is a popular leet-like exclamation, usually of happines
[/quote]

Alright folks, let's not n00b l33t. The word is w00t. WOOT is a program here. w00t is a rather l33t expression of pwnage. :-D

oo pwnage, such an unfortunate word and i can say that cuz ive played all the online games that came from and it was just a very common typo that became has just become become so overused and abused that i just hope it dies, just like 1337 is now only for ppl who actually dont know where it came from and for some reason think they are cool for using, w00t a classic though and n00b will never die

now anyway, i personally always liked the AoPS skin, the mathlinks one just seems really dull

[quote="Spoon"]w00t w00t, I am 1337 cuz I pwnz0rs j00 and r0x0rs t3h b0x k thx

sigh, cs will always hold a place in my heart.[/quote]Ok, enough chatspeak. Only w00t is allowed :D

I use Mathlinks because I couldn't see some of the 'math' emoticons when I clicked 'View more emoticons'  (black on dark color)  with the AOPS skin.  I liked it so I haven't checked to see if it's still like that.

I use the MathLinks skin. I like the colors a lot better.

[quote="paladin8"][quote="Valentin Vornicu"]A small notice: it's Math[b]L[/b]inks not Math[b]l[/b]inks :) Capital [b]L[/b] that is :D[/quote]

Gee, don't be so (case) sensitive! HAHA... :D[/quote]A good one :P  :lol: But still ...  :blush:  :roll:


Find the ordered quadruple of positive integers (a,b,c,d) such that 2^a + 3^b + 4^c + 5^d = 2006.

[quote="tjhance"][quote="junggi"][hide]take the equation mod 3.
$ 2^{a}+3^{b}+4^{c}+5^{d}=(-1)^{a}+1^{c}+(-1)^{d}= 2006=2\mod 3$
$ (-1)^{a}+(-1)^{d}=1\mod3$
which isn't quite possible.
No solutions.

[/hide][/quote]

Why isn't that possible? If a and d are odd then $ -1-1\equiv-2\equiv 1\mod 3$[/quote]
because i was sleepy


did anyone else get one?

a kid in my class got one cause he took answers off dan :ninja:

im scared

i don't know if i can pay for it anyways

jorian

You don't need to be sent a brochure to apply, why do you care?

[quote="jhredsox"]why do you have to post something negative? :D

i'm not good at geometry, so i couldn't do it anyways

jorian[/quote]
He posted something positive, he said that it didn't make a difference that the other guy got one you were still on even footing.
Next time [b]read[/b] what others say before commenting on it.

umm...i was kidding in the first and third post

-jorian

just kidding about the other person, not the topic.

xxxxxxxx, xxxx'x xxx xxxxxxx ...

jorian

[color=red]Asking for help with problems is to be avoided.
--- Thomas/ MathPath[/color]

[quote="13375P34K43V312"]Then why bother? You know that's spam, right?[/quote]

umm that's spam too. saying the other post was spam.  :rotfl:

guys don't be all oooh that's spam! don't spam! most people do it once in a while. after all, it is a forum to discuss and comment...
it's like there are spam police out here

oh and does anyone know what fun stuff people do at mathpath? other than math...

maybe since leet doesn't get so many pms, he can show you the explanation he gave me

jorian

[quote="jhredsox"]just kidding about the other person, not the topic.

xxxxxxxx, xxxx'x xxx xxxxxxx ...

jorian
[size=75]
[color=red]Asking for help with problems is to be avoided.
--- Thomas/ MathPath[/color][/size][/quote]
DUDE THAT'S CHEATING

if you learned factoring you get it after a while

well, i know the formula, just how to get the exact fraction is my problem

jorian

use what you know i can't help anymore but just

[quote="jhredsox"]well bpms, i don't know every formula, and i don't know how to search for it. i still have to do the problem. fine, i'll go searching.

jorian[/quote]
excuse me, but asking for help on an admission test is as bad as copying someone's answers on Mathcounts.

first off, it's a qualifying quiz.

secondly, well, i'm sorry.

but i figured it out

i just need one last thing which i shall not post

-jorian

it's still a quiz you shouldn't cheat on. They look at it to see your true ability, not searching for answers ability with other people.   :fool:  :oops:


In a 4 x 4 grid, three coins are randomly placed in different squares. Determine the probability that no two coins lie in the same row or column

Maybe...

[hide]
Place a coin in an arbitrary square. This eliminates 6 squares.

Placing the 2nd coin eliminates another 4 squares.

Therefore, the probability is:

$ \frac{3}{5}\cdot\frac{2}{7}\equal{}\frac{6}{35}$[/hide]

[quote="Carbon57"]Maybe...

[hide]
Place a coin in an arbitrary square. This eliminates 6 squares.

Placing the 2nd coin eliminates another 4 squares.

Therefore, the probability is:

$ \frac {3}{5}\cdot\frac {2}{7} \equal{} \frac {6}{35}$[/hide][/quote]

[hide="Loong method"]

Yours surely does look like a neat way to do it , but i didn't quite get it.

My method

(( 16c1 * 9c1 * 4c1 ) / 6 )/16c3 ==> 6/35

[/hide]

Yeah, carbon's would have been my preferred method.

The first coin can be placed anywhere. (P=1)

The first coin will eliminate 7 squares no matter where it is placed so the are 9 places to put the 2nd coin (out of a possible 15. Remember, one space has already been taken. P=9/15=3/5)

Following the same technique for the third coin. (P=4/14=2/7)

Multiplying the probabilities gives us 6/35.

It is interesting to note that every time a coin is placed, the number of available squares decreases to the next  highest square number. (16,9,4 in this example.)

If you have any questions, I'd be more than happy to answer them.

Can someone please explain carbon's method more clearly?
and shouldn't this be in the MC or Moems forum?

after putting first , the coin locks 6 squares other than itself, so req. probability of putting second coin is 9/15 or 3/5 .. second coin locks 4 more squares other than itself so, probability of putting third coin is 4/14=2/7
combinin both prob. =3/5*2/7 =6/35

Draw yourself a grid. Make it 4 x 4. You can place the first dot anywhere. Do so. Cross out any spaces in the same row or column. Count the number of places not blocked now. The second coin can be placed in 15 spaces, but only 9 are available. So the probability of placing the second one in an unblocked space is 9/15 = 3/5.

Draw in the second dot in any available space. Cross out all spaces in the same row and column. Count how many places are available now. The third coin can be placed in 14 places, but only 4 are available. The probability of placing the third coin in one of these spots is 4/14 = 2/7.

Multiply the individual probabilities to get 1*(3/5)*(2/7) = 6/35.

i liked the "looong  method" more cause it was more straight forward  :D  :D

It was just the first solution, except reduced to just the combinatorics.


summer program
PROMYS
HCSSiM
number theory

Does anybody have a good experience with a summer program that was more focused on research than math contest preperation.  Since I am currently a junior, I don't really see the point in attending an intensive preperation program.  I am also applying to summer science program and an internship at lawrence labs.

RSI is research-oriented, but the deadline passed...

rsi, hah.  Considered applying but realized there was no way I could get in.

Well I asked rrusczyk the same question about a week ago and he told me to check out MathCamp. Based on their website, it seems pretty good, though I've never been before so I can't vouch for it personally. However, people who have gone that I asked about it generally said that it was a good experience.

[quote="K81o7"]PROMYS is a good choice here as well if you haven't seen much number theory.[/quote]
Even if you have seen number theory it is still a good program for
working on proof skills. They are most decidedly not contest oriented, and you can do research as a first-year, but they are focused more on teaching number theory to first years.

Even though HCSSiM isn't billed as a research-oriented program, some visiting faculty and some recent students have described it as a valuable research or research-like experience.


algebra
polynomial
Vieta
trigonometry
algebra proposed

Find all integers $a,b,m,n$, with $m>n>1$, for which the polynomial $f(X)=X^n+aX+b$ divides the polynomial $g(X)=X^m+aX+b$.

[i]Laurentiu Panaitopol[/i]

Someone can give a solution or at least a hint about this problem. :(

There are 7 solutions (I remember this because when we were grading the problem, it was basically 1 point for each solution found :D) and it's pretty computational, nothing extremely smart about this problem.

[hide]Let $ m \equal{} n \plus{} k$ for some $ k \in \mathbb{N}$.

$ f(x)$ divides $ g(x)$ iff $ f(x)$ divides $ g(x) \minus{} f(x) \equal{} x^m \minus{} x^n \equal{} x^n(x^k \minus{} 1)$.

Since the roots of $ x^n(x^k \minus{} 1)$ are all of magnitude $ 0$ or $ 1$, the roots of $ f(x) \equal{} x^n \plus{} ax \plus{} b$ must also be of magnitude $ 0,1$.

Thus, $ b \equal{} 0$ OR $ b \equal{} \pm 1$.

----------------------------------------------------------------------------------------------------

Case I: $ b \equal{} 0$:

$ x^n \plus{} ax | x^n(x^k \minus{} 1) \Longleftrightarrow x^{n \minus{} 1} \plus{} a | x^{n \minus{} 1}(x^k \minus{} 1)$

Since the roots of $ x^{n \minus{} 1}(x^k \minus{} 1)$ are all of magnitude $ 0$ or $ 1$, the roots of $ x^{n \minus{} 1} \plus{} a$ must also be of magnitude $ 0,1$. Thus, $ a \equal{} 0$ OR $ a \equal{} \pm 1$.

$ (a,b) \equal{} (0,0)$ yields: $ x^n|x^m$ which is true for all integers $ m > n > 1$.
Thus, $ (a,b,m,n) \equal{} (0,0,m,n)$ is a solution for all positive integers $ m > n > 1$.

$ (a,b) \equal{} ( \minus{} 1,0)$ yields: $ x^{n \minus{} 1} \minus{} 1|x^k \minus{} 1$.
$ x^{n \minus{} 1} \minus{} 1|x^k \minus{} 1$ iff the $ (n \minus{} 1)$th roots of unity are a subset of the $ k$th roots of unity.
Therefore, $ n \minus{} 1|m \minus{} n \Leftrightarrow m \minus{} n \equal{} t(n \minus{} 1) \Leftrightarrow m \equal{} (t \plus{} 1)n \minus{} t$ for any $ t \in \mathbb{N}$.
Thus, $ (a,b,m,n) \equal{} ( \minus{} 1,0,(t \plus{} 1)n \minus{} t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$.

$ (a,b) \equal{} (1,0)$ yields: $ x^{n \minus{} 1} \plus{} 1|x^k \minus{} 1$.
The roots of $ x^{n \minus{} 1} \plus{} 1$ are the $ 2(n \minus{} 1)$th roots of unity that are not $ (n \minus{} 1)$th roots of unity.
$ x^{n \minus{} 1} \plus{} 1|x^k \minus{} 1$ iff the $ 2(n \minus{} 1)$th roots of unity are a subset of the $ k$th roots of unity.
Therefore, $ 2(n \minus{} 1)|m \minus{} n \Leftrightarrow m \minus{} n \equal{} 2t(n \minus{} 1) \Leftrightarrow m \equal{} (2t \plus{} 1)n \minus{} 2t$ for any $ t \in \mathbb{N}$.
Thus, $ (a,b,m,n) \equal{} (1,0,(2t \plus{} 1)n \minus{} 2t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$.

----------------------------------------------------------------------------------------------------

Case II: $ b \equal{} \pm1$:

$ x^n \plus{} ax\pm1 | x^n(x^k \minus{} 1) \Longleftrightarrow x^n \plus{} ax\pm1 | x^k \minus{} 1$ (Since $ 0$ is not a root of $ x^n \plus{} ax\pm1$).

Let the roots of $ x^n \plus{} ax \pm 1$ be $ r_i$ for $ i \in \overline{1,n}$.

If $ n > 2$, by Vieta's:
$ \sum r_i \equal{} 0$, $ \prod r_i \equal{} \pm1$, and thus, $ \sum \frac {1}{r_i} \equal{} \sum \frac {\pm r_1 r_2 \cdots r_{n \minus{} 1} r_n}{r_i} \equal{} \pm a$

However, since $ x^n \plus{} ax \pm 1$ has all real coefficients, if $ r_i$ is a root, then $ \overline{r_i}$ must also be a root.

Note that since all the $ r_i$ are $ k$th roots of unity: $ \overline{r_i} \equal{} \frac {1}{r_i}$.

Thus, $ r_i \plus{} \overline{r_i} \equal{} \frac {1}{r_i} \plus{} \frac {1}{\overline{r_i}}$ for all non-real $ k$th roots of unity $ r_i$. The other two possible $ k$th roots of unity are, $ \pm1$ which satisfy $ r_i \equal{} \frac {1}{r_i}$.

Therefore, if $ n > 2$, then $ \sum \frac {1}{r_i} \equal{} \sum r_i \Longleftrightarrow \pm a \equal{} 0 \Rightarrow a \equal{} 0$.

$ (a,b) \equal{} (0, \minus{} 1)$ yields: $ x^n \minus{} 1 | x^k \minus{} 1$.
$ x^n \minus{} 1|x^k \minus{} 1$ iff the $ n$th roots of unity are a subset of the $ k$th roots of unity.
Therefore, $ n|m \minus{} n \Leftrightarrow m \minus{} n \equal{} tn \Leftrightarrow m \equal{} (t \plus{} 1)n$ for any $ t \in \mathbb{N}$.
Thus, $ (a,b,m,n) \equal{} (0, \minus{} 1,(t \plus{} 1)n,n)$ for all positive integers $ n,t$ such that $ n > 1$.

$ (a,b) \equal{} (0,1)$ yields: $ x^n \plus{} 1 | x^k \minus{} 1$.
The roots of $ x^n \plus{} 1$ are the $ 2n$th roots of unity that are not $ n$th roots of unity.
$ x^n \plus{} 1|x^k \minus{} 1$ iff the $ 2n$th roots of unity are a subset of the $ k$th roots of unity.
Therefore, $ 2n|m \minus{} n \Leftrightarrow m \minus{} n \equal{} 2tn \Leftrightarrow m \equal{} (2t \plus{} 1)n$ for any $ t \in \mathbb{N}$.
Thus, $ (a,b,m,n) \equal{} (0,1,(2t \plus{} 1)n,n)$ for all positive integers $ n,t$ such that $ n > 1$.

$ n \equal{} 2$ yields: $ x^2 \plus{} ax\pm1 | x^k \minus{} 1$.

Since the roots of $ x^2 \plus{} ax\pm1$ are $ k$th roots of unity, either the roots are $ \pm 1$ (in which case $ f(x) \equal{} x^2 \minus{} 1$ which has already been covered above with $ (a,b) \equal{} (0, \minus{} 1)$), or in the form $ cis\left(\pm\theta\right)$.

Therefore, $ x^2 \plus{} ax\pm1 \equal{} (x \minus{} cis(\theta))(x \minus{} cis( \minus{} \theta)) \equal{} x^2 \minus{} 2\cos(\theta)x \plus{} 1$.
Thus, $ b \equal{} 1$ and $ a \equal{} \minus{} 2\cos(\theta)$ must be a non-zero integer, the only possibilities are $ \pm1$ and $ \pm2$.

If $ a \equal{} \pm2$, then $ f(x)$ has a double root, which is a contradiction since $ x^k \minus{} 1$ does not have a double root.

$ (a,b) \equal{} ( \minus{} 1, \minus{} 1)$ yields $ x^2 \minus{} x \plus{} 1|x^k \minus{} 1$.
The roots of $ x^2 \minus{} x \plus{} 1$ are $ cis\left(\pm\frac {\pi}{6}\right)$ which must be $ k$th roots of unity.
Therefore, $ 6|m \minus{} n \Leftrightarrow m \minus{} n \equal{} 6t \Leftrightarrow m \equal{} n \plus{} 6t$.
Thus, $ (a,b,m,n) \equal{} ( \minus{} 1, \minus{} 1,n \plus{} 6t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$.

$ (a,b) \equal{} (1, \minus{} 1)$ yields $ x^2 \plus{} x \plus{} 1|x^k \minus{} 1$.
The roots of $ x^2 \minus{} x \plus{} 1$ are $ cis\left(\pm\frac {\pi}{3}\right)$ which must be $ k$th roots of unity.
Therefore, $ 3|m \minus{} n \Leftrightarrow m \minus{} n \equal{} 3t \Leftrightarrow m \equal{} n \plus{} 3t$.
Thus, $ (a,b,m,n) \equal{} ( \minus{} 1, \minus{} 1,n \plus{} 3t,n)$ is a solution for all positive integers $ n,t$ such that $ n > 1$.

----------------------------------------------------------------------------------------------------

Therefore, the solutions are:
$ (0,0,m,n)$
$ ( \minus{} 1,0,(t \plus{} 1)n \minus{} t,n)$
$ (1,0,(2t \plus{} 1)n \minus{} 2t,n)$
$ (0, \minus{} 1,(t \plus{} 1)n,n)$
$ (0,1,(2t \plus{} 1)n,n)$
$ ( \minus{} 1, \minus{} 1,n \plus{} 6t,n)$
$ ( \minus{} 1, \minus{} 1,n \plus{} 3t,n)$
Where $ m,n,t \in \mathbb{N}$ such that $ m > n > 1$.[/hide]

Thanks for your answer Xantos C. Guin :)


calculus
derivative
function
calculus computations

For the function

$f(x,y) = \frac{xy}{x^{2}+y^{2}}$if$(x,y) \neq (0,0) f(x,y) = 0 if (x,y) = (0,0)$

show that $f_{x}(0,0), f_{y}(0,0)$ both exist but $f$ is not differentiable at $(0,0)$

($f_{x}(0,0), f_{y}(0,0)$ are partial derivatives of the function)

Both derivatives are equal to zero, but the function is not even continuous at $(0,0)$ (consider its restriction to the line $y=x$).


ARML HMMT email geometry AMC USA(J)MO USAMO \/closed

I am now in a math-specialized class in a Vietnamese high school, and I am going to the US to continue my studies.  I wonder if  there are any classes of the same kind in US high schools.  If not, how can I get access to and be prepared for regional or national contests in mathematics.

I would be very happy if anyone could give me any helpful information.

Are you still in high school in the USA? It depends greatly on what school you're going to. Most contests are organized by the math team coach or a math teacher at each school. If your school doesn't really have that stuff, you can try to join with regional teams for some major contests, like ARML and HMMT and PUMaC (;)).

As for classes, that also depends on your school. If your school doesn't have any challenging math classes for you, you can take college courses (online or at a local univeristy) or you could consider AoPS classes.

Thank you very much for your help Xevarion :)
I am still in Vietnam and I am looking for a school that fits me (somewhere in San Jose, CA)
But how can I know which school tend to have students participating in those contests? :?:

[quote="humanbeing"]But how can I know which school tend to have students participating in those contests? :?:[/quote]

Well, you can ask the principal or teachers via email if they offer those contests as well as an average on how many people participate.

I know Mission San Jose high school does well in National Science Bowl. I don't know anything else about the school, but you might want to consider them.

You can also look at results for major math and science contests online and see if you can find any records of the school's participation or performance.

Are you looking for a public or private high school?

The San Jose area has a lot of schools which do very well on contests- look at the list of USAMO qualifiers. Most of those schools are public.

A more specific version of what jmerry just said and a note about what Xevarion said:

Mission San Jose HS is actually in Fremont, not San Jose. That's a few miles to the east around the bay.

Here are a few schools in San Jose, Cupertino, Mountain View, or Palo Alto. The number in the brackets is the number of AIME qualifiers in 2007. Some of these schools had over 100 AIME qualifiers in 2006 when the AMC12 was easier.

Henry M. Gunn HS, Palo Alto [37]
Homestead HS, Cupertino [6]
Leland HS, San Jose [6]
Los Alto HS, Mountain View [8]
Lynbrook HS, San Jose [55]
Monta Vista HS, Cupertino [8]
Mountain View HS, Mountain View [5]
Palo Alto HS, Palo Alto [30]

And I left out several others. There are also several significant private schools in the area, and there are middle schools (grades 6-8) with significant numbers of AIME qualifiers.


If 0º < x  < 90°, and , compute log10(sin x) + log10(cos x) + log10(tan x).

The value is not uniquely defined  :huh:

Try for example 56º and 60º.....isn't a value for each angle measure????

Well, so jhcreinhardt didn't write the problem well...........

I accidentally double posted this topic.  I edited the other one so that cosx was unique.


symmetry
analytic geometry
graphing lines
slope
geometry
perpendicular bisector
algebra

Can someone post the solutions for numbers 6 and 7 on mandelbrots 1 and 2. (2008)(Regional)(November and December)

here are the answers, ill post solutions later:

number 1, 6. dont know
number 1, 7. -9/2 or -4.5

number 2, 6. 3/5
number 2, 7. 1/(-x+2)

sorry for not posting solutions.

I am extremely doubtful that my #6 round 1 is right, so I won't say that. Also, I'm too lazy to actually write out #6 and #7 for round 2, so I'll just give hints.

[hide="#7, round 1"] We notice that $ \frac{6x}{x^2\plus{}2x\minus{}3}\equal{}\frac{\frac{9}{2}}{x\plus{}3}\plus{}\frac{\frac{3}{2}}{x\minus{}1}$ (or something like that, I forget). Hence, to be left over with just $ \frac{B}{x\minus{}1}$, we'll need to get rid of the fraction over x+3, which happens when A is -9/2.[/hide]

[hide="hint for #6, round 2"]Notice that the perpendicular bisector of a line segment contains all points equidistant from both of the endpoints of the line segment. So you should draw a perpendicular bisector...[/hide]

[hide="hint for #7, round 2"] Uh, basically substitute in three terms in the sequence, multiply out, and solve the resulting system of equations? 3 variables isn't that bad.[/hide]

Thanks.....but please post the solution if you can.

I got 11 on the first mandelbrot, and 8 on the second.......

Number 6:
[hide]
There are 6 ways to choose what two colors go into the top row (6C2).  After picking any combination, there are 3 possibilities for the row immediately below it.  For example, if the top row was AB, the next could be BC, BA, CA.  Therefore, there are also 3 possibilities for each of the 5 bottom rows.  Our desired answer is $ 6\ast 3^5\equal{}1458$  [/hide]

I believe this is correct, I remember solving it this way and had it correct, but I may have made a mistake somewhere....

I think I had the same answer, but I did casework with the first two rows, then had some sort of a weird recursion thing.

It was something like there are 3 ways to choose the first car, then 2 ways to choose the one right next to it. Now, if the car below the first is the same as the second, there are 2 ways for the fourth. If it's not, there is just one care for the fourth.

The first case gives 3*2*2=12, the second gives 3*2=6. So 18 for the first two rows. Now I think there are 3 ways for the third row and then by symmetry, 3 for every row after, so $ 18\cdot3^4 \equal{} 6\cdot3^5 \equal{} 1458$ like your answer.

EDIT: Wait, this is pretty much like your method, because I count 6 ways for the first row, then find the number of ways for the others.

[hide="Round 2, Problem 6"]The midpoint of CD is $ (\frac{5}{2},1)$.
Looking at O to this point, we have the equation of the line as $ y\equal{}\frac{2}{5}x$

This intersects $ AB$ at $ (1,\frac{2}{5})$. Clearly, for the line $ OP$ to intersect $ CD$ at a point closer to $ C$, we need $ P$ to be greater than $ \frac{2}{5}$. So the answer is $ 1\minus{}\frac{2}{5}\equal{}\boxed{\frac{3}{5}}$[/hide]

Problem 7 is just an equation in three variables...
[hide]
We're given that $ g(x)$ has the form $ \frac{a}{bx\plus{}c}$.

So plugging in the first three values: $ \frac{a}{\frac{1}{2}b\plus{}c}\equal{}\frac{2}{3}$
$ \frac{a}{\frac{2}{3}b\plus{}c}\equal{}\frac{3}{4}$
$ \frac{a}{\frac{3}{4}b\plus{}c}\equal{}\frac{4}{5}$

In the first equation, we have $ 3a\equal{}b\plus{}2c$
Second gives us: $ 4a\equal{}2b\plus{}3c$
Third: $ 5a\equal{}3b\plus{}4c$

$ b\equal{}3a\minus{}2c\rightarrow 4a\equal{}6a\minus{}c \rightarrow c\equal{}2a$

So $ b\equal{}\minus{}a$

Then, $ g(x)\equal{}\frac{a}{\minus{}ax\plus{}2a}\equal{}\minus{}\frac{1}{x\minus{}2}$[/hide]

Haha I remember on #6 round 2 how I thought the perpendicular bisector actually intersected the origin...argh, epic phail.

Yeah, those are the solutions I have.

I dont quite understand...
that second number 6..
I thought the distance to the line to a point is measured by its perpendicular distance...

not just finding a midpoint, as the line CD is NOT perpendicular to line OP if you just found the midpoint...(so the shortest distance from C to line OP is not measured on the distance between C and P on the line CD)

If you try that formula somehting like $ \frac {|Ax \plus{} By \plus{} C|}{\sqrt {a^2 \plus{} b^2}}$, you get a totally different answer...like 1/2...

Actually, you'll notice that it doesn't matter in this case.

[hide="Solution"]So let's find the distance from C to the line $ y\equal{}\frac{2}{5}x$ and the distance from D.

The only lengths we need to find are those from the midpoint of CD to the feet of the perpendiculars from D and C. Then we need to prove this distance is the same so that the distances to C and D are the same.

From C: Since the perpendicular line must have slope $ \frac{\minus{}5}{2}$, we take $ y\equal{}\frac{\minus{}5}{2}x\plus{}7$ to contain C on the line.

Intersecting this with $ y\equal{}\frac{2}{5}x$, we have that $ 7\equal{}\frac{29}{10}x\rightarrow x\equal{}\frac{70}{29}$ which is $ |\frac{\minus{}5}{58}|\equal{}\frac{5}{58}$ away from the midpoint of CD (x-coordinate wise).

From D: Similarly, the equation of the line passing through D is $ y\equal{}\frac{\minus{}5}{2}x\plus{}\frac{15}{2}$.

Intersecting this with the other line yields $ x\equal{}\frac{10}{29}*\frac{15}{2}\equal{}\frac{75}{29}$

But we see the difference in x-coordinates is $ \frac{75}{29}\minus{}\frac{5}{2}\equal{}\frac{5}{58}$.
Thus, the distances are the same on this line, so the point on AB must lie above this line.[/hide]

Yea, my friend assumed that the line from the origin to the midpoint met the line at a right angle, and in this case (actually this math fact is ALWAYS TRUE, as proven below), he got lucky. I however, used the analytic formula posted above for distance from point to line and did it the hard way

easy way (that I figured out after I took the test):
Let G and H be the feet of the altitudes from C and D to line OP, respectively. Also, let the intersection of OP and CD be X. Triangles CGX and DHX are similar by AA similarity. In fact, since we need CG = DH, they are congruent by SAS congruence. Thus we know that the line that is equidistant to C and to D through the origin O passes through the midpoint of segment CD, and this problem becomes trivial after this observation.

kinda cool that the nice solution involves euclidean geometry when the problem has coordinates.

out of curiousity, which forum is this in? The label of "Other U.S. contests and programs" doesnt seem to have a mandelbrot subforum.....and the only reason I found this post today was because it was a recent post

lol... Click on the other us contests and programs. and then scroll down. below the subforums section. xD

Is it ok to start talking about this recent round?

I know that some people are taking it tomorrow. Best to not discuss until they put the scores up online, at least.


This is just to get a Penn State AoPS gathering going w/o the mixup in the other thread.

I'm going. :D

This is my first time going to ARML! I'm very excited.

bah, i'd be there... I qualified for that Eastern Massachusetts A team but its the same weekend as my graduation so i cant go, oh well.

[quote="JackPo"]I'll be there if Joel is there.[/quote]

Jack is actually a parasitic fungus who lives in my armpit and types on this forum at night while I'm asleep.  :D

Yet he's in Baltimore, and you're in Cambridge.

I'm there, as one of the three canadian teams (two from our school and one from somewhere in the abyss ;)

I'll (probably) be there; my team's organizers want me to take one more test so they can be sure.

[quote="cats..."]i'll be there if i can get my SAT situation resolved...[/quote]

SAT 2C math?  If so, I'm annoyed about that, too, as I'm learning most of what's on it this year.. but I'm a freshman, so I could study over the summer and take it in October instead.. =/

I will be there on the NC team.

cyall there.

I'll be there w/ Western Mass.

Anyone got a good idea for the song contest?

I am going.  yay! :)  Look for young NYC coaches with curly hair ;)

i was told that the whole baltimore team isn't going.  Is that true?

Very simple you can either meet at the giantantic commons area. Get killed by flying frisbees. (Oh I shall play). Or even better yet meet at the local Hooters.

Does anyone have definitive location yet?

I am especially interested in meeting JBL.

Will the AoPS instructors be here? Any IMOers? Famous people from this forum?

I'm none of the above, but I will be there as one of the coaches of the South Carolna team.

I think you qualify as a "famous person from this forum."

For anyone interested in meeting me, I suggest waiting until everyone is in the big auditorium and finding NYC people and asking them, "Do you know where Joel Lewis is?"  :)

I saw JBL-- YAY (for about 3 seconds, and then I had to leave for the bus). There was no way I would have known who you were except that I heard someone say your name as I came down the stairs, and there weren't many "Joel"s at ARML, as far as I knew. I am pretty sure I saw gauss202, but you were talking to some people and I didn't want to interrupt. I was also worried that you wouldn't be gauss202, and I would look like an idiot.

Mathnovice and Gauss, it was great to meet you!  Sorry there wasn't more time to talk.  Perhaps next year we will organize somewhat better.

And Alison and Jen, it is of course always a pleasure to see you.

Nukular, good to see you as well ... 65*66! + 16, eh?

And congratulations to NY!  Yay!  Good job! :)

It was nice to meet you too JBL. Sorry we didn't get a chance to talk more. Maybe next year.

It was also nice to meet Alison and Jen.

And I'm sorry I didn't get to meet you mathnovice. I can't say if that was me or not. I was with the SC team and we had on the red shirts with a SC logo on the back. I was one of two African-American coaches on our team - me being the younger and shorter of the two.

I was on South Carolina B... as well as being the person in the lowest grade on either of the SC teams.

I had a great time and hope to go back next year.

I didn't really meet many people from AoPS at Penn State.  I saw JBL in the cafeteria but I never went up to him.  I met two other people from AoPS there, trickdacy  and gauss202 who was one of my coaches.  It was really cool being on his team.  I really learned a lot from him on the bus ride back.   I look forward to going next year and I also hope to meet more people.

Maybe those of us who are interested in meeting other AoPS people should wear name tags like Joel did, except with usernames. I wanted to meet some people here at the HMMT Tournament also, but I had no way of identifying who was who unless I had already seen him before.


The digits 1, 2, and 5 can be scrambled to form three-digit numbers. How many of the numbers are prime?

Clearly, neither $ 2$ nor $ 5$ can be the units digit. I'm sure all of us have memorized that $ 2008\equal{}2^3\cdot251$, so $ 251$ is prime, and $ 521$ is also prime after a check. So $ \boxed2$.

Uh, where did it ask for a probability?

[b]Edit:[/b] Now that's correct. :)


hi,
prove
\[ \sqrt[n]{n} < 1 \plus{} \sqrt[k]{\frac{n\minus{}1}{\binom{n}{k}}},\]
where $ k,n \in \mathbb{N}, 2 \leq k \leq n$.


Let $\triangle ABC$ a triangle and debote $H,N,R$ the orthocenter, the centre of nine-point circle, the circumradius respectively. Prove that

\[
AN^2  + BN^2  + CN^2  + HN^2  = 3R^2
\]

Using Stewart theorem in triangle $NAO$ and $AG$
$AG^2 = 2/3AN^2+1/3AO^2 - 2/9ON^2$

Summing and multiplying by 3:

$3 %Error. "sumAG" is a bad command.
^2 = 2\sum AN^2 +3R^2 - 2ON^2$

Using $HN = ON^2$

$ 3(\sum AG^2) + 4ON^2 = 2\sum AN^2 + 2ON^2 +3R^2$

$(a^2+b^2+c^2) + OH^2 = 2(\sum AN^2 + HN^2) +3R^2$

$9R^2 = 2(\sum AN^2 + HN^2) +3R^2$ or $3R^2 = (\sum AN^2 + HN^2)$ c.t.d.

See also Theorem 3 of http://www.mathlinks.ro/Forum/viewtopic.php?p=185878 post #3.

  Darij


$1.$  Let 7 denote system of numeration. If $8^{26}.125^{48}=(A)_{7}$ what is the last two digit of $A$?
$2.$  Let 4 denote system of numeration. If $10.3^{195}.49^{49}=(B)_{4}$ what is the last three digit of $B$?

[quote="mayhem"]K81o7, Answers are not correct. Why did you use mod 49?[/quote]

I assuming you meant that you want the first number converted into base 7, and then you want the last two digits?  I forgot to finally convert the base, so that the final answers should be 21 and 102.  I used base 49 because we want the last two digits of a base 7 number, and mod 64 because we want the last three digits of a base 4 number.  But did you mean something else?


Yep, it's simple but i'd like to know

There are 3 coloured balls that can be spun in a lotto motion with a funnel for the ball to dropout
The red ball comes out which is a one third probabilty.
What's the formulae for the red ball to come out a 2nd time and a 3rd and 4th and so on.
Is it correct that 2 times would be 1/2 x1/3 equals 1/6 or 0.16 ththe
3 times 1/3 X 1/6 equals 1/18 or .05
4 times 1/4 x 1/18 equals 1/72 0.013 etc etc
Whats the formalue? wcosmos@dodo.com.au

Well I think that the color of the first ball to come out is irrelevant to what the color of the next ball is, so it is always a/b, where a is the number of red balls, and b is the number of balls.


geometry
3D geometry
sphere
algebra
function
domain
conics

In $ xyz$ space with the origin $ O$, take a point $ A(0,\ 1,\ 1)$ and donote by $ l$ the straight line $ OA$. Suppose that the sphere $ S$ with radius 1 is in the domain $ z\geq 0$ and moves with touching to both the $ xy$ plane and $ l$, then find the locus of the tangent point $ P$of $ S$ and the $ xy$ plane, then sketch the locus.

S can be parametrized by $ 1\equal{}(h(t)\minus{}x)^2\plus{}(j(t)\minus{}y)^2\plus{}(k(t)\minus{}z)^2$
with $ A(t)\equal{}(h(t),j(t),k(t))$ we know $ A(0)\equal{}(0,0,1)$
and we can say at time t, it is touching $ (0,t,t)$ on OA, so
until time $ t\equal{}2$ after which it is impossible, so $ A(2)\equal{}(0,2,1)$

$ 1\equal{}h(t)^2\plus{}(j(t)\minus{}t)^2\plus{}(k(t)\minus{}t))^2$
to be tangent to the xy plane, we need to have only one solution $ (x,y)$ to

$ 1\equal{}(h(t)\minus{}x)^2\plus{}(j(t)\minus{}y)^2\plus{}k(t)^2$
or, $ (x\minus{}h(t))^2\plus{}(y\minus{}j(t))^2\equal{}1\minus{}k(t)^2$ so $ k(t)\equal{}1$ for all t
now we get

$ h(t)^2\plus{}(j(t)\minus{}t)^2\equal{}1\minus{}(1\minus{}t)^2\equal{}t(2\minus{}t)$
which is satisfied by $ j(t)\equal{}t, h(t)\equal{}\sqrt{t(2\minus{}t)}$
since the center moves here, the xy tangents move along the same path with z=0 instead of 1,
$ x(t)\equal{}\sqrt{t(2\minus{}t)}, y(t)\equal{}t$
which traces out on the xy plane to look like the right half of the circle $ x^2\plus{}(y\minus{}1)^2\equal{}1$

I assume that line $ l$ is the line passing through $ (0,0,0)$ and $ (0,1,1)$, not merely the line segment $ \overline{OA}$.

Consider an equivalent situation in which a unit sphere whose center travels in the plane $ y \equal{} z$, is tangent to the $ z$-axis.  Clearly, such a sphere is confined within the cylinder $ x^2 \plus{} y^2 \equal{} 4$, and within the parallel planes $ |z \minus{} y| \equal{} \sqrt {2}$.  Furthermore, the center of this sphere lies on the surface of the cylinder $ x^2 \plus{} y^2 \equal{} 1$.  The intersection of this cylinder with the plane $ y \equal{} z$ is an ellipse, which we can parameterize as

$ C(t) \equal{} (\cos\theta, \sin\theta\, \sin\theta)$.

The locus of the point of tangency $ T$ of the sphere to the lower plane $ z \equal{} y \minus{} \sqrt {2}$ has the same shape as the curve $ C$, but is translated by $ (0,1/\sqrt {2}, \minus{} 1/\sqrt {2})$:

$ T(t) \equal{} \left( \cos\theta, \frac {1}{\sqrt {2}} \plus{} \sin\theta, \minus{} \frac {1}{\sqrt {2}} \plus{} \sin\theta \right)$.

The major axis of the ellipse has length $ |C( \minus{} \pi/2) \minus{} C(\pi/2)| \equal{} \sqrt {2^2 \plus{} 2^2} \equal{} 2\sqrt {2}$, and the minor axis is easily seen to be 2.

Thus, in the original coordinate system, the locus of the point of tangency is an ellipse with vertices at

$ \left(0, \minus{} \cot \frac {3\pi}{8}\right) \equal{} (0, 1 \minus{} \sqrt {2})$,

$ (0, 1 \minus{} \sqrt {2} \plus{} 2\sqrt {2}) \equal{} (0,1 \plus{} \sqrt {2})$,

$ (1,1)$,

$ ( \minus{} 1,1)$.

This ellipse has parametric equation

$ E(t) \equal{} \left(\cos\theta, 1 \plus{} \sqrt {2} \sin\theta\right)$.

That's correct, which is same as my idea.

I have another solutions in using 1. Discriminant 2. Scalar Product.

By the way, doesn't straight line $ OA$ mean the line passing though the points the origin and the point $ A$?


Let $ X=(a_{ij})$ be $ n \times K$ matrix, $ A = \left(\begin{array}{ccccc}{cc...c}E_{t_1} & 0 & 0 & ... & 0 \
0 & E_{t_2} & 0 & ... & 0 \
0& 0 & E_{t_3}& ... & 0 \
\cdots \
0 & 0 & 0 & ... & E_{t_k}\end{array}\right)$ where $ E_{t_k}$ is $ t_k \times t_k$ matrix with all entries equal $ 1$, and $ \sum t_k =n$. Find $ \text{tr}((X^TX)^{-1}X^T A X)$

..please help.. thx.

NB: We see that for the case $ X$ is invertible then the trace equals trace $ A$..

[hide]
Then, if $ X$ is not invertible, then $ X^T$ is also not invertible, then the inverse of $ XX^T$ is undefined.
Sorry if my answer is false, Kak Ajat...
[/hide]

The interesting case here is when $ X$ is not [i]square[/i].

@QRUNI, for this problem $ XX^T$ is invertible.

I've tried to prove it, but i fail, to get the feeling about the problem, by the help of computer, i generated the full rank matrix $ X$ and compute the trace directly, i smell the interesting relation.

btw, i got this problem on my statistic class, from a statistic paper..it is stated there by the language of statistic (expectation,variance,covariance...) and no proof is given there.

Beginner, what exactly do you want to show about the trace? This is not to say I suspect to find much of interest, but I wouldn't rule it out either. But I let Mupad compute the trace for $ X \equal{} \left(\begin{array}{cc} a & a' \\
b & b' \\
c & c'\end{array}\right)$ and $ A \equal{} \left(\begin{array}{ccc} 1 & 1 & 0 \\
1 & 1 & 0 \\
0 & 0 & 1\end{array}\right)$ and obtained

$ \left(a^{2}(b^{\prime })^{2} \minus{} 2aca^{\prime }c^{\prime } \minus{} 2bcb^{\prime}c^{\prime } \minus{} 2aba^{\prime }b^{\prime } \plus{} b^{2}(a^{\prime })^{2}\right.$
$ \left. \plus{} a^{2}(c^{\prime })^{2} \plus{} c^2 (a^{\prime})^{2} \plus{} b^{2}(c^{\prime })^{2} \plus{} c^{2}(b^{\prime })^{2}\right)^{ \minus{} 1}$
$ \cdot\left( 2ab(c^{\prime })^{2} \minus{} 4aca^{\prime }c^{\prime } \minus{} 2acb^{\prime }c^{\prime } \minus{} 2bca^{\prime}c^{\prime } \minus{} 4bcb^{\prime }c^{\prime } \minus{} 4aba^{\prime }b^{\prime} \plus{} 2c^{2}a^{\prime }b^{\prime } \plus{} 2a^{2}(b^{\prime })^{2}\right.$
$ \left. \plus{} 2b^{2}(a^{\prime})^{2} \plus{} 2a^{2}(c^{\prime })^{2} \plus{} 2c^{2}(a^{\prime })^{2} \plus{} 2b^{2}(c^{\prime})^{2} \plus{} 2c^{2}(b^{\prime })^{2}\right)$.

This cannot be factored further - but maybe it can be simplified in some other way.

I feel that if anything can be said, the proof would begin by rewriting $ \text{Tr}\left(\left(X^TX\right)^{ \minus{} 1}X^TAX\right)$ as $ \text{Tr}\left(X\left(X^TX\right)^{ \minus{} 1}X^TA\right)$ and studying the matrix $ X\left(X^TX\right)^{ \minus{} 1}X^T$. This matrix is idempotent, symmetric and nonnegative definite (and every idempotent and symmetric matrix can be written this way!). This yields that $ \text{Tr}\left(\left(X^TX\right)^{ \minus{} 1}X^TAX\right)\geq 0$, because for any two nonnegative definite $ n\times n$ matrices $ A$ and $ B$, we have $ \text{Tr}\left(AB\right)\geq 0$. But it is not clear to me how to derive more about the trace...

  darij


A man was asked what price per 100 he paid for some apples.
His reply was as follows: "If they had been 4 cents more per 100 I should have got five less for 1.20 dollars." Can you say what the price was per 100?

Well...

Let x be the price per 100 (the price for one apple x/100)
and y be the number of apples the man got for 1.20 dollars (120 cent)

then y * x/100 = 120 and y = 120 * 100 / x

If the price was higher

then (y-5) * (x+4)/100 = 120

y * x/100 = (y-5) * (x+4)/100 ⇒

4y - 5x = 20 ⇒

x² + 4x - 9600 = 0 ⇒ x= 96

i.e. the price per 100 was 96 cent?  :?:

Sorry, I'm not very smart and my English is not good enough to give a proper explanation...  :blush:


simple question
we do electolytic reduction of alumina. but can we do electrorefining of alumina?

[hide="Sodium"]Sodium (and the other alkali metals) "burn" in water because in their reaction,

$ 2M \plus{} 2H_2O \longrightarrow 2MHO \plus{} H_2$

it is evolved not only hydrogen gas but also heat, which is enough to ignite the hydrogen. Of course, as we go down the group the react gets more violent.[/hide]

What do you mean by electrorefining of alumina? Purification of aluminium by electrolitic methods?

tx for sodium
in electrolytic reduction , we use carbon anodes for co2 to liberate.
but i was asking electrolytic refining, like for copper where anode is impure metal and it settles on cathode on passing electricity

We do not practically need pure alumina, do we?? We are interested in its electrolysis, not because of obtaining pure alumina, but for obtaining the aluminium metal. Because, both pure alumina and impure alumina will result in the same quality of aluminium. So, after all purification of alumina is of no use!
But we use [b]Hope's process[/b] for purification of aluminium electrolytically.
 :)

If you are interested in electrorefining of alumina, just because you are interested, then I'll say it's not impossible.....maybe you will be able to do it some day!!! :D


trigonometry
calculus
calculus computations

Find
\[ \sum_{n\equal{}1}^{\infty} \frac{\cos{n}}{n^2 \plus{} 1}\].

By using Fourier series, we find that

$ \sum_{n\equal{}1}^{\infty} \frac{\cos(nx)}{n^2\plus{}a^2} \equal{} \frac{\pi}{2a}[\coth(\pi a) \cosh(ax) \minus{} \sinh(ax)] \minus{} \frac{1}{2a^2}$.

Setting x = a = 1, we get $ \sum_{n\equal{}0}^{\infty}\frac{\cos n}{n^2\plus{}1} \equal{} \frac{\pi}{2}[\coth(\pi) \cosh(1) \minus{} \sinh(1)] \minus{} \frac{1}{2}$.


If $N=\frac{n-1}{n+1}$ in base $10$, express $N$ in base $n$.

edit: lack of sleep causes mistakes

bad mistakes  :lol:

Sorry, I meant to ask to find the [i]decimal [/i]representation of $N$ in base $n$. I believe there's more than one possible fractional answer...

Also, you forgot to subtract $1$ from your answer.

[hide="division algorithm"]
You can see pretty easily that the greatest integer multiple of $\frac{1}{n}$ less than $N$ is $\frac{n-2}{n}$.
$N-\frac{n-2}{n}=\frac{2}{n^2+n}$
Then subtract $\frac{1}{n^2}$ to get $\frac{N}{n^2}$.
Therefore $N=\frac{n-2}{n}+\frac{1}{n^2}+\frac{n-2}{n^3}+\frac{1}{n^4}+\cdots=0.\overline{(n-2)1}_n$.[/hide]


A 1000 kg boat is traveling at 90 km/h when its engine is shut off.  The magnitude of the kinetic frictional force between boat and water is proportional to the speed, v, of the boat: kinetic frictional force = 70v, where v is in meters per second and the kinetic frictional force is in newtons.  Find the time required for the boat to to slow to 45 km/h.

[hide]$ F_{net} \equal{} \minus{} 70v \equal{} m\dfrac{dv}{dt}$. Rearrange and separate with variables:
\[ \minus{} dt \equal{} m\dfrac {dv}{v}\]
Then we integrate with limits:
\[ \int_0^{t_f} \minus{} dt \equal{} \int_{90}^{45} m\dfrac{dv}{v}\]
Evaluating the integral, we get $ \minus{} t_f \equal{} m\ln \dfrac{45}{90}$ so $ t_f \equal{} \minus{} m\ln \dfrac 12 \equal{} \minus{} 1000\ln \dfrac 12$.[/hide]

I think the units should be fixed.
90km/h=25m/s and 45km/h=12.5m/s
and $ \minus{}70dt\equal{}m\frac{dv}{v}$

(The problem states that F=70v has v in m/s)


hameye $x\in\mathbb{R}$ ra biyabid, ke $\{(x+1)^3\}=x^3$.
$(\{x\}=x-\lfloor{x}\rfloor)$


Let  O  be  a  point  in  the  interior  of  an equilateral  triangle  ABC  with  sides  of  lenght  a . The  lines  AO , BO ,CO  intersect  the  edges  in  the points  A1 , B1 , C1 . Prove  that
    OA1 + 0B1 + 0C1<a

(VIII  Nordic  Mathematical  Contest , 1994)
MERRY  XMASS!!

Here's a solution which matches the text of the problem (nice and short :D ):

Let D, E, F be the projections of O on BC, CA, AB, respectively. It's obvious that angles DOA1, EOB1 and FOC1 are <30 deg. this means that OA1=OD/cos(DOA1)<OD/cos30 and the analogous relations. This means that OA1+OB1+OC1<(OD+OE+OF)/cos30=h/cos30=a (h is the altitude of ABC).


Ring Theory
superior algebra
superior algebra solved

Does there exist infinite rings such that all their subrings are finite?(the rings have unity)

Let $K=\mathbb{Z}_2$. Let $K\subset \Omega$ be an algebraic closure of K.

Let $K_n=K[\xi_n]$ where $\xi_n$ is a primitive $2^n$-th root of unity in $\Omega$.
$K_n$'s are fields and $K_1=K$.
It can be proven that $K_n$ is the field generated by the $2^n$ -th roots of unity.

It is easy to see that $K_n\subset K_{n+1}$ and that $[K_{n+1}:K_n]=2$.

Take $L=\bigcup_{n\geq 1}K_n$. It is easy to see that $L$ is an infinite field.

Define $f:L\to \mathbb{N}^*$ to be the map sending $x\in L$  to $\min\{n\in \mathbb{N}^*| x\in K_n\}$.

Let A be a subring of L. Consider the set $B=f(A)\subset \mathbb{N}^*$.

[b]First case[/b]: B is bounded.  This means that there exists $M\in \mathbb{N}^*$ such that $f(x)<M\  \forall x\in A$. This, by the definition of f implies $A\subset K_M$ and so A is finite.

[b]Second case[/b]: B is unbounded. This means that for all $n\in \mathbb{N}$ there exists $x_n\in A$ such that $f(x_n)\geq n$. This means that A contains an $f(x_n)$-th root of unity so it contatins $K_{f(x_n)}$ and so also $K_n$. This implies $A\supset L$ and so A=L.

This shows that all subrings of L not equal to L are finite.


geometry
perimeter
pigeonhole principle
number theory
arithmetic sequence
relatively prime
geometry unsolved

This problem is 2008  Germany Olympiad problem 5. Can anyone from Germany Help me?

[quote="tfy"]We put finite circles in the interior of a unit square ( two circles maybe have joints ), if the sum of all the perimeters of the circles is 10, then try to prove there exists a line which has joints with at least 4 circles.

Sorry for my poor English, I hope I've expressed the problem clearly. :help:  :gathering:[/quote]
Project all circles to the side AB of unit square.
The sum of all diameters is $ \frac {10}{\pi} > 3$
But AB=1 then applying Dirichlet's theorem, there exists a point M which lies on at least 4 projections.
A line d which passes through M and perpendiculars to AB cuts at least 4 circles.

[quote="tfy"]livetolove212 has mentioned the Dirichlet Theorem, but I don't know what it is. Would you say it more clearly?[/quote]

If I am not mistaken, livetolove212 actually refers to what is usually callled "Dirichlet's pigeonhole principle", which states in the general form that, if there are $ N$ elements divided into $ k$ categories, such that $ N>qk$ for some integer $ q$, then there is at least one category with at least $ q\plus{}1$ elements.  It is called the "pigeonhole principle" because one of its formulations is "If there are $ N$ pigeonholes and $ N\plus{}1$ pigeons, then there is at least one pigeonhole shared by $ 2$ angry pigeons".

In the present problem, the idea is that the sum of the lengths of the projections of the diameters onto one of the sides of the square being larger than $ 3$, but the side having length $ 1$, there is at least one segment that needs to belong to at least $ 4$ projections (otherwise the sum of the lengths of the projections would be at most $ 3$).  The perpendicular to the side of the square through one of the points inside this segment therefore cuts the four circles projected onto that point.

NB: "Dirichlet's Theorem", if I am not mistaken, usually refers to a number theory result: given a nonconstant arithmetic sequence $ a_n\equal{}a_0\plus{}dn$, such that $ a_0$ and $ d$ are relatively prime integers, there exist infinitely many primes that belong in the sequence.

[quote="daniel73"][quote="tfy"]livetolove212 has mentioned the Dirichlet Theorem, but I don't know what it is. Would you say it more clearly?[/quote]

If I am not mistaken, livetolove212 actually refers to what is usually callled "Dirichlet's pigeonhole principle", which states in the general form that, if there are $ N$ elements divided into $ k$ categories, such that $ N > qk$ for some integer $ q$, then there is at least one category with at least $ q \plus{} 1$ elements.  It is called the "pigeonhole principle" because one of its formulations is "If there are $ N$ pigeonholes and $ N \plus{} 1$ pigeons, then there is at least one pigeonhole shared by $ 2$ angry pigeons".

In the present problem, the idea is that the sum of the lengths of the projections of the diameters onto one of the sides of the square being larger than $ 3$, but the side having length $ 1$, there is at least one segment that needs to belong to at least $ 4$ projections (otherwise the sum of the lengths of the projections would be at most $ 3$).  The perpendicular to the side of the square through one of the points inside this segment therefore cuts the four circles projected onto that point.

NB: "Dirichlet's Theorem", if I am not mistaken, usually refers to a number theory result: given a nonconstant arithmetic sequence $ a_n \equal{} a_0 \plus{} dn$, such that $ a_0$ and $ d$ are relatively prime integers, there exist infinitely many primes that belong in the sequence.[/quote]

Sorry fly. In Viet Nam, I understand that "Dirichlet's theorem" is "Dirichlet's pigeonhole principle"  :blush:
Thanks you very much, daniel73. :D

[quote="tfy"]I've known the "pigeonhole principle" before. But in China, we usually use it in problems about finite elements, could it be also used in infinite-elements conditions? For there exists infinite points in a segment, besides, a point has no length.[/quote]

The principle is the same as in problems about finite elements, except in this case you break the side of the square into segments, such that each segment belongs to the projection of a constant number $ k$ of circles - if the number of circles is finite, and each one of them has finite length, this is possible and the side of the square is neatly expressed as the disjoint union of a finite number of such segments (plus maybe a finite number of isolated points with $ 0$ length where the perpendicular to the side of the square is tangent to some of the circles).  The idea is that the sum of all constants $ k$, weighted by the length of the corresponding segment, must add to the sum of the projections of the circles onto the side, which is larger than $ 3$.  So at least one such constant is $ 3$, hence it is at least $ 4$.  You could view it as a "weighted version" of the pigeonhole principle: if $ N$ elements are assigned each an integer weight $ w$, and a real value $ v$, and the sum of the weighted values $ wv$ exceeds $ qV$ for some integer $ N$, and where $ V$ is the sum of the values $ v$, then there is at least one of the weights that is at least $ q\plus{}1$ - otherwise the total sum of the weighted values would not exceed $ q\sum v\equal{}qV$.  In our case, $ v$ is the length of each segment, $ w$ is the number of circles that project onto the segment, $ V\equal{}1$ is the total length of the side of the square, and the sum of the weithed values is $ \frac{10}{\pi}>3$...


function
algebra
rational function
geometric series
calculus
calculus computations

I'm looking to find the ilt of $\frac{1}{s(e^s-1)}-\frac{1}{s^2(e^s-1)}$.

I realize that I can combine this into a rational function whose ilt is [tex]1-t[/tex] or something like that and so the main part I'm interested in is $\frac{1}{e^s-1}$.  I'm not finding it in any table except when I expand it as a geometric series; then it's a sum of delta functions or something but I was wondering if there was a better closed form.  Once I get the ilt of the rational expression and the inverted exponential, I can convulte the two, right?

Any help appreciated.

Oh, btw, what I'm trying to do is find the asymptotic behavior of $\sum_{k=1}^{t-1}(1-t/k)^k$.  I took the lt of this and I *think* the first two terms in its Laurant series expansion are given above, the third term being really small.  I'm then hoping that the ilt of the first two terms in the lt will give me the asymptotic behavior of the original.

Thanks.

I don't think it has a nice form

You could try to use the complex inversion formula to see if it can be written as a nice integral...

Otherwise a power series expansion is your best bet

Thanks for replying. I will work on it expanding $\frac{1}{e^s-1}$ as a geometric series which I believe inverts back to a sum of delta functions...  We'll see what happens.


function
integration
real analysis
real analysis solved

Find all continuous functions f:[0, 1}->R s.t.  \int {0, 1} f(x)dx=1/3+ \int{0, 1} (f(x^2))^2dx.

I particularly like this one.

we have that:
[0,1]-->g-->[0,1]-->f-->R so that x=g(t)-t^2 and then we have that : \int {g(0),g(1)} f(x)dx =  \int {0,1} (fog)(t)*g'(t)dt from which we have that g is derivable (hurray) and with its derivable continuos on [0,1] and g(0)=0 and g(1)=1. so we have that:
 \int {0,1} f(x) dx=  \int {0,1} f(t^2)2t dt=  \int {0,1} 2f(x^2)x dx and we also know that  \int {0,1} x^2 dx=1/3 and we get that:

 \int {0,1} (x-f(x^2))^2 dx =0 [1]
we also have that: (x-f(x^2))^2<=0

From [1] and [2] we get that (x-f(x^2))^2=0 for any x in [0,1] so f(x^2)=x for any x in [0,1] so we have that f(k)=sqrt(k) , for any k in [0,1]

cheers! :D

I wonder why the soln in the book takes that odd turn. I mean it's easy to prove  \int {0, 1} f(x)dx= 2 \int{0, 1} xf(x^2)dx by changing the variable x=t^2.

Indeed grobber .. I found that observation on those papers! :D
it's a lot shorter and easier!! :D

BTW if we stick to the 1st solution I guess it's x=k^2  :D :D


I know that Contest stopped for a long time ago. Is there a chance that it comes back? I am sure that there are people which will compete (including me) and people who make problems...

I saw many people ask this... Come on let's start it again!

And the Contest section is unavailable...

No one responded, no admins, no mods, no users (OK, it hasn't been a day yet, but this was asked even a year ago by some people and no answers, also I sent a PM to Valentin Vornicu a long time ago asking this).

Why does no one answer?

Of course, it's not necessary that the problems be open. But IMO style problems are nice. I mean, how many people on this forum want to practice for IMO, and competitive style could be good (of course, lots of problems in Olympiad section are good, but everyone feels better competing).

Why does no one respond? I see the topic had 125 views, is anyone here who has to say something about this? Pros/Cons? Anyone?

I agree with Bugi.

At leat one admin have to reply and tell the reasons that mathlinks contests are ended.

Come on Admins, someone reply to Bugi's nice suggestion.
Why cant the Mathlinks contest again start?

I know that it is not so easy to arrange such a competition, or at least it takes a enourm amount of time; I assure it is hard to find suitable and original problems, especially for a online one; and if people (like in the lasts MLC round) do not partecipate, also the autors and mods in general can be discouraged (sorry for my english, but i think you can understand what i am meaning..).
As Bugi said, a solution can be made of getting easier problem, just for training, at level of some imo problem (ok, not the first, and not the fourth  :D ). But i think that a lot of student will appreciate this idea..It is possible also to assign correcting work to some users.

Ps, thanks Bugi for putting a link to contest of mine in your signature  :wink:

Why not first and fourth also? It can encourage users which aren't so trained for hardest problems (like me, I usually do 1/4, sometimes 2)...

You are welcome

[quote="Raúl"]Can we, users, start a competition by ourselves? I'm interessted in it, so if them does nothing about it, we can do :D

Agree? Some ideas about how we can do that? :maybe:[/quote]

It was the Algebra Tournament ,but for unknown reason it stoped  :(  .It was started by [b]ghjk[/b] ,and not many users was active to submit their solutions.I'm quite sure that mathlinks contests are over for similar reasons.

I also agree  with Bugi ,it's another feeling when you're competing

ghjk stopped the tournament because he wasn't active. If you followed his posts, you wold see that after his last post in the tournament, he had only 2 posts up to today.

Well, no admin response?

If you really want the attention that this Contest deserves, PM our admin Valentin Vornicu, he is the founder of Mathlinks and also the Contest.

I'm really busy with some personal projects right now, at this moment I do not have the time for this. But I will think about a solution for the near future. Thanks for bringing this to my attention.


calculus
real analysis
real analysis unsolved

I think this has already been discussed. Since $ f^m(x) \equal{} x$ for all $ x \in [0,1]$, it follows $ f^m \equal{} \textrm{id}_{[0,1]}$, which is one-to-one (bijection).

Lemma. [color=blue]$ f\circ g$ bijective implies $ f$ surjective AND $ g$ injective.[/color]
(Proof. Trivial, from definitions.)

But $ f^m \equal{} f\circ f^{m\minus{}1} \equal{} f^{m\minus{}1}\circ f$, so according with Lemma, $ f$ is both surjective AND injective, hence bijective (one-to-one).

[quote="debanik2"]Which book is this excerpt from?[/quote]

Problems in Real Analysis: Advanced Calculus on the Real Axis

[quote="mavropnevma"]I think this has already been discussed. Since $ f^m(x) \equal{} x$ for all $ x \in [0,1]$, it follows $ f^m \equal{} \textrm{id}_{[0,1]}$, which is one-to-one (bijection).

Lemma. [color=blue]$ f\circ g$ bijective implies $ f$ surjective AND $ g$ injective.[/color]
(Proof. Trivial, from definitions.)

But $ f^m \equal{} f\circ f^{m \minus{} 1} \equal{} f^{m \minus{} 1}\circ f$, so according with Lemma, $ f$ is both surjective AND injective, hence bijective (one-to-one).[/quote]

Thanks very much !

[quote="debanik2"]can you provide a link where i can download it from ?[/quote]

google it !


Is there an if..then function I can use in Asymptote? I have the diagram posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1101217#1101217]here[/url]

and the code:

[code][asy]
unitsize(1cm);
for(int i=0;i<6;++i){
pair A=(0,0), B=(dir(60+60i)), C=(shift(dir(40+60i))*B), D=(shift(dir(60(i-1)))*C), E=(dir(60i));
path unit=D(A--B--C--D--E--cycle);
path unit2=D(rotate(-100,D)*unit);
D(rotate(140,C)*unit2);
}
dot("X",shift(dir(40))*(1,0),SE);
dot("Y",dir(60),SE);
dot("Z",shift(dir(160))*dir(120),NE);

D(shift(-7,-.5)*((0,0)--dir(60)--shift(dir(40))*dir(60)--shift(dir(40))*(1,0)--(1,0)--cycle));
MP("A",shift(-7,-.5)*(0,0),SW);
MP("B",shift(-7,-.5)*dir(60),NW);
MP("C",shift(-7,-.5)*shift(dir(40))*dir(60),N);
MP("D",shift(-7,-.5)*shift(dir(40))*(1,0),E);
MP("E",shift(-7,-.5)*(1,0),SE);
[/asy][/code]

I feel that I'm being way too inefficient there at the end, especially because I'm including redundant code to redraw "unit"

I was looking for something like this (syntax nonexistent :lol:) within the _for_ loop:

[code]if(i=0){
  D(shift(-7,-.5)*unit);
  }[/code]

meh it's not much..

also, how do I shift the labels around with the figure? Just using _shift_ only seems to move the shape but not the labels.

Thanks!

Asymptote has an if statement.  The syntax is "if (i == 0)".  Take a look at the Asymptote manual.


This problem was confusing me, so any help would appreciated.

Use row reduction to show that the determinant of the n by n matrix symbolically represented by
[A C
 O B]

is |A||B|, where

A is an m by m submatrix,
B is an (n-m) by (n-m) submatrix
C is an m by (n-m) submatrix, and
O is an (n-m) by m zero submatrix.

[quote="Kent Merryfield"]We can row reduce $A$ to $T_{1},$ which is upper triangular, and whose determinant we can read off. That means we can find $P_{1}$ such that $P_{1}A=T_{1}.$ (And if $P_{1},$ which is just record of the row steps we used, contains any row swaps or scalings which change the determinant, then we kept track of that.) Similarly, by row reduction, $P_{2}B=T_{2}.$

Then

$\begin{bmatrix}P_{1}&0\\0&P_{2}\end{bmatrix}\begin{bmatrix}A&C\\0&B\end{bmatrix}= \begin{bmatrix}T_{1}&P_{1}C\\0&T_{1}\end{bmatrix}$

Note that $P_{1}C$ is just some junk that we're not even going to look at. We have an upper triangular matrix and we can read off the determinant.[/quote]

Thanks!  I was wondering - which submatrices are you using to notate matrices A, B, C, and O?  I guess I had trouble notating them, which is why I was having trouble with the problem.

Sorry, you can disregard that previous question, but I have another question.

 I know where you got P1 and P2 from in that first matrix, but how did you get 0's for the top right hand corner and bottom left hand corner?

If we had anything other than zero in the lower left block, we'd be taking some linear combinations of the top rows and adding them to the bottom rows; if we had anything other than zero in the upper right block, we'd be taking some linear combinations of the bottom rows and adding them to the top rows. Since we don't want to do either of those things, we leave those blocks as zeros.


[quote="Ulanbek_Kyzylorda KTL"]x+y+z=1
x^2+y^2+z^2=3
x^3+y^3+z^3=7
x^4+y^4+z^4=?
for real[/quote]

The answer is 11

Let, $ x\plus{}y\plus{}z\equal{}p$, $ xy\plus{}yz\plus{}zx\equal{}q$, $ xyz\equal{}r$.
So,
$ x\plus{}y\plus{}z\equal{}p\equal{}1$,
$ x^2\plus{}y^2\plus{}z^2\equal{}p^2\minus{}2q\equal{}3 \iff q\equal{}\minus{}1$,
$ x^3\plus{}y^3\plus{}z^3\equal{}p^3\minus{}3pq\plus{}3r\equal{}7 \iff r\equal{}1$.

Therefore, $ P(x)\equal{}x^3\minus{}x^2\minus{}x\minus{}1$.

[b]1st[/b]
$ x^4\plus{}y^4\plus{}z^4\equal{}p^4\minus{}4q(p^2\minus{}2q)\minus{}6q^2\plus{}4pr\equal{}\boxed{11}$

[b]2nd[/b]
$ x^4\plus{}y^4\plus{}z^4\equal{}(p^2\minus{}2q)^2\minus{}2q^2\plus{}4pr\equal{}\boxed{11}$

[b]3rd[/b]
$ x^4\plus{}y^4\plus{}z^4\equal{}p(p^3\minus{}3pq\plus{}3r)\minus{}q(p^2\minus{}2q)\plus{}pr\equal{}\boxed{11}$

[b]4th[/b]
Let,
$ S_1\equal{}x\plus{}y\plus{}z$,
$ S_2\equal{}x^2\plus{}y^2\plus{}z^2$,
$ S_3\equal{}x^3\plus{}y^3\plus{}z^3$,
$ S_4\equal{}x^4\plus{}y^4\plus{}z^4$.

So,
$ S_1\minus{}1\equal{}0$,
$ S_2\minus{}S_1\minus{}2\equal{}0$,
$ S_3\minus{}S_2\minus{}S_1\minus{}3\equal{}0$,
$ S_4\minus{}S_3\minus{}S_2\minus{}S_1\equal{}0$.

Thus, $ S_4\equal{}x^4\plus{}y^4\plus{}z^4\equal{}\boxed{11}$.

cf. [url=http://www.mathlinks.ro/Wiki/index.php/Newton_sums]Newton's sums[/url]


Let $ a,b,c$ be the sides of a triangle such that: $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\equal{}3a^2b^2c^2$.Prove that:
$ \frac{a}{3ab\plus{}1}\plus{}\frac{b}{3bc\plus{}1}\plus{}\frac{c}{3ca\plus{}1} \le \frac{4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2}{4}$

[quote="quykhtn-qa1"]Let $ a,b,c$ be the sides of a triangle such that: $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} 3a^2b^2c^2$.Prove that:
$ \frac {a}{3ab \plus{} 1} \plus{} \frac {b}{3bc \plus{} 1} \plus{} \frac {c}{3ca \plus{} 1} \le \frac {4(ab \plus{} bc \plus{} ca) \minus{} (a \plus{} b \plus{} c)^2}{4}$[/quote]
It's really a hard inequality but just uses two lemmas:
1.  $ \frac {b}{{ab \plus{} 3}} \plus{} \frac {c}{{bc \plus{} 3}} \plus{} \frac {a}{{ca \plus{} 3}} \le \frac {3}{4}$  for all $ a,b,c \ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$  (Vasile Cirtoaje).
2.  $ \left( {2ab \plus{} 2bc \plus{} 2ca \minus{} a^2 \minus{} b^2 \minus{} c^2 } \right)\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right) \ge 9$  with a,b,c are three sides of a triangle. (IRAN_1996)

[quote="leedt26"][quote="quykhtn-qa1"]Let $ a,b,c$ be the sides of a triangle such that: $ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \equal{} 3a^2b^2c^2$.Prove that:
$ \frac {a}{3ab \plus{} 1} \plus{} \frac {b}{3bc \plus{} 1} \plus{} \frac {c}{3ca \plus{} 1} \le \frac {4(ab \plus{} bc \plus{} ca) \minus{} (a \plus{} b \plus{} c)^2}{4}$[/quote]
It's really a hard inequality but just used twos lemma:
1.  $ \frac {b}{{ab \plus{} 3}} \plus{} \frac {c}{{bc \plus{} 3}} \plus{} \frac {a}{{ca \plus{} 3}} \le \frac {3}{4}$  for all $ a,b,c \ge 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$  (Vasile Cirtoaje).
2.  $ \left( {2ab \plus{} 2bc \plus{} 2ca \minus{} a^2 \minus{} b^2 \minus{} c^2 } \right)\left( {\frac {1}{{a^2 }} \plus{} \frac {1}{{b^2 }} \plus{} \frac {1}{{c^2 }}} \right) \ge 9$  with a,b,c are three sides of a triangle. (IRAN_1996)[/quote]
Yes,you are right :wink: .It very nice and not nice :wink:   :)


In general, $ f_{n\plus{}4} \equal{} 3f_{n\plus{}2} \minus{} f_n$. Therefore, $ g_n \equal{} 3g_{n\minus{}1} \minus{} g_{n\minus{}2}$.

I don't have a good answer for the second part at the moment.

[quote="bigman"]The sequence G_0, G_1, G_2, . . . consists of every other Fibonacci number. Show that
there is a linear recursion (e.g. of the form G_n = aG_{n-1}+bG{n-2}). (Follow-up: How
about a sequence consisting of every tenth Fibonacci number. How do you know
there's a recursion?[/quote]Note the term 'linear'. If you've taken a course in linear algebra, you can prove that there exists a recursion by picking every $ k^{th}$ term of a sequence with linear recursion, e.g. Fibonacci. More general propositions can be proved for general linear sequences and subsequences but that I won't go further, leaving that for researching purposes.

If you want some information to generalize your problem, here are some facts you would like to use (and prove them for some more fun):

The characteristic polynomial for the Fibonacci sequence is $ P(t) \equal{} t^2 \minus{} t \minus{} 1$. Let $ r$ be a root of $ P(t) \equal{} 0$.

Lemma 1: $ r^j \equal{} f_jr^1 \plus{} f_{j \minus{} 1}$ for all $ j$.

Lemma 2: $ \frac {f_{2j}}{f_j}$ is an integer for all $ j$.

Proposition: There exists a linear recurrence relation of the sequence $ g_n$ by taking every $ k^{th}$ term of the Fibonacci sequence.


Two side lengths of a right triangle are $ 2\sqrt{11}$ meters and $ 6\sqrt{3}$ meters. The number of meters in the length of the third side is a whole number. How many meters long is the third side?


Does anyone know if a school can hold the Mandelbrot individual test at two different points during a day? (e.g. in the morning and afternoon.) It says that all the tests should be taken the same day, but nothing about taking it at different times.

Thanks!

I'm pretty sure it can be done since it's not mentioned in the policy, but obviously something you'd want to aviod.

You should e-mail Sam Vandervelde, the contest director/writer, and ask him.

I don't recolect it saying that all tests had to be taken the same day, where does it say this?

~lyra

What do you mean by that?  The rules clearly indicate that the National and Regional tests should be given concurrently.  In fact, that also answers the original question -- you should only administer the test at one time during the day.  My previous advice still holds, however: talk to Sam Vandervelde if you want an exception.


If ${a_{1}}<{a_{2}}<...<{a_{n}}$ are integers, show that

$\prod\limits_{1\leq{i} <j\leq{n}}\frac{a_{i} - a_{j}}{i - j}$

is an integer.

Its a nice problem, but i think its been posted a few times before.

This problem is Turkish TST(for 2005 IMO) 's fourt question.

 I remembered you need to look power of primes.