Anyone go to ARML last year or previously (preferably the PA location)? I'm planning to go this year and I want to find out all I can about it. What is the competition/experience like? What are the problems like? (the ones on the webpage look something like AIME problems but more painful... fantastic...) What do you recommend to learn/study/brush up on?

Also, if anyone knows any links to practice problems I'd be grateful, the official website doesn't help that much...

There are at least a couple of ARML problem books.  You might want to check out the ARML homepage for info on those.

Also, I think ARML problems are typically a bit easier than AIME problems.  At the very least, hard AIME problems are definitely more involved than hard ARML problems.  I think that AIME problems (especially the first 10) are a great way to prepare for ARML.

The ARML power question is proof-based.  You might take a look at easier Olympiad problems or any Mandelbrot team problems for something that will help you develop those skills.

I've been on the all PA team since freshman year. I'd suggest you brush up on some easier AIME problems, which may appear on the ARML. Also, HMMT (web.mit.edu/hmmt) also has a team round that you may consider practicing on. I suggest the 2002 team, which is one of the harder ones and more characteristic of the ARML.

The individual questions on HMMT are also good practice. They are fairly difficult for the later years...especially after Gabe Carroll began writing problems, lol. Anyway, I hope to see you there.

wow, the easy AIME problems? o.O That works out rather well then, I can practice for AIME and ARML at the same time...  :twisted:

thanks for the suggestions :D It's a good thing I still have a lot of time left to practice.

masamune2387:
yes, I hope so too :mrgreen: it would be funny if I practiced a lot for ARML and didn't make it. ha, ha.


geometry
rectangle
circumcircle
projective geometry
geometry unsolved

Let a point $ M$ be out of circle $ (O)$. Let $ MA$, $ MB$ tangen with $ (O)$ at points $ A,B$ respectively. Let $ AO$ meet $ (O)$ at point $ C$, let $ BH \perp AC$ with $ H$ lie on $ AC$. Let $ BH$ meet $ CM$ at point $ N$, let $ AB$ meet $ MO$ at point $ K$. Prove that $ NK // AC$.

Dear Mathlinkers,
1. let P, Q, R the second points of intersection of MC, BO, BH with (O), and Tb the tangent to (O) at B.
2. According to Pascal's theorem, MO is the Pascal's line of AB Tb QPCA ; then P, K and Q are collinear.
3. AC and QR are parallel.
4. According to Pascal's theorem applied to ABRQPCA, KN is parallel to QR.
Sincerely
Jean-Louis

Thank you very much. Hi every One, I have a solution for this problem by use pole and polar. But i think this problem can be done by an easy way. Can you help me?

[quote="dgreenb801"]Here is my solution:
Let MO intersect BH at P and BC intersect AM at Q. Then as $ AB \perp BC$ (AC is a diameter),  PMQB is a parallelogram.
$ \frac {MN}{NC} \equal{} \frac {MP}{BC} \equal{} \frac {BQ}{BC} \equal{} \frac {MK}{KO}$.
So since $ \frac {MN}{NC} \equal{} \frac {MK}{KO}$, $ NK \parallel AC$.
EDIT: No Menelaus needed![/quote]

Yes! it's a nice solution. Thank you very much.

Let $ BC\cap KN \equal{} \{L\}$. Then $ L$ is on circle circumscribed to $ BKOH$, since $ \angle LBK \equal{} 90^{\circ}$.
Then $ BKOL$ is rectangle since $ OB \equal{} KL \equal{} r$. Now $ \angle LKB \equal{} \angle OBK \equal{} \angle OAB$ thus $ AC\parallel{}KN$.

Edit: it is not O.K.

[quote="Number1"]Let $ BC\cap KN \equal{} \{L\}$. Then $ L$ is on circle circumscribed to $ BKOH$, since $ \angle LBK \equal{} 90^{\circ}$.
Then $ BKOL$ is rectangle since $ OB \equal{} KL \equal{} r$. Now $ \angle LKB \equal{} \angle OBK \equal{} \angle OAB$ thus $ AC\parallel{}KN$.[/quote]

Numeber1, why $ L$ is on circle circumscribed to $ BKOH$?

[quote="Number1"] Then $ L$ is on circle circumscribed to $ BKOH$, since $ \angle LBK \equal{} 90^{\circ}$.
[/quote]
I don't get this either. $ \angle CBK\equal{}90$ but C is not on the circumcircle of BKOH.

[quote="dgreenb801"][quote="Number1"] Then $ L$ is on circle circumscribed to $ BKOH$, since $ \angle LBK \equal{} 90^{\circ}$.
[/quote]
I don't get this either. $ \angle CBK \equal{} 90$ but C is not on the circumcircle of BKOH.[/quote]

No, the point $ L$ must lie on the circle $ (BKOH)$. But, why it's true? Number1!


[quote="SalmanMalik"]hello moderators I  am of the view of creating a Pakistani community here.So can anyone help me?[/quote]

I guess a national subforum requires some other Pakistani members with more than one posting.  ;)

[quote="SalmanMalik"][quote="Valentin Vornicu"][quote="SalmanMalik"]nice thinking.... :lol: .....but more Pakistani people will soon join in after the formation of the Paki subforum.[/quote]
well how can you be so sure that more pakistani will join after the creation of the sub-forum?[/quote]

Because I know many of them who are willing to join in.[/quote]Have them joining and posting in this topic. We'll see then ;)

We are very happy to create local and national communities, but we want to be sure that we are not spending our time creating something that will not be used.  There are some communities with 100 or more members on the site that haven't shown a lot of activity (though some have).  If we see a good number of requests from active AoPSers who wish to participate in a new community, we will create a new community.

[quote="chess64"]hey, i'm from pak... though I wouldn't really waste my time in some pak community forum. I like math![/quote]

But some people might.  When we created local communities, there were some students who participated immediately in those communities who did not often (or at all) participate in the regular forums.  Different people have different interests.  After a few years I suspect we'll have communities for many nations.

[quote="Valentin Vornicu"][quote="MCrawford"]After a few years I suspect we'll have communities for many nations.[/quote]Mathew ... we already have many national communities :P[/quote]

Yes, but I meant "many" as in "looks like the list of IMO participating nations."

Hi - Isn't is the first time Pak is going to take part in IMO (2005, Mexico). Any  info on who is in the team?
I heard that  Becheanu   is/was going to help as a  coach to  the Pak team.

Salman - BTW  are  you the  Salman Malik ( Multan)  who is a member of the team? Just curious.

[quote="SalmanMalik"]yes I am the same Salman.But who are you and how do you know so much about the pakistani team????[/quote]It is well known in Romania that the former team and deputy leaders Mircea Becheanu and Bogdan Enescu trained the first time participant team of Pakistani ...  :roll:

[quote="SalmanMalik"]yes I am the same Salman.But who are you and how do you know so much about the pakistani team????[/quote]
There was a news report in some newspaper  which gave the name of the students  and as Valentin said Becheanu is well known,  but other than that  I do not know much. So you may like to add some info here.

hey..salman..didnt know ppl knew us in romania...  :o  ..

OK, your wish has come true :P You have now your own Pakistani community! Enjoy :)

assalamalikum all  iam suffering from imo fever .could any body help me.


calculus
integration
trigonometry
function
logarithms
algebra
rational function

Could someone teach me the t-method of integration with an example. Unfortunately I do not have a good book which teaches this method and the internet websites are difficult to understant

regards bos1234

The words "t-method of integration" do not mean anything to me. They're not standard teminology, at least as far as I know. If you mean "integration by substitution" (for American textbooks, the favorite letter is $u$ rather than $t$), then that's the main topic of a substantial proportion of a first-semester calculus course.

t- substitution is standard terminology for the substitution t= tan(x/2). It is popular because it makes an trig integral a rational function to integrate, which can be easier.

$\int{R(\sin{x},\cos{x})\,dx}$ where $R$ is ratioanal function, with substitution $\tan\frac{x}{2}=t$ becomes integral of rational function.
$\sin{x}=\frac{2t}{1+t^{2}}, \cos{x}=\frac{1-t^{2}}{1+t^{2}}, dx=\frac{2\,dt}{1+t^{2}}$
After that you have $\int{R(\frac{2t}{1+t^{2}},\frac{1-t^{2}}{1+t^{2}})\frac{2\,dt}{1+t^{2}}}$ where $R$ is rational function.

Here is an example.It's maybe to easy but it is an example.
$\int{\frac{dx}{sin{x}}=\int{\frac{1+t^{2}}{2t}\cdot\frac{2}{1+t^{2}}\,dt=\int{\frac{dt}{t}=\ln\left|t\right|+C=\ln\left|\tan\frac{x}{2}\right|+C, \tan\frac{x}{2}=t.}}}$

You should use the substitution tan(1/2)x when the denominator of the intagrand is of the form a + bcosx or a + bsinx. That is the only time I ever use it.

[quote="Kent Merryfield"][b][color=red]The words "t-method of integration" do not mean anything to me.[/color] [/b]They're not standard teminology, at least as far as I know. If you mean "integration by substitution" (for American textbooks, the favorite letter is $u$ rather than $t$), then that's the main topic of a substantial proportion of a first-semester calculus course.[/quote]

OK, that was pretty funny !  :rotfl:

i've also [u]never [/u]heard of a "t-method of integration" par se, sure i have used it
many times before, but it is just referred to over here as being a clever method
of "integration by substitution" and nothing else.  :|

The name differs from place to place. Here we call it the t-substitution but often when I use it I just say let $t= \tan \frac{x}{2}$.

@bos1234 you can try this as an exercise: $\int \frac{dx}{ 1+\sin{x}+\cos{x}}$.


Find $ n$ possitive integer least such for any $ n$ triples $ (a_i,b_i,c_i)\in (1,2,. . .,7)^3$ then there exits 2 triples $ (a_i,b_i,c_i), (a_j,b_j,c_j)$ : $ a_i\leq a_j,b_i\leq b_j, c_i\leq c_j$.

Same with [url]http://www.mathlinks.ro/viewtopic.php?t=307268[/url]
with the difference the values here range $ 1\minus{}7$ rather than $ 0\minus{}7$, therefore the answer is $ \min(n) \equal{} 36 \plus{} 1 \equal{}37$.


geometry
inradius
quadratics
algebra
geometry proposed

Hi, I've got a problem:

[color=green][b]Problem. [/b]Given $\triangle ABC$ such that $m\angle ABC=\pi /2$, let $Y \in \overline{AC}$ such that the inradius of $\triangle ABY$ and $\triangle YBC$ are equal. Prove that $BY^2=[ABC]$.[/color]

 :D  Capitan Gelatina!

Let $\triangle ABC$ be any triangle divided by a cevian BY so that the inradii of the triangles $\triangle ABY,\ \triangle CBY$ are equal. Let $r, s, \triangle$ be the inradius, semiperimeter, and area of the triangle $\triangle ABC,$ and x the common inradius of the triangles $\triangle ABY,\ \triangle CBY.$ Let $s_A, \triangle_A$ be the semiperimeter and area of the triangle $\triangle ABY,$ and $s_C, \triangle_C$ the semiperimeter and area of the triangle $\triangle CBY.$ Denote y = BY and we want to calculate y.

$\triangle = \triangle_A + \triangle_C = x(s_A + s_C) = x(s + y)$

Let h be the B-altitude of the triangle $\triangle ABC.$ According to the "infamous" :yup:  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19468]inradii problem[/url], $\frac 2 h = \frac 2 x - \frac{r}{x^2},$ which is a quadratic equation for the inradius x.

$x^2 - hx + \frac{rh}{2} = 0,\ \ x = \frac{h \pm \sqrt{h^2 - 2rh}}{2}$

Since $x < \frac h 2,$ we have to take the root with the minus sign. Therefore,

$x = \frac h 2 \left(1 - \sqrt{1 - \frac{2r}{h}}\right) = \frac h 2 \left(1 - \sqrt{1 - \frac b s}\right)$

$\frac{bh}{2} = \triangle = \frac h 2 \left(1 - \sqrt{1 - \frac b s}\right) (s + y)$

$b = \left(1 - \sqrt{1 - \frac b s}\right) (s + y)$

which is an equation for y. Solving,

$b \left(1 + \sqrt{1 - \frac b s}\right) = \left[1 - \left(1 - \frac b s\right)\right] (s + y) = \frac b s\ (s + y)$

$s \left(1 + \sqrt{1 - \frac b s}\right) = s + y$

$y = s\ \sqrt{1 - \frac b s} = \sqrt{s(s - b)}$

$y^2 = s(s - b) = \triangle\ \sqrt{\frac{s(s - b)}{(s - a)(s - c)}}$

This holds for any triangle $\triangle ABC.$ It is clear that $y^2 = \triangle$ iff $s(s - b) = (s - a)(s - c)$ or iff

$(a + b + c)(c + a - b) = (b + c - a)(a + b - c)$

$c^2 + 2ca + a^2 - b^2 = b^2 - c^2 + 2ca - a^2$

$c^2 + a^2 = b^2$

i.e., iff the angle $\angle ABC = 90^\circ$ is right.


Yeah I started with that and it turned out really, really ugly. Then I came up with a bunch of equations relating to real and imaginary parts, like Re(a)+Re(b)=1, Im(a)+Im(b)=0, and others, I can't remember exactly. This was on the Complex Number final, and the solutions haven't been posted yet.

hah, wow. How did you simplify the square root, deMoivre?

You just have to guess...it's like simplifying square roots of radicals.

Yea, you can set up a+bi to both sides and have equation in a, b...but almost all the time on contests they're simple enough to guess at, especially this one.

Another way that helps guessing is to take the square root of the norm.  For example, $N(21-20i) = 21^{2}+20^{2}= 841 = 29^{2}$, so the square root has components $a, b$ such that $a^{2}+b^{2}= 29$.  Assume they're integer and there's only one way to go from there :)

By the way, you should also notice that $20, 21, 29$ is a Pythagorean triple.

[quote="13375P34K43V312"]You mean like set it equal to a+bi then square both sides? Thanks.[/quote]
Yes, if you do this you get two equations by comparing real and imaginary parts. If you also remember that $a^{2}+b^{2}$ is known it will be very easy to solve the equations. This is what you should do, not guess.

[quote="mathgeniuse^ln(x)"][quote="13375P34K43V312"]Find the product of the real parts of the roots of the equation $z^{2}-z=5-5i$.

I came up with a long, somewhat ugly solution to this. I'm wondering if there's anything better.[/quote]

So $z=\frac{1%Error. "pmsqrt" is a bad command.
{1-20+20i}}{2}$

This is going to be ugly.[/quote]

Wow you are stupid, you didn't even realize that it should've read

$z=\frac{1 \pm \sqrt{21-20i}}{2}$.

[hide="The Solution"]
By purely using the Quadratic formula, one will get that:
\[z=\frac{1\pm\sqrt{1-20(\imath-1)}}{2}\]

Then, by simplifying that radical, one will get:
\[\frac{1\pm\sqrt{21-20\imath}}{2}\]

And then by simplifying and getting rid of that big radical, but leaving the $i$, I got, after tons of conjecture:

\[\frac{1\pm\left(5-2\imath\right)}{2}\]

and then one can find the roots of that quadratic from that, thus leading to the answer.

So:

The roots must be:

\[\frac{1\pm\left(5-2\imath\right)}{2}\longrightarrow\{3-i,-2+i\}\]

And then finding the product would be something like:

\[\Re\{3-\imath\}\times\Re\{-2+\imath\}\]

and then, I believe, everything should pretty much be self explanatory :)
[/hide]

Wow I thought your spam in high school forums died down a bit, but are you starting again?
why post a solution that was already posted and discussed?
(same with this place here [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=717424#717424[/url])

What? I just posted a solution.

Is that considered spam.

Uhh, yes, because that solution has already been posted.

But, at least he wasn't doing it two months later.

In all honesty, did you come up with that yourself or did you look at paladin8's first?

He probably came up with it himself because it was on the complex numbers final and he is in trig/complex numbers...

all the posts between this and anirudh's solution are quite unnecessary

even if it was a repost, you guys should not use the thread to talk about spamming ways or anything, especially since he hid the solution;

using private means to communicate with anirudh or a moderator would have been much better

Also, I don't copy off of other's solutions. :wink:


This is my schools first year of doing mathcounts. I also help start up the AMC 8 at our school. I was wondering, for mathcounts, what is best for the team. If we have two members will we be penalized in any way? such as not being able to answer all the questions because we have to have more pairs or can we do them all still? Thx for replies.

The more people on your team, the better the team score will be.  This is because:

\[ \text{Team Score} \equal{} \dfrac{\text{Sum of Individual Scores}}{4} \plus{} 2 \times \text{Team Round Score}\]

so a short-handed team is severely disadvantaged.

Or is it the average? If it was the average, then no, you don't have a disadvantage. If it is the sum divided by four, then yes, you do.

Sum divided by four; AIME15's right. Non-existent team members count as having a 0/46 individual.


inequalities trigonometry calculus derivative AMC USA(J)MO USAMO

Let $ \alpha,\beta$ and $ \gamma$ be the angles in any acute triangle. Demonstrate that:

$ 2<\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}$

[hide="What I have till now"]

\begin{eqnarray*}
2&<&\sin \alpha + \sin \beta + \sin \gamma \le \dfrac{3\sqrt{3}}{2}\\
\dfrac{2}{3}&<&\dfrac{\sin \alpha + \sin \beta + \sin \gamma}{3}\le \dfrac{\sqrt{3}}{2}
\end{eqnarray*}

Let $ f(x)=\sin (x)$, since $ f''(x)\le 0$ in the interval $ [0,\dfrac{\pi}{2}]$, $ f(x)$ is concave down.

Using [url=http://www.artofproblemsolving.com/Wiki/index.php/Jensen's_Inequality]Jensen's inequality[/url] for $ f(x)$ where $ a_1=a_2=a_3=\dfrac{1}{3}$.

The inequality is:

\begin{eqnarray*}
\dfrac{\sin \alpha+ \sin \beta + \sin \gamma}{3}&\le& \sin \left(\dfrac{\alpha+\beta+\gamma}{3}\right)\\
\dfrac{\sin \alpha+ \sin \beta + \sin \gamma}{3}&\le& \sin \left(\dfrac{180^{\circ}}{3}\right)\\
\dfrac{\sin \alpha+ \sin \beta + \sin \gamma}{3}&\le& \sin \left(60^{\circ}\right)\\
\dfrac{\sin \alpha+ \sin \beta + \sin \gamma}{3}&\le& \dfrac{\sqrt{3}}{2}
\end{eqnarray*}

So I have half of it done, I am missing to prove that it is bigger than $ \dfrac{2}{3}$.
[/hide]
Any help or hints will be great.  :roll:

I'm not sure if this is in the right forum.

EDIT: I forgot the part about it being acute. But the inequality is still equivalent to a+b+c>4R

Okay, I think I have a proof using this.
Let O be the center of the circumcircle. Let AO (extended) intersect BC at H. Let DE be the diamater perpendicular to AO. Since ABC is acute, we can assume B and C are on the other side of DE as A. Let DE intersect AB and AC at F and G, respectively. Drop perpendiculars BJ and CK to DE. Then, AF>AO=R. Similarly AG>R. Also, FB+BH>BJ+JO>BO=R. Similarly HC+CK>R. So AF+FB+BH+HC+CG+GA=a+b+c>4R.

[quote="tonypr"]Let $ \alpha,\beta$ and $ \gamma$ be the angles in any acute triangle.[/quote]
Self explanatory.

I think I found a solution to the second part, a friend of mine gave me it but I'm not sure if it's correct:

Second derivative is negative, so its increasing at a decreasing rate. Now imagine having to distribute the 180 degrees worth of angles among a, b, and c. And because it's increasing at a decreasing rate in the worst case scenario (where the sum of the sine's is as low as possible) you would have the angles be as high as possible without passing 90, and this means: $ a\equal{}90^{\circ},b\equal{}90^{\circ},c\equal{}0^{\circ}$ which gives $ 2$. But since this triangle is clearly impossible the sum is strictly bigger than $ 2$.

[quote="tonypr"]Second derivative is negative[/quote]

I read up to there and stopped. This is not a calculus forum.

But nice solution, though.

There's a USAMO question:

$ \minus{}2\leq \sin{3A}\plus{}\sin{3B}\plus{}\sin{3C}\leq \frac{3\sqrt{3}}{2}$

where $ A$, $ B$, and $ C$ are the angles of a triangle. I think that if we solve one then we have a solution for the other.

Testing convexity and concavity with a second derivative is one of the few legitimate uses of calculus on olympiads. (Incidentally this probably does not go in intermediate)

The solution two posts above is not precise; don't expect that to get 7 on an olympiad. You should prove that it is a "worst case scenario."

[quote="enndb0x"]Why you can not use every method you know in an Olympiad ? (calculus etc )[/quote]

You can but it's discouraged on this specific section of the site because it's considered too advanced for the types of problems that are posted.

But, that's just an easier way of showing the convexity of $ \cos x$, it can be done by drawing the graph and drawing the segment which joins two points on the graph in a closed interval (in this case from 0 to $ \dfrac{\pi}{2}$) and checking if it lies above or below the graph, so calculus is not required but it's just a tool to make it easier.

[quote="dgreenb801"]Since ABC is acute, we can assume B and C are on the other side of DE as A.[/quote]

Sorry for not clearing this up, my internet was down. WLOG, assume A is the smallest angle, and B the second smallest. Then, since ABC is acute, we can assume $ \angle B > 45$, or else $ \angle C > 90$. But $ \angle AOC \equal{} 2\angle B$, so $ \angle AOC > 90$, so C is on the opposite side of DE as A. Similarly for B.
Also, we can assume B and C are on opposite sides of line AO or the triangle would be obtuse.

[quote="dgreenb801"] FB+BH>BJ+JO[/quote]
What is the proof for this? (This doesn't look right in my drawing  :huh: )

FB>BJ because BJ is the perpendicular to EF, and BH>the distance from B to line OH=JO.

Remember, we are assuming A is the smallest angle.

Whoops I just re-read your solution and I realized for some strange reasons last time I mis-read it as "drop perpendicular BJ to AC, CK to AB"... really strange :o  :huh:
By the way I don't think we have to assume A is the smallest angle (all angles are acute), but very nice proof anyways... :D


\mathcal is not working approtiately

see this $ \mathcal{F_3} , \mathcal{F_a} , \mathcal{F_i}$

What is wrong with this? or am i missing somethig?

You are trying to use undefined things. Mathcal was not made for any kind of symbol, for example _. And I think you wanted $ \mathcal{F}_a$.

[quote="cgyao15"]hmm.. maybe this should be moved to the latex forum?[/quote]Good idea, done!


Relate the positive solutions of the differential equation

\[ \frac{d^{2}y(x)}{dx^{2}} \plus{} f(x)y(x) \equal{} 0, \,\,\,\, (a \le x \le b)\]

to those of the differential equation

\[ \frac{dy(x)}{dx} \plus{} y^{2}(x) \plus{} f(x) \equal{} 0, \,\,\,\, (a \le x \le b).\]

You sure it's not $ y''\plus{}y^2\plus{}f(x)y\equal{}0$?


group theory
abstract algebra
superior algebra
superior algebra solved

find group $G$ and a set $H\subseteq G$ so that $H^{g}=H$ for all $g \in G$, but $H$ is not a normal subgroup of $G$.

does $H^{g}=H$ mean that $ghg^{-1}=h$?, would any group of integers work with subset being $(0,1)$?

$H^g=H$ means $gHg^{-1}=H$, I think.
Then $H$ is normal. :?

That example is fine. The key is to make sure that $H$ is not a subgroup.

Another example: Let $G$ be any group with at least two elements. Let $H$ be the set of all elements other than the identity.

I didn't notice that $H$ is just a [b]set[/b].  :blush:  :blush:

another canonical example build all subsets of an abelian group


So how many do you know.  I was reading a post involving lots of Bulgarian, and well I was confused.  Anyways I understand some kids speak um different languages like UK kid speaking Russian?  Anyway I am American, but I speak Dutch and I'm learning spanish in school 3rd year...

Anyways...
discuss (be sure to list what languages you may speak and perhaps why).

I speak Romanian (born in Romania huh :) ) and English fluently without any major grammatical flaws (except maybe when I'm typing :D).
I also speak French (but can't write that well), Italian (cuz I studied a little bit of Latin in high-school and also because Italian is very similar to Romanian) and I understand a few words in German :)

I used to know Marathi, but then I forgot it after I started going to school  :)

But I'm learning Spanih in school, and can obviously speak English since I'm lving in America.  And I kind of know a couple of words in Tamil, but I don't count that as knowing the language.

I sorta speak (Mandarin)Chinese, used to speak Shanghainese, have studied French but can't speak very well, obviously speak English, and have tried unsuccessfully to learn Russian and slightly more successfully to learn Esperanto.

I am fluent in english, and I also know some German and i am getting quite good at it.  I have taken some spanish in school, but nothing serious and I can somewhat get what someone is saying in spanish.  As for why I know german.  I was born in germany and I was interested in it so I just messed around with it am I am picking up on a lot of the terms and such.

I am a native speaker of English. I am conversant and literate in Modern Standard Chinese and have worked many years as a translator or interpreter, or as a language teacher of Chinese in English-speaking places or of English in Chinese-speaking places. I studied German off and on for years, beginning in elementary school, and can read German for meaning but can't really speak it. I studied Russian in high school and began college as a Russian major and can still read most Slavic languages for the gist of a passage (thus I knew what was going on with the Bulgarian thread). I have some conversational ability in Taiwanese (Taiwan dialect of Minnan Chinese) and rather less in Japanese and Cantonese. I have read the whole Bible in its original languages (Hebrew, Greek, and Aramaic). I have also had formal instruction in some other languages, and self-study of a few more too. It used to be both my hobby and my occupation to learn languages.

I speak and write English fluently (being a resident of the US, this isn't too uncommon).  I also speak Mandarin, but I don't write it that well.  I can understand some Taiwanese but can't speak it.

I can speak English fluently, (I was born in the U.S.) and can speak/understand Mandarin Chinese, though I can only read/write a few words.  (I quit chinese school in 5th or 6th grade-4 years ago.)  I am also taking my 3rd year of Spanish at school.

I speak English well and French weakly.  In addition, I know the word for "ice cream" in more than a dozen languages (although I can only write it in the European languages on that list).  I think the poll needs to go into fractions for me.

[quote="tokenadult"] I have also had formal instruction in some other languages, and self-study of a few more too. It used to be both my hobby and my occupation to learn languages.[/quote]

Wow. I must say I am really impressed. Have you been working as a translator at university, school or industry ? Were that any specific texts, e.g. mathematics ? And what is your relation to math and college question if your coming from a language background ?

Can you elaborate what other languages you meant in the quotation above ?

You stated that you know several languages. How many words (just approximately) do yo know in each of these languages. How about your grammar ? Do you have your knowledge rather from books or do you also know some daily-life stuff in these languages ? How long does it take you to reach which level of profiency ? Which languages do you find easiest/hardest to learn ?

P.S.: Don't hesitate to ask me for proposing more questions... :)

[quote="somedude"]like UK kid speaking Russian?
[/quote]

 :D

[quote="tetrahedr0n"]I speak English and Romanian. My english is good - according to standardized tests :) but my romanian is decaying quickly. My romanian vocabulary's half life is about a year. I take French in school, so I can read and get the gist of most books, but speaking or having a conversation is not feasible.[/quote]You should perhaps pay attention in the Romanian Forums then? :D :P

I speak Russian and English. I used to know hebrew but forgot it.

Perhaps you all already know it but  last time when I was in Europe,  someone asked:

What do you call a person who speaks many languages?
[hide]multilingual[/hide]
And what do  you call someone who speaks two languages?
[hide]bilingual[/hide]
And what do you call someone who speaks only one language?
[hide]Yes ... I knew what he was going to say .....[hide] :D American  :D [/hide][/hide]

Toeknadult - Impressive.  I tried to teach my sons   my native language (Hindi and Rajasthani) but did not have success in the begining.. (  my wife - who is Eastern_ Europiann and/or (from upstate New York )-  was much faster learner  than my sons) . .. But  both have taken  4 years of Latin in High School, which was  very helpful in kindling their interest in other languages. One son is  majoring in Physics/Math with minor in classics and he has/is taken/taking  Spanish, Russian  and a  little Sanskrit (one semester) .. in college.. and yes now he is better  even in Hindi than he used to be..

But ...I do envy my brother a lot,  he speaks (fluently I may add) French (He is French), Hindi, English, Spanish, Russi an, Italian, German, Japanese,  Mandrin (and can read written chinese ),  Marathi, Rajasthani,  Bengali (and a few more indian languages since his Sanskrit is  *very* good) .. and these,  he is very good at.. he can get by  in  a few more languages....(almost all Indian languages and Europian  languages).

I am a native speaker of English, and am learning French in school.

[quote="djimenez"][quote="EUCLA"]i speak italian (i wonder why :D )..english not very good, german, and chinese :lol: ...latin is not a language, is it? :P[/quote]

Of course it is! that it is not spoken as a live language is other thing, but it is a language, as well as Persa, Sanskrit, and many others that are no longer spoken, but it was the way people talked long time ago![/quote]

so i know 6 languages italian..english..german...chinese..latin...and greek. That' s it! :D

[quote="djimenez"][quote="Valentin Vornicu"][quote="djimenez"]I want to take some language while I am a Grad Student. I wanted it to be either Italian, Portuguese or Romanian, but non is offered in this university. From the ones that they offer and I would like to take are German, French (to relearn) and Russian, but I haven't decided yet.

Best,[/quote]Romanian wasn't offered? :mad: ...

Just kiddin' :lol: ... who would be so nuts as to offer a Romanian language class in grad school in US?! :) :P[/quote]

Nooo, let me explain. I have taken enough courses for my Ph.D. (Here you have to take 17 courses at least, in grad level), after that, you can take whatever you want. Languages, at least here (GATech) are not courses of Grad Level, but given that I have to take at least 3 courses a semester regardless the level, I thought that to learn some language would be a good option, even when they are absolutely not required, and not taken in consideration as completition of program when I present my Thesis.

I though about Portuguese, Italian and Romanian because I believe that they would be easy to learn (or Sardinian, but this is very hard to find a place to learn it, except Sardinia of course!!), but unfortunatelly, they are not offered, then, I still want to find anotherone to learn, and German seems a good option.

Best,[/quote]
you think italian is easy to learn? :?

[quote="Algebra 1 ©"]I speak Urdu, and I'm from Pakistan.  English is my second language, and currently I take French 2 in school since we're required to take a language course in High School.[/quote]

Interesting. I thought, that in Pakistan all people speak arabic or at least can understand it  :D (I don't know, what made me think so  :? ).  Are there any people in your school, who learn arabic?

[quote] I thought, that in Pakistan all people speak arabic..[/quote]
Urdu script has some common features with Persian (and  VERY slght relation to Arabic) the spoken part is very similar to Hindi. My written urdu is a little rusty now but I can speak and understand it fluently. (Reading speed is a little slower though having no practice for last 20-30 years)

Other languages spoken there, among others  are Punjabi, Sindhi etc.... If you are good in Hindi (and have taken a few years of Sanskrit) and have interest you can easily understand many of these languages....Arabic is not very common but Persian (If you  read really read old literature)  - this was also the official "royal" language for law etc a few hundred yeas ago in courts) has more influence. ..

... 1234... Bhojpuiri  ? ehh .. Did  you see "Lagaan" .. lot of the language there is.. well a lot of is in Avadhi, I guess .. but you may still like the movie..it was nominated for oscars but missed the best foreign film award...
(I can understand  these languages .. but can not speak very well)

[quote]you think italian is easy to learn? [/quote]

I don't know but my son (who had never taken Italian in school  and only knew Latin except  he has taken a few semesters of Spanish and may be knew very little French) did ok in Italy generally understanding what was going around.

My wife says it was very easy to learn (As she grew up with people who spoke Italian)...

[quote="diophantus"]Anyone speak German.  I have a teacher who is particularly offended if you refer to her as the frow.  My friends tell me that frow means "woman".  Is this insulting in German?[/quote]

frau, bitte :D i do think that's not offending

[quote="djimenez"][quote="EUCLA"]you think italian is easy to learn? :?[/quote]

I think it should be for me. I mean, I have a bunch of italian friends, and when I listen to them I can understand fairly well the meaning of what they are talking about. It is not that I understand each word, but at least the ideas (Spanish is not that different from Italian!)[/quote]
i know spanish isn't very different from italian, but in italian it isn't easy to put verb into correct form...also italian kids have some difficulties when they learn it at school...

but i didn't want to discourage you :?

native speaker of Korean.
fluent in English
almost-fluent in spanish(talk with hispanic employees at my mom's workplace
 :D )
2-3rd yr in Chinese
Some Portuguese....

so about 4.25-lingual :)

[quote="Gyan"] ... 1234... Bhojpuiri  ? ehh .. Did  you see "Lagaan" .. lot of the language there is.. well a lot of is in Avadhi, I guess .. but you may still like the movie..it was nominated for oscars but missed the best foreign film award...
(I can understand  these languages .. but can not speak very well)[/quote]

Great movie, Avadhi is similar to Bhojpuri, reminded me of my village in India.

And gyan, your'e right- Indian languages are very related. My mom speaks Hindi, Bhojpuri, Mahrati, Gujrati, Sanskrit (not really related). She also has some Punjabi and Bengali skills, too (from friends). But yes, they are all closely related -  the nouns and adjectives are about the same; it is just how the verbs are expressed and idioms that are different (though still similar).

My mother tongue is German, I speak Englisch and some French. I'd like to learn Japanese, but school is enough, and learning something all by myself is sooooo boring...

Native in Chinese

also know 2 dialect of Chinese...didn't count them for they are not languages...

then of course i know English

German at school...pretty good at that...

Some Latin

[quote="Zratonx"]Native in Chinese

also know 2 dialect of Chinese...didn't count them for they are not languages...
[/quote]
which two? I think that one of them is Cantonese ;)


geometry
function
induction
algorithm
complex numbers
algebra open
algebra

I want to know more about propositions wich are true, but not provable.Can someone give some examples?
Is there a criteria for seeing that a statement is not provable?

Statements which are true, but not provable from other statements, are called axioms. They are statements which are taken to be true on faith or from experience.

One such statement from Euclidean Geometry is that the sum of the interior angles of a plane triangle always equals 180 degrees. This statement is seen to be "true" from experience, but it cannot be proven from other statements.

All systems of logic, including mathematics, are based on axioms (or basic truths) and on well-formed definitions. From those axioms and definitions, you can go on to prove theorems using logic. This is how Euclid wrote and proved things in The Elements.

A more unsettling fact, however, was proven by Godel in the early 1900s. Godel showed that in any system of logic that includes arithmatic, there will either be some theorems which cannot be proven either true or false; or there will be some theorems which can be proven both true and false!! Simply put, any mathematical system will either be incomplete or inconsistent. Furthermore, Godel went on to show that you can never prove that any mathematical system is either complete or consistent.

This was very unsettling to mathematicians when it was discovered, but most mathematicians went on to have faith that our current mathematical system was consistent - even if they could never prove it. And they thought that any such theorems which showed incompleteness would have to be highly pathological.

An example of such a theorem was found, however, in the 1930s by Cohen. It's now called the Continuum Hypothesis. The theorem states  that there do not exist a subsets of the real numbers with cardinality greater than that of the rational numbers, and less than that of the entire real set. While it's not obvious whether this statement is either true or false in any real sense, it must be either taken or rejected as an axiom. From the standpoint of the rest of mathematics the theorem is both true AND false!

I think some of the above is quite inexact and may lead to the misunderstanding of critical facts. I'll try to clarify it a bit.

First of all, it's not necessarily true that no axiom of a theory can be proven from other statements. An axiom set is simply any set (finite of not) of logical statements, and a model of a theory is a structure (i.e a set endowed with some functions and relations...) that satisfies all the axioms of this theory. If an axiom can be proven using just the other axioms of the same theory, then it's redundant. It can be removed without changing the models of the theory, but it's OK also if we keep it. The axioms might even contradict one another, and in this case we say that the theory is inconsistent or contraditory. An inconsistent theory has no models, so any statement is both true and false in such a theory.

Then, it's not true that the statement of Euclidean Geometry "the sum of the interior angles of a plane triangle always equals 180 degrees" is an axiom. It's actually a theorem, in fact Euclidean Geometry has just 5 axioms, that are a lot "simpler" than this one, and can be combined to prove it. Of course, nothing changes if we take this one as an axiom as well (together with the classical 5), but it's just redundant, so Mr. Euclid wisely decided not to take it, and to prove it later on.

Moreover, Gdel did NOT prove that there are some theorems which cannot be proven true or false. A theorem of a theory is by definition a sentence that CAN be proven in that theory. First, we must say that Gdel's theorems are related just to First Ordel Logic (FOL), that is a simpler and less expressive logic in which we can use quantifiers just on variables, and not on relations. An example of a theory in Second Order Logic is the famous 2nd Order Peano Arithmetics, whose axioms are the classical ones related to the definition of successor, and of +, * and < (which are all First Order axioms, as they only quantify on variables), plus the induction axiom. This last axiom is a Second Order one, as it quantifies on unary relations (or, in other words, subsets of the structure). This theory is actually cathegoric, as it has just one model (up to isomorphisms...), that is obviously N (Dedekind's theorem).

On the other hand, we are not happy with 2nd Order Peano Arithmetics, as there's a theorem stating that we cannot define a formal system of logical inference to prove theorems in this theory, as the set of true sentences in N is not recursively enumerable. That is, there's no algorithm that can decide in a finite time if any given sentence is true in N. On the contrary, in FOL we can indeed define a formal mean of proving theorems (modus ponens, and so on...), and moreover there's a nice theorem (Gdel's completeness theorem) stating that a sentence of FOL is true in every model of a theory if and only if it can be proven formally from the axioms of that theory.

So, to get rid of the induction axiom of 2nd Order Peano Arithmetics, we introduce an infinite set of new First Order induction axioms, one for every sentence in FOL. So, instead of a big induction axiom, we have many small First Order induction axioms. Since the FOL sentences are countable many, and the subsets of N are not, it's obvious that these infinite axioms cannot substitute the big one without losing some subsets of the models, and in fact it happens that the so formed theory (1st Order Peano Arithmetics) is not cathegoric. Of course, there's N among its models, but there are also MANY non-standard models (i.e. models that are not isomorphic to N).

There are many interesting theorems about non-standard models of 1st Order PA, one of wich stating that there's no algorithm to compute the function + or * in any given non-standard model (i.e. they're not recursive: Tennenbaum's theorem). But the most famous result is Gdel's incompleteness theorem, basically stating that there's a sentence that is true in some models of 1st Order PA, and false in some other models (Gdel also showed how to "build" such a sentence). It immediately follows from this theorem and from Gdel's completeness theorem that there are some First Order sentences which are true in N, but not provable from the axioms of 1st Order PA, i.e. they're true sentences, but not theorems.

Then, it's inexact that Gdel proved the impossibility to prove the completeness or consistency of a theory. This is just what he did in his incompleteness theorem: he proved that 1st Order PA is incomplete! However, he did prove (in his 2nd incompleteness theorem) that any theory in any FOL language that can "encode" statements and proofs cannot prove its own completeness. For example, suppose you have an encoding for sequences of natural numbers with natural numbers (such an encoding is easily defined using the Chinese remainder theorem). Using this encoding, you can define another encoding for sentences and proofs. And then, you can say by Gdel's 2nd incompleteness theorem that the sentence "for any sentence s, there's a proof of s or a proof of not(s)" cannot be proven in 1st Order PA. We need encodings for sentences and proofs so that we can form the sentence above just using quantifiers on numbers, and so getting a FOL sentence.

Finally, the Continuum Hypothesis is not an example of Gdel's incompleteness theorem, as its statement is not in FOL. In fact, it quantifies on subsets of R, so forming at least a 2nd Order statement. Actually, the FOL theories of real numbers, complex numbers and Euclidean Geometry are all complete, meaning that they can formally prove or disprove any FOL sentence.

Note that Gdel's completeness and incompleteness theorems have similar names, and this is sometimes source of much confusion, as they're not mutually contradictory, and actually cohexist in 1st Order PA. The completeness theorem states that in First Order Logic, a sentence is true in every model of a theory if and only if it's provable from the axioms of this theory (i.e. it's a theorem). The incompleteness theorem states that there are true sentences in N that cannot be proven from the 1st Order axioms of Arithmetics.

Of course, whether proving both a theorem called "completeness" and one called "incompleteness" just in a few years can drive a man to madness, I can't tell.   ;)

How can we prove that the set of sentences of A (A be an Axiom System)
that are true in all models of A is an r.e.(recursively enumerable) set.

This is true in First Order Logic, and it's easy to prove indeed, using Godel's Completeness Theorem.
By this theorem we know that the set of sentences that are true in all models of a theory is precisely the set of sentences that are provable in that theory. By provable, we mean that these sentences can be derived from the axioms (and tautologies) by applying a finite set of inference rules (for example modus ponens). By applying again and again these inference rules to the axioms and to the theorems obtained so far, we obtain all provable sentences.
Now, we clearly see that this set is r.e.: take a sentence, and start building the tree of proofs within the given theory, comparing each theorem with the given proposition. Provided that the nodes of the tree are built in the correct order (for example in breadth-first order), then this algorithm stops if and only if the given proposition is a theorem.

Example of an undecidable proposition:

'I am an idiot.'   :D

(

Also, have a look at this:

http://www.geocities.com/robert.milleker

)


I have this 3'rd degree equation i could need some help with.

I've been told that 1 is one of the roots.
(x^3)-(6x^2)+(11x)-6=0

By putting 1 into x's place i get this:
(x^2)+5x+16+(10/x-1)

How do i proceed from here? As i've read i will now have to use this:
x=(-b+-root(d)/(2a)
But 'd' is negative. Can this really be true?

(For my own use: 307)

And here's another one in the same chapter as the above stated:

h(x)=(x^3)+(2x^2)-5x+2

What are the roots?

If i'm correct there are none... :?

[hide]

Use latex like $h(x)=x^3+2x^2-5x+2$. Well using factor theorem you plug in $1$ for $x$ and see if it = 0. If it does then (in this case it does) you use synthetic division or long division like $\frac{x^3+2x^2-5x+2}{x-1}$ and u get $(x-1)(x^2+3x-2)$ thus the roots are $1,\frac{-3\pm\sqrt17}{2}$.  :rotfl:[/hide]

P.S. i think this belongs in like the lower levels.

I have to correct myself. For the question in the second post it was:

h(x)=(x^3)+(2x^2)+x+2

What are the roots?

If i'm correct there are none...

For the first question i would need to see you go through the division of this equation:

(x^3)-(6x^2)+(11x)-6 : x-1

...or you can use the rational zeros theorem, put the equation in your calculator, and plug your potential rational zeros into "x= "....

if 1 is one of the zeros for sure,

and you cannot find any other zeros using the rational zeros theorem,

then you might have 2 complex zeros...

1 does not give me a zero. With no minusses in the equation you can't get anywhere near zero. Anyway i would like to do the most in my head instead of leaving everything to the calculator ;)

We have $h(x)=x^3+2x^2+x+2=x^2(x+2)+(x+2)=(x+2)(x^2+1)$ Now it's obvious that its roots are $x_1=-2, x_{2,3}=\pm i$.

Maybe you should practice polynomial factoring?

I will try and take a look at it. Could you help me with this dividend then?

(x^3)-(6x^2)+(11x)-6 : x-1

Plz not just the result. I would much rather see how you do the division line by line.

Well, it's much like long division of numbers. First you write

$x^3-6x^2+11x-6 : x-1$

Then you take the most significant terms in both polynomials and divide them. In the first it's $x^3$, in the second it's $x$. The quotient's $x^2$. You write that down after the equal sign:

$x^3-6x^2+11x-6 : x-1=x^2$

Now multiply the divisor and the quotient and underwrite the result underneath the corresponding terms of the first polynomial. Here $x^2\cdot(x-1)=x^3-x^2$:

$x^3-6x^2+11x-6 : x-1=x^2$
$x^3-x^2$

Now subtract what's underwritten from the corresponding terms of the upper polynomial. Here $(x^3-6x^2)-(x^3-x^2)=-5x^2$:

$x^3-6x^2+11x-6 : x-1=x^2$
$x^3-x^2$
$---$
$=\quad -5x^2$

Now lower down the next term from the upper polynomial, in this case $11x$:

$x^3-6x^2+11x-6 : x-1=x^2$
$x^3-x^2$
$---$
$=\quad -5x^2+11x$

and repeat the procedure (divide the most significant terms, write the quotient etc.) Keep doing it until you get the remainder, which generally should be a polynomial of a degree lower than the degree of the divisor, but in this case we know it will be zero.

Thx for this so far, but as you mention yourself this one should end with nothing remaining. However, i do end with remainings in this. I have tried look at it several times but can't see where i do it wrong.

This is what i get:
(x^2)+5x+16 and 10 remaining.

Can you show me how you would finish this one plz.

Hey, why don't you try dividing ${{-5x^2} \over {x^2}}$ correctly... no more on this from me. Feed your head :)

I've tried using latex but i really can't. I don't find it too easy to handle, sorry :( Might learn it in the future though.

Ok, now i've done the first one. So finding a solution to an equation is actually the same as finding the roots to the equation. You do it all the same way. And when you are told to factorize the equation you also need to find a solution to the equation before factorizing.

Yes! I can't believe your teacher hasn't taught you that. Consider a polynomial $P(x)$. Then we have that $P(\alpha) = 0$ if and only if $(x-\alpha)$ is a factor of $P(x)$.

I am my own teacher, so no help there :D

Well, i have learnt that, but just sometimes knowledge is jumping out of your head as fast as you receive it.

Now i thought i had it all right, but then a new problem seems to have appeared.

I need to solve this equation - plz tell me where i do something wrong.
(x^4)+(1,5x^3)-(1,5x^2)-x
x=1
(x^3)+(2,5x^2)+x
x=1

If i do this 3rd degree equation i end with (x^2)+3,5x and remainings set at 4,5x.

Note how easily you can factor $x$ out of that third degree one, reducing the problem to a quadratic.

I'm not sure how to do so. Is this the way to do it?

(x^3)+(2,5x^2)+x --> x((x^2)+2,5x+1)

and then:

x=0, x=(-b+-root(d))/(2a)

Am i doing it correctly, or...?

I seriously suggest you buy the AoPS books. They will be mighty handy to you.

wtf man, this is really easy, the factors are 1 and 5...

[u]@ Treething[/u]
1,353 posts where at least half of them are about if ppl are stupid or not. I just don't get you type of guys. Why the damn hell don't you just go somewhere else - is it really that funny to tell ppl when they are stupid or not? Or do you just have to have your finger in all threads. OK, i'm totally stupid, so what, this is not fascism so i still have the right to live and try to improve.

[u]@ Everyone else[/u]
As said, i'm stupid, so can anyone (with pathience) tell me why the hell i can't make this work.

(x^4)+(1,5x^3)-(1,5x^2)-x
I divide it with x-1
(x^3)+(2,5x^2)+x
--> x((x^2)+2,5x+1)
x=0^x=(-b+/-root((b^2)-4ac))/(2a)

x={-2;-0,5;0;1}

But this obviously isn't right.

[quote="Sporally"](x^4)+(1,5x^3)-(1,5x^2)-x
I divide it with x-1
(x^3)+(2,5x^2)+x
--> x((x^2)+2,5x+1)
x=0^x=(-b+/-root((b^2)-4ac))/(2a)

x={-2;-0,5;0;1}

But this obviously isn't right.[/quote]

Why not? The roots -2, -1/2, 0, and 1 are all roots in the original polynomial. That should mean it obviously [b]is[/b] right.

[quote="Sporally"][u]@ Treething[/u]
1,353 posts where at least half of them are about if ppl are stupid or not. I just don't get you type of guys. Why the - - don't you just go somewhere else - is it really that funny to tell ppl when they are stupid or not? Or do you just have to have your finger in all threads. OK, i'm totally stupid, so what, this is not fascism so i still have the right to live and try to improve.

[u]@ Everyone else[/u]
As said, i'm stupid, so can anyone (with pathience) tell me why the - i can't make this work.

(x^4)+(1,5x^3)-(1,5x^2)-x
I divide it with x-1
(x^3)+(2,5x^2)+x
--> x((x^2)+2,5x+1)
x=0^x=(-b+/-root((b^2)-4ac))/(2a)

x={-2;-0,5;0;1}

But this obviously isn't right.[/quote]

It is right, as scorpius119 said. Don't doubt every answer you get! :huh: And yes, AoPS textbooks might come in handy, too.

I calculated it too after i replied to this thread and found out, to my big surprice, it obvious was right. But Treething said it was 1 and 5, and for that reason i was confused.

I believe you have noticed that button to the left called 'LaTex'. I have taken a look at it several times and tried writing it the way it says, but still no big luck. I just need to take a greater look at it whenever i get some time to do so. Right now the math in the school is more important, but of course i would have been better learning it the other way around :oops:

[quote="sen"]hey guys

this is totally off topic

but i dont use latex similar to sporally

however, my reason is because i don't know how to create a document with latex..

like seriously.. ive installed miktex.. and i dont know.. ?![/quote]

You don't need to install anything or create a document to use LaTeX on this site. Just put the LaTeX code inbetween dollar signs. :)

the correct answers are 1,2, and 3 and these can be all found by using the solve function on yuor ti-89  :ninja:

yep ok i see.. let me see

$today$ $is$ $wednesday$

$5^2 = 25$

ahaha okk got it guys  :D

But i have to say it becomes quite harder when you try something more difficult, that is the reason i still don't use LaTex.


[quote="Arvind_sn"]Billy Bob has fourteen different pairs of socks in his drawer. They are just thrown around randomly in the drawer. Billy Bob once woke up in a hurry and had to get his socks quickly. Without switching the light on, he pulled out enough socks to know that he had at least one pair, and then he ran out of the room. How many socks did Billy Bob pull out?[/quote]

[hide]Using the Pigeonhole Principal, $\boxed{15}$[/hide]

[hide="Using what SplashD used,"]You get 15, which is 14+1[/hide]

[hide] Pigeonhole: If there are $kn+1$ objects in $kn$ boxes, there must be one box with at least $k+1$ objects in it.

You have $14$ boxes for the 14 pairs of socks. We want at least $2$ socks in one of the boxes. So, in this case, $k=1$, $n=14$, $kn+1=\boxed{15}$. [/hide]

[quote="Arvind_sn"]Billy Bob has fourteen different pairs of socks in his drawer. They are just thrown around randomly in the drawer. Billy Bob once woke up in a hurry and had to get his socks quickly. Without switching the light on, he pulled out enough socks to know that he had at least one pair, and then he ran out of the room. How many socks did Billy Bob pull out?[/quote]
[hide]After you pick fourteen, if they are all the same type, then the fifteenth will be different.[/hide]

[hide]By the pigeonhole property, 15.[/hide]

the answer is 15 because if bob picks 14 different socks, the 15th one is obviously going to be the same as one of the 14 different socks.
By the way, what's the pigeonhole property? (Sorry I'm kind of new to this)

[quote="Arvind_sn"]Billy Bob has fourteen different pairs of socks in his drawer. They are just thrown around randomly in the drawer. Billy Bob once woke up in a hurry and had to get his socks quickly. Without switching the light on, he pulled out enough socks to know that he had at least one pair, and then he ran out of the room. How many socks did Billy Bob pull out?[/quote]

[hide]15? or 3[/hide]


Given $ a,b,c\in R$
[b]Problem 1[/b]:Prove that  $ a^4\plus{}b^4\plus{}c^4\plus{}ab^3\plus{}bc^3\plus{}ca^3\geq2(a^3b\plus{}b^3c\plus{}c^3a)$

[b]Problem 2[/b]:True or false:$ a^4\plus{}b^4\plus{}c^4\plus{}2(ab^3\plus{}bc^3\plus{}ca^3)\geq3(a^3b\plus{}b^3c\plus{}c^3a)$
[b]
Problem 3[/b]: Find the best constant: $ a^4\plus{}b^4\plus{}c^4\plus{}k(ab^3\plus{}bc^3\plus{}ca^3)\geq(k\plus{}1)(a^3b\plus{}b^3c\plus{}c^3a)$

[quote="dduclam"]Given $ a,b,c\in R$
[b]Problem 1[/b]:Prove that  $ a^4 \plus{} b^4 \plus{} c^4 \plus{} ab^3 \plus{} bc^3 \plus{} ca^3\geq2(a^3b \plus{} b^3c \plus{} c^3a)$
[/quote]
It's very old problem. I didn't find the link.
$ \sum_{cyc}(a^4 \plus{} ca^3 \minus{} 2a^3b ) \equal{} \frac {1}{2}\sum_{cyc}(x^2 \minus{} y^2 \minus{} xy \plus{} yz)^2.$
[quote="dduclam"]
Problem 2:True or false:$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(ab^3 \plus{} bc^3 \plus{} ca^3)\geq3(a^3b \plus{} b^3c \plus{} c^3a)$[/quote]
It's wrong: $ a \equal{} b \equal{} 1.3$ and $ c \equal{} 0.$

[quote="akai"][quote="dduclam"]
Problem 2:True or false:$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2(ab^3 \plus{} bc^3 \plus{} ca^3)\geq3(a^3b \plus{} b^3c \plus{} c^3a)$[/quote]
It's wrong: $ a \equal{} b \equal{} 1.3$ and $ c \equal{} 0.$[/quote]

Arqady, i think that you were wrong  :wink:[/quote]
I mean $ a \equal{} 1.3,$ $ b \equal{} 1$ and $ c \equal{} 0.$ :D  Thank you! :)


When written as a decimal, the fraction m/n (with m<n) contains the consecutive digits 2,5,1. Find the smallest possible n.

Just checking to see if I was right...

This question has been posted before.  Try a search for it.


function
ratio
algebra
polynomial
calculus
calculus computations

Let $ f(x)$ be a polynomial of degree five. Determine an example of $ C,\ L,\ M$ such that the graph of $ C: y \equal{} f(x)$ is tangent to two parallel lines $ L,M$ at two points respectively, hence for the example, find the ratio of two line segments cut by three points which comprise of the intersection point of $ C,\ L$ and the two tangent points.

[color=green]Moderator edit: added correction as requested[/color]


floor function
logarithms
function
irrational number

Find, with proof, the 2008th decimal digit of
\[ \sum_{k \equal{} 1}^\infty\frac {1}{10^k \minus{} 1}
\]
[hide="Answer"]
8
[/hide]
How about the $ n$th decimal digit?
[hide="Answer"]
\[ \left(\sum_{k \equal{} n}^{n \plus{} \lfloor\log_{10}n\rfloor}{\left\lfloor\frac {\psi(k)}{10^{k \minus{} n}}\right\rfloor}\right)\mod 10
\]
where $ \psi(m)$ denotes the number of factors of $ m$ and $ \lfloor\cdot\rfloor$ is the usual floor function.
[/hide]
Does the sum converge to an irrational number?
[hide="Answer"]
Hell if I know, I posted this in hoping someone would help me prove it :P.  Though it seems that it should be possible to prove using elementary methods, since the sum being rational would clearly force very tight constraints on the distribution of primes.
[/hide]

Well..
 [hide] $ 2008 \equal{} 2^3*251$, so that makes $ 8$ factors, which makes the digit $ \boxed{8}$[/hide]

[quote="jimhu"]Well..
 [hide] $ 2008 = 2^3*251$, so that makes $ 9$ factors, which makes the digit $ \boxed{9}$[/hide][/quote][hide="No"]
First, we look at the first few terms of the sum:
$ \frac{1}{9}+\frac{1}{99}+\cdots$. This happens to also be ${ .\overline{1}+.\overline(01}+.\overline{001}+\cdots$. Now, we see that a $ 1$ is added to each place for each factor it contains. Thus, $ 2008=2^3\cdot251$ has 8 factors and thus that digit is $ 8$. But we must also check the next spot to make sure it does not trade over to this one when summing up the places - it's $ 2009$, which is prime, so the answer is $ 8$. (Beyond the next place the place after would not have enough factors to trade enough over).[/hide]

[hide="By the way, the value of this sum is..."] $ \zeta(10)^2$.  See [url=http://en.wikipedia.org/wiki/Dirichlet_series]Dirichlet series[/url]. [/hide]
Edit:  Misread.   :blush:  What this actually is is a [url=http://mathworld.wolfram.com/LambertSeries.html]Lambert series[/url].

t0rajir0u, if by $ \zeta$ you meant the usual Zeta function, then certainly not.  For one thing, the sum is clearly less than 1, whereas $ \zeta(x)$>1 for real $ x > 1$.

[quote="t0rajir0u"][hide="By the way, the value of this sum is..."] $ \zeta(10)^2$.  See [url=http://en.wikipedia.org/wiki/Dirichlet_series]Dirichlet series[/url]. [/hide][/quote]

It isn't a Dirichlet series, it is $ \sum{\frac{d(n)}{10^n}}$, where d is the number of divisors, not $ \sum{\frac{d(n)}{n^{10}}}$. But the sum of the second series can be found from $ d=u*u$ to be what you gave.


algebra
polynomial
function
Euler
national olympiad

[b]INDIAN NATIONAL MATHEMATICAL OLYMPIAD 1992[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340072#p340072][b]Problem 1 [/b][/url]
In a triangle $ABC$ , $\angle A = 2 \times \angle B$. Prove that $a^2 = b (b+c)$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340073#p340073][b]Problem 2[/b][/url]
If $x , y, z \ \in \ \mathbb{R}$ such that $x+y +z =4$ and $x^2 + y^2 +z^2 = 6$ , then show that each of $x, y, z$ lies in the closed interval $[ \dfrac{2}{3} , 2 ]$. Can $x$ attain the extreme value $\dfrac{2}{3} \ or \ 2$ ?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340074#p340074][b]Problem 3[/b][/url]
Find the remainder when $19^{92}$ is divided by 92.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340076#p340076][b]Problem 4[/b][/url]
Find the number of permutations $( p_1, p_2, p_3 , p_4 , p_5 , p_6)$ of $1, 2 ,3,4,5,6$ such that for any $k , 1 \leq k \leq 5$ , $(p_1, ... p_k)$ does not form a permutation of $1 , 2, ... k$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340077#p340077][b]Problem 5[/b][/url]
Two circles $C_1$ and $C_2$ intersect at two distinct points in a plane. Let a line passing through $P$ meet the circles $C_1$ and $C_2$ in $A$ and $B$ respectively. Let $Y$ be the midpoint of $AB$ and let $QY$ meet the cirlces $C_1$ and $C_2$ in $X$ and $Z$ respectively. Show that $Y$ is also the midpoint of $XZ$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340078#p340078][b]Problem 6[/b][/url]
Let $f(x)$ be a polynomial in $x$ with integer coefficients and suppose that for five integers $a_1, \ldots a_4$ one has $f(a_1) = f(a_2) = \ldots = f(a_5) = 2$. Show that there does not exist an integer $b$ such that $f(b) = 9$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340079#p340079][b]Problem 7[/b][/url]
Find the number of ways in which one can place the numbers $1, 2, 3, \ldots n^2$ in the $n^2$ squares of a $n \times n$ chesboard , one on each , such that the numbers in each row and in each column are in Arithmetic Progression. ( $n \geq 3$)

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340081#p340081][b]Problem 8[/b][/url]
Determine all pairs $(m,n)$ of positive integers for which $2^{m} + 3^{n}$ is a perfect sqaure.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340082#p340082][b]Problem 9[/b][/url]
Let $A_1, A_2, \ldots , A_n$ be an $n$ -sided polygon. If $\dfrac{1}{A_1 A_2} = \dfrac{1}{A_1 A_3} + \dfrac{1}{A_1A_4}$ , find $n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=340084#p340084][b]Problem 10[/b][/url]

Determine all functions $f : \mathbb{R} - [0,1] \mapsto \mathbb{R}$ such that
\[ f(x) + f ( \dfrac{1}{1-x} ) = \dfrac{2(1-2x)}{x(1-x)} \]

This problem is a direct application of Eulers theorem.
Euler function of 92 = 44
Therefore 19^44 is congruent to 1 modulo 92,
which implies, 19^88 is congruent to 1 modulo 92,
19^4 is congruent to 49 modulo 92 (by calculation)
Therefore, 19^92 shall leave a remainder 49 when divided by 92.

[quote="white_horse_king88"]Why did you repost these, all of them, again in Intermediate?[/quote]Each problem must be posted in its own thread. This is just a collection of links (which is now slowly replaced by the [url=http://www.mathlinks.ro/Forum/resources.php]resources pages[/url]) ;)


function
algebra
functional equation
algebra unsolved

Let $a,b$ be two constants within the open interval $(0,\frac{1}{2})$. Find all continous functions $f(x)$ such that \[f(f(x))=af(x)+bx\] holds for all real $x$.

This is much harder than the problems we had in the 1st TST...

Suppose some k so that f(k) = k.

Now k = ak + bk, so that a+b = 1, contradiction.

---

Case 1:  f(x) - x > 0 for all x.

Then f(f(x)) > f(x), whereby (a-1)f(x) + bx > 0 for all x.

We have bx < b f(x), whereby (a+b)f(x) > f(x), or a+b > 1, contradiction.

---

Case 2:  f(x) - x < 0 for all x.

Then f(f(x)) < f(x), where again by a similar process to Case 1, we get a+b > 1, contradiction.

====
Thus, there is no such f.

*note: i didnt check my proof but it appears right.

Your proof is wrong. Firstly you can have $f(0)=0$ so you cannot assume that for all $x$ you have $f(x)>x$.

Secondly there exists at least two such functions:

$f(x)=cx$ where $c$ is the root of $x^2-ax-b=0$

But the good thing about your post above is that it can be used to prove $f(0)=0$.

This has been incredibly frustrating. This problem was used in a very old Romanian NMO, I distinctly remembered solving  a very long time ago, but couldn't nail it :).

Let $t>0,t'<0$ be the roots of $x^2-ax-b=0$, put $u=\frac 1t,u'=\frac 1{t'}$, and let $g=f^{-1}$ ($f$ is continuous and injective, and thus bijective, and also strictly monotonic). Notice that $|t|>|t'|\Rightarrow|u|<|u'|\ (*)$. I'll use the fact that $f(0)=0$, which can indeed be shown by employing the method used by singular.

We now write the recurrences for both $f$ and $g$. We get $f^{(n)}(x)=\frac{f(x)-t'x}{t-t'}\cdot t^n+\frac{tx-f(x)}{t-t'}\cdot t'^n\ (1)$ and $g^{(n)}(x)=\frac{g(x)-u'x}{u-u'}\cdot u^n+\frac{ux-g(x)}{u-u'}\cdot u'^n\ (2)$. Now, $(*),(1),(2)$ imply that for a fixed $x$ and large $n,\ f^{(n)}(x)$ will have the same sign as $\frac{f(x)-t'x}{t-t'}\cdot t^n\ (\#)$, while $g^{(n)}(x)$ will have the same sign as $\frac{ux-g(x)}{u-u'}\cdot u'^n\ (\#\#)$ (unless $f(x)=t'x$ and $g(x)=ux$, respectively).

Suppose first that $f$ is strictly increasing. In this case, we will have $f(x)>0,\ \forall x>0$, which, in turn, means that $g^{(n)}(x)>0,\ \forall x>0,\ \forall n$. If we could find some $x$ for which $g(x)\ne ux$, then for large $n,\ (\#\#)$ would change sign infinitely often, and so would $g^{(n)}(x)$, according to the last observation in the previous paragraph, thus yielding a contradiction. It means that in this case $g(x)=ux\iff f(x)=tx$.

Assume now that $f$ is strictly decreasing. In this case $f^{(n)}(x)$ has alternate signs as $n$ increases (for every $x$). If we could find an $x$ for which $f(x)\ne t'x$, for large enough $n,\ (\#)$ would have constant sign, and then so would $f^{(n)}(x)$, leading to contradiction. This means that when $f$ is decreasing we have $f(x)=t'x$.

I hope it's correct. It would be very dissapointing if it wasn't :).

I'm not sure if your solution is correct or not, though I can't find any mistakes... I must note that none of the 19 students writing this problem got it right, though.

I thing that this kind of functional equation could be solved by using this sequence:
$x_{0}=x$
$f(x_{n})=x_{n+1}$
Reporting this in the original equation gives:
$x_{n+2}=ax_{n+1}+bx_{n}$
And thus there exist two reals $A$ and $B$ such that:
$x_{n}=A(\frac{a+\sqrt{a^2+4b}}{2})^n+B(\frac{a-\sqrt{a^2+4b}}{2})^n$
Now ( The tricky part) by using the propreties of $f$ we can find an argument to eliminate alternatively $A$ or $B$.
And conclude by putting$x_{1}=f(x_{0})$

Nothing new here :?. That's exactly what I did in order to get $(1)$ and $(2)$.

[quote="grobber"]This has been incredibly frustrating. This problem was used in a very old Romanian NMO, I distinctly remembered solving  a very long time ago, but couldn't nail it :).

[/quote]
In fact, the problem proposed at the Romanian NMO (author M. Piticari) only asked to prove that $f(0)=0$ ;)


modular arithmetic
combinatorics unsolved
combinatorics

Let $ (a_i)$ for $ 1\leq i\leq 37$ be a finite sequence of [b]positive[/b] integers, that $ \displaystyle\sum_{i \equal{} 1}^{37} a_i \leq 60$.
Show that there exists subsequence of consecutive elements of $ (a_i)$, such that its sum is exactly $ 13$.

There must be $ 13$ of $ 1$s in the sequence, otherwise $ \sum_{i \equal{} 1}^{37}a_{i}$ would be greater then $ 60$.

Oh of course, sorry, i should have mentioned that we want to show there exists a subsequence of consecutive elements of sequence $ (a_n)$ that it's sum is $ 13$ (I've just edited it).

Consider $ 37$ sums of the form $ S_k \equal{} \sum_{i \equal{} 1}^{k} a_i,\forall 1 \le k \le 37$, we have that $ 1 \le S_1 < S_2 < ... < S_{37} \le 60$. Thus we have $ S_k \ge k,\forall k \equal{} \overline{1,37}$.

Consider the sums $ S_9,S_{10},...,S_{37}$ (there are 28 sums), by pigeon-hole theorem, there exist $ 9 \le a < b < c \le 37$ such that $ S_a \equiv S_b \equiv S_c \pmod{13}$. If three differences of them $ |S_a \minus{} S_b|,|S_b \minus{} S_c|,|S_c \minus{} S_a|$ are all greater than $ 13$ then they must be at least $ 26$, thus $ S_c \ge 26 \plus{} S_b \ge 52 \plus{} S_a$, but $ S_a \ge a \ge 9$ so $ S_c \ge 52 \plus{} 9 > 60$, contradiction.

Thus, one of those differences is $ 13$, this yields a subsequence of consecutive elements of $ (a_n)$ whose sums is $ 13$, and the problem is solved.

[b]Comment:[/b] We may just need $ 36$ numbers :lol:


oopss edited... mom was calling : again dinner... bye

sigh, i believe this has been issued more than once. keep your mistakes, cause people might think that people never make mistakes if its always edited

yeah, I have no clue as to [b]what[/b] you guys are talking about :)

Tare, why do you post so much when most of the time you don't even post the answer? It's really bugging me because I have to wade through a million meaningless posts before getting to a solution.

Also, are you sure this is the only information given? Is it a right triangle? I'm having a bit of trouble proving anything without anymore info.

A couple things
a.) I'm [b]trying [/b]not to post answers since if the comoderator answers all the questions, there won't be any fun
b.) I do know the answer (well, at least most of the time :) ), fyi. I mean, even if I can't solve any Intermediate Topics Question and act like a total idiot, that doesn't mean I'm still a stupid idiot compared to other middle-schoolers. I'm always comparing myself to high schoolers when I'm in middle school so I'll be fully prepared for Colledge/University.
...because it sounded like you were doubting my skilz (Now [b]that[/b] sounds Gangsta:) )

besides, it's not actually my job to solve the problems. My job is to post questions, help people out, and make the whole process fun. So acting stupid and fooling around can be considered a part of my job, especially in an easy forum  :)

c.) If you just want posts deleted, all you've got to do is PM me. I mean, you don't have to call them "meaningless"...that's pretty harsh, I think
d.) I was asking what in the world was going on between Syntax and Chinaboy

and e.) Yes, I think it's enough. (unless I screwed up :) )

Syntax Error wrote:there is a triangle ABC.  F is a point on AB, E is a point on AC, and D is a point on BC.  BDEF is a rhombus with vertices on ABC.  Given AB=10 and BC=15, find DE

[hide]

Ok... proportions... i think. yahhh...

x:15 = 10-x:10

EF:BC = AE:AB

25x = 150

x = 6[/hide]

lol, i dont care. just like, explain um, i dunno. like your thinking, and how you like arrived at your calculations and stuff. like the proportion part i guess. hmmmmm...sure, that could do. whatever. i dont care


What is your score (Sprint/Target) of the mathcounts school test?
Anybody got a perfect score?

My coach nvr told me my School score, she just told me if i made it or not.

Are the school tests easier than chapter?

I think so, last year I got a 42 on the School test, but I did too horrible on the Chapter test to even mention my score (it was less than I got at State!)

[quote="plokoon51"]i never knew of a school mathcounts test. is it standarized? in 7th grade our teacher just gave us 10 questions to see who would qualify and in 8th grade we got a 30 question mathcounts-style sprint round test that was as hard as the State round of 2004.[/quote]

MathCounts has a specific 'school round,' but it isn't always used by the coaches. It's ultimately the coach's choice who they want on the team, and some don't use the school round. For some schools, the school round is too easy and they have too many ties for perfects or high scores so they make their own problems or use old state or national competitions.

Every school that registers gets a packet containing ribbons, certificates, and the "School Competition Booklet".  The School Competition Booklet contains competitions in the same format as the Chapter Competition -- Sprint, Target, Team, and Countdown Rounds.  Whether and when a coach uses the School Competition Booklet is entirely up to your individual coach.

Some coaches use this resource to choose their team.  Some don't use it at all.  [Our District holds an intramural competition, prior to the Chapter Competition, using the rounds from the School Competition.]

Up to this point, everything in this thread looks appropriate to me.  But, I want to remind everyone to refrain from making comments about specific questions (or difficulty of specific questions), as some schools will be using this right up through their Chapter Competitions, which run through the end of February.

-- Coach from Northern Colorado

I got all right on my target and missed one (possibly 2, can't remember) on my sprint... It was a dumb error though...REALLY DUMB.

all target right... and 4 wrong on sprint (maybe I should learn how to answer to right question and/or actually read the entire question... ust might help.)
but our coach uses the school round for the simple reason: gives most accurate depicition of everyone's ability.

our coach has only been giving us sprints...uh oh.......we done the test, but she never tells us who got what and it's really nervewracking.  are all of u guys so good that even one mistake is bad? :blush:

is it just me or has no one posted their score for this year?  I got 39.  I think its the same for all mathcounts test for the school...

elvenchamp777

Is there any place online where they have a sample of the school test? I don't think I've ever seen one.

I don't know, but don't worry about it. It's extremely basic... And I mean extremely...If you want an example really bad, I might be able to scan one up for you.  E-mail me if it matters that much, and I'll put it up on my website.

The school round I'm talking about is the one from last year.  So I won't be giving anything away...

I got perfect on school last year, and stuff, but somehow i got cocky and my school rival (#2) beat me on countdown round.  and he didn't do that great on written either!  Ya I'm kindof wierd at MathCounts.  I got first at state (written and after countdown), but got 6th at chapter!  And at state I won by like 2 points off 2nd place.  I guess I was just my ideal test.

well...target is always a lot easier than sprint... so it would help ur score

yea... all of the smart 9th graders ar gone

I got something around, 38.

You think that is bad...

Our school has been state champs for 4 years in a row, and we're probably make states again this year.

Only one problem. NEXT YEAR!!!

Next year, all the people that are anywhere near smart will be in 9th grade, so there will be only 3 returing next year, and only one (me) was on the team, or even an individual! And next year I want to win states again!

What am I going to do?... :(

You're going to start working extra hard 20 hours a week including vacations and summer so that you get a 46 at State and do all 10 team problems so you can carry your team :)

In case you're wondering, I'm dwx's coach.

I got 21/7 (35). But since I'm in seventh grade, I have a chance next year.

I came tied for fourth in my school (Hunter, NY) with a score of 39.
[i]Mod's note: let's not talk about the chapter competition yet, okay?[/i]

I had a pathetic 33 on the school competition, and I was the one who beat Azn in the chapter.

Thats funny. Did u make it to state?  :P

Not sure yet. They might try to say that I was "on the winning team." Nice way to pull out a technicality.  :roll:

I AM MATHELTE...I GOT 2nd IN CHAPTER COMPETITION RAWR :D

I AM BABA MATHLETE OF DOOM :D  :D  :D  :D  :D  :D  :D  :blush:  :blush:  :blush:  I AM A GOOD SCORING TOP MATHLETE PEANUT BUTTER JELLY TIME OF DOOM MY NAME IS DANE HENNESEY

[quote="LuCy4EvA"]wait, does every school take the same test for the school quilifying?[/quote]I think most of the schools with MATHCOUNTS teams are sent the same in-school tests because my MATHCOUNTS teacher said that she talked to a teacher in Indiana and 2 of her kids got perfects on the test that we took.  She said that most people use the same test for the in-school competition if they are registered.

27 on sprint (mostly dumb mistakes)

7 on target (dumb mistake)

7 on team (dumb mistakes)

41 total (team doesn't count and targets double, or 1 point for each)

Edit: First at school!

actually, metzfan got lucky on the last 10 sprint questions. I panicked toward the end and got most of those wrong. So the graders used a tiebreaker and that's the only reason. lol :blush:

I made three very stupid arithmetic mistakes on Sprint, landing me a 43.  I'm happy though.

I got a 40 or 41 for my school thingy. Our school is really dumb so this is like a record besides this girl in our school. I think she got like a 44 or a 45. Our school is so dumb.[/hide]


calculus
integration
function
trigonometry
calculus computations

hello, setting $ t\equal{}\sqrt[3]{\tan(x)}$ we get $ x\equal{}\arctan(t^3)$ and from here $ dx\equal{}\frac{3t^2}{1\plus{}t^6}\,dt$
so our integral is now $ 3\int\frac{t^3}{1\plus{}t^6}\,dt$, this is an integral of a rational function. Note that
$ \frac{t^3}{1\plus{}t^6}\equal{}\frac{1}{3}\frac{t^3\plus{}t}{t^4\minus{}t^2\plus{}1}\minus{}\frac{1}{3}\frac{t}{t^2\plus{}1}$.
Sonnhard.


For $a,b,c,x,y,z$ positive numbers prove

$\frac{x}{y+z}a+\frac{y}{z+x}b+\frac{z}{x+y}c +\frac{1}{2}(a+b+c) \geq \sqrt{ab} + \sqrt{bc} +\sqrt{ca} $

It's not very hard.

By noting that
\[
\frac{ax}{y+z} + a = \frac{x+y+z}{y+z}a
\]
We can rearrange the ineq into
\[
\left( \frac{a}{y+z} + \frac{b}{x+z} + \frac{c}{x+y}\right) ((y+z) + (x+z) + (x+y) ) \geq (\sqrt{a} + \sqrt{b} +\sqrt{c})^2
\]
which is clear with Cauchy-Schwarz.


here is a 9th grade problem which I found in the short list of this years national mathematical competition ..

Let a,b,c be real numbers strictly pozitive and n is natural. Compare the numbers

A= a/(na+b)+b/(nb+c)+c/(nc+a)  and B=3/(n+1)

so... we have that

if 0<=n<=1/2 we have A>=B
if n>=2 we have A<=B

what happens for 1/2<n<2 ?

waiting for further comments.

cheers! :D

if n is natural and 1/2<n<2 => n=1 we have to campare:
a/(a+b)+b/(b+c)+c/(c+a) with 3/2 if x=a+b y=b+c z=c+a we have:
a=(x+z-y)/2 b=(x+y-z)/2 c=(y+z-x)/2
A=1/2[(x+z-y)/x+(x+y-z)/y+(y+z-x)/z] so we have to compare
C=(x+z-y)/x+(x+y-z)/y+(y+z-x)/z with D=3
C=(z-y)/x+(x-z)/y+(y-x)/z+3 D=3
if E=(z-y)/x+(x-z)/y+(y-x)/z>0 then A>B
(z-y)/x+(x-z)/y+(y-x)/z=(-1)*(x-y)(y-z)(x-z)/xyz =>
if x>y>z  E<0 => B<A
if z<y<z E>0 => A>B
if x=y=z => A=B
hope it'z corect


I want to learn Fridrich F2L. Do you know any simple, easy websites where I can learn it?

[quote="Treething"]Third one down.

PG-13 warning.
http://tinyurl.com/362w2l[/quote]

I have but all I've found is expert stuff.

I like this one: [url]http://www.cubewhiz.com/f2l.html[/url]

Use the algs that say FR (front right) (remember this is cross down)

This site is awesome:
[url]http://cubefreak.net/F2L.html[/url]


Chris and Mike take turns, with Chris going first. Chris tosses a fair coin at each of her turns, while Mike rolls a fair die at each of his turns. The game ends when either Chris tosses a head, in which case Chris wins, or Mike rolls a 3, in which case Mike wins. What is the probability that Mike wins?

Since Chris goes first, he needs to make tails, in order for Mike to go. Therefore he has a $1/2$ chance of making tails. But, then Mike needs to get a $3$, in which there is a $1/6$ probability, therefore, on the first try, Mike has a $1/12$ chance of winning. But say that Mike doesn't roll a 3, then the whole process keeps continuing until someone wins.
Then, though, it becomes an infinite geometric series, in which $a$ is $1/12$ and $r$ is $1/12$ therefore, it is equal to $1/11$, which is the probability that Mike wins.

[quote="Palytoxin"][quote="4everwise"]Chris and Mike take turns, with Chris going first. Chris tosses a fair coin at each of [b]her[/b] turns, while Mike rolls a fair die at each of his turns. The game ends when either Chris tosses a head, in which case Chris wins, or Mike rolls a 3, in which case Mike wins. What is the probability that Mike wins?[/quote]

i didn't know Chris can be a female name...  :blush:[/quote]

Christine, or Christina

Most of the female Chris's I know are spelled Kris...but anyways, wer're gettting a bit off topic here  :D

[quote="Iversonfan2005"]Hey is $1/11$ correct?  ;)[/quote]

No, I don't think $r=\frac 1{12}.$

[hide="Solution 1"]
Doing the infinite progression, like Iversonfan2005 explained, we get:

\[\left(\frac 12\right)\left(\frac 16\right)+\left(\frac 12\right)^2\left(\frac 56\right)\left(\frac 16\right)+\left(\frac 12\right)^3\left(\frac 56\right)^2\left(\frac 16\right)+\cdots = \left(\frac 12\right)\left(\frac 16\right)\left(1+\frac 12\cdot\frac 56+\cdots\right)=\boxed{\frac 1{7}}.\][/hide]

[hide="Solution 2"]
Let $p$ be the probability that he wins.  Consider every possibility for him winning.  He can either win on the first turn or on any turn after the first.  But since the game can go on forever, there is no way to distinguish the first turn from the second for a game going on indefinitely :!: Thus

\[p=\frac 12 \cdot \frac 16 + \frac 12\cdot \frac 56\cdot p\Rightarrow p=\boxed{\frac 1{7}}.\][/hide]

Oh, I see what I did wrong, yes Joe, you're right, I should've known that $1/12$ was not the ratio, because Mike has a $5/6$ chance of not rolling a $3$ which would bring it to an infinite series, not $1/6$. That was my mistake right there.

[quote="math92"][hide]Mike has a 5/6 chance of not rolling a 3.  Cris has a 1/2 chance of flippin a head.  16*1/2=[b]1/12[/b][/b][/hide][/quote]

Try explaining that again.  Your argument didn't make sense.

The probability of Mike winning the game is equal to the probability of Mike winning on his first turn plus the probability of Mike winning on his second turn plus the probability of Mike winning on his third turn, and so on.

For Mike to win on his first turn, Chris must toss tails and Mike must roll a three.  We will call the event of Chris tossing tails $A$.  We will call the event of Mike rolling a three $B$.

\[A=\frac{1}{2}\]
\[B=\frac{1}{6}\]

The probability of Mike winning on his first turn is $A\times B=\frac{1}{12}$.

For Mike to win on his second turn, Chris must toss tails, Mike must roll a number other than three, Chris must toss tails, and Mike must roll a three.  We will call the event of Mike rolling a number other than three $C$.

\[C=\frac{5}{6}\]

The probability of Mike winning on his second turn is $A\times C\times A\times B=\frac{5}{12}\times\frac{1}{12}$.

Each time we "add a turn", we multiply the probability of Mike winning on the previous turn by $A\times C$.  Therefore, the probability of Mike winning the game is the sum of an infinite geometric series with first term $A\times B=\frac{1}{12}$ and rate $A\times C=\frac{5}{12}$.  Another way to look at the problem is that each time $A\times C$ happens, the game effectively starts over.

To find the sum of the series, we use the formula for the sum of an infinite geometric series $\frac{a}{1-r}$.  The sum is

\[\frac{\frac{1}{12}}{1-\frac{5}{12}}=\frac{\frac{1}{12}}{\frac{7}{12}}=\frac{1}{7}\]

Therefore, the probability of Mike winning the game is $\frac{1}{7}$

O man!  I've really been off this week.  I messeup my arithmetic but I get 1/7 now also  :blush:

yeah, $\frac{1}{7}$ is the correct answer.


linear algebra
matrix
superior algebra
superior algebra unsolved

Let $a_i,x_i \in R, a_i,x_i \neq 0 (i=1,2,\cdots,n)$ and $m \geq n $ be natural numbers.
$B$ be the matrix such that:$b_{ij}=\sum_{k=0}^na_kx_k^{m+i-j}$
Find $det(B)$

Suppose $x_1,x_2,...,x_{n-1}$ are roots of $P(x)=x^{n-1}+c_{n-2}x^{n-2}+...+c_1x+c_0=0$. Then replace the last row $b_n$ with $b_n+b_{n-1}c_{n-2}+...+b_1c_0$. We see that $b_n$ becomes \[(a_nx^{m}P(x_n),a_nx^{m-1}P(x_n),...,a_nx^{m-n+1}P(x_n)).\] So we can divide it into $a_nx^{m+1-n}P(x_n)=a_nx^{m+1-n}(x_n-x_1)...(x_n-x_{n-1})$, then $b_n$ becomes $(x^{n-1},...,x,1)$. Now we easily exclude $x_n$ from all other rows. Performing the same thing with $b_{n-1}$, ..., $b_2$ we obtain a Vardermonde determinant.

Therefore, answer is $(-1)^{n(n-1)/2}a_1...a_n(x_1...x_n)^{m+1-n}\prod_{i\neq j} (x_i-x_j)^2$.

Sorry.
Find $det(B)$ with $b_{ij}=(\sum_{k=0}^na_kx_k^{m+i-j})^n$

Are you sure?  :huh:
I even don't want to think about...


basically, one person posts a fact for people to find, and everyone else has to try and find it, using the internet.

[u]Finding the answer[/u]
you may use ANYTHING on the internet.
when you find it, post
*answer*
*URL*
when you find the answer, you may post a question, or you may ask someone else to post a question
and we will keep score. everything you answer correctly is 1 point.

[u]Posting a question[/u]
KEEP IT CLEAN!!
try to make it hard so someone cant just google it. dont make it impossible either
if someone cant figure it out, then post FORFEIT, and the poster of the question gets 1 point.

i will start(easy)

name 1 person born in the year 1337 and something they accomplished.

[hide="click to reveal scores"]
none yet :(
[/hide]

Louis II, Duke of Bourbon

Launched a crusade against Hafsids of Tunis

New:

Find out how much money the richest man in the world had (not adjusted for inflation.

remember to post the URL so people can make sure your not making stuff up.

John D. Rockefeller
200 billion dollars
http://www.askmen.com/toys/top_10/11b_top_10_list.html

Martine75

[url=http://www.artofproblemsolving.com/Forum/index.php]http://www.artofproblemsolving.com/Forum/index.php[/url]

How many finger does Tiktali (sp?) fossil have?

four fingers

http://tiktaalik.uchicago.edu/photos.html

New

What is the effect of photo receptors becoming desensitized to a color called?

Afterimage effects

[url]http://www.houman.org/color.htm[/url]

Which came first, the chicken or the egg?  Not joking, the answer is so obvious.

The egg.

[url]http://www.en.wikipedia.org/wiki/Dinosaur[/url]

Why did Alfred Nobel make the Nobel Prize and to whom was it first awarded to?[/url]

According to Nobel's will, the Peace Prize should be awarded "to the person who shall have done the most or the best work for fraternity between nations, for the abolition or reduction of standing armies and for the holding and promotion of peace congresses."

Frédéric Passy

http://en.wikipedia.org/wiki/Nobel_peace_prize

New question

[img]http://www.glyphweb.com/esky/_images/illustrations/betelgeuse.gif[/img]

What star is this?

Betelgeuse.

http://www.glyphweb.com/esky/_images/illustrations/betelgeuse.gif

(lol, I just used the image location)

How many legs do millipedes have?

Though the name clues that they have 1000 legs, most millipedes have between 80 and 400 legs.
http://en.wikipedia.org/wiki/Millipede

Who became Prince of Turov on 1134?

Also, I think urbandictionary should not EVER be used as a resource.

http://encarta.msn.com/related_761561349_42/smallest_mammal.html
Kitti’s Hog-Nosed Bat

[quote="james4l"]Iziaslav Vladimirovich or Viacheslav Vladimirovich, depending on which portion of 1134 you mean (beginning of 1134 or end of 1134?)

http://en.wikipedia.org/wiki/Prince_of_Turov

What is the maximum number of units/buildings can there be in one time in a game of Starcraft?[/quote]

Heh, he said who became Prince of Turov, which is obviously the second guy.

Fritz Platten
Lenin
Ture Nerman
Otto Grimlund
Leon Trotsky
Zinoviev
Bukharin
Leonard Shapiro
Marx
Mao
Friedrich Engels
Rosa Luxemburg
Joseph Stalin
Hegel
Kim Il-sung
Peter Kropotkin
Josip Broz Tito
Daniel De Leon
Anton Pannekoek
Herman Gorter

http://en.wikipedia.org/wiki/Left_communism
http://en.wikipedia.org/wiki/Titoism
http://en.wikipedia.org/wiki/De_Leonism
http://en.wikipedia.org/wiki/Communism
http://en.wikipedia.org/wiki/Juche
http://en.wikipedia.org/wiki/Lenin

[quote="hunter34"]200.

My Starcraft game.

Where is the smallest mammal located.[/quote]

you're supposed to use the web, hence the name [b][u]"the internet game"[/u][/b]

don't use another source like your Starcraft game in this case

[hide="click for scores"]Ihatepie-1
computingthis-4
pianoforte-1
AIMEman-1
hm29168-1
alkjash-1
MustItAlwaysBeMe-1
james4l-1
hunter34-1(use the internet)
abacadaea-1[/hide]

What is the TV show that has aired the longest? It does not have to still be on air.

Meet the Press.

http://en.wikipedia.org/wiki/Meet_the_Press

Who wrote Les Aventures du Capitaine Hatteras au Pôle Nord?

Jules Vernes

[url]http://www.google.com/search?hl=en&q=Who+wrote+Les+Aventures+du+Capitaine+Hatteras+au+P%C3%B4le+Nord%3F[/url]

the name pops up everywhere you you search it on google  :D

who invented  banana puddings?

Recyclable plastic number five is polypropylene and its strong, tough, resistant to heat, chemicals, grease and oil, versatile, barrier to moisture.

[url]http://en.wikipedia.org/wiki/Plastic_recycling[/url]

Make it a rule we cant use wikipedia?

How much waste does the average american produce in a lifetime?

[quote="distracted523"][quote="dingzhou"]Jules Vernes
[/quote]

You mean [b]Jules Verne[/b].

Some random Southern US cook.
http://www.joyofbaking.com/BananaPudding.html

What is recyclable plastic #5?[/quote]

stop buggin mi abuot spelin misdakes...plese!!!!!!

123,662 lbs (61.8 tons) of solid waste
http://answers.yahoo.com/question/index?qid=1005123003199

Woot, Yahoo! Answers.  Also, wasn't there a rule that THEY SHOULD BE CLEAN!!!

Anyway, name 5 insecurities with PHP.

PHP remote file inclusion vulnerability in quiz/common/db_config.inc.php in the Online FlashQuiz (com_onlineflashquiz) 1.0.2 component for Joomla! allows remote attackers to execute arbitrary PHP code via a URL in the base_dir parameter.

PHP-Nuke Platinum 7.6.b.5 allows remote attackers to obtain configuration information via a direct request to maintenance/index.php, which reveals settings such as magic_quotes_gpc.

Directory traversal vulnerability in index.php in Sava's Link Manager 2.0 allows remote attackers to include and execute arbitrary local files via directory traversal sequences in the q parameter. NOTE: the provenance of this information is unknown; the details are obtained solely from third party information.

Directory traversal vulnerability in admin/login.php in EasyNews 4.0 allows remote attackers to include and execute arbitrary local files via a .. (dot dot) in the lang parameter.

SQL injection vulnerability in dynamicpages/index.php in EasyNews 4.0 allows remote attackers to execute arbitrary SQL commands via the read parameter in an edp_Help_Internal_News action.

http://nvd.nist.gov/nvd.cfm?advancedsearch , then type in "KEYWORD SEARCH" PHP

Was there a code in the PERL Obfuscation Contest which had no letters (any year), and if so, post the code.

[quote="7h3.D3m0n.117"][quote="computingthis"]you're supposed to use the web, hence the name [b][u]"the internet game"[/u][/b]

don't use another source like your Starcraft game in this case[/quote]

abacadaea, any reason why you're creating multiple accounts?[/quote]

thats not my account. i was supposed to be on my account. i was at a study session. this is my friends. he is using it for school purposes(not giving away what.)

anyway

all of them have letters
http://www.foo.be/docs/tpj/issues/vol3_2/tpj0302-0012.html

the dstance(in inches), that mt. Everest gains ever year.

and starcraft works, as long as you can find it on the internet.

0.16 inches (4 mm):
http://en.wikipedia.org/wiki/Mount_Everest

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

What is the product of the hit points of a prismatic dragon great wyrm, all the ability scores of a prismatic dragon ancient, the spell resistance of a wyrm prismatic dragon, and the sum of all the skill modifiers of the sample old prismatic dragon (given in the monster entry) as well as each entry of hit points for force dragon (that is, wyrmling * very young * young * juvenile ... * ancient * wyrm * great worm)?  (All numbers from D&D version 3.5)

In maximum ride: school's out -forever, what is the name of the evil genetic scientist that max looks up on the internet?


[quote="silouan"]Maybe  $cos x=\frac{e^{ix}+e^{-ix}}{2}$  and de Moivre  formula to find  $cos 4x$  and $cos 5x$  ?
I am in a wrong way Arqady ?[/quote]
I mean something another.

if you plug in the complex number substitution $z=e^{\frac{2\pi i}{21}}$, multiply both sides by $4$, subtract $1$, square, then subtract $z^{10}+z^{9}+...+z^{-10}=0$ four times, then return to the trig, you want to show that:

$\cos\frac{18\pi}{21}-\cos\frac{14\pi}{21}+\cos\frac{12\pi}{21}+\cos\frac{6\pi}{21}=0$

reduce the fractions, and evaluate the second one, to get

$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{1}{2}$

then let $w=e^{\frac{2\pi i}{7}}$, and the expression becomes

$w^{3}+w^{2}+...+\frac{1}{w^{3}}=1$

which is true by factoring $w^{7}-1=0$

... perhaps there is a more elegant solution?

I seem to recall a similar problem where the solution was to times LHS by a $sin \theta$ for some value of $\theta$ and then Product-to-Sum to solve the problem. It might work here.

[quote="Altheman"]
reduce the fractions, and evaluate the second one, to get

$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{1}{2}$

quote]

This is easy to show if you multiply the expression by $\sin \frac{\pi}{7}$

[quote="M4RI0"][quote="Altheman"]
reduce the fractions, and evaluate the second one, to get

$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{1}{2}$

[/quote]

This is easy to show if you multiply the expression by $\sin \frac{\pi}{7}$[/quote]
Easy to show that   $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.$ :wink:
[quote="Altheman"]
... perhaps there is a more elegant solution?[/quote]
I think yes: by M4RI0's hint and quadratic equation. :)


geometry
geometric transformation
reflection
number theory proposed
number theory

A problem proposed by me for Mathematical Reflections 4:

Let  $a,b,c,d$ be positive integers such that
$ab+cd=(a+b)(c+d)$.

a) Prove that equation has infinitely many of solutions
such that $a,b,c,d$ are pairwise not equal.

b) Prove that $ \max(a,b,c,d)\geq \frac{\sqrt{3}+1}{4\sqrt{3}}(a+b+c+d)$

For (a), taking (a,b,c,d) = (n, 2n, 10n, 4n) will do.


Epeidi exoun psiloksexastei askiseis tetoiou tipou dinw mia gia zestama...

Na eksetastei an iparxei sinartisi $f: \mathbb{Z}\longrightarrow\mathbb{Z}$ h opoia na epalitheuei tis sinthikes:

(I) $f\big(f(n)\big)=n, \ \ \forall \ n\in\mathbb{Z}$
(II) $f\big(f(n+2)+2\big)=n, \ \ \forall \ n\in\mathbb{Z}$
(III) $f(0)=1.$

Alexandros

Έχουμ ότι αν $f(n_{1})=f(n_{2})$ tote $f(f(n_{1}))=f(f(n_{2}))$ ara $n_{1}=n_{2}$
Συνεπώς η f είναι ''1-1 " .Είναι
$f(f(n))=n=f(f(n+2)+2)$ άρα $f(n+2)+2=f(n)$
Τώρα από αυτή και από το γεγονός ότι $f(0)=1$ επαγωγικά βρίσκουμε ότι $f(n)=1-n$ που πράγματι ικανοποιεί τις συνθήκες του προβλήματος .

Διορθώστε με αν κάνω λάθος.....

Βρήκα το ίδιο .  :)

Silouan mia paratirisi mono, oti i sinartisi f einai orismeni sto $\mathbb{Z}$ sinepws den mporeis na efarmoseis epagwgi gia na deikseis oti $f(n)=1-n$ (pou pragmati auti einai i sinartisi). Den einai omws kati idiaitero auto, arkei na theseis opou $n$ to $-n$ kai na efarmoseis meta epagwgi diakrinontas periptwseis gia $n=2k$ kai $n=2k+1$.

Etsi kataligeis sto epithimito apotelesma... (kai exeis tautoxrona deiksei oti i monadiki sinartisi pou ikanopoiei tis sinthikes tou provlimatos einai i parapanw).

Alexandros

Αυτό έκανα  :)  Απλά παράλειψή μου που δεν έγραψα λεπτομέρεις...... :(


Salut! Are cineva subiectele de la locala Bucuresti 2005?

[quote="Leliaa"]Locala Bucuresti? Ce clasa? Eu am numai a 12-a[/quote]
   In ce clasa esti si cum te numesti?Sunt si eu tot din Slatina, dar mai fac deplasari si pe la Valcea...


A basketball player is on the line to shoot a one-and-one free throw. (In a one-and-one situation, a second shot is allowed only if the first shot is successful.) If the player's free throw shooting average is .750, what is the probability that she will score exactly one basket? Express your answer as a common fraction.

This is $ \left (\frac{3}{4} \right ) \left (\frac{1}{4} \right )\equal{}\boxed{\frac{3}{16}}$.


calculus
integration
real analysis
function
real analysis unsolved

Let $\psi_{[n,n+1)}(x)=\left\{\begin{array}{cc}1,&\mbox{ if }x\in{[n,n+1)}\\0, & \mbox{ if }x\in\mathbb{R}\backslash[n,n+1)\end{array}\right.$

1) Let $f_{n}: \mathbb{R}\rightarrow\mathbb{R}, f_{n}=\frac{1}{2n}\psi_{[n,n+1)}(x) , n=1,2,3,...$
 Solve $\int_{\mathbb{R}}(\sum_{n=1}^{\infty}f_{n}(x))dm(x)$, $m=$Lebesgue measure

2) Solve $\int_{\mathbb{R}}s(t)dm(t)$, where $s(t)=\psi_{[1,2]}(t)+\frac{1}{2}\psi_{[2,3]}(t), t\in\mathbb{R}$

For the first one, since the simple functions are measurable and then by monotone convergence theorem, you can switch limit and integral directly.

For the second one, just notice $\int_{R}\chi_{E}dm = m(E)$, where $E$ is measurable and $\chi$ is the characteristic function.

I'd like to know the final result. Thanks


A1=(3)^1/2

An=(3+2an-1 )^1/2
n≥2

Show the sequences converges and estimate the limit of the sequence.

thanks

[url=http://www.math123xyz.com]Multimedia Math Resources[/url]

[hide]when a_n is less than three than a_n < a_(n+1) so the sequence is increasing.  Now suppose a_n > 3 for some n, there must be aleast such n, but a little arithmetic shows that a_(n-1) is greater than three. So the sequence is bounded above by three.  So its monotone and bounded which implies convergence.

To find the value you can just set an=an+1 and see that it must converge to 3[/hide]

We claim that the sequence converges to 3.

It is thus sufficient to prove that for all a_(n-1) is non-negative in the interval $ [\sqrt{3}, 3)  U  (3, 6\minus{}\sqrt{3}]$ that

$ |a_{n\minus{}1} \minus{} 3| > |a_{n} \minus{}3|$

Setting a_(n-1) = x and substituing the given equation for a_n, we square both sides to obtain:

$ x^2 \minus{}8x \minus{}3 \plus{} 6\sqrt{3\plus{}2x} > 0$

Using derivatives, one find that the function is positive for all x in the interval $ [\sqrt{3}, 3) U (3, 6\minus{}\sqrt{3}]$. Therefore, the sequence will get closer and closer to 3.