So who will be there?

I'll be there for the South Carolina team.

I will be there as a captain of New York City A team.

In looking to see what ARML's up to this year, I discovered this little gem in Penn State's application.  Great idea, not so great implementation.

Hint: Any time someone feels like defending themselves after one sentence into the description of the event, there's a problem.

Convoluted description of how a song ends up winning this competition? CHECK.

Bizarre, inexplicable 30-second song length requirement? CHECK.

No prizes at all to get people to actually show up to this thing? CHECK.

I could launch into a diatribe about how this is the type of thing you get when schools simply participate in the same competitions year after year [i]without even thinking about it[/i], or taking time to ask "is this something we really ought to be spending money on?" But I'll save that for another day...ARML needs to stop reading from Mu Alpha Theta's [u]How Not To Run A National Convention[/u] playbook, though.

[b]ARML Song Contest [/b]
At Penn State University

Teams are encouraged to write a song about the ARML competition, their trip to Penn
State, their ARML team, or math in general. This is a great team building activity for
teams to work on during their trip to Penn State or during the day on Friday.
A sign up sheet will be posted outside the registration booth in Findlay Commons and
teams are encouraged to register for the competition upon arrival. All teams wishing to participate must register by 7:00 pm Friday evening.

The preliminaries will be held on Friday evening in Findlay Commons at 9:35 pm. The
audience, along with ARML judges, will select five teams who will perform at the main
competition on Saturday after the Super Relay. The winning song will be selected by
audience applause.

Teams are allowed to have as few, or as many performers as they want (1 or 2 up to 15), but each team is limited to one entry. Alternates may compete on the team that they came to Penn State with.
Songs will be judged based on originality, creativeness, and overall presentation. Songs must be limited to thirty seconds and any entries with inappropriate content or dress will be disqualified.

I don't know if you're aware, but the song contest is about 4 years (maybe 5?) old, and is solely for the purpose of filling time to allow them to finish grading power, coordinate among the various sites, get awards ready, etc.  The final voting is audience applause.  As there are no prizes, there is in fact 0 cost, and yet people do participate.  Your message is the equivalent of writing a nasty letter about the ultimate frisbee games that take place, or something -- it's just there for fun to fill up some time.

If this is a tradition, as well as something that serves a functional purpose within their tournament schedule, so be it.

I don't have any problem with people having fun, either.

[quote]it's just there for fun to fill up some time.[/quote]

Instead of just "filling time," and occupying a captive audience, I wish math competitions (not just ARML, by any means, as its a great event) would MARKET themselves toward new groups of students and new schools, not just the same ones year after year.  [b]Putting a little bit more thought - [u]and a lot more incentive[/u] - behind participation in the recreational events that go on at math competitions would be a tremendous step forward in this regard.[/b]

...  :?

but anyway, getting away from this diatribe against (of all things) the song contest, North Carolina has not chosen it's ARML team yet (the contest will be this thursday), but hopefully I will be there.

I'll also be there for south carolina. but after my bad performance at practice, it looks like i'll be on the B team.  :?

I'll be there for Vermont. :D

Maybe I can go for a 7 or 8 this year-that would be nice.  question-if you're on a team that qualifies for the A division, how many years do you have to stay in that division?

I'll be there for Georgia, definitely on the B team.

EDIT: or alternate ... forgot they did those  :(

mkkool: 2 years i think, but i could be making that up.

North Carolina A team... I wonder how many other 8th graders are going? And yes, mysticterminator is on there again (getting top two as usual).

Im going to be there representing the NYC A Team...
                                                                          Stan

I'll be there, giving the Friday night talk...

NYC A

[quote="Bictor717"]If you expect me to be at all alive, you'll be sorely disappointed because my prom is on Thursday night-Friday morning.[/quote]

do you really want to miss out on the manhattan bus antics? :D

Hi.  I'm coming, as a coach.  For NYC C.  That's right, I found this place too.

There is too much to miss on the Manhattan bus.  You know that I will try to get so many games running...

Hmm...Sarah-Marie is running the Friday talk this year...I doubt if she remembers me though.  HCSSiM 2001...and I was too little then.

So much to do! I can't not sleep for 2 days and expect to do well on the third, after roughly 5-6 hours of sleep. I love Manhattan bus antics, but I also don't want to miss Sarah-marie's talk. I was in HCSSiM 2003. Will there be a training session as well?
And we'll meet fellow AoPSers, right? Just to say hello.
I am so looking forward to this. Must go pack now.

i don't know if i want to meet any of you. i might have to put a mask on to conceal my identity.  :ninja:

Bictor717 and other HCSSiM alumns, you've likely seen this talk before:  it's "Seventeen plus Seven Flavors of Wallpaper."  (I think we've done it as a prime time almost every year since 2000.)  You may recall it as the one involving some jumping around.

[quote="Midnight"]
Hmm...Sarah-Marie is running the Friday talk this year...I doubt if she remembers me though.  HCSSiM 2001...and I was too little then.[/quote]

I remember everyone from 2001 (though not all the last names).  Given the number of 2001 alumns who are not in NYC, you're at most one of 11 people.  If by "little" you meant that you were one of the younger students (rather than one of the shorter students) then you're likely to be one of three people.    Of course, you might look different now---most people change quite a bit in four years of teenage-ness---so maybe I won't recognize you!  Be sure to say hi to me.

[quote="qweretyq"]quick question:
does anyone know the answer to number 8 team round? (the parametric equations one)
i got 0 as an answer 5 seconds after time was up.
Is that right?[/quote] No- it's 100.  Look at a=b.

Congratulations Arnav!

Great job.

Same to all the top scorers!

Edit: And to all the people on our team, including hmmm, who scored really high!

Go Lehigh Fire!

Edit 2: That guy at the podium! OUCH. He screwed up everybody's name - Wow...

yea I know that was funny I was like hahaha that's funny

w00t I actually know some people on lehigh fire go lehigh fire!!!

LOL WHAT
NAH NO WAY
And im doing arml next year btw

yes way, it happened.
I did the impossible
19 year, 6 months, and 12 day bump.
I think... if im doing the math right.


The inside wheels of a car traveling on a circular path are rotating half as fast as the outside wheels. The front two wheels are six feet apart. What is the number of feet in the path traced by the inside front wheel in one trip around the circle?

[hide="long way"]If $x$ is the distance of the inner wheel from the center of the circle, then $x+6$ is the distance of the outer wheel from the center. For the inner wheel to be traveling at $\frac{1}{2}$ the speed of the outer wheel, the circumferences need to be in a ratio of $1: 2$ (inner to outer). Then you set $2*\pi*(x+6)$ equal to $2*(2*\pi*x)$
$2\pi(x+6)=4x\pi$
$x\pi+6\pi=2x\pi$
Divide by pi: $x+6=2x$
$x=6$ and the circumference of the inner cirle is $12\pi$[/hide]
[hide="short way"]The distances from the center of the circle are $x$ and $2x$ or $(x+6)$ so then $2x=x+6$ and $x=6$
circumference is $12\pi$[/hide]

[quote="jhredsox"]wait, since when is there an, "Inside," wheel for a car?

-jorian[/quote]
Inside is referring to the wheel that is closer to the center of the circular path.

Oh, I think I get it now!

I stink at figuring out stuff like this, and have attempted the problem, but I got it wrong.

-jorian


Calculate
\[ \exp{
\begin{bmatrix}
t&1&\cdots&0\\
&t&1&0\\
&0 & \ddots &1\\
& & &t\\
\end{bmatrix}}, ( t\in R)\]
Help me!

Let's start by letting $ B\equal{}
\begin{bmatrix} 0 & 1 & \cdots & 0 \\
& 0 & 1 & 0 \\
& 0 & \ddots & 1 \\
& & &0 \\
\end{bmatrix}.$

Its powers can be explicitly computed (it is nilpotent) and put into the definition of the exponential. The result is

$ e^B\equal{}\begin{bmatrix} 1 & 1  &\frac12&\cdots & \frac1{(n\minus{}1)!} \\
0 & 1 & 1 &\cdots &\frac1{(n\minus{}2)!} \\
0 & 0 &1 &\cdots & \frac1{(n\minus{}3)!} \\
\vdots &\vdots&\vdots&\ddots&\vdots\\
0&0 & 0&\cdots&1 \\
\end{bmatrix}.$

The matrix whose exponential we want is $ A\equal{}tI\plus{}B.$ Since $ tI$ and $ B$ commute, the result is $ e^A\equal{}e^{tI}e^B\equal{}e^{t}Ie^B\equal{}e^te^B$ where $ e^B$ is computed as above.


[hide]
AM-HM:
$\sqrt[3]{abc} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} $

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{\sqrt[3]{abc}} \geq \frac{3}{abc}$

From the question:

$a+b+c \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{3}{abc}$
[/hide]

[hide="Not complete"]
Why should  $\frac{3}{\sqrt[3]{abc}} \geq \frac{3}{abc}$ ?
[/hide]

[quote="devious_"][hide]
If a,b,c are positive, then:

$abc \geq \sqrt[3]{abc}$

$1 \geq \frac{\sqrt[3]{abc}}{abc}$

$\frac{1}{\sqrt[3]{abc}} \geq \frac{1}{abc}$

Is that correct?
[/hide][/quote]No. Same mistake.

\[
abc(a + b + c) \geqslant bc + ca + ab \geqslant \sqrt {3abc(a + b + c)}
\]
so,
\[
abc(a + b + c) \geqslant 3
\]

[quote="devious_"]siuhochung, can you please explain how you arrived at $\sqrt {3abc(a + b + c)}$?[/quote]
It is just $(x+y+z)^2\geq 3(xy+yz+zx)$.

I was just coming to edit my post and say that I figured it out, then I noticed two replies. :D

I don't know whether it has a name, but you see that it is $x^2+y^2+z^2\geq xy+yz+zx$, which is very well known and widely used.

Its most ancient ancestor is called the trivial inequality.   :lol:

[quote=Valentin Vornicu]Let $a,b,c$ be positive numbers such that $a+b+c \geq \dfrac 1a + \dfrac 1b + \dfrac 1c$. Prove that
\[ a+b+c \geq \frac 3{abc}. \][/quote]

easy!

my solution =

the given condition is $abc(a+b+c)\ge ab+bc+ca$

but $(ab+bc+ca)^2\ge 3abc(a+b+c)\ge 3(ab+bc+ca)\Longrightarrow ab+bc+ca\ge 3\Longrightarrow abc(a+b+c)\ge 3$

so we are done :D

Let $a,b,c$ be positive real numbers  such that: $abc(a+b+c) =3.$ Prove that
$$a+b+c \geq \frac{3}{ab+bc+ca}+\frac{2}{abc}$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{3}{a+b+c}+\frac{2}{abc}$$
Let $a,b,c$ be positive real numbers, such that: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$ [url=https://artofproblemsolving.com/community/c6h148342p838560]Prove that[/url]
$$a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}$$
[url]https://artofproblemsolving.com/community/c6h1490860p8749593[/url]
[quote=Valentin Vornicu]Let $a,b,c$ be positive numbers such that $a+b+c \geq \dfrac 1a + \dfrac 1b + \dfrac 1c$. Prove that
\[ a+b+c \geq \frac 3{abc}. \][/quote]
[url]https://artofproblemsolving.com/community/c6h1184601p5748463[/url]


Does anybody know where to find results from USAMOs before 2002? The 2002-2005 results are on the AMC website.

Here is to 98: http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/ .

I meant statistics (_% of people got _, ...); I guess I wasn't very clear.


[hide]$\sqrt{x+\sqrt{x+74}}+\sqrt{x-\sqrt{x+74}}=10$
Squaring both sides, we get $x+\sqrt{x+74}+2\sqrt{(x+\sqrt{x+74})(x-\sqrt{x+74})}+x-\sqrt{x+74}=100$.  Simplifying, we get $x+\sqrt{x^{2}-x-74}=50$.  Rearranging and then squaring both sides, we see $x^{2}-x-74=2500-100x+x^{2}$, so $99x=2574$ and $x=26$.[/hide]


It's conservation of momentum in inertial and non-inertial frames.

[quote="climskywell"]A student on a frictionless skateboard at rest has a box of 20 baseballs.  Each ball has a mass of 175 grams.  She has a mass of 49 kg.  The box is 100 grams and the skateboard is 900 grams.

1.  first she threw all of the baseballs horizontally at once away from her from the box.  She held on to the box.   They were going 33m/s with respect to her, not the ground.  How fast was her recoil speed across the ground?  How fast were the baseballs going relative to the ground?

2.  Next, she threw the 20 baseballs one at a time.  Each time they traveled away from her horizontally 33m/s with respect to her.  How fast was her recoil speed across the ground after the last ball was thrown?[/quote]

Momentum conservation as ^ said.

Let the girl's absolute velocity be 'v'. abs velocity of baseball is 33-v (moving in opp dir)
(20x175)x(33-v) = (49000+100+900)xv

solve for v.

2. first ball:

(175)x(33-v)=(m-175)xv
v1=(175)x33
    -----------
     m

again:

(m-175)xv1=(m-350)xv2 -175(33-v2)

solve. It follows a geometric progression. get the general form and substitute for n=20/

P.S: Does this even belong in advanced physics?


I just recently downloaded both Latex programs and I'm trying to type something for my math assignment but I seem to be caught in a lot of problems.  Can someone help me take a look at this?

Thanks

[code]2.  Determine the length of each side, ie.  AB , BC , AC

AB = $\sqrt{{(2-8)}^2+{(5+1)}^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2}

BC = $\sqrt{{(-4-8)}^2+{(3+1)}^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}

AC = $\sqrt{{(2+4)}^2+{(5-3)}^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}

3a.  The triangle is scalene because all the lengths are different
b.  The triangle is obtus because it satisfies ${c^2+b^2<a^2}.
$Thus, $\triangle ABC is obtuse[/code]

The compilation will tell you about missing dollar signs. And that's the problem. Math mode must start [i][b]and[/b][/i] end with a dollar sign.
[I can't use $\$$ here easily as it is misinterpreted, so will use [i]dollar sign[/i] instead]

You have the dollar sign starting in the middle of an equation but not finishing anywhere - hence the errors.
You need to use something like:
[code]2.  Determine the length of each side, ie.  $AB , BC , AC$

$\\ AB = \sqrt{{(2-8)}^2+{(5+1)}^2}=\sqrt{36+36}=\sqrt{72}=6\sqrt{2} \\
BC = \sqrt{{(-4-8)}^2+{(3+1)}^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\\
AC = \sqrt{{(2+4)}^2+{(5-3)}^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}$ \\
\\
3a.  The triangle is scalene because all the lengths are different \\
b.  The triangle is obtuse because it satisfies ${c^2+b^2<a^2}$.
Thus, $\triangle ABC$ is obtuse[/code]
Thus the three long equations start with a dollar sign and end with another dollar sign with \\ signifying a new line.

Similarly the formula on the next to last line starts and ends with a dollar sign. This should then compile.

Thanks stevem.  I understood a bit more about Latex and it works real nice now.

Ok I got a problem again now...

After I added a header to the Tex File, I seem to have this 2.5 inch margin from the top of the sheet before I start my document.  How do I change this?

P.S. I have already set \topmargin to 0in

Thanks

I'm just upping this topic because I'm in urgent need of help.  Thanks.

Again, the problem is...

After I added a header to the Tex File, I seem to have this 2.5 inch margin from the top of the sheet before I start my document. How do I change this?

Great care is needed with margins. Have a look at http://www.tex.ac.uk/cgi-bin/texfaq2html?label=changemargin and http://www.image.ufl.edu/help/latex/margins.shtml


[quote="maokid7"]Find all positive $n$ such that $5^5-5^4+5^n$ is a perfect square.

[hide="i have gotten this so far"]
$5^4(5^{n-4}-4)=z^2$ for some integer $z$
so since $5^4$ is already a perfect spuare then $5^{n-4}-4$ has to be a perfect square.
but i dont know how to find any values other than $n=5$.
i also know that $n=7$ works but i found this through guess and check. [/hide]
could someone please give me a hint or point me in the right direction.[/quote]

Eh, it factors into $5^{n-4}+4$, not $5^{n-4}-4$...

Yes, that is how I would have done it  :)

maokid, $n = 7$ does not work  :huh: $5^5 - 5^4 + 5^7 = 80625 \approx (283.9454...)^2$

yeah i got $n=7$ because i factored wrong.
man i should check my work for stupid mistakes like that.
well thanks for the help.

Just some thought

When $n>4$ , we have

$5^4(4+5^{n-4}$

Let $4+5^{n-4}=z^2$

$\implies 5^{n-4}=(z+2)(z-2)$

Since they diff by 4 , it must be $n-4=1$ or $n=5$ .

For $n\leq 4$ , it should be easy to check that there are no solution .


Let $ABC$ be a triangle and $w$ it's circumcircle. Denote $AD$ the bisector of $\angle{BAC}$, where $D \in (BC)$, and consider the circles $k_{1}$ tangent to $w, BD,AD$ and $k_{2}$ tangent to $w,CD,AD$. Prove that $k_{1}$ and $k_{2}$ are tangent in $I$, where $I$ is the incircle of $\triangle{ABC}$.

[hide="Remark"]See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=146656[/url][/hide]

Let $E$ the intersection of $AD$ and $w$. We know that $EB = EC = EI$.
Do one inversion $(E, EB^{2}= EC^{2}= EI^{2})$.
The circle w is mapped to the line $BC$, because $B = B'$ and $C = C'$ (and, obviously, $BC$ is mapped to $w$)
See that the line $AD$ is fixed, because $E\in AD$. So, $k_{1}'$ will be tangent to $B'C'$, $w'$ and $(AD)'$, and this implies that $k_{1}' = k_{1}$.
So, The intersection $X$ of $k_{1}$ and $AD$ will be fixed, and $EX^{2}= EI^{2}\Longrightarrow X = I$.

(The same for $k_{2}$)

See, for example, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=307]here[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=46342]here[/url]. Check Darij's post!

Oh, thanks. I knew that it is a nice particulary case of the Thebault theorem, but I didn't know that something like that was given in a contest before; I was just generalizing the problem proposed by stergiu here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=146656[/url]


What is the probability that a positive integer less than or equal to 24 is a factor of 24? Express your answer as a common fraction.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24, making a total of 8 out of 24 numbers, or a probability of 1/3.


calculus inequalities derivative algebra function domain parameterization

Find the maximum value of the expression
$A = 13\sqrt{x^2-x^4}+9\sqrt{x^2+x^4}$, where $0\leq x\leq 1$.

this is a hard one....i dunno i think it might be 162, ie when $x=1$ ...lol...like i simplified the expresion to:

$A^2 = 250x^2 - 88x^4 + 234\sqrt{x^4 - x^16}$

Using Calculus, I have found that [i][b]maxiumum of A = 16[/b][/i] when $x = \frac 4 5$.
But I want to know if there is any other way to find the maximum without using Calculus.

Edit: maximum of A is attained when $x^2=\frac 4 5$.

I think it can be solved by BCS .See here in posts #52  and #54 for a very similar problem

http://www.mathlinks.ro/Forum/viewtopic.php?p=352548#p352548

Sorry, I don't understand the language in the link provided.
Do you mind explaining the detail in English?  :lol:

In the post #54 it says <<At first (2A)^2=...... . after use BCS >>

At the post #52 the problem tells <<if   0<=x<=2 then find the maximum of A=....>>

I suppose BSC = Cauchy-Schwarz inequality ?
I tried it but the result doesn't follow.  :P

You should find numbers $k,m$ such that   the root of the equation

$\frac{m}{\sqrt{x^2-x^4}}=\frac{k}{\sqrt{x^2+x^4}}$  has root $x=\frac{4}{5}$

Sorry ! Why root $x=\frac 45$ ? At first you find this root with calculus ... and then You painted proof !

luimichael can you post your solution using calculus?
how did you find the roots of the derivative?

First of all, I have to apologize for telling an incorrect value of x for the maximum of A.
In fact, A is maxiumum when $x^2=\frac 4 5$.
Here is my calculations.
Let $t=x^2$.
Then $A=13\sqrt{t-t^2}+9\sqrt{t+t^2}$, where $0\leq t\leq 1$.
$\frac{dA}{dt}=\frac{13(1-2t)}{2\sqrt{t-t^2}}+\frac{9(1+2t)}{2\sqrt{t+t^2}}$.
$\frac{dA}{dt}=0$ if and only if $\frac{13(2t-1)}{2\sqrt{t-t^2}}=\frac{9(1+2t)}{2\sqrt{t+t^2}}$----(*)
.......
$2t(500t^3-375t+44)=0$
$t=0, \frac 4 5, -\frac{3\sqrt{3}+4}{10}, \frac{3\sqrt{3}-4}{10}$.
t = 0 corresponds to A = 0 , which is clearly the absolute minimum of A.
The third root is out of range.
While the fouth root does not satisfy (*).
Checking second derivative, $t=\frac 4 5$ corresponds to a maxiumum.
As there is only one turning point in the given domain and this value corresponds to a maxiumum, $t=\frac 4 5$ corresponds to the absolute maximum of A.
When $t= \frac 4 5$, $A = 16$.
[b]
And, Virgil, I agree with your comment[/b] :!:

Aren't you missing $\sqrt{t-t^2}+\sqrt{t+t^2}$ in the derivative of A?
I mean by the product rule, the derivative of $13\sqrt{t-t^2}$ is $\sqrt{t-t^2}+\frac{13(1-2t)}{2\sqrt{t-t^2}}$
Am I wrong?

[quote="manuel"]Am I wrong?[/quote]

Yes.  You're taking the derivative of $t\sqrt{t - t^2}$ or something.

oops, true. the derivative of 13 is 0 not 1. sorry  :oops:

This problem reminds me of why in the United States graphing calculators are recommended. With a graphing calculator, you can find the maximum in one minute.

Cautiong: Just an outline.
I don't like using x as a parameter, and since the problem seeks the max of A and does not care for x, we change:
$0 \le x \le 1, x^2 = t \Rightarrow 0 \le t \le 1$
$A = 13\sqrt{t - t^2} + 9\sqrt{t + t^2}$
We have x and y parametrically defined:
$x = \sqrt{t - t^2}$ and $y = \sqrt{t + t^2}$
These give two equations (both of which must be simult. true):
$x^2 + y^2 = 2t$ and $y^2 - x^2 = 2t^2$
$\Rightarrow 0 = 2(0) \le x^2 + y^2 \le 2(1) = 2$

We also have a family of vertical hyperbolas. We can find our maximum from the points that lie in the circle with radius $\sqrt{2}$ and on on of the hyperbolas. We have the line $13x + 9y = k$ for some $k$. The greater the y-intercept, the greater the $k$. So, we know if the line is tangent to the outer circle, we will have our max, except the tangency point may not lie on one of the hyperbolas. If it doesn't, we must find the closest points that do. Gotta run.


inequalities
integration
function
limit
logarithms
geometry
real analysis

For natural number $ n$, prove that

$ \sum_{i\equal{}1}^{n}\frac{1}{\sqrt{n^2\plus{}i^2}} < \ln(1\plus{}\sqrt{2})\minus{}\frac{1}{2(2\plus{}\sqrt{2})n}.$

We have to show that $ \frac {d_{n} \plus{} D_{n}}{2} < \int^{1}_{0} \frac {1}{\sqrt {1 \plus{} x^{2}}}$ where $ d_{n}$ is the lower and $ D_{n}$ the upper Darboux sum of the function $ f(x) \equal{} \frac {1}{\sqrt {1 \plus{} x^{2}}}$ on $ [0,1]$. However, since the function is convex, I would say that we have $ \frac {d_{n} \plus{} D_{n}}{2} >  \int^{1}_{0}\frac {1}{\sqrt {1 \plus{} x^{2}}}$ for all $ n$-s?

I didn`t have time before to explain my thoughts, so I will be clearer now.

$ \sum_{i \equal{} 1}^{n}\frac {1}{\sqrt {n^2 \plus{} i^2}} \equal{} \sum_{i \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}}$, which is the Riemann sum of the function $ f(x) \equal{} \frac {1}{\sqrt {1 \plus{} x^{2}}}$ for the divison of the interval $ [0,1]$ on $ n$ equal segments and choosing the right most point on each of this segments.

Now we know that $ \lim_{n \to \infty}\sum_{i \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}} \equal{} \int^{1}_{0} \frac {1}{\sqrt {1 \plus{} x^{2}}} \equal{} \ln{(1 \plus{} \sqrt {2})}.$

Now we have to show that $ \sum_{i \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}} < \ln(1 \plus{} \sqrt {2}) \minus{} \frac {1}{2(2 \plus{} \sqrt {2})n} \qquad (\ast)$.

We have to somehow find $ A, B$, such that:
\[ \ln(1 \plus{} \sqrt {2}) \minus{} \frac {1}{2(2 \plus{} \sqrt {2})n} < A\leq \sum_{i \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}}\leq B < \ln(1 \plus{} \sqrt {2})
\]
.
Therefore we can study the upper and the lower Darboux sum of the function $ f$ for the same division.

Since $ f^{\prime}(x) \equal{} \minus{} \frac {x}{\sqrt {(x^{2} \plus{} 1)^{3}}} \Rightarrow f^{\prime}(x) < 0$ on $ (0,1]$, the function $ f$ is strictly decreasing on $ [0,1]$. Therefore we take the right most point on each interval of our division to get the lower Darboux sum $ d_{n}$ and the left most point on each interval to get the upper Darboux sum $ D_{n}$.
Therefore
\[ d_{n} \equal{} \sum_{i \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}} \quad,\quad D_{n} \equal{} \sum_{i \equal{} 0}^{n \minus{} 1}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}}
\]
We can see that
\[ D_{n} \minus{} d_{n} \equal{} \frac {2}{2n(2 \plus{} \sqrt {2})}
\]
.
Using this in $ (\ast)$, we have to show
\[ \frac {D_{n} \plus{} d_{n}}{2} < \ln{(1 \plus{} \sqrt {2})}.
\]
Now we know, that on the intervals, where $ f$ is convex, the mean value of the upper and the lower Darboux sum is greater than the area below $ f$ and vice versa for concave parts of $ f$. Hence $ \frac {D_{n} \plus{} d_{n}}{2}$ partitions into $ \frac {D_{n} \plus{} d_{n}}{2} \equal{} \frac {D^{\prime}_{n} \plus{} d^{\prime}_{n}}{2} \plus{} \frac {D^{\circ}_{n} \plus{} d^{\circ}_{n}}{2}$, where $ D^{\prime}, d^{\prime}$ are from convex parts of $ f$ and $ D^{\circ}, d^{\circ}$  are from concave parts of $ f$. If there wouldn`t be any concave parts of $ f$ on $ [0,1]$, we would have $ \frac {D_{n} \plus{} d_{n}}{2} > \ln{(1 \plus{} \sqrt {2})}$, but if there wouldn`t be any convex parts of $ f$ on $ [0,1]$, we would have $ \frac {D_{n} \plus{} d_{n}}{2} < \ln{(1 \plus{} \sqrt {2})}$.

But the problem appears, since $ f^{\prime\prime}(x)$ gives us, that $ f$ is concave on $ [0,\frac {\sqrt {2}}{2}]$ and convex on $ [\frac {\sqrt {2}}{2},1]$. Now it remains to decide which part prevails. I am stuck here.

I have made a mistake in the upper proof in the line:
\[ \ln(1 \plus{} \sqrt {2}) \minus{} \frac {1}{2(2 \plus{} \sqrt {2})n} < A\leq \sum_{i \equal{} 1}^{n}\frac {1}{n}\frac {1}{\sqrt {1 \plus{} \left(\frac {i}{n}\right)^2}}\leq B < \ln(1 \plus{} \sqrt {2}).\]
It is not needed, so just ignore it.

However, does anybody have any hint on how to show that the concave part of $ f(x)$ prevails over convex in the upper reasoning?


trigonometry
geometry
circumcircle
incenter
geometry unsolved

Help me to solve this problems(can you solve them using trigonometry).

[b]1.[/b] On the sides $AB$, $BC$, $CD$, and $DA$  of quadrilatera $ABCD$ are given points $M$, $N$, $P$ and $Q$ and for them is $AM=BN=CP=DQ$. If $MNPQ$ is square then $ABCD$ is also square.

[b]2.[/b] Let $k$ be circumcircle of triangle $ABC$ and $D$ be an arbitrary point on side $AB$. Circle $k_1$ touches Circle $k$ inside, $AD$ and $CD$, and circle $k_2$ touches $k$ inside, $BD$ and $DC$. Prove that $k_1$ touches $k_2$ iff $CD$ is bisector of $\angle ACB$.

[b]3.[/b] The incircle $s$ of triangle $ABC$ touches $BC$ in $K$. Let $M$ be the midpoint of the altitude $AD$. Let $N$ be the second intersection point line $KM$ with circle $s$. Prove that $N$ is the common point of circle $s$ and circumcircle of triangle $BCN$.

[quote="X-man"][b]3.[/b] The incircle $s$ of triangle $ABC$ touches $BC$ in $K$. Let $M$ be the midpoint of the altitude $AD$. Let $N$ be the second intersection point line $KM$ with circle $s$. Prove that $N$ is the common point of circle $s$ and circumcircle of triangle $BCN$.[/quote]

The point N is the tangency point of the incircle of $\triangle ABC$ and circumcircle of $\triangle BCN$. For a proof, see [url=http://www.mathlinks.ro/Forum/post-141487.html]posted before? [a circumcircle tangent to the incircle][/url], with a significant contribution from pestich = armpist:

[quote="pestich"]:?[/quote]
(no kidding).

[quote="X-man"][b]2.[/b] Let $k$ be circumcircle of triangle $ABC$ and $D$ be an arbitrary point on side $AB$. Circle $k_{1}$ touches Circle $k$ inside, $AD$ and $CD$, and circle $k_{2}$ touches $k$ inside, $BD$ and $DC$. Prove that $k_{1}$ touches $k_{2}$ iff $CD$ is bisector of $\angle ACB$.[/quote]

To use such a valuable tool as trigonometry for a problem that can be solved synthetically at the first glance would be a waste.

Assume that CD bisects the angle $\angle C$. Then it passes through the midpoint M of the circumcircle arc AB opposite to the vertex C. Invert the figure in a circle (M) with radius MA = MB. The line AB is carried into the circumcircle k, the circumcircle k into the line AB, and the line CD passing through the inversion center M is carried into itself. This means that the circles $k_{1}, k_{2}$ are carried into themselves, which is possible only if they are perpendicular to the inversion circle. But the line CD is perpendicular to the inversion circle (M), hence, the circles $k_{1}, k_{2}$ have to touch this line at the same point, i.e., they have to touch each other.

Conversely, assume that the circles $k_{1}, k_{2}$ touch each other. Let o, m be tangents to the circumcircle k and to the circle (M) at their intersection A. We have $\angle (AB, o) = 180^\circ-\angle C$ and $\angle (AB, m) = \frac{\angle AMB}{2}= \frac{180-\angle C}{2}$. This means that the circle (M) bisects the angle formed by the line AB and the circumcircle k. Hence, any circle $k_{i}$ centered on the same side of AB as the vertex C, touching the line AB, and touching the circumcircle k internally is perpendicular to the circle (M). Consequently, any 2 such circles $k_{1}, k_{2}$ touching each other have their tangency point on the circle (M), which means that their single common internal tangent must pass through the center M of this circle. If this tangent also passes through the vertex C, it is  the bisector of the angle $\angle C$.

[quote="X-man"][b]1.[/b] On the sides $AB$, $BC$, $CD$, and $DA$  of quadrilateral $ABCD$ are given points $M$, $N$, $P$ and $Q$ and for them is $AM=BN=CP=DQ$. If $MNPQ$ is square then $ABCD$ is also square.[/quote]

Assuming that the points $M, N, P, Q$ lie on the segments $AB, BC, CD, DA.$ The vertices $A, B, C, D$ lie on 4 congruent circles $(M), (N), (P), (Q)$ with radius $r = AM = BN = CP = DQ$, respectively. The line $AM$ is either tangent to the circle $(N)$ at a point $B_0$ or it meets it at 2 points $B_1, B_2$ on different sides of the line $B_0N$, such that $AB_1 < AB_2$. The line $B_0N$ is tangent to the circle $(P)$ and the line $B_1N$ does not intersect it. Hence, the quadrilateral vertex $B$ is identical either with the tangency point $B_0$ or with the more distant point $B_2$. Assume that $B \equiv B_2$ is different from $B_0$ and let $F$ be the midpoint of the chord $B_1B_2$. Since the angles $\angle MNP = \angle AFM = 90^\circ$ are both right, it follows that the angles $\angle(NP, FN) = \angle BMN$ are equal and $\angle CNP \equiv \angle(NP, BN) < \angle(NP, FN) = \angle BMN$. In exactly the same way, we can show that $\angle DPQ < \angle CNP, \angle AQM < \angle DPQ, \angle BMN < \angle AQM,$ etc. Combining,

$\angle BMN > \angle CNP > \angle DPQ > \angle AQM > \angle BMN$

which is a contradiction. Hence, the quadrilateral vertex $B$ must be identical with the tangency point $B_0$. But the line $BN$ is then tangent to the circle $(P)$ and the single tangency point must be identical with the quadrilateral vertex $C$. Likewise, the line $CP$ is then tangent to the circle $(Q),$ the tangency point identical with the quadrilateral vertex $D,$ the line $DQ$ is tangent to the circle $(M)$, the tangency point identical with the quadrilateral vertex $A$, and finally, the line $AM$ is tangent to the circle $(N)$, the tangency point identical with the quadrilateral vertex $B$, thus closing the square $ABCD$.

[color=darkred][b]The proof of the problem II.[/b] The circles $k_{1}$ and $k_{2}$ are tangent (in the point $T\in CD$) $\Longleftrightarrow$ the rays $[AT$, $[BT$ are the bisectors of the angles $\widehat{BAC}$, $\widehat{ABC}$ respectively $\Longleftrightarrow$ the point $T$ is incentre of the triangle $ABC$ $\Longleftrightarrow$ the ray $[CD$ is the bisector of the angle $\widehat{ACB}\ .$[/color]

[b]EDIT:[/b] This is wrong, I am sorry ! (It doesn't follow that the rays $[AT$, $[BT$ are the bisectors of the angles $\widehat{BAC}$, $\widehat{ABC}$ respectively!) I wanted to apply the following lemma, but I made a great mistake confusing the letters. I shall come back soon for correcting of proof.

[color=darkred][b]Lemma.[/b] The circle $k_{1}$ is interior tangent to the circle $k$ in the point $T$. The tangent to the circle $k_{1}$ in a point $M\in k_{1}$ meet the circle $w$ in the points $N$ and $P$. Then the ray $[TM$ is the bisector of the angle $\widehat{NTP}$.[/color]

Try to apply this lemma to the circles $k_{1}$ and $k_{2}$ for the their common tangents $AB$ and $CD$ !


Hi!

Sorry for my non-mathematic description of my problem. But here we go...

Let's say I have an ordered sequence of objects. Let's take N letters for examples.
How many ways can I create k groups of letters without changing the orders.

Example : ABCDE

With k=4 (I want 4 groups), I can divide it 4 different ways :
[list][*]{AB},{C},{D},{E}
[*]{A},{BC},{D},{E}
[*]{A},{B},{CD},{E}
[*]{A},{B},{C},{DE}[/list]With k=3 (I want 3 groups), I can divide it 6 different ways :
[list][*]{ABC},{D},{E}
[*]{A},{BCD},{E}
[*]{A},{B},{CDE}
[*]{AB},{CD},{E}
[*]{AB},{C},{DE}
[*]{A},{BC},{DE}[/list]With k=2, I can only divide is two ways :
[list][*]{ABCD},{E}
[*]{A},{BCDE}
[*]{AB},{CDE} (Edit)
[*]{ABC},{DE} (Edit)[/list]Is there a simple formula/theorem that can give me the total number of partition I can create for a give N and k ?

Mat

[quote="BleuBleu"]With k=2, I can only divide is two ways :
[list][*]{ABCD},{E}
[*]{A},{BCDE}[/list][/quote]  What about AB-CDE and ABC-DE?  Assuming you just overlooked these (and so there aren't any other restrictions you haven't mentioned), the answer is $ \binom{n \minus{} 1}{k \minus{} 1} \equal{} \frac {(n \minus{} 1)!}{(k \minus{} 1)!(n \minus{} k)!}$ -- from among $ n \minus{} 1$ gaps between the letters you must choose $ k \minus{} 1$ to be the dividers between the different groups.

Cool, I never tough about looking at the problem in terms of "dividers".
That makes perfect sense!
I just edited my original post to include those 2 missing combinations.

[code]Binomial[4, 3] = 4
Binomial[4, 2] = 6
Binomial[4, 1] = 4[/code]

It works! Thank you sooo much!

Mat


algebra
polynomial
calculus
derivative
induction
number theory
relatively prime

Prove that the remainder of the polynomial
p(x)=x^2007+2x^2006+3x^2005........+2006x^2+2007x+2008

is the same upon division by x(x+1) as upon division by x(x+1)^2

It's shorter than it looks at first glance, but it still bothers me that I ended up using calculus.

[hide="Solution"]There exists a unique combination $ P(x)$, $ C_1$, $ C_2$, $ C_3$, such that: \[ x^{2007} + \ldots + 2007x+2008 = P(x) \cdot x (x+1)^2 + C_1 x(x+1) + C_2 x + C_3 \] We must show that $ C_1 = 0$ in the above formulation.

[b]1.[/b] The instance $ x=0$ gives $ C_3 = 2008$. Subtract that out and divide by $ x$ to get: \[ x^{2006} + \ldots + 2006x + 2007 = P(x) \cdot (x+1)^2 + C_1 (x+1) + C_2 \]
[b]2.[/b] The instance $ x=-1$ gives $ C_2 = 1004$. Subtract that out to get: \[ x^{2006} + \ldots + 2006x + 1003 = P(x) \cdot (x+1)^2 + C_1 (x+1) \]
[b]3.[/b] We want to show that $ (x+1)^2 \; | \; x^{2006} + \ldots + 2006x + 1003$, or equivalently, [i]both[/i] of the following:  \[ x+1 \; | \; x^{2006} + \ldots + 2006x + 1003 \] \[ x+1 \; | \; \frac{\partial}{\partial x} ( x^{2006} + \ldots + 2006x + 1003 ) \]However, the first is already asserted by the right-hand side of step 2, so we must only prove the second.

[b]4.[/b] Note that all non-constant terms of the polynomial on the right-hand side take the form $ n x^{2007 - n}$, for $ n \in \{ 1 \ldots 2006\}$. We can sum the terms as pairs $ n x^{2007 - n} + m x^{2007 - m}$ such that $ n+m = 2007$. Observe that one of $ n,m$ is even and the other is odd.
\[ \frac{\partial}{\partial x} n x^{2007 - n} = n(2007-n) x^{2006-n} = nm x^{m-1} \] \[ \frac{\partial}{\partial x} m x^{2007 - m} = n(2007-n) x^{2006-n} = nm x^{n-1}\]
Testing divisibility by $ x+1$ is equivalent to testing if the instance $ x=-1$ results in a root of the polynomial. Plug in $ x=-1$ to find that: \[  \; \frac {\partial}{\partial x}  | _{x=-1}  ( n x^{2007 - n} + m x^{2007 - m} )  \; = nm (-1)^{m-1}+nm (-1)^{n-1} = 0 \]
Thus, the sum of the derivatives at each pair $ n,m$ evaluates to $ 0$. Similarly, the derivative of the constant term evaluates to zero, and thus at $ x=-1$, the entire derivative is $ 0$, establishing divisibility as desired, and we are done.[/hide]

[hide]You can prove using induction that \[ \frac{p(x)}{x(x\plus{}1)} \equal{} x^{2005} \plus{} x^{2004} \plus{} 2x^{2003} \plus{} 2x^{2002} \plus{} \ldots \plus{} 1003x \plus{} 1003 \plus{} \frac{1004x}{x(x\plus{}1)}.\] The quotient is divisible by $ x\plus{}1$ and the result follows.[/hide]

[quote="TZF"]it still bothers me that I ended up using calculus.[/quote]
It shouldn't; the formal derivative is a powerful and perfectly reasonable way to tease out the behavior of a polynomial modulo the power of an irreducible polynomial.  One of the things I didn't elaborate on in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237669]this post[/url] is a certain asymmetry between how Lagrange interpolation and CRT are usually presented:  whereas CRT is stated to apply to any set of pairwise relatively prime ideals of $ \mathbb{Z}$, Lagrange interpolation tends to be restricted to the prime ideals of $ \mathbb{C}[x]$.  On the other hand, it makes just as much sense to replace questions modulo linear factors with questions modulo "powers of primes," that is, questions such as

$ P(x) \equiv a \bmod x$
$ P(x) \equiv b \plus{} c(x \plus{} 1) \bmod (x \plus{} 1)^2$

which is equivalent to asking

$ P(x) \equiv a \bmod x$
$ P(x) \equiv b \bmod (x \plus{} 1)$
$ P'(x) \equiv c \bmod (x \plus{} 1)$

which is an equally sensible interpolation problem; asking to specify the behavior of a polynomial $ \bmod (x \plus{} 1)^2$ specifies both its value at $ x \equal{} \minus{} 1$ and its slope.  This is a special case of the more general problem of defining [url=http://en.wikipedia.org/wiki/Intersection_number]intersection multiplicity[/url]; in my opinion, the generic methods of defining intersection multiplicity are actually more difficult to understand than the use of the derivative in this special case.


inequalities
geometry
circumcircle
inequalities unsolved

I had a proof for this problem , it's very nice :
Let ABC is a triangle .Prove that
$R\leq \frac{4abc\sqrt{3}.\sum a^2b^2}{18a^2b^2c^2-\sum a^2.\sum a^2b^2}$.

[quote="Dr Sonnhard Graubner"]hello, is R the radius of the circumcircle?
Sincerely,Sonnhard.[/quote]
Yes .This inequality equivalent with :
$ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq \frac{18}{a^2+b^2+c^2+4S_{ABC} \sqrt{3}}$. :D

[quote="dhthstn"]nice inequality.For all positive real numbers x,y,:
 [tex]\frac{1}{x^2+y^2+xy}+\frac{1}{y^2+z^2+yz}+\frac{1}{z^2+x^2+zx}\geq\frac{9}{(x+y+z)^2}[/tex]
In triangle ABC,let M be toroxelli point,apply this inequality for [tex]MA=x;MB=y;Mc=z..[/tex],we obtain reyes"s inequality[/quote]
Ok. :) .This's my friend's solution.I have a question for you .Ok?How do you find this solution ?


Given the numbers $ 1,2,2^2, \ldots ,2^{n\minus{}1}$, for a specific permutation $ \sigma \equal{} x_1,x_2, \ldots, x_n$ of these numbers we define $ S_1(\sigma) \equal{} x_1$, $ S_2(\sigma)\equal{}x_1\plus{}x_2$, $ \ldots$ and $ Q(\sigma)\equal{}S_1(\sigma)S_2(\sigma)\cdot \cdot \cdot S_n(\sigma)$. Evaluate $ \sum 1/Q(\sigma)$, where the sum is taken over all possible permutations.

Cute.  Here's an inductive solution:

[hide]Forget for a moment that we know some values for the $ x_i$.  I claim that the given expression is equal to $ \frac {1}{x_1\cdot x_2 \cdots x_n}$.  The proof is inductive:  it's clearly true for $ n = 1$.  Let $ S(k)$ be the set of permutations of $ \{x_1, \ldots, x_n\}$ that end in $ k$ and let $ T(k)$ be the set of permutations of $ \{x_1, \ldots, x_n\} \setminus \{x_k\}$.  Then for $ n \geq 2$,
\begin{eqnarray*} \sum_\sigma 1/Q(\sigma) & =& \sum_k \sum_{\sigma \in S(k)} 1/Q(\sigma) \\
& = &\sum_k \sum_{\sigma \in T(k)} \frac {1}{Q(\sigma)(x_1 + \ldots + x_n)} \\
& = &\frac {1}{x_1 + \ldots + x_n} \sum_{k} \sum_{\sigma \in T(k)} 1/Q(\sigma) \\
& = &\frac {1}{x_1 + \ldots + x_n} \sum_{k} \frac {x_k}{x_1 \cdots x_n} \\
& = &\frac {1}{x_1 \cdots x_n} \end{eqnarray*}
where we apply the inductive hypothesis going from the third to the fourth line.
[/hide]

That's nice. Of course, the final expression can be simplified to $ 2^{\minus{}{n\choose 2}}$.

Yes.  I wonder if there is a more direct solution, or something interesting with generating functions?  (I suppose there could also be something clever for the specific case of powers of 2, but my guess is that anything nice will apply in the general case.)

The general/normal solution is by induction.  As far as I know, this problem was posted before and solved by towerfreak with the use of induction.

Indeed, Towersfreak2006's solution at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125603 is essentially the same as mine.


Factor completely:

$ 6x^2\plus{}2x\minus{}6y^2\minus{}5xy\plus{}23y\minus{}20$

[quote="Smartguy"]Factor completely:

$ 6x^2 \plus{} 2x \minus{} 6y^2 \minus{} 5xy \plus{} 23y \minus{} 20$[/quote]

$ 6x^2 \plus{} 2x \minus{} 6y^2 \minus{} 5xy \plus{} 23y \minus{} 20\equal{}6x^2\plus{}(2\minus{}5y)x\minus{}6y^2\plus{}23\minus{}20\equal{}0$,

by solving this for $ x$ we have

$ x\equal{}\frac{\minus{}(2\minus{}5y)\pm \sqrt{(2\minus{}5y)^2\minus{}4\cdot 6(\minus{}6y^2\plus{}23y\minus{}20)}}{12}$

$ \equal{}\frac{5y\minus{}2\pm (13y\minus{}22)}{12}\equal{}\frac{3y\minus{}4}{2},\frac{\minus{}2y\plus{}5}{3}$,

so finally we have $ 6x^2 \plus{} 2x \minus{} 6y^2 \minus{} 5xy \plus{} 23y \minus{} 20\equal{}6(x\minus{}\frac{3y\minus{}4}{2})(x\minus{}\frac{\minus{}2y\plus{}5}{3})$
$ \equal{}\boxed{(2x\minus{}3y\plus{}4)(3x\plus{}2y\minus{}5)}$

using the quadratic formula would make it very messy--> easy to make a mistake

can you do guess and check?

First $ 6x^2\minus{}5xy\minus{}6y^2\plus{}2x\plus{}23y\minus{}20$
Factor--> $ (3x\plus{}2y)(2x\minus{}3y)\plus{}2x\plus{}23y\minus{}20$

Then look at the -20 and find factors to fill the following spaces:

$ (3x\plus{}2y\pm$___$ )$$ (2x\minus{}3y\pm$___$ )$

Comments?

This solution is easier, and finding factors is not to cumbersome here. But how did (or would) you find (3x+2y)(2x-3y) without looking at the previous post?

you can factor the $ 6x^2\minus{}5xy\minus{}6y^2$
right?

Okay, but I guess I'm just wondering if there was a reason that you should factor that part first... :|

it does make the problem simpler, because once you factor that part, you can look at the other part of the equation and try to factor that, less terms are easier to factor unless its 2x+23y-20 where trial and error is necessary with the exception of the quad. for.

Yes,in this problem,I believe your solution is the best.

With quadratic formula way we can reach the answers automatically.If the constant term has many dividers,it could be onerous to find the pairs.

Ok. I just didn't know the general steps you should do when facing a quadratic in 2 variables like this.

Now I get it.


A sequence $ (a_n)$ satisfies $ a_0\equal{}0,0\leq a_{k\plus{}1}\minus{}a_k\leq 1$. Prove that

$ \sum_{k\equal{}0}^na_k^3\leq (\sum_{k\equal{}0})^na_k)^2$

Correct version:

A sequence $ (a_n)$ satisfies $ a_0 \equal{} 0,0\leq a_{k \plus{} 1} \minus{} a_k\leq 1$. Prove that

$ \sum_{k \equal{} 0}^na_k^3\leq (\sum_{k \equal{} 0}^na_k)^2$

[quote="Carlez Tevos"]
A sequence $ (a_n)$ satisfies $ a_0 \equal{} 0,0\leq a_{k \plus{} 1} \minus{} a_k\leq 1$. Prove that

$ \sum_{k \equal{} 0}^na_k^3\leq (\sum_{k \equal{} 0}^na_k)^2$[/quote]
For $ n \equal{} 0$ it's true.
If $ \sum_{k \equal{} 0}^{n}a_k^3\leq \left(\sum_{k \equal{} 0}^{n}a_k\right)^2$ then $ \sum_{k \equal{} 0}^{n \plus{} 1}a_k^3\leq \left(\sum_{k \equal{} 0}^{n}a_k\right)^2 \plus{} a_{n \plus{} 1}^3.$
Hence, it remains to prove that $ \left(\sum_{k \equal{} 0}^{n}a_k\right)^2 \plus{} a_{n \plus{} 1}^3\leq\left(\sum_{k \equal{} 0}^{n \plus{} 1}a_k\right)^2,$
which is equivalent to $ a_{n \plus{} 1}^2\leq2\sum_{k \equal{} 0}^{n}a_k \plus{} a_{n \plus{} 1},$ which is still true for $ n \equal{} 0.$
If $ a_{n \plus{} 1}^2\leq2\sum_{k \equal{} 0}^{n}a_k \plus{} a_{n \plus{} 1}$ then $ a_{n \plus{} 2}^2\leq a_{n \plus{} 2}\left(a_{n \plus{} 1} \plus{} 1\right) \equal{} a_{n \plus{} 2}a_{n \plus{} 1} \plus{} a_{n \plus{} 2}\leq$
$ \leq \left(a_{n \plus{} 1} \plus{} 1\right)a_{n \plus{} 1} \plus{} a_{n \plus{} 2}\equal{} a_{n \plus{} 1}^2 \plus{} a_{n \plus{} 1} \plus{} a_{n \plus{} 2}\leq2\sum_{k \equal{} 0}^{n \plus{} 1}a_k \plus{} a_{n \plus{} 2}.$
Thus, $ a_{n \plus{} 1}^2\leq2\sum_{k \equal{} 0}^{n}a_k \plus{} a_{n \plus{} 1}$ is true for all $ n\geq0.$
Id est, $ \sum_{k \equal{} 0}^{n}a_k^3\leq \left(\sum_{k \equal{} 0}^{n}a_k\right)^2$ is true for all $ n\geq0.$

Nice...

What about if $ a_0\equal{}0$ and $ a_{k\plus{}1}\geq a_k\plus{}1$. Prove that

$ \sum_{k\equal{}1}^n a_k^3\geq (\sum_{k\equal{}1}^na_k)^2$.

[quote="Carlez Tevos"]

What about if $ a_0 \equal{} 0$ and $ a_{k \plus{} 1}\geq a_k \plus{} 1$. Prove that

$ \sum_{k \equal{} 1}^n a_k^3\geq (\sum_{k \equal{} 1}^na_k)^2$.[/quote]
I think, this is the same.


Given  S = ( 1/2 , 1/3 , 2/3, 3/4 , 5/8 , 5/16 , 15/32 )

What is the least positive difference between two distinct numbers from set S?

A.   1/48  B.    1/32  C.    1/24  D.   1/12 	E.  1/8

[hide="my solution"]Ok. So you have to pick the 2 lowest numbers...which would give you the answer as E..I don't quite know if this is right [/hide]

What about $ \frac12\minus{}\frac{15}{32}\equal{}\frac{1}{32}$? :wink:


A $9 \times 9$ square consists of $81$ unit squares. Some of these unit squares are painted black, and the others are painted white, such that each $2 \times 3$ rectangle and each $3 \times 2$ rectangle contain exactly 2 black unit squares and 4 white unit squares. Determine the number of black unit squares.

Hmm... what if we just triple-cover every square, then we can find a perfect covering, and have $3B = 3\frac{2}{6}81 = 81 \Rightarrow B=27$ so there are 27 black squares. Did I miss something? :?

The answer is correct, but I don't have a clue what you mean. Please try to write a clear solution. For example, what do you mean by "triple-cover"? And "perfect covering"?

Ah well, won't make a computer drawing for this. :)

But even easier: use the rectangles to cover all but the middle 3x3 square. We have covered 24 so far.

So the 3x3 must either have 3 black squares, or 4 in the pattern
b w b
w w w
b w b

If 4, taking the rectangle $A_{5-7,4-5}$ tells that either $A_{7,4}$ or $A_{7,5}$ is black, which gives 3 blacks in $A_{6-7,4-6}$, contradiction. So we have 3. And 24+3=27.

And such a coloring exists: put $A_{i,j}$ black iff $i=j \pmod 3$.


Let $a \in R$ s.t. $0\leq a\leq \frac{1}{n}\forall n \in Z^{+}$ then $a=0$, please explain the proof.

The Archimedian axiom tells us that for every $x>0$, there exists an integer $n$ with $n>x$.  Taking reciprocals, we see that for every $y>0$, there exists $n$ such that $1/n<y$ (since $f(x)=1/x$ maps $(0,\infty)$ to itself).

In view of the condition, we conclude that $a\leq 0$.  By the condition, $a\geq 0$.  So we must have $a=0$, and it's a trivial matter to see that $a=0$ satisfies the condition.


Let $ a_i$ be positive real numbers such that $ a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2\equal{}n$ prove that
($ S\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n$)
\[ \frac{a_1^2}{S\minus{}a_1}\plus{}\frac{a_2^2}{S\minus{}a_2}\plus{}...\plus{}\frac{a_n^2}{S\minus{}a_n}\ge \frac{n}{n\minus{}1}\]

By Chebyshev's inequality we directly get:
$ \sum_{i\equal{}1}^n \frac{a_i^2}{S\minus{}a_i}\geq \sum_{i\equal{}1}^n \frac{1}{S\minus{}a_i}$
by AM-HM this is greater than $ \frac{n^2}{(n\minus{}1)S}$
So all we need is $ S\le n$ which is true because $ n(a_1^2\plus{}a_2^2\plus{}\cdots \plus{}a_n^2)\geq S^2$


geometry
rectangle
symmetry
floor function
combinatorics unsolved
combinatorics

[hide]
hint colorise like a chessboard and make an inequality[/hide]

Continue the discussion here and lock my topic.

We remove $2(i-1)$ squares if $i\leq n$, and $2(2n-i)$ squares if $i>n$. Symmetry with respect to the line that separates $n$th row and $n+1$th  row.

We remove the corresponding number of unit squares from every line
or just from one?

Sorry for taking back this topic !
But if anyone has a solution please post it here.
Thanks...

[quote="Arberi1"][quote="shobber"]Consider a $2n \times 2n$ board.  From the $i$th line we remove the central $2(i-1)$ unit squares. What is the maximal number of rectangles $2 \times 1$ and $1 \times 2$ that can be placed on the obtained figure without overlapping or getting outside the board?[/quote]
I have tried to solve this problem for a long time.
I want to ask you if the solutions are :if $n$ is odd we have $\frac{(n-1)^2}{2}$
if $n$ is even then $\frac{(n-1)^2+1}{2}$[/quote]

The solution is certain higher:

we find for $n=1,2,3,4$ resp. $2,6,12,20$ $(=n(n+1))$
, because here we can full in all squares (there are there for each $n$ $2n(n+1)$ when we calculate that.

The way to the solution:

Divide the $2n*2n-square$ in $4$ $n*n-$suares.
In this squares, we have $\frac{n(n+1)}{2}$ litle squares, when we colour this in the chesscoloring (black, white,...),
we have ${\frac{(n+1)}{2}^2}$ black (the corner is black) and $\frac{n^2-1}{4}$ white squares.
This mean we have $\frac{n+1}{2}$ black ones more, we can only use $2$ with an other litle square of the $n*n-squares.$

So we have $2(n-3)$ litle squares we can't fill in.
We get $n^2+3$ rectangles we can set in the picture.
One remark; this formula is other for only $3.$

Use we the same strategy, we find if $n$ is even, we can fill in the whole picture
and then the answer is $n^2+4.$
Here the formula doesn't work for $2$

Edit: We can write the answer as $n(n+1)-2\lfloor{\frac{n-3}{2}}\rfloor$ where the floorfunction is 0 if the number is negative
(wording of Alberi1)

Problem assumes that we remove $2(i-1)$ squares if $i\leq n$, and $2(2n-i)$ squares if $i>n$.

Divide the entire board into 4 quadrants each containing $n^2$ unit squares.

First we note that the $2$ squares on the center on each of the $4$ bordering lines of the board can always be completely covered by a single tile, so we can count in the first and last unit squares (which are diagonally opposite) of each quadrant as being filled in completely by a tile.

So in each quadrant we have:

if $n$ is even, there are exactly $(n-4)/2$ unit squares which cannot be filled by the tiles and if $n$ is odd, there are exactly $(n-3)/2$ unit squares which cannot be filled by the tiles.

Above can be seen by drawing a diagram and noticing that alternate columns have even and odd number of unit squares (leaving a unit square uncovered by tiles in odd numbered blocks of columns).

Also, note that the total number of unit squares which were removed from each quadrant = $(1 + 2+ 3 + ... n-1) = n(n-1)/2$

Let us consider the 2 cases for parity of $n$:

$Case 1: n$ is even

for $n = 2$, it can be seen easily that we can use a maximum of $6$ tiles.

for $n \ge 4:$
Total number of squares that cannot be filled in each quadrant is:
$n(n-1)/2 + (n-4)/2 = (n^2 - 4)/2$
So total number of squares that cannot be filled on the entire board = $2(n^2 - 4)$
So total number of squares that CAN be filled completely by the tiles = $4n^2 - 2(n^2 - 4) = 2n^2 + 8$
So the maximum number of tiles that can be used = $(2n^2 + 8)/2 = n^2 + 4$

$Case 2: n$ is odd

for $n = 1$, it can be seen easily that we can use a maximum of $2$ tiles.

for $n \ge 3:$
Total number of squares that cannot be filled in each quadrant is:
$n(n-1)/2 + (n-3)/2 = (n^2 - 3)/2$
So total number of squares that cannot be filled on the entire board = $2(n^2 - 3)$
So total number of squares that CAN be filled completely by the tiles = $4n^2 - 2(n^2 - 3) = 2n^2 + 6$
So the maximum number of tiles that can be used = $(2n^2 + 6)/2 = n^2 + 3$


1!+1+2!+2+3!+3.....8!+8+9!+9 what is the units digit? this is a countdown question so its not exactly effective if i write it all out is it?
thanks

Pick at the little pieces you know.  First, break in into pieces you like.  Then, think about what you know about last digits of factorials.

[hide]8

Since 5! on ends in a 0, then the units digit for 1! + 2! + 3! + 4! is 3

1+...+9's unit digit is 5... so 8[/hide]

Is that supposed to be 1!+1+2!+2+..., or should it be 1!*1+2!*2+3!*3+...? The latter occurs quite frequently on math contests.


Thank you, Andreas.
However, can I have
3^105 + 4^105 = 5Q + R, where R is a product of prime factors except 5,
as the answer?
Also, I don't know what is "mod". Could you tell me what it is? Thanks. =]

Besides, how can I put the LATEX in my posts instead of in the form of an attachment?

Sorry for being troublesome. =]

It's clear that $p+q\geq 0$.  So $p^3+q^3=(p+q)(p^2-pq+q^2)=(p+q)((p+q)^2-3pq)\geq \frac{1}{4}(p+q)^3$ and the rest is trivial.

[quote="M4RI0"]In number 2, you just have to use the power mean inequality. :P[/quote]

What if $p$ and $q$ are not both positive?  Then it requires some more work.  My method avoids that difficulty.


Prove that every positive rational number can be represented in the form \[\frac{a^{3}+b^{3}}{c^{3}+d^{3}}\] for some positive integers $a, b, c$, and $d$.

For a positive rational $\frac{m}{n}$, find another positive rational $\frac{s}{t}$ such that

$\frac{t^3m}{s^3n}=\frac{p}{q}$ lies between $\frac12$ and $2$. Such a rational clearly exists.

Then $\frac{p}{q}=\frac{(p+q)^3+(2p-q)^3}{(p+q)^3+(2q-p)^3}$, and all four cubes are positive.

Then just multiply back, giving $\frac{m}{n}=\frac{(s(p+q))^3+(s(2p-q))^3}{(t(p+q))^3+(t(2q-p))^3}$.


I have students that are preparing for the Pythagoras Contest which is a nation wide Grade 6 math challenge, Do you have any Past tests?

There is some prepatory material on the sponsor's site:

[url]http://www.mathematica.ca/eng/prep.htm[/url]


Hi!

I need help to solve this problem:

Given: Two Circles k1 and k2, that meet in A and B. The tangent to k1 in A meets k2 in C2. The tangent to k2 in A meets k1 in C1. The line C1C2 meets k1 in D.

Proof that the line BD bisects the chord AC2.

Greetings,
Ulysses

STOP !

The problem is exactly problem 3 of the ONGOING 2nd Round of the BWM 2004 (BWM = Bundeswettbewerb Mathematik = German national mathematical competition).

DON'T POST SOLUTIONS !

  Darij

Well, cool, i am from italy. I got this problem from an old book called "mathematical problems". But in the book, there're no solutions for the problems.

Could you give me the name, authors and publishing house of the book, please?

[b]PS.[/b] Your first posting (before you edited it) contained a link to a figure for the problem on an Austrian provider (I'm sad I didn't save the link); so I may assume that you are an Italian with some close relationship to German-speaking countries. Maybe you can even read German? If so, I am delighted to announce you that on 2/9/2004 a detailed solution of the problem will appear on [url=http://de.geocities.com/darij_grinberg/Dreigeom/Inhalt.html#bwm]my website[/url]. I am almost sure your patience will suffice for this one week.

  Darij

Sure, i can: "problemi matematici" by "Damiano Scarponi" published 1951 in italy.

You're right... i live close to austria, but my german is quite bad, it is as bad as my english :-)
It would be great if you offered a detailed solution on your website :-)
Thank you,

Ulysses

PS: you're quite good in math, aren't you?
[quote]
A brief note about me: I am a school student (gymnasiast) in Karlsruhe, Germany and participant of several mathematical olympiads like the Bundeswettbewerb Mathematik, the Landeswettbewerb Mathematik Baden-Wrttemberg and the Deutsche Mathematik-Olympiade. Have won first prizes in the 1st Rounds of the Bundeswettbewerb Mathematik 2001, 2002, 2003 and 2004, as well as in the 2nd Round of the Bundeswettbewerb Mathematik 2003 (where I gained third prizes in 2000 and 2002), and became Bundessieger 2003. Participated in the "Deutsche Mathematik-Olympiade" since 2002, and gained a 1st prize in the 41th olympiad 2002 (Hamburg), an honorable mention in the 42th olympiad 2003 (Bremen), and a 2nd prize in the 43th olympiad 2004 (Essen). Silver medal in the 45th IMO 2004.[/quote]
congratulations! :-)

[quote="Ulysses"]It would be great if you offered a detailed solution on your website :-)[/quote]

It will be detailed, actually it will be the solution I submitted to the olympiad. I will also post my solution on the forum (in English). But I'll do this all on 2 September, when the deadline will be over.

What is really strange is that this problem is really identic with the 3rd problem of the BWM 2004, 2nd round. The wording is almost equal, and the notations are completely the same. Maybe the BWM guys have their secret sources...

Thanks for the congratulations  ;)

  Darij

I'm going to lock this thread; you can remind a mod to unlock it once the contest is over.

I think it would be best to lock this thread until the contest is over (just in case someone solves the problem without looking at Darij's post, thinking it would contain a solution). Concur?

Okay, now the contest is over, let me now announce that [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16201]the "right" thread contains some solutions to this problem[/url].

  Darij


modular arithmetic
number theory
prime numbers
number theory unsolved

6. Find all prime numbers p,q for which pq divides (5^p-2^p)(5^q-2^q).

A solution is given to this problem, but there is a part of it that I don't understand:

"If p|5^p-2^p, then p|5-2 by Fermat's theorem, so p=3. Suppose p,q!=3; then p|5^q-2^q and q|5^p-2^p. Without loss of generality, assume p>q, so that (p,q-1)=1." Continuing, "Then if a is an integer such that 2a==5(mod q), then the order of a mod q divides p as well as q-1, a contradiction."

The proof continues to give the solutions. But I do not understand this statement: how can you justify that the order of a mod q divides p and q-1?

By the way, the symbol "!=" means "not equal to" and the symbol "==" means "is congruent to."

Could anyone please tell me what kind of theorem about orders justifies this statement?

The order of $a \mod q$ must divide $q-1$ since $a^{q-1} \equiv 1 \pmod{q}$ by Fermat's Little Theorem. By our definition of $a$ as $2a \equiv 5 \pmod{q}$, raising both sides of the congruence by a power of $p$, we get $(2^p)(a^p) \equiv 5^p \pmod{q}$, but we earlier have $q \mid 5^p-2^p \Leftrightarrow 5^p \equiv 2^p \pmod{q}$, thus we have $a^p \equiv 1 \pmod{q}$, thus the order of $a \mod q$ must divide $p$, and consequently we have the contradiction as desired, since $a \neq 1$.


Hi Guy's ... Did anybody have a 4x4x4 Rubik's Revenge Cube solution ? ( In PDF file or u can email me at kofxi123@gmail.com ) Thank's a million.... :blush:  :P

You could always search for it. Try this:

http://www.jaapsch.net/puzzles/cube4.htm


[quote="ReyBoB"]Two regular polygons have sides of 6 inches and 8 inches respectively. Find the ratio of their perimeters.

I am totally confused on how to answer this. :([/quote]

Is there some info missing? i don't think  you can get a definite answer with that much info. unless, of course, you want to take the number of sides as $x$ and $y$.

I guess the polygons have the same number of sides?

could always assume one has n sides and the other m sides and then multiply by the 6 and 8 respectively and divide.  If they have the same number of sides then its trivial


Let $ ABC$ be a triangle such that $ \angle ABC \equal{}120$.
The foot of the bisector of $ \angle ABC$ is $ L$.
$ K$ is a point on $ BC$ such that $ \angle KAC \equal{} 30$.
Prove that $ AK$ is a bisector of $ \angle BKL$.

Let $ BM$ be $ \angle LBC$ bisector and $ P$ the intersection point of $ BM$ and $ AK$. We have $ \angle PBL\equal{}\angle PAL$ hence points $ B,P,L,A$ lie on a circle and $ \angle KPL\equal{}120^{\circ}$. We also see that $ BKMA$ is inscribed quadrilateral hence $ \angle KMC\equal{}120^{\circ}$ which means that $ PKML$ is inscribed => $ \angle BKA\equal{}\angle BMA\equal{}\angle PKL$ QED

[quote="Ligouras"][quote="Nak"]Let $ ABC$ be a triangle such that $ \angle ABC \equal{} 120$.
The foot of the bisector of $ \angle ABC$ is $ L$.
$ K$ is a point on $ BC$ such that $ \angle KAC \equal{} 30$.
Prove that $ AK$ is a bisector of $ \angle BKL$.[/quote]

___________
Can I add the following question (NAK agrees?): Prove or disprove that $ CK \equal{} CL$[/quote]

$ CK \neq CL$ becouse if $ AB < BC$ then there is a point $ T$ on $ [AB]$ such $ AL\equal{}AT$ but $ \angle TCA < 30$.

Next problem (harder than first)
Let $ ABC$ be an equilateral triangle with center $ O$.
Line $ l$ passes throught $ O$ intersects $ AB$ and $ AC$ at $ P$ and $ Q$ respectively.
$ P'$, $ Q'$ are points on $ BC$ such that $ BP'\equal{}BP$, $ CQ'\equal{}CQ$.
Prove that there is a point $ K$ on $ l$ such that $ P'K\equal{}P'B$, $ Q'K\equal{}Q'C$.

this problem is solving with harmonic conjugate. D is intersection of BL and AK. E is intersection of KL and AB. F is intersection ED and AC. therefore A,F,L,C is harmonic conjugate. if CB is exterior bisector, BF is right to BC. and BDFA is cyclic quadrilateral.<FBA=<FDA=30 and <EFC=60=<CBE. so that EBFC is also cyclic quadrilateral.and EC is right to EF. if <CEF=90, EF is bisector of <LEA. in triangle EBK, ED is interior bisector, BD is exterior bisector, KD have to exterior bisector.
so that AK is a bisector of <BKL.


[color=darkblue]Slove this equation: $ \sqrt[3]{x^2\plus{}7}\plus{}3\sqrt{2x\minus{}1}\plus{}\sqrt{x}\equal{}6$[/color]

[quote="bunhiacovski"][quote="thanhnam2902"][color=darkblue]Slove this equation: $ \sqrt [3]{x^2 \plus{} 7} \plus{} 3\sqrt {2x \minus{} 1} \plus{} \sqrt {x} \equal{} 6$[/color][/quote]
determination $ x$ be  $ x\ge \frac {1}{2}$

the equation equivalence to:

$ (\sqrt [3]{x^2 \plus{} 7} \minus{} 2) \plus{} 3(\sqrt {2x \minus{} 1} \minus{} 1) \plus{} (\sqrt {x} \minus{} 1) \equal{} 0$

<=> $ \frac {x^2 \minus{} 1}{(x^2 \plus{} 7)^{2/3} \plus{} 2(x^2 \plus{} 7)^{1/2} \plus{} 4} \plus{} \frac {6(x \minus{} 1)}{(2x \minus{} 1)^{1/2} \plus{} 1} \plus{} \frac {x \minus{} 1}{x^{1/2} \plus{} 1} \equal{} 0$

<=> $ (x \minus{} 1)[\frac {x \plus{} 1}{(x^2 \plus{} 7)^{2/3} \plus{} 2(x^2 \plus{} 7)^{1/2} \plus{} 4} \plus{} \frac {6}{(2x \minus{} 1)^{1/2} \plus{} 1} \plus{} \frac {1}{x^{1/2} \plus{} 1}] \equal{} 0$

$ \frac {x \plus{} 1}{(x^2 \plus{} 7)^{2/3} \plus{} 2(x^2 \plus{} 7)^{1/2} \plus{} 4} \plus{} \frac {6}{(2x \minus{} 1)^{1/2} \plus{} 1} \plus{} \frac {1}{x^{1/2} \plus{} 1} > 0$ with $ x\ge \frac {1}{2}$

thus, $ x \minus{} 1 \equal{} 0$ equivalence to $ x \equal{} 1$[/quote]
[color=darkblue]It's nice. Thank you very much.[/color]

$ x\ge 1/2$. The L.H.S. is clearly strictly increasing. Therefore, only one solution exists. Guess x=1. Q.E.D.


geometry
3D geometry
algebra
polynomial
group theory
abstract algebra

Find all perfect squares that can be written as the sum of three consecutive perfect cubes.

[hide="Partially somewhere?"]
$ b^2 \equal{} (c\minus{}1)^3\plus{}c^3\plus{}(c\plus{}1)^3 \equal{} 3c^3 \plus{} 6c \equal{} 3c(c^2\plus{}2)$, so $ b$ must be divisible by three.  Let $ b\equal{}3b'$, where $ b'$ is an integer.
$ 9b'^2 \equal{} 3c^3\plus{}6c$
$ 3b'^2 \equal{} c^3\plus{}2c$.
[/hide]

My approach:

[hide]In general, if the first of the consecutive cubes is k^3, we have k^3 + (k+1)^3 + (k+2)^3.  Since these are consecutive, this can be written with the summation powers of cubes: ((k+2)^2*(k+3)^2)/4 - ((k-1)^2*k^2)/4.  Simplifying yields a polynomial function.

Now we cheat and use a calculator to find values for which this holds true.

If we consider 0 and 1 as perfect cubes, 9 and 36 can be written as this sum.

23, 24, and 25 yield 204

I went farther on my calculator but the values all seemed to be rounded.[/hide]

[quote="mysmartmouth"][hide]
Let the perfect cubes be $ (x \minus{} 1)^3$, $ (x)^3$, and $ (x \plus{} 1)^3$.

Then $ (x \minus{} 1)^3 \plus{} (x)^3 \plus{} (x \plus{} 1)^3 \equal{} k^2$ for some integer k.

This simplifies to $ 3x(x^2 \plus{} 2) \equal{} k^2$

I had a mistake in there... it's fixed now but I don't know where to go from here.
[/hide][/quote]

It might help to analyze $ 3x$ and $ x^2 \plus{} 2$ in terms of their factors...

But I did that for a while and only got two values for x.  The third I found using a calculator.

Moral: Diophantine equations= calculator abuse.

Once you expand and combine like terms you'll have $ n^2 \equal{} 3k(k^2 \plus{} 2)$. After a little musing I subtracted $ 9k^2$ from both sides so I could factor the thing. That gave me
\[ (n \minus{} 3k)(n \plus{} 3k) \equal{} 3k(k \minus{} 1)(k \minus{} 2)
\]
Obviously the RHS is zero for $ k \equal{} \{0,1,2\}$. Using that in conjunction with $ n^2 \equal{} 9k^2$, the three ordered pairs $ (k,n)$ : $ (0,0)$, $ (1,9)$, $ (2,36)$ are easily found. I also performed some subsitution letting $ \pm k \equal{} \pm \frac {n}{3}$. I completed a square and got
\[ n((n \minus{} 9)^2 \minus{} 63) \equal{} \minus{} 81k^2
\]
Looking at that thing we can say exactly where the LHS is negative and a not so bad approximation is $ n\in (2,16)$. It's equivalent to finding integer solutions to
\[ (u \plus{} 9)(u^2 \minus{} 63) \equal{} \minus{} 81k^2
\]
with $ u$ ranging the integers from 0-7. Doesn't look like there are any.

That's all I've got I don't know how to extract n=204 from that.
[b]edit:[/b] Actually I made a mistake. If someone wants, plug and chug the u-k equation again. I forgot that the u on the outside isn't squared so that butchered all of my computations.  :D

If $ k$ is odd, then $ (k,k^2 \plus{} 2) \equal{} (k,2) \equal{} 1$.

Thus $ 3k \equal{} m^2$, $ k^2 \plus{} 2 \equal{} p^2$

(easy to factor and solve)

OR

$ k \equal{} m^2$, $ 3(k^2 \plus{} 2) \equal{} p^2$

$ m^4\plus{}2\equal{}3p'^2$... $ (m^2\plus{}1)(m\plus{}1)(m\minus{}1)\equal{}3(p'\minus{}1)(p'\plus{}1)$

not much help...

if $ k$ is even there are even more cases...

The reason elementary techniques aren't working is that this is a nonsingular [url=http://planetmath.org/encyclopedia/ArithmeticOfEllipticCurves.html]elliptic curve[/url].  In particular, it has Weierstrass form $ y^2 \equal{} k^3 \plus{} 18k$.  If you'd like, I can give a rather computational solution using those tools.

Edit:  Well, not really a solution.  The techniques to exhaustively find all the integer points on an elliptic curve aren't very elementary.  What I can tell you is that Siegel's Theorem guarantees that the number of integer points on a cubic curve is finite.  A short calculation demonstrates that the above curve has [url=http://en.wikipedia.org/wiki/Torsion_subgroup]torsion subgroup[/url] $ \{\mathcal{O}, (0, 0) \}$ and [url=http://planetmath.org/encyclopedia/RankOfAnEllipticCurve.html]rank[/url] at least $ 1$ (since it has the non-torsion point $ (3, 9)$).  I believe the point $ (72, 612)$ is independent, which would give this curve rank at least $ 2$.  (The point $ (6, 18)$ is not independent; $ (3, 9) \plus{} (0, 0) \equal{} (6, \minus{}18)$.)

Thank you all for your replies, and t0rajir0u for your explanation :) And my applogies for inaccurately guaging the difficulty of the problem.


Has any pianist here played a piece from Prokofieff's Romeo and Juliet, the score he transcribed from his ballet? All of the pieces are just phenomenal music. My favorite is the Montagues and Capulets.

Cool, I played Capulet and Montague...It's a great piece...

Difficult though. I love it, been playing it for about six months and it's almost perfected. I wish I had more time to spend at the piano bench.

I like #10!

I've played 2, 3, 4, and 10.


geometry
circumcircle
trigonometry
geometric transformation
reflection
geometry proposed

Let $MNP$ be an acute-angled triangle. We denote by $O$ the circumcenter of the triangle given.
 We suppose that $MO$ and $NP$ intersect in $K$ and we also suppose that on the edges $[MN]$ and $[MP]$ of the
  triangle, we draw the points $A$ and $B$ such that $KA=KN$ and $KB=KP$. Show that $ABNP$ is a trapezoid. ;)

Caesar, the your proposed problem is nice, but easily and wellknown !

[hide="The first (metrical) method."]
$\frac{KN}{KP}=\frac{c\sin \widehat {NMK}}{b\sin \widehat {PMK}}=\frac{c\cos P}{b\cos N}$,

$AN=2KN\cos N$, $BP=2KP\cos P$ $\Longrightarrow$

$\frac{AN}{BP}=\frac{KN}{KP}\cdot \frac{\cos N}{\cos P}=\frac{c\cos P}{b\cos N}\cdot \frac{\cos N}{\cos P}=\frac cb=\frac{MN}{MP}$ $\Longrightarrow$

$\frac{MN}{MP}=\frac{AN}{BP}\Longrightarrow AB\parallel NP.$ [/hide]
[hide="The second (syntetical) method."]
I note the reflection $M'$ of the point $M$ w.r.t. the centre $O$

and the middlepoints $U$, $V$ of the segments $AN$, $BP$ respectively.

$KA=KN$, $KB=KP$ $\Longrightarrow$ $KU\perp AN$, $KV\perp BP$ $\Longrightarrow$

$KU\parallel M'N$, $KV\parallel M'P$ $\Longrightarrow$

$\frac{MU}{MN}=\frac{MK}{MM'}=\frac{MV}{MP}$ $\Longrightarrow$ $\frac{MU}{MN}=\frac{MV}{MP}\Longrightarrow$

$\frac{MA+MN}{2MN}=\frac{MB+MP}{2MP}\Longrightarrow\frac{MA}{MN}=\frac{MB}{MP}$ $\Longrightarrow$ $AB\parallel NP.$
[/hide]


given a positive integer, consider the set
$ A_n$={$ 1 \le a \le n: gcd(a,n)\equal{}gcd(a\plus{}1,n)\equal{}1$}
prove that the product of all the element of $ A_n$ congruent 1 modulo n

il prove that if $ a$ satisfies$ gcd(x(x\plus{}1),n)\equal{}1$, then so does the multiplicative inverse of $ a$ mod n
Let b be the inverse of a, $ ab\equal{}1(modn)$. Then, obviously $ gcd(b,n)\equal{}1$. Now observe tht $ a(b\plus{}1)\equal{}1\plus{}a (modn)$. Since $ gcd(a\plus{}1,n)\equal{}1$ it follows that $ gcd(b\plus{}1,n)\equal{}1$. Hence $ gcd(b(b\plus{}1),n)\equal{}1$

Now, if a number which is its own inverse satisfies the condition, say $ t$, then $ t^2\equal{}1(mod n)$. Since $ gcd(t\plus{}1,n)\equal{}1$ which forces $ n$ to divide $ t\minus{}1$. Since $ t<n$, we must have $ t \equal{} 1$

We're done


Scuze, de fapt:
calculati limita sirului $ \displaystyle\lim_{n\to\infty}\sum_{k \equal{} 1}^n \frac {1}{k2^k}$.
$ \displaystyle\sum_{k \equal{} 1}^n \frac {k}{2^k}$ se poate calcula prin inmultire cu 1/2.

Sirul tinde la infinit.Poti sa observi ca $ \frac {2^k}{k} > 1$ , $ \forall k\geq 1$

Dar ce zici de $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \frac{k}{2^k}$ ?  :lol:

[quote="vador"]
calculati limita sirului $ \displaystyle\lim_{n\to\infty}\sum_{k \equal{} 1}^n \frac {1}{k2^k}$.
.[/quote]

Pentru $ x \in (0,1)$ avem:

$ \frac { \minus{} 1}{x \minus{} 1} \equal{} \sum_{k \equal{} 0}^{\infty} x^k$ $ \Rightarrow$ $ \int \frac { \minus{} 1}{x \minus{} 1}dx \equal{} \int \sum_{k \equal{} 0}^{\infty}x^kdx$ , deci

$ \int \frac { \minus{} 1}{x \minus{} 1}dx \equal{} \sum_{k \equal{} 0}^{\infty} \frac {x^{k \plus{} 1}}{k \plus{} 1}$, sau scrisa sub alta forma

$ \lim_{n\to\infty} \sum_{k \equal{} 1}^n \frac {x^k}{k} \equal{} \minus{} \ln|x \minus{} 1|$, si pentru $ x \equal{} \frac {1}{2}$ rezulta ca limita cautat e $ \ln 2$.

Sau poti sa folosesti formula "de-a gata", ca sa zic asa dezvoltand dupa formula lui Taylor. vezi [url=http://en.wikipedia.org/wiki/Taylor_series]aici[/url].(parca se cere formula lui Taylor la cls 11 pt olimpiada, eu am incercat sa fac o justificare destul de naturala pentru formula.)


[quote="Albanian Eagle"]Show that if $a,b,c$ are any three positive numbers
$(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\geq (\sqrt{ab}+\sqrt{bc}+\sqrt{ca})^{3}$[/quote]
$(a^{2}+a+1)(b^{2}+b+1)\geq\left(a+b+\sqrt{ab}\right)^{2}.$
Thus, $(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)\geq(a+b+\sqrt{ab})(a+c+\sqrt{ac})(b+c+\sqrt{bc})\geq$
$\geq\left(\sum_{cyc}\sqrt[6]{ab}\cdot\sqrt[3]{a}\cdot\sqrt[3]{b}\right)^{3}=\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)^{3}.$


trigonometry
function
quadratics
algebra
quadratic formula
trig identities
Law of Cosines

A teacher asks his pupil to find the greates angle of a triangle with sides 21, 41 and 50. The pupil "solves" the problem as follows :

[i]Let C be the vertex that corresponds with the greatest angle.
Then sin(<C) = 50/41 = 1,21951...
1 corresponds to sin 90 and 0,21951 corresponds to sin 1240'48".
Hence <C = 90 + 1240'48" = 10240'48".[/i]

Of course this method is ridiculous ... but ... (surprise) :):
THE ANSWER IS CORRECT !

(1) Explain why this method has to give the correct answer, although (mathematically) it does not make sense.
(2) Find out whether there exist other triangles, not similar to the given triangle, for which this method works.

What an interesting problem!  It turns out there are many other triangles with this property, including the very similar 20-41-49 triangle.  Thanks!

I appear to have thrown out the scraps of paper I worked on yesterday, but I can give you the gist of the solution, but I didn't get a general form for all applicable triangles.  Basically,

[hide]use the law of cosines to find cosC as a function of a, b, and c.  For an obtuse triangle, you want cosC to be equal to (1 - c/b), because of a cool trigonometric identity that turns up, in which arcsin(x) + arccos(x) = 90.  This was a reasonably complicated equation, which is a quadratic in terms of all of a, b, and c, so I ended up solving for a (using the quadratic formula) in terms of b and c, then made an Excel spreadsheet to try a bunch of cases to find another solution.  There are lots of solutions!  I thought they'd be less common.[/hide]


Find all n such as n divide 2^n-1.(n is possitive integer.) :)

[quote="Rust"]Only n=1.
If n>1, let p is minimal prime divisor p|n, and T(p) minimal natural period ($T(p)=min(m| \ p|(2^{m}-1))$. Then $T(p)\not =1$ (p>2) and $T(p)|n$ give contradition.[/quote]

can you explain clearly what is $T(p)$ ? I cant get what you mean . :maybe:

T(p) is minimal natural m, suth that $p|(2^{m}-1).$ For example $T(7)=3$. T(p) is divisor p-1, therefore $T(p)\le p-1$.


geometry
perimeter
algebra
polynomial
trapezoid
modular arithmetic
quadratics

No calculators.

1. [4] Determine $x$ if \[x-\frac{x}{x-\frac{x}{x-\frac{x}{x-\cdots}}}= \frac{x}{x-\frac{x}{x-\frac{x}{x-\cdots}}}.\]

2. [6] If the perimeter and hypotenuse of a right triangle are 14 and 6, respectively, determine its area.

3. [8] The roots of the polynomial $ax^{3}+bx^{2}+cx+a$ form a geometric series, with $a\ne 0$. Find $b/c$.

4. [9] $O$ is the center of the inscribed circle in $30^\circ-60^\circ-90^\circ$ $\triangle ABC$ with the right angle at $C$. If the circle is tangent to $AB$ at $D$, find $\angle COD$.

5. [10] Define a sequence of numbers $a_{1},a_{2},\ldots,a_{n}$ by $a_{1}=9$ and $a_{n}= a_{n-1}^{2}-1$, $n\ge 2$. Find the smallest $n$ such that $N$ is the multiple of 10, where \[N = \sum_{k=1}^{n}a_{k}.\]

6. [11] In isosceles trapezoid $ABCD$, the sum of the bases $AB$ and $CD$ is 3, and $\angle ACD=60^\circ$. Find the length of diagonal $AC$.

7. [12] Find all integer pairs $(m,n)$ such that \[(m+n)^{4}-(m-n)^{4}= 256.\]

8. [13] Twelve points are equally distributed on the circumference of a circle. How many right triangles have three of these points as their vertices?

9. [13] Alex writes the number 16 on a blackboard. He then performs several "replacements" in which he erases the current number and replaces it with one of its factors. In how many ways can Alex end up with the number 1 after five replacements?

10. [14] $a,b,c,d,x,y,z$ are real numbers such that

\begin{eqnarray*}ax+by+cz &=& d,\\ bx+cy+dz &=& a,\\ cx+dy+az &=& b,\\ dx+ay+bz &=& c. \end{eqnarray*}

If $a\ne c$, find all possible values of $y$.

[hide="#8"]

An angle inscribed in a circle is a right angle if and only if it subtends a semicircle.  Thus the opposite side is a diameter.  There are 6 diameters among the 12 points.  For each there are 10 other points to be the right angle.  So $6*10 = 60$ right triangles.

[/hide]

[quote="E^(pi*i)=-1"][quote="7h3.D3m0n.117"][hide="#1"]

$x=4$ lol after some explanation from chess.....thought it was 0 the first time   :oops:
[/hide][/quote]

Please post a solution.

[hide="#5"]
Note that $9 \equiv-1 \pmod{10}$.  $(-1)^{2}-1 = 0$, so $a_{2}\equiv 0 \pmod{10}$.  Similarly, $0^{2}-1 =-1$, so $a_{3}\equiv-1 \pmod{10}$.  Thus the sequence is $-1, 0,-1, 0,-1, 0, \cdots \pmod{10}$.  We need $10$ $-1$'s to get down to $0 \pmod{10}$.  Since the sequence starts with $-1$, we need 19 terms and $N = 19$.
[/hide][/quote]

[hide]$x=4$ isn't a solution?[/hide]

:rotfl: Sorry.
[hide]$x=4$ [/hide]
is the answer to the problem.  I meant that you should post how you solved the problem instead of just the answer.

The answer for 2 is correct but it isn't necessary to invoke the quadratic equation. Take sides $a,b,6$ and let $A$ denote its area. Then

\begin{eqnarray*}a+b+6&=&14,\\ a+b&=&8,\\ a^{2}+b^{2}+2ab&=&8^{2},\\ 6^{2}+4A&=&64,\\ A&=&\fbox{7}. \end{eqnarray*}

[quote="wonki_noh"]Apologies for not using Latex. my computer has some problems, and i cannot install latex. I tried few times and i kind of gave up installing it.
Anyway,
[hide="#2"]
Hypotenuse=6, so the other sides are x and (8-x).
(1) Using Pythagoras,
   x^2+(8-x)^2=36
   x^2-8x+14=0
Using quadratic formula,  x=4+Square root(2) or 4-Square root(2)
       two x-values will give the same answer.

(2) the area of the triangle=1/2*(x)(8-x)
                                      =1/2*(4-square root(2))(4+square root(2))
                                      =1/2*(16-2)
                                      =7 square units.
[/hide][/quote]

you do not need to install LaTeX in order to program it on forums like AoPS. just use the $\$$ signs for programming

[quote="wonki_noh"]serious??

can you do the latex writing,, without actually installing it?

could you please tell me how??[/quote]

The forum is LaTeX enabled.  Just type as you normally would in LaTeX and it will work.

[hide="Here is my solution to #1"]Define

\[y = \frac{x}{x-\frac{x}{x-\frac{x}{x-\cdots}}}.\]
Then
\[y ={x\over y}\ \ \Rightarrow\ \ y^{2}= x.\]
From the original equation:
\[x-y = y\ \ \Rightarrow\ \ x=2y.\]
So we have $2y = y^{2}\ \ \Rightarrow\ \ y=2.$  This gives
\[x=4.\]
[/hide]

[hide="#4"]
We know that a tangent to a circle is tangent to the radius of the circle, so $m\angle ODB=90^{\circ}$.  Say that the circle is tangent to $BC$ at $E$ and to $AC$ at $F$.  Clearly $\angle COD=\angle COE+\angle EOD$.

In quadrilateral $EODB$, $m\angle OEB=m\angle ODB=90^{\circ}$ and $m\angle DBE=60^{\circ}$.  Thus $m\angle EOD=120^{\circ}$.

In quadrilateral $CFOE$, $m\angle CFO=m\angle ECF=m\angle OEC=90^{\circ}$, thus the quadrilateral is a rectangle.  Since $EO=FO$, both being radii of the circle, the quadrilateral is in fact a square. Since $CO$ is a diagonal, $m\angle COE=45^{\circ}$.

Thus $m\angle COD=m\angle COE+m\angle EOD=120^{\circ}+45^{\circ}=165^{\circ}$.
[/hide]

[hide="#7"]
So expanding we see that this is $(m^4 + 4m^3n + 6m^2n^2 + 4mn^3 + n^4) - (m^4 - 4m^3n + 6m^2n^2 - 4mn^3 + n^4) = 256$, so $8m^3n + 8mn^3 = 256$, and $mn(m^2 + n^2) = 32$.  Now we test factors and casework.  We find that since $m^2$ and $n^2$ must be positive, $mn$ must be positive and all we need to look for are positive factors.

If $mn = 1$, then $m^2 + n^2 = 32$, so adding $2mn + m^2 + n^2$ yields $34$ which is not a perfect square.  So we can throw out this case.

If $mn = 2$ then $m^2 + n^2 = 16$, so adding yields $m^2 + 2mn + n^2 = 20$ which is not a perfect square.  So we can throw out this case.

If $mn = 4$ then $m^2 + n^2 = 8$, so adding yields $m^2 + 2mn + n^2 = 16$ which is a perfect square!  Hence $m + n = \pm 4$.  If $m + n = 4$ then $m = 4 - n$, so $4n - n^2 = 4$ and $(n - 2)^2 = 0$, hence $n = 2$ and $m = 2$.  $\boxed{(2,2)}$.  If $m + n = -4$ then $m = -4 - n$, so $-4n - n^2 = 4$, hence $n^2 + 4n + 4 = 0$ and $n = -2$, so $m = -2$.  $\boxed{(-2,-2)}$.

If $mn = 8$ then $m^2 + n^2 = 4$, so adding yields $m^2 + 2mn + n^2 = 20$ which is not a perfect square.

If $mn = 16$ then $m^2 + n^2 = 2$, so adding yields $m^2 + 2mn + n^2 = 34$ which is not a perfect square.

If $mn = 32$ then $m^2 + n^2 = 1$, so adding yields $m^2 + 2mn + n^2 = 65$ which is not a perfect square.

So we are done :)
[/hide]


conics
parabola
analytic geometry
linear algebra
matrix
algebra unsolved
algebra

Consider all parabolas of the form $ y\equal{}x^2\plus{}2px\plus{}q$ for $ p,q \in \mathbb{R}$ which intersect the coordinate axes in three distinct points. For such $ p,q$, denote by $ C_{p,q}$ the circle through these three intersection points. Prove that all circles $ C_{p,q}$ have a point in common.

The points of intersection are $ A(0,q), \ B(x_1,0), \ C(x_2,0)$, where $ x_1, x_2$ are the roots of the equation $ x^2\plus{}2px\plus{}q\equal{}0$

The equation of the circle passing through A, B, C is

$ \begin{vmatrix}x^2\plus{}y^2 & x & y & 1\\
q^2 & 0 & q & 1\\
x_1^2 & x_1 & 0 & 1\\
x_2^2 & x_2 & 0 & 1\end{vmatrix}\equal{}0\Rightarrow (x_1\minus{}x_2)(q^2y\plus{}qy\minus{}q(x^2\plus{}y^2)\minus{}q^2\minus{}2px)\equal{}0$

For $ x\equal{}0, \ y\equal{}1$ the equality is true $ \forall p,q$.

    I claim that the point $(0, 1)$ is on all circles $C_{p, q}$.

    The circle $C_{p, q}$ must pass through the three points
    $$
    (-p-\sqrt{p^{2}-q}, 0), \quad (-p+\sqrt{p^{2}-q}, 0), \quad \text{and} \quad (0, q).
    $$

    Some computation yields that the circle has the center $O$ and radius $r$ with
    $$
    O = \left(-p, \frac{1 + q}{2}\right), \quad \text{and} \quad r = \frac{1}{2} \sqrt{4 p^{2}+(q-1)^{2}}.
    $$
    giving the equation of the circle
    $$
    (x + p)^2 + \left(y - \frac{1 + q}{2}\right)^2 = \frac{4p^2 + (q - 1)^2}{4}.
    $$

    Letting $x = 0$ and $y = 1$ on the LHS gives
    $$
    (p)^2 + \left(1 - \frac{1 + q}{2}\right)^2 = \frac{4p^2 + (q - 1)^2}{4},
    $$
    so the point $(0,1)$ lies on all circles $C_{p, q}$.


[quote="DaReaper"]Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...

i hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think...  :oops:

If ${a}\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\oplus{b}=-3$ and $a=1$

i know the answer...have no solution....and no clue how to do this (other than guess and check) :P[/quote]
[hide]Just make b a variable and solve for it. So you have $(1+b)^{2}-6b=-3$. Foil and simplify until you get $b^{2}-4b+4=0$. Factoring gets you $(b-2)^{2}$. So b=2.[/hide]

wow..i guess that was simple thanks... :blush:  :P

[quote="DaReaper"]Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...

i hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think...  :oops:

If ${a}\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\oplus{b}=-3$ and $a=1$

i know the answer...have no solution....and no clue how to do this (other than guess and check) :P[/quote]
$(a+b)^{2}-6b=a^{2}+2ab+b^{2}-6b$
Plug in $a=1$ to get $1+2b+b^{2}-6b=-3$ so $b^{2}-4b+4=0$ and $(b-2)^{2}=0$ so $b=2$.

[quote="myspacesucks"]Ckck, you are completely wrong.  You cannot simplify until b^2-4b +4, you can simplify to b^2-6b+4=0, in which there is no solution.
This is how it SHOULD be done:
(a+b)^2-6b=-3
(1+b)^2-6b=-3
  1+b^2-6b=-3
 +3             +3  (To get rid of the stuff on the right)
b^2-6b+4=0

There is no way for x to be true so it is no solution.[/quote]
$(1+b)^{2}\ne1^{2}+b^{2}$, in fact, in general $(a+b)^{2}\ne(a^{2}+b^{2})$.  Instead, $(a+b)^{2}=a^{2}+2ab+b^{2}.$

[quote="myspacesucks"]Ckck, you are completely wrong.  You cannot simplify until b^2-4b +4, you can simplify to b^2-6b+4=0, in which there is no solution.
This is how it SHOULD be done:
(a+b)^2-6b=-3
(1+b)^2-6b=-3
  1+b^2-6b=-3
 +3             +3  (To get rid of the stuff on the right)
b^2-6b+4=0

There is no way for x to be true so it is no solution.[/quote]

I've seen you post things like that a few times now.  Note that

$(5+5)^{2}\ne 5^{2}+5^{2}$
$(5+5)^{2}= 5^{2}+2(5)(5)+5^{2}$

In general, $(a+b)^{2}=a^{2}+2ab+b^{2}$

[quote="DaReaper"]Well, the test was multiple choice and i just plugged in the possible answers until the problem worked out...

i hope thats not the way to these kinds of problems...but its worth a try...and its the easiest way....i think...  :oops:

If ${a}\oplus{b}=(a+b)^{2}-6b$, then find $b$ if ${a}\oplus{b}=-3$ and $a=1$

i know the answer...have no solution....and no clue how to do this (other than guess and check) :P[/quote]

[hide]$(a+b)^{2}-6b=-3$

plug in the given $a$ value...

$(1+b)^{2}-6b=-3$

$1+2b+b^{2}-6b=-3$

$b^{2}-4b+4=0$

$(b-2)(b-2)=0$

$\boxed{b=2}$[/hide]

hee hee, stupid me!!!!!!!!!!!!!!!!
sorry, i'm wacko
 :blush:


calculus
trigonometry
complex numbers
\/closed

Will you have these classes available over the summer? pre-calc+calculous?

We'll have Intermediate Algebra and Intermediate Trig/Complex numbers in the spring.  These are essentially pre-calculus courses.  The Algebra course will run through the summer.

[quote="dbkarp"]If I wanted to take pre-calc over the summer, would I need to take the trig one? (I already learned algebra II in school)[/quote]

I'm not sure what you're asking.  Our trig class is a lot harder than your typical pre-calc class.  (And our algebra is a lot more challenging than Algebra II in school.)  If your target is preparing for AMC12/AIME, you'll want to take both of these Intermediate classes (assuming you aren't already easily passing both).

If you want school credit, you will need ask the people at your school if they will accept the course.  My guess is that they will not because the AoPS courses do not cover exactly the same material that your state likely wants you to have covered.  However, the courses are still good to learn material for contests and for more knowledge in general.

My precalc course covered only the very basics of what was covered in the Intermediate Trig and Complex Numbers course...

[quote="dbkarp"]I admit my questions were a bit vague, here is what I need to know:

I'm in eight grade, but taking what they call Algebra III Honors. It's a course with me and a bunch of 11th graders to learn all the hard parts of Algebra II+Some Trig. I'm going into pre-calc next year, but I want to know if I can make that calculous instead. Are there any courses on AoPS I can take to skip pre-calc over the summer.[/quote]

If you want to replace your precalc with AoPS work, then I would recommend the Interm Algebra and Interm Complex/Trig classes.

ok, maybe i should just learn pre-calc at school... Thanks anyway!


geometry
3D geometry
calculus
calculus computations

this is not the exact question because it was from a test, and i didnt have to time to copy it down, but this is basically what we dealing wit here:

theres a mountain that resembles a cone (treat it like a cone), and theres a cloud surrounding top of the mountain, (if u cut the cone side ways parallell to the base, the cloud would cover the top small cone) and the bottom of the cloud is moving up at 0.5 m/s (we r assuming the bottom of cloud is a straight line, lol). the height of the mountain, or cone, is 1350 m, diameter of base is 1200, and the question is, how fast is the area not covered by the cloud increasing when the bottom of the cloud is 800 m above the ground (meaning when the bottom part has height 800 m, and small cone on top has 550 m)? in the question it also gave us the equation to find the surface area of the cone, which i think y'all know, but i'll jsut put it here anyways...A= (pi) (r) (s), where r is radius, and s is the slanted height

if anyone doesnt understand it, can some1 else make a diagram somehow to explain it? just make it simple, assume we looking at it horizontally, so just draw a triangle, the cone in 2d, representing the mountain, and a line in the mid paralell to the base representing the bottom of the cloud...

i dunno about u guys, but i think that question was quite hard...so...if any1 can solve it for me, that'd be great...

Isn't this just a straightforward shadow on a cone problem that is to be solved using "related rates"?  By the way, very nice avatar. :blush:

thx...

ya i think it is, i think ur supposed to find the rate of area w.r.t rate of height, by using chain rule, rate of area / rate of radius x rate of radio / rate of height, but then somehow the slanted height have to go in somewher, i dunno...i was rushed during the tset, so i wasnt thinking straight, i have no idea how to solve it now, and i just dun feel like thinking about it now...so i was hoping some ppl would solve it and post  the solutions

i just tried it, and i got 840. something, i dunno if its right...can some1 verify the answer plz???


Define a sequence by $a_1=a_2=1$ and $a_n a_{n-2} = a_{n-1}^2+2$ for $n>2$. Show that all terms of this sequence are odd integers.

It is easy to verify that the sequence is also defined by $a_{n+2} = 4a_{n+1} - a_n$.
The result is now trivial by induction.

Pierre.

[quote="pbornsztein"]It is easy to verify that the sequence is also defined by $a_{n+2} = 4a_{n+1} - a_n$.
The result is now trivial by induction.[/quote]

That's the solution I saw too. But how exactly did you manage to reduce the non-linear recurrsive relation to $a_{n+2} = 4a_{n+1} - a_n$? Just by guess-and-check?

The first thing which came to my mind is that it smelt like a Fibonacci property ($F_{2n+3}F_{2n-1} = F_{2n+1}^2 + 1$).
Thus I looked over a recurrence of order 2, and that one is easy to find.

Pierre.


are we supposed to receive an email/mail saying AMC has already received our faxed/mailed permission slip?

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamo-cklist.html

Steve Dunbar
AMC Director


conics
ellipse
Euler
geometry proposed
geometry

Hi all!
I'd like to presant my new problem :) .
I think it is rather unusual problem and also a beautiful result. :P

[color=blue]Problem.Let $ABC$ be an arbitrary triangle in the plane.${\varepsilon}_{A}$ is an elipse with focuses $B$ and $C$ and passing trough $A$.The elipses ${\varepsilon}_{B}$ and ${\varepsilon}_{C}$ are defining similary.Let $A'$ be the intersection point of ${\varepsilon}_{B}$ and ${\varepsilon}_{C}$ that lies in the angle $A$.The points $B'$ and $C'$ are defining similary.Prove that the segments $AA'$,$BB'$ and $CC'$ are intersecting at the same point $E$,which lies on the $Euler$'s line of the triangle.[/color]

Enjoy it! :wink:

Dear users don't forget that you have an unsolved problem here :wink: .

I was playing with some test cases, and I found that $AA'$, $BB'$, and $CC'$ do not concur.

But the same test cases strongly suggest that the common chords of the ellipses, taken in pairs, concur at the [url=http://mathworld.wolfram.com/deLongchampsPoint.html]de Longchamps point[/url] of triangle $ABC$, which does lie on the Euler line.  No proof yet, though.

But I agree, it is a very nice problem.

[quote="nsato"]I was playing with some test cases, and I found that $AA'$, $BB'$, and $CC'$ do not concur.

But the same test cases strongly suggest that the common chords of the ellipses, taken in pairs, concur at the [url=http://mathworld.wolfram.com/deLongchampsPoint.html]de Longchamps point[/url] of triangle $ABC$, which does lie on the Euler line.  No proof yet, though.

[/quote]
It sounds like a contradiction,isn't it?
How can two of them intersect in a fixed point,but all three do not? :maybe:
Maybe there is something wrong with the test cases? :blush:

I think you misunderstand me.  The ellipses $\epsilon_{A}$ and $\epsilon_{B}$ have a common chord, as do $\epsilon_{A}$ and $\epsilon_{C}$, and $\epsilon_{B}$ and $\epsilon_{C}$.  My claim is that all three common chords concur.

A related problem was proposed for the 1985 IMO:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=29201[/url]
[url]http://www.kalva.demon.co.uk/short/soln/sh859.html[/url]

Ooops :oops: ,I see what you meant.
Then can you show me a drowing that will show that the main problem is incorrect?In my drowings it is always true :huh: .

I found a reference for my conjecture:
[url]http://tech.groups.yahoo.com/group/Hyacinthos/message/458[/url]


symmetry
function
geometry
3D geometry
algebra proposed
algebra

[hide="Step 1"]$ \frac{x^3\plus{}1}{2}\equal{}\sqrt[3]{2x\minus{}1}$
Do you notice something?[/hide]
[hide="Step 2"]$ f(x)\equal{}\frac{x^3\plus{}1}{2}$ $ \to$ $ f^{\minus{}1}(x)\equal{}\sqrt[3]{2x\minus{}1}$
The equation translates to
$ f(x)\equal{}f^{\minus{}1}(x)$[/hide]
[hide="Step 3"]Using symmetry, $ f(x)\equal{}f^{\minus{}1}(x)$ $ \to$ $ f(x)\equal{}x$[/hide]
[hide="Step 4"]$ f(x)\equal{}x$ $ \to$ $ \frac{x^3\plus{}1}{2}\equal{}x$
$ x^3\minus{}2x\plus{}1\equal{}0$ $ \to$ $ (x\minus{}1)(x^2\plus{}x\minus{}1)\equal{}0$
The solutions are:
$ x\equal{}1$ , $ x\equal{}\frac{\minus{}1 \pm \sqrt{5}}{2}$[/hide]

Hmm, does $ (f(x)\equal{}f^{\minus{}1}(x))\Rightarrow (f(x)\equal{}x)$ hold for all continuous functions? If yes, could you post a short explanation about that?

The solution

$ x_0\equal{}\minus{} \frac {1 \plus{} \sqrt {5}}{2}$

doesn't hold for the equation.

The equation has 2 real and 2 complex solutions. The two reals are already written, and the other two are the complex solutions with negative real parts to the following equation:

$ x^6\plus{}2 x^4\plus{}2 x^3\plus{} 4 x^2 \plus{}2 x\plus{}9\equal{}0$

[quote="polskimisiek"]Hmm, does $ (f(x) \equal{} f^{ \minus{} 1}(x))\Rightarrow (f(x) \equal{} x)$ hold for all continuous functions? If yes, could you post a short explanation about that?[/quote]

I did not read the other posts, but I can said that $ (f(x) \equal{} f^{ \minus{} 1}(x))\Rightarrow (f(x) \equal{} x)$ does not hold for all continuous functions.

Example of continuous functions different from $ f(x)\equal{}x$ and such that $ f(x)\equal{}f^{\minus{}1}(x)$ :

$ f(x)\equal{}a\minus{}x$

$ f(x)\equal{}\sqrt[3]{a\minus{}x^3}$

$ f(x)\equal{}g^{\minus{}1}(\minus{}g(x))$ for any bijective function $ g(x)$, $ \mathbb{R}\rightarrow\mathbb{R}$

Or, a simple one:

$ f(x)\equal{}\frac{1}{x}$

[quote="milin"]Or, a simple one:

$ f(x) \equal{} \frac {1}{x}$[/quote]

which, unfortunately, is not continuous at $ x\equal{}0$

True.. But adding $ f(x)\equal{}1, x\equal{}0$ should fix the problem.

The complex roots aren't even particularly well-defined since it's not clear which cube root to take.  Anyway, I'd suggest looking at the graph of $ f(x) \equal{} \frac {x^3 \plus{} 1}{2}$ and then reflecting it across the line $ y \equal{} x$.  If there are intersection points off of $ y \equal{} x$, we can find them graphically and then show them algebraically.

[quote="milin"]True.. But adding $ f(x) \equal{} 1, x \equal{} 0$ should fix the problem.[/quote]

Not really.

It allows $ f(x)\equal{}\frac{1}{x}$ to be defined at $ x\equal{}0$ but not to be continuous.

OK, so if that does not hold for every continuous function, than what are the required assumptions? And why [b]lordWings[/b] could use that here?

Although many functions can have $ a$ such that $ f(f(a))\equal{}a$ without $ f(a)\equal{}a$, only solution to $ f(x)\equal{}x$ will solve $ f(f(x))\equal{}x$, so you can know what kind of factor to pull out.

In this case, $ x^3\plus{}1\equal{}2\sqrt[3]{2x\minus{}1}$ can be written as
\[ x^3\minus{}2x\plus{}1\equal{}2\left( \sqrt[3]{2x\minus{}1}\minus{}x\right) \equal{}\minus{}2\cdot \frac{x^3\minus{}2x\plus{}1}{x^2\plus{}x\sqrt[3]{2x\minus{}1}\plus{}(2x\minus{}1)^{2/3}}\]
So any solution other than a root of $ x^3\minus{}2x\plus{}1$ must have
\[ x^2\plus{}x\sqrt[3]{2x\minus{}1}\plus{}(2x\minus{}1)^{2/3}\equal{}\minus{}2\]
But this has no real solutions, since $ y\equal{}\sqrt[3]{2x\minus{}1}$ implies $ x^2\plus{}xy\plus{}y^2\equal{}\left(x\plus{}\frac{y}{2}\right)^2\plus{}\frac{3}{4}y^2\geq 0$.


calculus
integration
inequalities
function
real analysis
induction
triangle inequality

This one is pretty tough and technical:
    Prove the existence of a sequence of positive numbers $c_k$ with the properties:
   1) For any $\lambda>0$, any $ a<b$ and  any function $f$ of class $C^{2}$  such that $ f'$ is monotonic and $|f'(x)|\geq 1$ for all $x\in [a,b]$ we have $|\int_{a}^{b}{e^{i\cdot\lambda\cdot\ f(x)}dx}|\leq\frac{c_1}{\lambda}$ and
  2) For any numbers $a<b$, any $k\geq 2$, any $\lambda>0$, any function $f$ of class $C^{k+1}$ such that $|f^{(k)}|\geq 1$ for all $x\in [a,b]$ we have ${|\int_{a}^{b}{e^{i\cdot\lambda\cdot f(x)}dx}|\leq\frac{c_k}{\sqrt[k]{\lambda}}}$.  This is the problem with the greatest number of "any" that I have ever seen. :D

I'll take care of the easy part here  :) , namely proving that there is such a $ c_1.$

This resembles the Riemann-Lebesgue lemma in the case when the function is $ \mathcal{C}^1,$ so the proof will be similar.

WLOG, we may assume that $ \forall x \in [a, b], \ \ f'(x) \geq 1.$

An integration by parts yields :
\begin{eqnarray*} \int_a^b e^{i \lambda f(x)} dx & = & \int_a^b \frac 1{i \lambda f'(x)} (e^{i \lambda f(x)})' dx \\
& = & \frac 1{\lambda} \left ( \left [\frac 1{i f'(x)} e^{i \lambda f(x)} \right ]_a^b + \int_a^b \frac 1i e^{i \lambda f(x)} \frac {f''(x)}{f'^2(x)} dx \right ) \end{eqnarray*}

Therefore we get, with a few applications of the triangle inequality :
\begin{eqnarray*} \left | \int_a^b e^{i \lambda f(x)} dx \right | & = & \frac 1{\lambda} \left |\frac 1{i f'(b)} e^{i \lambda f(b)} - \frac 1{i f'(a)} e^{i \lambda f(a)} + \frac 1i \int_a^b e^{i \lambda f(x)} \frac {f''(x)}{f'^2(x)} dx \right | \\
& \leq & \frac 1{\lambda} \left (\frac 1{f'(a)} + \frac 1{f'(b)} + \int_a^b \frac {|f''(x)|}{f'^2(x)} dx \right ) \\
& = & \frac 1{\lambda} \left (\frac 1{f'(a)} + \frac 1{f'(b)} + \left | \int_a^b \frac {f''(x)}{f'^2(x)} dx \right | \right ) \\
& = & \frac 1{\lambda} \left (\frac 1{f'(a)} + \frac 1{f'(b)} + \left | \frac 1{f'(a)} - \frac 1{f'(b)} \right | \right ) \\
& \leq & \frac 2{\lambda} \left (\frac 1{f'(a)} + \frac 1{f'(b)} \right) \\
& \leq & \frac 4{\lambda} \end{eqnarray*}

Hence, we may take $ \boxed{c_1 = 4}.$

Well, it's actually almost all the work. A proof by induction using the same arguments will prove the statement. Note however that the proof will not be completely trivial, since there will be some cases to discuss. It's a good exercise.


Let $MAB$ be a triangle with constant base $AB$. Assume that the point $M$ moves

on to its plane such that the external squares $MAKC$, $MBHZ$, $MBLI$ that figures

fulfil the condition $[AMCK]+[BMZH]+[ABLI]+4[MAB]=2002$ where with $[...]$ we

symbolise the areas. Find the locus of the point $M$.

Noone has a solution to this nice problem ?Thanks.

BTW, the 34 views in two months saw that my problem didn't interest you :(


Supose that the refractive index of a region in the space is proportional to the inverce of the distance from the point $ (0,0)$ (working in 2 dimensions for simplicity).

If a ray of light start at the point $ (r,\theta)$ for $ \frac {\pi} {2} > \phi> 0 > r$ , $ \frac {\pi} {2} < \theta < \pi$ ,  The ray will describe a Arc of a circunference and will ever end in $ (0,0)$. ($ \phi$ is the tangent angle needed to solve this problem

(Theorically speaking, 'cause at the point $ (0,0)$ $ n$ would be infinity, and at very far points $ n$ would be zero.)

[url=http://www.mathlinks.ro/viewtopic.php?search_id=682405916&t=167520]here[/url] is  a similar problem posted by me


Factorise:

 a^3 - 2b^3 + (a+b)^3

Thank you.

I think it can't be factored into 2 polynomials of $ a,b$ with integer coefficients.