quadratics
abstract algebra
vector
group theory
algebra proposed
algebra

a) Find two sets $X,Y$ such that $X\cap Y =\emptyset$, $X\cup Y = \mathbb Q^{\star}_{+}$ and $Y = \{a\cdot b \mid a,b \in X \}$.

b) Find two sets $U,V$ such that $U\cap V =\emptyset$, $U\cup V = \mathbb R$ and $V = \{x+y \mid x,y \in U \}$.

b) Let $U$ be the set of all numbers which in base three have a 1 after the decimal point.

a) we'll see...

a) Let $p$ be an odd prime.  Let $X$ be the set of all $\frac{a}{b}$ (in simplest form) where one of $a, b$ is a quadratic residue $(\mod p)$ and the other isn't, or one of $a, b$ is divisible by $p$ and if the second is a residue then the part of the first which we get by fully dividing out $p$ is a nonresidue, or vice versa. That $X$ and $Y = \{xy|x, y \in X\}$ are disjoint results from the fact that the multiplication among residues and non-residues follows the same rules as $+$ and $-$ signs follow in multiplying real numbers.

That $X$ and $Y$ together make up all positive rationals results from the fact that any non-residue can be written as the product of a residue and a non-residue--namely, as the product of itself and 1--and any residue can be written as a product of two residues--namely, itself and 1, again.

To put them both in the same bucket:

Both $(\mathbb Q_+^*,\cdot)$ and $(\mathbb R,+)$ are free monoids with infinite bases: the former over $\mathbb Z$ (it's the free abelian group generated by the primes), and the latter over $\mathbb Q$ (i.e. a vector space). For a free module having an infinite basis over a ring take any basis $(a_j)_{j\ge 1}$, and take $X$ to be the set of elements which have an odd number of non-zero coefficients when written in the basis $(a_j)_j$, and $Y$ the set of elements with an even number of non-zero coefficients.


Hello everyone,

I'm from Italy (I know you can guess it from the flag  :D ) and I wanted to know some little informations about studying in the USA (I mean at college) cause here we've got a great confusion  :blush: .

First of all is there a fixed period of time to graduate? Here in Italy is of five years.

When do you need to apply (how long before) to get admitted?

What sort of admission tests do you need to pass?

If anyone can write more pecise infos about any university in particular he or she attends I would be very grateful.

Thanks a lot!

Bye  ;)

Most United States courses for an undergraduate (bachelor's) degree take four years to complete. But if you need more time, you are usually allowed to take more time (I did). The two main college entrance examinations are both privately administered (not set up by the government). They are the [url=http://www.collegeboard.com/student/testing/sat/about.html]SAT[/url] and the [url=http://www.actstudent.org/index.html]ACT[/url]. Applicants who did not receive their education in an English-speaking country usually have to take a test of English language proficiency, such as the [url=http://www.ets.org/toefl/]TOEFL[/url].

Always, your secondary school records will be evaluated by college admission officers. Routinely, you will need to submit letters of recommendation from former teachers. Often, you will be interviewed by an alumnus of the college as part of the application process. United States college admission is much less based on a single selection factor than in most other countries.

Separately from applying for admission, you will apply for financial aid. Sometimes very good financial aid is available for international students. Usually both admission and financial aid are more competitive for international students than domestic students.

Usually students apply during their last year of secondary education, with application deadlines coming in November (for the early round of application decisions) or January (for the late round) before the next school year begins. (United States school years for most colleges begin in August or September.) Financial aid applications usually must be completed by late January, just after the student's family has income tax information for the preceding calendar year. Be careful to check deadlines for the colleges you like--some application deadlines are earlier, and other colleges give highest priority to applicants who apply the earliest (what is called "rolling admission").

I hope this helps answer your first questions. Please feel free to ask follow-up questions as needed.

Buona fortuna.

Thanks for the answers!!

Just a few more questions... :blush:

Is it possible to achieve the bachelor taking less than four years? (I think it's not so but I ask you as a confirmation!)

You said that SAT and ACT are administered privately, you mean they are set up by universities right? I've read that you can take these tests more than one time, is it so? In any case after sending the application I will be contacted for the dates?

Sorry for my bad english and thanks again for the help!!  :D

Bye

[quote="XT"]Is it possible to achieve the bachelor taking less than four years?[/quote]

That's possible. It's unusual, but possible.

[quote="XT"]You said that SAT and ACT are administered privately, you mean they are set up by universities right?[/quote]

The organizations that run the tests are independent of the universities, but each university decides to accept those test results as one part of how it evaluates a candidate for admission.

[quote]I've read that you can take these tests more than one time, is it so?[/quote]

Yes. That is quite common.

You can go to the Web sites of the testing organizations, linked to in my last reply, to find the test dates.

Just to add to  the above:

SAT tests can be taken in Italy (see collegeboard.com),  you  can search for sites in Italy for any particular date..

Also look at the individual schools websites , you will find that they are quite useful.

Again, a little question!!  :D

Surfing through the pages of some universities and looking for example at the tests they require I often find something like "For the class of 2011..." or "..students enrolling in 2008...". Does it mean that admssions to the previous years have already been done?

I know it may seem a stupid question but I'm a little confused  :blush:

Many thanks to anyone who's answering!!

 :lol:

Class of 2011 for instance means that you are supposed to graduate in 2011.

:blush:  :blush:

I should have guessed it...

Thanks !

I'm sorry again but I have some more questions.. :D

Does anyone know which arguments covers the Mathematics Sat (I and II)? And I suppose the best to prepare for the test is to buy apposite made books, isn't it? And what about the Physics Sat? Do you know anything?

Thanks a lot boys (and girls)!!

Bye!  :D

by the way, do you think that medals in the national MO will help?

Medals in national mathematical olympiads help, but they do not guarantee admission. Essentially NOTHING guarantees admission to the top United States universities, because they have "holistic" admissions, which means they can accept or reject a candidate for any reason or no reason at all other than lack of space for all applicants.

Replying to the earlier question in this thread, both the SAT II Level 1 and SAT II Level 2 are tests strictly of precalculus mathematics, rather easier than olympiad mathematics.


These questions are in Section 9 of the RocketReview "Ultimate Guide to the New SAT" practice SAT.

Questions:
7. Having planned the event far in advance, [u]that the weather forced its cancellation disappointed the reunion committee greatly[/u].
(A) that the weather forced its cancellation disappointed the reunion committee greatly
(B) the weather's forcing its cancellation was a great disappointment to the reunion committee
(C) the reunion committee's disappointment at the weather's forcing its cancellation was great
(D) the reunion committee found the weather's forcing its cancellation greatly disappointing
(E) the reunion committee was greatly disappointed in the weather forcing its cancellation

I chose (D), because I thought "disappointed in" was incorrect. Shouldn't "by" follow "disappointed"? The answer is apparently (E).

14. [u]Although Ireland's government censors had long stifled artistic expression, the country is now discovering the tremendous talents of its native poets and novelists.[/u]
(A) Although Ireland's government censors had long stifled artistic expression, the country is now discovering the tremendous talents of its native poets and novelists.
(B) Artistic expression that was once being stifled by Ireland's government censors, but now they are discovering the tremendous talents of their native poets and novelists.
(C) The government censors had long stifled artistic expression, but Ireland is now discovering the tremendous talents of their native poets and novelists.
(D) Artistic expression was long stifled by Ireland's government censors, and so now they are discovering the tremendous talents of their native poets and novelists.
(E) Now discovering the tremendous talents of its native poets and novelists, Ireland's government censors had stifled artistic expression for a long time.

I think it is (A). The answer is (C). But isn't the use of "their" wrong in (C)? Shouldn't it be "its" if the sentence is referring to Ireland? and if it's referring to the government censors, isn't the sentence really unclear?

In 7D, "its" is ambiguous, referring to the weather or the event. In 14A, "the country" cannot "discover." People discover things, not places.

Okay I guess I see the argument for #14.
But for #7, choice (E) has the same problem, and (E) is the correct choice..


Anyone know a good way to learn all of the weird grammar things that might show up on the SAT Writing?

Being a homeschooler, grammar is not really a part of the English courses I take so that's probably a problem. The grammar from public middle school usually gets me a 75-78 on the PSAT, so I'm basically just looking for something to brush up the extra stuff that I might've missed.

EPGY has a very useful grammar course, [url=http://epgy.stanford.edu/courses/english/EG20/]EG20 Grammar and Style of the Sentence, Part 1[/url]. There may be a follow-up course to that pretty soon. And EPGY has good financial aid for middle-class families, so don't let the list price scare you.

There are some very readable and funny books about English grammar and style, such as [url=http://www.amazon.com/exec/obidos/tg/detail/-/1592400876/learninfreed]Eats, Shoots & Leaves: The Zero Tolerance Approach to Punctuation[/url], which was a national best-seller in Britain.


find all [b]m,[/b][b]n[/b] in natural number such that [b]m 3+n[/b] and [b]n 3+m[/b] divisible by [b]m 2+n 2 [/b].

Asumme that $m,n$ satisfy the statement of the problem.
Let $m=dx$ and $n=dy$ where $x,y$ are coprime positive integers.
Then $\frac {d^2x^3+y} {d(x^2+y^2)}$ is an integer. It follows that $d$ divides $y$. Similarly $d$ divides $x$, so that $d=1$, which means that $m,n$ are coprime.

Now :
$\frac {m^3+n} {m^2+n^2} = m - \frac {n(mn-1)} {m^2+n^2}$ which forces $\frac {n(mn-1)} {m^2+n^2}$ to be an integer. Since $\gcd (m,n) = 1$ it follows that $\frac {mn-1} {m^2+n^2}$ is a non-negative integer.
But sinc $2mn \leq m^2+n^2$ we deduce that $0 \leq \frac {mn-1} {m^2+n^2} < 1$.
Then $0 = \frac {mn-1} {m^2+n^2} $, which leads to $mn=1$ and $m=n=1$.

Conversely, it is clear that $(1,1)$ is a solution.

Pierre.


If $a,b,c$ be nonnegative real numbers such that $a+b+c=1$ then
\[\frac{a}{\sqrt{\frac{1}{b}-1}}+\frac{b}{\sqrt{\frac{1}{c}-1}}+\frac{c}{\sqrt{\frac{1}{a}-1}}\le \frac{3\sqrt{3}}{4}\sqrt{(1-a)(1-b)(1-c)}\]
:)

I have a nice solution for it :) Hope you so.

It is nice problem :P
By expanding expression we have:
\[\sum \sqrt{\frac{a^{2}b}{(a+c)^{2}(a+b)(b+c)}}\le \frac{3\sqrt{4}}{4}\]
Without loss of generally, assume that: $a=min(a,b,c)$ and therefore we will prove with two case:
[color=green]Case 1:[/color] $a\le b\le c$
[color=green]Case 2:[/color] $a\le c\le b$
If you want to prove above two case, you will use
\[\frac{ab}{(b+c)(c+a)}\le \frac{ca}{(a+b)(b+c)}\le \frac{bc}{(a+b)(c+a)}\]
Aplly in case 1. And similar to case 2.

[quote="zaizai-hoang"]It is nice problem :P
By expanding expression we have:
\[\sum \sqrt{\frac{a^{2}b}{(a+c)^{2}(a+b)(b+c)}}\le \frac{3\sqrt{3}}{4}\]
Without loss of generally, assume that: $a=min(a,b,c)$ and therefore we will prove with two case:
[color=green]Case 1:[/color] $a\le b\le c$
[color=green]Case 2:[/color] $a\le c\le b$
If you want to prove above two case, you will use
\[\frac{ab}{(b+c)(c+a)}\le \frac{ca}{(a+b)(b+c)}\le \frac{bc}{(a+b)(c+a)}\]
Aplly in case 1. And similar to case 2.[/quote]
Show the proof, Hoang! My solution is much difference and I think it's very nice! :)


How many different triangles are in the problem?
[asy]size(70);
pair A = (0,0), B = (7,0), C = (3,6), D=(3,0), E=(5,3),F=(1.5,3);
draw(A--B--C--A,linewidth(0.6));
draw(D--E--F--D,linewidth(0.6));[/asy]

We see $ 1$ large triangle and $ 4$ small triangles, so there are a total of $ \boxed{5}$ triangles.


Find the equation of the tangent line to the graph of $ f(x)\equal{}\frac{4x\minus{}2}{x\plus{}1}$ at the point at which $ x\equal{}0$

[quote="ben_b"]Find the equation of the tangent line to the graph of $ f(x) = \frac {4x - 2}{x + 1}$ at the point at which $ x = 0$[/quote]

$ y=\frac{4x-2}{x+1}
\\
\\
\; \text{General Equation of line is: } (y-y_1)=m(x-x_1)
\\
\\
\;
m=\frac{dy}{dx}|_{x=0}=\frac{4(x+1)-(4x-2)}{(x+1)^2}=\frac{6}{(x+1)^2}|_{x=0} = 6
\\
\\
\\
\;\text{Substituting in the abouve general equation: }
\\
\\
\boxed{\color{red}6x-y-2=0}$


Is it a^2+b^2+c^2=3abc? There is a "/" there which confused me.

Do you mean  \sum ab-2c-2 \geq 9/(a + b + c) ?

I think that the fact that there is a "/" is just that insead of typing "." he just typed "/". Those keys are really close! :D :D

cheers! :D

We rewrite it like this:

 \sum (a^3/a^2b^2c^2) >= 9/ \sum a iff ( \sum a^3)( \sum a) >= 9a^2b^2c^2=( \sum a^2)^2, which is true by Cauchy.


Let $ x,y,z>0$ satisfies $ (x\plus{}y\minus{}z)(\frac{1}{x} \plus{} \frac{1}{y} \minus{} \frac{1}{z}) \equal{} 4$. Find the minimum value of $ P \equal{} (x^4 \plus{} y^4 \plus{} z^4)(\frac{1}{x^4} \plus{} \frac{1}{y^4} \plus{} \frac{1}{z^4})$.

hello, we have $ P\geq2304$ and the equal sign holds for $ x\equal{}1,y\equal{}\frac{1}{2}(7\minus{}3\sqrt{5}),z\equal{}\frac{1}{2}(3\minus{}\sqrt{5})$
Sonnhard.

[quote="Mashimaru"]Let $ x,y,z > 0$ satisfies $ (x \plus{} y \minus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \minus{} \frac {1}{z}) \equal{} 4$. Find the minimum value of $ P \equal{} (x^4 \plus{} y^4 \plus{} z^4)(\frac {1}{x^4} \plus{} \frac {1}{y^4} \plus{} \frac {1}{z^4})$.[/quote]

\[ 4 \equal{} \left( {a \plus{} b \minus{} c} \right)\left( {\frac{1}{a} \plus{} \frac{1}{b} \minus{} \frac{1}{c}} \right) \equal{} \left( {a \plus{} b} \right)\left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \minus{} \left( {\frac{{a \plus{} b}}{c} \plus{} c\left( {\frac{1}{a} \plus{} \frac{1}{b}} \right)} \right) \plus{} 1\]
\[ \le \left( {a \plus{} b} \right)\left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \minus{} 2\sqrt {\left( {a \plus{} b} \right)\left( {\frac{1}{a} \plus{} \frac{1}{b}} \right)}  \plus{} 1\]
\[ \Rightarrow \left( {a \plus{} b} \right)\left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \ge 9,\]
by Cauchy:
\[ \left( {a^4  \plus{} b^4  \plus{} c^4 } \right)\left( {\frac{1}{{a^4 }} \plus{} \frac{1}{{b^4 }} \plus{} \frac{1}{{c^4 }}} \right) \ge \left( {\frac{{a^2 }}{{b^2 }} \plus{} \frac{{b^2 }}{{a^2 }} \plus{} 1} \right)^2\]
\[ \equal{} \left( {\left( {\left( {a \plus{} b} \right)\left( {\frac{1}{a} \plus{} \frac{1}{b}} \right) \minus{} 2} \right)^2  \minus{} 1} \right)^2\]
\[ \ge \left( {\left( {9 \minus{} 2} \right)^2  \minus{} 1} \right)^2\]\[ \equal{} 2304\]

[quote="Mashimaru"]Let $ x,y,z > 0$ satisfies $ (x \plus{} y \minus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \minus{} \frac {1}{z}) \equal{} 4$. Find the minimum value of $ P \equal{} (x^4 \plus{} y^4 \plus{} z^4)(\frac {1}{x^4} \plus{} \frac {1}{y^4} \plus{} \frac {1}{z^4})$.[/quote]
It's an old Vasc's problem.


geometry ratio Princeton college LaTeX logarithms

So what's the chemistry olympiad like?  the regional competition's early in april.  Anyone familiar with the competition know a good way to prepare?

no the regional is NOT in early april!!
the dead line for the local is end of march ;)

do you mean USNCO;)?  If so, I can tell you a bit about what I did last year (I made it to the camp, so I must have done something right).  Make very good friends with the tests posted on the USNCO website.  Practice them under timed conditions and with the (scientific) calculator you will be using during the real deal.  I studied out of both a gen. chem textbook and the AP Princeton Review book, and both helped enormously.  the textbook was great to learn new concepts, and the Princeton Review book helped refresh old ones.  Oh and of course if you come across any problems that you just don't know how to do, go ahead and ask your chemistry teacher or post them on here.

My local section exam is set for March 29th.

well i have already taken the exam and on the bottom of the page of questions it says:

"Not valid for use as an ACS Olympiad Local Section Exam after March 28, 2006."

so i don't know how you are taking it that late!

because the exam we take is [i]not an ACS exam[/i] :!:  :mad:  :rotfl: They probably just make it up.  How many different tests are there before nationals and how much do we need to know?

btw, my local is April 7th, the day before the ACT (priorities...?)

really?? i thought the ACS (American Chemistry Society) was the only place responsible for USNCO (wait you are talking about United States Chem Olympiad right?!?)

anyways, my apologies

after the Local exam there is the National Exam (about 8 ppl from each region take it), after the Nationals they select people for the summer camp, and then they select the ICHO team

for prepration i would recomend reading both volumes of General Chemistry by Charles Mortimer, as they are very very help full and similar kind of questions that they ask.

In terms of the ACS local exams, they are only one way to select a region's qualifiers for the national exam.  Other possible methods include recomendations or interviews.  A section may send as many qualifiers to write the national exam as the ACS permits, based on the section's student population.

:blush: my bad
[url=http://www.cbu.edu/sciences/chemistry/Exams.htm]http://www.cbu.edu/sciences/chemistry/Exams.htm[/url]

well, i didn't remember "ACS" on our exam last year, apparently, but it is the memphis chapter of it that is responsible.  Still, I think i was partly misled because of the exam date you guys are talking about.  8 ppl [i]will[/i] advance from this one, but it is being held the 7th.

http://www.chemistry.org/portal/a/c/s/1/acsdisplay.html?DOC=education/student/co01.html

[quote]Local sections select their nominees by various means including: the USNCO local section exam, a locally-prepared exam, laboratory practicals, teacher recommendations, [b]or[/b] other regional events with competitive activities among school teams.[/quote]

Weeeeeeeeeeeeeeeeeeeeeeee....
I took it today... it was fun  :)

most of the results are in
so who made it to nationals??

i got first place in my region out of 115 ppl. :)

Wow, that is awesome  :)  What was your score? Anyways, my school didn't get our scores yet. I am very excited though.

i think i got like a 56

i know i missed that question on solution that dissolves gold
and i know i missed that one question about which organic substance is used to flavorise stuff
i liked the questions in general, although i liked last years better, mainly because i got a higher score (59 :) ), but i think some of the questions this year weren't at standard level (not saying that they were hard, cause they weren't at all, but i just didn't like some of them)

Here's what I (tried) to do for that HCN problem:

Kp = 7.8 = [HCN]/ P_HCN
The main problem with this is that I don't know if this is the correct expression for the Kp. This really doesn't match up with the Kp = Kc * (RT) ^ n

At pH of 7, H+ = 1.0 * 10^-7, so [HCN] = ~ 0.10 M or 0.099 M (they gave us that [HCN] + [CN-]  = 0.10 M)

Plugging this in, we get P_HCN = 0.013 atm

0.013 atm / (300. * 10^-6 atm) = ~ 43:1 ratio

I know of only one other kid in my region who did it this way (or at least got this answer).
Hopefully, it was close to the right answer though, or something...
Anyways, here's a picture of the actual graph:
http://www.nizkor.org/features/techniques-of-denial/images/weightp.JPG

So, 0.10 moles of HCN in approximately a 1-Liter solution = 0.10 * (27 g/ mol) = 2.7 g
~55 * 18 = 990 g of water

2.7/992.7 * 100% = 0.27%by weight?

I think the answer to the allotrope question was silicon (I don't remember the question entirely, but I do recall putting down tin, then going home and finding out that tin wasn't right 'cause apparently it exists as some sort of allotrope at that temperature range).

I'm pretty sure the Ce(IV), Fe(II) acid redox does yield Ce(III) and Fe(III) (from looking at the oxidation numbers of Ce, anyway. hehe I put Ce(II) but oh well).

And um, about the H3NO4 vs. H3PO4 question, I put that the nitrogen atom is too small to accomodate 4 oxygen atoms surrounding it-- not sure if that's right or not, but that was the only reason I could think of.

*sigh*.

For the Cerium one though, why acidic solution?  I couldn't find a role for the acid to play... catalyst maybe?

Hm, maybe the H+ became water? I don't remember what I wrote...

It appears that all four (silicon, tin, phosphorus, and sulfur) elements have allotropes in that temperature range.

I hope every single one of us gets a phone call tomorrow!  :P

Congrats!  ;) What time did you get the call?

thanks!

at lunch... haha... that is... lunch for me... i'm on pacific standard time

Ooh, that's awesome. Three people from my section made it to camp this year. (I wasn't one of them though =\) Congratulations to everyone that makes camp!

THREE people?  Do you happen to be in the New Jersey area or are you in the Southern California section?... lol... i guess i could be wrong with both... but those seem most probable to me.

Oh man, congratulations!

Just wondering... did ACS call you, or did your local coordinator do it?

Yep, I'm in the North Jersey Section.  ;)


Prove that: x+y+z<= 2 + xyz for all real number x,y,z satisfy: x2+y2+ z2 =2

This is an IMO 1987 shortlist problem. Here is a hint:

xy<=1, xz<=1 and yz<=1. Hence

(x+y+z-xyz)^2 - 2= 2(xy+xz+yz)-2(x+y+z)xyz+(xyz)^2<=2.


linear algebra
matrix
Gauss
algebra unsolved
algebra

From the second equation, $ x\equal{} \pm \sqrt{2y^2 \plus{} 5}$, and we can take the positive root since if $ (x, y)$ is a solution then so is $ (\minus{}x, \minus{}y)$.

Substituting into the first equation, we get

$ (2y^{2} \plus{} 5)\sqrt{2y^{2} \plus{} 5} \minus{} 8\sqrt{2y^{2} \plus{} 5} \equal{} y^3 \plus{} 2y$

$ (2y^{2} \minus{}3)\sqrt{2y^2\plus{}5} \equal{} y^3 \plus{} 2y$

$ (4y^{4} \minus{} 12y^{2} \plus{} 9)(2y^{2} \plus{} 5) \equal{} y^{6} \plus{} 4y^{4} \plus{} 4y^{2}$

$ 7y^{6} \minus{} 8y^{4} \minus{}46y^{2} \plus{} 45 \equal{} 0$.

It seems that, by Gauss' theorem on rational roots, the real roots are irrational.

Using an equation solver, I get the real roots as $ y\equal{}\pm1.64192$ and $ y\equal{}\pm0.975671$.

Solving for $ x$ in the same way, I get $ 11x^{6} \minus{} 185x^{4} \plus{} 757x^{2} \plus{} 5 \equal{} 0$ with roots $ x \equal{} \pm3.125466$ and $ x\equal{}\pm2.656358$.

So the (approximate) solutions are $ (\minus{}2.656358, 0.975671)$, $ (\minus{}3.125466, \minus{}1.64192)$, $ (2.656358, \minus{}0.975671)$ and $ ((3.125466, \minus{}0.975671)$.

hello, i have obtained only one real solution.
Sonnhard.

you are right Dr Sonnhard Graubner this equation has only one root


AMC AIME geometry circumcircle USA(J)MO USAMO inradius

In triangle $ ABC$, $ AB\equal{}13, BC\equal{}14,$ and $ AC\equal{}15$. Circle $ D$ is drawn such that it is internally tangent to the circumcircle at $ A$ and externally tangent to the incircle. Given that $ AD\equal{}m/n$, where $ m$ and $ n$ are relatively prime, find $ m\plus{}n$

Okay, first let's clear up some notation.

Let the circumcircle of $ ABC$ be $ \Omega$ and let the incircle be $ \omega$.  This is what I'm reading.

Circle $ \delta$ with center $ D$ is drawn so that it is internally tangent to $ \Omega$ at $ A$ and externally tangent to $ \omega$.  Given that $ AD \equal{} \frac{m}{n}$ blah blah blah find $ m\plus{}n$.

Is that what you're saying?

[quote="Altheman"]Use the same inversion in USAMO 2007 #6...The computation is still a bit messy...[/quote]
What's that?

[hide="BIG HINT"] Draw a triangle with vertices at $ A$, the circumcenter, and the incenter. Then draw a cevian to $ D$. [/hide]

Whats your solution? That still seems like a mess...

[hide="BIG HINT (ctd.)"] Then apply Stewart's theorem, and simplify. It all clears up at the end[/hide]

[hide="what's wrong with this? bad answer"]
Designate $ O$ as the center of the incircle. Drop a perpendicular from $ O$ to $ AC$ and call the intersection $ R$. Since the inradius is $ 4$, $ OR\equal{}4$. Taking a walk, we have that $ AR\equal{}7$. Since $ A$, $ D$, $ O$ and the point of tangency between the two circles, $ P$, are collinear (because of the tangency), by Pythagoras we have $ AD\equal{}\frac{AP}{2}\equal{}\frac{AO\minus{}OP}{2}\equal{}\frac{\sqrt{65}\minus{}4}{2}$.
[/hide]

[quote="worthawholebean"][hide="what's wrong with this? bad answer"]
Designate $ O$ as the center of the incircle. Drop a perpendicular from $ O$ to $ AC$ and call the intersection $ R$. Since the inradius is $ 4$, $ OR \equal{} 4$. Taking a walk, we have that $ AR \equal{} 7$. Since $ A$, $ D$, $ O$ and the point of tangency between the two circles, $ P$, are collinear (because of the tangency), by Pythagoras we have $ AD \equal{} \frac {AP}{2} \equal{} \frac {AO \minus{} OP}{2} \equal{} \frac {\sqrt {65} \minus{} 4}{2}$.
[/hide][/quote]

I don't think we can say that A, D, and O are collinear. The circle that's tangent to incircle, circumcircle is kind of up, to the left/right of the incircle; A,D,O collinear only if the incenter and circumcenter are the same.

[quote="dnkywin"]And, the answer is a fraction in the form of $ m/n$, where $ m$ and $ n$ are relatively prime positive integers[/quote]

Actually, I got the same answer as worthawholebean, and it's not in that form of the fraction m/n.  :huh:

Oh, and, I don't think it's AIME #15 level difficulty  :P

I just tried this problem quickly, and I got $ \frac {5249}{2592}$.  If this is correct, I'll post my solution, but don't AIME problems usually have answers from 0 to 999?  This gives me $ 7841$...

EDIT: Actually, I just saw my mistake.  My answer is wrong...

let me just post my solution, and see if anything is wrong with it  :huh:
my answer's the same as worthawholebean's.

[hide]
Using the equation $ A \equal{} rs$, we find the incircle radius is $ r \equal{} 4$ ($ A$ being the area of the triangle $ ABC$ and using heron's to find it.) Let the center of the incircle be $ O$. $ AO$ is the angle bisector of $ \angle A$. Using $ a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc*\cos A \Longrightarrow \cos A \equal{} 33/65$, applying half angle formula, $ \sin A \equal{} \frac {4}{\sqrt {65}}$.
$ AO \equal{} 4 * \sin A \equal{} \sqrt{65}$. It follows that $ AD \equal{} \frac {\sqrt {65} \minus{} 4}{2}$
[/hide]

NOTE: A,D, and the incenter are NOT collinear!!!

[quote="Teki-Teki"]The use of Stewart's Theorem is correct.  However, you still need several trigonometric identities, some of which are nontrivial, as well as knowledge of triangle invariants to come up with a non-inversive solution.  The answer does end up being of the form m/n, however, because $ a,b,c,s,r,K,R$ are all rational in this triangle.

Edit: Also, this is most definitely not an AIME #15.  It's more like a ripoff of, as Altheman said, the 2007 USAMO #6, which was certainly not AIME difficulty.  For example, take a look at the 2007 AIME II #15 [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2007]here[/url].  You will find that it is [i]significantly[/i] easier than this problem.[/quote]
Actually, you can use coordinates to find the length of the third side of the triangle I was referring to. No trig is involved. That makes it much easier. So no, it is not much harder than last year's #15. Try it yourself.

A few preliminaries: $ a \equal{} 14$, $ b \equal{} 15$, and $ c \equal{} 13$. $ A \equal{} 84$, $ R \equal{} \frac {65}{8}$, $ r \equal{} 4$, $ OI^2 \equal{} R^2 \minus{} 2r \equal{} \frac {65}{64}$.

Okay so call the tangency point $ D$. Lets start with $ AO$ and $ AI$. We can compute these lengths: $ AO \equal{} \frac {65}{8}$, and $ AI \equal{} \sqrt {7^2 \plus{} 4^2} \equal{} \sqrt {65}$.

Now $ D$ is on $ AO$ and $ J$ is on $ ID$ such that $ IJ \equal{} r \equal{} 4$, and $ AD \equal{} JD \equal{} p$. Then we have to find $ x$. Suppose that $ IO \equal{} y$. Then by stewart's theorem on $ AIO$, cevian $ AD$,
\[ \frac {65}{64}p \plus{} (\sqrt {65})^2\left(\frac {65}{8} \minus{} p\right) \equal{} (4 \plus{} p)^2\cdot \frac {65}{8} \plus{} (p)\left(\frac {65}{8} \minus{} p\right)\left(\frac {65}{8}\right)
\]
so $ p \equal{} \frac {49}{24}$.

Okay so I will admit, this isn't that hard. I just didn't take the time to draw out a diagram. It is pretty intensive in terms of formula knowledge. The most important simplification is knowing the value of $ OI^2$.

In general, it is easy to see that the radius is
\[ \frac {Rx^2}{r^2 \plus{} 4rR\plus{} x^2}
\]
where $ x \equal{} s \minus{} a$.

Yeah, that's how I did it, and your answer is correct. I noticed that even if you used $ B$ or $ C$ instead of $ A$ $ m \plus{} n$ still equals $ 73$


How many functions $f: \{1,2,...,2005\}\longrightarrow\{0,1\}$ can be found such that the value of $\sum^{2005}_{i=1}f(i)$ is an even number?
How many functions like $f: \{1,2,...,2005\}\longrightarrow\{0,1\}$ can be found such that:
$3$ $|$ $\sum^{2005}_{i=1}f(i)$   ?


abstract algebra
modular arithmetic
superior algebra
superior algebra unsolved

Are groups
G=<x,y,z|x^5=y^5=z^4=[x,y]=1,x^z=x^2,y^z=y^2> and H=<x,y,z|x^5=y^5=z^4=[x,y]=1,x^z=x^2,y^z=y^3>  isomorphic? How to prove?

What is $ x^z$? Does it mean conjugation?

They are not isomorphic.
We have $ P\equal{}\langle x,y\rangle \cong C_5\times C_5$ the unique normal Sylow 5-subgoup of both $ G,H$.
In $ G$ we have $ g^z\equal{}g^2$ for all $ g\in P$.
So if $ G\cong H$ we must find $ w\in H$ with $ g^w\equal{}g^2$ for all $ g\in P$.
But $ H\equal{}P\rtimes \langle z\rangle$ with $ P$ abelian, so WLOG $ w\equal{}z^k$ for some $ k\in\mathbb{N}$.
But then $ x^2\equal{}x^w\equal{}x^{2^k}$ and $ y^2\equal{}y^w\equal{}y^{3^k}$ in $ H$ gives
$ 2\equiv 2^k\equiv 3^k\pmod{5}$ for some $ k\in\mathbb{N}$. Contradiction!


geometry
trigonometry
cyclic quadrilateral

What is the maximum area of a quadrilateral with sides 1,4,7,8.

[hide="a hint"]
If we have a convex quadrilateral $ ABCD$, with sides $ a,b,c,d$, and $ \angle A+\angle C=2\beta$, then:
$ S=\sqrt{(p-a)(p-b)(p-c)(p-d)-abcd\cos^{2}\beta}$, where $ 2p=a+b+c+d$[/hide]

Since it's possible to find a cyclic quadrilateral with side lengths 1, 4, 7, and 8, by Brahmagupta's, $ \sqrt{(10-1)(10-4)(10-7)(10-8)}=18$.


[hide="Solution"]
Let $ S_n \equal{} a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n$, next notice that $ a_{n} \equal{} S_{n} \minus{} S_{n\minus{}1}$ which is equivalent to,
\[ a_{n} \equal{} \dfrac{n(n \plus{} 1)(n \plus{} 2)}{6} \minus{} \dfrac{(n\minus{}1)n(n \plus{} 1)}{6}\implies a_{n} \equal{} \dfrac{n(n \plus{} 1)}{2}\]
Hence,
\[ \dfrac{1}{a_n} \equal{} \dfrac{2}{n(n \plus{} 1)} \equal{} 2\left(\dfrac{1}{n} \minus{} \dfrac{1}{n \plus{} 1}\right)\]
From where we get a telescoping series,
\[ \sum_{i \equal{} 1}^n{\dfrac{1}{a_i}} \equal{} 2\left(\sum_{i \equal{} 1}^n{\dfrac{1}{i} \minus{} \dfrac{1}{i \plus{} 1}}\right) \equal{} 2\left(1 \minus{} \dfrac{1}{n \plus{} 1}\right) \equal{} 2 \minus{} \dfrac{2}{n \plus{} 1} \equal{} \dfrac{2n}{n \plus{} 1}\]
Now just substitute $ 95$ for $ n$ and we have that,
\[ \dfrac{1}{a_1} \plus{} \dfrac{1}{a_2} \plus{} \cdots \plus{} \dfrac{1}{a_{95}} \equal{} \dfrac{2\cdot 95}{96}\equal{}\boxed{\dfrac{95}{48}}\]
[/hide]


geometry
3D geometry
sphere
tetrahedron
circumcircle
IMO Shortlist
IMO Longlist

Prove this proposition: Center the sphere circumscribed around a tetrahedron which coincides with the center of a sphere inscribed in that tetrahedron if and only if the skew edges of the tetrahedron are equal.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let $ABCD$ be the tetrahedron. $I$, its incenter, is projected on the four planes of the faces in the circumcenters of the faces. Let $O_A,O_B,O_C,O_D$ be the circumcenters of $BCD$ and so on. We clearly have $d(O_D,BC)=d(O_A,BC)\Rightarrow \angle BAC=\angle BDC$. From this equality and the like we find that the sum of the palne angles around each vertex is $\pi$ (call this assertion $(*)$).

Now take the faces $BCD,CAD,ABD$ and rotate them around $BC,CA,AB$ until they are on the same plane as $ABC$. It is now easy to see from $(*)$ that $ABC,CBD,ADB,DAC$ are congruent, and the conclusion follows.


Hi everyone im doing a survey for some school project, so i was hoping u guys could help me out. All u hav to do is tell me what u do most of the time(ex: gaming,reading, tennis) and take this simple IQ test
[url=http://www.ivillage.co.uk/ivillageuk/articles/0,,161171_182561,00.html]IQ test[/url]

plz post ure answer and ure IQ results if u dont mind :wink: Thanks!

Well , this forum does look fine for it.Anyway , I spend most of time reading something or the other.
As for the IQ test , I'll take it a little later and then post the result.
[hide]EDIT : IQ as per your site : 160.[/hide]

I like gaming and sports

IQ [hide="150"]139 :blush: [/hide]

That was a fun one.

I tend to hang out online alot on different forums.  I also like reading.
[hide="Score"]157[/hide]

Uh, I'm on the computer a lot.. or do the Rubik's cube.. or read, or do art, or play games, or play guitar. I used to play tennis.

[hide="score"]166[/hide]

I dont read usually, but i do art, spend a lot of time online and watch a lot tv. i havent played tennis in a year, but i used to play quite frequently.
score: 159

Usually I'm either doing something outside (sports, running) or reading or playing video games or something like that.

I hate IQ tests because they're inaccurate, but fine.
[hide]160[/hide]

[quote="ambierona"][quote="yif man12"]I hate IQ tests because they're inaccurate, but fine.[/quote]

The only thing IQ tests test is your ability to take IQ tests.  :D[/quote]

Hehe, so true.  :)

[quote="ambierona"]The only thing IQ tests test is your ability to take IQ tests.  :D[/quote]

Haha...I could make similar statements about the SAT :-) but oh well.

Anyways, I'm changing the title on this.

Score [hide]170[/hide]

but honestly, "People who brag about their IQ are losers."

--Stephen Hawking, most important theoretical physicist since Einstein, as quoted in the New York Times Magazine.

Oh and I play tennis, I ski during the winter, I play the piano, I do math, and I do homework.  I don't play too many video games.  I also read sometimes.  I also eat.  I also sleep.

[quote="bleumoose"]um... reading, hiking, and my job (which i dont really consider a "job") as a veterinary assistant. i got a 143. but that's loads too high, i think im reallly more like 125... and no offense to anyone here, but i think most of yours are probably a tad on the high side, as well. Mensa members only have to be over like, 150, is it? and that's pretty prestigious, isnt it? besides, i think the purpose of sites like these is to give you a really high IQ so that you will spend the money for their "detailed report" because you are so goshdang proud of yourse.f[/quote]

yesssss...i finally don't feel thaaat stupid

Yeah, I took this other IQ test on the internet a couple of years ago when I was smarter and I got like 30 less on my IQ.. and this one had questions that were a lot easier.

Woohoo a 162. Although this is not very trustworthy, since my official IQ is in the range 136-155.

[quote="paladin8"]I got 158... I've taken various IQ tests though, with scores ranging from 130-180, so I've learned not to trust them very much :P[/quote]

Hehe, not as big as my range.

90-190 Yay! Even though if you throw out the highest and lowest, you'd get 120-180. I don't even remember what happened the time I got 90.  :)  :?

you can't really trust IQ tests online... :D BTW nice range

activeites: gaming, reading, math, posting on AoPS  :lol:
score: 146. apperently that is low compared to the others ppl recieved. i think my real IQ is somewhere around 130.

That was a stupid test.

Hobbies: basketball, baseball, trivia, (yes, trivia is a hobby), math, uh... spending time with friends, card games(poker, bridge, euchre, spades, hearts, yeah I play a lot uh... cribbage as well)

Score: 164 <--- sooooo inaccurate.

You're right. They inflate your score. My real IQ lies somewhere in the 135-155 range.

[quote="GoBraves"]I got a 60 once because i stopped after 6 questions and pressed finished. It said people that get 69 or under are mentally retarded[/quote]

or they are smart enough not to waste anymore time on the test :D

yup they inflate the score way too much on this one, and the questions were pretty easy.. [hide]155[/hide] ..IQ tests are never accurate.. on the others i usually get from like 130-150.

I play basketball, football, baseball.. I bike, I game on computer & xbox, i read, i go to school, i listen to music, i post on AoPS.

Oh yeah I forgot, I play card games too. Like.. spades, hearts, communism, euchre, gin, etc.. And magic. with cards. or other stuff.


recurrence relation
factorial
number theory
binomial coefficients
IMO
IMO 1972

use the lagrange formula for the max power of $p$, p prime, dividing $n!$

and $\left\lfloor 2m/p\right\rfloor+\left\lfloor 2n/p\right\rfloor\geq \left\lfloor m/p\right\rfloor+\left\lfloor n/p\right\rfloor+\left\lfloor m/p+n/p\right\rfloor$

using that: $\left\lfloor 2m/p\right\rfloor-\left\lfloor m/p\right\rfloor=\left\lfloor m/p+1/2\right\rfloor$

and $\left\lfloor m/p+1/2\right\rfloor+\left\lfloor n/p+1/2\right\rfloor \geq \left\lfloor m/p+n/p\right\rfloor$

let $f(m,n) = \frac{(2m)!(2n)!}{m!n!(m+n)!}$.

we have $f(m,n) = 4f(m,n-1) - f(m+1,n-1)$.

see that we can use this many times to represent $f(m,n)$ by $\sum{f(m+x,0)}$,
but $f(a,0) = C^a_{2a} \in Z$

so, the problem is solved.  :)

[quote="orl"]Prove that $(2m)!(2n)!$ is a multiple of $m!n!(m+n)!$ for any non-negative integers $m$ and $n$.[/quote]

Suppse that $p$ be an arbitrary prime then we should prove that $[\frac{2m}{p}]+[\frac{2n}{p}]\geq [\frac mp]+[\frac np]+[\frac{m+n}{p}]$ (where [$x$]  denote the greatest integer less than or equal to $x$)
And we know $[\frac{2m}{p}]-[\frac mp]=[\frac mp+\frac 12]$ so that $[\frac mp+\frac 12]+[\frac np+\frac 12]\geq [\frac mp+\frac np]$ and we are done. :D

To prove $\lfloor \frac{2m}{p} \rfloor + \lfloor \frac{2n}{p} \rfloor \geq \lfloor \frac{m}{p} \rfloor + \lfloor \frac{n}{p} \rfloor + \lfloor \frac{m+n}{p} \rfloor$, it suffices to prove that $\lfloor 2a \rfloor + \lfloor 2b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor + \lfloor a+b \rfloor$ for arbitrary real $a,b \in [0,1)$, which is obviously true by considering the four cases: $a \geq \frac{1}{2}$ & $b \geq \frac{1}{2}$, $a \geq \frac{1}{2}$ & $b < \frac{1}{2}$, $a < \frac{1}{2}$ & $b \geq \frac{1}{2}$, and $a < \frac{1}{2}$ & $b < \frac{1}{2}$.

From the nature of the solution, I feel this problem should be in the Number Theory section instead  :D

This is the only one case of Kummer's Theorem.

True. I'm just proving this particular case here, which is sufficient for this problem.

[hide="Another solution"]
To prove \[ \sum^{\infty}\left\lfloor \frac{2m}{p^k}\right\rfloor\plus{}\left\lfloor \frac{2n}{p^k}\right\rfloor\geq \sum^{\infty}\left\lfloor \frac{m}{p^k}\right\rfloor\plus{}\left\lfloor \frac{n}{p^k}\right\rfloor\plus{}\left\lfloor \frac{m\plus{}n}{p^k}\right\rfloor\]
we establish that for any $ a$, \[ \left\lfloor \frac{2m}{a}\right\rfloor\plus{}\left\lfloor \frac{2n}{a}\right\rfloor\geq \left\lfloor \frac{m}{a}\right\rfloor\plus{}\left\lfloor \frac{n}{a}\right\rfloor\plus{}\left\lfloor \frac{m\plus{}n}{a}\right\rfloor\]
Let $ m\equal{}r_1a\plus{}q_1$ and $ n\equal{}r_2a\plus{}q_2$ with $ q$'s non-negative integers less than $ a$. Then we see the inequality is equivalent to
\[ \left\lfloor \frac{2q_1}{a}\right\rfloor\plus{}\left\lfloor \frac{2q_2}{a}\right\rfloor\geq \left\lfloor \frac{q_1\plus{}q_2}{a}\right\rfloor\]
But one of the quantities on the left is greater than the entire right hand side.
[/hide]

[quote="e.lopes"]we have $ f(m,n) \equal{} 4f(m,n \minus{} 1) \minus{} f(m \plus{} 1,n \minus{} 1)$.[/quote]
Very nice!  Let

$ G(x, y) \equal{} \sum_{m, n \ge 0} f(m, n) x^m y^n$.

Then the recurrence you gave is equivalent to $ 4G(x, y) \equal{} \frac{G(x, y) \minus{} G(0, y)}{x} \plus{} \frac{G(x, y) \minus{} G(x, 0)}{y}$, or $ (x \plus{} y \minus{} 4xy) G(x, y) \equal{} y G(0, y) \plus{} x G(x, 0)$.  But of course

$ G(x, 0) \equal{} \sum_{m \ge 0} {2m \choose m} x^m \equal{} \frac{1}{ \sqrt{1 \minus{} 4x} }$

and similarly, which gives

$ G(x, y) \equal{} \frac{ \frac{y}{ \sqrt{1 \minus{} 4y}} \plus{} \frac{x}{ \sqrt{1 \minus{} 4x} } }{x \plus{} y \minus{} 4xy}$.

Questions:

1.  Can anyone come up with a natural combinatorial description of $ f(m, n)$?  Does it lead to the above in a more general way?  (Such a description would trivially solve the original question.)

2.  What other identities of the form "a quotient of factorials is an integer" follow from [url=http://en.wikipedia.org/wiki/Hermite%27s_identity]Hermite's identity[/url]?  (Here several of you have used Hermite's identity with $ n \equal{} 2$.)

3.  The [url=http://www.google.com/search?q=hook-length+formula&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]hook length formula[/url] is the most general statement I know of this form; for example, it implies that the Catalan numbers are integers.  Is there a generalization of the hook length formula that reduces to this question as a special case?

4.  What is the most general statement of this form we can make?  I made an obvious conjecture some time ago (the terms in the numerator should majorize the terms in the denominator) that is quickly verified to be wrong, so what is the appropriate generalization?  For example, is

$ \frac{(3m)!(3n)!}{m!n!(m\plus{}n)!^2}$

an integer?

By factorization, f(m,n) = (2m  m)*(2n  n)/(m+n   m). I don't know if it will be useful(it isn't so natural), but I got a counting solution trough this form. It's a little hard to explain, and I'm very tired. Tomorrow I will try to explain my solution.

I'm not sure if this works, but could someone check it? Thank you in advance.

[hide="Solution"]
\[ \frac{(2m)! (2n)!}{m!n!(m\plus{}n)!} \equal{} \frac{\binom{2m}{m}\binom{2n}{n}}{\binom{m\plus{}n}{n}}\]

We have 2m+2n people in total. Arbitrarily split them into two groups of size 2m and 2n. From a group of 2m people we want to pick m people to join our club. From another group of 2n people, we want to choose n people to get preferential treatment in our club. Later the rules were changed, and we decided not to give preferential treatment to n people from the group of n+m people. This expression counts the number of ways you can do this, and thus is an integer.
[/hide]

[quote="bokagadha"]We have 2m+2n people in total. Arbitrarily split them into two groups of size 2m and 2n. From a group of 2m people we want to pick m people to join our club. From another group of 2n people, we want to choose n people to get preferential treatment in our club. Later the rules were changed, and we decided not to give preferential treatment to n people from the group of n+m people. This expression counts the number of ways you can do this, and thus is an integer.[/quote]

Could you be more precise about it? The group of $ n\plus{}m$ people chosen can still be split uniquely into those with preferential treatment and those without, because the initial splitting is known.

I'm not very sure about my solution, and I can see why that last part might be sketchy at best. Here's another stab at a combinatorial interpretation:

Represent the quantity as such:
\[ \frac{m! \binom{2m}{m} n! \binom{2n}{n}}{(m\plus{}n)!}\]

So take a group of 2m boys and 2n girls. The numerator counts the way we can take m boys and order them in terms of height and do the same with the girls. The denominator says that we don't care about the order and just want a group of m+n people in no arrangement whatsoever.

Again, I'm not very sure about this argument, and any help/comments would be greatly appreciated.

[quote="bokagadha"]

 The numerator counts the way we can take m boys and order them in terms of height and do the same with the girls.
The denominator want a group of m+n people in no arrangement whatsoever.

[/quote]

I don't think it is good, because the denominator don't say sth about the gene/ they may be all men for example.

[quote=e.lopes]
we have $f(m,n) = 4f(m,n-1) - f(m+1,n-1)$.
[/quote]

How you got $f(m,n) = 4f(m,n-1) - f(m+1,n-1)$ ?

WLOG let $n > m$; if the two variables are equal, the expression simply equals ${2m \choose m}$.

Taking $\nu_p$, it suffices to show that $$\sum_{i=1}^\infty \left \lfloor \frac{2m}{p_i} \right \rfloor + \sum_{i=1}^\infty \left \lfloor \frac{2n}{p^i} \right \rfloor \geq \sum_{i=1}^\infty \left \lfloor \frac m{p^i} \right \rfloor + \sum_{i=1}^\infty \left \lfloor \frac n{p^i} \right \rfloor + \sum_{i=1}^\infty \left \lfloor \frac{m+n}{p^i} \right \rfloor.$$ In other words, it suffices to show that $$\lfloor 2a \rfloor + \lfloor 2b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor + \lfloor a+b \rfloor$$ for all real numbers $a, b$; note that $b>a$. This follows by using the inequality $\lfloor a+b \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor$ on the following three expressions and summing:
\begin{align*}
\lfloor 2b \rfloor - \lfloor b-a \rfloor &\geq \lfloor a+b \rfloor \\
\lfloor 2a \rfloor &\geq 2\lfloor a \rfloor \\
\lfloor b-a \rfloor &\geq \lfloor b \rfloor - \lfloor a \rfloor.
\end{align*}
So we are done.


so any indians preparing or giving ACT this year

is this act thing same as ACTURY???one of my frnds was enquiring??

meanwhile congrats 4 ur score :oops:

ACTUARY? What is that?
And thanks! :) I wish there was someone else here who could tell me if I should retake to increase my score :(

ofc u shud take more trials

u hav plenty of time...almost a full year

but it wud b better if u retake SAT first :)

I'm retaking my SAT in a week. The problem with retaking is that its so costly and you need to travel to another city for the test+ miss IIT-JEE coaching for a couple of days.

yes,u r rite :P

finally i see u hav a piece of ur mind interested in JEE :rotfl:

Whether I go to IIT or not, JEE prep is THE BASE for anything, SAT subject tests, advance placement exams, boards, Olympiads, intellectual growth, everything! I can't miss JEE coaching even if there;s a school exam the next day and I'm going to fail(which almost happened!  :rotfl: )

good going ..............the ghost of JEE is all around me nowadays......... :rotfl:  :rotfl:

[quote="nuclear_alchemist"]if there;s a school exam the next day[/quote]
i cudn't  hav missed nething 4 that (even sm movie which i dont like :rotfl: )

not now though :(


function
combinatorics proposed
combinatorics

Consider a set $D$  with [b]Card[/b]$(D)=10$.
[b]Question:[/b]  [i] Find the number  $m$ of all   nonidentical functions $f : D \to D$ such that  [/i]
\[ \left(f\circ f\circ f\circ f\circ f \right)(x)=x \; \; \; ,\; \; (\forall x\in D)\; \; . \]
[b]( $m=73008$ ? ) Generalization ? [/b]

It is just the classification of permutations by order.  I do not know if there is a general formula, but counting small cases is not hard at all using cycles.  Especially this case, as 5 is a prime number.  I think your answer is a little too small, though.  I get m=78624.

I found o recurence of $\ x_n$ which is the number of permutatin with cicles of length 1 and 5 of n elements(the functions which have your desired property). $\ x_n= x_{n-1}+\displaystyle\binom{n-1}{4} x_{n-2}$. To prove it we split the set of functions of n elements in two subsets: first whith the functions with f(n)=n and the second with the functions in witch n is in a cicle of length 5 and the ather 4 elements can be chosen in $ \displaystyle\binom{n-1}{4}$ ways. This recurence take us to the formula for the exponential generating function of $\ x_n$   $ F(x)=\sum_{n \ge 0} x_n\frac{t^n}{n!} = exp(t+ \frac {t^5}{5})$. From these we can calculate easily $\ x_n$ for small values of n.


A bush pilot wants to fly her plane to a lake that is 250 km [N 30 E] from her strating point. The plane has an air speed of 210 km and a wind is blowing from the west at 40 km/h.

In what direction should she head the plane to fly directly to the lake?

It's a three-part question but I only need help drawing the vector diagram.

I have a vector going west at 40 km/h and another going 250 km [N 30 E] from the reference point. However, I can't add/subtract the two because they have different units, right?

Can someone show me how to draw the vector diagram? Much thanks.

[/img]

You don't need a vector representing the displacement. What you want is only the direction of that displacement.

First of all, you have your wind velocity vector backwards. It says [i]from[/i] the west, not towards the west. Second, the other vector you want is the other velocity vector (that of the plane itself). It has magnitude 210 km/h (you must have made a typo), but you don't know its direction, so don't draw it in yet. Hold that thought.

Now, the net velocity of the plane will be the resultant vector resulting from the addition of the wind and plane velocity vectors. You already know the direction that you want that resultant to be. So, draw it in. For your purposes, the length (in the diagram) is unimportant at the moment. You now have two vectors in your diagram. The third is the direction in which you want to fly the plane. So draw it in, but in a position so that you are adding it to the wind velocity vector to produce the resultant.

You now know two magnitudes and can easily determine (with a little geometry) at least one angle in your triangle. Use trig to find the other useful information.

I hope that was clear. The explanation may have been a little bit awkward.


algebra
polynomial
IMO Shortlist
algebra unsolved

Let be given a polynomial $ P(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ satisfy $ |P(x)|\leq 1$ for every$ |x|\leq 1$.
Prove that $ |a|\plus{}|b|\plus{}|c|\plus{}|d|\leq 7$

[quote="zaya_yc"][quote="QuylaoKame"]Let be given a polynomial $ P(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ satisfy $ |P(x)|\leq 1$ for every$ |x|\leq 1$.
Prove that $ |a| \plus{} |b| \plus{} |c| \plus{} |d|\leq 7$[/quote]

IMO Shortlist 1996, A5

http://www.mathlinks.ro/viewtopic.php?p=1218556#1218556[/quote]
Did you have any other solution???


Hey Ray (that rhymes!), if you want to suggest some problems for my mock mathcounts competition (preferably geometry), go ahead.

YES! now i feel special. i made like 5 good ones and then i left tem somewhere. once i find them i willl post them

wow not a lot of people here live in PA.  But why do you cry because of that? You can just post in other forums...

there are no other forums to post it on...

Are we the only few people living in PA on this whole website???

I know at least ten others who live here, but apparently they don't use the PA forum all that much.

[quote=abacadaea]nobody posts in the PA forum. :([/quote]

I did

We can make graph theory problems where the points are cities in Pennsylvania (Pittsburgh, Erie, Greensburg, Altoona, Wilkes Barre, etc.)

[quote=mamis511]I live in PA! How many of you guys are on an ARML team?[/quote]

I'll probably do the Lehigh Valley team next year.

What grade are you in?  I'll be in 7th grade next year and i might do it but it might be too hard.

I'm in 8th grade.  Well there are past Lehigh Contests online and if you can get 15+ on the later ones, you'd probably make one of the teams.

Could i have the link to these?  Thanks!


topology
search
advanced fields
advanced fields unsolved

Ive searched and have found nothing for this problem-studying for a prelim soon.  Hopefully someone can help:

Suppose $ f: X \rightarrow Y$ is a continuous onto map and $ X$ is a compact metric space.  Show that $ Y$ is metrizable.

Any suggestions?

Are you sure if this is true? If I take X = [0,1] with usual metric, Y = [0,1] with finite complement topology, then the identity map is continuous but Y isn't even first countable.

A more trivial example is to take X to be any compact space and Y to be a finite set with the trivial topology.

No, I am not sure that this is true.  It would be really nice if $ f$ were an open-then $ Y$ would be normal (hence regular) and
be 2nd countable.  I haven't been able to figure out a way around the open condition.

There's a proof [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=904893872&t=117480]here[/url] for the stronger, similar assertion which assumes only that $ X$ is a separable metric space.

You certainly need another condition either on $ Y$ or on the map- the linked post had as a condition that $ Y$ is compact and Hausdorff. You can get compactness from compactness of $ X$, but that's not enough.

This is from a prelim that tacitly assumes all spaces are $ T_2.$  I believe there is one other question on this that needs such a hypothesis, as well.

-Jon


modular arithmetic
induction
number theory unsolved
number theory

Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^{n}$ all of whose digits are odd

Let $M_{n}$ be a $n$-digit number recursively defined as

$M_{n+1}= a_{n}\cdot 10^{n}+M_{n}$

for $n \geq 1$ where $M_{1}=5$ and $a_{n}$ is an odd digit which is a solution of the congruence

$(1) \;\; a_{n}\cdot 10^{n}+M_{n}\equiv \pmod{5^{n+1}}$.

In other words, $M_{k}$ is a natural number consisting of $k$ odd digits which is divisible by $5^{k}\; (*)$. This statement can be easily verfied by induction: Clearly $(*)$ is true for $k=1$. Next assume $(*)$ is true for all $k \leq n$. Hence $M_{n}= 5^{n}\cdot K_{n}$ where $K_{n}$ is a natural number. Then by (1)

$5^{n}(a_{n}\cdot 2^{n}+K_{n}) \equiv \pmod{5^{n+1}}$,

i.e.

$(2) \;\; a_{n}\cdot 2^{n}\equiv-K_{n}\pmod{5}.$

Since $GCD(2^{n},5)=1$, the congruence (2) has a solution $a_{n}\equiv r_{n}\pmod{5}$ where $r_{n}\in \{0,1,2,3,4\}$. Then by choosing $a_{n}= r_{n}$ if $r_{n}$ is odd and $a_{n}= r_{n}+5$ if $r_{n}$ is even, we obtain a number $M_{n+1}$ which satisfies the conditions given in the statement $(*)$.


[size=150]
Let C1: $y=x - \frac{1}{2}x^2$ and C2: $x= y - \frac{1}{2}y^2$ be curves on the xy plane.

1.) Find the equation of the tangent to the curve C1 at x=k.
2.) Suppose the line obtained in number 1 is also tangent to the curve C2. Find
all values of k and the equations of the tangent.[/size]

using calc
[hide]
i didn't simplify but for part a:
$y=(1-k)(x-k)+(k-\frac{1}{2}k^2)$

use this to solve the second part ;) [/hide]

I didn't get it sir. Please post step by step procedure because I am a little bit confuse. Thank you.

This isn't the place to post this, but you use calculus (as the above poster stated).  Take the derivative and use $\left(k,f\left(k\right)\right)$ as your point.


If the roots of x^3+bx+cx+d=0 are -4,-1 and 4, then what is b+c+d?

Hint: Plug in those $ x$-values to form a system of equations.

By the factor theorem, $ (x\minus{}c)$ is a factor of a polynomial $ f(x)$ iff (if and only if) $ c$ is a factor of $ f(x)$.

[hide="Another Solution"]
We have $ f(x) \equal{} x^3 \plus{} bx \plus{} cx \plus{} d \equal{} (x \plus{} 4)(x \plus{} 1)(x \minus{} 4)$ by the Factor Theorem. We have $ f(1) \equal{} 1 \plus{} b \plus{} c \plus{} d \equal{} (5)(2)(\minus{}3) \equal{} \minus{}30\implies b \plus{} c \plus{} d \equal{} \boxed{\minus{}31}$.
[/hide]


Let $ f: R \to R$ be a function satisfies $ |f(x\plus{}y)\minus{}f(x)\minus{}f(y)|\leq 1 \forall x,y \in R$.Prove that there exists a function $ g$ satisfying $ |f(x)\minus{}g(x)|\leq1$ and $ g(x\plus{}y)\equal{}g(x)\plus{}g(y)$ for all $ x,y \in R$

I wonder if just Zorn's Lemma could prove its existence ...


Show that if a polynomial $f(x)$ satisfies the following property:"$f(x)$ is rational if and only if $x$ is rational", then the coefficients of $f$ are rational.

If deg f=n , we have f(0),f(1),...,f(n) are in Q. Therefore by  Lagrange Interpolating Polynomial
  http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

we have done.

Am I wrong?

The problem to find all such polynomials was already posted three times, one of them lying three days back  :roll:


calculus
integration
inequalities
trigonometry
real analysis
real analysis unsolved

Prove the following inequality.

\[\int^\pi_0 e^x\sin^2xdx>8\]

Taking by part it easy to prove that

$ \int^\pi_0 e^x\sin^2xdx = \frac{2(1-e^{pi})}{5} $ which is approximately equal to 8.2

Taking by parts it easy to prove that

$ \int^\pi_0 e^x\sin^2xdx = \frac{2(e^{\pi}-1)}{5} $ which is approximately equal to 8.2

To eugene, thank you for your reply.

What we should prove is that $e^\pi>21, by the way ,how can you obtain your vallue?

kunny

eugene, thank you!

It is very kind of you to explain in detail.What I want to know is how you are  induce the value of $e^\pi$.

kunny


modular arithmetic
number theory
relatively prime

Sorry, what does the broken divisibility bar mean?

If the bar means what I think it does, I can show for 120 but not 240.

Looking at the LHS mod 3, (somecrap)^5 is (somecrap) mod 3 by Fermat's Little. Summing the terms mod 3, we obtain 0.

Looking at the LHS mod 8, (somecrap)^5 is (somecrap) mod 8 by Euler's Theorem since phi(8) = 4. Similarly summing, we obtain 0.

Looking at the LHS mod 5, (somecrap)^5 is (somecrap) mod 5 by FLT. We obtain a sum of 0 mod 5.

So 120 | the LHS.

Now that I look back at it, I think I've made several dangerous overlookings of the constraints of FLT and Euler's; the modulus has to be relatively prime with (somecrap) which is not necessarily true. Yufei mentioned something about being able to fix this. How? I am very much a noob problem solver.

[quote="DavidL"]Ithe modulus has to be relatively prime with (somecrap) which is not necessarily true. Yufei mentioned something about being able to fix this. How? I am very much a noob problem solver.[/quote]
The FLT may be writen as
$ a^{p}\equiv a\pmod{p}$, and it's not neccessary for a to be coprime with prime p.
(if a IS coprime, the statement follows from the famous form $ a^{p\minus{}1}\equiv 1\pmod{p}$, if a is not coprime, $ a^{p}\equiv0\equiv a\pmod{p}$.
So your congruences with somecrap are valid for mod 5 and mod 3.
Mod 8 it doesn't work.
But it's easy to do the mod 16 part of the problem:
if a+b+c=2k, we have a+b-c=2k-2c etc, so all the terms are even and thir fifth powers are divisible by 16.
if a+b+c=2k+1, a+b-c=2k-2c+1 etc, so all the terms are odd. Fourth power of odd numbers are 1 (mod 16), so $ (SomeOddCrap)^{5}\equiv SomeOddCrap\pmod{16}$. And the answer follows :)

[quote="silouan"][quote="arqady"]$ \{a,b,c\}\subset\mathbb Z.$ Prove that
$ (a\plus{}b\minus{}c)^{5}\plus{}(a\plus{}c\minus{}b)^{5}\plus{}(b\plus{}c\minus{}a)^{5}\minus{}(a\plus{}b\plus{}c)^{5}\vdots240.$[/quote]

It means <<is divided by>>

Here is my solution.
$ (a\plus{}b\minus{}c)^{5}\plus{}(a\plus{}c\minus{}b)^{5}\plus{}(b\plus{}c\minus{}a)^{5}\minus{}(a\plus{}b\plus{}c)^{5}\equal{}\minus{}80abc(a^{2}\plus{}b^{2}\plus{}c^{2})$

So we need to prove that $ 3|abc(a^{2}\plus{}b^{2}\plus{}c^{2})$ .
That's true why :
if 3|a or 3|b or 3|c we are done .
If 3 doesn't divide either a, either b ,or c then $ a^{2}\equiv b^{2}\equiv c^{2}\equiv 1\ (\mod 3)$
so $ 3|(a^{2}\plus{}b^{2}\plus{}c^{2})$ so again we are done[/quote]

Where did you get that simplification from?  I am talking about the that power of 5 one.


geometry
rectangle
inequalities
Asymptote
triangle inequality

I just thought of this problem while messing around in Inkscape, and how would I solve it?  :huh:

What is the maximum area of the bounding rectangle of a triangle with side lengths 7,8,9?

And what would be the general method?'

Even worse, is the area always the same? If so, why?

what do u mean bounding rectangle? if u mean the rectanlge that touches all the vertices of the triangle then i would think u figure out which combination of a side of the triangle along with its corresponding altitude will yield the greatest product.

Yeah, i don't understand that part either.

Bounding Rectangle refers to the possible surrounding rectangles of the triangle, yes.

Hmm.
Let's make a triangle like this:

Select one vertex to be $ (0, 0)$ and one on the X axis to be the point $ (X_{1}, 0)$.
(This fixes the location of this side, but we can still consider the triangle to be free rotating because fixing one side does not limit the orientations of a triangle.)

Then we need to find the locus of points $ (x_{2}, y_{2})$ such that the triangle inequality is not violated. This is clearly any point in space NOT on the X axis ( for obvious reasons, i.e degenerate triangle).

Therefore, we have the restriction that $ X_{2}\neq 0$

Now, because the rectangle must (optimally) pass through at least one pair of vertices in a straight line, we have three cases for each triangle:

I'll try and draw a picture (sorry for not using Asymptote, I could never get the hang of it):
  A                              B
-*----------------------*
|00*000000000000000*00|
|00000*0000000000*0000|
|00000000*0000*0000000|
|------------*----------|
                  C
where vertices A and B are connected by the rectangle.

The other two cases are where AC is connected (i.e. the rectangle is on a slant) or BC is connected.

So because the rectangle ideally passes through a pair of these points, at least one side has to go through there.

So we can look at the cases like this:
For the first case, where it goes through $ (0,0)$ and $ (x_{1}, 0)$ one side of the rectangle is x_1, and the other is the altitude to C (the y coord of $ (x_{2}, y_{2})$) and the area is $ x_{1}\cdot y_{2}$

For the other two cases, we use the distance formula to derive the length of the base and altitude.

I'm not sure how you would prove maximality, but I THINK that this approach is the right general method.

Comments are welcome :0

I don't think that there is a rectangle of maximum area for the following reason.
the triangle is $ ABC$ where $ A$ $ B$ $ C$ are the vertices of the triangle.
Now let line $ x$ pass through $ AB$, no select the line, $ y$, through $ C$ that is parallel to $ x$. For the two remaining sides all you need to do is select almost any two lines that are orthogonal to $ x$ and $ y$.

are you sure that there are no other restrictions for this problem?  :maybe:

[asy]pair A,B,C,D,E,F,G; A=origin; B=(9,0); C=(5.333333333,7.824463063); D=(-2,0); E=(11,0); F=(11,7.82446306); G=(-2,7.824463063); Draw (A--B--C--A); Draw (D--E--F--G--D); Dot A; Dot B; Dot C;[/asy]

If this is considered bounded, then clearly, the area of the rectangle can be infinitely large.

[asy]pair A,B,C,D,E; A=origin; B=(9,0); C=(5.333333333,7.824463063); D=(9,7.824463063); E=(0,7.824463063); Draw (A--B--C--A); Draw (A--B--D--E--A);[/asy]

If this is what is meant, then the area is always twice the area of the triangle because they contain a common base and common height, so its area would be $ 2\sqrt{12(5)(4)(3)}=24\sqrt5$.


calculus
integration
trigonometry
logarithms
limit
complex numbers
calculus computations

Evaluate $ \int_{0}^{\infty} \frac{x}{x^{6}\plus{}1}\ dx$.

hello, use that
$ \frac{x}{x^6\plus{}1}\equal{}\frac{1}{3}\frac{x}{1\plus{}x^2}\minus{}\frac{1}{3}\frac{x(x^2\minus{}2)}{x^4\minus{}x^2\plus{}1}$
Sonnhard.

[quote="AndrewTom"]Evaluate $ I = \int_{0}^{\infty} \frac {x}{x^{6} + 1}\ dx$.[/quote]

$ I = \mathop {\lim } \limits_{ n \to \infty}(\int_{0}^{n}\frac {x}{x^6 + 1}dx)$

 $ = \mathop {\lim } \limits_{ n \to \infty} ( \frac {\pi\sqrt {3}}{36} + \frac {\ln(1 + n^2)}{6} - \frac {\ln(n^4 - n^2 + 1)}{12} + \frac {\sqrt {3}}{6}arctan(\frac {2n^2 - 1}{\sqrt {3}}))$

$ = \frac {\pi\sqrt {3}}{9}$

How about this ????

[quote]Evaluate $ \int_{0}^{\infty} \frac {x}{x^{10} + 1}\ dx$.[/quote]

;)

Suppose the real part of $ \alpha$ lies inside $ \left(0, 1\right)$.
Show that:
$ I(\alpha) \equal{} \int_{0}^{\infty} \frac{x^{\alpha \minus{} 1}}{1 \plus{} x} dx \equal{} \pi \cdot \csc \pi \alpha$

Then the two problems given here will be special cases corresponding to appropriate $ \alpha$ in $ I(\alpha)$.

hello, we have
$ \int_{0}^{\infty}\frac{x}{x^{10}\plus{}1}\,dx\equal{}\frac{\pi}{10\sin\left(\frac{\pi}{5}\right)}$
Sonnhard.

Dr. Sonnhard Graubner can you show your solution?

With the substitution $ x \equal{} y^{\frac{1}{10}}$, one can easily write down the answer that you have from the formula provided in my previous thread,

\begin{align*}
\int_0^\infty \frac {x^{\alpha - 1}}{1 + x} dx
& = B\left(\alpha, 1 - \alpha\right) \\
& = \Gamma\left(\alpha\right)\Gamma\left(1 - \alpha\right) \\
& = \frac {\pi}{\sin \pi \alpha} \\
& = \pi \csc \pi \alpha
\end{align*}

Another derivation uses complex analysis.


factorial
number theory unsolved
number theory

Solve the equation $ a_1 ! \cdot a_2 ! \cdot ... \cdot a_{n \minus{} 1} ! \equal{} a_n !$  for  $ n \ge 3\,,\,a_i  \in N\,,\,\left\{ {i \equal{} 1,2,...,n} \right\}$.

$ a_{n\minus{}1}\equal{}a_n$ and $ a_1\equal{}...\equal{}a_{n\minus{}2}\equal{}1$

[quote="Rust"]$ a_{n \minus{} 1} \equal{} a_n$ and $ a_1 \equal{} ... \equal{} a_{n \minus{} 2} \equal{} 1$[/quote]

Quite trivial solution, but certainly not unique. For example : $ 1!\cdot 1!\cdot 1! ... 1!\cdot (a!\minus{}1)! \cdot a! \equal{} (a!)!$

just an estimated answers.
try to figure out solution.

let me estimate another solution: $ 3! \cdot 5! \equal{} 6!$.

@theSA : Your "another" solution is the one I gave (choose $ a\equal{}3$ and you obtain $ 3!\cdot 5!\equal{}6!$, as you could obtain $ 4!\cdot 11!\equal{}12!$, or $ 5!\cdot 119!\equal{}120!$, ...).

More generally : $ a_1!a_2!...a_{n \minus{} 2}!(a_1!a_2!...a_{n \minus{} 2}! \minus{} 1)! \equal{} (a_1!a_2!...a_{n \minus{} 2}!)!$

But I'm rather sure there are other solutions.

[quote="pco"]@theSA : Your "another" solution is the one I gave (choose $ a \equal{} 3$ and you obtain $ 3!\cdot 5! \equal{} 6!$, as you could obtain $ 4!\cdot 11! \equal{} 12!$, or $ 5!\cdot 119! \equal{} 120!$, ...).

More generally : $ a_1!a_2!...a_{n \minus{} 2}!(a_1!a_2!...a_{n \minus{} 2}! \minus{} 1)! \equal{} (a_1!a_2!...a_{n \minus{} 2}!)!$

But I'm rather sure there are other solutions.[/quote]

this problem was proposed for our National Olympiad.
And solution was to find out the solution form that  [b]pco[/b] gave:  $ (n! \minus{} 1)! \cdot n! \equal{} (n!)!$.

Only one student solved(my student :D).

is there any necessary steps to show with solution???
or just giving 3-4 examples and showing the form is enough???
 :oops:

Excuse me everyone.
there is some mistake in question.
[b]shoki[/b] mentioned me the problem so i try to edit the question, but I can't, MathLinks not allowing.

the original question was:

SHOW THAT THERE EXIST DIFFERENT NATURAL NUMBERS(infinitely many) SATISFYING THE GIVEN EQUATION.

Sorry again... :blush:


geometry
geometric transformation
dilation
ratio
parallelogram
Euler
geometry unsolved

Let $A,B,C$ be fixed points in the plane , and $D$ be a variable point on the circle $ABC$, distinct from $A,B,C$ . Let $I_{A},I_{B},I_{C},I_{D}$ be the Simson lines of $A,B,C,D$ with respect to triangles $BCD,ACD,ABD,ABC$ respectively. Find the locus of the intersection points of the four lines $I_{A},I_{B},I_{C},I_{D}$ when point $D$ varies.

One for all! :D
[b]Problem[/b]:Given triangle $ABC$ inscribed $(O)$, $M$ lie on $(O), d$ is Simson's line of $M$ with triangle $ABC$ ,$d'$ is Simson's line of $A$ with triangle $MBC$. Find the locus of intersections of $d$ and $d'$
[b]Proof[/b]
[b]Lemma[/b] (Well know):Steiner's line is dilation of Simson's line $d$ center $M$ ratio $2$, $s=D_{M}^{2}(d)$ and Steiner's line is through orthocenter
$H$ of $ABC$.Thus $d$ is through midpoint $K$ of $MH$ $(1)$

Called $H'$ is othorcenter of triangle $MBC$ easily seen $AHH'M$ is paralelogram, thus $K$ is midpoint of $AH$.Apply the lemma we have $d'$ is through $K$ $(2)$.
From $(1),(2)$ we have intersection $K$ of $d$ and $d'$ is midpoint of $MH$ thus $K=D_{H}^\frac{1}{2}(M)\Leftrightarrow \{K\}=D_{H}^\frac{1}{2}(\{M\})=D_{H}^\frac{1}{2}((O))$=Euler circle.
@N.T.TUAN If you don't post all Taiwan Olympiad yet, you don't want stop
Please post all them in one box of olympiad box and post solution for it if you have  :D

[quote="gemath"]
@N.T.TUAN If you don't post all Taiwan Olympiad yet, you don't want stop
Please post all them in one box of olympiad box and post solution for it if you have[/quote]

I don't understand. I haven't got solution of these problems. I only problem and NOT solution!

First, I post all them in box National Olympiad. Second, I post all them in boxes : Algebra, Combinatoric,...

That is LAW of this forum.

I will check your solution later. Now, I have to sleep :D

I know it has been more than 13 years since this post was published by @N.T.TUAN, but at the moment I only knew how to do basic calculations (addition only). However, now I have a solution for this via complex plane geometry:

Let us have quadrilateral $ABCD$ with a unit circle around it. It is known (if anybody needs proof, I can write it) that Simson's lines (from $A$ to $BCD$, $B$ to $ACD$, $C$ to $ABD$  and $D$ to $ABC$) all intersect in one point, and that point is (I can write the proof of this as well if needed): $x = \frac{a+b+c+d}{2}$. So the geometric place of all points $x$ is circle with center in $\frac{a+b+c}{2}$ and radius $1$, because $|x - \frac{a+b+c}{2}| = |\frac{d}{2}| = \frac{1}{2}$.


A man goes to a stream with a 9–pint container and a 16–pint
container. What should he do to get 1 pint of water in the 16–pint container?

If $ \gcd(a,b)\equal{}1$, then there exists integers $ x,y$ such that $ ax\plus{}by\equal{}1$. Does this help?

[hide]
Fill nine pint, pour into sixteen pint, fill nine pint, pour into sixteen pint, empty sixteen pint, pour remaining two pints from the nine pint into the sixteen pint, fill and pour the nine pint into the sixteen pint, fill and pour in the same manner, empty the sixteen pint, pour the fours pints in the nine pint into the sixteen pint, fill and pour same manner, fill and pour same manner, empty the sixteen pint, pour the six pints from the nine pint into the sixteen pint, fill and pour same manner, fill and pour same manner, pour the eight pints in the nine pint into the other, fill and pour, and done.
[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Define a subgroup $H$ of a group $G$ to be permutable if $HK=KH$ whenever $K$ is a subgroup of $G$. It is easily seen that every normal subgroup is permutable. However, not every permutable subgroup is normal.

I have not been able to find any trivial examples of permutable non-normal subgroups. Are there any? Also, is there any general class of subgroups that is always permutable?

Vipul

I guess it's a big surprise that such an example should arise from permutation groups and from a quite simple one $S_3$.

It is known that $S_3=<a,b| a^2=b^3=1, ab=b^2a>$ (i.e. $S_3$ is generated by a (a transposition) and b (a cycle) and we have some relations which in this case are $a^2=b^3=1$ and $ab=b^2a$)

Take $H=<a>$. Then $<b>H=H\cup bH \cup b^2H=\{1,a,b,ba,b^2,b^2a \}=S_3$
Also $H<b>=S_3$.
But $bH=\{b,ba\}$ and $Hb=\{b,ab\}$ and these are distinct, so H is not normal in G.

For your second question: aren't normal subgroups general enough for you? If not, just take the class of permutable subgroups.

That example doesn't work, amfulger.

Consider the subgroup $<ab>$ $=\{1,ab\}.$

Then $H<ab>$ $=\{1,a,ab,b\}$, while

$<ab>H = \{1,a,ab,b^2\}.$

Thus $H$ is not permutable.


Hi, folks !!!   Here is a nice function problem..

Function [tex] f [/tex] defined on [tex] 0 \preceq x \preceq 1 [/tex], and satisfied the following, with all [tex]a,b [/tex] and [tex]a, b[/tex] are not the same.

and..  (a,b) is not (1,0) or (0,1).

Following : [tex]f(0)=f(1)=1, \mid {f(a)-f(b)} \mid < \mid {a-b}\mid [/tex]

Then show that [tex] \mid {f(a)-f(b)} \mid < 1/2 [/tex].

_____________________________________________

Very nice problem.....    :D

Have a nice day all![tex][/tex][tex][/tex][tex][/tex][tex][/tex][tex][/tex][tex][/tex][tex][/tex][tex][/tex]

I do not understand this problem! If we take a=0, b=1 is the last relation verified?

[quote]and.. (a,b) is not (1,0) or (0,1).
[/quote]


I am running MiKTeX on Win XP directly from CD.  I want to add some new packages to $\LaTeX$ texmf tree.  I have tried to place the new packages in localtexmf folder but no use.  I think there may be a way of adding packages by putting them in another directory and adding this directory to search path of MiKTeX. Please, help me.

I don't understand why would you want to run them from the CD in the first place anyway? I doubt that in our days there is someone which doesn't have 100-200 MB free on his HDD to put the packages ...  :?

Portable to use on computers that don't have TeX installed?

[quote="DPatrick"]You can add directories to MiKTeX's search path by clicking on the "Roots" tab in the MiKTeX options window.  (I don't know if this works if you're running it off a CD.)

You may also need to click on "Refresh Now" button on the "General" tab of the options window, after adding a new package.[/quote]

This process doesnot working when running from CD.


[quote="shyong"]There are $m$ distinct positive even integer and $n$ distinct positive odd integer which sum up to $2005$ . For these values of $m,n$ , what is the maximum value of $3m+4n$ ?[/quote]
Let $x_1<\dots<x_m$ are distinct even integers and $y_1<\dots<y_n$ are distinct odd integers, so
$x_1 + \dots + x_m + y_1 + \dots + y_n = 2005$

Let $x_i = 2x'_i + 2i$ and $y_i = 2y'_i + 2i - 1$ then
$0\leq x'_1 \leq \dots \leq x'_m$ and $0\leq y'_1 \leq \dots \leq y'_n$ and
\[
x'_1 + \dots + x'_m + y'_1 + \dots + y'_n = (2005 - m(1+m) - n(1+n))/2.
\]
This equation has non-zero integer solutions if and olny if $2005 - m(1+m) - n(1+n)\geq 0$.

So the original problem is equivalent to maximizing $3m+4n$ with the constraint $m(1+m) + n(1+n)\leq 2005$.
The maximum value 220 is attained for $m=28, n=34$.

How do u get from that contrait to the result...i think is the non trivial part of the problem and is the one that you missed...can you explain it please?


consider a 2k*2k array of dots. Color the dots red or blue in such a way that
1) the number of red and blue dots in a row are equal.
and the same holds for the columns.
join two adjacent red dots by a red segment and two adjacent blue dots with a blue segment.
Prove that the number of red and blue segments are equal..
easy problem it seems but i seem to be missing something.

waiting for replies

cheers

Let S the difference of the number of black segments and white segments!

i)See that you can change two adjacent dots of different colors and the value of S is the same!

ii) Using i) a lot of times, you can compute the value of S for the table with the first k dots of each line black and the last k dots of each line white (easily we find S = 0)

Unfortunately, I don't think (i) quite works:  If we have switches near the corner of a grid, S may change after all (for example, if one corner of the grid is

[code]RBB
BBB[/code]

switching the R and B in the top row decreases the number of blue segments by 1 while leaving the number of red segments unchanged.

Here's another method:

Define the degree of a dot to be the number of adjacent dots in our grid (so the degree is 2 at the corners, 3 at the sides, and 4 elsewhere in the grid).  The sum of the red degrees is equal to twice the number of red edges plus the number of uncolored edges, and similarly for blue.  Thus to show the number of red and blue segments are equal, it suffices to show that the sum of the degrees of the red dots is equal to that of the blue dots.  However, the sum of the red degrees is just four times the number of red dots in the grid minus the number of red dots on each side of the grid (since this subtracts one for points on the sides, 2 for points on the corners).  Since the number of red dots in the grid is equal to the number of blue dots, and each of the four sides of the grid is equally distributed between red and blue, so we are done.


Find p,q>0 naturals if p,q,p*q+1,p*q-1 simultaneous primes.

obviously one of $p,q$ must be $2$ , else $pq+1$ will be even .

WLOG let $p=2$ then $q,2q+1,2q-1$ must be all prime . one of $q,2q-1,2q+1$ must be divisible by $3$ . So the only answer is $p=2$ , $q=3$ or vice versa


Gauss
arithmetic sequence
geometric sequence

I have a scenario where I would like to have the sum of all numbers between:
X....Y without adding each incrementally. How to do this? I remember there was a much easier formulated method of doing this.

Similarly would it be possible to do the same thing in none incremental scenario? say if I am dealing with numbers involving ^-1 or -^2  or say even ^2, etc.?

Thanks in advanced.

You mean all integers, and Gauss' method.

The formula is just the Arithmetic Sequence formula: $ S\equal{}\frac{1}{2}y[2x\plus{}(y\minus{}1)d]$, with the difference $ d$ being $ 1$. $ S\equal{}\frac{1}{2}y[2x\plus{}y\minus{}1]$, where the sequence you wish to add up is $ x,x\plus{}1,x\plus{}2....x\plus{}y\minus{}1$.

Many thanks for the response... that is exactly what I was looking for.  :)

However what if I am not dealing with integers and say the difference was varying on each interval.
To note if I can something like => Sum( (x+1)^-2 )     where x is 0...100.
If that makes any sense?

Thanks again.

If it's a geometric sequence, see [[Geometric sequence]].

Else, there is no simple universal way. Actually, there may be for Fibonnaci and Harmonic sequences, but if there are, then I on't know them.


vector
geometry
3D geometry
trigonometry

In general, how do you find the angle between a line and a plane or two planes? Specifically, the problem is how to find the angle between the space diagonal of a cube and an adjacent face? I would have guessed 45 but the textbook says 54.7.

I would suggest using the dot product:

$\vec a . \vec b=|\vec a| \times |\vec b| cos \theta$.

54.7 degrees is the angle between the diagonal and an edge with a common vertex. To find that, construct a right triangle whose bases are that edge and the diagonal of one of the faces. So $\tan{\theta}=\sqrt{2}$.


Natural number n divides (3^n)+(4^n). Prove that 7 divides n.

Nice one, I suppose that $n > 1$ as steven128 mentioned.

So, let $n>1$ be natural number such that $n|3^n + 4^n$. Because $(3,4)=1$ it's easy to see that $(n, 3)=1$ and $(n, 4)=1$.
Now, let $p$ be the smallest prime divisor of $n$. We will prove that $p=7$. Ofcourse, we have:
$3^n + 4^n \equiv 0 \mod{p}$
which is equivalent to
$(\frac{3}{4})^n \equiv -1 \mod{p}$
and this implies
$(\frac{3}{4})^{2n} \equiv 1 \mod{p}$
Let $t = ord_p(\frac{3}{4})$. It's clear that $t \not |n$ and $t|2n$ so $t$ is even, let's say $t=2k$ for some positive integer $k$.
Then $k|n$ and from the FLT $t=2k|p-1$ which means that $k|\frac{p-1}{2}$. But this implies $k \leq \frac{p-1}{2} < p$ so $k$ has to equal $1$, because otherwise $n$ would have a smaller prime divisor than $p$ which is impossible, because of the assumption that $p$ is minimal.
So $k=1$ and $t=2$, also $(\frac{3}{4})^2 \equiv 1 \mod{p}$, or equivalently $7 \equiv 0 \mod{p}$. $p=7$ as desired, QED.

[quote="TomciO"]It's clear that $t|n$ and $t \not | 2n$ [/quote]
How is this possible ?
We have $t | n$ and $n|2 n$ therefore by transitivity $t | 2 n$.
I guess you meant $t \not | n$ and $t | 2n$... (which is OK)
Otherwise it's a very nice proof :)

Let $t = ord_p(\frac{3}{4})$. What does that mean?

See the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Formulary[/url]  :)
(fractions $\mod n$ can be treated as integers as long as the denominator is coprime to $n$)

Sorry, I was in a hurry and couldn`t find it.Could you please write the answer of my question right here?

The third post with the conventions etc. is the correct one for such things...


Let $ ABC$ be a triangle such that $ AB\equal{}2AC$ and angle $ \hat{A}\equal{}\alpha$; $ \hat{B}\equal{}\beta\minus{}\alpha$; $ \hat{C}\equal{}\beta\plus{}\alpha$.
Find the value of $ \alpha$ and $ \beta$

hello, easily we get $ \alpha=\pi-2\beta$ and from the theorem of sine we get $ \frac{\sin(\beta-\alpha)}{\sin(\beta+\alpha)}=\frac{1}{2}$ or equivalent with ${ \frac{\sin(3\beta)}{\sin(\beta)}}=-\frac{1}{2}$
Sonnhard.


calculus
integration
trigonometry
logarithms
calculus computations

$\int \left( \frac{1}{x^4}+ \frac{16}{3 x} \right) dx=$

$\int \frac{e^{6 x}}{e^{12 x} + 64} dx=$

$\displaystyle{ \int \frac{x}{\sqrt{x+6}}\,dx} =$

$\int 27\cos^2(51 x)\,dx=$

These are the problems which I can't figure out on the test, any help would be great!  Thanks!

1] Write the integral in the form $\int x^{-4} + \frac{16}{3x} = - \frac{x^{-3}}{3} + \frac{16}{3} \ln x +C$
2] take $u=e^{6x}$ then $du=6e^{6x} \, dx$ then the integral becomes $\int \frac{1}{6(u^2+64)} \, du = \frac{1}{48} \arctan \frac{u}{8} = \frac{1}{48} \arctan \frac{e^{6x}}{8}$

I think you forgot a '+' sign for the first problem.

for #2, shouldn't du= ((e^6x)/6 dx) ?

What about #3 and #4, anyone have any ideas?

#3: That square root in the denominator looks like trouble- why not substitute so it's easier to deal with ($u=x+6$).
#4: This is basically $\cos^2$; remember your double-angle identities.


vector
function
linear algebra
linear algebra unsolved

let C(R) be a vector space of all continuous functions from R( reals) to itself.

$ C_{\plus{}}$ is the collection of all even functions

and

$ C_{\minus{}}$ is the collection of all odd functions.

How do i show that C(R)$ \equal{} C_{\plus{}}\oplus C_{\minus{}}$ ?

Construct it. Find the unique functions $ f_{e}$ and $ f_{o}$ (even and odd respectively) such that $ f_{e}\plus{}f_{o}\equal{}f$.

[quote="mstrip"]So should i just say simply that $ f_{e}$ and $ f_{o}$ are both continuous functions. Thus, the sum of these two continuous functions gives a continuous function in $ C(\mathbb R)$? And then conclude the direct summand from that or what?[/quote]

He means find $ f_{e}$ and $ f_{o}$ in terms of f. I don't believe there's a way to derive those expressions, just another case of inspired guessing

The continuity will be obvious from the expressions.
A hint on finding them- the definition of even/odd involves the relation between $ f(x)$ and $ f(\minus{}x)$...

Of course, it also proves the version where we don't include continuity.

should i say that by definition of an even function $ f(\minus{}x) \equal{} f(x)$ for all x and an odd function where $ f(\minus{}x) \equal{}\minus{}f(x)$, and since $ C_{\plus{}}, C_{\minus{}}$ are both vector spaces that $ C_{\plus{}}\oplus C_{\minus{}}$ form $ C(\mathbb R)$ analogous to the positive and negative numbers that make up the $ \mathbb R$ line?

No, you haven't proved anything. You've just restated the problem.

ok, so I guess i started out by constructing a formula for arbitrary $ f_{\minus{}}$ in $ C_{\minus{}}$ and $ f_{\plus{}}$ in $ C_{\plus{}}$. The whole purpose of this construction ( to me at least ) was to get this sum $ f_{\plus{}}\plus{}f_{\minus{}}\equal{} f$.

So after playing around, I get

$ f_{\minus{}}\equal{}\frac{1}{2}(f(x)\minus{}f(\minus{}x))$

and

$ f_{\plus{}}\equal{}\frac{1}{2}((f(x)\plus{}f(\minus{}x))$

when these two eqns are summed together i get $ f_{\plus{}}\plus{}f_{\minus{}}\equal{}f$. is this enough to show it's direct? I guess, i should really be asking, how does one go about showing two even and odd functions are direct?

To show that $ V$ is the direct sum of $ A$ and $ B,$ you must show two things. One is that $ V\equal{}A\plus{}B$ - in other words, that anything in $ V$ can be written as a sum of something in $ A$ and something in $ B.$ You've just done exactly that. The second thing is to show that this decomposition is unique, which amounts to showing that $ A\cap B\equal{}\{0\}.$

So, assume that $ f$ is both even and odd. What do the definitions of even and odd tell you about $ f?$

That really explains a lot, thank you so much for your help.


Can GeoGebra shade the area between two circles (two circles.png )?

Can GeoGebra shade the area between the two lines AB, AC, and arc BC ( arc two lines.png) ?

Thanks!

Sort of:  [geogebra]fa8466191ec6baa0c54282290d3cd5770d315c97[/geogebra]  Unfortunately, even this imperfect solution requires more programming than you would use in Asymptote. On the other hand, Geogebra allows you to fill both circles with color easily (from the Properties menu) and the color of the intersection is deeper than each of the colors for the circles, so you can see it quite well: [geogebra]42bb149974f56b41426f1d413dbbcf37a83c15e1[/geogebra]
Also, you can fill the intersection with color filling 2 arcs:
 [geogebra]8367cb2c64302b1ad48f761e1691d1c906db9a7a[/geogebra]

Thank you very much!

How to shade the area inside one circle but outside the other?

How to shade the area $ y \geq x^2$   or  $ y  \geq x^2 (x  \geq 0)$ ?

also, how to colour the segment $ y \equal{} x^2 (x  \geq 0)$?

Thank you!

[quote="yzymat"]
How to shade the area inside one circle but outside the other?[/quote]
This gives you the general idea:
 [geogebra]f3510d4b0e74e0c4067c31b0f80acf9fa24409aa[/geogebra]
Note that it almost disables interactivity of the picture (it just takes too long to recalculate the list) and that the idiotic kk+kk-kk in the function list definition is intentional: if you just do kk, the file won't save correctly (don't ask me why).

I don't know any color filling solution :(.


geometry
AMC
AIME
AMC 12
AMC 10
Pythagorean Theorem

omg... the AMC was so much harder this year! How is it that #19, 20, 21, 22, 23, and 25 were all geometry! Drat...

I've noticed the same thing.  I got all of those last problems right and I got a 130 for my score, so I qualified for the AIME.  It seemed quite difficult for an AMC 10 I have to say to be honest.  I think most people that took the AMC 10 and qualified for the AIME [b][i]barely[/i][/b] qualified for it.

Gah... I got a 118... I answered 21 questions and got three of them wrong because I didn't get a chance to check them over.. I got all the really easy and stupid questions wrong such as #3, where I didn't realize that the answer was in cents... oh well.

I haven't taken any other AMCs but the general consensus seems to be that this years was harder. I got a 125.5 - BARELY making it, as most of my friends did. I doubt, however, that since this was a hard to test, that the top 1% will be higher than 120, so if you got a 118, I'm pretty sure you'll still make it.

On a side note, I answered 18, got all of them right. Thought I'd play it safe since I suck at geometry.

PS how do you do the last 5 or so geometry problems?

Show up at the Math Jam in about 90 minutes - we'll be starting with the last 5 problems on each exam (then taking requests).

ya, way too much geometry, but none of it was over the edge... as long as you know how to apply all the theorems and such it was pretty easy (with the exception of 21 which was brutal if you did not have a good formula, aka me) and overall it was definitely harder than last year's though i had a 21.5 pt improvement because i got owned last year... i hope the AMC writers decide to make the B easier.

P.S. I have to say, #25 was the coolest problem :D cuz i knew how to do it rather quickly and #19 was a trick question and if you knew the easy way it was an excuse for a question. but #3 was SOOOOOO stupid i just had to put it in dollars... grr.

:(
119...
I answered like too many.  I got pwned because I forgot that if you get a question wrong it counts back on your score.

[quote="darkquantum"]naw man
the geometry was awesome
i LOVED this AMC, cuz im such a geo freak

but sadly, my stomach started convulsing around question 2, and then becasue i solved all teh geo problems, i didn't get to look at some of the other ones. SO i only answered 18 correctly and left 7 blank

125.5

Drats[/quote]
Wow, exact same score as me

[quote]now i get to feel nervous/stupid...i ended with a 116[/quote]
i ended with a 99   :cry: . i made 4 silly mistakes... 4!!!

Don't feel bad; I know a guy who took AMC 12 and got 6 questions wrong......

Omg, that was way harder than most of the practices and ones from the past. The sucky thing is i havent taken geo. I'm only in 8th grade, so i struggled on the last few. I prolly only got a 117. Even worse than my 120.5 last year....

hmm..... this test is certainly harder than most of the past AMC 10s. However, it's not that incredibly hard. I looked at the test just a few days ago and the problems seemed much easier than before. Geometry problems are not as intricate as you described; a factor that made the test appear to be harder was the change of topics (from similar polygons and quadrilaterals to triangles and clever manipulation of "pythagorean theorem!!"ex: #25!!?)

Haha... You guys need to come to a school like mine.  6 wrong on the AMC12?  That's not bad compared to the kids who only get a couple of questions total.  One of my peers was sleeping during the test!  No one in 10th grade will probably go on to the AIME except for maybe me.

believe me, i know what you mean, last year i was in 8th grade and no one in the high school made it to AIME... so this year i transferred and already like 10 made it :D

uh are teachers supposed to open the AMC package the day when the AMC is held?

Thank goodness there's still AMC 10B because I bombed AMC 10A.


geometry
parallelogram
circumcircle
geometry proposed

A parallelogram ABCD with an acute angle BAD is given. The bisector of Angle BAD intersects CD at point L, and the line BC at point K. Let O be the circumcenter of Triangle LCK. Prove that quadrilateral DBCO is inscribed in a circle.

I don't know if this is a noob question or not, but how can you say that O lies on KD? The solution I read just assumed that. Again sorry if this is a dumb question.[/code]

Well I checked my solution with the book's, but the book (AOPS V2) just assumed that O was on KD since the beginning of the problem. I don't know how I can prove O is on KD, but knowing that makes the answer easy to get. I know O is outside Triangle LCK, because LCK is obtuse.

[quote="modularmarc101"]Actually this problem is a tough cookie (to be put in intermediate level, cuz this should be a piece of cake for olympiad competitors). But it shouldn't lie on KD, since KDC is obtuse. It should lie outside the triangle...correct me if im wrong..[/quote]

Note: the problem talks about the circumcenter of triangle $ LKC$ not $ KDC$.

Oh that's true, thanks. But still, the point is somewhere outside LKC.

Just post it on Olympiad Geo. They'll solve it.  :wink:

[quote="JITTT"]Thanks bud, but I still don't get how to prove O on KD and L on OB, because that is how you got your solution.[/quote]
You are confused JITTT because neither O on KD or L on OB is true .The upper proof in post number 7 is ok and does not use this wrong assertion as you said , read more carefully  :) . See my sketch below

[geogebra]b1fee98a668f853812465af3a5dad11802d48d36[/geogebra]

[quote="Kenny O"][quote="game.guy"]How can BAD's bisector intersect both BC and CD? That would only be possible if it intersected at the intersection which would be the opposite vertex. What did I do wrong?[/quote]

gameguy this happens when the parallelogram is a rombus , in this case ABCD is an arbitrary paralllelogram[/quote]

I still don't understand how it can intersect two sides of it. How is that possible.


For what value of $ n$ is the four-digit number $ 712n$, with units digit $ n$, divisible by 18?

divisible by 18 means divisible by 9 and 2. Therefore, $ 7\plus{}1\plus{}2\plus{}n$ is a multiple of 9, and it is an even integer. This is the case when $ n \equal{} \boxed{8}$


Let $a,b$ be integers greater than 2. Prove that there exists a positive integer $k$ and a finite sequence $n_1, n_2, \dots, n_k$ of positive integers such that $n_1 = a$, $n_k = b$, and $n_i n_{i+1}$ is divisible by $n_i + n_{i+1}$ for each $i$ ($1 \leq i < k$).

[quote="paladin8"]Call two integers $ a$ and $ b$ "linked" if there exists a sequence as described in the question. We wish to show that all integers $ a,b > 2$ are linked. If we can show that $ m$ and $ m \plus{} 1$ are linked for any integer $ m > 2$ in a finite number of terms (links), the problem is solved. Consider the following sequence:

$ m$, $ m(m \minus{} 1)$, $ m(m \minus{} 1)(m \minus{} 2)$, $ 2m(m \minus{} 1)$, $ 2m(m \plus{} 1)$, $ m(m \plus{} 1)(m \minus{} 1)$, $ m(m \plus{} 1)$, $ m \plus{} 1$,

which can easily be shown to satisfy the property in the question. Thus $ m$ and $ m \plus{} 1$ are always linked for $ m > 2$. Hence, by induction, any integers $ a, b > 2$ are linked, as desired.[/quote]

A similar approach by my friend:
Case 1:$ a$ is odd
$ a\rightarrow a(a\minus{}1)\rightarrow a(a\plus{}1)\rightarrow a\plus{}1$
Case 2:$ a$ is even
$ a \rightarrow a(\frac{a}{2}\minus{}1) \rightarrow a(\frac{a}{2}\plus{}1)\rightarrow a\plus{}2 \rightarrow (a\plus{}2)(a\plus{}1) \rightarrow a(a\plus{}1) \rightarrow a\plus{}1$
QED


Let $ a,b,c,d,e,f,g\in R\plus{}$
find the minimum of
$ \frac{a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2\plus{}f^2\plus{}g^2}{ab\plus{}bc\plus{}cd\plus{}de\plus{}ef\plus{}fg}$
 :D

[quote="Rose_joker"]Let $ a,b,c,d,e,f,g\in R +$
find the minimum of
$ \frac {a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2}{ab + bc + cd + de + ef + fg}$
 :D[/quote]
Let $ A(a,b,c,d,e,f,g) = \frac {a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2}{ab + bc + cd + de + ef + fg}$
Then we have $ A(ka,kb,kc,kd,ke,kf,kg) = A(a,b,c,d,e,f,g)$
So we can assume that $ a^2 + b^2 + c^2 + d^2 + e^2 + f^2 + g^2 = 1$
Then we just have to find the maximum of $ ab + bc + cd + de + ef + fg$ with the above given condition.
We know that $ ab + bc + cd + de + ef + fg +ga \le \frac{a^2 + g^2}{2} + b^2 + c^2 + d^2 + e^2 + f^2$ or $ ab + bc + cd + de + ef + fg +ga \le1 - \frac{a^2 + g^2}{2}$ Then i guess we can say that $ a,g \le b,f \le c,e \le d$.

I am stuck.... -.-

the minimum is $ \sqrt{4\minus{}2\sqrt{2}}$  :D

How??! :maybe:
Have u solved it apollo? Please post it here then!


Let k be arbitrary natural number . Find all f(m,n):N/{0}->{-1;1} satisfying:
1) $\sum_{1<m;n\leq k} f(m,n) = 0$
2)$\forall m |\sum_{n=1}^k f(m,n)| >\displaystyle\frac{k}{2}$
$\forall n  |\sum_{m=1}^k f(m,n)| >\displaystyle\frac{k}{2}$

I haven't solved this problem :( .Can anybody help me?

Yesterday,I asked my teacher about the solution and he told me this problem is combinatorics problem not algebra.
Admin, please move it!


Let $a,\ b,\ c$ be positive real numbers such that $\frac{1}{a^{2}b^{2}}+\frac{1}{b^{2}c^{2}}+\frac{1}{c^{2}a^{2}}=12.$

Prove that $4(ab+bc+ca)+12\geq 3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right).$

I know a lil bit about Muirhead :D

but I think $(2ab-1)^{2}(ab+1)\geq 0$ :D

The problem can be restated as:$a^{2}+b^{2}+c^{2}=12$,prove that \[4(\frac1a+\frac1b+\frac1c)+12\geq3(a+b+c)\] .Let $S=a+b+c$.We know that \[4(\frac1a+\frac1b+\frac1c)\geq\frac{36}{S}\] therefore we only need to prove \[\frac{12}{S}+4\geq S\iff\]  \[\iff S^{2}-4S-12\leq0\iff\]  \[\iff(S+2)(S-6)\leq0\] ,which is true since $a+b+c\leq\sqrt{3(a^{2}+b^{2}+c^{2})}$


Find all the monotone functions$f: R\Rightarrow{R}$ st:
$f(x^{2007}+y^{2007})=f^{2007}(x)+f^{2007}(y) \forall{x,y\in{R}}$
:)

Double post: http://www.mathlinks.ro/Forum/viewtopic.php?t=121710 .
Please don't answer here, but there.
Mods: please delete this thread.


algebra
polynomial
calculus
calculus computations

Let $ f(x)\in\mathbb R(x)$ be a polynomial of degree $ 3$.
If there exist a real constant $ c$ such that $ f'(x)$ divide $ f(x)\plus{}c$,then show that there exist just one real $ a$ such that $ f(a)\equal{}0$.

[hide="after trying the hard way"]
$ h(x)\equal{}f(x)\plus{}c, h'(x)\equal{}f'(x)$ divides $ h(x)$

we know exists at least one real a which is a zero of h,

 $ h(a)\equal{}h'(a)\equal{}0 \to h(x)\equal{}m(x\minus{}a)^2(x\minus{}r) \to h'(x)\equal{}m(x\minus{}a)[3x\minus{}(r\plus{}2a)]$

must divide h, so $ \frac{r\plus{}2a}{3}\in \{ a,r \}\to r\equal{}a$ for which our claim holds trivially.

now we can assume $ h'(a) \ne 0$

let $ a_k$ be another zero of h, then

$ h(a)\equal{}h(a_k)\equal{}0 \to \exists a_{k\plus{}1}\in (a,a_k ),  h'(a_{k\plus{}1})\equal{}0$ by rolles

since h' divides h, $ h(a_{k\plus{}1})\equal{}0$  hence we can construct an infinite sequence of zeros of h, which is a polynomial of degree 3 (contradiction)[/hide]


inequalities
logarithms
function
inequalities unsolved

Let $x_1, x_2 , x_3, .... x_n$ be $n$ real numbers such that $0 < x_j < \frac{1}{2}$. Prove that \[ \frac{ \prod\limits_{j=1}^{n} x_j } { \left( \sum\limits_{j=1}^{n} x_j \right)^n} \leq \frac{ \prod\limits_{j=1}^{n} (1-x_j) } { \left( \sum\limits_{j=1}^{n} (1 - x_j) \right)^n}  \]

Write our inequality equivanelntly:

\[ \ln \frac{\prod x_j}{\prod (1-x_j)}\leq n\ln \frac{\sum x_j}{\sum (1-x_j)} \]
\[ \sum \ln \frac{x_j}{1-x_j} \leq n\cdot \ln \frac{\sum x_j}{\sum (1-x_j)} \]

Function $f(x)=\ln \frac{x}{1-x}$ is concave ( f''<0 :D) in $\left(0,\frac{1}{2}\right)$, so from Jensen's Inequality we have:

\[ \sum \ln \frac{x_j}{1-x_j}=\sum f(x_j) \leq n\cdot f\left(\frac{\sum x_j}{n}\right) = n\ln\frac{\sum x_j}{\sum 1-x_j}, \] QED

Btw.: It's Fan-Tsa's inequality.