inequalities
calculus
derivative
function
real analysis
real analysis unsolved

First observe that,

$x_{k+1}=\frac{1}{3}\left(x_{k}+x_{k}+\frac{a}{x_{k}^{2}}\right)\geq \sqrt[3]{a}$.

On the other hand,

$x_{k+1}-x_{k}=\frac{a-x_{k}^{3}}{3x_{k}^{2}}$, hence the sequence decreases and since it is bounded bellow it converges, the limit is $L=\sqrt[3]{a}$.

The second part of the problem, namely  the properties of $z_{k}$, follow simply from the properties of $x_{k}$.

[quote="didilica"]

The second part of the problem, namely  the properties of $z_{k}$, follow [b]simply[/b] from the properties of $x_{k}$.[/quote]

Can you post your solution?

[quote="didilica"]$z_{n}=x_{n}-L>0$,  $x_{n}\geq L$ since the sequence decreases.

$z_{k+1}-L<\frac{2}{3}\left(x_{k}-L\right)$. It follows that

$x_{k+1}-\frac{2}{3}x_{k}=\frac{a}{3x_{k}^{2}}<\frac{L}{3}$.
For the last inequality I think that it should be

$z_{k+1}<\sqrt[3]{a}z_{k}$, which is equivalent to

$\frac{2}{3}x_{k}+\frac{1}{3}\frac{L^{3}}{x_{k}^{2}}-Lx_{k}<L-L^{2}$
and now study the variation of $x\rightarrow (2/3-L)x+\frac{L^{3}}{3x^{2}}$ on the interval $[L, \infty)$.
Study the derivative of the function and I think that you should make a discussion according to $2-3L$ is less than or equal to 0 and greater than 0.[/quote]

is this symply?  :P

Comment: this is Newton's method applied to finding a root of the equation $x^{3}-a=0.$ As is typical of Newton's method, the convergence of the sequence is "quadratic." That is, once $x_{k}$ is within a certain neighborhood of $L,$ we have for some positive constant $C$ that $|x_{k+1}-L|\le C|x_{k}-L|^{2}.$


geometry
geometric transformation
induction

Consider an infinite strip of unit squares. The squares are numbered "1", "2", "3", ... A pawn starts on one of the squares and it can move according to the following rules:
(1) from the square numbered "$n$" to the square numbered "$2n$", and vice versa;
(2) from the square numbered "$n$" to the square numbered "$3n + 1$", and vice versa.
Show that the pawn can reach the square numbered "$1$" in a finite number of moves.

I am not so happy about my translation of this problem. Is it clear enough, and can anyone improve the way I stated the problem?

How I would word it:
Consider an infinite graph $G$ with vertices $v_1,v_2,v_3,...$. For every $n$, connect $v_n,v_{2n}$ and  $v_n,v_{3n+1}$. Show that $G$ is connected.

My solution:

Consider the remainder modulo 3.
If 0 go $3n \rightarrow 9n+1 \rightarrow 36n+4 \rightarrow 12n+1 \rightarrow 4n \rightarrow 2n \rightarrow n$
If 1 go $3n+1 \rightarrow n$
If 2 go $3n+2 \rightarrow 6n+4 \rightarrow 2n+1$

Every time you decrease, so in the end, you must get on 1 :)

[quote="Peter VDD"]How I would word it:
Consider an infinite graph $G$ with vertices $v_1,v_2,v_3,...$. For every $n$, connect $v_n,v_{2n}$ and  $v_n,v_{3n+1}$. Show that $G$ is connected.[/quote]

Obviously that's what the problem is about but I'm sure you know that it's not an acceptable translation for this problem! You even changed the problem, and lots of people don't know what a connected graph is :huh:

Connected graph = graph in which there is a path between each couple of vertices? Anyway I just gave you an idea for the wording. You can alter it for high school like this:

Number the squares of an infinite chessboard (1,2,3,...) Call 2 squares "connected" if their indices can be written as a pair $(n,2n)$ or a pair $(n,3n+1)$ for a postive integer $n$. A pawn may move from $a$ to $b$ iff those squares are connected. Prove that we can reach square 1 through a finite number of moves.

I didn't say that [b]I[/b] don't know what a connected graph is!

That's better, but when translating you should always try to follow the structure of the original statement, and you're still not doing that. Never mind. This discussion doesn't make sense.

If $n$ is even, we go from $n$ to $n/2$.  If $n$ is odd, we go from $n$ to $3n+1$.  That this will always eventually lead to 1 is left as an exercise to the interested reader  :lol:

(For those who haven't seen it before, whether the above procedure always yields 1 or not is a famous open problem called the Collatz conjecture.  It's been verified for all starting values less than about 10^53 so far).[/hide]

[quote="kevinatcausa"](For those who haven't seen it before, whether the above procedure always yields 1 or not is a famous open problem called the Collatz conjecture.  It's been verified for all starting values less than about 10^53 so far).[/hide][/quote]Do you mean my proof is wrong? Or what do you mean by this?

I know about the Collatz problem, but this problem is fine. The clue is that you can go "backwards" as well: from $3n + 1$ to $n$ and from $n$ to $2n$.

Peter VDD, no that was just a joke about the Collatz conjecture, I guess. Don't worry about it, your proof looks fine to me.

prove by induction from $n$ we can obtain a number less than $n$
if $n\equiv 1(mod3)$
$n\rightarrow\frac{n-1}{3}$
if$n\equiv 2(mod3)$
$n\rightarrow\frac{2n-1}{3}$
if$3|n$
$n\rightarrow 3n+1\rightarrow 2(3n+1)\rightarrow 4(3n+1)\rightarrow 4n+1\rightarrow\frac{4n}{3}\rightarrow\frac{2n}{3}$
QED


inequalities function logarithms trigonometry

These inequalities are copied from p.6 of Kiran Kedlaya's MOP notes on inequalities. #3 has been proved by tetrahedr0n using the concavity of the function [tex]f(x) = \ln\sin(x) [/tex]  and Jensen's Inequality. (See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12640]here[/url]) although an elementary proof is possible.

Let $A,B,C$ be the angles of a triangle. Prove that

1. $\sin A + \sin B + \sin C \leq 3\sqrt{3}/2$;

2. $\cos A + \cos B + \cos C \leq 3/2$;

3. $\sin A/2 \sin B/2 \sin C/2 \leq 1/8$;

4. $\cot A + \cot B + \cot C \geq \sqrt{3}$.

I believe that all these problems have been previously posed and solved on this site.  But they still make fun problems.

Solution(s): Use Jensen (with a few difficulties here and there)

Alright, show that #4

[tex]\cot A + \cot B + \cot C \geq \sqrt{3}[/tex]

is equivalent to:

[tex]a^2+b^2+c^2\geq4\sqrt{3}S[/tex]

(The 2nd inequality was posted by Bottema, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12639]here[/url])


Quite easy, but good excersize.

Let $ x_1, x_2, \cdots , x_n > 0$ and $ m \in \mathbb{N}$

$ \sqrt[m]{\frac{\left(x_1\right)^m \plus{} \left(x_2\right)^m \plus{} \cdots \plus{} \left(x_n\right)^m}{n}} \ge \frac{x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n}{n}$

[quote="Mathias_DK"]Quite easy, but good excersize.

Let $ x_1, x_2, \cdots , x_n > 0$ and $ m \in \mathbb{N}$

$ \sqrt [m]{\frac {\left(x_1\right)^m \plus{} \left(x_2\right)^m \plus{} \cdots \plus{} \left(x_n\right)^m}{n}} \ge \frac {x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n}{n}$[/quote]
It 'll be kill by Holder or AM-GM
With Holder
$ [ \left(x_1\right)^m \plus{} \left(x_2\right)^m \plus{} ... \plus{} \left(x_n\right)^m]n^{m \minus{} 1} \ge\ (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^m$

[quote="Mathias_DK"][quote="onlylove_math"]
It 'll be kill by Holder or AM-GM
With Holder
$ [\frac {\left(x_1\right)^m + \left(x_2\right)^m + ... + \left(x_n\right)^m]n^}{m - 1} \ge\ (x_1 + x_2 + ... + x_n)^m$[/quote]
-.-[/quote]
Sorry Mathias_DK ; I have edited.Thank!

it follows immediately from jensen's inequality, since $ x^t$ is convex for all $ t\geq 1$.

this problem is too easy for this subforum in general, and it doesn't really belong in the proposed section since it's very, very classical.

It can also be killed using $ Chebycheve.$ :wink:


integration
logarithms
number theory unsolved
number theory

Let $ a,k,N$ be positive integers.

Let $ P(N)$ be the number of $ d \le N$ such that $ a,a \plus{} d,a \plus{} 2d,...,a \plus{} k*d$ are all primes.

By the PRIME NUMBER THEOREM , can we say

$ P(N) \sim \int_1^N \frac {dt}{\prod_{i \equal{} 0}^k \log (a \plus{} i*t) }\ \ ?$

Not by the prime number theorem as it is known today. It would be great to know some generalization of this kind, of course, but so far the best result in this direction is the Green-Tao theorem, which merely asserts that the set of primes contains arbitrarily long arithmetic progressions.

Also, you should be careful with how you state this problem. In order to have it meaningful, you have to assume that $ a$ is a prime not dividing $ k!$ at the very least and introduce a correcting factor that takes into account that certain values of $ d$ have no chance to satisfy the condition for trivial divisibility reasons.


geometry
incenter
inradius
circumcircle
trigonometry
power of a point
radical axis

A triangle $ \bigtriangleup ABC$ is given with incenter $ I,$ orthocenter $ H$ and let $ M$ be, the midpoint of its side-segment $ BC.$ Prove that $ (HM)^{2} \minus{} (HI)^{2} \equal{} \frac {a^{2}}{4} \minus{} 2r^{2},$ where $ a \equal{} BC$ and $ r,$ is the inradius of $ \bigtriangleup ABC.$

Kostas Vittas.

We use the well-known formulae:

$ IH^2 \equal{} 4R^2 \minus{} \frac {_1}{^2}(a^2 \plus{} b^2 \plus{} c^2) \plus{} 2r^2 \ \ (1)$

$ OH^2 \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2) \ \ (2)$

Power of $ H$ WRT the circle with diameter $\overline{ BC}$ is $ \frac {_1}{^4}a^2 \minus{} HM^2 \equal{} k^2$ and $ k^2$ is the power of the inversion taking the 9-point circle $(N)$ of $\triangle ABC$ into its circumcirle $(O).$ From inversion properties, we get:

$ \frac {k^2}{OH^2 \minus{} R^2} \equal{} \frac {R}{2R}  \Longrightarrow  k^2 \equal{} \frac {_1}{^2}(R^2 \minus{} OH^2)$

$\Longrightarrow k^2 \equal{}  \frac {_1}{^2}(a^2 \plus{} b^2 \plus{} c^2) \minus{} 4R^2 \equal{} \frac {_1}{^4}a^2 \minus{} HM^2 \ \ (3)$

Combining the expressions $(1)$ and $(3)$ gives $ HM^2 \minus{} IH^2 \equal{} \frac {_1}{^4}a^2 \minus{} 2r^2$

Thank you dear [b][size=100]Luis[/size][/b], for your interest and nice solution.

I like better your approach with the argument of the inversion power that takes the circucircle of $ \bigtriangleup ABC,$ to its nine-point circle.

Since I am not good in metrical solutions, I had only an ugly approach, used some ready formulas from the bibliography I have in mind.

(1)	- $ (HI)^{2} \equal{} 2r^{2} \minus{} 4R^{2}\cos A\cos B\cos C$

(2)	- $ (HM)^{2} \equal{} \frac {2[(HB)^{2} \plus{} (HC)^{2}] \minus{} a^{2}}{4}$

(3)	- $ HB \equal{} 2R\cos B,\ HC \equal{} 2R\cos C$

(4)	- $ \cos^{2}A \plus{} \cos^{2}B \plus{} \cos^{2}C \equal{} 1 \minus{} 2\cos A\cos B\cos C$

From the wanted equality, $ (HM)^{2} \minus{} (HI)^{2} \equal{} \frac {a^{2}}{4} \minus{} 2r^{2}$, combining the above formulas, we obtain by calculations the equality $ a \equal{} 2R\sin A,$ which is true.

$ \bullet$ As an interesting remark, we can say that the radical axis of the circles $ (M),$ with diameter $ BC$ and $ (I'),$ centered at incenter $ I,$ with radius $ r' \equal{} \sqrt {2} r,$ passes through the orthocenter $ H,$ of the given triangle $ \bigtriangleup ABC.$

Thank you again, Kostas Vittas.


how to prove that the angle bisectors of the angles of a quadrilateral ABCD with AB + CD = AD + BC

ARE CONCURRENT .
 :maybe: i worked out a contradiction but there's a shorter method too. :!:

We know that $ ABCD$ is a circumscribed quadrilateral therefore  the angle bisectors of the angles concur at the center of the incircle of $ ABCD$.

Let's call $ I$ the intersection of the bisectors of the angles $ \angle BAD$ and $ \angle ABC$, and call $ M, N$ and $ P$ on $ AB, BC$ and $ AD$. Obviously, $ AM\equal{}AP, BM \equal{} BN$ and, from the given relation we get $ PD \plus{} NC \equal{} CD$, that is, there exists the point $ Q \in [CD] | PD \equal{} DQ$ and $ CQ \equal{} CN$. From the isosceles triangles $ \triangle AMP, \triangle BNM, \triangle CQN$ and $ \triangle DPQ$ we shall notice that $ \measuredangle MPQ \plus{} \measuredangle MNQ \equal{} 180^\circ$, i.e. $ MNQP$ is cyclic.
Next, we have to show that there is no other point on $ CD$ and circle $ (MNP)$.
Taking $ Q'$ on the circle such that $ DQ'$ be tangent to the circle, we get $ PD \equal{} PQ \equal{} PQ'$, that is, there are three points on the circle $ (MNP)$ at the same distance from $ D$, impossible, hence $ Q \equiv Q'$.

Best regards,
sunken rock


There seems to be a problem being able to see the forum and posts with internet explorer 8. I don't really know anything about it besides that everything seems to be shifted when you view a thread.

It's definitely a bug - it should show an "install Firefox" link.

[size=59]SCNR[/size]

  darij

[quote="darij grinberg"]It's definitely a bug - it should show an "install Firefox" link.

[size=59]SCNR[/size]
[/quote]

HAHAHA niice one.

Anyway, because IE8 isn't out yet (other than betas I'm assuming), the admins wouldn't have updated the site so that it looks the way it should.

So, alternatively, instead of waiting for the admins, you could [url=http://www.mozilla.com/en-US/firefox/]click here to update IE8 with Microsoft's latest content[/url].

With all due respect for the awesomeness of Firefox, I found [url=http://www.stanford.edu/~pgbovine/geek-behaviors.htm]this article on geek behaviors[/url] to be quite interesting. Specifically, the last: evangelizing your favorite technology by degrading (whether justly or not) what someone else thinks is [i]their[/i] favorite technology or general technology of choice. It's somewhat amusing and a bit off-turning to see Firefox being force-fed someone.

{But I do believe, once upon a time, I tried to get jhredsox to get Firefox too. Firefox is decently good and easier for me to use than IE :).}

haha yes, actually, I usually use firefox (thanks to you haha) but decided to give ie8 a try...it's actually pretty nice...this seems to be the only problem I've encountered

Firefox is great because of the thousands of addons, I don't consider it the ultimate browser. Without the addons, I wouldn't even consider it better than any other popular browser; it certainly has a number of annoyances IE doesn't have. But with [url=https://addons.mozilla.org/de/firefox/addon/722]NoScript[/url] (for safety and against annoying scripts that take longer to load than the rest of the page), [url=https://addons.mozilla.org/de/firefox/addon/953]RefControl[/url] (against people tracking where you come to their site from), [url=https://addons.mozilla.org/de/firefox/addon/684]FireFTP[/url] (self-explanatory, I think), [url=https://addons.mozilla.org/de/firefox/addon/201]DownThemAll[/url] (ever wanted to download all files starting with 001.pdf until 593.pdf? this is the solution; even if you know wget this will make some things easier), [url=https://addons.mozilla.org/de/firefox/addon/966]TamperData[/url] (for, uhm, special needs...) and many more, I have no idea how I could work on any other browser.

But this is pretty much offtopic; Solafidefarms, did you notice the "[url=http://en.wiktionary.org/wiki/SCNR]SCNR[/url]" in my post? ;)

  darij

Oh, I didn't know what SCNR meant. Thanks, and apologies for going so off-topic. :) [I'm a FlashBlock user; my chief pleasure in Firefox though is that I have my tabbar at the top, and at the bottom I have the Edit menu, Bookmarks menu, address bar, and search bar - and there's nothing else on there! 1440x770 room for the page I'm looking at, instead of fluff! :)]

[quote="solafidefarms"]Oh, I didn't know what SCNR meant. Thanks, and apologies for going so off-topic. :)[/quote]

The off-topic began with my post :D

[quote="solafidefarms"][I'm a FlashBlock user; my chief pleasure in Firefox though is that I have my tabbar at the top, and at the bottom I have the Edit menu, Bookmarks menu, address bar, and search bar - and there's nothing else on there! 1440x770 room for the page I'm looking at, instead of fluff! :)][/quote]

FlashBlock is new to me, but judging from its description, everything it does NoScript does too (with default settings).
If you are trying to waste as little space on the display as possibly, I guess you are familiar with [url=http://lifehacker.com/392452/shrink-firefox-3s-supersized-back-button]this[/url]? (Given that going back just requires pressing backspace while focussed on the website, I don't see any point to a back button at all...)

  darij

I know we have some problems with MSIE8, but it's in BETA stages right now, and things are still changing. I'll enhance the support for MSIE8 soon though.

Awesome to know how much the people at this site care, thank you


My coach wanted me 2 make a "year" of MC comps 4 her(as thet're more challenging than b4, and she thinks I'm good at that stuff-not so much...) and I can put them on here if anyone wants-they're not great, but they're not AWFUL. reply if at all interested. they're a little harder than normal 1s, but not much.

Yeah.

The message is too small. Please make the message longer before submitting.


Do all posts have to be a problem to figure out? Cause this one isnt.

I noticed that 243, which is 3^5, is 100 less than 343, which is 7^3. 100 is also 10^2. Cool! Theyre all a^b! Heres what it looks like:

10^2 + 3^5 = 7^3

split them into just bases and exponents, and change the operators, and:

2 = 5 - 3
10 - 3 = 7

Actually, this could be a problem (maybe too advanced for this forum).

Is there any other equation where a,b,c ; d,e,f ; and a^d, b^e, c^f can be arranged into correct equations?

I don't know, Fermats last theorem (proved by Andrew Wiles) stated that for the any value of n>2 the equation a^n+b^n=c^n has no solutions.  It would be interesting if someone proved that a^x+b^y=c^z has an  :inf: solutions, or even that it dosn't.

It wouldn't be very interesting actually. So here's a problem that appeared on a math contest once:

Find a solution to [tex]x^3+y^4=z^5[/tex] or prove that there aren't any solutions. (x, y, and z are positive integers.)

[quote="Hexaditidom"]Do all posts have to be a problem to figure out? Cause this one isnt. [/quote]

Welcome to the forum.

As for your comment, no, they don't have to be a problem. Feel free to make as many interesting comments or conjectures as you wish. I can speak at least for this forum.

[quote]Fermats last theorem (proved by Andrew Wiles) stated that for the any value of n>2 the equation a^n+b^n=c^n has no solutions.[/quote]

You mean no trivial solutions. There are solutions like (a,0,a) and (0,a,a).

Can anyone explan this Frey curve used in Fermat's last theorem?

No. Why?

We don't really know how hard the original open problem is. Simon's problem probably belongs here (though I haven't gotten the solution yet :)). As for the proof of FMT, lets just wait until someone does answer masamune2387's question. Is the Frey curve a modular elliptic curve?

Ack. I didn't get to Frey curves yet in my study of elliptic curves. Perhaps I'll go to the bookstore and read some of the book on elliptic curves that I don't understand tomorrow.

The way I understand it Frey's curve is a hypothetical elliptic equation derived from Fermat's Last theorem so that if Fermant was wrong (aka there WAS a solution to A^n+B^n=C^n where n>2, therefore this modular equation existed) then the Taniyama-Shimura conjecture, (that all elliptic equations are modular) would be FALSE.  So, if someone could prove the Taniyama-Shimura conjecture Fermat would be vindicated.
Edit: this is the nontechnical synopsis of course

[quote="NAS"][quote="tetrahedr0n"]You mean no trivial solutions. There are solutions like (a,0,a) and (0,a,a).[/quote]
Yeah, let me clarify any 3 distinct positive integers have no solution when n>2.  I did a report on Fermat in math class so I know a bit about it.[/quote]

..when n is a positive integer > 2.

;)

There's an interesting conjecture called the Beal Conjecture that has to do with the equation a^x+b^y=c^z. Im not sure about the details. I suggest a google search. I do know, though, that FMT is an immediate consequence of this conjecture if it happened to be true. I also know that there is a prize for a proof or counterexample :D

Ok here it is: [quote]BEAL'S CONJECTURE: If Ax +By = Cz , where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.[/quote]
[url]http://www.math.unt.edu/~mauldin/beal.html[/url]

[quote="TripleM"]
..when n is a positive integer > 2.

;)[/quote]

Is there any NONpositive integer that is greater than two?

;) ;)

integer was the key word. that was missed out before.


probability
geometry
3D geometry
tetrahedron

Find a combinatorial proof of the following:

$\sum_{k=1}^n k^2 = \frac{1}{4} \binom{2n+2}{3}$

Unfortunately, the end requires just a tiny bit of algebra - but the argument is combinatorially motivated:

[hide]Take the set $\{x,y,z\}$ such that $z>x$ and $z>y$. In how many ways can we pick $x,y,z$ from the set $\{1,2,...,n\}$?

First, if $z=1$ we have $0$ ways. If $z=2$ we have $1^2$ ways. If $z=j$ we have $j^2$ ways. Total, we have $1^2+2^2+...+n^2 = \sum_{k=1}^n k^2$ ways to do it.

Or, if $x=y$ we pick two numbers and order them, in $\binom{n+1}{2}$ ways. If $x>y$ or $x<y$, we pick three numbers and order them, so total $2\binom{n+1}{3}$ ways. So total we have $\binom{n+1}{2} + 2\binom{n+1}{3}$ ways.

Thus,

$\sum_{k=1}^n k^2 = \binom{n+1}{2} + 2\binom{n+1}{3} = \frac{2n^3+3n^2+n}{6} = \frac{1}{4}\binom{2n+2}{3}$ and we are done. [/hide]

Now, as a challenge, find an expression and give a proof for $\sum_{k=1}^n k^j$

Er, I don't know what constitutes a combinatorial proof, but I have a proof.

[hide]
A formula for the sum of the first n squares (to rewrite the left side) is $\frac{n(n+1)(2n+1)}{6}$
We write out the choose to get:
$\frac{n(n+1)(2n+1)}{6}=\frac 1 4 \cdot \frac{(2n+2)!}{(2n-1)!3!}$
$4n(n+1)(2n+1)=\frac{(2n+2)!}{(2n-1)!}$
$4n(n+1)(2n+1)=\frac{(2n+2)(2n+1)(2n)(2n-1)!}{(2n-1)!}$
$2(n+1)=2n+2$
[/hide]

Right - but you've actually skipped the hard part of the problem - to prove that fomula for the sum of the first $n$ squares -  that's really the problem, not the bit of algebra at the end to get to the binomial coefficient.

Haha I could use one of them flash thingeys on the side of the page to prove it  :D

But I have seen it proved... maybe it was in one of the AoPS books.

Then check out my proof - it proves the formula in the combinatorical method (altough it only gives the non-factored polynomial.)

Just in case deej isn't familiar with the terminology: a combinatorial proof is one which shows a result by a counting (or probability) argument of some sort.  It's not an "official" term, but it makes for some very lovely proofs.  Usually, when you have two algebraic expressions to prove equal, you show that the two are just different ways of counting the same thing as Tetrahedron did in this question.


...but I can't seem to do the following question:
Show that for all integer polynomials p(x) with degree > 1, there exists an integer polynomial q(x) such that p(q(x)) is reducible in Z[x].

Jack


logarithms
quadratics
conics
parabola
special factorizations

Let  $ a\neq 1$ be the positive constant number and $ x,y$ be postive numbers such that $ \log_{\sqrt{a}}\ x\equal{}1\plus{}\log_a y$.
Find the maximum value of $ x\minus{}y$.

[hide="solution"]

The given identity is equivalent to $ \log_a{x^2} \equal{} 1 \plus{} log_a{y}$.  It's easy to rearrange it into $ x^2 \equal{} a*y$.  Then $ x\minus{}y \equal{} 1/a(ax \minus{} ay) \equal{} 1/a(ax \minus{} x^2) \equal{} x \minus{} x^2/a$.  Completing the square, $ x\minus{}x^2/a \equal{} \minus{}1/a(x\minus{}a/2) \plus{} a/4$.  So $ a/4$ is the maximum possible value that can be obtained.

[/hide]

What's the values of $ x,\ y$ for which $ x \minus{} y$ is maximized?

[quote="kunny"]Let  $ a\neq 1$ be the positive constant number and $ x,y$ be postive numbers such that $ \log_{\sqrt {a}}\ x \equal{} 1 \plus{} \log_a y$.
Find the maximum value of $ x \minus{} y$.[/quote]
I don't know if this is correct, it seems a little wierd to me. nvm, I typed most of this before dinner then I came back to finish after.
[hide]
Let's first convert to log base a. We have:
$ \log_a x^2 \equal{} 1 \plus{} \log_a y$
$ \log_a x^2 \minus{} \log_a y \equal{} 1$
$ \log_a {\frac {x^2}{y}} \equal{} 1$
$ \frac {x^2}{y} \equal{} a$
$ y \equal{} \frac {x^2}{a}$
We want to find the maximum of x-y, and we have obtained y, so we can write the expression:
$ x \minus{} \frac {x^2}{a}$
Notice that this is a quadratic, so we want to find the y value vertex of the parabola if we were to graph the equation (it is facing down).
The vertex is at a/2, and plugging a/2 into our equation we get that the maximum value of x-y is a/2-a^2/4a=a/2-a/4=a/4[/hide]


Hey im in highschool and I really hate to do mathematical proofs.  I was wondering and this may sound odd if anyone knows a proof that proves that mathematical proofs don't need to be done?


Let $ f: \mathbb R\longrightarrow \mathbb R$ be a continuous function such that
i) $ f(2010)=2009$, and
ii) $ f(x)\cdot f^{4}(x)=1$, $ \forall x\in \mathbb R$.

Find $ f(2008)$.

Notice that in this problem, we denote by $f^4(x)$ the composition of $f(x)$ with itself four times. That is, $f^4(x)=f(f(f(f(x))))$.

If we add, as mathVNpro suggested, the condition "$ f(x)$ continuous", then :

We have $ f(f(f(x)))\equal{}\frac{1}{x}$ $ \forall x\in A\equal{}f(\mathbb R)$ and so we also have $ f(\frac{1}{x})\equal{}\frac{1}{f(x)}$ $ \forall x\in A$

Notice that $ 0\notin A$ and $ 2009\in A$, so $ A\subseteq\mathbb R^{\plus{}*}$

So $ f(x)$ is an injective function from A$ \to$ A, so monotonous, so decreasing (else $ f(f(f(x)))$ would be increasing).

Let $ a\in A$
Suppose $ f(a)>\frac{1}{a}$ : Then $ \frac{1}{a}\equal{}f(f(f(a)))>f(f(\frac{1}{a}))$ (since $ f(f(x))$ is strictly increasing). So  $ f(a)>\frac{1}{a}>f(f(\frac{1}{a}))$  and so $ f(a)>f(f(\frac{1}{a}))$.

But $ f(a)>f(f(\frac{1}{a}))$ implies $ a<f(\frac{1}{a})\equal{}\frac{1}{f(a)}$ (since $ f(x)$ is strictly decreasing) and so $ f(a)<\frac{1}{a}$
Similarly $ f(a)<\frac{1}{a}$ $ \implies$ $ f(a)>\frac{1}{a}$

So $ f(x)\equal{}\frac{1}{x}$ $ \forall x\in A$

Since $ 2009\in A$, $ \frac{1}{2009}\equal{}f(f(f(2009)))\in A$, so $ [\frac{1}{2009},2009]\subseteq A$, and so $ 2008\in A$

So $ f(2008)\equal{}\frac{1}{2008}$

==========================
Notice that such a function exists :

$ f(x)\equal{}2009$ $ \forall x< \frac{1}{2009}$

$ f(x)\equal{}\frac{1}{x}$ $ \forall x\in[\frac{1}{2009},2009]$

$ f(x)\equal{}(2009\minus{}\frac{1}{2009})(x\minus{}2010)\plus{}2009$ $ \forall x\in(2009,2010)$

$ f(x)\equal{}2009$ $ \forall x\geq 2010$


Find the smallest positive integer $n$ such that $\sqrt{n+2}-\sqrt{n}<\frac1{2004}$.

umm after simplifying it out u get the smallest possible integer to be like 4016016 ...
is that right?

Yeah, you treat it as an equality, then add 1 to your answer.


geometry
perimeter
videos
area of a triangle
Heron\u0027s formula

Find the area of a triangle with sides $ 7, 9, 15$.

This is practice for Heron's formula!

As you say, you use Heron's formula here. Pretty staraightforward-

$ 7 \plus{} 9 \plus{} 15 \equal{} 31$
semi perimeter is $ 15.5$

now plug, with s, s-a, s-b, and s-c being muliplied under the radical-

$ sqrt(15.5*.5*6.5*8.5)$
$ sqrt(428.1875)$
area is approx. 20.6927

If three sides of a triangle a, b, c are given, we can use Heron's formula:
    Area = [i]sqrt[/i]((s)(s - a)(s - b)(s - c)) where s= (a+b+c)/2

Here,
a = 7
b = 9
c = 15

s = (7 + 9 + 15) / 2
s = 15.5

Area = [i]sqrt[/i](15.5 * (15.5 - 7) * (15.5 - 9) * (15.5 - 15))
       = [i]sqrt[/i](15.5 * 8.5 * 6.5 * .5)
       = [i]sqrt[/i](428.1875)
       = 20.6927

Answer: Area of a triangle with sides 7, 9, 15 is [b]20.6927[/b].

To learn more you may click the link below and go through video lesson on ‘Area of a triangle':
http://www.winpossible.com/lessons/Geometry_Area_of_a_Triangle_-_Areas_of_Polygons_and_Circles.html


What did you get?

I got 17, which was around my expectations, so I need a 14 for bronze-- seems feasible.

This one was the best round I had so far

55535

5 5 5 4 [b]5[/b]

24...again.   :roll:
Well, if this continues, I'll get gold. :lol:

congratulations on those great scores :coolspeak:

unfortunately, I learned about USAMTS about three days before round 3 was due, so my score was only 25500. I was actually surprised I got that much.

Yehaaa! I got my first 2! commendations and I got a score 2 more than I expected, because my good grader gave me 3 points instead of 1 for a completely wrong solution. And I almost got another commendation, but the second grader of that solution noticed that I counted some things twice and so they gave me a 3 instead.

35[b]5[/b]3[b]5[/b]

[b]5[/b] 5 5 3 5

Yay for my first commended!  Heh, I don't know how it slipped my mind to show that $n+k-2$ didn't beat $k$ players...

Surprisingly another 25, although I expected to lose points for either a silly mistake or for not considering the trivial $k=0$ case on #4.  Still no commendations, though.

A stupid dumb 52333, when I was expecting considerably better, meaning that I need 1 point for bronze, and 26 for silver :mad: .

25, with #5 commended.

I really doubt I'll manage to get a 100, but I think I have a very good chance of getting gold.

Wtf I had a 50 for the last two rounds and I submitted everything and I got an 11 this round??????????? :(

45555

They docked me a point on #1 because they couldn't find any diagram, but I included a diagram among the files I sent, and explained that it was the jpeg file in the email.  I now hope I can get that score changed, because it seems that was the only reason they didn't give me a 5...

5 [b]5[/b] 5 5 5

so i finally got one commended.  and my total is now 75.  i'm excited.

3+[b]5[/b]+[b]5[/b]+4+5 = 22

Now I just need to get 26 points on round 4 to get gold.  :mad:

5 5 5 5 5 = 25.

That makes me happy, but I didn't get any commendations.

I seem to get more commendations when I type up the solutions the night of the due date than when most of them are typed up a week in advance.

Total score: 72
Score needed for Silver: 3
Score needed for Gold: 24

5 5 [b]5 5[/b] 5

gold is within my reach.

eh 5 5 5 [b]5[/b] 5

not too bad, but i've had #2 commended round 1, #3 round 2, #4 round 3... i guess #5 round 4!?

lol

have 72 now :( got 22 1st round

[quote="ehehheehee"]A 22 is pretty good.  Some people would love to have a 22...

And how is a 25 not too bad?  It couldn't be better.[/quote]

Well you could get 5 commended  :rotfl:

[quote="ehehheehee"]That does not help your score.

Would you rather have [b]5 5 5 5[/b] 4 or 5 5 5 5 5?[/quote]
I'm not sure, I wouldn't get either.
But if I knew that I could get the same thing everytime then I'd go with first because a 96 is gold too.

5 5 5 [b]ZOMG 5[/b] 5

I love the grader's comment to my #1:  "BAD SOLUTION! - 5"  Oh the irony of brute force...

[quote="lowest_reynolds"]
I love the grader's comment to my #1:  "BAD SOLUTION! - 5"  Oh the irony of brute force...[/quote]

That frustrates me, considering my grader wrote "Good solution, but no diagram - 4" when there was a diagram...

EDIT - I am happy: score got changed, and we got a snow/ice day today!

To calc rulz: So you interpreted #1 as coloring the vertices?

[quote="calc rulz"]Yay, it got corrected. I got 1 5 5 [b]5 5[/b]. Pretty ironic set of scores and too bad I misread #1, but at least I'm doing well and I can still probably get a gold  :D[/quote]

I always read each problem like 10 times.

Do you do that in a real competition?  Doesn't that take time?

I am glad you got 10 of your points back, calc rulz.

I think the poll is kind of skewed.  Only people in AOPS could vote.  And mostly only people who thought they did well did.

i don't do it that many times during a timed competition (thought if i did, it might help) unless I have time at the end to check my answers.

Woww..

I got [b]5[/b] 2 5 - 2

There is a good new and a bad news for me.

14 is my worst score ever. :mad:
This is my first - (aka "you got an F") :mad: to the 100th power

BUT, I got my first commended... with a pretty diagram  :maybe:

I hope I do well on the last round and get at least bronze!  :roll:

[quote="vishalarul"]What would you expect an 8th grader to get?[/quote]
Well it depends on how much time is spent.

How does being in 8th grade have that much to do with how good at math you are?

I know the seniors in my school think they are the only ones that know Calculus, while in reality, there are freshmen and sophmores that actually do competitions and probably know more Calc by just flipping through a Calc book from time to time (not that Calc is important).

it doesn't. 365 days difference in lifespan ( assuming an 8th grader is about 13.5 and a 9th grader is about 14.5 ). Some can work hard, and then there are the lazy ones.

[quote="ehehheehee"]What about "Good work, but you do not need to give so many details."  (on #3)[/quote]
You forgot to quote the initials "EJS" at the end. I initial all my comments, because sometimes I get snarky and I wouldn't want anyone else to take the blame for my comment.

There was nothing wrong with your solution, but meticulous detail is tiring to read. Also, I wouldn't want you to labor under the misconception that we require that much detail.

Erin J. Schram
USAMTS grader

Sorry, I did not see the initials.  It is just that I rather be too detailed then leave holes where the grader takes off.

I guess I'll go look at other people who got 5's to see what it should look like under the thread where people post their solutions.


Let $ m; n$ be positive integers, and let G be a connected graph with $ mn$ vertices. Show that the vertices can be partitioned into m sets of size n such that any two vertices in
the same set are at distance · $ \leq 2n-2$ from each other.   :D

Take a longest path, suppose it has $ k$ vertices. We observe from the last $ n$ vertices (if there aren't $ n$ vertices in the path, the problem is solved), any (other) path beginning from vertex $ k-i$ can have at most $ i$ length, so in this set of $ n$ vertices and the vertices from their paths we have that the distance between $ 2$ points is $ \leq 2n-2$ (this can be reached between $ 2$ endpoints of $ 2$ paths from vertex $ k-n+1$(the last one)). We select $ n$ points from these so that the graph remains connected(for example we choose and erase the endpoint from the initial longest path, after the endpoints of the longest path containing the initial $ k-1$ vertices, after this we erase the $ k-1th$ vertex and so on, untill we erased $ n$ points), and these form a set. We repeat the procedure until all $ m$ sets are formed.


[quote="mdk"]How many ordered pairs of positive integers (x,y) with x>y are there such that 1/x+1/y=1/2007?[/quote]
[hide="hint"]SFFT
[size=75][edit]: in other words, simon's favorite factoring trick[/size][/hide]

[quote="SM4RT"]okay so using that i get..

$\frac{x+y}{xy}=\frac{1}{2007}$
$2007x+2007y = x\cdot{y}$
$x\cdot{y}-2007x-2007y=0$ add $2007^{2}$ to both sides
$\left(x-2007\right)\left(y-2007\right)=2007^{2}$

$2007^{2}=3^{4}223^{2}$..i dont think 223 has any factors..does it??

so $2007^{2}$ has $\left(4+1\right)\left(2+1\right) = 15$ factors..however 2007 cannot work so there are $\frac{14}{2}=\boxed{7}$ pairs of factors that work..

is this right??[/quote]
With this math, none of the factors of $2007^{2}$ work.

[quote="SM4RT"]okay so using that i get..

$\frac{x+y}{xy}=\frac{1}{2007}$
$2007x+2007y = x\cdot{y}$
$x\cdot{y}-2007x-2007y=0$ add $2007^{2}$ to both sides
$\left(x-2007\right)\left(y-2007\right)=2007^{2}$

$2007^{2}=3^{4}223^{2}$..i dont think 223 has any factors..does it??

so $2007^{2}$ has $\left(4+1\right)\left(2+1\right) = 15$ factors..however 2007 cannot work so there are $\frac{14}{2}=\boxed{7}$ pairs of factors that work..

is this right??[/quote]

Yep that is correct  Good JOb


parameterization
probability
function
advanced fields
advanced fields unsolved

Can you tell me a method to produce  Xn random variables with Laplace distribution and parameter μ, from samples with uniform distribution in [0,1]?

And how can i calculate theoretically the probability P{x>10}  for μ=1?

thx a lot...

I don't know what the Laplace distribution is, but here's a general procedure for constructing a real-valued random variable $\lambda$ from a uniform distribution:

Step 1: Find the cumulative distribution function $f(x)=P(\lambda \le x)$ for all $x$.
Step 2: Find $f^{-1}$ as a function. Given random input $z$ uniform on $[0,1]$, $f^{-1}(z)$ has the same distribution as $\lambda$.


geometry
parallelogram
3D geometry
tetrahedron
geometry unsolved

How to construct a parallelepiped(a parallelepiped is a three-dimensional figure formed by six parallelograms) by a given Tetrahedron so that every vertices of the Tetrahedron is a vertices of the Parallelepiped and no two vertices of the Tetrahedron lies on the same edge of the parallelepiped.

Choose any corner of the parallelepiped as one vertex of the tetrahedron, and the three faces diagonals that converge to this vertex as three edges of the tetrahedron . The "free' extremes of those diagonals define the tetrahedron base.

yes, but the problem is that a Tetrahedron is given and you should construct with compass and ruler a parallelepiped with that properties


Two opposite sides of a rectangle are each divided into $n$ congruent segments, and the endpoints of one segment are joined to the center to form triangle $A$. The other sides are each divided into $m$ congruent segments, and the endpoints of one of these segments are joined to the center to form triangle $B$. [See figure for $n = 5, m = 7$.] What is the ratio of the area of triangle $A$ to the area of triangle $B$?

[asy]
int i;
for(i=0; i<8; i=i+1) {
dot((i,0)^^(i,5));
}
for(i=1; i<5; i=i+1) {
dot((0,i)^^(7,i));
}
draw(origin--(7,0)--(7,5)--(0,5)--cycle, linewidth(0.8));
pair P=(3.5, 2.5);
draw((0,4)--P--(0,3)^^(2,0)--P--(3,0));
label("$B$", (2.3,0), NE);
label("$A$", (0,3.7), SE);[/asy]

$\text{(A)} \ 1 \qquad \text{(B)} \ m/n \qquad \text{(C)} \ n/m \qquad \text{(D)} \ 2m/n \qquad \text{(E)} \ 2n/m$

[hide="Solution"]
Let the side being divided into n segments be called X. Let the side being divided into m segments be called Y.

The long leg of triangle A is Y/2, and the long leg of triangle B is X/2.

The area for A is $\frac{XY}{4n}$.
The area for B is $\frac{XY}{4m}$.

That makes the ratio $\frac{\frac{XY}{4n}}{\frac{XY}{4m}}=\boxed{\frac{m}{n}}$.
[/hide]


For a fixed positive number x, what is the minimum value of the expression
  $\frac{1}{(1+xa)^2}+\frac{1}{(1+bx)^2}+\frac{1}{(1+cx)^2} $ when a,b,c are positive with product 1?

For x<sqrt(3)-1 minimal value doesn't exist. For other x I have an ugly solution using derivatives.

Sorry, I wanted to say the infimum of the expression. And I have a very nice solution without derivatives. I hope you didn't apply that method as in ineq 3, since I have real doubts that it's totally corect. Manlio can confirm that.

Three years ago, I proved the following problem:
Let $x_1,x_2,...,x_n$ be positive numbers such that $x_1x_2...x_n=1$.  If $n\geq3$ and $k\geq \sqrt{n}-1$, then

$\frac{1}{(1+kx_1)^2}+\frac{1}{(1+kx_2)^2}+...+\frac{1}{(1+kx_n)^2} \geq \frac{n}{(1+k)^2}$.

I will send this problem either to Crux or to GM-A.
The proof is very difficult.

Upsss, sorry, I didn't know you also treated this problem. I apologise. I think, but I'm not sure, that I also managed to prove this generalized inequality. Again, I apologise, but I didn't know
.

I think I solved the easy one(n=3)
We know that 1/(1+x)^2+1/(1+y)^2>=2/(1+sqrt(xy))^2 if xy>=1/2
(it's not so hard, just clear denominators)
If x>sqrt(3)-1 then we can find a,b s.t. ax*bx>=1/2( as sqrt(3)-1>1/sqrt(2))
So we can replace a,b by sqrt(ab).
Then we get the inequality in 2 variables.
Again open the brackets and the result will come out.


What is the probability on a toss of a fair coin 10 times, there is no 3 consecutive heads or tails?
For the probabiliy of no 3 consecutive heads only, I got the recurrence [hide]$ P(n)\equal{}(1/2)P(n\minus{}1)\plus{}(1/2)^2P(n\minus{}2)\plus{}(1/2)^3P(n\minus{}3),P(0)\equal{}P(1)\equal{}P(2)\equal{}1$ with P(10) as the answer.[/hide]

how did you get that recurrence? it doesn't seem right.

here's what i would do. let $ k_n$ be the number of sequences of $ H$'s and $ T$'s of length $ n$ with no three consecutive heads or tails. let $ t_n$ be the number of such sequences ending in $ T$ and $ h_n$ the number ending in $ H$. obviously, $ h_n \equal{} t_n$, so $ k_n \equal{} 2t_n$. so we'll find a recurrence for $ t_n$. if the sequence ends in $ ...TT$, obviously it must end in $ ...HTT$, so the number of sequences in this case is $ h_{n \minus{} 2}$. if the sequence ends in $ ...HT$, then either it ends in $ ...THT$ or $ ...HHT$. in the first case, there are $ t_{n \minus{} 2}$ possible sequences; in the second, there are $ t_{n \minus{} 3}$ possible sequences. so there are $ t_{n \minus{} 2} \plus{} t_{n \minus{} 3}$ sequences ending in $ ...HT$. combining with the $ ...TT$ case, our recursion is therefore $ t_n \equal{} h_{n \minus{} 2} \plus{} t_{n \minus{} 2} \plus{} t_{n \minus{} 3} \equal{} 2t_{n \minus{} 2} \plus{} t_{n \minus{} 3}$. multiplying by 2, we get $ k_n \equal{} 2k_{n\minus{}2} \plus{} k_{n \minus{} 3}$. the probability is $ \frac {k_{10}}{2^{10}}$

My recurrence is on the probability of no 3 consecutive heads ONLY - 1/2 prob. of it ending in tails times $ P(n\minus{}1)$; 1/2 * 1/2 of it ending in tail+head times $ P(n\minus{}2)$; 1/2 ^3 of it ending in tail+head+head times $ P(n\minus{}3)$. Is that right?

Here's a new problem (don't want to create an extra new topic for more space):
Candidate A receives 1032 votes, and candidate B receives 971 votes. If the votes are counted one at a time in a random manner, what is the probability that candidate A was never behind at any point in the counting?


geometry
trigonometry
LaTeX
area of a triangle

If the sides of a triangle are $2,3,4$, what is the radius of the circle circumscribing the triangle?

[quote="Iversonfan2005"][quote="warut_suk"]Formula: $\frac{abc}{4A}$
where $A$ is the area of the triangle[/quote]

Is that the formula to obtain the radius?[/quote]

Of course! The circumradius.

[quote="cezar lupu"]Hey,Iversanfan2005 this problem is trivial.Here's a difficult problem: Find the maximal area of a triangle $ABC$ with an angle $60$ degrees such that there is a point $P$ such that $PA=1,PB=2,PC=3$[/quote]

Sorry if it was trivial :o , I mean, this was the last and hardest problem on a math competition here in SC. Very well, I'll try your problem then.

[quote="Iversonfan2005"]Sorry, I never knew that formula :blush:[/quote]

That's OK. I suggest you to remember it  :)

[quote="cezar lupu"] Find the maximal area of a triangle $ABC$ with an angle $60$ degrees such that there is a point $P$ such that $PA=1,PB=2,PC=3$[/quote]

Hmm... I've only seen something like "When $PA=1, PB=2, PC=3$ and $ABC$ is equilateral then find the side length". Your problem is quite interesting.

And it is very difficult but also very beautiful.

Let me have a guess. Is it when $ABC$ is equilateral?

[quote="furious"][quote="Iversonfan2005"] this was the last and hardest problem on a math competition here in SC.[/quote]
which contest?[/quote]

It was one of the older ones from the USC, maybe from the late 90s? I was going through them and I found this one, which was the last problem.

[quote="cezar lupu"]can you post some other inetrresting problems? i assume that thre're problems harder than the one you've posted.[/quote]

Sure, but don't blame me if they're not tough enough :P

Just go [url=http://www.math.sc.edu/contest/]here[/url] to see past USC's.

The problem is not too bad if you know some formulas.  And 'trivial' when used in a math context can mean a variety of different things...and is relevant to only the person using the term a lot of the time.

Anyway,  the formulas you would need for this problem are mostly a bunch of different formulas for the area of a triangle:

[b]Heron's (aka Hero) Formula[/b]

The area, $A$, of a triangle with sides of length $a$, $b$, and $c$, is

\[A=\sqrt{s(s-a)(s-b)(s-c)}\]

where $s$ is the semiperimeter, i.e. $s=\frac{a+b+c}{2}.$

The next formula for the area of a triangle is \[A=\frac{abc}{4R}\]

where $R$ is the circumradius.

Have you done any trig, Shobhit?  If so proving this formula is a good problem.  Try it.

Nope, I even skipped trig in the AOPS book, but thanks for the formulas Joe :)

EDIT: Never mind, I totally screwed up my logic.  Ignore this post. :blush:

[quote="K81o7"]You might use this formula for area for that problem: $\frac{1}{2}\sin{a}bc$.  I think that's it.  But if it is, you could probably find the maximum product for two of the sides given that the angle between them is 60 degrees.[/quote]

Well, by looking at the $\LaTeX$ the formula you [i]intended[/i] to write is correct (except you should use $A$ instead of $a$). It should look more like this: $\frac 12bc\sin A.$

[quote]Here's a difficult problem: Find the maximal area of a triangle $ABC$ with an angle $60$ degrees such that there is a point $P$ such that $PA=1,PB=2,PC=3$[/quote]

Could you post the solution?


What's the sum of the digits of this number:

$ \sqrt{2004\times2002\times1998\times1996\plus{}36}$


function
counting
distinguishability
modular arithmetic
number theory open
number theory

First, define a function
$ \varsigma(n) \equal{} \sum{\pi(n)}$,
where $ \pi(n)$ means the permutations (including "indistinguishable" ones) of the digits of $ n$. Single-digit numbers are defined to be two digit numbers with the first digit 0 (i.e, 7 = 07).
For example, $ \varsigma(23) \equal{} 23 \plus{} 32 \equal{} 55$, $ \varsigma(11) \equal{} 11 \plus{} 11 \equal{} 22$, $ \varsigma(7) \equal{} \varsigma(07) \equal{} \varsigma(70) \equal{} 07 \plus{} 70 \equal{} 77$,
and $ \varsigma(125) \equal{} 125 \plus{} 152 \plus{} 215 \plus{} 251 \plus{} 512 \plus{} 521 \equal{} 1776$.

Next, for all $ k,j \in \mathbb{N}$, define a function $ D_{k,j}: \mathbb{N} \rightarrow \mathbb{N}$ with $ D_{k,j}(0) \equal{} k$,

$ D_{k,j}(n) \equal{} \left\{\begin{array}{rl} \varsigma(D_{k,j}(n\minus{}1)) \pmod{10^j} & \text{if } 2 \mid n \\
\sum_{i \equal{} 0}^{n \minus{} 1}D_{k,j}(i) \pmod{10^j} & \text{if } 2 \nmid n \end{array}\right.$

The sequence $ D_{k,j}(0),D_{k,j}(1),D_{k,j}(2),\dots$ is said to be generated by $ k$ and $ j$. For example, the sequence generated by $ k \equal{} 12, j \equal{} 2$ is $ 12, 33, 45, 99, 89, 87, 65, \dots$. The sequence generated by $ k \equal{} 12, j \equal{} 3$ is $ 12,33,45,99,189,996,374\dots$.

1.) Call a sequence irrational if it has no infinitely repeating parts, and rational otherwise (like with irrational and rational numbers). Prove that there are no irrational sequences generated by any $ k,j$.
2.) Put a maximum and minimum bound on the repeating part of the sequence generated by $ k,j$ in terms of $ j$.
3.) Find the number of distinguishable repeating parts of the sequence generated by $ k,j$ in terms of $ j$ (A repeating part is distinguishable if it can't be shifted to the right or left to become the repeating part of some sequence. For example, $ 17,89,43,57$ is considered the same as $ 43,57,17,89$, but different from $ 17,43,89,57$).

[quote="akessler"]First, define a function
$ \varsigma(n) \equal{} \sum{\pi(n)}$,
where $ \pi(n)$ means the permutations (including "indistinguishable" ones) of the digits of $ n$. Single-digit numbers are defined to be two digit numbers with the first digit 0 (i.e, 7 = 07).
For example, $ \varsigma(23) \equal{} 23 \plus{} 32 \equal{} 55$, $ \varsigma(11) \equal{} 11 \plus{} 11 \equal{} 22$, $ \varsigma(7) \equal{} \varsigma(07) \equal{} \varsigma(70) \equal{} 07 \plus{} 70 \equal{} 77$,
and $ \varsigma(125) \equal{} 125 \plus{} 152 \plus{} 215 \plus{} 251 \plus{} 512 \plus{} 521 \equal{} 1776$.

Next, for all $ k,j \in \mathbb{N}$, define a function $ D_{k,j}: \mathbb{N} \rightarrow \mathbb{N}$ with $ D_{k,j}(0) \equal{} k$,

$ D_{k,j}(n) \equal{} \left\{\begin{array}{rl} \varsigma(n) \pmod{10^j} & \text{if } 2 \mid n \\
\sum_{i \equal{} 0}^{n \minus{} 1}d_i \pmod{10^j} & \text{if } 2 \nmid n \end{array}\right.$

The sequence $ D_{k,j}(0),D_{k,j}(1),D_{k,j}(2),\dots$ is said to be generated by $ k$ and $ j$. For example, the sequence generated by $ k \equal{} 12, j \equal{} 2$ is $ 12, 33, 45, 99, 89, 87, 65, \dots$. The sequence generated by $ k \equal{} 12, j \equal{} 3$ is $ 12,33,45,99,189,996,374\dots$.

1.) Call a sequence irrational if it has no infinitely repeating parts, and rational otherwise (like with irrational and rational numbers). Prove that there are no irrational sequences generated by any $ k,j$.
2.) Put a maximum and minimum bound on the repeating part of the sequence generated by $ k,j$ in terms of $ j$.
3.) Find the number of distinguishable repeating parts of the sequence generated by $ k,j$ in terms of $ j$ (A repeating part is distinguishable if it can't be shifted to the right or left to become the repeating part of some sequence. For example, $ 17,89,43,57$ is considered the same as $ 43,57,17,89$, but different from $ 17,43,89,57$).[/quote]

It looks like a very nice problem indeed, but you've probably made some mistakes (Or I'm too stupid to understand...)

In your function:
$ D_{k,j}(n) \equal{} \left\{\begin{array}{rl} \varsigma(n) \pmod{10^j} & \text{if } 2 \mid n \\
    \sum_{i \equal{} 0}^{n \minus{} 1}d_i \pmod{10^j} & \text{if } 2 \nmid n \end{array}\right.D_{k,j}(n) \equal{} \left\{\begin{array}{rl} \varsigma(n) \pmod{10^j} & \text{if } 2 \mid n \\
\sum_{i \equal{} 0}^{n \minus{} 1}d_i \pmod{10^j} & \text{if } 2 \nmid n \end{array}\right.$

What is $ d_i$ ..?

You say that $ D_{12,2}(2) \equal{} 45$, but obviously $ D_{12,2}(2) \equal{} 02 \plus{} 20 \equal{} 22$.
Please enlightent me :)

[quote="Mathias_DK"]
It looks like a very nice problem indeed, but you've probably made some mistakes (Or I'm too stupid to understand...)

In your function:
$ D_{k,j}(n) \equal{} \left\{\begin{array}{rl} \varsigma(n) \pmod{10^j} & \text{if } 2 \mid n \\
\sum_{i \equal{} 0}^{n \minus{} 1}d_i \pmod{10^j} & \text{if } 2 \nmid n \end{array}\right.$

What is $ d_i$ ..?

You say that $ D_{12,2}(2) \equal{} 45$, but obviously $ D_{12,2}(2) \equal{} 02 \plus{} 20 \equal{} 22$.
Please enlightent me :)[/quote]
Ah, there was a slight mistake in the first part. Below is the correct version, and I'm going to replace the formula in the original post:
$ D_{k,j}(n) \equal{} \left\{\begin{array}{rl} \varsigma(D_{k,j}(n \minus{} 1)) \pmod{10^j} & \text{if } 2 \mid n \\
\sum_{i \equal{} 0}^{n \minus{} 1}D_{k,j}(i) \pmod{10^j} & \text{if } 2 \nmid n \end{array}\right.$

As for the sequence generated by $ k \equal{} 12, j \equal{} 2$, we have:
$ \boxed{12}, 12 \plus{} 21 \equal{} \boxed{33}, 12 \plus{} 33 \equal{} \boxed{45}, 45 \plus{} 54 \equal{} \boxed{99}, 12 \plus{} 33 \plus{} 45 \plus{} 99 \equal{} \boxed{89},...$

(I meant to start the sequence at $ D_{k,j}(1),D_{k,j}(2),...$, not at $ D_{k,j}(0)$)


Although ethers are weakly polar why are they used as solevnts for grignard's reagents?

[hide="My Answer"]

Grignard's reagent  has both lipophillic and ionic part so as to get it solvated we need a solvent which has both organic(hydrocarbon) and polar part so probably ethers are the best choice to do so [/hide]

What you have said is valid.  Also, of at least equal importance is the fact that ethers don't react with Grignards.  If you used something like water as your solvent instead, you could have some problems...

A Grignard reagent is extremely basic. Much more basic than hydroxide ion. So basic it could pull a proton off of an alcohol. So basic it could pull a proton off of ammonia. So basic it might even pull a proton off of a ketone. And that proton would neutralize the reagent and prevent its reactivity with the desired substrate. Hence you need an [b]aprotic[/b] polar solvent. The most readily available of such solvents are ethers such as diethyl ether or tetrahydrofuran.

Now I'm wondering if I still have my Morrison and Boyd textbook from the organic class I took 30 years ago. I remember a chain of homework problems involving a hapless graduate student who was always having his attempted syntheses, usually involving Grignard reagents, go bad - the problem was always to explain what went wrong. As I recall, the last problem in the series ended with the student giving up chemistry and embarking on a political career in which he was reasonable about most things but had an irrational hatred of the French.

thank you all
[hide="Kent Merryfield"]
 this is a question from Morrison&Boyd  :wink:

any idea who the 'hapless graduate student' is....[/hide]

[quote="pardesi"]any idea who the 'hapless graduate student' is....[/quote]
I always assumed he was a completely fictional character onto whose head the authors heaped every mistake they had heard about or could imagine.

Hmm...

[quote] always assumed he was a completely fictional character onto whose head the authors heaped every mistake they had heard about or could imagine.[/quote]

These lines are tougher than those in my bio text books to correlate :rotfl:

[quote="Kent Merrifield"]So basic it might even pull a proton off of a ketone.[/quote]

If that were true then the reactions of Grignard reagents weren't certainly be so usefull. Its true that the alpha hydrogens of aldehydes and ketones are "specially" acidic, but when they react  with Grignard reagents addition across the carbon-oxygen double bond is the preferred path. Why? Because when you mix the carbonyl compound with the Grignard the first reaction that occurs is complexation of one molecule of Grignard by the oxygen of a molecule of carbonyl compound. This makes the carbonyl carbon even more electrophilic and in the very next step of the mechanism it is attacked by the nucleophilic carbon of another molecule of Grignard.
Even when there are bulky groups around the carbonyl, the preferred pathway is still not an acid-base reaction, but very probably a reduction to the alcohol.

[quote="Pardesi"]Although ethers are weakly polar why are they used as solevnts for grignard's reagents?[/quote]

Because of two reasons. One of them already stated: because ethers (but not all ethers) don't react with Grignard reagents. But another very important reason is because the oxygen of ethers is able to coordinate with the magnesium atom of the Grignard that is forming, thus stabilizing it. In fact, when Et2O or THF are used, every Grignard molecule coordinates with two molecules of ether. As a final note, ethyl ether is commonly used when we want to prepare alkyl Grignard reagents; for, vinyl, alkynil, and aryl Grignard reagents THF is the solvent of choice.


In an arithmetic sequence the $ 113^{\text{th}}$ term is $ 786$ and the $ 125^{\text{th}}$ term is 870. Find the $ 150^{\text{th}}$ term.

The common difference is $ \frac{870\minus{}786}{125\minus{}113}\equal{} 6$.  Therefore, the $ 150$th term is $ 870\plus{}(150\minus{}125) \times 6\equal{} \boxed{1020}$.

[hide="The long way"]

Let a be the first term and d the difference:

$a+112d=786$
$a+124d=870$

Thus, $12d=84 \implies d=7$.

This gives $a=2$, and thus $a+149d=2+149(7)=\boxed{1045}$.

Problem with solution above is that $\frac{870-786}{25-13}$ is not 6.[/hide]


geometry
analytic geometry
graphing lines
slope
trigonometry
inequalities
quadratics

[quote="worthawholebean"]Let $ ABC$ be a triangle with $ AB \equal{} 5$, $ BC \equal{} 4$ and $ AC \equal{} 3$. Let $ \mathcal P$ and $ \mathcal Q$ be squares inside $ ABC$ with disjoint interiors such that they both have one side lying on $ AB$. Also, the two squares each have an edge lying on a common line perpendicular to $ AB$, and $ \mathcal P$ has one vertex on $ AC$ and $ \mathcal Q$ has one vertex on $ BC$. Determine the minimum value of the sum of the areas of the two squares.

[asy]import olympiad;
import math;
import graph;

unitsize(1.5cm);

pair A, B, C;

A = origin;
B = A + 5 * right;
C = (9/5, 12/5);

pair X = .7 * A + .3 * B;
pair Xa = X + dir(135);
pair Xb = X + dir(45);

pair Ya = extension(X, Xa, A, C);
pair Yb = extension(X, Xb, B, C);

pair Oa = (X + Ya)/2;
pair Ob = (X + Yb)/2;

pair Ya1 = (X.x, Ya.y);
pair Ya2 = (Ya.x, X.y);
pair Yb1 = (Yb.x, X.y);
pair Yb2 = (X.x, Yb.y);

draw(A--B--C--cycle);
draw(Ya--Ya1--X--Ya2--cycle);
draw(Yb--Yb1--X--Yb2--cycle);

label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, N);

label("$\mathcal P$", Oa, origin);
label("$\mathcal Q$", Ob, origin);[/asy][/quote]
[hide]
Label $ A(0,0)$ and $ B(5,0)$ and let the intersection of $ \mathcal P$ and $ AC$ be $ D$ and the intersection of $ \mathcal Q$ and $ BC$ be $ E$. Also let $ \mathcal P$ have length $ p$ and $ \mathcal Q$ have length $ q$.

The slope of line $ AC$ is $ \tan\angle CAB \equal{} \frac43$. Then label $ D\left(\frac {3}{4}p, p\right)$. Then label $ E\left(\frac {7}{4}p \plus{} q,q\right)$.

The slope of line $ BC$ is $ \tan \angle CBA \equal{} \frac34$. Write the equation of line $ BC$ as $ y \equal{} \minus{} \frac {3}{4}x \plus{} \frac {15}{4}$. Then $ q \equal{} \minus{} \frac {3}{4}\left(\frac {7}{4}p \plus{} q\right) \plus{} \frac {15}{4}$. Then $ 21p \plus{} 28q \equal{} 60$. By the Cauchy-Schwarz inequality, $ (p^2 \plus{} q^2)(21^2 \plus{} 28^2)\ge (21p \plus{} 28q)^2 \equal{} 60^2$. So $ p^2 \plus{} q^2\ge\frac {60^2}{21^2 \plus{} 28^2} \equal{} \frac {60^2}{35^2} \equal{} \frac {12^2}{7^2} \equal{} \boxed{\frac {144}{49}}$.

Edit: Equality when $ \frac{p}{21}\equal{}\frac{q}{28}$, or $ p\equal{}\frac{36}{35}, q\equal{}\frac{48}{35}$. [/hide]

Let x be a side of square P, y be a side of square Q. The triangle with point A containing a side of square P and the one with B containing a side of Q are both similar to the whole triangle by AA. Thus, the four lengths that add up to the hypotenuse are 4x/3, x, y, and 3y/4. Since these add up to 5, we can deduce that $ y \equal{} 20/7 \minus{} 4x/3$. So we want to minimize $ x^2 \plus{} (20/7 \minus{} 4x/3)^2$. This can easily be done by expanding it, finding the x coordinate of the vertex, and plugging it back in. And the answer, as before, is 144/49.

Umm i just saw some old AMC problem that told you to find the maximum of something like 3x+4y or something like that and some guy used Cauchys...so
when you get to 7x/3+7y/4=5 and you want to minimize x^2+y^2 (area of the two squares)
you can use Cauchys (instead of substitution and then vertex on a quadratic)

(x^2+y^2)((7/3)^2+(7/4)^2)=>(7x/3+7y/4)^2

So we have S=x^2+y^2, (S){[(25)(49)]/[(16)(9)]}=>(25).  So S=>144/49 so the minimum is at 144/49.

There is a nice purely geometrical solution.


Prove that no integer in the sequence

$ 11, 111, 1111, 11111, ...$

is a perfect square.

[hide="Solution"]
Note that the sequence can be defined has $ a_n\equal{}10a_{n\minus{}1}\plus{}1$ with $ a_1\equal{}11\equiv 3\pmod{4}$, if we evaluate this in $ \pmod{4}$ we get, $ a_n\equiv 2a_{n\minus{}1}\plus{}1$. From which we can see that $ a_2\equiv2\cdot 3 \plus{} 1\equiv 3$, and notice that every $ a_n$ is equivalent to $ 3\pmod{4}$. Since $ 3$ is not a quadratic residue in $ \pmod{4}$ we are done. :ninja:
[/hide]


geometry
3D geometry
algebra
polynomial
parameterization
AMC
USA(J)MO

What are some useful factorizations in algebra that you might see on a math contest but not in a standard high school algebra class?

During math team, an upperclassman used a factorization I hadn't seen before in order to relate $ x^3\plus{}y^3\plus{}z^3$ to $ xyz$ and $ xy\plus{}xz\plus{}yz$ or something like that in a factorization step...

This might just be [url=http://www.artofproblemsolving.com/Wiki/index.php/AM-GM]AM-GM[/url] instead of a factorization.

I know what you're saying and, yeah, someone did do another problem involving what i wrote earlier using AM-GM.

Maybe I wrote some of the parts incorrectly, but it was a factorization involving cubes of some of the variables and the product of the three variables.

If $ x \plus{} y \plus{} z \equal{} 0$, what is $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz$

Maybe you mean this:

$ x^{3}\plus{}y^{3}\plus{}z^{3}\minus{}3xyz\equal{}(x\plus{}y\plus{}z)(x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}xy\minus{}zx\minus{}yz)$

Last post was for aufha but it is also the reponse of the last question.
So if $ x \plus{} y \plus{} z \equal{} 0$

$ x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} zx)$
$ x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz \equal{} (0)(x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} xz \minus{} zy) \equal{} 0$

Or maybe it was this:

$ (x\plus{}y\plus{}z)^{3}\equal{}x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}2yz\plus{}2xy\plus{}2xy$

The discussed factorization can also be written

$ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz \equal{} \frac {1}{2} (x \plus{} y \plus{} z)( (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2 )$;

in other words, the right factor is zero iff $ x \equal{} y \equal{} z$.  This is occasionally a more useful form.

[b]Exercise:[/b]  Exactly one set of three points on the graph of $ x^3 \plus{} y^3 \plus{} 3xy \equal{} 1$ forms an equilateral triangle.  What is its area?

Another interesting factorization is the [i]Sophie Germaine identity[/i]

$ a^4 \plus{} 4b^4 \equal{} (a^2 \plus{} 2ab \plus{} 2b^2)(a^2 \minus{} 2ab \plus{} 2b^2)$.

[b]Exercise:[/b]  Find all values of $ n$ such that $ n^4 \plus{} 4^n$ is prime.

In general, though, the skill of finding factorizations is more useful than the skill of memorizing them.  A few tips:

- The factors of a symmetric expression are symmetric.  To actually find them, however, it is useful to treat one of the variables as the only variable and to treat the other variables as constants; in other words, polynomial factorization.
- Some problems involve polynomials whose coefficients have parameters in them; for example, a polynomial in $ x$ with a constant $ t$ determining some of its coefficients.  To factor these, it can be useful to treat them as polynomials in $ t$ with a constant $ x$ determining some of its coefficients.
- The AoPS book discusses the "incomplete" factorization $ x^2 \plus{} y^2 \equal{} (x \plus{} y)^2 \minus{} 2xy$.  This becomes a "complete" factorization if $ x, y$ are replaced with expressions such that $ 2xy$ is a square, and then we can use difference of squares.  This technique can be used to derive the Sophie Germaine identity and to solve [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=825508#825508]USAMO 2007 #5[/url].

[quote="Patterns_34"][quote="moogra"]Or maybe it was this:

$ (x + y + z)^{3} = x^{3} + y^{3} + z^{3} + 2yz + 2xy + 2xy$[/quote]

Actually: $ (x + y + z)^{\color{red}2\color{black}} = x^{\color{red}2\color{black}} + y^{\color{red}2\color{black}} + z^{\color{red}2\color{black}} + 2yz + 2xy + 2x\color{red}z{\color{black}}$

$ (x + y + z)^3 = x^3 + y^3 + z^3 + 3x^2(y + z) + 3y^2(x + z) + 3z^2(x + y) + 6xyz$

 :)[/quote]

lol i meant squared. Hmmm then that wouldn't work then for the thread starter's problem


This year Kazakhstan team has such results:
2 gold medals (Yessenov Kuat -37, Andrey Kim -32 - lucky man :) ), 2 bronze (Yeleussizov Damir -23, Kelbetov Daniyar -21).
We are 21st (with Canada, Brazil and Serbia and Montenegro).
Comparing to our past IMO results, this is not good for us since we were in top 20 for last 3 years.

Kuat, thank you for this information.
Congratulation to you. You have done a new record for Kazakhstan:
1 silver IMO2002, 2 gold medals IMO2003 and IMO2004.

Not bad for Andrey KIM.
First time participating and winning a gold medal! And he has 2 more posibilities.

Congratulations to everybody, who has received medals, especially to Andrey Kim, a lucky man like me.

[quote="Valentin Vornicu"][quote="billzhao"]It's weird that Kazakhstan tied Canada this year as well as last year.[/quote] is this good or bad for Canada? :)[/quote]

It is bad for Kazakhstan.

Congratulations to the team from Kazakhstan! They are great mathematicians and also great guys! I was their guide this year, so i am very glad for them!  :) I am not sure but i think that Andrey Kim is the youngest gold-medalist of 2004... He is just 14! Anyway, congratulations!

:D Hi man here U R!!!
Everything was great there in Greece, but unfortuately I coudn't win any medal this year. I believe everything gonna be great next year in Mexico!!!!!!

[quote="Nikolas"]Nikos man how R U?
how about your lessons and where is Sri LAnka guide :D


go here: and press f5 gett hundreds of thousands of views here if u have no life!!!

enjoy! and feel free to go emo
 :D

[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=963[/url]

dude, this is bordering spam. don't advertise your blog on your state forum. geez.
i mean, at least i advertised somebody else's identity  :wink:


Well done Valentin,looks good.What technical problems did arise?

it's a beta. there's still much to be done, but hopefully by the next week I will make it.

the latex installation went wrong, and therefore there is not latex yet on the site ...

I have a suggestion for the beginners corner.A sticky in which is explained how to post a problem and eventually requiring the users to post their tryings on the exercise/problem.Because,we dont want that some just post a problem they didnt even try and then just copy the solution.Right?

[b]Very nice[/b] indeed !

Perhaps there are a lot a topics->subtopics->subtopics->subtopics->subtopics->subtopics->subtopics  :D but it's only a very minor remark.

I've just had a look at the beginner section, do you think problems from national contests are of beginner's level  :? Your beginner's section is the advanced section of some other forums :D

you can make that post yourself, give it a try. in the beginner's corner forum directly ... :D

[quote="belenos"] I've just had a look at the beginner section, do you think problems from national contests are of beginner's level  :? Your beginner's section is the advanced section of some other forums :D[/quote]

beginners can also post any problem they would like comments on or solutions :)

Another thing:in the upper right corner,the text appears a bit blury,maybe you could choose a different font or colour or something :)

Upper right corner of the page:
News|Links|About|Home

those are not functional yet. they are not blurry to me  :?

I agree with Belenos about the a lot of forums -> subforums -> subtopics thing...it's a lot to click just to check out each forum, especially as now we don't know which particular subforum the new post is in.  I noticed that there are more posts on each page now, though; that might be a good thing because 10 posts per page (as it used to be) is relatively little actually.

Valentin, you really need to do something about everybody's status; we're all showing up as "Moderator" :D

[quote="Valiowk"]Valentin, you really need to do something about everybody's status; we're all showing up as "Moderator" :D[/quote]
except him!  :) He's "admin".

sorry guys it seems something went wrong while implementing the source code for the source line in the old post ... I have to work at it.

anyway don't worry the database is backed-up every hour so your posts won't be lost.

and remember for now this is a beta :) (not fully featured & functional site ) :D

Just a tiny note: I'm not sure if it's a good idea to have separate "Theorems and Formulas" (actually, it should be [i]formulae[/i] :D ) sub-forums for each category for both Advanced and Beginners...it occurred to me that the same ones might be frequently used for both, so to have 2 such sub-forums might cause some confusion when searching for it...

[quote="Valiowk"](actually, it should be [i]formulae[/i] :D ) [/quote]
I just looked up in the dictionary. it said that both forms are ok.

"data" has been wrongly used for many years. It is the plural form of "datum"...  but many information says "the data is corrupt"...

I have to agree with the other members,the number of subcategories is confusing me...

Hmmm... Me too (kind of :)), but I guess we'll get used to it. You know what they say: "Man is an adaptable animal".

So true  :)  (But some die while trying to adapt :D)

you will see in time that this is a much more better arrangement than before. I think that all of the moderator's can appreciate the new indexing, as in the old index one would have to be carefull when moving a solved algebra problem into the "Algebra - Solved" or "Solved Algebra Problems" sections (the first one was AS and the latter CP).
:D

Just wanted to mentioned that i appreciate the effort and time you're putting in this project.Thanks go to all users,for keeping this forum alive.

"But some die while trying to adapt "

I'm sure it's not the case here :D:D

Small request: can the flags be made bigger?  I think they looked better when they were slightly bigger...

I thought (and tried) about different size versions, but I have reached the conclusion that this is the best size (fits in nicely with the rest of the topic display) ...


Welcome to DMAT, a new competition created by gfour84 and his friend Caelestor. [b]This is Round 1, taking place over the week of 7/7-7/19. [/b]You can read all about the rules, prizes, problems, points, etc in the attached pdf. Listed below are the goals:

(1) We wish to hone our problem solving and creating skills via writing and solving our questions (and testing their difficulty with your help!).

(2) By creating this friendly competition, we wish that everyone will take pride in, and have fun doing these questions over the summer holidays and perhaps continue to do them on into the school year.

The first two questions count for points, but, they are pretty easy because we are taking this first week as sort of a practice round.  Be sure to read the attachment!

Here you go, let's start then with 2 rather simple problems (sorry that this was posted one day later than the other place)! (Try not to miss the first problem :maybe: )

[b]1-A.[/b] (2 pts) Find the product of the four smallest prime factors of $ 62^{32} \minus{} 29^{32}$.

[b]1-B.[/b] (6 pts) Consider the following functions:
$ f(x \minus{} y) \equal{} g(x)g(y) \plus{} f(x)f(y)$
$ g(x \minus{} y) \equal{} g(x)f(y) \minus{} f(x)g(4)$ where $ x \not \equal{} 4$.

Given $ f( \minus{} 4) \equal{} 15$ and $ g(4)$ is a non-zero real, compute the numerical value of $ \left(f(2)\right)^{2} \plus{} \left(g(2)\right)^{2}$. Partial credit will be awarded to those that submit nontrivial statements (for example, equalities which are useful in solving the problem).

[b]Please submit answers to gfour84 or Caelestor by 12:01 A.M. EDT, Thursday 7/9 (9:01 P.M. PDT, Wednesday 7/8)[/b][/quote]

Hello, a reminder that answers to the first 2 problems are due in about 24 hours. Since quite a few have already solved these questions, you may continue on and solve the next problem.

[b]1-C.[/b] (4 pts) How many integers between 1 and 1000 inclusive do not satisfy $ x\minus{}\lfloor\frac{x}{3}\rfloor \equal{}\lfloor\frac{3x}{4}\rfloor\minus{}\lfloor\frac{x}{12}\rfloor$

Please feel free to discuss the DMAT on this thread! Anyone got any comments, opinions, or suggestions for improvement?

The reason why I decided to also post gfour's competition here is that there are a lot of tourneys (and people) active here. Know where I can find an intermediate touney forum?

Of course, if you desire, I can also give easy questions. (Perhaps a second version).

Eh I think it's fine as there is no highschool tourney forum,and aops is probably the only site where people will take a contest

Hello everybody, let's take a look at the scorings so far.

[hide="1-A"]
Scoring Rubric: 1 point awarded for 3 correct prime factors; 2 points awarded for all correct prime factors. [hide="Solution:"]
Yes, it's just factoring of squares: $ (62^{16} \plus{} 29^{16})(62^8 \plus{} 29^8)(62^4 \plus{} 29^4)(62^2 \plus{} 29^2)(62 \plus{} 29)(62 \minus{} 29)$.

We notice $ 91 \equal{} 7*13$ and $ 33 \equal{} 11*3$ are factors.
[hide="Trick Answer:"]
$ 3*7*11*13 \equal{} 3003$. Did you forget some prime factors?
[/hide]
[hide="Correct Answer:"]
$ 3*5*7*11 \equal{} 1155$. Also note that $ 5|(62^2 \plus{} 29^2)$.
Note: this is not mean at all; you should've expected it.
[/hide]

[/hide]
[/hide]
[b]1-B Announcement[/b]
Apparently there's a lot of confusion (in fact, I realized a typo, so sorry about that), so I've decided to extend the deadline for this problem by two days. Try it again:

Consider the following equations:
$ f(x \minus{} y) \equal{} g(x)g(y) \plus{} f(x)f(y)$, where $ f( \minus{} 4) \equal{} 15$
$ g(x \minus{} y) \equal{} g(x)f(y) \minus{} f(x)g(4)$, which is not defined when $ x \equal{} y \equal{} 4$

If $ g(4)$ is a nonzero real, compute the numerical value of $ \left(f(2)\right)^{2} \plus{} \left(g(2)\right)^{2}$. [b]Now due 12:01 A.M. EDT Saturday 7/11 (9:01 P.M. PDT Friday 7/10).[/b]

[hide="Current Scores"]
abcak: +1
Bugi: +2
Brut3Forc3: +2
cyanredyellow: +1
mercenary90: +2
narik: +1

[/hide]
Question 1-D to come soon.

[b]1-D.[/b] (4 pts) There exist quite a few positive integers m and n such that $ \frac {2}{m} \plus{} \frac {3}{n} \equal{} \frac {1}{16}$. What is the sum of all possible odd values of n?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1556318#1556318]Just go here for updates from now on.[/url] This link will be updated daily.

[quote="rd5493"]since this is on the middle school tourneys forum, i think you should maybe make the problems a little easier.  i mean, there's no need to post the same problems as on the intermediate forum for 6-8th graders[/quote]

There were tries to make High School Tournaments, but it left on, put them in MST. So no need for easier problems

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288549]Sign up for Round 2![/url] (Use the link given in a previous post to access Round 1.

DMAT Round 2 is now up! Click the link in my signature for the problems.

Edit: Sorry about that.

um you probably have to click on attach signature before you post


Enter here ____________ an algebraic expression that gives the area of a circle with a circumference .

Thanks that makes sense now that i see it

:huh:  I'm pretty sure what I did was right.

[quote="tjhance"][quote="1337h4x"][hide]
$A=\pi\left(\frac{C}{2\pi}\right)^{2}=\frac{C^{2}}{4\pi}$[/hide][/quote]

[hide="should be"]
$\frac{C^{2}}{4\pi^{2}}$
[/hide][/quote]

Wtf?

oh i'm sorry! :wallbash: :wallbash: :wallbash:

i didn't notice the other $\pi$

That is a very important formula! Thanks!

But...If you were given the area $A$, would the circumference be $\frac{2A}{r}$?

um
yeah

according to the weird people up above, this message is too small for their picky minds
go figure

This should be in the middle school forum...


What prize rank did you earn? Gold, silver, bronze, honorable mention, or none? Vote here.

I got an honorable mention with a 55. :( Darn I missed bronze by 5 points, which I shoulda gotten if it weren't for the stupid mistakes I made on the three #1 counting and probability problems from Rounds 1-3 which were ridiculously easy for me. :(

so close...73

so i guess i'm stuck w/bronze again :(

eye gots no prises, cuz i got a 4 (1/0/1/0)  :D

no but seriously, i got a 37 21/16/0/0. I forgot to do round 3 and i totally bsed round 2. I had a solution for #1 on round 4, but i forgot to submit it on time. Dang, i would've gotten a 42. oh well.

Anyone know what the exact list of prizes will be?

[quote="Yongyi781"]Hmmmmm.... silver with 81.......... yeah.[/quote]
what does this mean precisely?
yes silver as well

19/13/15/22=bronze
if i had lost round 2, i would be 1 point from silver

[quote="funtwo"]19/13/15/22=bronze
if i [b]had [/b]lost round 2, i would be 1 point from silver[/quote]

i assume you mean hadn't?

no, i didnt lose round 2, and i didnt do good on it. so im saying, if i had lost round 2, i would have done better

I lied. The round 2 debacle pushes me up to a silver!

20 16 21 = 57

20/13/14/15=bronze.

Yay?

r1+r3+r4=49 still bronze.

0/0/0/0 (never took it)

btw how hard are the problems

Well, I got 100 usamts /6 aime last year...

The thing is that you can use anything for usamts with essentially no time limit whereas aime is time- and resource- limited. So it's much easier to make stupid mistakes on aime than on usamts (though this year I made 3 mistakes on usamts versus 0 mistake on aime).

Despite the differences between USAMTS and AIME as pointed out above, I think there's still a pretty strong positive correlation between scores in the two contests. I had 56 USAMTS (three rounds)/4 AIME last year and 98 USAMTS/10 AIME this year.

Freshman year: 96 USAMTS 4 AIME
Sophomore year: 97 USAMTS 12 AIME

Of course, if you score better on the USAMTS, you're probably generally better at math. I just thought my scores were an isolated case until now.

[quote="hurdler"][quote="123456789"]Gold with 96 (and 4 on AIME). yay[/quote]

How do you do so well on usamts but not as well on aime?[/quote]

I just messed up really bad this year. I got a 7 on AIME last year. This year, I scored 112.5 on AMC 12...a very down year.

That's what happens if you do too many olympiad problems. :(

60 silver in three rounds, lost round two

wow i feel like my score sucks compered to my AIME score of 11, maybe i should stop putting it off until the last day and actually use resources like you're allowed to

That wasn't nice.

I missed the bronze by one point.

0 on USAMTS(never took it) first year AIME: 8 or 9(not sure). But I made USAMO.

[quote="BOGTRO"]That wasn't nice.[/quote]
well he was saying that to feel better about his USAMTS score vs. his skill level so idt he minded

How do you prepare for USAMTS and is it all proofs or some applied math problems?

It's all proofs.  If you want to see the types of problems that come up, check out some of the previous years contests on the USAMTS website.  That's a good way to prepare, too.

Of course, preparation isn't all that necessary because you have a month to do each set...


linear algebra
matrix
geometry
geometric transformation
rotation
algebra
polynomial

"A square matrix A is called nilpotent if A^n =0 for some positive integer n.  What can you say about the eigenvalues of a nilpotent matrix?"

The eigenvalues of A are all zero.  A matrix is nilpotent iff its eigenvalues are zero.

Am I right?

Almost- be careful on the "if" in that last statement. Over some fields, a matrix which isn't nilpotent may have no nonzero eigenvalues.

Jmerry, forgive me but can you give such an example of a nonnilpotent matrix with only zero as eigenvalue

ooow, or do you mean : the polynomial only has zero as roots in this field, but over a splitting field there are others

what a nasty subtlety in that case :(