geometry
geometric transformation
reflection

hey guys,
i had an argument with my friends about grades. So I decided to ask here. Are grades important(if u r an international student) (grades in high-school) when you apply to good universities and does it matter if you have 3.7 or 3.8? Is it all about grades?

I am not an international student but I can tell you that grades are important.  Different countries have different systems but yes, they're very important.  So are SAT's and if you do not speak english as a first language, your TOEFL scores.  You need the whole package, not just grades, but yes, they are important.

Everything is important, grades reflect that you work hard in school, and care about it.  An IMO participant who fails all his classes is probably not someone colleges want to admit  :wink: :maybe:

If grades are not important, then why do you have to spent almost the whole day every day at school just to study? Hmm...Sharpen your critical thinking if you really want to get admitted to a good school. But, remember that learning for learning's sake is still the no. 1 priority here. Grades are important to reflect your comprehension/understanding (to a certain "standard") of the subject matter.


When will the "official" AMC scores be out?

Also, would my teacher already have received an e-mail of my scores?

Your adviser should have received an e-mail for your school's scores.

As of 10:30 am CDT on March 11, 2009:

All email reports for the A-date (February 10, 2009) contests have been sent, with the last being sent approximately February 25, 2009.  All A-date postal reports for schools with AIME qualifiers have been sent, with the last being sent approximately February 27, 2009.   The postal mailing contains the AIME I information.  The postal report is a printout of the emailed report, both reports contain the same information

All email reports for the B-date (February 25, 2009) contests have been sent, with the last being sent approximately March 9, 2009.  All B-date postal reports for schools with AIME qualifiers have been sent, with the last being sent approximately March 11, 2009.   The postal mailing contains the AIME I information.  The postal report is a printout of the emailed report, both reports contain the same information.

Postal reports for schools *without* AIME qualifiers are now being packaged and sent.  This will take several more days.  Those postal reports for schools without  AIME qualifiers will be reaching those schools in the next week to 10 days depending on mail delivery.  In almost all cases, your teacher or Contest Manager already has the result by email.

Check first with your teacher or Contest Manager. your Teacher or Contest Manager should have the emailed report.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="AMCDirector"]All B-date postal reports for schools with AIME qualifiers have been sent, with the last being sent approximately March 11, 2009. [/quote]
Got my mailing today (March 11), in Southern California. So the mail works, at least from my end, and you must have started the mailings some time last week.


[quote="manjil"]Show that every multiple of $ 3$ can be expressed as the sum of four cubes.[/quote]

hi. do you mean $ \forall  x\equiv0[3]$ then $ x\equal{} a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}d^{3}$

take for example  $ x\equal{}6$ !! :!:

Well, then you can have $ a \equal{} 2, b \equal{} c \equal{} \minus{} 1, d \equal{} 0$

Also after some fun (and referring to the numbers I chose) it seems
$ 6n \equal{} (n\plus{}1)^3\plus{}(n\minus{}1)^3\minus{}n^3 \minus{} n^3 \equal{} a^3\plus{}d^3\minus{}1^3\minus{}1^3.$

This seems Ok.Now my question is can we replace the question by: Show that every multiple of $ 3$ can be expressed as the sum of four $ distinct$ cubes.


So... has anyone taken the Spanish SAT? If so, how was it? I've taken 4 years of Spanish; currently taking AP Spanish Lit, though not particularly aceing it. Anyway, if I get a decent score I might be able to get out of foreign language requirements in college, so that'd be really nifty.

[quote="aznphatso"]I might do this too just to get out of that requirement (because it's way too late for chinese now :( )
i have 4 years of spanish in high school, no AP, but year 4 second semester was taught from the AP book. what's the score stanford wants, a 650 or something? it's definitely doable for you. for me, not so sure.[/quote]

Heh, my Spanish is awful though... but I heard from one of my friends at Stanford that the placement test they have isn't so bad; she placed out of the requirement even though she didn't take Spanish her junior/senior years :!:

[quote="arjunlandes"]I took the Spanish SAT II after 6 years of Spanish class and got an 800.  Since the curve is unforgiving, it's hard to get a perfect, but the difficulty of the test makes it easy to do well nevertheless.  You'll do fine.  Congrats on Intel by the way.  :wink:[/quote]

Haha ok, thanks for your response. Congrats to yourself, I recognized the name and I was like huh, he's on AoPS :wink:.

I took five years of spanish in high school in which my classmates and I learned basically nothing (briefly covered present, past, and subjunctive tense) since most of our teachers seemed to have been chosen soley based on being native speakers and in their early twenties.  If you talk to any of my classmates, including the ones who got A's, you'd think they were just starting to learn the language.
This past summer, after graduating, I went to Uruguay for a month where I thought I'd learn Spanish but anyone who I met who spoke any English refused to talk in Spanish so they could practice their English so it wasn't as helpful as I thought it would be.  Still, Uruguay is a great place; you should visit their one time.

The point of my rambling is that I ended the summer knowing just a little bit more Spanish and I signed up to take the SAT in November overseas (I deffered for a year). What I did was take a Barron's SAT Spanish book and study from that as well as all the online exercises I could find.  The scale on the test to get a 650 means you can get almost 25 wrong out of 85.  I ended up with a 730 but by now I've forgotten most of what I had learnt.

Just learn all the grammer tenses, memorize basic vocab, and take a lot of practice tests. Good luck with your test. Hope what I wrote helped at all.

I did the same as Basketball4Life :)  That SAT spanish book was extremely helpful. I think that after you learn all the grammar u can not encounter problems...Although I have never taken spanish classes (I learned it from TV and books) I managed to take a 750. So u need not worry paladin8  :)

[quote="Elemennop"]I heard the spanish SAT had the same problems as the Chinese SAT in that native speakers are a huge population of test takers and thus skew the scale...

Though it isn't as bad as the Chinese, where you miss one question and get a 230.[/quote]

WTF now?


calculus
integration
logarithms
Euler
linear algebra
matrix
real analysis

Let $ n\geq 0$ be a nonnegative integer. Calculate the following integrals:

1) $ \int\limits_{0}^{1}x^{n}\ln(1\minus{}x)\ln x \ln(1\plus{}x)dx.$

2) $ \int\limits_{0}^{1}\frac{\ln(1\minus{}x)\ln x\ln(1\plus{}x)}{1\plus{}x}dx.$

3) $ \int\limits_{0}^{1}\frac{\ln(1\minus{}x)\ln x\ln(1\plus{}x)}{1\minus{}x}dx.$

(3) can also be  written as $ \sum_{n \equal{} 1}^\infty \textbf H_n\sum_{m \equal{} 1}^\infty \dfrac{( \minus{} 1)^{m \plus{} 1}}{m(m \plus{} n \plus{} 1)^2}$

but i don't   see how that helps me !  :?:

Can we use?

$ \sum _{j \equal{} 1}^{\infty } \sum _{i \equal{} 1}^{2 j \minus{} 1} \frac {( \minus{} 1)^{i \plus{} 1} x^{2 j}}{i (i \minus{} 2 j)} \equal{} \log (1 \minus{} x) \log (1 \plus{} x)$

To simplify?

For the first, this leads to:

$ \sum _{j\equal{}1}^{\infty } \sum _{i\equal{}1}^{2 j\minus{}1} \frac{(\minus{}1)^i}{i (i\minus{}2 j) (2 j\plus{}n\plus{}1)^2}$

Since it is very similar to misan post, I hope it is rigth

1/ m=1 was posted in March 2021
[url= https://artofproblemsolving.com/community/c7h2490132_nice_integral] see here [/url]
https://artofproblemsolving.com/community/c7h2490132_nice_integral


inequalities
trigonometry
geometry
circumcircle
geometry proposed

A triangle ABC is inscribed in a circle with center $ O$ and radius $ R$. If the inradii of the triangles $ OBC, OCA, OAB$ are $ r_{1}, r_{2}, r_{3}$ , respectively, prove that $ \frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}\geq\frac{4\sqrt{3}+6}{R}.$

Maybe this problem need triangle is acute. I want to change notation $ r_{1},r_{2},r_{3}$ into $ r_{a},r_{b},r_{c}$. we get
\[ r_{a}=\frac{S_{BOC}}{p_{BOC}}=\frac{2\sin 2AS}{(\sum\sin 2A)(2R+2R\sin A)}=\frac{S\cos A}{2R\sin B\sin C(1+\sin A)}=\frac{2R^{2}\prod\sin A\cos A}{2R\sin B\sin C(1+\sin A)}=\frac{R\cos A\sin A}{1+\sin A}\Leftrightarrow \frac{R}{r_{a}}=\frac{1+\sin A}{\sin A\cos A}\]
thus we need to prove
\[ \sum\frac{1+\sin A}{\sin A\cos A}\ge 4\sqrt{3}+6\Leftrightarrow\sum\frac{2}{\sin 2A}+\sum\frac{1}{\cos A}\ge 4\sqrt{3}+6 \]
but we can prove easily $ \sum\frac{1}{\cos A}\ge 6$ and $ \sum\frac{2}{\sin 2A}\ge 4\sqrt{3}$, note we assume triangle is acute, but forall triangle I think it isn't true.  :roll:

[quote="gemath"]note we assume triangle is acute, but forall triangle I think it isn't true.  :roll:[/quote]
Please give an example if it is false!   :D

Of course here is counter example, please see the figure  :D

Extension maybe It was proved
Given triangle acute $ ABC$ and $ P$ is an arbitrary point inside it and cirumradius $ R$. If the inradii of the triangles $ PBC, PCA, PAB$ are $ r_{a}, r_{b}, r_{c}$ , respectively, prove that \[ \frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}\geq\frac{4\sqrt{3}+6}{R}\].


[b]If [size=150]g_{i}>=0[/size] you can use Lagrange multiplier method
 (g=0 IFF -g^2>=0)[/b]

Maybe You wanted to post a REPLY [hide="here"]http://www.mathlinks.ro/viewtopic.php?t=164304[/hide] instead of opening a new topic!
Anyway I don't understand any thing in your post... :huh: [/url]

I think that he asks if the Lagrange Multipliers method can be used when the given condition is an inequality instead of equality (for example $ a\plus{}b\plus{}c\leq 3$ instead of $ a\plus{}b\plus{}c\equal{}3$). Anyway, often in such inequalities it's sufficient to prove it when the condition is an equality (I mean that equality case is "worst", but of course not always).


geometry
circumcircle
ratio
rhombus
parallelogram
angle bisector
perpendicular bisector

Let $ABC$ be a triangle .$M$ be midpoint of $BC$.Let $D$ be a point on $BC$ such that $\hat{BAD}=\hat{DAC}$.the circumcircle of the triangle $ADM$ meets $AB;AC$ at $E;F$,respectively.Prove that $BE=CF$

I have a solution that is has a lot of computation...sorry I cannot offer a better solution; here is an outline.

WLOG $b>c$. Let $EF\cap BC=X$ and $AD\cap EF=Y$.

We can easily determine $BD=\frac{ac}{b+c}$, $CD=\frac{ab}{b+c}$, $BM=MC=\frac{a}{2}$, so we can determine $BE=\frac{a^{2}}{2(b+c)}\implies AE=c-\frac{a^{2}}{2(b+c)}$ and $AF=b-\frac{a^{2}}{2(b+c)}$ by power of a point.

$\bullet$ We want $DX: MX$

-Use menelaus' theorem on $\triangle ABC$, transversal $EFX$. Consequently, one can determine $CX$, then $DX: MX$ in terms of the sides of $\triangle ABC$

$\bullet$ We want to determine $YN: YX$

-$EN: NF=1$

-Use menelaus on $\triangle BEX$, line $AFC$ to determine $EX: FX$

-$EY: YF=AE: AF$ by the angle bisector theorem

Consequently, we can determine the ratio $YN: YX$

It should follow that $YX: NX=DX: MX$. So $YD//NM$, as desired.

[quote="Virgil Nicula"][color=darkred]Given are the triangle $ABC$ and the points $\{\begin{array}{c}{c}M\in (BC)\ ,\ MB=MC\\ D\in (BC)\ ,\ \widehat{BAD}=\widehat{DAC}\end{array}$.
The circumcircle of  $\triangle ADM$ meets the lines $AB$, $AC$ in the points $E$, $F$ respectively.
Denote the middlepoint $N$ of the segment $[EF]$. Prove that $MN\parallel AD$.[/color][/quote]
[color=darkblue][b]Proof.[/b] We have: \[\{\begin{array}{c}BE\cdot BA=BM\cdot BD\ CF\cdot CA= CM\cdot CD\end{array}\|\implies\frac{BE}{CF}\cdot\frac{BA}{CA}=\frac{BM}{CM}\cdot\frac{BD}{CD}\implies \frac{BE}{CF}=1\]
Take $X\in AB,\,Y\in AC: \,AX=AY=1.$ Construct the rhombus $AXPY$ and denote $BE=CF=k$
We have: $\overrightarrow{NM}=\frac{1}{2}(\overrightarrow{EB}+\overrightarrow{FC})=\frac{k}{2}(\overrightarrow{AX}+\overrightarrow{AY})=\frac{k}{2}\cdot\overrightarrow{AP}$
Thus $MN\parallel AP\implies MN\parallel AD$[/color]

[quote="sunken [size=150][/size]rock"][quote="Virgil Nicula"][color=darkblue]Prove easily that $BE=CF=\frac{a^{2}}{2(b+c)}$. Here is a more interesting problem :[/color]
[quote="chien than"][color=darkred]Given are the triangle $ABC$ and the points $\{\begin{array}{c}M\in (BC)\ ,\ MB=MC\\\ D\in (BC)\ ,\ \widehat{BAD}=\widehat{DAC}\end{array}$.
The circumcircle of  $\triangle ADM$ meets the lines $AB$, $AC$ in the points $E$, $F$ respectively.
Denote the middlepoint $N$ of the segment $[EF]$. Prove that $MN\parallel AD$.[/color][/quote][/quote]

After having seen that $BE=CF$, we know that the midpoint $P$ of the arc $\widehat{BAC}$ lies on the perpendicular bisector of $EF$; indeed, $\Delta DEP\equiv\Delta CFP\implies PE=PF$ and $AEFP$ cyclic.
If $O, O'$ are the circumcenters of $\Delta ABC$ and $\Delta AEF$, from $\Delta PEF\sim\Delta PBC$ we get $\frac{PO'}{PN}=\frac{PO}{PM}\iff MN\parallel OO'$, but $OO'\bot AP$ as line joining the circumcenters. And since $AP\bot AD$ as external bisector, and we are done.

Best regards,
sunken rock[/quote]

It seems I useless complicated the solution:
The property, [b]valid for any convex quadrilateral $BCFE$ with $BE=CF$: the line joining the midpoints $M, N$ of $BC, EF$ respectively, is parallel to the angle bisector of $\widehat{BE,CF}$.[/b]

Proof: construct the parallelograms $BENP, CFNQ$, we get $BPCQ$ parallelogram, hence $PM=MQ\ (\ 1\ )$ and $\Delta NPQ$ issocelex at $N$, consequently the median $NM$ is angle bisector of $\angle PNQ$, having the sides parallel to $BE, CF$ respectively and we are done.

[color=darkred][b][u]Lemma[/u].[/b] Let $\triangle ABC$ and $X\in AB$ , $Y\in AC$ so that $AX=CY$ . Denote the midpoints $M$ , $L$ of $[BC]$ , $[XY]$
respectively. Prove that: if $BC$ doesn't separate $X,Y$, then $ML$ is parallel with the ([b][u]interior[/u][/b]) $A$-bisector of $\triangle ABC$;
 if $BC$ separates $X,Y$, then $ML$ is parallel with the [b][u]exterior[/u][/b] $A$-bisector of $\triangle ABC$.[/color]

[b][u]Proof[/u].[/b] Suppose w.l.o.g. that $Y$ belongs to the ray $(CA$ . Denote the symmetrical point $U$ of $Y$ w.r.t. $M$ and the midpoint $V$ of $[XU]$ .
Observe  that $\triangle XBU$ is $B$-isosceles and $XU\parallel ML$ . Prove easily that: if $BC$ doesn't separate $X$ , $Y$ , then $XU$ is parallelly with
(interior) $A$-bisector; if $BC$ separates $X$ , $Y$ , then $BV$ is parallelly with $A$-bisector, i.e. $XU$ is parallely with the exterior $A$-bisector.
In conclusion, if $BC$ doesn't separate $X$ , $Y$ , then $ML$ is parallelly with (interior) $A$-bisector and if  $BC$ separates $X$ , $Y$ , then $ML$
is parallely with the exterior $A$-bisector.

Dear Mathlinkers,

1. the circumcircle of the triangle ADM goes throug the midpoint of the arc BAC
2.the triangle ABF and ACE are equal... and we are done.
Sincerely
Jean-Louis


HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:


[quote="bellahsayn"]
Source: book
$a$,$b$,$c$>$0$and $abc=1$
show that $a^{3}+b^{3}+c^{3}+15$$\geq$6$(a^{-}1+b^{-}1+c^{-}1)$[/quote]
It's my inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72290
Which book do you mean? Thank you.


function
trigonometry
integration
real analysis
real analysis unsolved

2 questions:

One of these transforms is not a transform which one? Why?

Find the function having fourier transform

a)
\[\frac{\cos\omega}{\omega^{2}+1}\]
b)
\[\frac{\omega\cos\omega}{\omega^{2}+1}\]
c)
\[\omega^{2}\frac{\cos\omega}{\omega^{2}+1}\]

[quote="color"]2 questions:

One of these transforms is not a transform which one? Why?

Find the function having fourier transform
[/quote]

It depends how you define Fourier transform.
The way I define it is like this:

$F(\omega)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t}f(t) dt$

Other authors do not write the $\frac{1}{\sqrt{2\pi}}$ factor in the Fourier transform pair.

[b]Theorem[/b] The fourier transform of a fast fourier transform of a well-behaved function returns the original function.

Therefore, the answer to your question is that you just need to find the Fouier transform of all those functions.

I tend to prefer putting the $2\pi$ in the exponent:

$\widehat{f}(\xi)=\int_{-\infty}^{\infty}e^{2\pi i\xi x}f(x)\,dx$

That amounts to measuring frequency in Hz rather than in radians/sec.

The Riemann-Lebesgue Lemma says that the Fourier tranform of an integrable function must tend to zero at infinity. Of your choices, (c) violates that, so it cannot be the transform of an integrable function.

For (a), I would star with the inverse tranform of $\frac1{\omega^{2}+1}.$ It should be something like $e^{-|x|},$ depending, of course, on where you put the $2\pi$ in your definition.

Now, write $\cos\omega=\frac12\left(e^{i\omega}+e^{-i\omega}\right).$ Make the function the sum of two functions, each with a complex exponetial. Make use of the theorem that says that translating the original function results in multiplying the transform by a complex exponential. You should get something related to $e^{-|x-1|}+e^{-|x+1|}.$

Note that the $\widehat{f}(\omega)$ in (a) is real and even, meaning that $f(x)$ will be real and even, and that $\widehat{f}\in L^{1},$ which means that $f$ will be continuous and bounded.

In (b), we're dealing with a real, odd transform, meaning that the orginal function will be odd and pure imaginary, and $\widehat{f}\not\in L^{1},$ meaning we have no expectation that $f$ should be continuous (and, in fact, $f$ will have jump discontinuities.}

Look to base the answer in (b) on something like $\text{sgn}(x)e^{-|x|}.$

Yes, thanks for your reply Prof. Merryfield.


inequalities integration algebra function domain trigonometry polynomial

Show that for all $ x \in W_0^{1,2} \left( {0,\pi } \right)$ the inequality
\[ \int\limits_0^\pi {\left| {x\left( t \right) } \right|^2dt} \leqslant \int\limits_0^\pi {\left| {\dot x\left( t \right)} \right|^2 dt}
\]
holds true.

Since such an $ x$ is zero at both ends of the domain, we should be able to represent it by a Fourier sine series:

$ x(t)\equal{}\sum_{n\equal{}1}^{\infty}b_n\sin nt.$

[More properly, we should speak of trigonometric polynomials of the form $ \sum_{n\equal{}1}^Nb_n\sin nt$ being dense in this Sobolev space.]

Then $ \int_0^{\pi}|x|^2\equal{}\frac{\pi}2\sum_{n\equal{}1}^{\infty}|b_n|^2$

and $ \int_0^{\pi}|x'|^2\equal{}\frac{\pi}2\sum_{n\equal{}1}^{\infty}n^2|b_n|^2$

The inequality is clear from that, as is also the status of $ x(t)\equal{}\sin t$ as the extremal function for the inequality.

If we repeat this proof on the equivalent space on an arbitrary interval $ [a,b],$ we get that

$ \int_a^b|x|^2\le\left(\frac{b\minus{}a}{\pi}\right)^2\int_a^b|x'|^2.$


linear algebra
matrix
group theory
abstract algebra
floor function
number theory proposed
number theory

Prove that the product:
$ \prod_{i\equal{}0}^{n\minus{}1}(2^{n}\minus{}2^{i})$ is divisible by $ n!$.
[hide]P.S:
To pbornsztein & ZetaX:
what about this title?[/hide]

It can be done with group theory: $ \prod_{k\equal{}0}^{n\minus{}1} (2^n\minus{}2^k)$ is the number of elements of the group $ GL(n,\mathbb{Z}_2)$ (we need $ n$ independent columns: there are $ 2^n\minus{}2^k$ ways to choose the $ k\plus{}1$st column so that it is not generated by the previous ones).

But this group contains a copy of the group of permutations of $ n$ elements, with each permutation $ \sigma$ being representable by the matrix with entries of 1 if the column number is the image of the row number under $ \sigma$ and 0 otherwise (i.e. for $ n\equal{}3$, the permutation $ (1\ 3\ 2)$ is represented by $ \left(
\begin{array}{ccc}
0&0&1\\
1&0&0\\
0&1&0\\
\end{array}\right)$).

So the group has a subgroup of $ n!$ elements and the divisibility follows.

Write the product as

$ P \equal{} 2^{n(n \minus{} 1)/2} \prod_{i \equal{} 1}^{n} (2^{k} \minus{} 1)$.

Now, the greatest power of $ 2$ that divides $ n!$ is $ \left\lfloor \frac {n}{2} \right\rfloor \plus{} ...$, which is at most $ n$; we can verify that this is not an issue for $ n \equal{} 1, 2$, and for $ n \ge 3$ we have $ \frac{n(n\minus{}1)}{2} \ge n$, so there's no problem here.

Consider an odd prime $ p$ less than or equal to $ n$.  The greatest power of $ p$ dividing $ n!$ is

$ \left\lfloor \frac {n}{p} \right\rfloor \plus{} \left\lfloor \frac {n}{p^2} \right\rfloor \plus{} ...$

whereas the greatest power of $ p$ dividing the desired product is at least

$ \left\lfloor \frac {n}{p \minus{} 1} \right\rfloor$

by Fermat's Little Theorem.  Since the above is equal to $ \left\lfloor \frac {n}{p} \plus{} \frac {n}{p^2} \plus{} ... \right\rfloor$, we are done.


function
geometry
geometric transformation
rotation
algebra solved
algebra

The graph of a function $f: \mathbb{R}\to\mathbb{R}$ has two has at least two centres of symmetry. Prove that $f$ can be represented as sum of a linear and periodic funtion.

It is well-known that the compose of two 180 degree rotation is a translation.
So the graph of f is invarable under a certain translation (a,b).
Clearly a=/=0, so we let b=ka.
we have f(x+a)=ka+f(x).
let g(x)=f(x)-kx, we have g(x+a)=g(x), so g is a periodic function.
Thus f(x)=g(x)+kx is sum of a linear function and a periodic function.


I encounter these questions often.

For example it says "What is the largest possible 5 digit product obtained by using the digits 5,6,7,8 and 9 only once ?"

Are there any useful tips?

[hide]
So clearly we'll do something like

$ abc$
$ \times de$
------------
stuff

So we look at the first column from the left.  We have only $ a$, so we let $ a\equal{}9$ since 9 is the biggest allowed digit.  Then, look at the next column.  We have $ b$ and $ d$, and since b is above d, $ b>d$, and b and d are the next 2 largest digits, so $ b\equal{}8$ and $ d\equal{}7$.  Similarly, $ c\equal{}6$ and $ e\equal{}5$.  So we have to evaluate $ 986\times 75$, and I don't feel like it.
[/hide]

[quote="alanchou"][hide]
So clearly we'll do something like

$ abc$
$ \times de$
------------
stuff

So we look at the first column from the left.  We have only $ a$, so we let $ a \equal{} 9$ since 9 is the biggest allowed digit.  Then, look at the next column.  We have $ b$ and $ d$, and since b is above d, $ b > d$, and b and d are the next 2 largest digits, so $ b \equal{} 8$ and $ d \equal{} 7$.  Similarly, $ c \equal{} 6$ and $ e \equal{} 5$.  So we have to evaluate $ 986\times 75$, and I don't feel like it.
[/hide][/quote]
No, I think I did that too, but it was wrong.

[hide="Answer"]The answer was $ 875\times 96\equal{}84000$[/hide]

Actually, 975 x 86 >  986 x 75. And 965 x 87 is 83955. You want the two factors to be as close together as possible, without too small.

Keep in mind, if a+b = c+d, then ab > cd, if a-b < c-d (assuming a>b, c>d).

So the largest possivle would be

84000 (875 x 96).

[quote="darkmatter47"]Actually, 975 x 86 >  986 x 75. You want the two factors to be as close together as possible.

Keep in mind, if a+b = c+d, then which ever is closer to a square is larger.

So the largest possivle would be

84000 (875 x 96). It's like adding 10 more 875, although each time, there's 100 x 86 less. And 8750> 8600, so naturally 875 x 96 is larger.[/quote]
Yes, correct.

So are you saying that the two-digit number should be the closest to the square number you can obtain in order to get the largest possible?

Sorry let me clarify.

if A+b = c+d, and a-b < c-b, then ab > cd (assume that a>b, c>d)

In other words, you want the smaller factor to be as close to the larger one as possible.

Here:

Represent digits by ABCDE.

ABC
xDE
----

Then, to exchange digits between ABC and DE, so that they are as close as possible. However, this is only true if the largest digit is in the largest place value of each individual number.


Find the number of integral $ x \in [3,3003]$ such that $ x \equiv 3 \pmod {6} \equiv 0 \pmod {33}$.

[hide]The conditions imply that
\[ x\equal{}6j\plus{}3\equal{}33k \implies 2j\plus{}1\equal{}11k.\]
If we turn this into a congruence, we have
\[ \implies 2j \equiv \minus{}1 \equiv 10 \pmod{11} \implies j \equiv 5 \pmod{11}.\]
Then we have $ j\equal{}11n\plus{}5$, and we have the bounds
\[ 3 \le x \le 3003,\]
so
\[ 3 \le 6\left(11n\plus{}5\right)\plus{}3 \le 3003 \implies \minus{}30 \le 66n \le 2970 \implies \minus{}\frac{5}{11} \le n \le \frac{495}{11}.\]
Since $ n$ is an integer, we have
\[ 0 \le n \le 45,\]
so there are $ 45\minus{}0\plus{}1\equal{}\boxed{46}$ solutions for $ x$.[/hide]

[hide]Meeting the first condition, we can say that $ x \equal{} 6k \plus{} 3$
since $ 3 \equiv \minus{} 3 \pmod {6}$ we can also say that $ x \equal{} 6k \minus{} 3$
considering 2nd condition we conclude that $ x$ is divisible by both $ 3$ and $ 11$
since $ 6k \minus{} 3$ is always divisible by $ 3$ it is enough to multiply it with $ 11$, so $ x \equal{} 33(2k \minus{} 1)$.

Consider a sequence $ (a_n)$ such that $ a_n \equal{} 33(2n \minus{} 1)$ and $ a_n \in [3,3003]$
then $ a_1 \equal{} 33(2 \minus{} 1) \equal{} 33$ is the 1st number of the sequence and satisfies both criteria.
Lets find the largest $ a_n$:
$ a_n \le 3003$
$ 33(2n \minus{} 1) \le 3003 \implies n \le 46 \implies n \equal{} 46$, $ a_{46} \equal{} 3003$

Since there are $ 46$ numbers in the sequence, the number of solutions for $ x$ is $ \boxed{46}$[/hide]


A set of positive integers has property P if it contains no 3 consecutive integers. The empty set has property P by definition. How many subsets of {1, 2, 3, ..., 10} have property P?

[hide]f(n)=f(n-1)+f(n-2)+f(n-3), where f(n) is the number of good subsets with property P of the first n integers, because either the last element is not in the subset, the last element is in the subset but the second-to-last is not, or the last two elements are in the subset (meaning that the antepenultimate is not). f(0)=1, f(1)=2, f(2)=4. f(3) is indeed 7, as a check, as f(4)=13. Anyway, we can now work our way up or find a closed form for f(n)[/hide]

is that anywhere close to correct?

Well, recurrence is a good way to solve it,
but I think your recurrence formula is wrong.

Hmm, I got the same recurrence formula that MysticTerminator got (for n :ge: 3), based on the same argument.  Working up, I found that f(10) = [hide]504[/hide].  It's not so easy to write a closed-form solution, because we would have to find the zeros of a cubic polynomial.

What's wrong with the recurrence or the argument?

I also found 504 with the same argument (Tribonacci sequence).

I wrote a small computer program to check and indeed it is 504  :?

Sorry
My original recurrence looks differently :$
But it in fact exactly the same recurrence.
So sorry :oops: I didn't look careful enough.


inequalities geometry inradius circumcircle trigonometry search

I was solving an inequality lately,and I have become to this inequality:
$ p^{2}\geq 2R^{2}+8Rr+3r^{2}$
where $ p,r,R$ are the semiperimeter,inradius,the circumradius of $ \triangle ABC$...
Im sure Ive seen this before and Im sure that this is a famous inequality,but I cant prove it right now,so probably Im missing something trivial,could some one post a proof of this one...?  :maybe:
this is driving me really crazy...  :(

the original problem is:
[b]let $ x,y,z$ be positive reals,s.t $ x^{2}+y^{2}+z^{2}+2xyz=1$ prove that:
$ xy+yz+zx\leq \frac{1}{2}+2xyz$[/b]

the solution I used is as follow:
[hide][b]we can assume that $ x=\cos A,y=\cos B,z=\cos C$ where $ \angle A+\angle B+\angle C=180$

so we have to prove that:

$ \cos A\cos B+\cos B\cos C+\cos C\cos A\leq \frac{1}{2}+2\cos A\cos B\cos C$

now in the other hand we have:

$ \cos A\cos B+\cos B\cos C+\cos C\cos A=\frac{p^{2}+r^{2}-4R^{2}}{4R^{2}}$

and also:

$ \cos A\cos B\cos C=\frac{p^{2}-(2R+r)^{2}}{4R^{2}}$

so its sufficient to show that:

$ \frac{p^{2}+r^{2}-4R^{2}}{4R^{2}}\leq \frac{1}{2}+2\frac{p^{2}-(2R+r)^{2}}{4R^{2}}$

$ \Rightarrow p^{2}+r^{2}-4R^{2}\leq 2R^{2}+2p^{2}-8R^{2}-2r^{2}-8Rr$

$ \Rightarrow 3r^{2}+8Rr+2R^{2}\leq p^{2}$

and as I said above,Im sure that this is a famous inequality in triangle,but I cant solve it right now  :( [/b][/hide]

Well, I'm not sure about this, it's so complicated ...

It's easy to prove $ x+y+z\leq \frac{3}{2}$
This gives $ x+y\leq \frac{3}{2}-z$
We can assume that $ z\leq y\leq x$ then $ 3z\leq x+y+z\leq \frac{3}{2}$ or $ z\leq \frac{1}{2}$ or $ 1-2z\geq 0$
And alos $ 0=x^{2}+y^{2}+z^{2}+2xyz-1 \geq 2xy+z^{2}+2xyz-1$
$ \Longleftrightarrow (1-z)(z+1) \geq 2xy(z+1)$
or $ 1-z\geq 2xy$ or $ xy+\frac{z-1}{2}\leq 0$
Now, we have
$ xy+yz+zx-2xyz-\frac{1}{2}= xy(1-2z)+z(x+y)-\frac{1}{2}$
$ \leq xy(1-2z)+z(\frac{3}{2}-z)-\frac{1}{2}= (1-2z)(xy+\frac{z-1}{2}) \leq 0$
Thus, $ xy+yz+zx \leq 2xyz+\frac{1}{2}$

I'm still finding solution for your first inequality

See Mr. Nicula`s post: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1583801365&t=126286


Source: Mathcounts Nationals 1984, Team Round #10

A function A(x,y) is defined by the following rules:

A(0,n)=n+1
A(m,0)=A(m-1,1)
A(m,n)=A(m-1,A(m,n-1))

Find A(2,3).

This is the well-known Ackermann Function:  http://en.wikipedia.org/wiki/Ackermann_function

I once solved a problem that required finding $A(4, 1994)$.  Consider this problem easy by comparison :)

[quote="236factorial"]Ummmm.... how would this problem be solved within 2-5 minutes?[/quote] You could memorize some values of the Ackermann function. :D I've done this problem before, and it took 15 minutes to get a wrong answer.  :mad:

I wouldn't have imagined that they did put Ackermann function into MATHCOUNTS, that's hard compared to other Team Round questions.

You could find a pattern that would help you solve, like $A(1,x)=x+2$. I believe that is true. :D

Yes that identity is true and makes the problem not so difficult at all.  I was working on typing up a quick solution but you guys seem to be talking too quickly :P I got [hide]9[/hide] if anyone was wanting to check that, luckily the numbers are small.

[quote="LordoftheMorons"]By brute force (I'm far too lazy to type up all of the calculations):
[hide]Edit: 9[/hide][/quote]

Incorrect
[hide="Similarly by Brute Force"]I got 9.[/hide]

[quote="Shogia"]I'm pretty sure that some explicit formulas are:
$A(1,n)=n+2$ and, using this fact, that $A(2,n)=2n+3$[/quote]

Yep.  Neither are particularly hard to prove, either.

For a challenge, and for those of you who haven't clicked on the link yet, try finding an explicit formula for $A(3, n)$.  :)

I had trouble coming up with the one for 3 let alone even trying 4 :?

[quote="t0rajir0u"][quote="Iron Fist"]... it can't be that hard to find A(2,3) :p

each A(n,x) is a recursively defined function...

A(0, x) = x + 1
A(1, x) = x + 2
A(2, x) = 2x + 3

[hide="Past this spoiler are a couple answers"]
A(3, x) = 2[A(3, x - 1)] + 3, A(3, 0) = 5... it's the first one you need a recursive definition for.

A(4, x) is ridiculously huge... it's A[3,  A(4, x - 1)] big... even the first term is
somewhere around 100,000...
[/hide][/quote]

[hide="You dont need a recursive definition for 3"]
$\boxed{ A(3, n) = 2^{(x+3)} - 3 }$

Before I discuss $A(4, n)$, I would like to diverge a little into generalized operators with function notation.  For these purposes I would like to designate

$\rho_0 (a, b) = b + 1$,

The "successor operator" (which is unary, so ignores $a$).  Now, define

$\rho_1(a, b) = a + b = \rho_0 (\rho_0 (\rho_0 ... \rho_0(a) ... ))$,

The "addition operator," or the repetition of $b$ successor operators on $a$.  Similarly recursively define

$\rho_2(a, b) = ab = \rho_1 (\rho_1 (\rho_1 ... \rho_1(a) ... ))$,

The "multiplication operator," or the repetition of $b$ addition operators on $a$ (starting with $\rho_2(a, 0) = 0$).  Then,

$\rho_3(a, b) = a^b$,

The "exponentiation operator," or repeated multiplication.  I'm not defining this function very well, but you get the idea, right?  We can just keep generalizing.

$\rho_4(a, b)$ is called the "ternary operator," and the ones higher than that have no name.  A "ternary operator" is repeated exponentiation - a power tower.

The Ackermann function can be described very simply using $\rho_m(a, b)$.  Notice the pattern

$A(0, n) = \rho_0(2, n+3) - 3$
$A(1, n) = \rho_1(2, n+3) - 3$
$A(2, n) = \rho_2(2, n+3) - 3$
$A(3, n) = \rho_3(2, n+3) - 3$

Now, it's not very hard to guess that

$\boxed{ A(m, n) = \rho_m(2, n+3) - 3 }$

And $\rho_m(a, b)$ is a pretty intuitive function, right?  :D

(In other words, $\boxed{ A(4, n) = 2^{ 2^{ 2^{ ... }}} - 3 }$ (there are $n+3$ twos).
[/hide][/quote]

From Wikipedia...of course. [hide="From Wikipedia, too"] $A(5,x)=(2\rightarrow(x+3)\rightarrow3)-3=(2\uparrow\uparrow\uparrow(x+3))-3=(2\uparrow\uparrow2\uparrow\uparrow...\uparrow\uparrow2)-3$ with $x+3$ 2s. :D Wikipedia can explain what that is.[/hide] If you want to see a REALLY big number (that actually has mathematical relevance), search Graham's number in Wikipedia (or Google).

[quote="Iron Fist"]Or $\frac{1}{Avogadro's number}$ :P[/quote] Graham's number is bigger. WAY BIGGER. :P

[quote="Hokkage"]Yeah, Avogadro is a particle compared to the universe of Graham's. He actually holds the world record for the largest number ever put to use.[/quote]

He also holds the record for the grossest probable overestimate.  Graham's number is an upper-bound he derived in a problem in Ramsey (?) theory, and most mathematicians think the actual number is about $6$.  :D


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let $ C$ and $ D$ be two intersection points of circle $ O_1$ and circle $ O_2$. A line, passing through $ D$, intersects the circle $ O_1$ and the circle $ O_2$ at the points $ A$ and $ B$ respectively. The points $ P$ and $ Q$ are on circles $ O_1$ and $ O_2$ respectively. The lines $ PD$ and $ AC$ intersect at $ H$, and the lines $ QD$ and $ BC$ intersect at $ M$. Suppose that $ O$ is the circumcenter of the triangle $ ABC$. Prove that $ OD\perp MH$ if and only if $ P,Q,M$ and $ H$ are concyclic.

Here is an image,that i've promised to upload earlier.
[img]http://img118.imageshack.us/img118/676/cwmo2007ul1.png[/img]

Remark:
The condition $ A,D,B$ are collinear is not necessary.I did not use it in my proof.
I'm writing this,because it make the problem more beautiful,and kinda easier,for someone,who is trying to solve the problem with using all the conditions.
Best regards.

$ OD\perp MH$
 equivalent$ {OM}^2\minus{}{OH}^2\equal{}{DM}^2\minus{}{OH}^2$......

What is it?If you really solved it,please post your solution,if you don't,then there is no reason,to post something like a hint,which is not true :rotfl:

[quote="Erken"]Let $ C$ and $ D$ be two intersection points of circle $ O_1$ and circle $ O_2$.A line,passing through $ D$,intersects with circle $ O_1$ and circle $ O_2$ at points $ A$ and $ B$ respectively.The points $ P$ and $ Q$ are on circles $ O_1$ and $ O_2$ respectively.The lines $ PD$ and $ AC$ intersect at $ H$,and the lines $ QD$ and $ BC$ intersect at $ M$.Suppose that $ O$ is the circumcenter of the triangle $ ABC$,prove that $ OD\perp MH$ is and only if $ P,Q,M$ and $ H$ are concyclic.
I'll add an image later,because there was an image on the problem list.[/quote]

Nice but easy!
We have:
$ P,Q,M,H$ are concylic $ \Leftrightarrow DP.DH\equal{}DM.DQ \Leftrightarrow HD.HP\minus{}HD^2\equal{}MD.MQ\minus{}MD^2$
$ \Leftrightarrow DM^2\minus{}DH^2\equal{}MD.MQ\minus{}HP.HD\equal{}MB.MC\minus{}HB.HA\equal{}(OM^2\minus{}R^2)\minus{}(OH^2\minus{}R^2)\equal{}OM^2\minus{}OH^2 \Leftrightarrow OD \perp MH$

[quote="dlt5"]Hey Erken my solution is the same 10maths_tp0609[/quote]
It is my proof too. :)
Yes,indeed the problem is very easy,but nice.

[quote]We have:
$ P,Q,M,H$ are concylic $ \Leftrightarrow DP.DH \equal{} DM.DQ \Leftrightarrow HD.HP \minus{} HD^2 \equal{} MD.MQ \minus{} MD^2$
$ MD.MQ \minus{} HP.HD \equal{} MB.MC \minus{} HB.HA$[/quote]
Very nice solution  :D  but  :maybe:
I think $ MD.MQ\minus{}HP.HD\equal{}MB.MC\minus{}HA.HC$

Different solution. Indeed $A-D-B$ is not required.

Let $T_1$ and $T_2$ be the circumcircles of $ABC$ and $PQD$. Let $O$ and $C_2$ be their centers, respectively.

I use a well known lemma: If $ABC$ is a triangle, $O$ its circumcenter, $X$ and $Y$ points in $AB$ and $AC$ then $BXYC$ is cyclic if and only if $AO \perp XY$. This is easy to prove.

From $HP \cdot DD = HA \cdot HC$ we get $H$ is in the radical axis of $T_1$ and $T_2$. Similarly $M$ is also on this radical axis. Hence we have $MH \perp C_1C_2$.

If $PHMQ$ is cyclic then $C_2D \perp MH$ by the lemma which implies $O - C_2 - D$ and $OD \perp MH$.

SImilarly if $OD \perp MH$ we have $D-O-C_2$ so $DC_2\perp MH$ so by lemma $PHMQ$ is cylic and we are done.


Suppose a projectile in projected into a headwind that pushes it back so as to reduce its horizontal speed by a constant amount w. Find the parametric equations for the path of the projectile.

The y direction is only affected by gravity, and since speed is reduced by a constant amount (not reduced at a constant rate, correct?) then
$ x\equal{}v_{0}t\cos\theta\ \minus{}w, y\equal{}v_{0}t\sin\theta\minus{}\frac{1}{2}gt^{2}$ with respect to t, time.

Shouldn't it bee

\[ x\equal{}(v_{0}\cos\theta\ \minus{}w)t,\; y\equal{}v_{0}t\sin\theta\minus{}\frac{1}{2}gt^{2}\]


search
graph theory
Ramsey Theory
combinatorics proposed
combinatorics

Let $ K_n$ be the complete graph with $ n$ vertices. What is the minimum $ n$ such that for any 3-edge coloring there exists a monochromatic cycle of length four?

I know that for any 2-edge coloring $ n \equal{} 6$ is the minimum number of vertices such that in $ K_n$ there exists a monochromatic cycle of length four...

Any suggestions?
 :(

You're doing Ramsey theory, so you're extremely unlikely to get exact answers without something akin to a complete enumeration.  However, there are in some cases theoretical results that can help get decent bounds -- see, for example, Ravi B's post here: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1637658511&t=8958

Maybe, i have found an article: "The ramsey number of $ r(C_4,C_4,C_4)$", but it seems unavailable...
Could anyone retrieve it?
 :oops:

[quote="JBL"]Some sort of reference (say, journal, volume and author) would be helpful  :roll:[/quote]

Sorry, the article is "The ramsey number of $ r(C_4,C_4,C_4)$", author "C. R. J. Clapham", journal "Periodica Mathematica Hungarica", Issue "Volume 18, Number 4 / December, 1987",...

[url=http://www.springerlink.com/content/q10v5275137u6181/]This is a link[/url].

Thanks in advance!
 :)

So, that paper is a two-page note that gives a simple three-coloring (with isomorphic color classes) of $ K_{10}$ with no $ C_4$.

(In particular, the coloring is as follows: the red edges are (1, 10), (4, 10), (7, 10), 12, 17, 25, 29, 34, 35, 38, 45, 67, 68, 69, 89, and we get the other classes by applying the permutation $ (123)(456)(789)(10)$ (given in cycle notation).  This can be made to look really nice if you draw the vertices 10, 7, 6, 8, 3, 4 in a hexagon.)

However, it does reference a paper showing that actually proves $ r(C_4, C_4, C_4) \equal{} 11$, namely A. BIALOSTOCKI and J. SCHONHEIM, On some Turan and Ramsey numbers for C4, [i]Graph theory and combinatorics[/i] (Cambridge, 1983), Academic Press, London, 1984; 29--33.  Are you able to access this paper?  (It seems that I am not so able via the internet, although it may be that my university library has a hard copy.)

Thank you for the reply!
So, the Clapham's paper merely shows a three-edge coloring of $ K_10$ with no $ C_4$, without a formal explanation of the problem?
I do not understand the meaning of the permutation $ (123)(456)(789)(10)$...
The red edges are [list]$ (1, 10), (4, 10), (7, 10), (1, 2), (1, 7), (2, 5), (2, 9), (3, 4), (3, 5), (3, 8), (4, 5), (6, 7), (6, 8), (6, 9), (8, 9)$[/list]... What about the green and blue ones? :blush:

[quote]
However, it does reference a paper showing that actually proves $ r(C_4,C_4,C_4) \equal{} 11$, namely A. BIALOSTOCKI and J. SCHONHEIM, On some Turan and Ramsey numbers for C4, Graph theory and combinatorics (Cambridge, 1983), Academic Press, London, 1984; 29--33. Are you able to access this paper? (It seems that I am not so able via the internet, although it may be that my university library has a hard copy.)
[/quote]
Unfortunately, I am not able to access it... Is that book a paper's collection or a handbook?

These four pages would be the answer at my curiosity... :wallbash_red:

[quote="Buddy Rich"]I do not understand the meaning of the permutation $ (123)(456)(789)(10)$...
The red edges are [list]$ (1, 10), (4, 10), (7, 10), (1, 2), (1, 7), (2, 5), (2, 9), (3, 4), (3, 5), (3, 8), (4, 5), (6, 7), (6, 8), (6, 9), (8, 9)$[/list]... What about the green and blue ones?[/quote]  Yes, those are the red edges.  To get the blue edges, replace everywhere 1 with 2, 2 with 3, 3 with 1, 4 with 5, 5 with 6, 6 with 4, 7 with 8, 8 with 9 and 9 with 7.  Then to get the green edges, perform this operation again.

I don't know anything about that reference, just that I copied it out of the paper by Clapham.  I'm not sure what you mean by "a formal explanation of the problem" -- he states that his main interest is in finding colorings in which the color classes are isomorphic, and then he states the coloring above.  There is a small amount of discussion and a reference to the paper of Bialostocki and Schonheim.

I think http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152703 is a proof?


Let $d,p,q$ be fixed positive integers, and $d$ is not a perfect square. $\mathbb{N}$ is the set of all positive integers, and $S=\{m+n\sqrt{d}|m,n \in \mathbb{N}\} \cup \{0\}$. Suppose the function $f: S \to S$ satisfies the following conditions for all $x,y \in S$:
(i) $f((xy)^p)=(f(x)f(y))^p$
(ii)$f((x+y)^q)=(f(x)+f(y))^q$
Find the function $f$.

[quote="Rust"]Only $f(x)=x$. If $S=a+b\sqrt d, \ a,b\in Z$ then $f(x)=x$ or $f(a+b\sqrt d)=a-b\sqrt d$.[/quote]
can you post your slution
i think it is not easy


inequalities
algebra
polynomial
inequalities unsolved

It's very tight...
Let $a\geq b\geq c$ and \[f(a,b,c): =\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\]
Then \[f(a,b,c)\ge f\left(\frac{2a+b}{3},\frac{2a+b}{3},c\right)\iff \\ \iff \frac{1}{(6b+3c)^{2}}+\frac{1}{(6c+3a)^{2}}\ge\frac{1}{(4a+2b+3c)^{2}}+\frac{1}{(6c+2a+b)^{2}}\]
Let $6b+3c=d, 6c+3a=e, a-b=f$ so all three are positive. Then this ineq. is equivalent to $\frac{1}{d^{2}}+\frac{1}{e^{2}}\ge \frac{1}{(d-2f)^{2}}+\frac{1}{(e-f)^{2}}$ which is obvious, since $d-2f$ and $e-f$ are positive.

So in general if the variables are sorted $a\geq b,c$ we can apply the transform $a\to \frac{2a+b}{3}, b\to \frac{2a+b}{3}, c\to c$ until after a possibly infinite process all three variables are equal, since for any $a>b$ we have that $a>\frac{2a+b}{3}>b$ is a number between the two.

I'm not quite sure where to go from here...

[color=#FF0000]Moderator Edit: do not quote the entire post above you.[/color]

how do you know that $10(ab+bc+ca)-a^{2}-b^{2}-c^{2}$ is positive?

and it still exists a big problem that why $x,y,z\ge 0$.

[quote="me@home"]I...
Let $a\geq b\geq c$
...[/quote]
Notice that $a,b,c$ are not symmetric!

[quote="Little Gauss"]$\frac{9}{10(ab+bc+ca)-a^{2}-b^{2}-c^{2}}\ge \frac{1}{ab+bc+ca}$   :)[/quote]

Nice idea to try to solve it with Iran96, but your last ineq isn't true. Look what happens if $a$ is very big  :(

The problem: your $x,y,z$ are not necessarily positive.
EDIT: Oops, didn't see the other posts below  :blush:

Such an useful but terrible method! I believe that it is true but I myself can not check it. It is so complicated!

It is a symmetric inequality :huh:
Why can't I assume a>=b>=c? Do you mean after the replacement of variables? I guess in that case, we only need $a\geq b$ so it doesn't really matter.

I've been trying to investigate the limit of this process of replacing by $2a+b/3$ until all three variables are equal. It's really interesting - for $(1,1,0)$ it yields $6/7$, for $(1,0,0)$ it yields $4/7$, for $(.5,1,1)$ it yields $10/11$, for $0,1,2$ it yields I think $10/7$ but possibly $16/11$...

[quote="me@home"]It is a symmetric inequality :huh:
[/quote] No. :wink:

if we apply CBS inequality...results the inequality...[/hide]

No. I dont think that the inequaality can be solved by classical inequality.

To Arquady: Can you solve this one by another method, different from tanpham's one?

Your last inequality is not symmetric and eighth degree. I don't think that $uvw$ helps here.

[quote="GacKiem"]If $x,y,z$ are positive real numbers then
\[\frac{1}{(2x+y)^{2}}+\frac{1}{(2y+z)^{2}}+\frac{1}{(2z+x)^{2}}\ge \frac{1}{xy+yz+zx}.\][/quote]

I found a different solution for this inequality.
Assume $z=min\{x,y,z\}$, then
\begin{eqnarray*} F&=&\sum_{cyc} \frac{1}{(2x+y)^2}-\frac{1}{xy+yz+zx}\\
&=&\left(\sum_{cyc} \frac{1}{(2x+y)^2}-\frac{1}{9xy}-\frac{8}{9(x+z)(y+z)}\right)+ \left(\frac{1}{9xy}+\frac{8}{9(x+z)(y+z)}-\frac{1}{xy+yz+zx}\right)\\
&=&A+B \end{eqnarray*}

\begin{eqnarray*} A &=&  \frac{1}{(2y+z)^2}+\frac{1}{(2z+x)^2}-\frac{8}{9(x+z)(y+z)} \geqslant -\frac{1}{(2x+y)^2}+\frac{1}{9xy}\\
\ &=& \text {which is true after expanding and use z = min}
\\
\\
B &=& \frac{1}{9xy}+\frac{8}{9(x+z)(y+z)} \geqslant \frac{1}{xy+yz+zx}\\
&\Leftrightarrow& \frac{xy+yz+zx}{9xy}+\frac{8 \left(xy+yz+zx+z^2-z^2 \right)}{9(x+z)(y+z)} \geqslant 1\\
&\Leftrightarrow&\frac{z(x+y)}{9xy} \geqslant \frac{8z^2}{9(x+z)(y+z)}  \\
&\Leftrightarrow& (x+y)(y+z)(z+x) \geqslant 8xyz \\ \end{eqnarray*}

Thus the original inequality is true

Urun, your proof is much simpler and elegent than Arqady's. Congratulation !

[quote=GacKiem]If $a,b,c$ are positive real numbers then
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}.\][/quote]

If If $a,b,c$ are positive real numbers then
\[\frac{1}{2a^2+b^{2}}+\frac{1}{2b^2+c^{2}}+\frac{1}{2c^2+a^{2}}\ge \frac{5}{3(ab+bc+ca)}.\]

[quote=GacKiem]If $a,b,c$ are positive real numbers then
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}.\][/quote]

Here is a 6th proof of this inequality. In fact, we can prove an even stronger one
For all positive $x,y,z$ prove that:
\[ \sum_{cyc}\frac{z}{2x+y} \geq \frac13 \cdot \frac{x^2+y^2+z^2}{xy+yz+zx}+\frac{2}{3}\]
\[\sum_{cyc}\frac{x(y-z)}{(2x+y)^2} +\frac13 \cdot \frac{x^2+y^2+z^2}{xy+yz+zx} \geq \frac13\]

Add up and get the original one.

[quote=GacKiem]If $a,b,c$ are positive real numbers then
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}.\][/quote]

Here is my 7th proof for this inequality. I briefly outlined the proof.
Multiply both sides by $9(a+b+c)^2$, and simplify the expression. We need to prove
\[ 2 \cdot \sum_{cyc} \frac{a+2b+3c}{2a+b} +\sum_{cyc}\frac{(a+2b+3c)^2}{(2a+b)^2} \geq 9 \cdot \frac{a^2+b^2+c^2}{ab+bc+ca}+15\]
Then prove this lemma
\[\sum_{cyc} \frac{(a+2b+3c)^2}{(2a+b)^2} \geq 5 + 7 \cdot \frac{a^2+b^2+c^2}{ab+bc+ca}\]

[quote=GacKiem]If $a,b,c$ are positive real numbers then
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}.\][/quote]
I think the following reasoning easy to implement.
Let $2a+b=x$, $2b+c=y$ and $2c+a=z$.
Hence, $a=\frac{4x-2y+z}{9}$, $b=\frac{4y-2z+x}{9}$ and $c=\frac{4z-2x+y}{9}$
and we need to prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{27}{5(xy+xz+yz)-2(x^2+y^2+z^2)}$.
Let $xy+xz+yz=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\leq0$, where $f$ is a convex function.
Id est, by $uvw$ it remains to prove one case only: $y=z=1$ for $\frac{1}{4}<x<\frac{5}{2}$, which gives $(x-1)^2(1+12x-4x^2)\geq0$, which is obvious.

[quote=arqady][quote=GacKiem]If $a,b,c$ are positive real numbers then
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}.\][/quote]
I think the following reasoning easy to implement.
Let $2a+b=x$, $2b+c=y$ and $2c+a=z$.
Hence, $a=\frac{4x-2y+z}{9}$, $b=\frac{4y-2z+x}{9}$ and $c=\frac{4z-2x+y}{9}$
and we need to prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{27}{5(xy+xz+yz)-2(x^2+y^2+z^2)}$.
[/quote]

Two problems are not equivalent when changing variables. It is not always true
\[ 5(xy+xz+yz)-2(x^2+y^2+z^2) \geq 0\]

[quote=bedwarmer]
Two problems are not equivalent when changing variables. It is not always true
\[ 5(xy+xz+yz)-2(x^2+y^2+z^2) \geq 0\][/quote]
It's true because $ab+ac+bc>0$. ;)

[quote=arqady]We can use  $uvw$ for proving of the general case.[/quote]

General case can be proved by assume  $z = min \{x,y,z\}$ and did similar estimation as Urun did.

[quote=MathPanda1]Urun, how do you come up with your fantastic proof? For example, how did you think of splitting $F$ into $A+B$[/quote]

I remember a proof of Iran 1996 by the same method, however used different coefficients. ;)

[quote=GacKiem]If $a,b,c$ are positive real numbers then
\[\frac{1}{(2a+b)^{2}}+\frac{1}{(2b+c)^{2}}+\frac{1}{(2c+a)^{2}}\ge \frac{1}{ab+bc+ca}.\][/quote]

There is an unique property of this inequality. If you take a partial derivative of the function with respect to $a,b,c$, you have a new and correct inequality. Please prove that
\[ \sum_{cyc} \frac{3}{(2x+y)^3} \geq \frac{x+y+z}{(xy+yz+zx)^2}\]
for all positive reals $x,y,z$.

I

[quote=bedwarmer]

Please prove that
\[ \sum_{cyc} \frac{3}{(2x+y)^3} \geq \frac{x+y+z}{(xy+yz+zx)^2}\]
for all positive reals $x,y,z$.

I[/quote]
Let $2x+y=a$, $2y+z=b$ and $2z+x=c$. Hence, $x=\frac{4a-2b+c}{9}$, $y=\frac{4b-2c+a}{9}$, $z=\frac{4c-2a+b}{9}$ and we need to prove that
$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\geq\frac{81(a+b+c)}{\left(\sum\limits_{cyc}(2a^2-5ab)\right)^2}$.
The following step for you, dear bedwarmer! :-D
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $\frac{27v^6-27uv^2w^3+3w^6}{w^9}\geq\frac{243u}{(2(9u^2-6v^2)-15v^2)^2}$ or $\frac{9v^6-9uv^2w^3+w^6}{w^9}\geq\frac{u}{(2u^2-3v^2)^2}$
or $f(w^3)\geq0$, where $f(w^9)=-uw^9+(2u^2-3v^2)^2w^6-9uv^2w^3(2u^2-3v^2)^2+9v^6(2u^2-3v^2)^2$.
But $f'(w^3)=-3uw^6+2(2u^2-3v^2)^2w^3-9uv^2(2u^2-3v^2)^2=-3uw^6-(2u^2-3v^2)^2(9uv^2-2w^3)\leq0$.
Thus, $f$ is a decreasing function.
Hence, $f$ gets a minimal value for maximal value of $w^3$, which happens, when two variables are equal.
Since our inequality is homogeneous, it remains to prove that $f(w^3)\geq0$ for $b=c=1$, where $\frac{1}{4}<a<\frac{5}{2}$, which gives
$(a-1)^2(8a^5-64a^4+56a^3+139a^2+22a+1)\geq0$, which is true for all $\frac{1}{4}<a<\frac{5}{2}$. Done!


Here in Georgia, there has been a "statewide recommendation" from the governor that all schools close this Monday and Tuesday due to gas shortages from Rita.  From what our bus driver said, we can expect other counties to follow the recommendation as well.  Is this happening in any other states?  What are your thoughts on this?

(And there was a math tournament scheduled for Monday.  Alas.)

[quote="236factorial"]That's an interesting fact  :o  I wish it would happen, but I dont' think it will happen up here in NY...[/quote]

Yes. NYS is so far up north that it probably won't happen. Does the gas shortage affect the whole nation? If it does, maybe it'll happen.

Problem is that the hurricane might prevent the import of fuel to Georgia, while, in NY, you could import from other sources.

gas shortage does affect the whole nation (especially in outrageous prices) , but not so bad in most places that school will stop.  though that [u]would[/u] be great....


Do professors that give you a letter of recommendation see your whole application after you submit it? Isn't this invasion of privacy?


blogs
inequalities unsolved
inequalities

The second see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66660
The third is Vasc's inequality.
The hint to nice proof: $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}\geq\frac{2}{ab+1}.$ :wink:

The three questions are from my blog: http://hi.baidu.com/shfdfzhjj
Arqady wrote:"The hint to nice proof: $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}\geq\frac{2}{ab+1}.$ :wink:[/quote]"

I think it isn't a nice proof.
$\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}\geq\frac{2}{ab+1}.$ :wink:[/quote]"
It holds iff ab=>1
We should  prove: $\frac{1}{c^{2}+1}+\frac{2}{ab+1}\geq\frac{3}{2}$
with ab=<(3+c^2)^(0.5)-c.
We can solve it by force.
But it is very ugly.
I'm waiting for a nice solution.

[quote="arqady"][quote="shfdfzhjj"]
I think it isn't a nice proof.
$\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}\geq\frac{2}{ab+1}.$ :wink:
[/quote]
It's not  proof. It's hint to very nice proof.
The hint for you, shfdfzhjj: let $a\geq b\geq c.$ :wink:[/quote]

I know,but how about the first?

[quote="shfdfzhjj"]
I know,but how about the first?[/quote]
If $a\geq b\geq c>0$ and $ab+ac+bc=3$ then $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}\geq\frac{2}{ab+1}$ is true. :wink:

So i wrote ab=>1.
The firt I mean is the fisrt inequality.
13.jpg

[quote="shfdfzhjj"]
The firt I mean is the fisrt inequality.
13.jpg[/quote]
I proved it, it's true, but my prove is very ugly. :(


MATHCOUNTS factorial geometry perimeter probability rectangle blogs

I looked over many of the posts on this forum.  Many have helped out and I thank everyone who participated in those threads greatly.  The thing I haven't seen mentioned are [size=150]formulas that are often used in States and Nationals[/size].  I think if we each posted a formula we know and how to use it, then we will be much more prepared.  It doesn't have to be some bizarre formula no one ever uses, even the simple ones are fine.  :lol:

[size=167] [color=red]If the above is too much to read, just look below.   :lol: [/size] [/color]

Let me start:

Pythagorean Theorem- very oftenly used in MATHCOUNTS
[color=darkred]Used for finding a side of a right triangle.[/color]
[color=green]Formula:  a^2 + b^2 = c^2 where c is the hypotenuse while a and b are the sides of the triangle.[/color]
[color=darkblue]Problem:  The base of a triangle is 5 cm, and the hypotenuse is 13 cm.  Find the length of the other side.
Solution using the formula: Let a= the base, b= the missing side, and c= the hypotenuse.
5^2+b^2=13^2
25+b^2=169
b^2=169-25=144
b=12[/color]

I know most people already knew that, but i just wanted to give an example.
  [color=red]Note:  Your explantion doesn't have to be as in depth as mine, you can just mention the formula and what its used for if you want.  [/color]
For example:
Formula: a^2 + b^2 = c^2 where c is the hypotenuse while a and b are the sides of the triangle (used for finding sides of a right triangle)

Simple as that.  I know its a lot to read but every little bit will help.  Thank you everyone.  I only have one week so type all the formulas you can!  :P

Please help me out! :blush:

[quote="lightspeed"]Sorry!  I accidenally clicked submit the first time, clicked preview, thought it didn't submit, then submitited it again.  I'm really sorry for any problems this might have caused. :([/quote]

All you have to do is to delete the other forum... and I think this thing is a really good idea... anyways I'll contribute my formula:

For any two relatively prime integers m and n, the greatest possible number that is impossible to attain by summing the numbers is [b]mn-n-m[/b]

1) Thanks for telling me that.  I'm new here (I joined today), so i really don't know much.
2)  I'm really glad that you think this is a good idea.
3)  I don't quite understand the formula.  Can you explain it?  Like where its used or how?

Hockey, I've noticed you've made like the most posts out of everyone here.  I think everyone should thank you emmenesly for all your contribution!  You should give yourself 5! (5 factorial) pats on the back!

Here's another formula:
[color=red]The sum of the first n odd natural numbers is n^2.[/color]

I think it would be a good idea if someone took all of the formulas that will hopefully come on this thread and made a sticky out of them.

Also...I have one more problem.  There's only 1 week before states.  I don't know how many formulas will arrive by then!  I hope there will be a lot!

Thanks to everyone who contributes! :lol:  :D

[quote="hockeyadi23"][quote="lightspeed"]3)  I don't quite understand the formula.  Can you explain it?  Like where its used or how?[/quote]
An example of this problem and its application is in the forum "Do the worm": I'll list the problem here.
[quote="Treething"]Every time Albatross does the worm, he gets pelted by thirteen rotten tomatoes. If he does the super-worm, he gets pelted by twenty-nine rotten tomatoes.

What is the most number of rotten tomato hits Albatross [b]cannot[/b] receive?[/quote]
Here, you would use the formula mn-m-n to find this answer. 29*13-29-13  = 335, which would be pretty difficult to get without the formula, and near impossible on the Sprints...[/quote]

Wow, that problem is super confusing.  How did you get so smart?  I'll have to ask you lots of questions that i don't get. Lol

Thanks  :lol:
P.S. Feel free to PM me anytime  :D

[quote="kidwithshirt"]i always loved hero's formula

this works good on linear triangle problem

Formula is:
A=sqrt[s(s-a)(s-b)(s-c)]

*s is the half of the perimeter of the triangle

Example

a triangle has side 3,4,5
s=2/(3+4+5)=6
A=sqrt[s(s-a)(s-b)(s-c)]
A=sqrt[6(6-3)(6-4)(6-5)]
A=sqrt[36]
A=6 sq.unit :P  :D  :D[/quote]

Thanks for the contribution  :lol:
I hope we get a lot more formulas in soon :D

[color=darkred]Here's the handshake one (it works only if each person shakes hands ONCE with another person):[/color]
[color=red]n(n-1)/2[/color]

Where n is the number of people.  Check the thread, "SOS..." for more info on this formula.

Thanks everyone who participates! :)

Here i ll be the genie for a millisec...

This might help a lot on probability

[b]basic: Probability[event]=(number of favorable outcomes)/(number of possible outcome)[/b]
example: A bag is filled with 3 yellow marble and 9 green marble. Probablity of yellow randomly: p[yellow]=3/12->1/4
[b]
Probablity of A and B if A and B are independent: P[A and B]=P[A]*P[B][/b]
Example: Roll a red die and a blue die. A red one and a blue 5 land face up. Probability:
P[red 1 and blue 5]=p[1]*p[5]
P[1 and 5]=(1/6)(1/6)=1/36

I ll get more later :P

Number of 0's at the end of n! is

[n/5] +[n/5^2] + [n/5^3] +........+[n/5^(infinity)]

where [x] is the greatest integer less than or equal to x

Thanks!  That could really help!  Oh yeah, thanks for being my genie for a short amount of time lol :D
Another request from me:  Could everyone post their formulas in red?  Thanks!

Here's something that might be of use:

[color=red](1/4)+(1/4)^2+(1/4)^3+....
=1/3[/color]

 :lol:

[quote="kidwithshirt"][b]Probability Of A and B both occur: p[A or B]=p[A]+p[B]-p[A and B] [/b]
example: select a 3 or a club from a deck of card
P[3]=4/52
P[Club]=13/52
P[3 and Club]=(4/52)*(13/52)=1/52
P[3 or Club]=4/52+13/52-1/52
                   =16/52or 4/13[/quote]

That's awesome thanks!  I could really use that as well!  I'm not really good with probability  :blush:

same here

btw all the formulas and examples are from a book called "algebra to go", its an awesome book

it comes down to in a problem like in how can you m,ake 1.00 dollar out of quarters dimes and nickels
$(\frac {25} {10} +1)+(\frac {50} {10}+1) + (\frac {75} {10}+1) + (\frac {100} {10} + 1)+1 =29$

Trust me once you do a few problems with it it's not as hard as the equation looks.

[quote="Press alt f4"]The Coin problem trick is a [b]fast[/b] way to solve coin problems.

$%Error. "display" is a bad command.
(  \lfloor  \frac {b_1}{LCM b_1, b_2, ..., b_n}  \rfloor+ \lfloor  \frac {2b_1}{LCM b_1, b_2, ..., b_n}  \rfloor+ ... +  \lfloor  \frac {\frac {V} {b_1} b_1}{LCM b_1, b_2, ..., b_n}  \rfloor + 1 $  $+$
$%Error. "display" is a bad command.
\lfloor  \frac {b_2}{LCM b_2, ..., b_n}  \rfloor+ \lfloor  \frac {2b_2}{LCM b_2, ..., b_n}  \rfloor+ ... +  \lfloor  \frac {\frac {V} {b_2} b_2}{LCM b_2, ..., b_n}  \rfloor + 1 $ $+$  $...$
$%Error. "display" is a bad command.
\lfloor  \frac {b_(n-2)}{LCM b_(n-2) ,b_(n-1) ,b_n}  \rfloor+ \lfloor  \frac {2b_(n-2)}{LCM b_(n-2), ..., b_n}  \rfloor+ ... +  \lfloor  \frac {\frac {V} {b_(n-2)} b_(n-2)}{LCM b_(n-2) ,b_(n-1) ,b_n}  \rfloor + 1 $ $%Error. "display" is a bad command.
\lfloor \frac {V} {LCM b_(n-1), b_n}  \rfloor  ) + 1$[/quote]

What's the simplified version of that equation?

Thats just it I forget how to symplify it  :blush: .

That formula seems way too complicated to be aplied to anything!  I think I would probably get the answer faster by doing it out the long way.

[quote="lightspeed"][quote="dwx314"][quote="lightspeed"]So much math... It's confusing my brain!  So many less people post now because many people lost at states :(  I know how it feels, for I was one of those people :(  Anyway, good luck to anyone who did make it to nationals!  Hope you have a great time w/o me :([/quote]
either that, or they were posting just because of states. (Notice the HUGE increase of posting the weeks before chapter and state) But nationals in MAY.
Doubtless the two weeks before nationals we'll see another surge. (Though it'll prob. be smaller :( )[/quote]
Yeah, you are probably right.

News update: [color=red][size=150]This thread now has the largest number of replies!  Yay us! When I first joined AoPS, I wanted to have the thread with the largest number of replies, and now I do!  Thanks to all who participated![/size][/color] Keep this thread going, if you can! :D
Thanks[/quote]

How does it matter?? There is no use in
having so many replies..its actually kind of annoying seeing the same thread over and
over again.

btw...this isnt the largest thread, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15270
has more replies than this.

Then I am glad to post to the largest thread :). How are you supposed to simplify something where practically every term is floored? I don't get how its possible. Anyway, you'd have to use it over and over again to actually get good enough to use it in the test! And see how long it takes when you add pennies or even another coin (perhaps a half-dollar).

gobraves, I was talking about on this Mathcounts Forum, not in the whole thing!  If I wanted to make the largest thread in all of AoPS,  I would choose a different topic.  As for the use, It makes me feel good and proud :D
To know that you accomplished your goal feels really good :D

[quote="DarkKnight"]If you really wanted the topic with the most posts, go to the word association game. 63 pages. :O

Anyway, I didn't participate in Mathcounts this year but all these formulas will help me next year. Thanks.[/quote]
63 pages?  That's crazy!  Yeah, my original purpose of this thread was to help as many people as possible, and for me to learn as many formulas as I could.  But somehow, it got kinda side tracked after the first few pages. hehe :D

[quote="PTheodora"][quote="LuCy4EvA"]PTheodora wrote:
LuCy4EvA wrote:
nope, but this dude almost almost made it.... at first they told him that he made it, but then they figured that the score was wrong... or was that in chapter? anyways, I knew the people who went to nationals last year, and they are incredible-io..... but they are in ninth grade now... why?

For some reason I thought a 7th grader from your school made it last year...but I'm probably wrong.

really? you sure she/he is from Pizitz? Kathy? Wayne? wait, Wayne made it to state but he wasn't on the team.... that's probably who.... and yes, he's on the team this year.... and he's really good at countdown..... [/quote] Kathy...maybe...is she an asian girl? Btw, the guy who got first place last year (before and after countdowns) was awesome. He was so quick!

And did you mean Wayne made it to nationals?

[quote="lightspeed"]Well, good luck to all at states!
*Cheers*
There are so many posts now!  This is awesome! :D:D:D:D
Drats, there was a MathJam yesterday?  I missed it!  When's the next one, and at what time?  How can I join?  Do I have to pay?

Sorry for all the questions in one post.  ;)[/quote] Math Jams are free. Click on the MathJams tab at the top of the page for a schedule. It's right next to the "Forum" tab.  :)

[quote="lightspeed"]*Update* *Update*

This thread has the second greatest number of replies! I remember in the beginning, it was just Hockey and me posting on this thread. I urged more people to post, but the posts didn't come in til a lot later! But now, it has the second greatest number of replies!!!!

Thanks to all those participated! Remember, keep posting! I want the GREATEST number of replies!  lol[/quote] To me, a lot of replies is great, but the quality of the posts in a thread matters more than the quantity...that's just me though.[/quote]

i know this is late and everything, but i came back to this topic today because Im thinking about making a formula sheet out of the stuff that u guys posted in here, and i saw this....

Kathy is Asian, she got 3rd at chapter this year, and got so close to making nationals this year, she got fifth after Countdown, although they made Countdown unofficial after they screwed up someone's scare. and the guy who got first last year's name was Wei, he's super duper smart.... he used to be my neighbor, and i used to ask him math questions, he's like good at everything... my coach said that Wei said that when he does countdown, he mainly read the numbers and the few words around it in the problem, so it saves time... i can never do that

mark zhang is fast in countdown too  :lol:

better watch out for him at nationals if you make the countdown


Results for the first round (the Slitherlink round). Here are the top 30:

$ \begin{tabular}{lll}Rank&Team&Score\\
1.&Henry M. Gunn HS&40\\
1.&Stuyvesant HS&40\\
3.&Lynbrook HS&38\\
3.&Sofia Math Circle&38\\
3.&State College Area HS (North)&38\\
3.&Western Washington ARML&38\\
7.&Wayzata HS&36\\
8.&Albany Area Math Circle&35\\
8.&Olimipiades Columbianas de Math.&35\\
8.&Oregon ARML&35\\
8.&Thomas Jefferson HS&35\\
12.&Georgia ARML&34\\
12.&NYC Math Team&34\\
12.&Univ. of Chicago Lab Schools&24\\
15.&Georgetown Day School&33\\
15.&Glenbrook North HS&33\\
15.&Palo Alto Math Team&33\\
18.&Colorado Math Circle&31\\
19.&Deerfield Academy&30\\
19.&The Dalton School&30\\
21.&Harvard\minus{}Westlake&29\\
21.&Iowa City West HS&29\\
23.&Central Academy&28\\
23.&Huron HS&28\\
23.&Long Beach Math Circle&28\\
23.&St. Louis Math Circle&28\\
27.&Mounds View HS&27\\
27.&Naperville North HS&27\\
27.&West Tennesee&27\\
27.&Windsor Math Enrichment&27\end{tabular}$

I really didn't like the first round. It wasn't mathematical at all, and I didn't find it remotely interesting. (Well, there was one proof, but it was the first question we solved, before I had even read it) Also people who had seen and played Slitherlink before had a big advantage.

[quote="AwesomeToad"]I thought ARML was Jun 4-5[/quote]

The Power Contest is independent of the ARML results in June. It's something additional to do over the year. Same with the ARML Local.

Final results for 2009-2010:

$ \begin{tabular}{lllll}Rank&Team&Round 1&Round 2&Total\\
1&Wayzata HS&36&39&75\\
2&Western Wash. ARML&38&35&73\\
3&Lynbrook HS&38&34&72\\
4T&Albany Math Circ.&35&36&71\\
4T&Stuyvesant HS&40&31&71\\
6&Sofia Math Circ.&38&30&68\\
7T&Colorado Math Circ.&31&36&67\\
7T&Georgia ARML&34&33&67\\
9&Gunn HS&40&23.5&63.5\\
10&State College HS (North)&38&25&63\\
11T&Long Beach Math Circ.&28&33&61\\
11T&Palo Alto&33&28&61\\
13&Iowa City West HS&29&31&60\\
14T&Georgetown Day School&33&26&59\\
14T&Oregon ARML&35&24&58\\
16&U. Chicago Lab School&34&24&58\\
17&Central Academy&28&26&54\\
18&Huron HS&28&25&53\\
19T&Hopkins School&26&25&51\\
19T&Windsor Math Enrich.&27&24&51\\
21&Utah ARML&24&24.5&48.5\\
22&Harvard\minus{}Westlake&29&19&48\\
23&Central HS&24&21&45\\
24&Western MASS ARML&15&29&44\\
25T&Deerfield Acad.&30&13&43\\
25T&Naperville North HS&27&16&43\\
27T&Dalton School&30&10.5&40.5\\
27T&University School&24&16.5&40.5\\
29&Evanston HS&19&20&39\\
30&Great Neck&21&17&38\end{tabular}$


You can also derive n(n-1)/2=120 by saying that there are n-1 triangles with a base of 1, n-2 with a base of 2, n-3 with a base of 3,..., and 1 with a base of n.

This gives us (n-1)+(n-2)+(n-3)+...+1=120.
Using the summation formula for the sum of all integers up to n (which can be altered to all the positive integers up to n-1) we have n(n-1)/2=120, which is what you achieved with an alternate method.

I just found that with two points, there is 1 triangle, with 3 points there are 3 triangles, with 4 points there are 6 triangles, and with 5 points, there are 10 triangles. Using this pattern I found that there had to be 16 points.

you just need to do algebra: recognizing that you must pick 2 vertices out of n, you have a combination...

[hide=Above Sol]As said above, you simply need to choose two of the $n$ possible vertexes in the form $A_n$. Thus;
$\binom{n}{2} = 120 \implies n(n-1) = 240 \implies  n = \boxed{16}$.[/hide]

i understand the solution but i cant picture how there will be 120 triangles if the last vertices or n is 16. Shouldn't it technically be 121?

[quote=arnolde1234][quote=haha0201][quote=Th3Numb3rThr33][hide=Above Sol]As said above, you [b][size=200]simply[/size][/b] need to choose two of the $n$ possible vertexes in the form $A_n$. Thus;
$\binom{n}{2} = 120 \implies n(n-1) = 240 \implies  n = \boxed{16}$.[/hide][/quote]

Yes, they overrated this. Level 18 this easy????[/quote]

Level 19 now for me![/quote]

me too level 19


Since dropping your calculator, two of the display digits act strangely.  After attempting to calculate $11^9$, your calculator display shows $\frame{\quad\tt 2357947\ddag 9\dag\quad}$.  Assuming your calculator is otherwise working, what is the actual value of the digit represented by the $\ddag$ symbol?

[size=75][Note:  Obviously, use of a calculator is not allowed.][/size]

[hide]the last digit of 11 is 1, so when you multiply it by itself, the last digit will always be 1.  so the wierd sign thingy=1[/hide]

[hide="to Technogram888"]You have found the units digit ($\tt\dag$) quite efficiently.  But the question asked about the hundreds digit ($\tt\ddag$).  Do you have any ideas for finishing the problem?[/hide]

[hide="hint"]
Divisibility rules
[/hide]


calculus
integration
limit
logarithms
function
real analysis
real analysis unsolved

Let $f(x)=\left\{ \begin{array}{cc} 1 & if\ x=0\\ \frac 1x\min\limits_{n\in N} |n-x| & if\ x>0\ \end{array}\right.$

and $a_n=\int^n_0 f(x)\ dx,\ n\in N^*$.

Prove that $\lim_{n\to \infty} a_n=\infty ,\ \lim_{n\to \infty} \frac{a_n}{\ln n}=\frac 14.$

Can you tell me what is the value of $a_1$?

That's irrelevant- you don't need to calculate any particular $a_k$ to solve this.

We're looking at the integral of $\frac1xg(x)$, where $g$ is the periodic function "distance to nearest integer". When you see an integral involving an oscillating function, you should often consider integration by parts, and integrate the oscillating part (it helps if the average value is zero). This is all about creating faster convergence.

Let $h(x)=\int_1^x g(t)-\frac14\,dt.$ $h$ is periodic and bounded. We have $f(x)=\frac1{4x}+\frac1x\cdot(g(x)-\frac14)$, so
$\int_1^M f(x)\,dx=\int_1^M\frac1{4x}+\frac1x\cdot(g(x)-\frac14)\,dx$ $=\frac14\ln(4M)+\frac1x\cdot h(x)|_{x=1}^{x=M}+ \int_1^M\frac1{x^2}\cdot h(x)\,dx$
$=\frac14\ln(4M)+\frac{h(M)}M+\int_1^M\frac{h(x)}{x^2}\,dx$

Since $|h(x)|<\frac14$ for all $x$, we have $|\frac{h(M)}M|<\frac14$ and $|\int_1^M\frac{h(x)}{x^2}\,dx|<\frac14$ for all $M>1$. We also have $|f(x)|\le1$ for $0<x<1$, so $|\int_0^1 f(x)\,dx|<1$ and $|\int_0^M f(x)\,dx-\frac14\ln(4M)|<\frac32$ for all $M>1$, proving the desired result.
The error estimates here could easily be improved. Slightly less precisely, we have $a_n=\frac14\ln(n)+O(1)$.


superior algebra
superior algebra unsolved

Inside R, look at the set K consisting of all numbers of the form a+b(sqr(2)), where a and b are rational. Assume you know that sqr(2) is not rational. Prove that K is a subfield of R.

Just construct addition, multiplication, and inverses.


calculus
derivative
function
vector
linear algebra
calculus computations

I'm studying partial derivatives and somethings confusing. In elementary maths by gradient of a function you mean the derivative at that point, and the derivative is always tangent to the function.So  how come gradient vector and tangent vector are two different things ?
What is the gradient vector??

Gradients and tangents live in different spaces. The tangent to a curve (a function from $\mathbb{R}$ to $\mathbb{R}^{n}$) at a point is an element of $\mathbb{R}^{n}$. The gradient of a function from $\mathbb{R}^{n}$ to $\mathbb{R}$ at a point is an element of the dual space $(\mathbb{R}^{n})^{*}$ of linear functionals.

The spaces $\mathbb{R}^{n}$ and $(\mathbb{R}^{n})^{*}$ are isomorphic, since they're both $n$-dimensional vector spaces. This isomorphism depends on the choice of basis, and the two spaces transform differently under linear transformations: the order in which we multiply matrices is reversed.

This may not make sense now, but linear algebra is important to really understand multivariable calculus. It's worth revisiting the subject later, when you have more of a background.


[quote="myyellowducky82"]Any advice??? Best is 2~3mins, but that time was simple.[/quote]

I think it would be really helpful if you had been more precise. ( What exactly do you mean by: "but that time was simple" ?)

Your frequent indulgence in "imprecise formulations" might lead to various interpretations of your query/queries. I interpret the first half of your query (also from the "title of the thread") as follows: [b] "How to solve a Rubik's cube more quickly? Can anyone offer some suggestions/tips?" [/b] The latter half of your query (or rather sentence) is ambiguous and horribly formulated.

Also, A more meaningful title for your thread could have been: "Rubik's Cube" with subject description as: "Can anyone offer some suggestions/tips on solving a Rubik's cube very quickly, say in under 2 minutes?"  (I think myyellowducky intends to say that the least time it ever took him to solve a Rubik's cube was 2 or 3 minutes)

Well the easiest way (for now) is just to practice and learn finger tricks. That'll bring your time down to 30 seconds. You can also learn more algorithms, but no one likes that right?

Hmm, would practicing piano help? :wink:

Unless your fingers are amazingly clumsy, no.

Not if you play slow songs. To start with, you need a good cube. Next, you need to use a good method and fingertricks.

The thing with Rubik's cube times is that it's really about how long you think between moves, not how fast you do moves. To get a faster time you need to be faster at pattern recognition. To see my point, think about how long it takes you to make a turn. You can probably make 3 in one second easily. Now, if your solving strategy makes you take 51 moves on average, that should be 17 seconds, right? The problem is that patern recognition makes you think for several seconds, or longer. Work on pattern recognition if you want to get faster.


Let $ 0<a_k<1$ for $ k=1,2,... .$ Give a necessary and sufficient condition for the existence, for every $ 0<x<1$, of a permutation $ \pi_x$ of the positive integers such that \[ x= \sum_{k=1}^{\infty} \frac{a_{\pi_x}(k)}{2^k}.\]

[i]P. Erdos[/i]

I think it should be
$ \inf a_k \equal{} 0$ and $ \sup a_k \equal{} 1$ but i'm not sure.

It is obvious that $\inf a_k = 0$ and $\sup a_k = 1$ are necessary conditions.

We will show that the these two conditions are also sufficient. That is, if $(a_k)_{k \geq 1}$ is a sequence such that $0 < a_k < 1$ and $\inf a_k = 0$ and $\sup a_k = 1$, then for any $x \in (0, 1)$ there exists a bijection $\pi: \Bbb Z_{\geq 1} \rightarrow \Bbb Z_{\geq 1}$ such that $x = \sum_{k = 1}^\infty \frac{a_{\pi(k)}}{2^k}$.
-----
Given any $x \in (0, 1)$, we will construct the map $\pi$ inductively.
We define $i_0 = 0$ and $s_0 = 0$. For $t \geq 0$, assume that we have defined $\pi(1), \dots, \pi(s_t)$ in such a way that $x_t = 2^{s_t}\left(x - \sum_{k = 1}^{s_t} \frac{a_{\pi(k)}}{2^k}\right)$ lies in the interval $(0, 1)$.
We define $i_{t + 1}$ as the smallest positive integer that doesn't belong to $\{\pi(1), \dots, \pi(s_t)\}$ and write $a$ for $a_{i_{t + 1}}$.

There are two cases: $x_t \leq \frac 1 2$ or $x_t > \frac 1 2$.
We first treat the case $x_t \leq \frac 1 2$.
In this case, we may choose $r \in \Bbb Z_{\geq 0}$ such that $\frac a 2 < 2^r x_t < \frac{a + 1}2$. In fact, if $x_t > \frac a 2$, then taking $r = 0$ suffices; otherwise, we can take $r$ to be the smallest integer such that $2^r x_t > \frac a 2$, so that $2^{r - 1}x_t \leq \frac a 2$ and hence $2^r x_t \leq a < \frac{a + 1}2$.
We choose $b_1, \dots, b_r$ such that all $b_j$ are different, none of the $b_j$ appears in $\{i_{t + 1}, \pi(1), \dots, \pi(s_t)\}$, and that $\sum_{j = 1}^r 2^{r - j}a_{b_j} < 2^rx_t - \frac a 2$. This is possible, thanks to the assumption $\inf a_k = 0$. It follows that $\frac a 2 < 2^r x_t - \sum_{j = 1}^r2^{r - j}a_{b_j} < \frac{a + 1}2$.
We define $\pi(s_t + j) = b_j$ for $1 \leq j \leq r$, $\pi(s_t + r + 1) = i_{t + 1}$, and $s_{t + 1} = s_t + r + 1$.

Let us verify that $x_{t + 1} = 2^{s_{t + 1}}\left(x - \sum_{k = 1}^{s_{t + 1}}\frac{a_{\pi(k)}}{2^k}\right)$ lies in the interval $(0, 1)$.
From the definitions, we compute $x_{t + 1} = 2\left(2^r x_t - \sum_{j = 1}^r 2^{r - j}a_{b_j}\right) - a$, which lies in the interval $(0, 1)$.

The case $x_t > \frac 1 2$ is treated similarly. For convenience, we write $\overline z = 1 - z$ for any $z \in (0, 1)$.
In this case, we have $\overline{x_t} < \frac 1 2$, thus we may choose $r \in \Bbb Z_{\geq 0}$ such that $\frac{\overline a}2 < 2^r \overline{x_t} < \frac{\overline a + 1}2$.
We choose $b_1, \dots, b_r$ such that all $b_j$ are different, none of the $b_j$ appears in $\{i_{t + 1}, \pi(1), \dots, \pi(s_t)\}$, and that $\sum_{j = 1}^r 2^{r - j}\overline{a_{b_j}} < 2^r \overline{x_t} - \frac{\overline a}2$, which is possible because $\sup a_k = 0$. This implies $\frac{\overline a}2 < 2^r\overline{x_t} - \sum_{j = 1}^r2^{r - j}\overline{a_{b_j}} < \frac{\overline a + 1}2$, which is equivalent to $\frac a2 < 2^rx_t - \sum_{j = 1}^r2^{r - j}a_{b_j} < \frac{a + 1}2$.
We define as before $\pi(s_t + j) = b_j$ for $1 \leq j \leq r$, $\pi(s_t + r + 1) = i_{t + 1}$, and $s_{t + 1} = s_t + r + 1$.

It is then verified in the same way that $x_{t + 1}$ lies in the interval $(0, 1)$.

This finishes the inductive construction of the map $\pi$. It remains to prove that $\pi$ is bijective and that $x = \sum_{k \geq 1}\frac{a_{\pi(k)}}{2^k}$.

The injectivity of $\pi$ is clear from the definition. To see that $\pi$ is surjective, we show that any integer $t \geq 1$ appears in $\{\pi(1), \dots, \pi(s_t)\}$ by induction on $t$.
Suppose this is true for $\leq t$ and we want to prove it for $t + 1$.
If $t + 1$ appears in $\{\pi(1), \dots, \pi(s_t)\}$, then it is already OK.
Otherwise, we must have $i_{t + 1} = t + 1$, because $1, \dots, t$ all appear in $\{\pi(1), \dots, \pi(s_t)\}$ by induction hypothesis. Thus $\pi(s_{t + 1}) = i_{t + 1} = t + 1$.

Finally, we know that $x_t = 2^{s_t}\left(x - \sum_{k = 1}^{s_t}\frac{a_{\pi(k)}}{2^k}\right)$ lies in the interval $(0, 1)$ for every $t$. This clearly implies that $x = \sum_{k = 1}^\infty\frac{a_{\pi(k)}}{2^k}$.


Let be $ 0< a\le b\le c\le d\le \sqrt [4]{7}$ . Show that :

$ a^5 \plus{} a(b^4 \minus{} 14) \plus{} b(c^4 \minus{} a^4) \plus{} c(d^4 \minus{} a^4) \plus{} d(14 \minus{} c^4) \minus{} d^5\le \frac {56\sqrt [4]{7}}{5}$

It's wrong! Try $ a\equal{}0$, $ b\equal{}c\equal{}d\equal{}\sqrt[4]{7}$.


I have a question about the current contest, problem 3.
"Let ABC be a triangle and let A' \in BC, B' \in CA and C' \in AB be three collinear points."
How is that possible? As far as I know collinearity means that all the points lie on a single straight line?

[quote="karsten bohlen"]As far as I know collinearity means that all the points lie on a single straight line?[/quote]Indeed that is the definition of collinearity. The problem is well stated. Read it again carefully :)

This is a problem of denotation, especially for German highschool students.

Many pupils don't know the differences of denoting lines and segments.

For instance I use $AB$ to denote a line and $\overline{AB}$ to denote a segment.

But various teachers use $(AB)$ to denote a line and $AB$ to denote a segment.

I think that helps.
Best regards.
D. F. G.

Ah, ok, I thought everyone knows that $(AB)$ is an open segment, $[AB]$ a closed segment, $\overline{AB}$ an orientated segment, $(AB$ a ray, $[AB$ a closed ray and finally $AB$ a line :)

Yes that was my problem Daniel. Now it is clear.
I indeed thought of a segment although I normally also use $\overline{AB}$ to mark it.. sorry :blush:

So it means that atleast one point lies of extended side of ABC ?


When I need to prove if it is impossible to have an integer solution to a equation with integers what mod do I consider everything in?

I know for cubes, mod 7 usually works.

What about squares,cubes,4th.....10th powers?

Here's what I usually try:
squares: 8 is [i]very[/i] useful (squares 0, 1, 4), sometimes 3 or 5 will also work well (0,1 mod 3, 0, 1, 4 mod 5)
cubes: 7 and 9 (both mods only have cubes of 0, +/- 1)
fourths: 5 and 16 (0 or 1 in each case)
kth powers: p if p is prime and equals k+1 (0 or 1) or 2k+1 (0 or +/- 1)

[quote="scorpius119"]kth powers: p if p is prime and equals k+1 (0 or 1) or 2k+1 (0 or +/- 1)[/quote]

FLT  for the first one, right? What's the proof for the second?Sorry, I am bad at number theory.

Both use Fermat's Little Theorem. For the first case, $a^p\equiv a\Rightarrow p|a(a^k-1)$. For the second, $a^p-a$ factors into $a(a^k-1)(a^k+1)$.


search
combinatorics unsolved
combinatorics

determine the number of subset S of the set T={1,2,...,2005} such that that the sum of elements in S is congruent to 2006 modulo 2048

A search for "2006 modulo 2048" brings up the following threads:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39550
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167883
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=228375


MATHCOUNTS
LaTeX
calculus
integration
number theory
least common multiple

I want to see how it is solved. Please show all your work.

By the way, I typed all the problems :) 154wpm

#28
When reading a book, Charlie made a list by writing down the page number of the last page he finished reading at the end of each day. (He always finished reading a page that he started) His mom thought his list indicated the amount of pages he had read on each day. At the end of the 8th day of reading, she added the numbers on his list and thought CHarlie had read 432 pages. IF Charlie started reading the book on page one, and he read the same amount of pages each day of this eight day period, how many pages did he actually read by the end of the 8th day?

#29 Suppose that each distinct letter in the equation MATH=COU+NTS is replaced by a different digit chosen from 1 through 9 in such a way that the resulting equation is true. If H=4, what is the value of the greater of C and N

   C   O   U
+ N   T   S
-------------
M A  T   H

(sorry for bad diagram :( didnt download latex yet)
( I got the right answer for this question, but I kinda guessed...)

#30
One gear turns 33 1/3 times in a minute. Another gear turns 45 times in a minute. Initially, a mark on each gear is pointing due north. After how many seconds will the two gears next have both their marks pointing due north?

Have fun solving them  :D

[hide="#30"]
The first gear turns $ \frac {33\frac {1}{3}}{60} \equal{} \frac {\frac {100}{3}}{60} \equal{} \frac {100}{180} \equal{} \frac {5}{9}$ times per second.  The second gear turns $ \frac {45}{60} \equal{} \frac {3}{4}$ times per second.  They will both be pointing due north when they turn an integral number of times.  Therefore, the number of seconds the gears turn for must be a number that can cancel out the $ 9$ in the denominator of $ \frac {5}{9}$, and a number that can cancel out the $ 4$ in the denominator of $ \frac {3}{4}$.  This number is the smallest when it is equal to $ \text{lcm}(4,9) \equal{} \boxed{36}$.[/hide]

[hide="#28"][hide="Hint"]Let $ x$ be the number of pages Charlie reads each day. In terms of $ x$, how many pages does his mother think that he has read? Use this equation, setting it equal to 432, to find his daily rate.[/hide]
[hide="Solution"]
Charlie reads $ x$ pages per day. On day one, he reads $ x$ pages, and he reads the same number of pages on each day until day 8. Thus, his marking for day one will show $ x$ pages, and each subsequent day will show $ x$ more pages than the previous day's number.

His mother thinks that he has read $ x \plus{} 2x \plus{} 3x \plus{} ... \plus{} 8x \equal{} 432$ pages, but Charlie has only read $ 8x$ total pages. We solve for the first equation to get $ 36x \equal{} 432\implies x \equal{} 12$.

Next, we see that Charlie has read $ 8x \equal{} 8(12) \equal{} \boxed{96}$ pages.[/hide][/hide]

[hide="#29"][hide="Hint"]Consider the fact that the sum of two 3-digit numbers, COU + NTS, gives a 4-digit sum. We now know immediately the value of M.[/hide]
[hide="Solution"]Using the info in the hint, we know that M = 1. Because O + T gives the digit T, we know that O must be either 0, or 9 with a regrouped 1 from U + S. No digit can be 0, and U + S has a maximum value of 17, so O = 9.

Next, we see that U + S must be 14. This sum can be 9 + 5 or 8 + 6, but the 9 is taken. Thus U and S are 8 and 6, interchangeably.

The fact that a T is in the tens place, with 9 + 1 + U, tells us that a 1 is regrouped to the hundreds place.

Our only remaining digits are in the one-to-one correspondence $ {2, 3, 5, 7} \equal{} {C, N, T, A}$. We see that the only way to achieve regrouping into the thousands is to use 7 as C or N. As 7 is the maximum value left, it must be the greater of C or N. The answer is $ \boxed{7}$.[/hide][/hide]

Thanks a lot guys :) :blush:
Now on to solving a 1000 more mathcount problems  :rotfl:

# 28.. im new at the hide stuff[hide]Since she adds all of the page numbers to get to 432,
you have to add x+2x+3x+4x+5x+6x+7x+8x, since it was at the end of the 8th day and
he read the same amount of pages each day. That equation = 36x which is =to 432.
So 36x=432 and x=12. But you have to multiply it by 8 because it was 8 days so the answer
would be 96 pages.[/hide]
#30[hide]First I would convert the number of times it turns into minutes. So (100/3)/60 which
is 5/9. Then the 2nd gear would be 45/60 or 3/4. Then you find the LCM of the denominators of 5/9 and 3/4 which ends up to be 36. So the answer would be after 36 turns.[/hide]


Given $ A_1A_2A_3A_4$ an arbitrary quadrilateral. $ H_1, H_2, H_3, H_4$ are the orthocenters of $ \triangle A_2A_3A_4, \triangle A_1A_3A_4, \triangle A_1A_2A_4, \triangle A_1A_2A_3$ respectively. Prove that $ S_{A_1A_2A_3A_4}\equal{}S_{H_1H_2H_3H_4}$
[img]http://img152.imageshack.us/img152/6118/sketch1wv0.png[/img]

http://www.mathlinks.ro/viewtopic.php?t=105181 yields $ \left[A_1A_2A_3\right]\equal{}\left[H_1H_2H_3\right]$ and $ \left[A_3A_4A_1\right]\equal{}\left[H_3H_4H_1\right]$, so that

$ \left[A_1A_2A_3A_4\right]\equal{}\left[A_1A_2A_3\right]\plus{}\left[A_3A_4A_1\right]\equal{}\left[H_1H_2H_3\right]\plus{}\left[H_3H_4H_1\right]\equal{}\left[H_1H_2H_3H_4\right]$,

qed.

I think this problem with exactly the same proof has been posted before, but it was easier to repeat it than to search for it...

  darij


LaTeX
pigeonhole principle
ceiling function
irrational number
absolute value
real analysis
real analysis unsolved

prove:

$ a$ is an irrational number $ \Leftrightarrow$ $ for$ $ all$ $ \epsilon$ $ >$ $ 0$ $ there$ $ exist$ $ integers$ $ x,$ $ y$ s. t. $ 0 < ax \minus{} y < \epsilon$

with absolute value for the last equation of course.  couldn't quite figure out how to use latex with absolute value

It's not quite a limit.  What the condition means is that it is possible to approximate $ a$ with rational numbers $ \frac {y}{x}$ such that $ \left| a \minus{} \frac {y}{x} \right| < \frac {\epsilon}{x}$; in other words, the degree of the approximation can be made arbitrarily stronger than would be expected based on the denominator of $ x$ alone.

It is in fact the case that $ a$ is irrational if and only if we can find an infinite sequence $ x_n, y_n$ of strictly increasing positive integers such that $ \left| a \minus{} \frac {y_n}{x_n} \right| < \frac {1}{x_n^2}$, which settles the matter.  There is a more elementary proof of the proposition as given (think Pigeonhole), but I always prefer generality; see [url=http://en.wikipedia.org/wiki/Continued_fraction#Theorem_5]Theorem 5 here[/url].

Edit:  On second thought, the elementary approach is instructive.  One way to interpret the statement is that the sequence of fractional parts $ a_n \equal{} \{ an \}$ contains points arbitrarily close to $ 0$.  Clearly this is not the case if $ a$ is rational.  Otherwise, for any $ \epsilon$ consider the first $ \left\lceil \frac{1}{\epsilon} \right\rceil$ terms of the sequence.  Since $ a$ is irrational, none of them can be equal to each other, so two of them must be within at least $ \epsilon$ of each other, say $ \{ ai \}$ and $ \{ aj \}$.  Then $ \{ ai \minus{} aj \} <  \epsilon$.  It is only slightly harder to show that in fact $ a_n$ is dense in $ [0, 1]$.

eh, I don't think it was supposed to be very hard.  I have done the problem before.  I wrote it on the diary, and last time I opened it, I couldn't figure out how to do it again!

$ \text{\LaTeX}$ note: don't put text in math mode. It looks bad, and it's an extra burden on the system here. If you need to use a word or two (such as in an array of some sort), there is \text{} for a better look.


if any of u has a nice problem in trigonometry u can send it it is just like a marathon

if any of u know how to find the area of a quadrilateral when only if its sides are given but not the diagonal and no side is parallel to other.[/b]

What you say it's possible, in fact, there is a formula: [b]Bretschneider's formula [/b](http://en.wikipedia.org/wiki/Bretschneider%27s_formula)
What it says is that the area for a quadrilateral is equal to:
$ \sqrt{(S-a)\cdot(S-b)\cdot(S-c)\cdot(S-d)-a\cdot b\cdot c\cdot d \cdot cos^{2}(\frac{A+C}{2})}$
Where a, b, c, d are the sides of a quadrilateral, S is the semiperimeter $ (\frac{a+b+c+d}{2})$ and A and C are oppossite angles of the quadrilateral.
When the quadrilateral is cyclic, $ A + C = 180$, so $ \frac{A+C}{2} = 90$ and cos 90 = 0, it takes the form of [b]Brahmagupta's formula [/b](http://en.wikipedia.org/wiki/Brahmagupta%27s_formula):
$ \sqrt{(S-a)\cdot(S-b)\cdot(S-c)\cdot(S-d)}$


logarithms
algorithm
combinatorics unsolved
combinatorics

Let $ n$ be a positive integer, and $ S$ an $ n$-element set of positive integers. Show that there is a subset $ A$ of $ S$ such that $ A$ has at least $ \log_3n$ elements and for every pair $ a,b$ of distinct elements in $ A$, $ a\plus{}b$ is not in $ S$.

actually you can improve $ \log_{3}n$ to $ 2\log_{3}n\minus{}1$ as [url=http://www.springerlink.com/content/m02g0814g70131t8/]this[/url] paper shows.

the proof goes along the lines of choosing a sequence of elements of our set by a greedy algorithm:
choose $ s_0$ to be the greatest element of the set and $ s_i$ to be the largest element so that the conditions are still valid (i.e. that $ s_i\plus{}s_j\not \in A$ for all $ j<i$) and prove this sequence contains $ >\log_{3}n$ elements.

[quote="Albanian Eagle"]

the proof goes along the lines of choosing a sequence of elements of our set by a greedy algorithm:
choose $ s_0$ to be the greatest element of the set and $ s_i$ to be the largest element so that the conditions are still valid (i.e. that $ s_i \plus{} s_j\not \in A$ for all $ j < i$) and prove this sequence contains $ > \log_{3}n$ elements.[/quote]

Can you show more?

[quote="randomgraph"]There are even better bounds, superlogarithmic, in [url=http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1121448866]this paper[/url]by Sudakov, Szemerédi, and Vu. I haven't read either paper but my guess is that they're far from easy.[/quote]

that's sort of why i didn't bring up that result. I've read that paper and though the proof is only ~30 pages long it includes most of the results found so far related to the problem, including some very ugly (heavy on computation) theorems of Freiman.

Umm.. can you show a bit more on my "worse" bound?  :)

Come on... you guys are posting links to papers which are unaccessible anyway..

Sorry, but I still want to see the solution  :(


group theory
abstract algebra
superior algebra
superior algebra theorems

Let $f:G\to G'$ be a surjective morphism.
Then $\ker f$ is a normal subgroup of $G$ and there exists an unique izomorphism $\overline f:G/\ker f\to G'$ s.t. $\overline {f}(p(x))=f(x),\forall x\in G$ where $p$ is the canonic projection morphism ($p:G\to G/\ker f,p(x)=xH$).

Actually, this is pretty straightforward: showing that $\ker f$ is a normal subgroup of $G$ consists of observing that if $f(x)=e_{G'}$, then $f(gxg^{-1})=f(g)f(x)f(g)^{-1}=f(g)f(g)^{-1}=e_{G'}$. As for the other part, you have defined $\bar f$, all we have to do is show that it's indeed an isomorphism, and show that it's unique.

We first show that it's well defined: $p(x)=p(y)\Rightarrow y^{-1}x\in \ker f\Rightarrow f(x)=f(y)$. It's clearly a morphism, and it's also pretty obvious that it's surjective, because $f$ is surjective. For the injectivity: $f(x)=f(y)\Rightarrow y^{-1}x\in\ker f\Rightarrow p(x)=p(y)$. This proves that $\bar f$ is indeed an isomorphism.

The unicity: take $\hat f$ satisfying those things. We have $\hat f(p(x))=\bar f(p(x))$, and since $p$ is surjective, it means that $\bar f=\hat f$ on the entire $G/\ker f$.


calculus
integration
logarithms
calculus computations

Help integral Dummy to learn Thanks

1. $ \int^e_1 (10/{x}) dx$

2. $ \int^3_0 (3x \plus{} 2)(4x \minus{} 1) dx$

hey man a lot of people never make it to calculus so don't even stress.
$ \boxed{1}$ so we've got
$ F(1,e)\equal{}10\int^e_1 \frac{1}{x}dx$
You should probably review logarithmic integration/differentiation, but recall that $ \frac{\partial}{\partial x}[\ln{x}]\equal{}\frac{1}{x}$ to find that $ F(e,1)\equal{}10(\ln{e}\minus{}\ln{1})\equal{}10$.

As for numero dos you might want to foil that.
 :)

2) Follow what igiul suggested:
By FOILing, we arrive at
$ \int_{0}^{3} 12x^{2}\plus{}5x\minus{}2\ dx\equal{}\left(4x^{3}\plus{}\frac{5}{2}x^{2}\minus{}2x\right)_{0}^{3}\equal{}79.5$
I hope I did these calculations right.

When I can't do basic arithmetic, it's a wonder I can do more advanced stuff.  :oops:

I think it should be 124.5.  Sorry for that error.

try using '\bigg|' for the integral evaluation, e.g
\[ \int^1_0 xdx \equal{} \frac {x^2}{2}\bigg|^1_0 \equal{} \frac 12
\]


algebra
polynomial
quadratics
Vieta
sum of roots
product of roots

So the sum of roots is $-b/a$ and the product of roots is $c/a$ do these $b$ and $c$ terms also include the [b]sign[/b] (pos/neg) of the number in the polynomial. For some reason, I am very confused. Thanks!

[quote="Drunken_Math"]$y=ax^{2}+bx+c$

Yes I do believe so. (in case you didn't know since you put polynomial, this only holds for quadratics)[/quote]

Right but in polynomials, it is the (- second coefficient/first) as sum of the roots, and (last coef/first) for the product ???

Yes-vietas theorm, theres more than that do.  Those are only the sum and product of the roots.  You can also find alternating sums and all that..

Look, you'll understand it more if you derive it yourself. If you let $r_{1}$ and $r_{2}$ be the roots of the quadratic, you have

$a(x-r_{1})(x-r_{2})= ax^{2}-a(r_{1}+r_{2})x+ar_{1}r_{2}$

Comparing this to $ax^{2}+bx+c$ (you always compare to this), you notice that $r_{1}+r_{2}$ is the sum of the roots, and so you have $-a(r_{1}+r_{2})=b$ and then $r_{1}+r_{2}=-\frac{b}{a}$. And something similar to the products.

Edit: Oh, I noticed you mentioned polynomials in general. You should look up Vieta's for more information. But you should just examine $a(x-r_{1})(x-r_{2})(x-r_{3})\cdots(x-r_{n})$. Note that if you expand it, the sum of the roots will always be negative and the product of the roots will differ (positive if $n$ is even and negative if $n$ is odd).

yes with the Eternica method you can generalize
you can find the sum and the product of roots of any polynom

plz clarify your question i dont indersand it well

It's

[hide]
$q$

A cubic polynomial would have three roots, so you have
$a(x-r_{1})(x-r_{2})(x-r_{3})$

Expanding it, you should get
$ax^{3}-a(r_{1}+r_{2}+r_{3})x^{2}+a(r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3})x-ar_{1}r_{2}r_{3}$

Comparing to
$ax^{3}+bx^{2}+cx+d$

So you have $-ar_{1}r_{2}r_{3}=d$ and then the product of the roots would be $-\frac{d}{a}$.

Your expression is $x^{3}-px^{2}+11x-q$. So you have $-q=d$. Substitute that into the product of the roots, you have  $-\frac{-q}{a}$ and because $a=1$, you have $q$.
[/hide]

[quote="Walk Around The River"][quote="jeez123"]Thanks for all your help so far!

One more question though:
Is the product of the roots of the following $q$ or $-q$, I said $-q$ but I was wrong...

$x^{3}-px^{2}+11x-q = 0$

and why?[/quote]

Isnt is 11? c/a?[/quote]

$\frac{c}{a}$ is ONLY for quadratic polynomials. In general, the product of the roots will be the constant term divided by the coefficent of the term with the highest degree... and maybe a $-$ sign if you feel like it.

But really. Look up Vieta's and try expanding them out for yourself. I had fun when I learned this.

Or read AoPS II Chapter 6. It's really very interesting, and you'll learn it better if you actually see where it's coming from.
In general, the product of the roots of polynomial with degree $n$ is ${(-1)^{n}}{a_{0}/{a_{n}}}$.

thanks for the formula it's very interesting

[quote="davidyko"]In general, the product of the roots of polynomial with degree $n$ is ${(-1)^{n}}{a_{0}/{a_{n}}}$.[/quote]

Yep, I went through it and that part was actually the one that I was really lost, but now with this problem and you guys help... it all fits together  :D .


Tangents of $ (O_1),(O_2)$ cut at $ K$.
Prove that $ K\in (O)$
See:

It's well-known (IMO 2008 problem 6).
http://www.mathlinks.ro/viewtopic.php?p=1191671#1191671